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© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Solutions Manual For

Water-Resources Engineering Third Edition By David A. Chin

Conventional solutions to all problems Also includes Mathcad® solutions to selected problems contributed by Dixie M. Griffin, Jr.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Preface This solutions manual contains the solutions to all end-of-chapter problems in WaterResources Engineering, Third Edition. This manual should be treated as confidential by course instructors and/or their trustees, such as teaching assistants and graders. Unauthorized use of this solutions manual by students would normally be considered as cheating. This solutions manual contains two sets of solutions: conventional solutions and Mathcad® solutions. The conventional solutions to all end-of-chapter problems were prepared by Dr. David A. Chin, using a calculator and/or electronic spreadsheet. Mathcad® solutions to selected problems were prepared by Dr. Dixie M. Griffin Jr. exclusively using Mathcad® software. Depending on the preference of the course instructor, students could be asked to solve problems in either format. The conventional solutions to all problems are presented first, and Mathcad® solutions to selected problems are presented thereafter.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Chapter 1

Introduction 1.1. The mean annual rainfall in Boston is approximately 1050 mm , and the mean annual evapotranspiration is in the range of 380–630 mm (USGS). On the basis of rainfall, this indicates a subhumid climate. The mean annual rainfall in Santa Fe is approximately 360 mm and the mean annual evapotranspiration is < 380 mm . On the basis of rainfall, this indicates an arid climate.

1

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Chapter 2

Fundamentals of Flow in Closed Conduits 2.1. D1 = 0.1 m, D2 = 0.15 m, V1 = 2 m/s, and π π A1 = D12 = (0.1)2 = 0.007854 m2 4 4 π 2 π A2 = D2 = (0.15)2 = 0.01767 m2 4 4 Volumetric flow rate, Q, is given by Q = A1 V1 = (0.007854)(2) = 0.0157 m3 /s According to continuity, A1 V1 = A2 V2 = Q Therefore V2 =

0.0157 Q = = 0.889 m/s A2 0.01767

At 20◦ C, the density of water, ρ, is 998 kg/m3 , and the mass flow rate, m, ˙ is given by m ˙ = ρQ = (998)(0.0157) = 15.7 kg/s 2.2. From the given data: D1 = 200 mm, D2 = 100 mm, V1 = 1 m/s, and π 2 π D = (0.2)2 = 0.0314 m2 4 1 4 π 2 π A2 = D2 = (0.1)2 = 0.00785 m2 4 4 A1 =

The flow rate, Q1 , in the 200-mm pipe is given by Q1 = A1 V1 = (0.0314)(1) = 0.0314 m3 /s and hence the flow rate, Q2 , in the 100-mm pipe is Q2 =

Q1 0.0314 = = 0.0157 m3 /s 2 2

3

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The average velocity, V2 , in the 100-mm pipe is Q2 0.0157 V2 = = = 2 m/s A2 0.00785 2.3. The velocity distribution in the pipe is

[ ( r )2 ] v(r) = V0 1 − R

and the average velocity, V¯ , is defined as 1 V¯ = A

(1)

∫ V dA

(2)

A

where A = πR2

and

dA = 2πrdr

Combining Equations 1 to 3 yields [∫ R [ ] ∫ R [ ∫ R 3 ] ( r )2 ] 1 r R4 2V0 2V0 R2 1 − V¯ = V rdr − − 2πrdr = dr = 0 2 πR2 0 R R2 R2 2 4R2 0 0 R =

V0 2V0 R2 = R2 4 2

The flow rate, Q, is therefore given by πR2 V0 Q = AV¯ = 2 2.4.

[ ] ∫ ∫ R 4 2r2 1 r4 2 2 v dA = V 1 − 2 + 4 2πrdr β = ¯2 R R πR2 V02 0 0 AV A [∫ R [ ] ∫ R 3 ∫ R 5 ] 2r 8 r R4 8 R2 R6 = 2 rdr − − dr + dr = 2 + 2 4 R R 2 2R2 6R4 0 R 0 0 R =

4 3

2.5. D = 0.2 m, Q = 0.06 m3 /s, L = 100 m, p1 = 500 kPa, p2 = 400 kPa, γ = 9.79 kN/m3 . D 0.2 R= = = 0.05 m 4 4 p1 p2 500 − 400 ∆h = − = = 10.2 m γ γ 9.79 (9.79 × 103 )(0.05)(10.2) γR∆h τ0 = = = 49.9 N/m2 L 100 πD2 π(0.2)2 A= = = 0.0314 m2 4 4 Q 0.06 V = = = 1.91 m/s A 0.0314 8τ0 8(49.9) f= = = 0.11 2 ρV (998)(1.91)2

4

(3)

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2.6. T = 20◦ C, V = 2 m/s, D = 0.25 m, horizontal pipe, ductile iron. For ductile iron pipe, ks = 0.26 mm, and ks 0.26 = = 0.00104 D 250 ρV D (998.2)(2)(0.25) Re = = = 4.981 × 105 µ (1.002 × 10−3 ) From the Moody diagram: f = 0.0202 (pipe is smooth) Using the Colebrook equation, 1 √ = −2 log f

(

2.51 ks /D √ + 3.7 Re f

)

Substituting for ks /D and Re gives 1 √ = −2 log f

(

0.00104 2.51 √ + 3.7 4.981 × 105 f

)

By trial and error leads to f = 0.0204 Using the Swamee-Jain equation, [ ] 1 ks /D 5.74 √ = −2 log + 3.7 f Re0.9 ] [ 5.74 0.00104 + = −2 log 3.7 (4.981 × 105 )0.9 which leads to f = 0.0205 The head loss, hf , over 100 m of pipeline is given by hf = f

L V2 100 (2)2 = 0.0204 = 1.66 m D 2g 0.25 2(9.81)

Therefore the pressure drop, ∆p, is given by ∆p = γhf = (9.79)(1.66) = 16.3 kPa If the pipe is 1 m lower at the downstream end, f would not change, but the pressure drop, ∆p, would then be given by ∆p = γ(hf − 1.0) = 9.79(1.66 − 1) = 6.46 kPa

5

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2.7. From the given data: D = 25 mm, ks = 0.1 mm, θ = 10◦ , p1 = 550 kPa, and L = 100 m. At 20◦ C, ν = 1.00 × 10−6 m2 /s, γ = 9.79 kN/m3 , and ks 0.1 = = 0.004 D 25 π π A = D2 = (0.025)2 = 4.909 × 10−4 m2 4 4 2 L Q 100 Q2 hf = f =f = 8.46 × 108 f Q2 2 D 2gA 0.025 2(9.81)(4.909 × 10−4 )2 The energy equation applied over 100 m of pipe is p1 V 2 p2 V 2 + + z1 = + + z2 + hf γ 2g γ 2g which simplifies to p2 = p1 − γ(z2 − z1 ) − γhf p2 = 550 − 9.79(100 sin 10◦ ) − 9.79(8.46 × 108 f Q2 ) p2 = 380.0 − 8.28 × 109 f Q2 (a) For Q = 2 L/min = 3.333 × 10−5 m3 /s, Q 3.333 × 10−5 = = 0.06790 m/s A 4.909 × 10−4 VD (0.06790)(0.025) = Re = = 1698 ν 1 × 10−6 V =

Since Re < 2000, the flow is laminar when Q = 2 L/min. Hence, 64 64 = = 0.03770 Re 1698 p2 = 380.0 − 8.28 × 109 (0.03770)(3.333 × 10−5 )2 = 380 kPa f=

Therefore, when the flow is 2 L/min, the pressure at the downstream section is 380 kPa . For Q = 20 L/min = 3.333 × 10−4 m3 /s, Q 3.333 × 10−4 = = 0.6790 m/s A 4.909 × 10−4 VD (0.6790)(0.025) Re = = = 16980 ν 1 × 10−6 V =

Since Re > 5000, the flow is turbulent when Q = 20 L/min. Hence, f=[

( log

0.25 ks /D 3.7

+

5.74

)]2 = [

0.25 ( 0.004 )]2 = 0.0342 5.74 log 3.7 + 16980 0.9

Re0.9 p2 = 380.0 − 8.28 × 109 (0.0342)(3.333 × 10−4 )2 = 349 kPa

Therefore, when the flow is 2 L/min, the pressure at the downstream section is 349 kPa .

6

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(b) Using the Colebrook equation with Q = 20 L/min, [ ] [ ] 2.51 2.51 1 ks /D 0.004 √ = −2 log √ √ + + = −2 log 3.7 3.7 f Re f 16980 f which yields f = 0.0337 . Comparing this with the Swamee-Jain result of f = 0.0342 indicates a difference of 1.5% , which is more than the 1% claimed by Swamee-Jain. 2.8. The Colebrook equation is given by 1 √ = −2 log f

(

ks /D 2.51 √ + 3.7 Re f

)

Inverting and squaring this equation gives f=

0.25 √ {log[(ks /D)/3.7 + 2.51/(Re f )]}2

This equation is “slightly more convenient” than the √ Colebrook formula since it is quasiexplicit in f , whereas the Colebrook formula gives 1/ f . 2.9. The Colebrook equation is preferable since it provides greater accuracy than interpolating from the Moody diagram. 2.10. D = 0.5 m, p1 = 600 kPa, Q = 0.50 m3 /s, z1 = 120 m, z2 = 100 m, γ = 9.79 kN/m3 , L = 1000 m, ks (ductile iron) = 0.26 mm, π 2 π D = (0.5)2 = 0.1963 m2 4 4 Q 0.50 V = = = 2.55 m/s A 0.1963 A=

Using the Colebrook equation, 1 √ = −2 log f

(

ks /D 2.51 √ + 3.7 Re f

)

where ks /D = 0.26/500 = 0.00052, and at 20◦ C Re =

ρV D (998)(2.55)(0.5) = 1.27 × 106 = µ 1.00 × 10−3

Substituting ks /D and Re into the Colebrook equation gives ( ) 1 0.00052 2.51 √ = −2 log √ + 3.7 f 1.27 × 106 f which leads to f = 0.0172

7

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Applying the energy equation p1 V12 p2 V22 + + z1 = + + z2 + hf γ 2g γ 2g Since V1 = V2 , and hf is given by the Darcy-Weisbach equation, then the energy equation can be written as p2 L V2 p1 + z1 = + z2 + f γ γ D 2g Substituting known values leads to 600 p2 1000 (2.55)2 + 120 = + 100 + 0.0172 9.79 9.79 0.5 2(9.81) which gives p2 = 684 kPa If p is the (static) pressure at the top of a 30 m high building, then p = p2 − 30γ = 684 − 30(9.79) = 390 kPa This (static) water pressure is adequate for service. 2.11. The head loss, hf , in the pipe is estimated by ( hf =

pmain + zmain γ

)

( −

poutlet + zoutlet γ

)

where pmain = 400 kPa, zmain = 0 m, poutlet = 0 kPa, and zoutlet = 2.0 m. Therefore, ( ) 400 hf = + 0 − (0 + 2.0) = 38.9 m 9.79 Also, since D = 25 mm, L = 20 m, ks = 0.15 mm (from Table 2.1), ν = 1.00 × 10−6 m2 /s (at 20◦ C), the combined Darcy-Weisbach and Colebrook equation (Equation 2.43) yields, ( ) √ gDhf ks /D 1.774ν 2 Q = −0.965D ln + √ L 3.7 D gDhf /L [ ] √ (9.81)(0.025)(38.9) 0.15/25 1.774(1.00 × 10−6 ) 2 √ = −0.965(0.025) ln + 20 3.7 (0.025) (9.81)(0.025)(38.9)/20 = 0.00265 m3 /s = 2.65 L/s The faucet can therefore be expected to deliver 2.65 L/s when fully open. 2.12. From the given data: Q = 300 L/s = 0.300 m3 /s, L = 40 m, and hf = 45 m. Assume that ν = 10−6 m2 /s (at 20◦ C) and take ks = 0.15 mm (from Table 2.1). Substituting these data

8

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into Equation 2.43 gives ) ks /D 1.784ν + √ Q = −0.965D 3.7 D gDhf /L √ ( ) −6 ) 0.00015 1.784(10 (9.81)D(45) 0.2 = −0.965D2 ln + √ (40) 3.7D D (9.81)D(45)/(40) √

2

gDhf ln L

(

This is an implicit equation in D that can be solved numerically to yield D = 166 mm . 2.13. Since ks = 0.15 mm, L = 40 m, Q = 0.3 m3 /s, hf = 45 m, ν = 1.00 × 10−6 m2 /s, the Swamee-Jain approximation (Equation 2.44 gives [

(

D = 0.66 ks1.25 {

LQ2 ghf

)4.75

( + νQ9.4

[

= 0.66 (0.00015)1.25

(40)(0.3)2 (9.81)(45)

L ghf

]4.75

)5.2 ]0.04

+ (1.00 × 10−6 )(0.3)9.4

[

40 (9.81)(45)

]5.2 }0.04

= 0.171 m = 171 mm The calculated pipe diameter (171 mm) is about 3% higher than calculated by the Colebrook equation (166 mm). 2.14. The kinetic energy correction factor, α, is defined by ∫ V3 v3 ρ dA = αρ A 2 A 2 ∫

or α=

3 A v dA V 3A

(1)

Using the velocity distribution in Problem 2.3 gives ∫

∫ 3

( r )2 ]2

1− 2πr dr R ∫ R[ ( r )4 ( r )6 ] ( r )2 3 +3 − r dr = 2πV0 1−3 R R R 0 ] ∫ R[ 3r3 3r5 r7 3 = 2πV0 r − 2 + 4 − 6 dr R R R 0 [ 2 ]R 4 6 3r r r8 3 r − + − = 2πV0 2 4R2 2R4 8R6 0 [ ] 3 1 1 2 3 1 = 2πR V0 − + − 2 4 2 8 πR2 V03 = 4 V03

v dA = A

[

R

0

9

(2)

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The average velocity, V , was calculated in Problem 2.3 as V0 V = 2 hence ( )3 V0 πR2 V03 3 V A= πR2 = 2 8 Combining Equations 1 to 3 gives α=

(3)

πR2 V03 /4 = 2 πR2 V03 /8

2.15. The kinetic energy correction factor, α, is defined by ∫ 3 v dA α= A 3 V A Using the given velocity distribution gives ∫ ∫ R ( r ) 37 3 2πr dr v dA = V03 1 − R A 0 ∫ R( r ) 37 3 = 2πV0 1− r dr R 0 To facilitate integration, let r x=1− R which gives r = R(1 − x) dr = −R dx

(1)

(2)

(3)

(4) (5)

Combining Equations 2 to 5 gives ∫ ∫ 1 3 v 3 dA = 2πV03 x 7 R(1 − x)(−R)dx A 0 ∫ 1 ∫ 1 3 3 10 2 3 2 3 x 7 (1 − x)dx = 2πR V0 = 2πR V0 (x 7 − x 7 )dx 0 0 [ ]1 7 10 7 17 = 2πR2 V03 x7 − x7 10 17 0 = 0.576πR2 V03

(6)

The average velocity, V , is given by (using the same substitution as above) ∫ 1 v dA V = A A ∫ R ( ∫ 1 r ) 17 2V0 0 1 V 1 − x 7 R(1 − x)(−R)dx = 2πr dr = 0 πR2 0 R R2 1 [ ] ∫ 1 1 8 7 8 7 15 1 = 2V0 (x 7 − x 7 )dx = 2V0 x 7 − x 7 8 15 0 0 = 0.817V0

(7)

10

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Using this result, V 3 A = (0.817V0 )3 πR2 = 0.545πR2 V03

(8)

Combining Equations 1, 6, and 8 gives α=

0.576πR2 V03 = 1.06 0.545πR2 V03

The momentum correction factor, β, is defined by ∫ 2 v dA β= A 2 AV

(9)

In this case, AV 2 = πR2 (0.817V0 )2 = 0.667πR2 V02

(10)

and ∫

∫

( r ) 27 2πr dr V02 1 − R 0 ∫ 0 ∫ 1 2 2 9 2 2 2 x 7 R(1 − x)(−R)dx = 2πR V0 = 2πV0 (x 7 − x 7 )dx 1 0 ]1 [ 7 16 7 9 = 2πR2 V02 x 7 − x 7 = 0.681πR2 V02 9 16 0 R

v 2 dA = A

(11)

Combining Equations 9 to 11 gives β=

0.681πR2 V02 = 1.02 0.667πR2 V02

2.16. The kinetic energy correction factor, α, is defined by ∫ 3 v dA α= A 3 V A Using the velocity distribution given by Equation 2.73 gives ∫ ∫ R ( r ) n3 v 3 dA = V03 1 − 2πr dr R A 0 ∫ R( r ) n3 3 = 2πV0 r dr 1− R 0 Let x=1−

r R

(1)

(2)

(3)

which gives r = R(1 − x) dr = −R dx

11

(4) (5)

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Combining Equations 2 to 5 gives ∫ ∫ 1 3 3 3 v dA = 2πV0 x n R(1 − x)(−R)dx A 0 ∫ 1 ∫ 1 3+n 3 3 2 3 2 3 n = 2πR V0 x (1 − x)dx = 2πR V0 (x n − x n )dx 0 0 [ ]1 3+n 3+2n n n = 2πR2 V03 x n − x n 3+n 3 + 2n 0 =

2n2 πR2 V03 (3 + n)(3 + 2n)

(6)

The average velocity, V , is given by ∫ 1 V = v dA A A ∫ R ( ∫ r ) n1 2V0 0 1 1 V0 1 − 2πr dr = 2 x n R(1 − x)(−R)dx = πR2 0 R R 1 [ ]1 ∫ 1 1+n 1+n 1+2n 1 n n n n n n )dx = 2V0 − = 2V0 (x − x x x 1+n 1 + 2n 0 0 [ ] 2n2 = V0 (1 + n)(1 + 2n)

(7)

Using this result, [ V 3A =

2n2 (1 + n)(1 + 2n)

]3 V03 πR2 =

8n6 πR2 V03 (1 + n)3 (1 + 2n)3

Combining Equations 1, 6, and 8 gives α=

2n2 2 3 (3+n)(3+2n) πR V0 8n6 πR2 V03 (1+n)3 (1+2n)3

=

(1 + n)3 (1 + 2n)3 4n4 (3 + n)(3 + 2n)

Putting n = 7 gives α = 1.06 , the same result obtained in Problem 2.15. 2.17. p1 = 30 kPa, p2 = 500 kPa, therefore head, hp , added by pump is given by hp =

p2 − p1 500 − 30 = = 48.0 m γ 9.79

Power, P , added by pump is given by P = γQhp = (9.79)(Q)(48.0) = 470 kW per m3 /s

12

(8)

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2.18. Q = 0.06 m3 /s, D = 0.2 m, ks = 0.9 mm (riveted steel), ks /D = 0.9/200 = 0.00450, for 90◦ bend K = 0.3, for the entrance K = 1.0, at 20◦ C ρ = 998 kg/m3 , and µ = 1.00 × 10−3 Pa·s, therefore π π A = D2 = (0.2)2 = 0.0314 m2 4 4 Q 0.06 V = = = 1.91 m/s A 0.0314 (998)(1.91)(0.2) ρV D Re = = = 3.81 × 105 µ 1.00 × 10−3 Substituting ks /D and Re into the Colebrook equation gives ( ) 1 0.00450 2.51 √ √ = −2 log + 3.7 f 3.81 × 105 f which leads to f = 0.0297 Minor head loss, hm , is given by hm =

∑

K

V2 (1.91)2 = (1.0 + 0.3) = 0.242 m 2g 2(9.81)

If friction losses, hf , account for 90% of the total losses, then hf = f which means that 0.0297

L V2 = 9hm D 2g

L (1.91)2 = 9(0.242) 0.2 2(9.81)

Solving for L gives L = 78.9 m For pipe lengths shorter than the length calculated in this problem, the word “minor” should not be used. 2.19. From the given data: p0∑ = 480 kPa, v0 = 5 m/s, z0 = 2.44 m, D = 19 mm = 0.019 m, L = 40 m, z1 = 7.62 m, and Km = 3.5. For copper tubing it can be assumed that ks = 0.0023 mm. Applying the energy and Darcy-Weisbach equations between the water main and the faucet gives p0 p1 v2 + z0 − hf − hm = + 1 + z1 γ γ 2g 2 2 480 f (40) v v 0 v2 + 2.44 − − 3.5 = + + 7.62 9.79 0.019 2(9.81) 2(9.81) γ 2(9.81) which simplifies to v=√

6.622 107.3f − 0.2141

13

(1)

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The Colebrook equation, with ν = 1 × 10−6 m2 /s gives [ ] 1 ks 2.51 √ = −2 log √ + 3.7D Re f f [ ] 2.51 1 0.0025 √ = −2 log + v(0.019) √ 3.7(19) f f 1×10−6 [ ] 1 1.321 × 10−4 −5 √ = −2 log 3.556 × 10 + √ f v f

(2)

Combining Equations 1 and 2 gives √ [ ] 1 1.995 × 10−5 107.3f − 0.2141 −5 √ = −2 log 3.556 × 10 + √ f f which yields f = 0.0189 Substituting into Equation 1 yields v=√

6.622

= 4.92 m/s 107.3(0.0189) − 0.2141 (π ) Q = Av = 0.0192 (4.92) = 0.00139 m3 /s = 1.39 L/s (= 22 gpm) 4 This flow is very high for a faucet. The flow would be reduced if other faucets are open, this is due to increased pipe flow and frictional resistance between the water main and the faucet. ∑ 2.20. From the given data: z1 = −1.5 m, z2 = 40 m, p1 = 450 kPa, k = 10.0, Q = 20 L/s = 0.02 m3 /s, D = 150 mm (PVC), L = 60 m, T = 20◦ C, and p2 = 150 kPa. The combined energy and Darcy-Weisbach equations give [ ] 2 p1 V12 p2 V22 fL ∑ V + + z1 + hp = + + z2 + + k (1) γ 2g γ 2g D 2g where V 1 = V2 = V =

Q = A

0.02 π(0.15)2 4

= 1.13 m/s

At 20◦ C, ν = 1.00 × 10−6 m2 /s, and Re =

VD (1.13)(0.15) = = 169500 ν 1.00 × 10−6

Since PVC pipe is smooth (ks = 0), the friction factor, f , is given by ( ) ( ) 1 2.51 2.51 √ = −2 log √ √ = −2 log f Re f 169500 f

14

(2)

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which yields f = 0.0162 Taking γ = 9.79

(3)

kN/m3

and combining Equations 1 to 3 yields [ ] 450 1.132 150 1.132 (0.0162)(60) 1.132 + + (−1.5) + hp = + + 40 + + 10 9.79 2(9.81) 9.79 2(9.81) 0.15 2(9.81)

which gives hp = 11.9 m Since hp > 0, a booster pump is required . The power, P , to be supplied by the pump is given by P = γQhp = (9.79)(0.02)(11.9) = 2.3 kW 2.21. (a) Diameter of pipe, D = 0.75 m, area, A given by A=

π 2 π D = (0.75)2 = 0.442 m2 4 4

and velocity, V , in pipe Q 1 = = 2.26 m/s A 0.442 Energy equation between reservoir and A: V =

7 + hp − hf =

pA VA2 + + zA γ 2g

where pA = 350 kPa, γ = 9.79 kN/m3 , VA = 2.26 m/s, zA = 10 m, and hf =

fL V 2 D 2g

where f depends on Re and ks /D. At 20◦ C, ν = 1.00 × 10−6 m2 /s and VD (2.26)(0.75) = = 1.70 × 106 ν 1.00 × 10−6 ks 0.26 = = 3.47 × 10−4 D 750

Re =

Using the Swamee-Jain equation, [ ] [ ] 1 ks /D 5.74 3.47 × 10−4 5.74 √ = −2 log + = −2 log + = 7.93 3.7 3.7 (1.70 × 106 )0.9 f Re0.9 which leads to f = 0.0159 The head loss, hf , between the reservoir and A is therefore given by hf =

fL V 2 (0.0159)(1000) (2.26)2 = = 5.52 m D 2g 0.75 2(9.81)

15

(1)

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Substituting into Equation 1 yields 7 + hp − 5.52 =

350 2.262 + + 10 9.81 2(9.81)

which leads to hp = 44.5 m (b) Power, P , supplied by the pump is given by P = γQhp = (9.79)(1)(44.5) = 436 kW (c) Energy equation between A and B is given by V2 pA VA2 pB + + zA − hf = + B + zB γ 2g γ 2g and since VA = VB , pB = pA + γ(zA − zB − hf ) = 350 + 9.79(10 − 4 − 5.52) = 355 kPa 2.22. From the given data: L = 3 km = 3000 m, Qave = 0.0175 m3 /s, and Qpeak = 0.578 m3 /s. If the velocity, Vpeak , during peak flow conditions is 2.5 m/s, then 2.5 = which gives

0.578 Qpeak = 2 πD /4 πD2 /4

√ D=

0.578 = 0.543 m π(2.5)/4

Rounding to the nearest 25 mm gives D = 550 mm with a cross-sectional area, A, given by A=

π 2 π D = (0.550)2 = 0.238 m2 4 4

During average demand conditions, the head, have , at the suburban development is given by have =

2 pave Vave + + z0 γ 2g

(1)

where pave = 340 kPa, γ = 9.79 kN/m3 , Vave = Qave /A = 0.0175/0.238 = 0.0735 m/s, and z0 = 8.80 m. Substituting into Equation 1 gives have =

340 0.07352 + + 8.80 = 43.5 m 9.79 2(9.81)

16

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For ductile-iron pipe, ks = 0.26 mm, ks /D = 0.26/550 = 4.73 × 10−4 , at 20◦ C ν = 1.00 × 10−6 m2 /s, and therefore Vave D (0.0735)(0.550) Re = = = 4.04 × 104 ν 1.00 × 10−6 and the Swamee-Jain equation gives ] [ [ ] 4.73 × 10−4 1 ks /D 5.74 5.74 √ = −2 log = −2 log + + 3.7 3.7 (4.04 × 104 )0.9 fave Re0.9 and yields fave = 0.0234 The head loss between the water treatment plant and the suburban development is therefore given by L V2 3000 0.07352 hf = f = (0.0234) = 0.035 m D 2g 0.550 2(9.81) Since the head at the water treatment plant is 10.00 m, the pump head, hp , that must be added is hp = (43.5 + 0.035) − 10.00 = 33.5 m and the power requirement, P , is given by P = γQhp = (9.79)(0.0175)(33.5) = 5.74 kW During peak demand conditions, the head, hpeak , at the suburban development is given by hpeak =

2 ppeak Vpeak + + z0 γ 2g

(2)

where ppeak = 140 kPa, γ = 9.79 kN/m3 , Vpeak = Qpeak /A = 0.578/0.238 = 2.43 m/s, and z0 = 8.80 m. Substituting into Equation 2 gives hpeak =

140 2.432 + + 8.80 = 23.4 m 9.79 2(9.81)

For pipe, ks /D = 4.73 × 10−4 , and Re =

Vpeak D (2.43)(0.550) = = 1.34 × 106 ν 1.00 × 10−6

and the Swamee-Jain equation gives [ ] [ ] 1 ks /D 5.74 4.73 × 10−4 5.74 √ = −2 log + = −2 log + 3.7 3.7 (1.34 × 106 )0.9 Re0.9 fpeak and yields fpeak = 0.0170

17

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The head loss between the water treatment plant and the suburban development is therefore given by L V2 3000 2.432 hf = f = (0.0170) = 27.9 m D 2g 0.550 2(9.81) Since the head at the water treatment plant is 10.00 m, the pump head, hp , that must be added is hp = (23.4 + 27.9) − 10.00 = 41.3 m and the power requirement, P , is given by P = γQhp = (9.79)(0.578)(41.3) = 234 kW 2.23. The energy equation applied between the reservoir and the outlet is given by [ ] 2 fl V V2 40 − Ke + + Kv − ht = D 2g 2g which can be put in the form ] 2 [ V fL + Kv + 1 ht = 40 − Ke + D 2g

(1)

For a sharp-edged entrance, Ke = 0.5, for an open globe valve, Kv = 10.0, and from the given data: D = 0.05 m, A = πD2 /4 = 0.001963 m2 , Q = 4 L/s = 0.004 m3 /s, V = Q/A = 2.038 m/s, L = 125 m, ν = 1.00 ×10−6 m2 /s, ks = 0.23 mm, Re = V D/ν = 1.02 × 105 , and using the Swamee-Jain equation, f=[

( log

=[

( log

0.25 ks 3.7D

+

0.23 3.7(50)

5.74

)]2

Re0.9 0.25

+

5.74 (1.02×105 )0.9

)]2 = 0.0308

Substituting into Equation 1 gives ] [ (2.038)2 (0.0308)(125) + 10.0 + 1 = 21.27 m ht = 40 − 0.5 + 0.05 2(9.81) Therefore, taking γ = 9.79 kN/m3 , the power extracted by the turbine is given by P = γQht = (9.79)(0.004)(21.27) = 0.833 kW A similar problem would be encountered in calculating the power output at a hydroelectric facility . 2.24. The head loss is calculated using Equation 2.78. The hydraulic radius, R, is given by R=

A (2)(1) = = 0.333 m P 2(2 + 1)

18

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and the mean velocity, V , is given by V =

Q 5 = = 2.5 m/s A (2)(1)

At 20◦ C, ρ = 998.2 kg/m3 , µ = 1.002 × 10−3 N·s/m2 , and therefore the Reynolds number, Re, is given by ρV (4R) (998.2)(2.5)(4 × 0.333) Re = = = 3.32 × 106 ν 1.002 × 10−3 A median equivalent sand roughness for concrete can be taken as ks = 1.6 mm (Table 2.1), and therefore the relative roughness, ks /4R, is given by ks 1.6 × 10−3 = = 0.00120 4R 4(0.333) Substituting Re and ks /4R into the Swamee-Jain equation (Equation 2.38) for the friction factor yields ] [ [ ] 0.00120 1 k /4R 5.74 5.74 √ = −2 log s = −2 log + + = 6.96 3.7 3.7 (3.32 × 106 )0.9 f Re0.9 which yields f = 0.0206 The frictional head loss in the culvert, hf , is therefore given by the Darcy-Weisbach equation as fL V 2 (0.0206)(100) 2.52 hf = = = 0.493 m 4R 2g (4 × 0.333) 2(9.81) 2.25. The frictional head loss is calculated using Equation 2.78. The hydraulic radius, R, is given by A (2)(2) R= = = 0.500 m P 2(2 + 2) and the mean velocity, V , is given by V =

Q 10 = = 2.5 m/s A (2)(2)

At 20◦ C, ρ = 998 kg/m3 , µ = 1.00 × 10−3 N·s/m2 , and therefore the Reynolds number, Re, is given by ρV (4R) (998)(2.5)(4 × 0.500) Re = = = 4.99 × 106 µ 1.00 × 10−3 A median equivalent sand roughness for concrete can be taken as ks = 1.6 mm (Table 2.1), and therefore the relative roughness, ks /4R, is given by ks 1.6 × 10−3 = = 0.0008 4R 4(0.500)

19

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Substituting Re and ks /4R into the Swamee-Jain equation (Equation 2.39) for the friction factor yields [ ] ] [ 1 ks /4R 5.74 5.74 0.0008 √ = −2 log + + = 7.31 = −2 log 3.7 3.7 (4.99 × 106 )0.9 f Re0.9 which yields f = 0.0187 The frictional head loss in the culvert, hf , is therefore given by the Darcy-Weisbach equation as fL V 2 (0.0187)(500) 2.52 hf = = = 1.49 m 4R 2g (4 × 0.500) 2(9.81) Applying the energy equation between the upstream and downstream sections (Sections 1 and 2 respectively), p1 V12 p2 V22 + + z1 = + + z2 + hf γ 2g γ 2g which gives p1 2.52 p2 2.52 + + (0.002)(500) = + + 0 + 1.49 9.79 2(9.81) 9.79 2(9.81) Re-arranging this equation gives p1 − p2 = 4.80 kPa 2.26. The Hazen-Williams formula is given by V = 0.849CH R0.63 Sf0.54

(1)

where

hf L Combining Equations 1 and 2, and taking R = D/4 gives ( )0.63 ( )0.54 hf D V = 0.849CH 4 L Sf =

which simplifies to hf = 6.82

L D1.17

(

V CH

(2)

)1.85

2.27. Comparing the Hazen-Williams and Darcy-Weisbach equations for head loss gives ( ) V 1.85 L V2 L =f hf = 6.82 1.17 D CH D 2g which leads to f=

134 1.85 D 0.17 CH

1 V

0.15

For laminar flow, Equation 2.36 gives f ∼ 1/Re ∼ 1/V , and for fully-turbulent flow Equation 2.35 gives f ∼ 1/V 0 . Since the Hazen-Williams formula requires that f ∼ 1/V 0.15 , this indicates that the flow must be in the transition regime .

20

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2.28. The Manning equation is given by 1 2 1 1 V = R 3 Sf2 = n n

(

D 4

)2 ( 3

hf L

)1 2

which re-arranges to give hf = 6.35

n2 LV 2 4

D3

2.29. Comparing the Manning and Darcy-Weisbach equations gives hf = 6.35

n2 LV 2 D

4 3

=f

L V2 D 2g

which leads to f = 125

n2 1

D3 For laminar flow, Equation 2.36 gives f ∼ 1/Re ∼ 1/V , and for fully-turbulent flow Equation 2.35 gives f ∼ 1/V 0 . Since the Manning equation requires that f ∼ 1/V 0 , this indicates that the flow must be fully turbulent or rough . 2.30. Equating the Hazen-Williams and Manning head loss expressions (

L 6.82 1.17 D which re-arranges to give

( n=

V CH

)1.85 = 6.35

n2 LV 2 4

D3

D0.082 1.04 0.075 V

)

1 0.93 CH

2.31. Choose the Darcy-Weisbach equation since this equation is applicable in all flow regimes. The Hazen-Williams and Manning equations are limited to particular flow conditions (transition and fully turbulent respectively). 2.32. (a) The Hazen-Williams roughness coefficient, CH , can be taken as 110 (Table 2.2), L = 500 m, D = 0.300 m, V = 2 m/s, and therefore the head loss, hf , is given by Equation 2.82 as ( ) ( ) L V 1.85 500 2 1.85 hf = 6.82 1.17 = 6.82 = 8.41 m D CH (0.30)1.17 110 (b) The Manning roughness coefficient, n, can be taken as 0.013 (approximation from Table 2.2), and therefore the head loss, hf , is given by Equation 2.85 as hf = 6.35

n2 LV 2 4

= 6.35

D3

(0.013)2 (500)(2)2 4

(0.30) 3

21

= 10.7 m

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(c) The equivalent sand roughness, ks , can be taken as 0.26 mm (Table 2.1), and the Reynolds number, Re, is given by Re =

VD (2)(0.30) = = 6.00 × 105 ν 1.00 × 10−6

where ν = 1.00 × 10−6 m2 /s at 20◦ C. Substituting ks , D, and Re into the Colebrook equation yields the friction factor, f , where [ ] [ ] 1 ks 2.51 0.26 2.51 √ √ √ = −2 log + = −2 log + 3.7D Re f 3.7(300) 6.00 × 105 f f Solving by trial and error leads to f = 0.0195 The head loss, hf , is therefore given by the Darcy-Weisbach equation as hf = f

L V2 500 22 = 0.0195 = 6.63 m D 2g 0.30 2(9.81)

It is reasonable to assume that the Darcy-Weisbach equation yields the most accurate estimate of the head loss. In this case, the Hazen-Williams formula gives a head loss that is 27% too high, and the Manning formula yields a head loss that is 61% too high. Problem 2.27 has demonstrated that the relationship between the friction factor, f , and the Hazen-Williams coefficient, CH , is given by 134

f=

1

1.85 D 0.17 CH

V

0.15

which can be re-arranged to give CH =

14.2 f 0.54 D0.092 V 0.081

=

14.2 (0.0195)0.54 (0.30)0.092 (2)0.081

= 126

which is 15% higher than the assumed value of CH (110). Problem 2.29 has demonstrated that the relationship between the friction factor, f , and the Manning coefficient, n, is given by f = 125

n2 1

D3 which can be re-arranged to give n = 0.0894f 0.5 D0.17 = 0.0894(0.0195)0.5 (0.30)0.17 = 0.010 which is 23% lower than the assumed value of n (= 0.013). 2.33. The pressure at the midpoint of the pipe as a function of time is shown in Figure 2.1.

22

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Figure 2.1: Problem 2.33. 2.34. From the given data: L = 100 m, V = 3 m/s, and T = 20◦ C. The critical time, tc , for valve closure is 2L tc = (1) c At 20◦ C, Ev = 2.15 × 109 Pa, and ρ0 = 998 kg/m3 (Appendix B, Table B.1), therefore √ √ Ev 2.15 × 109 c= = = 1470 m/s ρ0 998 Substituting into Equation 1 gives 2(100) = 0.14 s 1470 The maximum water-hammer pressure that can occur is given by tc =

∆p = ρcV = (998)(1470)(3) = 4.40 × 106 Pa = 4400 kPa If the water temperature is 10◦ C, Ev = 2.10 × 109 Pa, ρ0 = 999.7 kg/m3 (Appendix B, Table B.1), and √ 2.10 × 109 c= = 1450 m/s 999.7 and the maximum water-hammer pressure is given by ∆p = ρcV = (999.7)(1450)(3) = 4.35 × 106 Pa = 4350 kPa 2.35. Yes , in cases of rapid valve closure. 2.36. From the given data: T = 20◦ C, L = 150 m, D = 50 mm = 0.050 m, V = 4 m/s, e = 1.5 mm = 0.0015 m, and E = 1.655 × 105 MN/m2 = 1.655 × 1011 N/m2 . Taking ρ0 = 998 kg/m3 and Ev = 2.15 × 109 Pa, the speed of the pressure wave, c, is given by Equation 2.96 as √ √ Ev /ρ0 (2.15 × 109 )/(998) c= = = 1226 m/s 9 1 + (Ev D/eE) 1 + (2.15 × 10 × 0.050)/(0.0015 × 1.655 × 1011 ) Hence, the pressure increase, ∆p is given by Equation 2.89 as ∆p = ρ0 cV = (998)(1226)(4) = 4.89 × 106 Pa = 4890 kPa

23

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2.37. From the given data: T = 20◦ C, L = 150 m, D = 50 mm = 0.050 m, V = 4 m/s, e = 2.0 mm = 0.0020 m, and E = 1.7 × 104 MN/m2 = 1.7 × 1010 N/m2 . Taking ρ0 = 998 kg/m3 and Ev = 2.15 × 109 Pa, the speed of the pressure wave, c, is given by Equation 2.96 as √ √ Ev /ρ0 (2.15 × 109 )/(998) c= = = 719 m/s 1 + (Ev D/eE) 1 + (2.15 × 109 × 0.050)/(0.0020 × 1.7 × 1010 ) Hence, the pressure increase, ∆p is given by Equation 2.89 as ∆p = ρ0 cV = (998)(719)(4) = 2.87 × 106 Pa = 2870 kPa 2.38. D = 0.85 m, A = πD2 /4 = π(0.85)2 /4 = 0.567 m2 , ks = 0.26 mm (ductile iron), ks /D = 0.26/850 = 0.000306, T = 20◦ C, ρ = 998 kN/m3 , µ = 1.00 × 10−3 N·s/m2 . Assuming fully turbulent flow, then ( ) ( ) 1 ks /D 0.000306 √ = −2 log = −2 log = 8.16 3.7 3.7 f which leads to f = 0.0150 for each pipe. Writing the energy equations, 100 − f

LAJ Q2AJ LBJ Q2BJ = 80 − f D 2gA2 D 2gA2

(1)

100 − f

LCJ Q2CJ LAJ Q2AJ = 60 + f D 2gA2 D 2gA2

(2)

and

and the continuity equation is QAJ + QBJ = QCJ

(3)

Equations 1 and 3 assume that the flow is out of reservoir B. Substituting known values into Equation 1 gives 100 − 0.0150

Q2AJ Q2BJ 800 900 = 80 − 0.0150 0.85 2(9.81)(0.567)2 0.85 2(9.81)(0.567)2

which leads to Q2BJ = 1.12Q2AJ − 8.93

(4)

Substituting known values into Equation 2 gives 100 − 0.0150

Q2AJ Q2CJ 900 700 = 60 + 0.0150 0.85 2(9.81)(0.567)2 0.85 2(9.81)(0.567)2

which leads to Q2CJ = 20.4 − 1.29Q2AJ

24

(5)

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Substituting Equations 4 and 5 into Equation 3 leads to √ √ QAJ + 1.12Q2AJ − 8.93 − 20.4 − 1.29Q2AJ = 0 Solving for QAJ by trial and error gives QAJ = 2.84 m3 /s and Equations 4 and 5 give QBJ = 0.32 m3 /s QCJ = 3.16 m3 /s Recalculating f for each pipe (using Re and ks /D) and repeating the process until the assumed f is equal to the calculated f leads to QAJ = 2.81 m3 /s QBJ = 0.33 m3 /s QCJ = 3.14 m3 /s These results do not differ by much from the initial result that assumes complete turbulence. 2.39. For ductile-iron pipe, L Q2 hf = f = D 2gA2

(

L 1 D 2gA2

) Q2 f = rQ2

Assuming fully turbulent flow: Pipe AD BC BD AC

ks /D 0.000650 0.000867 0.000743 0.00104

f 0.0177 0.0190 0.0183 0.0198

r 143 517 345 1173

From the given data: hA = 25 m, and hB = 20 m. Since the flow directions in the pipe network are not known in advance, flow directions can be assumed and validated. Assumed flow directions are validated when the calculated flows are real and positive. Assuming that the flow in pipe AC is from C to A and that the flow in pipe BC is from C to B (i.e. hA < hC > hB ), the Darcy-Weisbach equation gives hC = 25 + 1173Q2CA

(1)

hC = 20 + 517Q2CB

(2)

and for flow from C to B,

25

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Combining Equations 1 and 2 gives 25 + 1173Q2CA = 20 + 517Q2CB which leads to Q2CA − 0.441Q2CB + 0.00426 = 0

(3)

The continuity equation requires that QCA + QCB = 0.2 or QCA = 0.2 − QCB

(4)

Combining Equations 3 and 4 gives (0.2 − QCB )2 − 0.441Q2CB + 0.00426 = 0 which yields 0.559Q2CB − 0.4QCB + 0.04426 = 0

(5)

Equation 5 has two real solutions that are given by QCB = 0.579 m3 /s

and QCB = 0.137 m3 /s

and substituting these results into Equation 4 gives QCA = −0.379 m3 /s

and

QCA = 0.063 m3 /s

Since only real and positive flows are acceptable, QCB = 0.137 m3 /s, QCA = 0.063 m3 /s, and the assumed flow directions are validated. [If invalid flow directions were assumed, then there would be no real and positive simultaneous solutions to the corresponding Darcy-Weisbach and continuity equations.] Assuming that the flow in pipe AD is from A to D and that the flow in pipe BD is from B to D (i.e. hA > hD < hB ), the Darcy-Weisbach equation gives hD = 25 − 143Q2AD

(6)

hD = 20 − 345Q2BD

(7)

and for the flow from B to D, Combining Equations 6 and 7 gives 25 − 143Q2AD = 20 − 345Q2BD which leads to Q2BD − 0.414Q2AD + 0.0145 = 0 The continuity equation requires that QAD + QBD = 0.2

26

(8)

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or QBD = 0.2 − QAD

(9)

Combining Equations 8 and 9 gives (0.2 − QAD )2 − 0.414Q2AD + 0.0145 = 0 which yields 0.586Q2AD − 0.4QAD + 0.0545 = 0

(10)

Equation 10 has two real solutions that are given by QAD = 0.495 m3 /s

and QAD = 0.188 m3 /s

and substituting into Equation 9 gives QBD = −0.295 m3 /s

and

QBD = 0.012 m3 /s

Since only real and positive flows are consistent with the assumed flow directions, QAD = 0.188 m3 /s, QBD = 0.012 m3 /s, and the assumed flow directions are validated. Therefore, based on these results, flow out of reservoir at A = QAD − QCA = 0.188 − 0.063 = 0.125 m3 /s flow into reservoir at B = QCB − QBD = 0.137 − 0.012 = 0.125 m3 /s The flows into and out of the reservoirs (0.125 m3 /s) are equal as expected. 2.40. From the given data: zE = zF = zG = 100 m, zA = zB = 0 m, and VAC = VBC = 2.5 m/s. Since pipes AC and BC are identical, they have the same flow. If Q is the flow through the pump, then Q π = VAC AAC = 2.5 (0.5)2 = 0.4909 m3 /s ⇒ Q = 0.9817 m3 /s 2 4 and so VCD = Q/ACD = 2.222 m/s. If hp is the head added by the pump, applying the energy equation gives 2.52 0.0055(100) 2.52 0.0050(300) (2.222)2 +0− − + hp − 2(9.81) 0.50 2(9.81) 0.75 2(9.81) Q2DE Q2DE 0.0060(500) [π ]2 = [ ]2 + 100 0.30 2(9.81) (0.3)2 2(9.81) π (0.3)2

0+

4

4

which simplifies to hp = 112.2Q2DE + 100.5

27

(1)

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For lines F and G, Q2DF Q2DF 0.0060(400) [π [ ]2 + 100 + ]2 = 0.25 2(9.81) 4 (0.25)2 2(9.81) π4 (0.25)2 Q2DG Q2DG 0.0060(500) + 100 + [ [ ]2 ]2 0.30 2(9.81) π4 (0.30)2 2(9.81) π4 (0.30)2 which simplifies to QDF = 0.7074QDG

(2)

Since E and G are identical pipes, then QDE = QDG

(3)

0.9817 = QDE + QDF + QDG

(4)

and by continuity Combining Equations 2 to 4 gives 0.9817 = QDE + 0.7074QDE + QDE which yields QDE = 0.3626 m3 /s. Substituting into Equation 1 gives hp = 112.2(0.3626)2 + 100.5 = 115.5 m Therefore, the pressure difference across the pump is 9.79(115.6) = 1128 kPa and the power consumed by the pump is γQhp /η = (9.79)(0.9817)(115.6)/0.76 = 1461 kW . 2.41. Using the Darcy-Weisbach head loss equation, r=

fL fL fL = = 0.0826 5 2 2 2 2gA D 2(9.81)(πD /4) D D

For ductile-iron, ks = 0.26 mm, and for fully turbulent flow the friction factor, f , is estimated using ( ) 1 ks /D √ = −2 log 3.7 f Using the given values of D and L in Equations 2 and 2 yields Pipe

f

AB BC CD DE EF FA BE

0.0190 0.0186 0.0210 0.0198 0.0190 0.0198 0.0183

L (m) 1000 750 800 700 900 900 950

28

D (mm) 300 325 200 250 300 250 350

r 646 318 4337 1172 581 1507 273

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Figure 2.2: Final Flows 2.42. The final flows using the Hardy Cross method are shown in Figure 2.2. The results of intermediate calculations will depend on initial flow assumptions. Since the pressure at P is 500 kPa, and the network is on flat terrain, then the head at other intersections can be calculated by accounting for the frictional head losses. For each pipe, the pressure, p, is given by p = 500 ± γhf kPa (1) where hf is the frictional head loss, and the ± accounts for whether the flow is towards P (+) or away from P (−). The frictional head loss, hf , is given by the Darcy-Weisbach equation as L Q2 hf = f D 2gA2 Since D = 0.3 m and A = πD2 /4 = π(0.3)2 /4 = 0.0707 m2 , then hf = f

L Q2 = 33.99f LQ2 0.3 2(9.81)(0.0707)2

(2)

where L is given for each pipe, Q has been calculated, and f can be calculated using the Swamee-Jain equation [ ] 1 k /D 5.74 √ = −2 log s + (3) 3.7 f Re0.9 where ks /D = 0.26/300 = 0.000867, and Re is Re =

ρV D ρQD (998)(Q)(0.3) = = = 4.23 × 106 Q µ Aµ (0.0707)(1.00 × 10−3 )

(4)

Combining Equations 3 and 4 gives the following expression for f in terms of the flow rate, Q, [ ] 1 0.000867 5.74 √ = −2 log + 3.7 (4.23 × 106 Q)0.9 f

29

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which simplifies to f={

0.25

[

log 2.34 × 10−4 +

6.22×10−6 Q0.9

]}2

(5)

Combining Equations 2 and 5 gives the following expression for the frictional head loss in each pipe 8.50LQ2 hf = { [ (6) ]}2 −6 log 2.34 × 10−4 + 6.22×10 0.9 Q The pressure drop, ∆p, in each pipe is equal to γhf , = 9.79hf kPa, and therefore 83.2LQ2 ∆p = { [ ]}2 −6 log 2.34 × 10−4 + 6.22×10 Q0.9 Calculating the pressure changes in each pipe, and adding/subtracting from the pressure at P (= 500 kPa) yields the following results: Pipe PB PF PH PE AB BC FI GH

L (m) 100 150 100 150 150 150 100 150

Q (m3 /s) 0.005 0.008 0.043 0.040 0.040 0.045 0.047 0.030

∆p (kPa) 0.02 0.08 1.28 1.67 1.67 2.10 1.52 0.96

Node B F H E A C I G

Pressure (kPa) 500.0 499.9 501.3 501.7 501.6 497.9 502.8 502.2

2.43. This problem can be solved using the Hardy Cross method. From the given data, ks = 0.05 mm and D = 1120 mm. Hence, f=[

0.25 ( )]2 = [

(

0.25

/D log ks3.7 log 4.464×10 3.7 π 2 π A = D = (1.12)2 = 0.9852 m2 4 4

−5

)]2 = 0.01033

and the Hardy Cross parameters corresponding to the Darcy-Weisbach equation are n = 2 and L 1 L 1 r=f = (0.01033) = 0.0004843L D 2gA2 1.12 2(9.81)(0.9852)2 Using this formulation with the given pipe lengths, the r values for each of the pipes are: CD = 2.42, DE = 5.81, EF = 4.36, FC = 2.91, FG = 3.39, GH = 3.87, and HC = 4.84. The intermediate flow values will depend on the initial assumed flow distribution. The final flow values are shown in Figure 2.3.

30

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48000 m3/d 8000 m3/d 8000 m3/d C

H

D 14700

8000 m3/d

8000 m3/d

4400

F

5700

3600

G

8000 m3/d

13700

11600

8000 m3/d E

2300

Figure 2.3: Flow distribution. 2.44. The specific speed in SI units is given by Equation 2.122 as 1

ns =

ωQ 2

3

(ghp ) 4

and the specific speed in English units is given by Equation 2.123 as 1

Ns =

ωQ 2 3

hp4 The unit conversions are as follows: 2π ω (rad/s) = ω (rpm) = 0.1047ω (rpm) 60 3.785 × 10−3 Q (m3 /s) = Q (gpm) = 6.308 × 10−5 Q (gpm) 60 hp (m) = 0.3048hp (ft) Taking g = 9.81 m/s2 gives ns in English units as 1

ns =

(0.1047ω)(6.308 × 10−5 Q) 2 3

3

1

= 0.000366

(9.81) 4 (0.3048hp ) 4

ωQ 2 3

= 0.000366Ns

hp4

which can also be written as Ns = 2730ns Therefore, the constant used to convert the specific speed in SI units to the specific speed in English units is 2730 . 2.45. The synchronous speed is given by Equation 2.124 as synchronous speed =

3600 No. of pairs of poles

The maximum speed occurs when there is 1 pair of poles, which gives a speed of 3600 rpm .

31

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2.46. From the affinity laws: ( ) ( ) Q Q = ωD3 1 ωD3 2

( and

hp ω 2 D2

)

( = 1

hp ω 2 D2

) 2

these equations can also be written as Q1 ω1 D13 = Q2 ω2 D23

and

hp1 ω2 D2 = 12 12 hp2 ω2 D2

Since the power, P is given by P = γQhp then γQ1 hp1 P1 Q1 hp1 = = · P2 γQ2 hp2 Q2 hp2 ω1 D13 ω12 D12 · = ω2 D23 ω22 D22 ω 3 D5 = 13 15 ω2 D2 The affinity law for power can therefore be written as P1 P2 = 3 5 ω13 D15 ω2 D2 2.47. From the given data: ω1 = ω2 = 1200 rpm, D1 = 500 mm, D2 = 250 mm, Q1 = 250 L/s, hp1 = 63.7 m, and η1 = 81%. Applying the affinity relationships given by Equation 2.125 requires that [ ] [ ] Q1 Q2 = (ω1 )(D1 )3 (ω2 )(D2 )3 [ ] [ ] 250 Q2 = (1200)(500)3 (1200)(250)3 which yields Q2 = 31 L/s. Also, [

] [ ] hp1 hp2 = (ω1 )2 (D1 )2 (ω2 )2 (D2 )2 [ ] [ ] hp2 63.7 = (1200)2 (500)2 (1200)2 (250)2

which yields hp2 = 15.9 m. Therefore, the best-efficiency operating point for a 250-mm pump in the given homologous series is at Q = 31 L/s and hp = 50.3 m . The efficiency at this operating point can be estimated using Equation 2.129 which gives ( )1 1 − η2 D1 4 = 1 − η1 D2 ( )1 1 − η2 500 4 = 1 − 0.81 250

32

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which yields η2 = 0.77. The efficiency can also be estimated by Equation 2.130 which gives ( )0.32 0.94 − η2 Q1 = 0.94 − η1 Q2 ( ) 0.94 − η2 250 0.32 = 0.94 − 0.81 31 which yields η2 = 0.69. Based on these results, it is estimated that the 250-mm pump will have an efficiency somewhere in the range of 69–77% . 2.48. Since the pump lifts the water over a height of 1.5 + 2 = 3.5 m, and the friction loss is given by the Darcy-Weisbach equation, then the energy equation is hp − f

L V2 = 3.5 D 2g

which can be written in the conventional form hp = 3.5 + f

L Q2 D 2gA2

(1)

The friction factor, f , can be approximated by the Swamee-Jain equation (smooth pipe, ks ≈ 0) ( ) 1 5.74 √ = −2 log10 (2) f Re0.9 where

VD QD = ν νA −6 In this case, D = 300 mm = 0.3 m, ν = 1.00 × 10 m2 /s (at 20◦ C), and A = π(0.3)2 /4 = 0.0707 m2 . Therefore, Re =

Re =

Q(0.3) = 4.24 × 106 Q (1.00 × 10−6 )(0.0707)

(3)

Combining Equations 2 and 3 gives [ ] [ ] 1 5.74 6.23 × 10−6 √ = −2 log10 = −2 log10 (4.24 × 106 Q)0.9 Q0.9 f or

[

(

f = 2 log10

Q0.9 6.23 × 10−6

)]−2 (4)

The system curve (Equation 1) can be written as hp = 3.5 + f

300 Q2 0.3 2(9.81)(0.0707)2

or hp = 3.5 + 1.02 × 104 f Q2

33

(5)

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where Q is in m3 /s. For Q in L/s, Equation 5 becomes

The pump curve is given as

hp = 3.5 + 1.02 × 10−2 f Q2

(6)

hp = 6 − 6.67 × 10−5 Q2

(7)

and solving Equations 6 and 7 simultaneously gives 6 − 6.67 × 10−5 Q2 = 3.5 + 1.02 × 10−2 f Q2 which can be re-arranged and put in the form √ 2.5 Q= 6.67 × 10−5 + 1.02 × 10−2 f

(8)

The flow is given by the simultaneous solution of Equations 4 and 8 which yields f = 0.0049 and Q = 146 L/s Since this flow is within 10% of the desired flow of 150 L/s, the pump is adequate . 2.49. The system curve is given by 0−f

L V2 L V2 + hp = 61 → hp = 61 + f D 2g D 2g

For DIP it can be assumed that ks = 0.12 mm, and for the system: π 2 D = 1.169 m2 4 Q = 1.711 m/s V = A ν = 1 × 10−6 m2 /s VD (1.169)(1.220) = Re = = 2.087 × 106 ν 1 × 10−6 0.25 f=[ ( )]2 = 0.0128 ks log 3.7D + 5.74 0.9 Re A=

Substituting the above results into Equation 1 gives 3200 (1.711)2 = 61 + 5.01 = 66.01 m 1.22 2(9.81) power delivered to water = γQhp = (9.79)(2)(66.01) = 1292 kW 1292 power consumed by pump = = 1520 kW 0.85 energy consumed in 8 h = 1520 × 8 = 12160 kWh hp = 61 + 0.0128

cost of electricity = 12160 × 0.06 = $730 per day

34

(1)

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When the system is operating in power-generation (turbine) mode, ht = 61 − 5.01 = 55.99 m power extracted from water = γQht = (9.79)(2)(55.99) = 1096 kW power delivered as electricity = 0.90(1096) = 986 kW energy delivered in 8 h = 986 × 8 = 7888 kWh revenue from sale of electricity = 7888 × 0.12 = $947perday Based on these results, the profit is $947 − $730 = $217/day = $79,200/year When the elevation difference is 65 m and hp = 80−3.5Q2 , then the combination of the pump curve and system curve gives 0−f

L Q2 + 80 − 3.5Q2 = 65 m D 2gA2

and for fully turbulent flow, 0.25 0.25 f=[ ( )]2 = 0.0119 )]2 = [ ( ks 0.12 log 3.7D log 3.7(1220) Substituting into the pump/system curve equation gives −(0.0119)

3200 Q2 + 80 − 3.5Q2 = 65 1.22 2(9.81)(1.169)2

which yields Q = 3.20 m3 /s hp = 80 − 3.5Q2 = 80 − 3.5(3.20)2 = 44.16 m hf = 11.92 m ht = 65 − 11.92 = 53.08 m γQhp cost of electricity = 8 × × 0.06 = $781 per day 0.85 revenue = γQht × 0.9 × 8 × 0.12 = $1437 per day Based on these results, the profit is $1437 − $781 = $656/day = $239,440/year . 2.50. From the given data: L = 300 m, ks = 0.26 mm, D = 800 mm, A = πD2 /4 = π(0.8)2 /4 = 0.503 m2 , ks /D = 0.26/800 = 0.000325, T = 20◦ C, ρ = 998.2 kg/m3 , γ = 9.789 kN/m3 , µ = 1.002 × 10−3 N·s/m2 , z1 = −5 m, zp = 0.5 m, z2 = 4 m, and pump characteristic curve given by hp = 12 − 0.1Q2 (1) Energy equation between well and reservoir hp = 9 + hf = 9 + f

35

L Q2 D 2gA2

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For fully turbulent flow, and ks /D = 0.000325, then f = 0.015 and the head loss equation becomes 300 Q2 hp = 9 + 0.015 = 9 + 1.13Q2 0.8 2(9.81)(0.503)2 Combining the energy equation with the pump equation gives 9 + 1.13Q2 = 12 − 0.1Q2 which gives Q = 1.56 m3 /s Check the assumption of f = 0.015, Re =

ρ(Q/A)D (998.2)(1.56/0.503)(0.8) = = 2.47 × 106 µ 1.002 × 10−3

The Swamee-Jain equation confirms that f = 0.015 and therefore Q = 1.56 m3 /s Specific speed, Ns , is given by

1

Ns =

ωQ 2

(2)

3 4

hp where

Q = 1.56 m3 /s = 24, 727 gpm hp = 12 − 0.1(1.56)2 = 11.8 m = 38.6 ft ω = 1200 rpm Substituting into Equation 2 gives 1

Ns =

(1200)(24727) 2 3

= 12,185

(38.6) 4

Therefore the type of pump required is an axial flow pump . 2.51. From the given data: ∆z = 3 m + 19.3 m = 22.3 m, L = 100 m, D = 50 mm = 0.05 m, A = π/4D2 = 0.00196 m2 , and Q = 370 L/min = 0.00617 m3 /s. For galvanized iron, ks = 0.15 mm. Taking Q = 370 L/min as the operating point gives V = Q/A = 3.15 m/s, Re = V D/ν = 1.58 × 105 (where ν = 10−6 m2 /s at 20◦ C) and ( ) ( ) 1 ks 2.51 0.15 2.51 √ = −2 log √ √ + = −2 log + 3.7D Re f 3.7(50) 1.58 × 105 f f which gives f = 0.0270. The energy equation for the system is given by [ ] fL Q2 Q2 0− + 1.8 + h = ∆z + p D 2gA2 2gA2 [ ] (0.0270)(100) Q2 Q2 0− + 1.8 + h = 22.3 + p 0.05 2(9.81)(0.001962 ) 2(9.81)(0.001962 )

36

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which simplifies to hp = 22.3 + 7.54 × 105 Q2 This equation is appropriate for hp in meters and Q in m3 /s. For hp in ft and Q in gpm, the system equation becomes ) ( hp = 22.3 + 7.54 × 105 (Q × 6.309 × 10−5 )2 3.281 which gives hp = 73.2 + 0.00985Q2 Plotting the system curve on the pump performance curve indicates that 7-inch pump will be required, with a maximum flow rate of around 112 gpm = 424 L/min. The next smaller pump size of 6.5-inch will yield a maximum flow of around 95 gpm = 360 L/min, which is slightly less than the minimum requirement of 370 L/min. Under the pump operating conditions, NPSHR = 15 ft = 4.57 m, pv = 2.34 kPa (at 20◦ C), p0 = 101 kPa, γ = 9.79 kPa, Q = 424 L/min = 0.00707 m3 /s, V = Q/A = 3.61 m/s, and the NPSH requirement is that f ∆zs V 2 pv p0 − ∆zs − − γ D 2g γ 101 0.0270∆zs 3.612 2.34 4.57 = − ∆zs − − 9.79 0.05 2(9.81) 9.79

NPSHR =

which gives ∆zs = 4.05 m. Since the water is 3 m below ground and the pump must be placed 4.05 m above the water, the pump must be placed a maximum of 4.05 m − 3 m = 1.05 m above ground . 2.52. The system curve is derived from the energy equation: 0−

V2 V2 − hf + hp − = 10 2g 2g

(1)

where the V 2 /2g terms account for the entrance and exit losses. The head loss, hf , is given by the Darcy-Weisbach equation as hf = f

L V2 L Q2 =f D 2g D 2gA2

The energy equation can therefore be written as ] [ L Q2 hp = 10 + 2 + f D 2gA2 For ductile iron, ks = 0.26 mm, therefore ks /D = 0.26/100 = 0.0026. Assuming the pipe is hydraulically rough, [ ] [ ] ks /D 0.0026 1 √ = −2 log = −2 log = 6.306 3.7 3.7 f

37

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which leads to f = 0.0251 For the pipeline, L = 104 m, D = 0.100 m, A = πD2 /4 = 0.00785 m2 , and therefore the system curve given by Equation 1 becomes [ ] [ ] L Q2 104 Q2 hp = 10 + 2 + f = 10 + 2 + (0.0251) D 2gA2 0.1 2(9.81)(0.00785)2 = 10 + 23250Q2 This equation is for Q in m3 /s. For Q in L/s, this equation becomes hp = 10 + 0.0233Q2 Solving the system curve and the pump curve yields 10 + 0.0233Q2 = 15 − 0.1Q2 which leads to Q = 6.37 L/s and hp = 10.9 m Under this operating condition, the (given) required net positive suction head, NPSHR , is 1.5 m. Putting NPSHR = NPSHA requires that 1.5 =

pv p0 − ∆zs − hL − γ γ

(2)

In this case, p0 = 101 kPa, γ = 9.79 kN/m3 , pv = 2.34 kPa (at 20◦ C), and ] [ L Q2 hL = 1 + f D 2gA2 where L = ∆zs + 1, f = 0.0251, D = 0.1 m, Q = 0.00637 m3 /s, A = 0.00785 m2 , which gives [ ] (∆zs + 1) 0.006372 hL = 1 + 0.0251 0.1 2(9.81)(0.00785)2 = 0.0420 + 0.00843∆zs (3) Combining Equations 2 to 3 gives 1.5 =

101 2.34 − ∆zs − 0.0420 − 0.00843∆zs − 9.79 9.79

which leads to ∆zs = 8.40 m 2.53. From the given data: L = 30 m, D = 0.15 m, A = πD2 /4 = π(.15)2 /4 = 0.01767 m2 , and specific speed = 3000.

38

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(a) Energy equation:

(

L zA + hp = zF + f + KA + KB + KC + KD + KE + KF D

)

Q2 2gA2

For PVC pipe, ks ≈ 0, and the friction factor is a function of the (unknown) Reynolds number according to the relation ( ) 1 2.51 √ = −2 log √ (1) f Re f Substituting known values into the energy equation gives ( ) 30 Q2 + 1 + 0.9 + 0.2 + 0.9 + 0.9 + 1 0 + hp = 10 + f 0.15 2(9.81)(0.01767)2 which simplifies to hp = 10 + 163.2Q2 (4.9 + 200f ) (b) The pump performance curve is given by hp = 20 − 4713Q2 Simultaneous solution of the energy equation and the pump performance curve gives 10 + 163.2Q2 (4.9 + 200f ) = 20 − 4713Q2 √

which leads to

10 (2) 5513 + 32640f This equation must be solved simultaneously with the friction factor equation (Equation 1), where Re is given by Q=

Re =

ρQD (998.2)(Q)(0.15) = = 8.46 × 106 Q Aµ (0.01767)(1.002 × 10−3 )

which combined with Equation 2 gives Re = 2.68 × 107 (5513 + 32640f )− 2 1

Substituting Equation 3 into Equation 1 gives the following implicit equation for f , √ [ ] 1 5513 + 32640f −8 √ = −2 log 9.37 × 10 f f which yields f = 0.014 from which Equation 2 gives Q = 0.0409 m3 /s and V =

Q 0.0409 = = 2.31 m/s A 0.01767

39

(3)

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(c) The available net positive suction head, NPSHA , is given by NPSHA =

p0 pv 101 2.34 − ∆zs − hL − = −3−0− = 7.08 m γ γ 9.79 9.79

Since the required net positive suction head, NPSHR at the pump operating point is given (by the pump manufacturer) as 3.0 m, and NPSHA (= 7.08 m) > NPSHR (= 3.0 m), cavitation is not expected to occur in the pump. (d) According to the affinity laws, Q1 = and hp1 =

ω1 800 Q2 = Q2 = 0.5Q2 ω2 1600

ω12 8002 hp2 = 0.25hp2 h = p2 16002 ω22

Using the affinity relations in the pump characteristic curve gives 0.25hp2 = 20 − 4713(0.5Q2 )2 which leads to hp2 = 80 − 4713Q22 2.54. (a) From the given data: D = 6 cm = 0.06 m, L = 107 m, ks = 0.01 mm, z1 = 10.00 m, z2 = 15.00 m, and ∆z = z2 − z1 = 15.00 m − 10.00 m = 5.00 m. The system curve is given by f L Q2 hp = ∆z + (1) D 2gA2 where π π (2) A = D2 = (0.06)2 = 0.002827 m2 4 4 and the Swamee-Jain formula gives f=[

( log

where

0.25 ks 3.7D

+

5.74

)]2

(3)

Re0.9

QD VD = ν Aν m2 /s (at 20◦ C) and combining Equations 2 to 4 yields Re =

Taking ν = 1.00 × 10−6

f=[

0.25

( log

0.01 3.7(60)

+

(

(4)

)]2 5.74

Q×0.06 0.002827×1.00×10−6

)0.9

which simplifies to f=

0.25 [log (4.505 ×

10−5

40

+ 1.461 × 10−6 Q−0.9 )]2

(5)

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Combining Equations 1 and 5, and substituting other known data yields hp = 5 +

107 Q2 [log (4.505 × 10−5 + 1.461 × 10−6 Q−0.9 )]2 0.06 2(9.81)(0.002827)2 0.25

which simplifies to hp = 5 +

2.84 × 106 Q2 [log (4.505 × 10−5 + 1.461 × 10−6 Q−0.9 )]2

(6)

These system curve is plotted on the pump curve in Figure 2.4 using the following tabulated values derived from Equation 6:

Figure 2.4: System Curve on Pump Curve. Q (gpm) 0 20 40 60 80 100 120

Q (m3 /s) 0 0.001262 0.002524 0.003785 0.005047 0.006309 0.007571

hp (m) 5.00 5.44 6.53 8.18 10.38 13.11 16.35

hp (ft) 16.4 17.9 21.4 26.9 34.1 43.0 53.7

The intersection of the system curve with the pump curve indicates that size A pump should be used. This yields an operating point of Q = 390 L/min (= 103 gpm) which is slightly greater than the desired flow rate of 380 L/min. The maximum flow rate can be throttled down with a valve. The head added by the pump, hp , at the operating point is 13.6 m (= 44.5 ft).

41

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(b) The efficiency, η, of the pump at the operating point is approximately 64%, and hence the power, P , of the required motor is given by P =

γQhp (9.79)(390 × 10−3 /60)(13.6) = = 1.35 kW η 0.64

(= 1.8 HP)

(c) The available net positive suction head, NPSHA , can be estimated by (neglecting pipeentrance losses) ( ) p0 pv p0 f V2 pv NPSHA = − ∆zs − hL − = − 1+ ∆zs − (7) γ γ γ D 2g γ From the pump curve at the operating point, NPSHA = 6.5 ft = 1.98 m, p0 = 101 kPa, γ = 9.79 kN/m3 , pv = 2.34 kPa (at 20◦ C), Q = 390 L/min = 0.00650 m3 /s, D = 0.06 m, A = 0.002827 m2 , V = Q/A = 2.30 m/s, and Equation 5 gives f = 0.0178. Substituting these data into Equation 7 gives [ ] 0.0178 2.302 2.34 101 − 1+ ∆zs − 1.98 = 9.79 0.06 2(9.81) 9.79 which yields ∆zs = 7.50 m. Therefore, the pump can be placed up to 7.50 m above the pond. 2.55. If Hp is the head added by the pump system, and Q is the flow through the system, then for n pumps in series Hp = 30 − 0.05Q2 n or Hp = 30n − 0.05nQ2 For n pumps in parallel,

( Hp = 30 − 0.05

Q n

)2

or Hp = 30 −

0.05 2 Q n2

2.56. The flow rate is given by the simultaneous solution of the system curve and the pump curve, hence 15 + 0.03Q2 = 20 − 0.08Q2 which gives Q = 6.74 L/s The pump curve for two identical pumps in series is given by Hp = 20 − 0.08Q2 2 or Hp = 40 − 0.16Q2

42

(1)

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Solving Equation 1 with the system curve gives 15 + 0.03Q2 = 40 − 0.16Q2 which leads to Q = 11.5 L/s The pump curve for two identical pumps in parallel is given by ( Hp = 20 − 0.08

Q 2

)2

or Hp = 20 − 0.02Q2

(2)

Solving Equation 2 with the system curve gives 15 + 0.03Q2 = 40 − 0.02Q2 which leads to Q = 10 L/s 2.57. From the given data: L = 20 km = 20000 m, D = 1120 mm = 1.120 m, ks = 0.05 mm, zA = zB = 5 m, hA = 2 m, zC = 15 m, Q = 48000 m3 /d = 0.556 m3 /s, and pC = 448 kPa. At T = 20◦ C, ν = 1 × 10−6 m2 /s and γ = 9.79 kN/m3 . (a) Using the given data, π 2 π D = (1.120)2 = 0.9852 m2 4 4 Q 0.556 V = = = 0.564 m/s A 0.9852 VD (0.564)(1.120) Re = = = 6.317 × 105 ν 1.0 × 10−6 ks 0.05 = = 4.464 × 10−5 D 1120 0.25 0.25 f=[ ( )]2 = [ ( )]2 = 0.01333 −5 /D 4.464×10 5.74 log ks3.7 + 5.74 log + 3.7 (6.317×105 )0.9 Re0.9 2 2 LV 20000 0.564 = (0.01333) = 3.858 m hf = f D 2g 1.120 2(9.81) A=

The system curve derived from the energy equation as follows: pA pC + zA + hp − hf = + zC γ γ 448 2 + 5 + hp − 3.858 = + 15 9.79

43

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which yields hp = 57.62 m. For N pumps in parallel the pump-system performance is given by ( )2 Q −8 hp = 65 − 7.6 × 10 N which gives

√ N=

7.6 × 10−8 Q= 65 − hp

√

7.6 × 10−8 (48000) = 4.87 ≈ 5 65 − 57.62

Hence 5 pumps are needed to deliver the required flow at the required pressure. (b) To determine the actual operating point, the system curve must be expressed in terms of the flow Q. Assuming fully turbulent flow, 0.25 ( )]2 = [

f=[ log hf = f

ks /D 3.7

( log

0.25 4.464×10−5 3.7

)]2 = 0.01033

L Q2 20000 Q2 = (0.01033) = 9.686Q2 2 D 2gA 1.12 2(9.81)(0.9852)2

The system curve is derived from the energy equation as follows: pA pC + zA + hp − hf = + zC γ γ 448 2 + 5 + hp − 9.686Q2 = + 15 9.79 which simplifies to hp = 53.76 + 9.686Q2 This equation is for Q in m3 /s. For Q in m3 /d, ( hp = 53.76 + 9.686

Q 86400

)2

= 53.76 + 1.298 × 10−9 Q2

(1)

For 5 pumps, the performance curve of the pump system is given by hp = 65 − 7.6 × 10−8

(

Q 5

)2

= 65 − 3.04 × 10−9 Q2

(2)

Combining Equations 1 and 2 gives 65 − 3.04 × 10−9 Q2 = 53.76 + 1.298 × 10−9 Q2 which yields 50,900 m3 /d . 2.58. (a) The energy equation between the supply reservoir and any delivery location, X, is given by 2 2 V2 fAB LAB VAB fBX LBX VBX pX 3− + hp − = + X + zX DAB 2g DBX 2g γ 2g

44

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which can be conveniently written as ] [ ] [ 2 2 VX2 fAB LAB VAB fBX LBX VBX pX hp = −3 + + + + zX DAB 2g DBX 2g γ 2g

(1)

where the first term in brackets on the right-hand side is the same for both pipe destinations. For pipe AB: QAB = 27 L/s = 0.027 m3 /s LAB = 1050 m DAB = 0.200 m π 2 π AAB = DAB = (0.200)2 = 0.03142 m2 4 4 0.027 QAB = = 0.859 m/s VAB = AAB 0.03142 VAB DAB (0.859)(0.200) ReAB = = = 171800 ν 10−6 0.25 0.25 fAB = [ ( )]2 [ ( )]2 = 0.02248 5.74 0.26 ks 5.74 + log 0.9 log 3.7DAB + 3.7(200) (171800) Re0.9 AB which gives

[

] [ ] 2 (0.02248)(1050) 0.8592 fAB LAB VAB −3 = − 3 = 1.439 m DAB 2g (0.200) 2(9.81)

(2)

For pipe BC: QBC = 12 L/s = 0.012 m3 /s LBC = 2800 m DBC = 0.150 m π 2 π = (0.150)2 = 0.01767 m2 ABC = DBC 4 4 0.012 QBC VBC = = = 0.679 m/s ABC 0.01767 VBC DBC (0.679)(0.150) ReBC = = = 101850 ν 10−6 0.25 fBC = [ ( )]2 = 0.0246 0.26 5.74 log 3.7(150) + (101850) 0.9 zC = 2 m which gives [

] [ ] 2 (0.0246)(2800) 0.6792 pC VC2 350 0.6792 fBC LBC VBC + + + zC = + + +2 DBC 2g γ 2g (0.150) 2(9.81) 9.79 2(9.81) = 48.567 m

45

(3)

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For pipe BD: QBD = 15 L/s = 0.015 m3 /s LBD = 2500 m DBD = 0.150 m π 2 π ABD = DBD = (0.150)2 = 0.01767 m2 4 4 QBD 0.015 VBD = = 0.849 m/s = ABD 0.01767 (0.849)(0.150) VBD DBD = = 127350 ReBD = ν 10−6 0.25 fBD = [ ( )]2 = 0.0242 0.26 5.74 log 3.7(150) + (127350) 0.9 zD = 5 m which gives [

] [ ] 2 fBD LBD VBD pD V2 (0.0242)(2500) 0.8492 350 0.8492 + + D + zD = + + +5 DBD 2g γ 2g (0.150) 2(9.81) 9.79 2(9.81) = 55.605 m

(4)

Assessing the results in Equations 1 to 4, it is apparent that the required conditions at location D will yield the maximum value of hp such that hp = [1.439] + [55.605] = 57.04 m and the required pump power, P , delivered to the water is given by P = γQAB hp = (9.79)(0.027)(57.04) = 15.1 kW (b) If n pumps are used in series and the pump size, D, can be adjusted to meet the desired operating conditions, then hp = 0.455D + 4000Q2AB n 57.04 = 0.455D + 4000(0.027)2 n which can be put in the form 125.4 − 6.4 n Assuming different values for n yields the following results D=

n 1 2 3 4

46

D (cm) 119 69 48 38

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Since the pump manufacturer requires that 40 cm < D < 50 cm, use 3 pumps with D = 48 cm . 2.59. At ω = 600 rpm, the operating point is determined by simultaneous solution of the pump and system curves, 6 − 0.05Q2 = 3 + 0.042Q2 which yields Q = 5.7 m3 /min . From the given data: ω1 = 600 rpm, ω2 = 1200 rpm, and the affinity laws (Equation 2.148) give that Q1 =

ω1 600 Q2 = Q2 = 0.5Q2 ω2 1200

h1 =

ω12 6002 h = h2 = 0.25h2 2 12002 ω22

Since the performance curve of the pump at speed ω1 is given by h1 = 6 − 0.05Q21 then the performance curve at speed ω2 is given by 0.25h2 = 6 − 0.05(0.5Q2 )2 which leads to h2 = 24 − 0.05Q22 The new operating point is determined by simultaneous solution of the pump and system curves, 24 − 0.05Q2 = 3 + 0.042Q2 which yields Q = 15.1 m3 /min .

47

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Chapter 3

Design of Water-Distribution Systems 3.1. (a) For geometric growth: ∫

dP = P

∫

dP = k1 P dt k1 dt

=⇒ ln P = k1 t + C

Which gives P = P0 ek1 t (b) For arithmetic growth: ∫

dP = k2 dt

∫ dP =

k2 dt

=⇒ P = k2 t + P0

(c) For declining growth: ∫

dP Psat − P

dP = k3 (Psat − P ) ∫ dt = k3 dt =⇒ ln(Psat − P ) = −k3 t − C ′

which gives P = Psat − Ce−k3 t ′

where C = e−C . 3.2. (a) By graphical extension P2030 = 100, 000 (b) For arithmetic growth: P = kt + P0 where k=

P1990 − P1980 61000 − 52000 = = 900 10 10

Therefore P2030 = 900t + P1990 = 900(40) + 61000 = 97, 000

49

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(c) For geometric growth: P = P0 ekt Therefore P1990 = P1970 ek(20) 61000 = 40000e20k k = 0.021 and hence P2030 = P1990 ek(40) = 61000e(0.021)(40) = 141, 298 (d) For declining growth:

P = Psat − Ce−kt

(1)

where Psat = 100,000 and 1970 : t = 0, P = 40000 1990 : t = 20, P = 61000 Substituting 1970 population into Equation 1 gives 40000 = 100000 − Ce−k(0) = 100000 − C which gives C = 60, 000 Substituting 1990 population into Equation 1 gives 61000 = 100000 − 60000e−k(20) which gives k = 0.0215 Equation 1 can now be used to predict the 2030 population (t = 60) as P2030 = 100000 − 60000e−0.0215(60) = 83, 483 (e) Equations 3.11 and 3.12 give the logistic-curve parameters a and b as Psat − P0 100 − 40 = = 1.5 P0 40 [ ] [ ] 1 P0 (Psat − P1 ) 1 40(100 − 52) b= ln = ln = −0.0486 ∆t P1 (Psat − P0 ) 10 52(100 − 40)

a=

The logistic curve for predicting the population is given by Equation 3.9 as P =

Psat 100, 000 = bt 1 + ae 1 + 1.5e−0.0486

t

In 2030, t = 60 years and the population given by Equation 3.19 is P = 92, 487 people

50

(2)

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3.3. From the given data: P0 = 13,000, P1 = 125,000, and P2 = 300,000. The saturation population, Psat , is given by Equation 3.10 as Psat =

2P0 P1 P2 − P12 (P0 + P2 ) 2(13)(125)(300) − (13)2 (13 + 300) = = −78 thousand people (13)(300) − (125)2 P0 P2 − P12

Since the calculated value of Psat is negative, estimation of the saturation population using Equation 3.10 is not possible . 3.4. average daily demand = 580(100000) = 58, 000, 000 L/d = 0.671 m3 /s maximum daily demand = 1.8(0.671) = 1.21 m3 /s = 1.04 × 108 L/d maximum hourly demand = 3.25(0.671) = 2.18 m3 /s = 1.89 × 108 L/d 3.5. The NFF can be estimated by Equation 3.20 as NFFi = Ci Oi (X + P )i where the construction factor, Ci , is given by Ci = 220F

√

Ai

For the 5-story building, F = 1.0 (Table 3.3, Class 2 construction), and Ai = 5000 m2 , hence √ Ci = 220(1.0) 5000 = 16000 L/min where Ci has been rounded to the nearest 1000 L/min. The occupancy factor, Oi , is given by Table 3.4 as 0.85 (C-2 Limited Combustible, Office), (X + P )i can be estimated by the median value of 1.4, and hence the needed fire flow, NFF, is given by NFFi = (16000)(0.85)(1.4) = 19000 L/min This flow must be maintained for a duration of 5 hours (Table 3.6), hence the required volume, V , of water is given by V = 19000 × 5 × 60 = 5.7 × 106 L = 5700 m3 3.6. The needed fire flow for any building should not exceed 45,000 L/min for a duration of 10 hours . 3.7. The (low-lift) supply pumps and the water-treatment plant should be designed for a capacity equal to the maximum daily demand (Table 3.7). With a demand factor of 1.8 (Table 3.2), the per capita demand on the maximum day is equal to 1.8 × 600 = 1080 L/day/capita. Since the population served is 200 000 people, the design capacity, Qdesign of the pumps and water-treatment plant is given by Qdesign = 1080 × 200000 = 2.16 × 108 L/d = 2.50 m3 /s

51

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The required fire flow, Qfire , is 28000 L/min = 0.467 m3 /s. According to Table 3.6 the fire flow must be able to be sustained for 7 hours . The volume, Vfire , required for the fire flow will be stored in the service reservoir, and is given by Vfire = 0.467 × 7 × 3600 = 11800 m3 The required flow rate in the distribution pipes is equal to the maximum daily plus fire demand or the maximum hourly demand, whichever is greater. maximum daily + fire demand = 2.50 + 0.467 = 2.98 m3 /s 3.25 × 2.50 = 4.51 m3 /s 1.80

maximum hourly demand =

where a demand factor of 3.25 has been assumed for the maximum hourly demand. The design capacity of the supply pipes in the distribution system should therefore be taken as 4.51 m3 /s . 3.8. According to Table 3.8, the minimum acceptable pressure under average daily demand conditions is 240 kPa . 3.9. From the given data: P = 200,000, per capita demand = 500 (L/d)/person = 5.79 × 10−6 (m3 /s)/person, and fire flow = 30,000 L/min = 0.5 m3 /s. (a) The required flow rate in the main distribution pipeline is the maximum daily demand plus fire demand or maximum hourly demand, whichever is greater. Taking a peaking factor of 1.8 for maximum daily demand and 3.25 for maximum hourly demand gives maximum daily demand + fire demand = 1.8(200000)(5.79 × 10−6 ) + 0.5 = 2.58 m3 /s maximum hourly demand = 3.25(200000)(5.79 × 10−6 ) = 3.76 m3 /s Therefore, the design capacity of the distribution pipeline should be 3.76 m3 /s . For a maximum allowable velocity of 1.5 m/s, the minimum required pipe diameter, D, requires that π 2 Q 3.76 D = = 4 V 1.5 which yields D = 1.78 m The pipe material to be used in ductile iron pipe (DIP) and a cover of 1.5 m would be appropriate. (b) For DIP with ks = 0.26 mm, D = 1.8 m, A = 2.544 m2 , Q = 3.80 m3 /s, V = Q/A = 1.494 m/s, and Re = V D/ν = (1.494)(1.80)/10−6 = 2.69 × 106 , [ f = −2 log

(

ks 5.74 + 3.7D Re0.9

)]−2

[ ( = −2 log

52

0.26 5.74 + 3.7(1800) (2.69 × 106 )0.9

)]−2 = 0.0134

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and, taking γ = 9.79 kN/m3 , the Darcy-Weisbach equation gives p1 p2 L V2 − =f γ γ D 2g 350 L (1.494)2 550 − = 0.0134 9.79 9.79 1.8 2(9.81) which yields L = 24.1 km For old DIP, take CH = 80 (Table 2.2) and the Hazen-Williams formula gives ( ) L V 1.85 p1 p2 − = hf = 6.82 1.17 γ γ D CH ( ) 550 350 L 1.494 1.85 − = 6.82 1.17 9.79 9.79 1.8 80 which yields L = 9.4 km The Hazen-Williams equation yields a significantly shorter distance. The Darcy-Weisbach equation is preferable since it covers all flow regimes, while the Hazen-Williams equation is restricted to a narrow range of flow conditions. Also, the value of CH might not be comparable to the value of ks used in the Darcy-Weisbach equation. 3.10. The required storage is the sum of three components: (1) volume to supply the demand in excess of the maximum daily demand; (2) fire storage; and (3) emergency storage. The volume to supply the peak demand can be taken as 25% of the maximum daily demand volume. Problem 3.7 gives the maximum daily flow rate as 2.50 m3 /s, hence the storage volume to supply the peak demand fluctuations over a day, Vpeak , is given by Vpeak = (0.25)(2.50 × 86400) = 54000 m3 The required fire flow, Qf , is calculated in Problem 3.7 as 0.467 m3 /s (= 28,029 L/min) and, according to Table 3.6, this fire flow must be maintained for at least 7 hours. The volume to supply the fire demand, Vfire , is therefore given by Vfire = (0.467 × 3600)(7) = 11768 m3 The emergency storage, Vemer , can be taken as the average daily demand, in which case Vemer = 200000 × 600 = 1.2 × 108 L = 120000 m3 The required volume, V , of the service reservoir is therefore given by V = Vpeak + Vfire + Vemer = 54000 + 11768 + 120000 = 185768 m3 The service reservoir should be designed to store about 185 800 m3 of water.

53

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3.11. (a) For line AB: population served = 50000 + 20000 = 70000 people average demand = (70000)(0.6) = 42000 m3 /d = 0.486 m3 /s maximum daily demand = (0.486)(1.8) = 0.875 m3 /s maximum hourly demand = (0.486)(3.25) = 1.58 m3 /s fire demand = 15000 + 10000 = 25000 L/min = 0.417 m3 /s maximum daily + fire demand = 0.875 + 0.417 = 1.29 m3 /s design flow = max(1.29, 1.58) = 1.58 m3 /s For line BC: population served = 20000 people average demand = (20000)(0.6) = 12000 m3 /d = 0.139 m3 /s maximum daily demand = (0.139)(1.8) = 0.250 m3 /s maximum hourly demand = (0.139)(3.25) = 0.452 m3 /s fire demand = 10000 L/min = 0.167 m3 /s maximum daily + fire demand = 0.250 + 0.167 = 0.417 m3 /s design flow = max(0.417, 0.452) = 0.452 m3 /s (b) For the steel transmission pipe (k = 1 mm, D = 1200 mm): π π A = D2 = 1.22 = 1.131 m2 4 4 1.58 = 1.40 m/s VAB = 1.131 VAB D (1.40)(1.2) ReAB = = = 1.68 × 106 ν ( 10−6 ) [ ] k/D 5.74 1/1200 5.74 1 √ = −2 log + = −2 log + 3.7 3.7 (1.68 × 106 )0.9 fAB Re0.9 AB fAB = 0.0191 0.452 VBC = = 0.400 m/s 1.131 VBC D (0.400)(1.2) = = 4.80 × 105 ReBC = ν ( 10−6 ) [ ] 1 k/D 5.74 1/1200 5.74 √ = −2 log + = −2 log + 3.7 3.7 (4.80 × 105 )0.9 fBC Re0.9 BC fBC = 0.0196 Applying the energy equation between A and B, neglecting minor losses, gives 2 V2 LAB VAB pB + hp = + AB + zB D 2g γ 2g 2 5000 1.40 550 1.402 0 − (0.0191) + hp = + + 20 1.2 2(9.81) 9.79 2(9.81)

0 − fAB

54

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which gives hp = 84.2 m. The required specific speed, ns , (where ω = 1200 rpm = 125.7 rad/s) is therefore given by 1

ns =

1

ωQ 2 3

=

(ghp ) 4

(125.7)(1.58) 2 3

(9.81 × 84.2) 4

= 1.03

This is a centrifugal pump . Applying the energy equation between B and C, neglecting minor losses, gives 2 V2 LBC VBC pC + hp = + BC + zC D 2g γ 2g 2 7000 0.400 480 0.4002 0 − (0.0196) + hp = + + 20 1.2 2(9.81) 9.79 2(9.81)

0 − fBC

which gives hp = 69.97 m. The required specific speed, ns , is therefore given by 1

ns =

1

ωQ 2

3

=

(ghp ) 4

(125.7)(0.452) 2 3

(9.81 × 69.97) 4

= 0.63

This is a centrifugal pump . (c) For the storage reservoir, taking the daily service storage as 25% of the maximum daily demand volume and noting that 10000 L/min fire flow is to be maintained for 3 h, and 15000 L/min for 4 h, gives service storage = (0.25)(1.8)(42000) = 18900 m3 fire storage = (10)(60)(3) + (15)(60)(4) = 5400 m3 emergency storage = 42000 m3 required storage = 18900 + 5400 + 42000 = 66300 m3 3.12. From the given data: Q = 4.67 L/s, L = 110 m, p1 = 380 kPa, and ∆z = 3 m. The pipe velocity is given by Q 4.67 × 10−3 0.00595 V = = = π 2 A D D2 4 Hence, for V < 2.4 m/s, √ D>

0.00595 = 0.0498 m = 49.8 mm 2.4

Take D = 50 mm and see if this is adequate for the pressure. For copper, ks = 0.0023 mm,

55

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and at 20◦ C, ν = 1.00 × 10−6 m/s2 , 4.67 × 10−3 = 2.38 m/s π 2 4 (0.050) vD (2.38)(0.050) Re = = = 119 × 105 ν 1 × 10−6 0.25 0.25 f=[ ( )]2 = [ ( )]2 = 0.0175 ks 0.0023 5.74 log 3.7D + 5.74 log 3.7(50) + (1.19×10 0.9 5 )0.9 Re L V2 110 2.382 hf = f = 0.0175 = 11.12 m D 2g 0.050 2(9.81) ( ) ( ) p1 380 p2 = γ − ∆z − hf = 9.79 − 3 − 11.12 = 241 kPa γ 9.79 V =

Since p2 > 240 kPa, a 50 mm copper line is (barely) adequate. 3.13. The design calculations for this problem are summarized in Table 3.1. 3.14. From the given data: Qref = 200 L/min = 0.00333 m3 /s, L1 = 20 m, L2 = 5 m, ∆z1 = 2 m, ∆z2 = 3 m, p0 = 380 kPa, and p2 = 240 kPa. For galvanized iron, ks = 0.15 mm = 1.5 × 10−4 . From the supply pipe to the first floor: p1 V12 L1 V12 p0 V02 + + z0 = + + z1 + f1 γ 2g γ 2g D1 2g where π 2 D = 0.7854D2 4 2Qref 2(0.00333) 0.008487 V1 = = = 2 A1 0.7854D D2 ( ) V1 D 0.008487 D 8487 = Re1 = = 2 −6 ν D 10 D 0.25 0.25 f1 = [ ( )]2 )]2 = [ ( 4.054×10−5 −4 0.9 5.74 log + 0.001671D log 1.5×10 + D 0.9 3.7D ( 8487 D ) A1 =

Combining the above equations and substituting known quantities yields p1 = 36.02 − [ γ

( log

0.25 4.054×10−5 D

+ 0.001671D

7.342 × 10−5 )]2 D5 0.9

From the first to the second floor, p1 V12 p2 V22 L2 V22 + + z1 = + + z2 + f2 γ 2g γ 2g D2 2g

56

(1)

Pipe AB BB’ B’C’ C’F’ C’D’ D’E’

Starting Head (m) 38.82 22.33 23.12 20.46 20.46 16.44

Flow (L/min) 409 144 144 45 45 45

Length (m) 17.0 2.7 2.5 46.0 4.0 46.0

Diam (mm) 64 51 51 25 51 25

Velocity (m/s) 2.12 1.17 1.17 2.62 0.37 1.52

Fitting Length (m) 5.12 4.66 3.05 1.52 2.13 1.52

Total Length (m) 22.12 7.36 5.55 47.52 6.13 47.52

Friction Loss (m) 1.37 0.21 0.16 5.15 0.02 5.15

Other Losses (m) 15.12 0 0 0 0 0

Elev Diff (m) 0 −1.0 2.5 0 4.0 0

Table 3.1: Head Loss in Standard Fittings in Terms of Equivalent Pipe Lengths Terminal Head (m) 22.33 23.12 20.46 15.31 16.44 11.29

Terminal Pressure (kPa) 216 226 200 146 161 109

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where π 2 D = 0.7854D2 4 Qref 0.00333 0.004244 V2 = = = 2 A2 0.7854D D2 ( ) V2 D 0.004244 D 4244 Re2 = = = 2 −6 ν D 10 D 0.25 0.25 f1 = [ ( )]2 )]2 = [ ( 4.054×10−5 −4 0.9 5.74 log + 0.003118D log 1.5×10 + D 0.9 3.7D ( 4244 D ) A2 =

Combining the above equations and substituting known quantities yields p1 = 27.51 + [ γ

( log

0.25 4.054×10−5 D

+ 0.003118D0.9

)]2

4.590 × 10−6 D5

(2)

Solving Equations 1 and 2 for D and taking the next larger available diameter yields D = 0.0508 m = 2 in. as the required pipe diameter. For this diameter, the actual pressure on the second floor (p2 ) is 262 kPa and the pressure on the first floor (p1 ) is 295 kPa .

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Chapter 4

Fundamentals of Flow in Open Channels 4.1. V = 1 m/s A = (b + my)y = (5 + 2 × 2)2 = 18 m2 Therefore Q = V A = (1)(18) = 18 m3 /s 4.2. Q = 8 m3 /s, w1 = 4 m, V1 = 1 m/s, and Q = w1 y1 V1 which leads to

Q 8 = = 2m w1 V1 (4)(1) w2 = 5 m, y2 = y1 − 0.5 m = 1.5 m, and y1 =

Q = w2 y2 V2 which leads to V2 =

Q 8 = = 1.07 m/s w2 y2 (5)(1.5)

4.3. The hydraulic radius, R, is defined by R=

A P

where, for circular pipes, A=

πD2 4

Hence R=

and P = πD πD2 /4 D = πD 4

or D = 4R

59

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4.4. The shear stress, τ0 , on the perimeter of the channel is given by τ0 = γRS0

(1)

From the given data b = 5 m, y = 1.8 m, m = 1.5, and the geometric properties of the channel are A = by + my 2 = 5(1.8) + 1.5(1.8)2 = 13.86 m2 √ √ P = b + 2 1 + m2 y = 5 + 2 1 + 1.52 (1.8) = 11.49 m A 13.86 R= = = 1.21 m P 11.49 From the given data, τ0 = 3.5 N/m2 , and since γ = 9790 N/m2 , Equation 1 gives the maximum allowable slope, S0 , as τ0 3.5 S0 = = = 0.00030 γR (9790)(1.21) For the excavated channel, ks = 3 mm = 0.003 m, and ν = 1.00 × 10−6 m2 /s at 20◦ C. Substituting these data into Equation 4.38 gives the flow rate, Q, as √ Q = −2A 8gRS0 log10

(

ks 0.625ν + √ 12R R 32 8gS0

)

√ Q = −2(13.86) 8(9.81)(1.21)(0.00030) log10

[

0.003 0.625(1.00 × 10−6 ) √ + 12(1.21) (1.21) 32 8(9.81)(0.00030)

]

= 17.2 m3 /s

Therefore, for the given flow depth restrictions in the channel, the flow capacity of the channel is 17.2 m3 /s . 4.5. From the given data: b = 8 m, S0 = 0.0001, ks = 2 mm = 0.002 m, and Q = 15 m3 /s. At 20◦ C, µ = 1.00 × 10−6 m2 /s, and, for a rectangular channel, A = by

and

R=

by 2y + b

Substituting into Equation 4.38 gives √ Q = −2A 8gRS0 log10

(

ks 0.625ν + 3√ 12R R 2 8gS0

)

[ ] √ 8y 0.002 0.625(1.00 × 10−6 ) 15 = −2(8y) 8(9.81)( )(0.0001) log10 + 8y 3 √ 8y 2y + 8 12( 2y+8 ) ( 2y+8 ) 2 8(9.81)(0.0001)

which yields y = 2.25 m Therefore, the uniform-flow depth in the channel is 2.25 m .

60

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4.6. Hydraulically rough flow conditions occur in open channels when u∗ ks ≥ 70 ν where u∗ =

√

gRSf

Equation 4.46 can be rearranged and put in the form ( n )6 = 2.84 × 108 n6 ks = 0.039

(1)

(2)

(3)

Substituting Equations 2 and 3 into Equation 1 and noting that ν = 1.00 × 10−6 m2 /s at 20◦ C and g = 9.81 m/s2 yields √ √ 9.81 RSf × 2.84 × 108 n6 ≥ 70 1.00 × 10−6 which simplifies to n6

√

RSf ≥ 7.9 × 10−14

(4)

From the given data: b = 5 m, S0 = 0.05% = 0.0005, and by definition: R=

by 5y A = = P 2y + b 2y + 5

(5)

Equation 4, can be combined with Equation 5 to give the following condition for fully turbulent flow, √( ) 5y 6 (0.0005) ≥ 7.9 × 10−14 (0.013) 2y + 5 This condition is satisfied when y ≥ 0.683 m . 4.7. The Darcy-Weisbach uniform-flow equation is given by Equation 4.38 as ( ) √ ks 0.625ν Q = −2A 8gRS0 log10 + 12R R 32 √8gS0 where the following variables are known: y = 2.20 m S0 = 0.0006 ks = 2 mm = 0.002 m ν = 1.00 × 10−6 m2 /s g = 9.81 m/s2 A = 3.6y + 2y 2 = 3.6(2.20) + 2(2.20)2 = 17.6 m2 R=

3.6y + 2y 2 3.6(2.20) + 2(2.20)2 √ = √ = 1.31 m 3.6 + 2 5y 3.6 + 2 5(2.20)

61

(1)

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Substituting these variables into Equation 1 yields Q = 34.0 m3 /s . Since y = 2.20 m corresponds to A = 17.6 m2 , then V = 34.0/17.6 = 1.93 m/s . The Manning’s equation gives the average velocity, V , as 1 2 12 R 3 S0 n Table 4.2 indicates that a mid-range roughness coefficient for concrete is n = 0.015. The average velocity given by the Manning equation is V =

2 1 1 (1.31) 3 (0.0006) 2 = 1.96 m/s 0.015 and the corresponding flow rate, Q, is

V =

Q = AV = (17.6)(1.96) = 34.5 m3 /s Hence, in this case, the Darcy-Weisbach and Manning equations give the similar results . 4.8. The Darcy-Weisbach uniform-flow equation is given by Equation 4.38 as ) ( √ ks 0.625ν Q = −2A 8gRS0 log10 + 12R R 32 √8gS0

(1)

where the following variables are either known or can be expressed in terms of the uniform-flow depth, y: S0 = 0.0001 ks = 1 mm = 0.001 m Q = 18 m3 /s ν = 1.00 × 10−6 m2 /s g = 9.81 m/s2 A = 5y + 2y 2 R=

5y + 2y 2 √ 5 + 2 5y

Substituting these variables into Equation 1 and solving for y yields y = 2.19 m . Check u∗ ks /ν and R/ks to determine the state of the flow and the validity of the Manning equation. Taking y = 2.19 m gives R = 1.39 m and √ √ (9.81)(1.39)(0.0001)(0.001) u∗ ks gRS0 ks = = = 37 ν ν 1.00 × 10−6 R 1.39 = = 1390 ks 0.001 Therefore, since 5 ≤ u∗ ks /ν ≤ 70 (i.e., 5 ≤ 37 ≤ 70) then according to Equation 5.19 the flow is in transition . Since u∗ ks /ν < 70 (i.e., 37 < 70) and R/ks > 500 (i.e., 1390 > 500), then the Manning equation is not valid .

62

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4.9. Comparing the Manning and Darcy-Weisbach equations √

which gives

1

8g R6 = f n

√ 1 1 1 1 1 f 2 R6 fR6 f 2 R6 n= √ = =√ 8.86 8g 8(9.81)

If the friction factor, f , is taken as a constant, the above relation indicates that n will be a 1 function of the depth (since R is a function of the depth). If f ∼ R− 3 , n would be a constant in the above equation. So the answer to the question is no . 4.10. Given: Q = 20 m3 /s, n = 0.015, S0 = 0.01

(a) Manning equation is given by 5 2 1 1 1 An3 12 Q = An Rn3 S02 = S n n P 23 0 n

where An = [b + myn ]yn = [2.8 + 2yn ]yn √ √ Pn = b + 2 1 + m2 yn = 2.8 + 2 5yn = 2.8 + 4.472yn Substituting into the Manning equation yields 5

1 1 [(2.8 + 2yn )yn ] 3 2 20 = 2 (0.01) 0.015 (2.8 + 4.472yn ) 3

or 5

[(2.8 + 2yn )yn ] 3 2

= 3.0

(2.8 + 4.472yn ) 3 Solving by trial and error yields yn = 0.91 m (b) Comparing the Manning and Darcy-Weisbach equations gives √

1

8g R6 = f n

which leads to f=

8gn2 1

R3

63

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In this case A = (2.8 + 2y)y = (2.8 + 2 × 0.91)(0.91) = 4.2 m2 P = 2.8 + 4.472(0.91) = 6.87 m A 4.20 R= = = 0.611 m P 6.87 therefore f=

8(9.81)(0.015)2 1

= 0.0208

(0.611) 3

For fully turbulent flow, where the Manning equation applies, ] [ 1 ks √ = −2 log 12R f [ ] ks 1 √ = −2 log 12(0.611) 0.0208 6.93 = −2 log[0.136ks ] which leads to ks = 0.00249 m = 2.5 mm 4.11. From the given information, 1

n = 0.039d 6 where d is in m. In this case, d = 30 mm = 0.030 m, and a 70% error in d is 0.7(0.030) = 0.021 m. Hence, d = 0.030 m ± 0.021 m. Hence, the “best estimate” of n, denoted by n ¯ , is given by 1 n ¯ = 0.039(0.030) 6 = 0.022 The lower estimate of n, nL , is given by 1

nL = 0.039(0.030 − 0.021) 6 = 0.018 and the upper estimate of n, nU , is given by 1

nU = 0.039(0.030 + 0.021) 6 = 0.024 The maximum percentage error in estimating n is therefore given by error =

0.022 − 0.018 × 100 = 18% 0.022

4.12. According to Equation 4.45, n 1

ks6

=

√1 8g

( )1 R ks

6

( ) 2.0 log 12 kRs

64

(1)

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Let y=

n 1

ks6 R x= ks and taking g = 9.81 m/s2 , Equation 1 can be written √

1 1 x6 8(9.81)

1

0.1129x 6 y= = 2.0 log(12x) 2.0(log 12 + log x)

(2)

1

1

0.1129x 6 0.1129x 6 = = 2.0(log 12 + 0.4343 ln x) 2.158 + 0.8686 ln x

(3)

1

The minimum value of n/ks2 (= y) occurs when dy/dx = 0, where (2.158 + 0.8686 ln x)( 61 × 0.1129x− 6 ) − (0.1129x 6 )(0.8686x−1 ) dy = =0 dx (2.158 + 0.8686 ln x)2 1

5

which yields x = 33.63 and substituting into Equation 3 yields y = 0.0389 1

Therefore, under fully-rough flow conditions, the minimum value of n/ks6 (= y) is 0.0389, or approximately 0.039 . 1

When n/ks6 differs by 5% from 0.039, n 1

= 1.05(0.039) =

ks6

√1 8g

( )1 R ks

6

( ) 2.0 log 12 kRs

or

1

0.1129x 6 0.04095 = 2.158 + 0.8686 ln x which yields x=6

or

281

1 6

Therefore, n/ks is within 5% of 0.039 when 6≤

R ≤ 281 ks

It is noteworthy that this range is narrower than suggested by Yen (1991) and Hager (1999). The reason for this is that the constant value they assumed is a bit higher than 0.039.

65

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4.13. From the given data: b =5 m, m = 3, L = 100 m, z1 = 24.01 m, z2 = 23.99 m, Z1 = 25.01 m, Z2 = 24.95 m, Q0 = 15 m3 /s, ∆Q = 2 m3 /s. (a) From the given data: y1 = Z1 − z1 = 25.01 m − 24.01 m = 1.00 m y2 = Z2 − z2 = 24.95 m − 23.99 m = 0.96 m √ √ P1 = b + 2 1 + m2 y1 = 5 + 2 1 + 32 (1.00) = 11.32 m √ √ P2 = b + 2 1 + m2 y2 = 5 + 2 1 + 32 (0.96) = 11.07 m A1 = by1 + my12 = (5)(1.00) + (3)(1.00)2 = 8.000 m2 A2 = by2 + my22 = (5)(0.96) + (3)(0.96)2 = 7.565 m2 ( 2) ∆y + ∆z + ∆ V2g Sf = − L Q2 ( 1 (0.96 − 1.00) + (23.99 − 24.01) + 2(9.81) − 7.5652 =− 100 A1 8.000 R1 = = = 0.7067 m P1 11.32 A2 7.565 R2 = = = 0.6834 m P2 11.07 ¯ = R1 + R2 = 0.7067 + 0.6834 = 0.6951 m R 2 2 A + A 8.000 + 7.565 1 2 A¯ = = = 7.783 m2 2 2

1 82

) = 0.006 − 9.422 × 10−6 Q2

According to the Manning equation, 2 ¯ 32 √ A¯R (7.783)(0.6951) 3 √ n= Sf = 0.006 − 9.422 × 10−6 Q2 Q Q

which gives n=

6.107 √ 0.006 − 9.422 × 10−6 Q2 Q

(1)

(b) Since Q = 15±2 m3 /s, Equation 1 gives Q (m3 /s) 13 15 17

n 0.0319 0.0254 0.0206

ks =

(

)6 n 0.039

× 1000

(mm) 299 76 22

Based on these results, Manning’s n is in the range 0.021–0.032 and the roughness height is in the range of 22–300 mm .

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(c) From the derived data, 2/15 = 0.52 (0.0319 − 0.0254)/0.0254 2/15 = = 0.71 (0.0254 − 0.0206)/0.0254 2/15 = = 0.045 (299 − 76)/76 2/15 = = 0.187 (76 − 22)/76

∆Q/Q ∆n/n ∆Q/Q ∆n/n ∆Q/Q ∆k/k ∆Q/Q ∆k/k

=

Based on these results, the flows are much more sensitive to to specification of Manning’s n than specification of the roughness height. The relative sensitivity to Manning’s n is in the range of 0.52–0.71 , while the relative sensitivity to the roughness height is 0.05–0.19 . 4.14. For fully-turbulent flow conditions, u∗ ks > 70 ν

(1)

where u∗ is given by Equation 4.30 as √ u∗ =

τ0 √ = gRS0 ρ

(2)

Combining Equations 1 and 2 gives √

or ks

gRS0 ks > 70 ν

√

70ν RS0 > √ g

Taking ν = 1.00 × 10−6 m2 /s (at 20◦ C), and g = 9.81 m/s2 yields the turbulence condition ks

√

RS0 >

which simplifies to ks

√

70(1.00 × 10−6 ) √ 9.81

RS0 > 2.2 × 10−5

For the given trapezoidal channel, ks = 3 mm = 0.003 m, S0 = 0.1% = 0.001, b = 3 m, m = 2, and for a flow depth y, R=

A by + my 2 3y + 2y 2 3y + 2y 2 √ √ = = = P 3 + 4.472y b + 2y 1 + m2 3 + 2y 1 + 22

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For turbulent flow, √ ks RS0 > 2.2 × 10−5 √( ) 3y + 2y 2 (0.001) > 2.2 × 10−5 0.003 3 + 4.472y which requires that y > 0.056 m Therefore, flow conditions are fully turbulent when the depth of flow exceeds 0.056 m = 5.6 cm . At this minimum flow depth, 3(0.056) + 2(0.056)2 = 0.0536 m 3 + 4.472(0.056) R 0.0536 m = = 17.9 ks 0.003 m R=

1

Since R/ks is within the range for n/ks6 to be assumed constant, using the Manning equation is appropriate . 4.15. From the given data: y = 4.00 m, b = 4 m, m = 3, and S0 = 0.0001. The Manning equation is valid under the following conditions, 3.6 < and ks

√

R < 360 ks

RS0 > 2.2 × 10−5

(1)

(2)

1

Assuming n = 0.013 and n/ks6 = 0.040, ( n )6 ( 0.013 )6 = ks = = 0.00118 m 0.040 0.040 and since R=

by + my 2 4(4) + 3(4)2 √ √ = = 2.18 m b + 2 1 + m2 y 4 + 2 1 + 32 (4)

then R 2.18 m = = 1847 ks 0.00118 m √ √ ks RS0 = (0.00118) (2.18)(0.0001) = 1.74 × 10−5 √ Since R/ks > 360 and ks RS0 < 2.2 × 10−5 , the flow is not fully turbulent and Manning’s equation is not applicable .

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4.16. From the given data: d = 15 cm = 0.15 m, Sx = 2% = 0.02, S0 = 1% = 0.01, and ks = 1 mm = 0.001 m. Assume ν = 10−6 m2 /s. The transverse slope corresponds to m = 50. (a) The following geometric characteristics of the channel can be derived from the given data: T = dm = (0.15)(50) = 7.5 m 1 1 A = dT = (0.15)(7.5) = 0.5625 m2 2 √ 2 √ P = d + (d2 + T 2 ) = 0.15 + (0.152 + 7.52 ) = 7.651 m A 0.5625 R= = = 0.07352 m P 7.651 Using these data, the Darcy-Weisbach equation gives ] [ √ 0.625ν ks + Q = −2A 8gRS0 log10 12R R 32 √8gS0 ] [ √ 0.001 0.625(10−6 ) = −2(0.5625) 8(9.81)(0.07352)(0.01) log10 + √ 12(0.07352) (0.07352) 32 8(9.81)(0.01) = 0.792 m3 /s (b) Using the Manning equation: 1

1

n = 0.039ks6 = 0.039(0.001) 6 = 0.0123 1 2 2 1 1 1 Q = AR 3 S02 = (0.5625)(0.07352) 3 (0.01) 2 = 0.803 m3 /s n 0.0123 (c) The following parameters are used to check the validity of the Manning equation: R 0.07352 = = 74 ks 0.001 √ √ ks RS0 = (0.001) (0.07352)(0.01) = 2.7 × 10−5 √ Since 4 < R/ks < 500 and ks RS0 > 2.2 × 10−5 , Manning’s equation is valid . (d) The discrepancy in Q calculated by the Darcy-Weisbach and Manning equations is 1

0.803 m3 /s − 0.792 m3 /s = 0.011 m3 /s. This is due to the approximation that n = 0.039ks6 . 4.17. From the given data: Q = 1.8 m3 /s, m = 2, n = 0.025, and S0 = 0.1% = 0.001. (a) Size the channel to accommodate the design flow under normal conditions. Assuming that the flow in the channel can be described by the Manning equation (i.e. fully turbulent) 1 2 1 Q = AR 3 S02 (1) n

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Since the lengths of the channel sides are equal to the bottom width, b, then the flow depth, y, is related to the bottom width by the relation y=√

b b =√ = 0.447b 2 1+m 1 + 22

(2)

The geometric properties of the channel are A = by + my 2 = b(0.447b) + (2)(0.447b)2 = 0.847b2 P = 3b R=

0.847b2 A = = 0.282b P 3b

Substituting into the Manning equation, Equation 1, gives 1.8 =

2 1 1 (0.847b2 )(0.282b) 3 (0.001) 2 0.025

which yields b = 1.67 m According to Equation 2 the depth of flow is given by y = 0.447(1.67) = 0.746 m The required channel is to have a bottom width of 1.67 m, side slopes of 2:1 (H:V), and a depth of at least 0.746 m. (b) Let y be the depth of flow when the average shear stress, τ , on the channel lining is equal to the critical shear stress, τc = 4.0 Pa. The channel lining then becomes unstable and the geometric properties of the channel are A = by + my 2 = 1.67y + 2y 2 √ √ P = b + 2 1 + m2 y = 1.67 + 2 1 + 22 y1.67 + 4.47y R=

A 1.67y + 2y 2 = P 1.67 + 4.47y

The average shear stress, τ , on the perimeter of the channel is given by τ = γRS0

(3)

where γ = 9790 N/m3 . The channel lining is unstable when τ = τc = 4.0 Pa, and Equation 3 gives 1.67y + 2y 2 4.0 = (9790) (0.001) 1.67 + 4.47y which yields y = 0.625 m

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Therefore, whenever the flow depth exceeds 0.625 m, the channel lining becomes unstable. In terms of flow, the Manning equation gives Q=

1 2 1 AR 3 S02 n

[ ]2 1 1 1.67(0.625) + 2(0.625)2 3 2 = [1.67(0.625) + 2(0.625) ] (0.001) 2 = 1.27 m3 /s 0.025 1.67 + 4.47(0.625) Therefore, whenever the flow rate exceeds 1.27 m3 /s , the channel lining becomes unstable. An alternative lining should be used if the channel is to accommodate the design flow of 1.8 m3 /s. 4.18. From the given data, b = 3 m, m = 2, S0 = 0.001, and n = 0.015. (a) For 4 < R/ks < 500, Manning’s n and the roughness height (= equivalent sand roughness), ks , are related by n 1 6

ks 0.015 1

= 0.039 = 0.039

ks6 which yields ks = 0.00324 m = 3.24 mm. (b) Manning’s n can be assumed to be approximately constant for 4 < R/ks < 500 4ks
0.0461 m

(2)

Equations 1 and 2 collectively indicate that n can be taken as a constant and the Manning equation is valid for 0.0461 m < R < 1.620 m For R = 0.0461 m, the (lower) flow depth, yL , satisfies the relation byL + myL2 √ = 0.0461 b + 2 1 + m2 y L 3yL + 2yL2 √ = 0.0461 3 + 2 1 + 2 2 yL

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which yields yL = 0.0479 m. Similarly, for R = 1.62 m, the corresponding flow depth, yU , is given by yU = 2.95 m. Therefore, the Manning equation can be applied with a constant value of n in the range 0.0479 m < y < 2.95 m . For R = 0.0461 m, A = 0.1483 m2 , and Q=

1 2 2 1 1 1 AR 3 S02 = (0.1483)(0.0461) 3 (0.001) 2 = 0.0402 m3 /s n 0.015

Similarly, for R = 1.62 m, A = 26.26 m2 , and Q=

1 2 2 1 1 1 AR 3 S02 = (26.26)(1.62) 3 (0.001) 2 = 76.36 m3 /s n 0.015

hence the range of flow for which the Manning equation can be applied with a (approximately) constant value of n is 0.0402 m3 /s < Q < 76.36 m3 /s . (c) The general equation for n under fully turbulent conditions is ( )1

1

n=

ks6 √ 8g

R ks

(

2 log

12R ks

1

6

)=

(0.00324) 6 √ 8(9.81)

2 log

(

(

R 0.00324

12R 0.00324

) 16

)

=

0.05644 1 = 3.569 + log R 63.24 + 17.72 log R

Hence, taking Q = 100 m3 /s, the Manning equation can be put in the form Q=

1 2 1 AR 3 S02 n 5

Q = [63.24 + 17.72 log R]

A3 2 3

1

S02

P ( [ )] 5 1 3y + 2y 2 (3y + 2y 2 ) 3 2 √ 100 = 63.24 + 17.72 log √ 2 (0.001) 3 + 2 5y (3 + 2 5y) 3 which yields y = 3.303 m and n = 0.0148. If n is assumed to be constant and equal to 0.015, then the Manning equation requires that 5

1 1 (3y + 2y 2 ) 3 2 100 = √ 2 (0.001) 0.015 (3 + 2 5y) 3

which yields y = 3.326 m. Therefore, the error in the calculated flow depth incurred by assuming a constant n (= 0.015) for Q = 100 m3 /s is 0.7% , which is a relatively small error. 4.19. From the given data: Q = 150 m3 /s, b = 10 m, m = 2.5, S0 = 0.1% = 0.001, y = 5 m, and A = by + my 2 = (10)(5) + (2.5)(5)2 = 112.5 m2 √ √ P = b + 2y 1 + m2 = 10 + 2(5) 1 + 2.52 = 36.93 m A 112.5 R= = = 3.046 m P 36.93

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(a) Using the Manning equation: 1 2 1 AR 3 S02 n 2 1 1 100 = (112.5)(3.046) 3 (0.001) 2 n

Q=

which yields n = 0.0498 . Assuming that 4 < R/ks < 500, n 1

ks6 0.0498 1

= 0.039 = 0.039

ks6 which yields ks = 4.354 m and R/ks = 3.046/4.354 = 0.7. Since R/ks < 4, n/ks cannot be taken as a constant (= 0.039) that is independent of R/ks . Assuming that the flow is fully turbulent, n

√1 8g

( )1 R ks

6

( ) 2.0 log 12 kRs ( )1 3.046 6 √ 1 ks 0.0498 8(9.81) ( ) = 1 2.0 log 12 3.046 ks6 ks 1

=

ks6

which yields ks = 1.577 m . This result is based on the assumption of fully turbulent flow, which requires that √ ks RSf > 2.2 × 10−5 √ (1.577) (3.046)(0.001) > 2.2 × 10−5 0.0870 > 2.2 × 10−5 Hence the fully-turbulent flow assumption is validated. (b) Under the flow conditions described here, the Manning equation is not valid since n is not a constant but depends on the flow depth. If n is expressed as a function of the flow depth, then the Manning equation can be used. (c) If the depth of flow increases by 50%, then y = 1.5(5.0) = 7.5 m A = by + my 2 = 10(7.5) + 2.5(7.5)2 = 215.6 m2 √ √ P = b + 2y 1 + m2 = 10 + 2(7.5) 1 + 2.52 = 50.39 m A 215.6 R= = = 4.279 m P 50.39

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Taking ks = 1.577 m and R = 4.279 m, Manning’s n is given by n

√1 8g

( )1 R ks

6

) ( 2.0 log 12 kRs ( 4.279 ) 1 6 √ 1 n 8(9.81) 1.577 ( ) 1 = 2.0 log 12 4.279 1.577 6 1.577 =

1

ks6

which yields n = 0.0475 . 4.20. The Horton (1933a) and Einstein (1934) formulae are based on the assumption that the total cross-sectional mean velocity is equal to the subarea mean velocity. According to the Manning equation 1 2 21 R 3 S0 ne 1 2 1 subarea mean velocity = Ri3 S02 ni

total cross-sectional mean velocity =

(1) (2)

Combining Equations 1 and 2, and using the Horton/Einstein assumption 1 2 12 1 2 1 R 3 S0 = Ri3 S02 ne ni ( )2 ( )2 1 A 3 1 Ai 3 = ne P ni Pi ( ) ( ) 1 A 1 Ai = 3 3 ne2 P ni2 Pi ( ) 3 1 A ni2 Pi = Ai 3 P 2 ne ( )∑ N N ∑ 3 1 A 2 (n P ) = Ai 3 i i P 2 ne i=1 i=1 Since A=

N ∑

Ai

i=1

Equation 3 simplifies to 1

∑N

i=1 (ni Pi )

3 2

P

ne which yields

 ne = 

2 3 3 2 P n i=1 i i  P

∑N

74

=1

(3)

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4.21. The Lotter (1933) formula is based on the assumption that the total discharge is the sum of the subarea discharges. According to the Manning equation, 1 2 1 AR 3 S02 ne 2 1 1 subarea discharge = Ai Ri3 S02 ni

total discharge =

(1) (2)

Combining Equations 1 and 2 and using the Lotter (1933) assumption ∑ 1 1 2 1 2 1 AR 3 S02 = Ai Ri3 S02 ne ni N

i=1

5

∑ 1 A3 1 A i = ne P 23 ni P 23 N

5 3

i=1

5 1 PR3 = ne

N ∑ i=1

i

5 1 Pi Ri3 ni

which simplifies to 5

ne =

PR3 ∑N

5

Pi Ri3 i=1 ni

4.22. From the given shape of the floodplain (Figure 4.5), the following geometric characteristics are derived: Section, i 1 2 3 4 5 6 7

Pi (m) 20.6 100.0 6.7 15.0 6.7 150.0 20.6 319.6

Ai (m2 ) 50 500 39 120 39 750 50 1548

Ri (m) 2.42 5.00 5.81 8.00 5.81 5.00 2.42 34.46

ni 0.040 0.030 0.015 0.013 0.017 0.035 0.060

yi (m) 2.50 5.00 6.50 8.00 6.50 5.00 2.50

The total perimeter, P , of the (compound) channel is 319.6 m, the total area, A, is 1548 m2 , and hence the hydraulic radius, R, of the compound section is given by R=

A 1548 = = 4.84 m P 319.6

Substituting these data into the formulae listed in Table 4.3 yields the following results:

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Formula Horton/Einstein Pavlovskii Einstein and Banks Lotter Krishnamurthy and Christensen Cox Yen Average

ne 0.034 0.035 0.034 0.026 0.029 0.032 0.033 0.032

A conservative (high) estimate of the composite roughness is 0.035, and the average composite roughness predicted by the models is 0.032 . 4.23. In the main channel: n = 0.016 and S0 = 0.005. When the main channel flows full: 1 A = [30 + 30 + 3(2) + 3(3)](3) = 112.5 m2 2 √ √ P = 30 + 3( 32 + 12 + 22 + 12 ) = 46.2 m A 112.5 R= = = 2.44 m P 46.2 The Manning equation gives the capacity, Q, of the main channel as 1 2 2 1 1 1 Q = AR 3 S02 = (112.5)(2.44) 3 (0.005) 2 = 900 m3 /s n 0.016 When flow is in the floodplain, use the Horton equation to calculate the equivalent Manning’s roughness, ne . If the depth of flow in the main channel is y: i

Section

1 2 3

Left Floodplain Main Channel Right Floodplain

Pi (m) 100(y − 3) 46.2 25(y − 3)

Ai (m2 ) 50(y − 3)2 112.5 + 45(y − 3) 12.5(y − 3)2

Ri (m) 0.5(y − 3) 2.44 + 0.974(y − 3) 0.5(y − 3)

n 0.040 0.016 0.050

Using Horton equation  2 [ ]2 3 3 ∑N 3 3 3 3 2 2 + 46.2(0.016) 2 + 25(y − 3)(0.050) 2 P n 100(y − 3)(0.040) i i=1 i  = n e =  ∑N 100(y − 3) + 46.2 + 25(y − 3) i=1 Pi which simplifies to

(

1.08y − 3.15 ne = 125y − 328.8 For the entire channel, including floodplains, A=

3 ∑

)2 3

Ai = 50(y − 3)2 + 112.5 + 45(y − 3) + 12.5(y − 3)2 = 62.5(y − 3)2 + 45(y − 3) + 112.5

i=1

P =

3 ∑

Pi = 100(y − 3) + 46.2 + 25(y − 3) = 125y − 328.8

i=1

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With Q = 1590 m3 /s, the Manning equation gives 5

1 2 1 1 A 3 12 AR 3 S02 = S ne ne P 23 0 ( )2 5 1 125y − 328.8 3 [62.5(y − 3)2 + 45(y − 3) + 112.5] 3 1590 = (0.005) 2 2 1.08y − 3.15 (125y − 328.8) 3

Q=

which yields y = 5.50 m Therefore, the encroachment on the left floodplain is (5.50 − 3.0)(100) = 250 m , and the encroachment on the right floodplain is (5.50 − 3.0)(25) = 62.5 m . 4.24. According to Equation 4.56,

( 1√ y) gdS0 1 + 2.3 log κ d The value of y where v(y) = V occurs when ( 1√ y) V =V + gdS0 1 + 2.3 log κ d which can be simplified as follows: ( 1√ y) 0= gdS0 1 + 2.3 log d (κ y) 0 = 1 + 2.3 log d y 1 log = − d 2.3 y = 10−1/2.3 d y = 0.368d v(y) = V +

4.25. According to Equation 4.56, v(y) = V +

( 1√ y) gdS0 1 + 2.3 log κ d

At y/d = 0.2, v(0.2d) = V +

1 √ 1 √ gdS0 (1 + 2.3 log 0.2) = V + gdS0 (−0.61) 0.4 0.4

(1)

At y/d = 0.8, 1 √ 1 √ gdS0 (1 + 2.3 log 0.8) = V + gdS0 (0.78) (2) 0.4 0.4 Combining Equations 1 and 2 gives [ 1 √ ] √ 2V + 0.4 gdS0 (0.78 − 0.61) v(0.2d) + v(0.8d) = = V + 0.21 gdS0 2 2 √ Therefore, assuming that the term 0.21 gdS0 is small, then the average velocity, V , can be estimated by [v(0.2d) + v(0.8d)]/2. v(0.8d) = V +

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4.26. The velocity distribution is given by Equation 4.57 as v(y) = Vmax

(y )1 7

d

Hence the average velocity, V , is given by 1 V = d

∫

d

Vmax 0

(y )1 7

d

1 Vmax dy = d d 17

[

8

y7 (8/7)

]d = 0

7 Vmax 8

Putting the v(y) = V in Equation 4.57 gives (y )1 7 7 Vmax = Vmax 8 d and solving this equation gives y/d = 0.393 or y = 0.393d 4.27. Q = 8.4 m3 /s, S0 = 0.001, L = 100 m. At Section 1: b1 = 2 m, m1 = 2, y1 = 1 m; and at Section 2: b2 = 2.5 m, m2 = 2, y2 = 1 m. The head loss is given by the energy equation, y1 +

V12 V2 − y2 − 2 = L(Sf − S0 ) 2g 2g

where A1 = (b1 + m1 y1 )y1 = (2 + 2 × 1)(1) = 4 m2 A2 = (b2 + m2 y2 )y2 = (2.5 + 2 × 1)(1) = 4.5 m2 Q 8.4 V1 = = = 2.10 m/s A1 4 Q 8.4 V2 = = = 1.87 m/s A2 4.5 Substituting into the energy equation gives 1+

2.102 1.872 −1− = (100)(Sf − 0.001) 2(9.81) 2(9.81)

which simplifies to Sf = 0.00147 and the head loss, hL , is given by hL = LSf = (100)(0.00147) = 0.147 m The power, P , dissipated is P = γw QhL = (9.79)(8.4)(0.147) = 12.1 kW where γw = 9.79 kN/m3 at 20◦ C.

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4.28. The Darcy-Weisbach equation can be written as hf =

f¯L V¯ 2 D 2g

Defining hf ¯=D and R L 4 and substituting into the Darcy-Weisbach equation gives S=

S=

f¯ V¯ 2 4R 2g

4.29. Q = 30 m3 /s, w = 5 m, and for a rectangular channel ( yc = where q= Hence

) 31

Q 30 = = 6 m2 /s w 5

( yc =

q2 g

62 9.81

) 13 = 1.54 m

Therefore, when the depth of flow is 3 m, yc < 3 m and the flow is subcritical . 4.30. From the given data: Q = 50 m3 /s, b = 4 m, and m = 1.5. Under critical flow conditions Q2 A3 = g T which gives 502 (4yc + 1.5yc2 )3 = 9.81 4 + 2(1.5)yc Solving by trial and error yields yc = 1.96 m When y = 3 m, the Froude number, Fr, is given by the relation Fr2 =

Q2 T (50)2 (4 + 2 × 1.5 × 3) = = 0.19 3 gA (9.81)(4 × 3 + 1.5 × 32 )3

hence Fr = 0.45 and the flow is subcritical .

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4.31. From the given data: w1 = 2 m, Q = 3 m3 /s, y1 = 1.2 m, and w2 = w1 − 0.4 m = 1.6 m. Conservation of energy requires that y1 +

V12 V2 = y2 + 2 2g 2g

where Q Q 3 = = = 1.25 m/s A1 w1 y1 (2)(1.2) Q Q 3 1.875 V2 = = = = m/s A2 w2 y2 1.6y2 y2 V1 =

Substituting into the energy equation gives 1.2 +

1.252 (1.875/y2 )2 = y2 + 2(9.81) 2(9.81) 0.179 1.28 = y2 + y22

Solving for y2 gives y2 = 0.47 m, 1.14 m These depths correspond to supercritical and subcritical flow conditions respectively. Since the upstream flow is subcritical, the flow in the constriction must also be subcritical, hence y2 = 1.14 m When choking occurs at the constriction, y = yc

V and Fr = 1 = √ gD

and the energy equation gives V12 V2 = yc + c 2g 2g ]2 [ ( 2 ) 13 q 1 q + 1.28 = 9.81 2(9.81) (q 2 /9.81) 31

y1 +

which yields q = 2.47 m2 /s and w2 =

Q 3 = = 1.21 m q 2.47

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4.32. From the given data: Q = 1 m3 /s, b = 1 m, and y1 = 1 m. The flow is choked when there is critical flow in the constriction. The upstream specific energy, E1 , is given by E1 = y1 +

V12 Q2 12 = y1 + = 1.0 + = 1.05 m 2g 2g(by1 )2 2(9.81)(1 × 1)2

At the constriction, Fr2c = 1 which leads to Q2 A3 = c g Tc Substituting given data (byc )3 12 = 9.81 b which leads to (byc )2 =

0.102 yc

(1)

The energy equation requires that yc + yc +

Q2 = 1.05 2gA2c

(1)2 = 1.05 2(9.81)(byc )2 0.0510 yc + = 1.05 (byc )2

(2)

Combining Equations 1 and 2 gives yc +

0.0510 = 1.05 0.102/yc

or yc = 0.70 m which leads to b = 0.55 m 4.33. From the given data: Q = 16 m3 /s, y1 = 2 m, b1 = 10 m, and m = 3. The following preliminary calculations will be useful, A1 = b1 y1 + my12 = (10)(2) + (3)(2)2 = 32 m2 T1 = b1 + 2my1 = (10) + 2(3)(2) = 32 m 32 A1 D1 = = =1m T1 32 Q 16 V1 = = = 0.5 m/s A1 32 V1 0.5 Fr1 = √ =√ = 0.160 gD1 (9.81)(1)

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(a) Denote the depth of flow and the bottom width of the contracted section as b and y, respectively. For critical flow in the contracted section: Q2 A3 = g T (by + 3y 2 )3 162 = 9.81 b + 2(3)y which yields (by + 3y 2 )3 = 26.10 b + 67

(1)

For conservation of energy, y1 + 2+

Q2 Q2 = y + 2gA2 2gA21

162 162 = y + 2(9.81)(32)2 2(9.81)(by + 3y 2 )2

which simplifies to y+

13.05 = 2.013 (by + 3y 2 )2

(2)

Both Equations 1 and 2 must be satisfied for choking to occur at the downstream section. (b) When b = 0, Equation 2 gives y+

13.05 = 2.013 (3y 2 )2

which yields y = 1.90 m or 1.10 m. Hence the depth in the contracted section will be 1.90 m and the flow will not be choked . 4.34. From the given data: b1 = 10.0 m, y1 = 1.00 m, Q = 8 m3 /s, b2 = 6 m, and L = 7 m. (a) According to the energy equation E1 = E2 +

V12 2g

where 8 Q = = 0.800 m/s b1 y1 (10.0)(1.00) V12 (0.800)2 = = 0.0326 m 2g 2(9.81) V2 E1 = y1 + 1 = 1.00 + 0.0326 = 1.0326 m 2g 82 0.0906 Q2 = y + = y2 + E2 = y2 + 2 2 2 2g(b2 y2 ) 2(9.81)(6y2 ) y22 V1 =

82

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Substituting into the energy equation, Equation 1, gives 1.0326 = y2 +

0.0906 + 0.0326 y22

which simplifies to 1.00 = y2 +

0.0906 y22

which yields the following positive solutions y2 = 0.383 m,

0.884 m

Since Fr21 =

V12 0.8002 = = 0.065 gy1 (9.81)(1.00)

the upstream flow is subcritical, and therefore the flow in the constriction must also be subcritical, and hence y2 = 0.884 m (b) To assess the effect of the energy loss, the depth of flow in the constriction must be calculated without including the energy loss. According to the energy equation E1 = E2

(2)

where E1 = 1.0326 m 0.0906 E2 = y2 + y22 Substituting into the energy equation, Equation 2, gives 1.0326 = y2 +

0.0906 y22

which yields the following positive solutions y2 = 0.371 m,

0.924 m

Since the upstream flow is subcritical, the flow in the constriction must also be subcritical, and hence y2 = 0.924 m Therefore, if energy losses are neglected the calculated flow depth is in error by (0.924 − 0.884)/0.884 × 100 = 4.5%. This effect is not very significant.

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(c) According to the energy equation E1 = E2 +

V12 2g

(3)

where V1 = 0.800 m/s, and E1 = 1.0326 m E2 = y2 +

82 0.1612 Q2 = y + = y2 + 2 2g(b2 y2 )2 2(9.81)(4.5y2 )2 y22

Substituting into the energy equation, Equation 3, gives 1.0326 = y2 +

0.1612 0.8002 + 2(9.81) y22

which does not have any positive solutions. Therefore, the flow is choked and critical flow exists within the constriction. Under critical flow conditions, Q2 A3 = g T 2 8 (4.5y2 )3 = 9.81 4.5 which yields y2 = 0.686 m (d) Since the flow is choked, the constriction influences the upstream flow depth. Under critical flow conditions, 3 3 E2 = y2 = (0.686) = 1.028 m 2 2 According to the energy equation E1 = E2 + or y1 +

V12 2g

V2 V12 = 1.028 + 1 2g 2g

which yields y1 = 1.028 m 4.35. From given data: b = 3 m, Q = 4 m3 /s, y1 = 1.5 m, and V1 =

Q 4 = = 0.889 m/s by1 3(1.5)

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Applying the energy equation, V22 + ∆z 2g (4/3y2 )2 + 0.15 2(9.81) 0.0906 1.54 = y2 + + 0.15 y22

V12 = y2 + 2g 0.8892 1.5 + = y2 + 2(9.81) y1 +

Solving this equation for y2 gives y2 = 1.34 m, 0.29 m Since the upstream flow is subcritical, select the subcritical flow depth, where y2 = 1.34 m When choking just occurs, ( y2 = yc =

q2 g

) 13

where q=

Q 4 = = 1.33 m2 /s b 3

and therefore ( yc =

1.332 9.81

) 31 = 0.565 m

and the energy equation can be written as (q/yc )2 V12 = yc + + ∆zm 2g 2g (1.33/0.565)2 1.54 = 0.565 + + ∆zm 2(9.81)

y1 +

which gives ∆zm = 0.69 m

85

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4.36. Flow in a rectangular open channel is choked when E1 = E2 + ∆zc V12 3 = yc + ∆zc 2g 2 ( )1 2 V1 3 q2 3 y1 + = + ∆zc 2g 2 g ( )1 3 Q2 3 V12 = y1 + + ∆zc 2g 2 gb2 [ ]1 3 (V1 by1 )2 3 V12 = + ∆zc y1 + 2g 2 gb2 y1 +

2

2

V2 3 (V13 y13 ) y1 + 1 = + ∆zc 2g 2 g 31 Dividing by y1 yields V2 3 1+ 1 = 2gy1 2

(

V12 gy1

) 13 +

∆zc y1

and defining V1 Fr1 = √ gy1 then Equation 1 can be written as ∆zc Fr2 3 2 = 1 + 1 − Fr13 y1 2 2 From Problem 4.35: b = 3 m, Q = 4 m3 /s, y1 = 1.5 m, and ∆zc = 0.15 m. Therefore Q 4 = = 0.889 m/s by1 (3)(1.5) V1 0.889 Fr1 = √ =√ = 0.232 gy1 (9.81)(1.5) V1 =

which yields Fr2 3 2 ∆zc = 1 + 1 − Fr13 y1 2 2 2 ∆zc (0.232)2 3 =1+ − (0.232) 3 1.5 2 2 and solving for ∆zc gives ∆zc = 0.69 m

86

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4.37. From the given data: Q = 4.3 m3 /s, b1 = 3 m, y1 = 1 m, and therefore the upstream specific energy, E1 , is given by E1 = y1 +

Q2 4.32 V12 = y1 + = 1 + = 1.10 m 2g 2g(b1 y1 )2 2(9.81)(3 × 1)2

At the constriction, y2 +

Q2 + 0.25 = 1.10 2g(b2 y2 )2

Since b2 = 3 − 0.75 = 2.25 m, then y2 +

4.32 = 0.85 2(9.81)(2.25 × y2 )2

which simplifies to y2 +

0.186 = 0.85 y22

There is no solution to this equation, so the flow is choked and the flow conditions in the constriction are critical. Under this condition, √ √ √ 2 2 2 3 (4.3/2.25) 3 q 3 (Q/b) y2 = yc = = = = 0.72 m g g 9.81 4.38. From the given data: Q = 18 m3 /s, b = 5 m, m = 2, y1 = 2 m, and there is a 0.50-m wide bridge pier placed in the channel. A1 = (b + my1 )y1 = (5 + 2 × 2)(2) = 18 m2 A2 = [(5 − 0.5) + 2y2 ]y2 = 4.5y2 + 2y22 Q 18 V1 = = = 1 m/s A1 18 and the energy equation gives y1 +

V12 V2 = y2 + 2 2g 2g

[ ]2 12 1 18 = y2 + 2+ 2(9.81) 2(9.81) (4.5y2 + 2y22 ) [ ]2 1 2.05 = y2 + 16.5 (4.5y2 + 2y22 ) with the solutions y2 = 0.59 m, 1.99 m which correspond to supercritical and subcritical flow conditions respectively. Since the upstream flow is subcritical, choose the subcritical downstream flow y2 = 1.99 m

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The maximum pier width produces critical flow at the constriction such that y2 = yc or Q2 A3 = c g Tw

(1)

where Ac = (5 − wp )yc + 2yc2 Tw = (5 − wp ) + 4yc Substituting into Equation 1 gives [(5 − wp )yc + 2yc2 ]3 182 = 9.81 (5 − wp ) + 4yc

(2)

The energy equation requires that V12 V2 = yc + c 2g 2g Q2 2.05 = yc + 2gA2c

y1 +

2.05 = yc +

182 2(9.81)[(5 − wp )yc + 2yc2 ]2

which simplifies to

[ ]1 2 1 16.51 wp = 5 + 2yc − yc 2.05 − yc Simultaneous solution of Equations 2 and 3 yields yc = 1.61 m

and

(3)

wp = 4.42 m

Therefore, the maximum width of the pier that will not cause a rise in the upstream water surface is 4.42 m . 4.39. From given data: Q = 15 m3 /s, b = 4.5 m, y1 = 1.9 m, ∆z = 0.15 m, m = 1.5. Let b′ be the bottom width at the step, then b′ = b + 2m∆z = 4.5 + 2(1.5)(0.15) = 4.95 m A1 = (b + my1 )y1 = (4.5 + 1.5 × 1.9)(1.9) = 13.97 m2 Q 15 V1 = = = 1.07 m/s A1 13.97 Q 15 15 V2 = = ′ = 2 A2 b y2 + my2 4.95y2 + 1.5y22 The energy equation gives V12 V2 = y2 + 2 + ∆z 2g 2g [ ]2 1.072 1 15 1.9 + = y2 + + 0.15 2(9.81) 2(9.81) 4.95y2 + 1.5y22 11.47 1.96 = y2 + + 0.15 (4.95y2 + 1.5y22 )2 y1 +

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Solving for y2 gives y2 = 1.73 m, 0.52 m These depths correspond to subcritical and supercritical flow conditions respectively. Since the upstream flow is subcritical, the flow at the step is also subcritical and y2 = 1.73 m At the maximum step height, y2 = yc and the energy equation gives ( )2 V12 1 Q y1 + = yc + + ∆zm 2g 2g Ac 11.47 1.96 = yc + + ∆zm A2c which can be written as ∆zm = 1.96 − yc −

11.47 [(4.5 + 3∆zm )yc + 1.5yc2 ]2

(1)

Under critical flow conditions, Q2 A3 = c g Tc which, in this case, can be written as 152 [(4.5 + 3∆zm )yc + 1.5yc2 ]3 = 22.94 = 9.81 4.5 + 3(yc + ∆zm )

(2)

Solving Equations 1 and 2 gives yc = 0.719 m

and

∆zm = 0.921 m

4.40. From the given data: Q = 20 m3 /s, y = 3 m, b1 = 3 m, m1 = 1, and the downstream section is rectangular with b2 = 3 m. Conservation of energy requires that Q2 Q2 = y2 + 2 2gA1 2gA22 202 202 3+ = y + 2 2(9.81)(3 × 3 + 1 × 32 )2 2(9.81)(3y2 )2 2.265 3.06 = y2 + y22 y1 +

which yields y2 = 2.76 m or y2 = 1.07 m. These depths correspond to subcritical and supercritical flow conditions respectively. The upstream conditions are subcritical (F1 = 0.25), so the flow depth in the constriction is subcritical and equal to 2.76 m . At the minimum allowable width to prevent choking, critical conditions occur, in which case, [ ]1 3 q2 3 E1 = 2 g

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Since E1 = 3.06 m, the above relation gives [ ]1 3 (20/b)2 3 3.06 = 2 9.81 which yields b = 2.19 m. Therefore, the minimum allowable width of the constriction to prevent choking in 2.19 m . 4.41. From the given data: S0 = 0.05% = 0.0005, m = 2, b = 5 m, and Q = 7 m3 /s. For a float-finished concrete channel, Table 4.2 gives n = 0.015. (a) The depth of flow, y, is given by the Manning equation as 5

1 A 3 12 Q= S n P 23 0 5

7=

1 [5y + 2y 2 ] 3 1 (0.0005) 2 √ 0.015 [5 + 2 1 + 22 y] 23

which yields y = 0.897 m For the Manning equation to be valid, the flow must be fully turbulent, which requires that √ n6 RS0 ≥ 9.6 × 10−14 (1) where R=

5(0.897) + 2(0.897)2 A √ = = 0.676 m P 5 + 2 1 + 22 (0.897)

Substituting into Equation 1 yields √ √ n6 RS0 = (0.015)6 (0.676)(0.0005) = 2.09 × 10−13 ≥ 9.6 × 10−14 Hence, the flow is fully turbulent. For the Manning’s n to be independent of the flow depth, R 4< < 500 (2) ks where ks ≈ (n/0.039)6 . In this case, R 0.676 = = 209 ks (0.015/0.039)6 which is within the range given by Equation 2. Based on the criteria given by Equations 1 and 2, the Manning equation is valid . (b) The head loss, hL , in the contraction can be estimated using the relation 2 V22 V 1 hL = C α2 − α1 2g 2g

90

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Since the contraction is abrupt, C = 0.6, and V1 and V2 are given by Q 7 = = 1.15 m/s A1 5(0.897) + 2(0.897)2 Q 7 V2 = = A2 4y2 + 2y22

V1 =

and α1 and α2 can be taken as unity. The energy equation requires that y1 +

V12 V2 = y2 + 2 + h L 2g 2g

(4)

Combining Equations 3 and 4 gives y1 + (1 + C)

V12 V2 = y2 + (1 + C) 2 2g 2g

(5)

Substituting known quantities gives 1.152 0.897 + (1 + 0.6) = y2 + (1 + 0.6) 2(9.81)

(

7 4y2 + 2y22

)2

1 2(9.81)

which simplifies to 1.005 = y2 +

3.996 (4y2 + 2y22 )2

which yields y2 = 0.815 m (subcritical)

or

0.609 m (supercritical)

At the upstream section, the Froude number is given by V1 1.15 Fr1 = √ =√ = 0.44 (subcritical) gD (9.81)(0.709) Therefore, the flow in the contracted section is subcritical and equal to 0.815 m . If the head loss is ignored, V12 V2 = y2 + 2 2g 2g )2 ( 2 1 1.15 7 0.897 + = y2 + 2 2(9.81) 2(9.81) 4y2 + 2y2 y1 +

which simplifies to 0.964 = y2 +

2.497 (4y2 + 2y22 )2

which yields y2 = 0.861 m (subcritical) Taking the head loss into account has a significant effect on the calculated flow depth in the contracted section (0.815 m vs. 0.861 m), with a depth difference of 5% when head loss is taken into account.

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4.42. Let Section 1 be the upstream section, Section 2 be the “throat” section, and Section 3 be the downstream section. From the given data: Q = 100 m3 /s, n = 0.025, S0 = 0.5%, y1 = 3.000 m, b1 = b3 = 30 m, b2 = 20 m, L12 = L23 = 40 m. Neglecting Energy Losses: The energy equation applied between sections 1 and 2 is given by [ ] [ ] V22 V12 y2 + = y1 + + (z1 − z2 ) (1) 2g 2g From the given data: y1 = 3.000 m A1 = b1 y1 = (30)(3.000) = 90 m2 Q 100 V1 = = = 1.111 m/s A1 90 z1 − z2 = L12 S0 = (40)(0.005) = 0.20 m P1 = b1 + 2y1 = 30 + 2(3.000) = 36.00 m A1 90 R1 = = = 2.500 m P1 36 A2 = b2 y2 = 20y2 Q 100 5 V2 = = = A2 20y2 y2 Substituting the calculated parameters into Equation 1 gives [ ] [ ] (5/y2 )2 (1.111)2 y2 + = 3.000 + + 0.20 2(9.81) 2(9.81) which yields y2 = 3.133 m . Between sections 2 and 3, the energy equation is [

] [ ] V32 V22 y3 + = y2 + + (z2 − z3 ) 2g 2g

where y2 = 3.133 m A2 = b2 y2 = (20)(3.133) = 62.66 m2 Q 100 = V2 = = 1.596 m/s A2 62.66 z2 − z3 = L23 S0 = (40)(0.005) = 0.20 m P2 = b2 + 2y2 = 20 + 2(3.133) = 26.27 m A2 62.66 R2 = = = 2.385 m P2 26.27 A3 = b3 y3 = 30y3 100 3.333 Q = = V3 = A3 30y3 y3

92

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Substituting the calculated parameters into Equation 2 gives [ ] [ ] (3.333/y3 )2 1.5962 y3 + = 3.133 + + 0.20 2(9.81) 2(9.81) which yields y3 = 3.414 m . Accounting for Energy Losses: The energy equation applied between sections 1 and 2 is given by [ ] [ ] V22 V12 y2 + = y1 + + (z1 − z2 ) − he − hf (3) 2g 2g where he and hf are the energy losses due to expansion/contraction and friction and are given by 2 V2 V12 (4) he = C − 2g 2g hf = Sf L (5) where C = 0.1 for contractions . Assuming that the friction slope, Sf , is approximately the same at all sections then Sf can be calculated at section 1 as [ Sf =

nQ 2

[

]2 =

AR 3

(0.025)(100) 2

]2 = 0.000227

(90)(2.500) 3

Between sections 1 and 2, Equations 4 and 5 give [ ] ( ) (5/y2 )2 1.1112 25 he = (0.1) − = 0.00510 − 1.234 2(9.81) 2(9.81) y22 hf = Sf L = (0.000227)(40) = 0.00908 m Substituting the calculated parameters into the energy equation (Equation 3) gives [ ] [ ] ( ) (5/y2 )2 (1.111)2 25 y2 + = 3.000 + + 0.2 − 0.00510 − 1.234 − 0.00908 2(9.81) 2(9.81) y22 which yields y2 = 3.116 m . Hence, A2 = (20)(3.116) = 62.32 m2 Q 100 V2 = = = 1.605 m A2 62.32 A3 = 30y3 Q 100 3.333 V3 = = = A3 30y3 y3 The energy equation applied between sections 2 and 3 is given by [ ] [ ] V32 V22 y3 + = y2 + + (z2 − z3 ) − he − hf 2g 2g

93

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where he and hf are the energy losses due to expansion/contraction and friction and are given by 2 V3 V22 (7) he = C − 2g 2g hf = Sf L (8) where C = 0.3 for expansions. Between sections 2 and 3, Equations 7 and 8 give [ ] ( ) (3.333/y2 )2 1.6052 11.11 he = (0.3) − = 0.01529 2.576 − 2(9.81) 2(9.81) y32 hf = Sf L = (0.000227)(40) = 0.00908 m Substituting the calculated parameters into the energy equation (Equation 6) gives [ ) ] [ ] ( (3.333/y3 )2 (1.605)2 1.11 − 0.00908 y3 + = 3.116 + + 0.2 − 0.01529 2.576 − 2 2(9.81) 2(9.81) y3 which yields y3 = 3.350 m . Without considering energy losses, the stage difference between the upstream and downstream sections is (3.000 m + 0.4 m) − 3.414 m = −0.014 m . Taking energy losses into account, the stage difference between the upstream and downstream sections is (3.000 m + 0.4 m) − 3.350 m = −0.050 m . Therefore an error of approximately (14-50)/50 = 72% is introduced by neglecting energy losses. 4.43. Taking Section 1 upstream of the bridge constriction, Section 2 at the bridge constriction, and Section 3 downstream of the bridge constriction (after expansion) then, from the given data, b1 = 10 m, b2 = 7 m, b3 = 10 m, Q = 20 m3 /s, and y1 = 2 m. (a) The specific energy, E1 , at section 1 is given by E1 = y1 +

Q2 V12 202 = y1 + =2+ = 2.051 m 2 2 2g 2(9.81)(10)2 (2)2 2gb1 y1

where it is noted that V1 = 20/(10 × 2) = 1 m/s. Applying the energy equation between sections 1 and 2 gives ) ( 2 V12 V2 E 2 = E 1 − Cc − (1) 2g 2g where Cc = 0.6 for an abrupt contraction, and it is assumed that V2 > V1 . Substituting the given and derived data into Equation 1 gives ( ) Q2 Q2 V12 y2 + = 2.051 − 0.6 − 2g 2gb22 y22 2gb22 y22 ( ) 202 202 12 y2 + = 2.051 − 0.6 − 2(9.81)(7)2 y22 2(9.81)(7)2 y22 2(9.81) 0.6657 y2 + = 2.082 y22

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which yields y2 =1.897 m , E2 = 2.013 m, and V2 = 1.506 m/s (confirming V2 > V1 ). Applying the energy equation between sections 2 and 3 gives ( E 3 = E 2 − Ce

V22 V32 − 2g 2g

) (2)

where Ce = 0.8 for an abrupt expansion, and it is assumed that V2 > V3 . Substituting the given and derived data into Equation 2 gives ( 2 ) V2 Q2 Q2 y3 + = 2.013 − 0.8 − 2g 2gb23 y32 2gb23 y32 ) ( 202 1.5062 202 y3 + = 2.013 − 0.8 − 2(9.81) 2(9.81)(10)2 y32 2(9.81)(10)2 y32 0.3670 y3 + = 1.968 y32 which yields y3 =1.862 m , E3 = 1.921 m, and V3 = 1.074 m/s (confirming V2 > V3 ). (b) If energy losses are neglected, the energy equation is given by E1 = E2 y2 +

Q2

= 2.051 2gb22 y22 202 y2 + = 2.051 2(9.81)(7)2 y22 which yields y2 = 1.941 m. When energy losses are taken into account, it was found that y2 = 1.897 m, and hence the error in the flow depth at section 2 associated with neglecting energy losses in the contraction is 2.3% . If energy losses are neglected in both the contraction and the expansion, then y3 = y1 = 2 m. When energy losses are taken into account, it was found that y3 = 1.862 m, and hence the error in the flow depth at section 3 associated with neglecting energy losses in the contraction and subsequent expansion is 7.4% . It is apparent that neglecting energy losses has a significant effect in this case. 4.44. From the given data: Q = 36 m3 /s, b = 10 m, n = 0.030, S0 = 0.001, and y = 3 m. Determine the normal depth, using the Manning equation 1 A3 √ Q= S0 n P 23 5

1 (10yn ) 3 √ 0.001 0.030 (10 + 2yn ) 23 5

36 = which gives

yn = 2.45 m

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Determine the critical depth, ( yc =

q2 g

) 13

( =

3.62 9.81

) 13 = 1.09 m

Since y > yn > yc , the water surface follows a M1 profile . Determine Sf using the Manning equation, ]2 [ ]2 [ ]2 [ [ ] 2 2 2 nQ 2 nQP 3 nQ(10 + 2y) 3 (0.030)(36)(10 + 2 × 3) 3 Sf = = = = = 0.00056 2 5 5 5 AR 3 A3 (10y) 3 (10 × 3) 3 The slope of the water surface is given by S0 − Sf dy 0.001 − 0.00056 = = ( )2 = 0.00046 2 dx 1 − Fr 1.2 1 − 9.81(3) where the velocity is taken as V = Q/A = 36/(10×3) = 1.2 m/s. If y = 2 m, then yn > y > yc and the water surface follows a M2 profile . Therefore the shape of the water surface would be different than when y = 3 m. 4.45. From the given data: Q = 30 m3 /s, b = 8 m, and n = 0.035. [ 2 ] 13 [ ]1 q (30/8)2 3 yc = = = 1.13 m g 9.81 [ ]2 [ ]2 [ ]2 2 2 2 nQP 3 nQ(8 + 2yc ) 3 (0.035)(30)(8 + 2 × 1.13) 3 Sc = = = = 0.016 5 5 5 A3 (8yc ) 3 (8 × 1.13) 3 Hence for Mild Slope: 0 < S0 < 0.016 and for Steep Slope: S0 > 0.016 . 4.46. From given data: b = 6 m, m = 2, n = 0.045, S0 = 0.015, Q = 80 m3 /s, and y = 5 m. Calculate normal depth using Manning equation, 1 A3 √ Q= S0 n P 32 5

1 (byn + 2yn2 ) 3 √ 1 (6yn + 2yn2 ) 3 √ S0 = 0.015 √ √ 2 0.045 (b + 2 5yn ) 3 0.045 (6 + 2 5yn ) 23 5

80 =

5

which gives yn = 2.21 m Calculate the critical depth, Q2 A3 = c g Tc 802 (byc + 2yc2 )3 = 9.81 (b + 4yc ) (6yc + 2yc2 )3 652.4 = (6 + 4yc )

96

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which gives yc = 2.07 m Since y > yn > yc , the water surface has a M1 profile and the depth increases in the downstream direction. The slope, Sf , of the energy grade line is given by the Manning equation as [ Sf =

2

nQP 3 5

A3

]2

[ =

]2 √ 2 nQ(b + 2 5y) 3 5

(by + 2y 2 ) 3

[ =

]2 √ 2 (0.045)(80)(6 + 2 5 × 5) 3 5

(6 × 5 + 2 × 52 ) 3

= 0.000508

Other hydraulic parameters are A = by + 2y 2 = (6)(5) + 2(5)2 = 80 m2 Q 80 V = = = 1 m/s A 80 T = b + 2my = (6) + 2(2)(5) = 26 m A 80 D= = = 3.08 m T 26 V2 12 Fr2 = = = 0.0331 gD 9.81(3.08) The slope of the water surface is therefore given by S0 − Sf dy 0.015 − 0.000508 = = = 0.0150 2 dx 1 − 0.0331 1 − Fr The depth, yu , 100 m upstream is given by yu = 5 − 100(0.0150) = 3.50 m and the depth, yd , 100 m downstream is given by yd = 5 + 100(0.0150) = 6.50 m 4.47. From the given data: H = 2.00 m, b = 3.00 m, m = 3, S0 = 0.005, and n = 0.025. Assuming α = 1 and that the slope is hydraulically steep (i.e., yn < yc ), Equations 4.129 and 4.130 require that Ac H = yc + 2Tc Substituting the geometric properties for a trapezoidal channel gives byc + myc2 2(b + 2myc ) 3yc + 3yc2 2.00 = yc + 2(3 + 2(3)yc ) H = yc +

97

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which yields yc = 1.52 m. The corresponding value of Q is then given by √ √ gA3c (9.81)(3yc + 3yc2 )3 Q= = = 35.3 m3 /s Tc 3 + 2(3)yc Determine the normal depth of flow, yn , corresponding to Q = 35.3 m3 /s by applying the Manning equation which requires that 5

5

1 1 1 An3 12 (byn + myn2 ) 3 2 Q= √ 2 S0 = 2 S0 nP3 n (b + 2yn 1 + m2 ) 3

n

5

1 1 (3yn + 3yn2 ) 3 2 35.3 = √ 2 (0.005) 0.025 (3 + 2yn 1 + 32 ) 3

which yields yn = 1.62 m. Since yn > yc , the slope is mild and so the initial assumption of a steep slope that was used in determining Q is not validated. Restart the problem with the assumption that the slope is mild. Equations 4.131 and 4.132 require that H = yn +

Q2 2gA2n

and Q =

2 1 1 An Rn3 S02 n

Eliminating Q from these equations yields the more convenient combined form as ( H = yn +

2 1 1 An Rn3 S02 n

)2

4

4

1 Rn3 S0 An3 S0 = yn + = yn + 4 2 2 2gAn 2gn 2gn2 Pn3

Substituting the geometric properties for a trapezoidal channel gives 4

(byn + myn2 ) 3 S0 H = yn + √ 4 2gn2 (b + 2yn 1 + m2 ) 3 4

(3yn + 3yn2 ) 3 (0.005) 2.00 = yn + √ 4 2(9.81)(0.025)2 (3 + 2yn 1 + 32 ) 3 which yields yn = 1.61 m. The corresponding value of Q is given by Manning’s equation, 5

5

1 1 An3 21 1 (3yn + 3yn2 ) 3 2 = 34.8 m3 /s Q= √ 2 S0 = 2 (0.005) 2 nP3 0.025 (3 + 2yn 1 + 3 ) 3

n

Determine the critical depth of flow, yc , using the critical-flow requirement Q2 g 34.82 9.81

= =

A3c Tc (3yc + 3yc2 )3 3 + 2(3)yc

which yields yc = 1.52 m. Since yn > yc (i.e., 1.61 m > 1.52 m), the slope is mild and so the revised assumption of a mild slope that was used in determining Q is validated. The discharge from the reservoir is 34.8 m3 /s .

98

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4.48. According to the momentum equation ∑

Fx = ρQ(V2 − V1 )

In this case, (

) Q Q γA1 y¯1 − γA2 y¯2 + γV sin θ = ρQ − A2 A1 ( ) Q Q A1 y¯1 − A2 y¯2 + V sin θ = Q − gA2 gA1 For a rectangular channel of width b, A1 = by1 A2 = by2 y1 y¯1 = 2 y2 y¯2 = 2 y1 + y2 5y2 b V= 2 sin θ = S0 Q = qb and substituting into the momentum equation gives b

y12 y 2 y1 + y2 q 2 b2 q 2 b2 −b 2 + 5y2 bS0 = − 2 2 2 gby2 gby1 y12 y22 y1 + y2 q2 q2 − + 5y2 S0 = − 2 2 2 gy2 gy1

or y12 q2 y2 q2 y1 + y2 + = 2 + − 5y2 S0 2 gy1 2 gy2 2 4.49. From given data: Q = 100 m3 /s, b = 8 m, y1 = 0.9 m, and q=

Q 100 = = 12.5 m2 /s b 8

and y1 y2 = 2 y1 = 2

(

)

√

−1 + 1 + √ ( −1 +

8Fr21

q2 1+8 3 gy1

)

0.9 = 2

(

99

−1 +

√

12.52 1+8 (9.81)(0.9)3

) = 5.52 m

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The energy loss, ∆E, is given by ∆E =

(y2 − y1 )3 (5.52 − 0.9)3 = = 4.96 m 4y1 y2 4(5.52)(0.9)

and the initial energy, E1 , is given by E1 = y1 +

V12 (12.5/0.9)2 = 0.9 + = 10.73 m 2g 2(9.81)

Therefore, the fraction of initial energy lost is 4.96 ∆E = = 0.462 E1 10.73 4.50. The head loss, hL , is defined by y1 + Dividing by y1 gives 1+

V12 V2 = y2 + 2 + h L 2g 2g

V12 y2 V2 hL = + 2 + 2gy1 y1 2gy1 y1

which can be put in the form hL y2 V2 V2 =1− + 1 − 2 y1 y1 2gy1 2gy1 Define Fr21 =

V12 gy1

(2)

Combining Equations 1 and 2 gives [ ] hL y2 Fr21 V22 =1− + 1− y1 y1 2 gy1 Fr21   y2 Fr21  V22 ( 2 ) + 1− =1− V1 y1 2 gy =1−

Fr21

y2 + y1 2

1

[ 1−

V22 V12

]

gy1

Since V1 = q/y1 and V2 = q/y2 , then [ ] hL y2 Fr21 (q/y2 )2 =1− + 1− y1 y1 2 (q/y1 )2 which simplifies to

[ ( )2 ] hL y2 Fr21 y1 =1− + 1− y1 y1 2 y2

100

(1)

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4.51. (a) From the given data: Q = 20 m3 /s, y1 = 1 m, b1 = 1 m, and m = 2, which gives A1 = b1 y1 + my12 = (1)(1) + (2)(1)2 = 3 m2 T1 = b1 + 2my1 = (1) + 2(2)(1) = 5 m A1 3 D1 = = = 0.6 m T1 5 Q 20 V1 = = = 6.67 m/s A1 3 V2 6.672 Fr21 = 1 = = 7.56 gD1 (9.81)(0.6) Since Fr1 > 1 a hydraulic jump will occur in the channel. (b) The upstream and downstream depths in a hydraulic jump are related by Q2 Q2 + A1 y¯1 = + A2 y¯2 gA1 gA2 where 1 A1 y¯1 = my12 · 2 my13 + = 3

y1 y1 1 y1 + by12 · + my12 · 3 2 2 3 by13 2(1)3 1(1)3 = + = 1.167 m3 2 3 2

Substituting the given and derived data: (y ) 202 202 2 + 1.167 = + (5y2 ) (9.81)(3) (9.81)(5y2 ) 2 8.15 14.76 = + 2.5y22 y2 which yields y2 = 2.08 m, 0.586 m, and −2.67 m. Hence the only feasible (subcritical) depth in the rectangular channel is 2.08 m . The energy loss in the hydraulic jump is equal to the change in specific energy, E1 − E2 , where V12 6.672 =1+ = 3.268 m 2g 2(9.81) 20 Q = V2 = = 1.92 m/s A2 2.08 × 5 V2 1.922 E2 = y2 + 2 = 2.08 + = 2.27 m 2g 2(9.81)

E1 = y1 +

Therefore, the energy loss is 3.628 m − 2.27 m = 1.358 m, and the power loss is given by Power loss = γQ∆E = (9.79)(20)(1.358) = 266 kW

101

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4.52. The general hydraulic jump equation is given by Q2 + A¯ y = constant gA For a trapezoidal channel,

( y¯ =

by 2 2

)

(

my 3 3 2 my

(1)

)

+

by +

(2)

Combining Equations 1 and 2 yields ( Q2 g(by + my 2 )

+ (by + my 2 )

which simplifies to Q2 + gy(b + my)

(

by 2 2

by 2 2

)

(

by +

)

( +

my 3 3 2 my

)

+

my 3 3

= constant

) = constant

which demonstrates that Q2 + gy1 (b + my1 )

(

by12 2

)

( +

my13 3

)

Q2 = + gy2 (b + my2 )

(

by22 2

)

( +

my23 3

)

4.53. From the given data: Q = 21 m3 /s, b = 2 m, m = 1, and y1 = 1 m. The momentum equations requires that Q2 Q2 + A1 y¯1 = + A2 y¯2 (1) gA1 gA2 where A1 = cby1 + y12 = 2(1) + 12 = 3 m2 Q 21 V1 = c = = 7 m/s A1 3 A1 = cby2 + y22 = 2y2 + y22 Q 21 V1 = c = A2 2y2 + y22 and ∑ (1)1 1 1 1(2)(0.5) + (1)1 A¯ y 2 3 + 2 3 y¯1 = ∑ = = 0.44 m A 3 ( ) [ ] 2y2 y22 + 2 12 y22 y32 y2 + y22 /3 y¯2 = = 2 + y2 2y2 + y22 Substituting into Equation 1 gives 212 212 y23 2 + y + + 3(0.44) = 2 9.81(3) 3 9.81(2y2 + y22 )

102

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which yields y2 = 2.59 m The energy equation gives the energy loss, ∆E, as ( ∆E = y1 +

V12 2g

− y2 −

V22 2g

=1+

72 2(9.81)

− 2.59 −

21 2×2.59+2.592

)2

2(9.81)

= 0.748 m

4.54. From the given data: m = 2, Q = 0.30 m3 /s, y = 15 cm, and A = my 2 = 2(0.15)2 = 0.045 m2 T = 2my = 2(2)(0.15) = 0.6 m 0.045 A = = 0.075 m D= T 0.6 Q 0.30 V = = = 6.67 m/s A 0.045 6.67 V =√ Fr = √ = 7.78 gD (9.81)(0.075) Since Fr = 7.78 > 1, the flow is supercritical.

The hydraulic jump equation is the same as for a trapezoidal channel with b = 0, hence my13 Q2 my23 Q2 + = + 3 3 gmy12 gmy22 m(0.15)3 (0.30)2 2y23 (0.30)2 + + = 3 (9.81)(2)(0.15)2 3 (9.81)(2)y22 0.00459 0.219 = 0.667y23 + y22 which yields y2 = 0.679 m

or

0.145 m

Since the downstream flow is subcritical, y2 = 0.679 m . 4.55. From given data: Q = 10 m3 /s, b = 5.5 m, S0 = 0.0015, n = 0.038, y2 = 2.2 m.

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(a) Using the direct-integration method, ( S0 − y1 = y2 −

2

nQP¯ 3 ¯ 35 A

)2

(x2 − x1 ) ¯2 1 − Vgy¯ ( )2 2 nQ(b+2¯ y) 3 S0 − 5 (b¯ y) 3 = y2 − (x2 − x1 ) 2 1 − gbQ2 y¯3 ( )2 2 (0.038)(10)(5.5+2¯ y) 3 0.0015 − 5 (5.5¯ y) 3 (100 − 0) = 2.2 − 2 10 1 − (9.81)(5.5) 2y ¯3

where

(1)

y1 + y2 y1 + 2.2 = 2 2

y¯ =

(2)

Solving Equations 1 and 2 gives y1 = 2.12 m (b) Using the standard-step equation [ y+ ∆L =

]1 V2 2g 2

(3)

S¯f − S0

This equation is solved iteratively until ∆L = 100 m, and the iterations are summarized in the following table: y2 2.2 2.2 2.2 2.2

A2 12.1 12.1 12.1 12.1

P2 9.4 9.4 9.4 9.4

R2 1.29 1.29 1.29 1.29

V2 0.826 0.826 0.826 0.826

S2 0.00070 0.00070 0.00070 0.00070

y1 2.20 2.10 2.11 2.12

A1 12.1 11.6 11.6 11.7

P1 9.4 9.2 9.22 9.24

R1 1.29 1.26 1.26 1.26

Therefore y1 = 2.12 m . Find the uniform flow depth, yn , using the Manning equation 1 A3 √ Q= S0 n P 23 5

which can be written as 1 (5.5yn ) 3 √ S0 n (5.5 + 2yn ) 23 5

10 =

1 (5.5yn ) 3 √ 10 = 0.0015 0.038 (5.5 + 2yn ) 23 5

104

V1 0.826 0.866 0.862 0.857

S1 0.00070 0.00080 0.00079 0.00078

S¯f 0.00070 0.00075 0.00075 0.00074

∆L 0 129 115 100

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which gives yn = 1.719 m Plugging this value of y into the direct-step equation, Equation 3 gives y2 2.2

A2 12.1

P2 9.4

R2 1.29

V2 0.826

S2 0.000704

y1 1.719

A1 9.455

P1 8.94

R1 1.00

V1 1.06

S1 0.00149

S¯f 0.001

∆L 1230

Therefore ∆L = 1230 m . 4.56. From the given data: Q = 5 m3 /s, b = 4 m, S0 = 0.04, n = 0.05, and y2 = 1.5 m. Let the given section be Section 2, then A2 = by2 = (4)(1.5) = 6 m2 P2 = b + 2y2 = 4 + 2(1.5) = 7 m A2 6 R2 = = = 0.857 m P2 7 Q 5 V2 = = = 0.833 m/s A2 6  2 [ ]2 (0.05)(5) nQ  = S2 =  = 0.00213 2 2 3 (6)(0.857) 3 A2 R2 Taking y1 = 1 m, then A1 = 4 m2 ,

P1 = 6 m,

R1 = 0.667 m,

V1 = 1.25 m/s,

S1 = 0.00670

and S1 + S2 0.00213 + 0.00670 S¯f = = = 0.00442 2 2 and ∆L = =

[y1 + V12 /2g] − [y2 + V22 /2g] S¯f − S0 [1 + 1.252 /2(9.81)] − [1.5 + 0.8332 /2(9.81)] = 12.9 m 0.00442 − 0.04

Therefore the depth is equal to 1 m at a location 12.9 m upstream . 4.57. Use the direct-step method where [ ]1 y + V 2 /2g 2 ∆L = S¯f − S0

105

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At Section 1, y1 = 1.6 m, A1 = [b + my]y = [2.8 + 2(1.6)](1.6) = 9.6 m2 √ √ P1 = b + 2 1 + m2 y = 2.8 + 2 5(1.6) = 9.96 m A1 9.6 R1 = = = 0.964 m P1 9.96 Q 20 V1 = = = 2.08 m2 A1 9.6  2 [ ]2 Qn 20 × 0.015  = Sf 1 =  = 0.00103 2 2 3 3 (9.6)(0.964) A1 R1 At Section 2, y2 = 1.4 m, A2 = [b + my]y = [2.8 + 2(1.4)](1.4) = 7.84 m2 √ √ P2 = b + 2 1 + m2 y = 2.8 + 2 5(1.4) = 9.06 m A2 7.84 R2 = = = 0.865 m P2 9.06 Q 20 V2 = = = 2.55 m2 A2 7.84 2 [  ]2 20 × 0.015 Qn  = Sf 2 =  = 0.00178 2 2 3 (7.84)(0.865) 3 A2 R2 Therefore

Sf 1 + Sf 2 0.00103 + 0.00178 = = 0.00141 S¯f = 2 2 Substituting data into direct-step equation gives [ ] [ ] 2.082 2.552 1.6 + 2×9.81 − 1.4 + 2×9.81 ∆L = = −10.5 m 0.00141 − 0.01 Therefore ∆L = 10.5 m downstream 4.58. From the given data: S0 = 0.01, m = 3, b = 3.00 m, n = 0.015, Q = 20 m3 /s, and y = 1.00 m. (a) The normal depth of flow is calculated using the Manning equation: 5

1 2 1 A 3 21 1 S Q = AR 3 S02 = n n P 23 0 5

1 1 [3yn + 3yn2 ] 3 2 20 = √ 2 (0.01) 0.015 [3 + 2 1 + 32 yn ] 3

106

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which gives yn = 0.82 m For the Manning equation to be valid, √ n6 RS0 ≥ 9.6 × 10−14 In this case, R=

A 3(0.82) + 3(0.82)2 √ = = 0.55 m P 3 + 2 1 + 32 (0.82)

which gives n6

√

RS0 = (0.015)6

√

(0.55)(0.01) = 8.45 × 10−13 ≥ 9.6 × 10−14

Hence, the Manning equation is valid, and yn = 0.82 m . (b) When flow conditions are critical, A3 Q2 = g T which requires that 202 [3yc + 3yc2 ]3 = 9.81 3 + 2(3)yc which gives yc = 1.15 m (c) Since yc > yn the slope is a steep slope . Since yn < y < yc , the water surface has a S2 profile . (d) In a S2 water surface profile, the depth decreases in the downstream section, so the depth of 1.1 m must occur upstream of the section where the depth is 1 m. At the location where the depth is 1.1 m: y1 = 1.1 m A1 = [3 + 3y1 ]y1 = [3 + 3(1.1)](1.1) = 6.93 m2 √ √ P1 = 3 + 2 10y1 = 3 + 2 10(1.1) = 9.96 m A1 6.93 R1 = = = 0.696 m P1 9.96 Q 20 V1 = = = 2.89 m/s A1 6.93 2 [  ]2 nQ (0.015)(20)  = Sf 1 =  = 0.00304 2 2 (6.93)(0.696) 3 A1 R 3 1

107

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and where the depth is 1.00 m: y2 = 1.00 m A2 = [3 + 3y2 ]y2 = [3 + 3(1.00)](1.00) = 6.00 m2 √ √ P2 = 3 + 2 10y2 = 3 + 2 10(1.00) = 9.32 m 6.00 A2 R2 = = = 0.644 m P2 9.32 20 Q = = 3.33 m/s V2 = A2 6.00 2 [  ]2 nQ (0.015)(20)  = Sf 2 =  = 0.00450 2 2 3 (6.00)(0.644) 3 A2 R2 Substituting the hydraulic parameters at sections 1 and 2 into the direct-step equation and taking α = 1 gives [ ] 2 1 y + α V2g 2 ∆L = ¯ Sf − S0 ( ) ( ) 2.892 3.332 1.1 + 2×9.81 − 1.00 + 2×9.81 ( 0.00304+0.00450 ) = − 0.01 2 = 6.26 m Hence, the depth in the channel increases to 1.1 m at a location that is approximately 6.26 m upstream of the section where the depth is 1.0 m. (e) The specific energy, E1 , at the gaging station is given by E1 = y1 +

V12 3.332 = 1.0 + = 1.565 m 2g 2(9.81)

Just downstream of the gaging station (at the hump), the specific energy, E2 , is given by E2 = E1 − 0.2 = 1.565 − 0.2 = 1.365 m Therefore, y2 + y2 +

Q2 = 1.365 2gA2

202 = 1.365 2(9.81)(3y2 + 3y22 )2 20.4 y2 + = 1.365 (3y2 + 3y22 )2

There is no solution to this equation. Therefore, critical depth will occur over the hump, the flow will be choked, and y2 = yc = 1.15 m

108

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4.59. Use the direct-step method where [ ]1 y + V 2 /2g 2 ∆L = S¯f − S0 At Section 1, y1 = 0.9 m, A1 = by = (6)(0.9) = 5.4 m2 P1 = b + 2y = 6 + 2(0.9) = 7.8 m A1 5.4 R1 = = = 0.692 m P1 7.8 Q 0.8 V1 = = = 0.148 m/s A1 5.4 5

1 A13 12 Q= S n1 P 23 0 1

5 2

n1 =

A1

5

5.4 3

1 2

2 3

S0 =

QP1

(0.8)(7.8)

1

2 3

0.005 2 = 0.374

At Section 2, y2 = 0.7 m, A2 = by = (6)(0.7) = 4.2 m2 P2 = b + 2y = 6 + 2(0.7) = 7.4 m A2 4.2 R2 = = = 0.568 m P2 7.4 Q 0.8 V2 = = = 0.190 m/s A2 4.2 5

n2 =

A23

5

4.2 3

1 2

2 3

QP2

S0 =

(0.8)(7.4)

1

2 3

0.005 2 = 0.254

Based on the information given in this problem, S¯f can be estimated using the following relations, n1 + n2 0.374 + 0.254 = = 0.314 2 2 A1 + A2 5.4 + 4.2 A¯ = = = 4.8 m2 2 2 ¯ = R1 + R2 = 0.692 + 0.568 = 0.630 m R 2 2 ( )2 ( ) n ¯Q 0.314 × 0.8 2 ¯ Sf = = = 0.00507 2 ¯ 23 A¯R 4.8 × 0.630 3 n ¯=

Substituting data into direct-step equation gives [ ] [ 0.1482 0.9 + 2×9.81 − 0.7 + ∆L = 0.00507 − 0.005

109

0.1902 2×9.81

] = 2850 m

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Therefore ∆L = 2850 m 4.60. From the given data: ∆L = 100 m, b = 5 m, y1 = 1 m, y2 = 0.9 m, Q = 2.5 m3 /s, and S0 = 0.5%. The energy equation requires that: [ ] 2 1 y + V2g 2 ∆L = (1) S − S0 The following parameters can be calculated from the given data: A1 = by1 = (5)(1) = 5 m2 P1 = b + 2y1 = 5 + 2(1) = 7 m 5 A1 R1 = = = 0.7143 m P1 7 Q 2.5 V1 = = = 0.5 m/s A1 5 A2 = by2 = (5)(0.9) = 4.5 m2 P2 = b + 2y2 = 5 + 2(0.9) = 6.8 m A2 4.5 R2 = = = 0.6618 m P2 6.8 Q 2.5 V2 = = = 0.5556 m/s A2 4.5 S0 = 0.005 Substituting the calculated parameters into Equation 1 yields [ ]1 V2 1 y+ S = S0 + ∆L 2g 2 [( ) ( )] 1 0.52 0.55562 S = 0.005 + 1+ − 0.9 + = 0.005970 100 2(9.81) 2(9.81) 2  [ ]2 2.5 nQ  = n2 = 0.3915n2 Sf 1 =  2 2 3 3 (5)(0.7143) A 1 R1 2  [ ]2 nQ 2.5 2  =n Sf 2 =  = 0.5352n2 2 2 (4.5)(0.6618) 3 A2 R23 1 1 S = (Sf 1 + Sf 2 ) = (0.3915 + 0.5352)n2 = 0.4634n2 2 2 Combining Equations 8.111 and 8.112 gives 0.005970 = 0.4634n2 which yields n = 0.114 .

110

(2)

(3)

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4.61. From the given data: Q = 11 m3 /s, b = 5 m, S0 = 0.001, n = 0.035, and y1 = 2 m. Calculate the uniform depth, yn , using the Manning equation 1 A3 √ Q= S0 n P 23 5

which gives 1 (5yn ) 3 √ 11 = 0.001 0.035 (5 + 2yn ) 23 5

and leads to yn = 2.19 m and therefore 95% of the normal depth is equal to 0.95 × 2.19 = 2.08 m. Using the standardstep method yields the following results: x (m) 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360

y2 (m) 2.00 2.01 2.01 2.02 2.02 2.03 2.03 2.04 2.04 2.05 2.05 2.06 2.06 2.06 2.07 2.07 2.08 2.08

A2 (m2 ) 10.0 10.0 10.1 10.1 10.1 10.1 10.2 10.2 10.2 10.2 10.3 10.3 10.3 10.3 10.3 10.4 10.4 10.4

P2 (m) 9.00 9.01 9.02 9.04 9.05 9.06 9.07 9.08 9.09 9.10 9.10 9.11 9.12 9.13 9.14 9.14 9.15 9.16

R2 (m) 1.11 1.11 1.11 1.12 1.12 1.12 1.12 1.12 1.12 1.13 1.13 1.13 1.13 1.13 1.13 1.13 1.13 1.14

V2 (m/s) 1.10 1.10 1.09 1.09 1.09 1.08 1.08 1.08 1.08 1.07 1.07 1.07 1.07 1.07 1.06 1.06 1.06 1.06

S2 0.0013 0.0013 0.0013 0.0013 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012

y1 (m) 2.01 2.01 2.02 2.02 2.03 2.03 2.04 2.04 2.05 2.05 2.06 2.06 2.06 2.07 2.07 2.08 2.08 2.08

A1 (m2 ) 10.0 10.1 10.1 10.1 10.1 10.2 10.2 10.2 10.2 10.3 10.3 10.3 10.3 10.3 10.4 10.4 10.4 10.4

P1 (m) 9.01 9.02 9.04 9.05 9.06 9.07 9.08 9.09 9.10 9.10 9.11 9.12 9.13 9.14 9.14 9.15 9.16 9.16

R1 (m) 1.11 1.11 1.12 1.12 1.12 1.12 1.12 1.12 1.13 1.13 1.13 1.13 1.13 1.13 1.13 1.13 1.14 1.14

V1 (m/s) 1.10 1.09 1.09 1.09 1.08 1.08 1.08 1.08 1.07 1.07 1.07 1.07 1.07 1.06 1.06 1.06 1.06 1.06

S1

S¯

0.0013 0.0013 0.0013 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012

0.0013 0.0013 0.0013 0.0013 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012

The upstream water surface profile is described by x and y2 . 4.62. From the given data: S0 = 0.0005, n = 0.040, Q = 250 m3 /s, y2 = 8 m, b2 = 12 m, m2 = 2, b1 = 16 m, and m1 = 3. Using these data A2 = b2 y2 + m2 y22 = 12(8) + 2(8)2 = 224 m2 √ √ P2 = b2 + 2y2 1 + m22 = 12 + 2(8) 1 + 22 = 47.8 m A2 224 = = 4.69 m P2 47.8 Q 250 V2 = = = 1.12 m/s A2 224 2 [  ]2 nQ (0.040)(250)  = Sf 2 =  = 0.000254 2 2 (224)(4.69) 3 A2 R 3 R2 =

2

111

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Use the standard-step equation [ y+ ∆L =

100 = 100 =

]1 V2 2g 2

= S¯f − S0

V12 2g − y2 Sf 1 +Sf 2 − 2

y1 +

V12

−

V22 2g

S0

1.122

2(9.81) − 8 − 2(9.81) Sf 1 +0.000254 − 0.0005 2 y1 + 0.05097V12 − 8.064

y1 +

0.5Sf 1 − 0.000373

(1)

where Q Q 250 = = 2 A1 b1 y 1 + m 1 y 1 16y1 + 3y12  2  2 nQ  nQ  = = 5 2 2 A1 R13 A13 /P13 [ ]2 (0.040)(250) = √ 2 5 (16y1 + 3y12 ) 3 /(16 + 2y1 10) 3

V1 =

Sf 1

(2)

(3)

Substituting Equations 2 and 3 into Equation 1 and solving for y1 gives y1 = 8.00 m Therefore the depth 100 m upstream is 8.00 m . 4.63. Let d be the depth of flow in the floodplain, and using the Horton equation, [ ]2 [ ]2 3 3 2(d + 45)(0.1) 2 + (16)(0.05) 2 3 3.023 + 0.0632d 3 ne = = 2d + 90 + 6 + 10 106 + 2d A = 30 + 100d P = 106 + 2d Substituting into the Manning equation gives 5

1 A 3 12 S n P 32 0 [ ]2 5 3 (30 + 100d) 3 1 106 + 2d 2 110 = 2 (0.0075) 3.023 + 0.0632d (106 + 2d) 3 Q=

which gives d = 0.842 m. Verify that the flow is non constrained in the main channel: 5

1 1 (10d) 3 100 = (0.0075) 2 0.05 (10 + 2d) 23

which gives d = 3.80 m. Therefore the flow is in the floodplain as calculated previously (d = 0.842 m).

112

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(a) If 15 m of the floodway is filled in, [

3

3

3

(d + 30)(0.1) 2 + (16)(0.05) 2 + (d + 45)(0.1) 2 ne = 2d + 75 + 6 + 10

]2 3

[

2.551 + 0.0632d = 91 + 2d

]2 3

A = 30 + 85d P = 91 + 2d Substituting into the Manning equation gives [

91 + 2d 110 = 2.551 + 0.0632d

]2

3

5

(30 + 85d) 3 (91 + 2d)

2 3

1

(0.0075) 2

which gives d = 0.904 m. Therefore, filling in 15 m of the floodplain causes the water-level to rise 0.904 m − 0.842 m = 0.062 m = 6.2 cm . (b) At the upstream section (Section 1), y1 = 3 + 0.904 = 3.904 m [ ]2 3.023 + 0.0632(0.904) 3 n1 = = 0.0935 106 + 2(0.904) A1 = 30 + 100(0.904) = 120.4 m2 V1 = Q/A1 = 0.914 m/s V12 = 0.0425 m 2g P1 = 106 + 2(0.904) = 107.8 m [ ]2 2 (110)(0.0935)(107.8) 3 S1 = = 0.006297 5 (120.4) 3 At the downstream section (Section 2), y2 = 54.50 − [50 − (0.0075)(150)] = 5.625 m ]2 [ 3.023 + 0.0632(2.625) 3 n2 = = 0.0937 106 + 2(2.625) A2 = 30 + 100(2.625) = 292.5 m2 V2 = Q/A1 = 0.376 m/s V22 = 0.00721 m 2g P2 = 106 + 2(2.625) = 111.3 m [ ]2 2 (110)(0.0937)(111.3) 3 S2 = = 0.0003424 5 (292.5) 3

113

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Combining the results, S1 + S2 0.006297 + 0.0003424 S¯ = = = 0.003320 2 2 ] [ 2 1 y + V2g [3.904 + 0.0425] − [5.625 + .00721] 2 = ∆L = ¯ 0.003320 − 0.0075 S − S0 which yields ∆L = 403 m . 4.64. From the given data: z1 = 102.05 m, Z1 = 105.27 m, m = 3, b1 = 20 m, S0 = 2%, n = 0.07, L = 100 m, b2 = 10 m, Q = 12 m3 /s, and y1 = 1.60 m. (a) From the given data: A1 = b1 y1 + my12 = 20(1.60) + 3(1.60)2 = 39.68 m2 √ √ P1 = b1 + 2y1 1 + m2 = 20 + 2(1.60) 1 + 32 = 30.1 m  2  2 ( )2 2 2 3 3 (0.07)(12)(30.1) nQ nQP 1   = Sf 1 =  = = 0.0003100 2 5 5 3 3 39.68 3 A1 R1 A1  2 4 4 2 3 3 P23 nQP2  2 P2  = (0.07 × 12) 10 = 0.07056 10 Sf 2 = 5 A23 A23 A23 4

1 P3 S¯f = (Sf 1 + Sf 2 ) = 0.0001550 + 0.3528 210 2 A3 2

The energy equation is ( 1.6 +

0.30242 2(9.81)

100 =

)

( − y2 +

1 2×9.81

(

12 A2

)2 )

S¯f − 0.02

Solving gives y2 = 3.586 m . Checking the Froude numbers of the upstream and downstream flows verifies that both are subcritical. (b) From the given data, the elevation of the floodplain at the bridge section is 105.27 − (0.02)(100) = 103.27 m. The elevation of the water surface at the bridge section is 102.05 − (0.02)(100) + 3.586 = 103.64 m. Therefore, since the water surface elevation is higher than the elevation of the floodplain, the floodplain will be flooded . 4.65. The usual energy equation that is used in the standard-step method is [ ] 2 B y + α V2g A ∆L = ¯ Sf − S0 which can be written as V2 V2 S¯f ∆L − S0 ∆L = yB + αB B − yA + αA A 2g 2g

114

(1)

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where

zB − zA (2) ∆L and zA and zB are the elevations of the bottom of the channel at stations A and B respectively. Denoting the water-surface elevations at A and B by ZA and ZB respectively, then S0 =

ZA = zA + yA

(3)

ZB = zB + yB

(4)

Combining Equations 1 to 4 gives V2 V2 S¯f ∆L = (zB + yB ) + αB B − (zA + yA ) − αA A 2g 2g 2 2 V V = ZB + αB B − ZA − αA A 2g 2g ) ( ) ( VA2 VB2 − ZA + αA = ZB + αB 2g 2g which can be written as

[ ]B V2 Z +α = S¯f ∆L 2g A

(5)

From the given data: ∆L = 140 m, Q = 280 m3 /s, n = 0.040, and ZA = 517.4 m. From the given cross-section information, AA = 144.98 m2 PA = 48.09 m 144.98 AA RA = = = 3.01 m PA 48.09 Q 280 VA = = = 1.93 m/s AA 144.98  2 [ ]2 (0.040)(280) nQ  = SA =  = 0.001373 2 2 (144.98)(3.01) 3 AA R 3 A

Taking SA + SB S¯f = ∆L and αA = αB = 1, the energy equation between sections A and B, Equation 5, can be written as [ ] VA2 SA V 2 SB ZB = ZA + + ∆L − B + ∆L 2g 2 2g 2 ] [ VB2 0.001373 SB 1.932 = 517.4 + + (140) − + 140 2(9.81) 2 2(9.81) 2

115

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which simplifies to ZB = 517.69 − 0.05097VB2 + 70SB where

 SB = 

2

(6)

2



nQ  (0.040)(280)  125.4 = = 2 2 4 AB RB3 AB RB3 A2B RB3

(7)

Combining Equations 6 and 7 gives the following form of the energy equation that is most useful for backwater computations, ZB = 517.69 − 0.05097VB2 +

8778 4

(8)

A2B RB3

Iterative calculations to determine ZB are indicated in the following table, where the initial estimate of ZB is 517.60 m. (1) ZB (m) 517.60 517.48 517.47

(2) AB (m2 ) 86.96 84.33 84.11

(3) PB (m) 32.77 32.45 32.42

(4) RB (m) 2.65 2.60 2.59

(5) VB (m/s) 3.22 3.32 3.33

(6) ZB (m) 517.48 517.47 517.47

The calculations begin with an assumption of ZB in Column 1, and the corresponding area, AB , in Column 2 and wetted perimeter, PB , in Column 3 are obtained from the given (tabular) data. The hydraulic radius, RB in Column 4 is obtained using RB = AB /PB , and the average velocity, VB , in Column 6 is obtained using VB = Q/AB . The values of AB , RB , and VB corresponding to the assumed value of ZB are substituted into the energy equation, Equation 8, to yield the calculated value of ZB shown in Column 6. If the calculated value of ZB in Column 6 is not equal to the assumed value of ZB in Column 1, then the calculations are repeated with assumed value of ZB equal to the calculated value of ZB . Based on the above calculations, the water-surface elevation at station B is 517.47 m . At Station B, the channel invert elevation is 515.10 m and hence the depth of flow, yB is given by yB = 517.47 − 515.10 = 2.37 m Since AB = 84.11 m2 , and the channel is approximately rectangular, the the width of the channel at section B is 84.11/2.37 = 35.49 m. Also, since the piers present an obstruction 2.50 m wide, the width of the channel adjacent to the piers is 35.49 m − 2.50 m = 32.99 m. The specific energy at section B, EB , is given by EB = yB +

VB2 3.332 = 2.37 + = 2.94 m 2g 2(9.81)

116

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Let yP be the depth of flow adjacent to the bridge pier, then neglecting energy losses requires that VP2 2g Q2 2.94 = yP + 2gA2P EB = yP +

2802 2g(32.99yP )2

2.94 = yP + which simplifies to

2.94 = yP +

3.672 yP2

(9)

which yields yP = 2.13 m

or

1.78 m

Since the flow at station B is subcritical (Fr < 1), the flow adjacent to the pier is probably subcritical also, and therefore the depth of flow at the pier is 2.13 m. The water surface elevation is 2.13 m + 515.10 m = 517.23 m . 4.66. From the given data: Q = 220 m3 /s, Z1 = 13.5 m, y1 = 1.00 m, Z2 = 21.5 m, b = 10 m, and n = 0.01. Based on these data, the downstream depth in the stilling basin is given by y2 = Z2 − (Z1 − y1 ) = 21.5 − (13.5 − 1.0) = 9 m Let L be the length of the stilling basin, so the hydraulic jump occurs at L/2. For y2 = 9 m, the conjugate depth equation gives y2 1 = y1 2 9 1 = y1 2

√

( −1 +

√

( −1 +

Q2 1+8 2 3 gb y1

)

2202 1+8 (9.81)(10)2 y13

)

1 = 2

( −1 +

√

394.7 1+8 3 y1

)

which gives y1 = 1.087 m. The next step is to find the distance, ∆L, over which the depth increases from 1.0 m to 1.087 m. This distance is given by [ y+ ∆L =

117

]1 V2 2g 2

S¯f − S0

(1)

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and the variables to be substituted into this equation are: y1 = 1 m V1 = 22 m/s A1 = 1 × 10 = 10 m2 P1 = 2 × 1 + 10 = 12 m 10 A1 R1 = = = 0.8333 m P1 12 )2 ( ( ) nQ 2 (0.01)(220) = 0.06172 Sf 1 = = 2 2 AR 3 (10)(0.8333) 3 y2 = 1.087 m A2 = 1.087 × 10 = 10.87 m2 220 V2 = m/s = 20.24 m/s 10.87 P2 = 2 × 1.087 + 10 = 12.17 m A2 10.87 R2 = = = 0.8606 m P2 12.17 )2 ( ) ( nQ 2 (0.01)(220) = 0.05004 Sf 2 = = 2 2 AR 3 (10.87)(0.8606) 3 Sf 1 + Sf 2 0.06172 + 0.05004 S¯f = = = 0.05588 2 2 S0 = 0 Substituting into Equation 1 gives [ 1+ ∆L =

222 2(9.81)

]

[ − 1.087 +

0.05588 − 0

20.242 2(9.81)

] = 66.3 m

Therefore, the length of the stilling basin should be 2 × 66.3 m = 133 m . 4.67. From the given data: b = 5 m, m = 2, n = 0.018, S0 = 0.001, T W = 1.00 m, and Q = 20 m3 /s. Use the standard-step method, where 2

2

Q Q [y1 + α1 2gA 2 ] − [y2 + α2 2gA2 ] 1 2 ∆L = S¯f − S0

where ∆L = 100 m, α1 = α2 = 1, and the objective is to find the depth y2 at the gate, given

118

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the depth y1 at a location 100 m upstream of the gate. In this case, y1 = 2.20 m − S0 ∆L = 2.20 − (0.001)(100) = 2.10 m A1 = by1 + = my12 = 5(2.10) + 2(2.10)2 = 19.32 m2 √ √ P1 = b + 2 1 + m2 y1 = 5 + 2 1 + 22 (2.10) = 14.39 m A2 = by2 + = my22 = 5y2 + 2y22 √ √ P2 = b + 2 1 + m2 y2 = 5 + 2 1 + 22 y2 = 5 + 4.472y2  2  2 2 3 nQ nQP 1   = Sf 1 =  2 5 3 3 A 1 R1 A1 [ ]2 2 (0.018)(20)(14.39) 3 = = 0.000234 5 (19.32) 3 2 [  ]2 2 2 4 3 3 (0.018)(20)(5 + 4.472y ) (5 + 4.472y2 ) 3 nQP 2 2   = = 0.1296 Sf 2 = 5 5 10 (5y2 + 2y22 ) 3 (5y2 + 2y22 ) 3 A3 2

4

(5 + 4.472y2 ) 3 S¯f = 0.5(Sf 1 + Sf 2 ) = 0.0001172 + 0.0648 10 (5y2 + 2y22 ) 3 Substituting into the standard-step equation yields [2.10 + 100 =

202 ] 2(9.81)(19.32)2

− [y2 +

202 ] 2(9.81)(5y2 +2y22 )2 4 3

) 0.0001172 + 0.0648 (5+4.472y2 210 − 0.001 (5y2 +2y2 )

3

which rearranges to 4

20.39 (5 + 4.472y2 ) 3 y2 + + 6.48 10 − 2.243 = 0 2 2 (5y2 + 2y2 ) (5y2 + 2y 2 ) 3 2

Viable solutions to this equation are y2 = 2.18 m and y2 = 0.73 m, which correspond to subcritical and supercritical flow conditions respectively (F2 = 0.26 and F2 = 1.75). Since the upstream flow is subcritical (y1 = 2.10 m, F1 = 0.28), the flow at the gate must also be subcritical, in which case y2 = 2.18 m. Under these conditions, Q = 20 m3 /s, HW = 2.18 m, T W = 1.00 m, and the gate discharge relationship requires that √ Q = 13.3h HW − T W √ 20 = 13.3h 2.18 − 1.00 which yields h = 1.38 m. Therefore, the gate must be opened at least 1.38 m to prevent the water elevation 100 m upstream of the gate from exceeding an elevation of 2.20 m.

119

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4.68. (a) y2 = 1.25 m: In this case the flow is contained entirely in the main channel and the flow variables are as follows: A = 23(1.25) = 28.75 m2 P = 180 + 2(1.25) = 28.04 m 5

5

1 A3 1 28.75 3 K= = 1169 m3 /s 2 = n2 P 3 0.025 28.04 23 α= 1 Fr∗ =

(1)(163)2 (23) αQ2 T = = 2.69 gA3 (9.81)(28.75)3

(b) y2 = 3.25 m: In this case the flow is contained in all three parts of the compound channel and the flow variables are as follows: A1 = 180(3.25 − 2.52) = 131.4 m2 P1 = 180 + (3.25 − 2.52) = 180.7 m 5

5

1 131.4 3 1 A13 = 1180 m3 /s K1 = 2 = n1 P 3 0.090 180.7 32 1

A2 = 23(3.25) = 74.75 m2 P2 = 23 + 2(2.52) = 28.04 m 5

5

1 74.75 3 1 A23 K2 = = 5749 m3 /s 2 = n2 P 3 0.025 28.04 32 2

A3 = 130(3.25 − 2.52) = 94.9 P3 = 130 + (3.25 − 2.52) = 130.7 m 5

5

1 A33 1 94.9 3 K3 = = 851.7 m3 /s 2 = n3 P 3 0.070 130.7 32 3

A = A1 + A2 + A3 = 131.4 + 74.75 + 94.9 = 301.1 m2 K = K1 + K2 + K3 = 1180 + 5749 + 851.7 = 7781 m3 /s ( ) A2 K13 K23 K33 α= 3 + 2 + 2 = 6.57 K A21 A2 A3 dα ∆α ≈ = 1.56 dy ∆y2 T = w1 + w2 + w3 = 180 + 23 + 130 = 333 m Fr∗ =

αQ2 T Q2 dα + = 0.246 gA3 2gA2 dy

An increment of ∆y2 = 0.01 m was used in estimating dα/dy. At y2 = 1.25 m the flow is supercritical , and at y2 = 3.25 m the flow is subcritical.

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4.69. The calculated flow variables at section BD are as follows: yD1 = 2.29 m AD1 = 11.4 m2 PD1 = 7.29 m KD1 = 309.1 m3 /s yD2 = 4.79 m AD2 = 239.4 m2 PD2 = 102.9 m KD2 = 14015 m3 /s yD3 = 2.29 m AD3 = 11.4 m2 PD3 = 7.29 m KD3 = 309.1 m3 /s AD = 262.3 m2 KD = 14633 m3 /s αD = 1.06 Sf D = 0.00042 VD = 1.14 m/s This gives the stage at BD as 4.79 m + 83.21 m = 88.00 m .

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Chapter 5

Design of Drainage Channels 5.1. For a rectangular section of width b and depth of flow y, A = by

(1)

and P = b + 2y or b = P − 2y

(2)

A = P y − 2y 2

(3)

Combining Equations 1 and 2 gives

The Manning equation can be put in the form ( A=

Qn √ S0

)3 5

2

P5

(4)

Therefore minimizing A minimizes P (i.e. dA/dy = 0 when dP/dy = 0). Differentiating Equation 3 with respect to y gives dA dP =P +y − 4y dy dy

(5)

At minimum value of A, dA/dy = dP/dy = 0, and Equation 5 becomes P − 4y = 0 or P = 4y

(6)

Combining Equations 2 and 6 gives b = 4y − 2y = 2y Therefore, the best hydraulic section is one in which the bottom width is twice the flow depth.

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5.2. For a triangular section with side slope m and flow depth y: T = 2my Ty A= 2√ P = 2y 1 + m2 T 2A/y A m= = = 2 2y 2y y According to the Manning equation ( A=

Qn √ S0

which can be put in the form P

2 5

)3

(√

5

S0 Qn

=A

2

P5 ) 35

Substituting the geometric properties of a triangular section and re-arranging gives (√ P = which simplifies to

(√

S0 Qn

S0 Qn

) 32

)3

( 5

A = 4y

(

A2 A = 2y 1 + 4 y 5 2

2

A2 1+ 4 y

) 21

) = 4y 2 + 4

A2 y2

Taking the derivative with respect to y and putting dA/dy = 0 gives 8y − 8

A2 =0 y3

which reduces to A = y2 and therefore T =

2A 2y 2 = = 2y y y

The best hydraulic triangular section is therefore one in which the top width is equal to twice the depth. √ 5.3. For the best trapezoidal section m = 3/3 = 0.577, and for the best triangular section m = 1, therefore the side slope is steeper in the trapezoidal section . 5.4. From the given data: Q = 2.4 m3 /s, m = 3, n = 0.020, and S0 = 0.8% = 0.008. According to Equation 5.11, for the most efficient trapezoidal section, the bottom width, b, is given by √ √ b = 2( 1 + m2 − m)y = 2( 1 + 32 − 3)y = 0.325y

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The area, A, and wetted perimeter, P , of the trapezoidal section is given by A = by + my 2 = (0.325y)y + (3)y 2 = 3.325y 2 √ √ P = b + 2 1 + m2 y = (0.325y) + 2 1 + 32 y = 6.649y The Manning equation requires that 5

Q=

1 A 3 12 S n P 23 0 5

2.4 =

1 1 (3.325y 2 ) 3 (0.008) 2 0.020 (6.649y) 23

which yields y = 0.600 m and b = 0.325(0.600) = 0.195 m. Therefore, the most efficient trapezoidal section has a bottom width of 0.195 m , side slopes of 3:1 (H:V) , and a flow depth of 0.600 m . 5.5. From the given data: b = 3 m, m = 2, ymax = 1 m, S0 = 0.005, n = 0.020, and τp = 25 Pa. It can be assumed that γ = 9790 N/m3 (at 20◦ C). Under limiting conditions, the bottom shear stress, τb , is equal to τp , in which case 25 = γRS0 = 9790

by + my 2 (3)y + (2)y 2 √ √ (0.005) = 9790 (0.005) b + 2 1 + m2 y 3 + 2 1 + 22 y

which yields y = 0.714 m. At this depth of flow, A = by + my 2 = (3)(0.714) + (2)(0.714)2 = 3.162 m2 √ √ P = b + 2 1 + m2 y = 3 + 2 1 + 22 (0.714) = 6.193 m 5

Q=

5

1 1 A 3 12 1 (3.162) 3 2 = 7.14 m3 /s 2 S0 = 2 (0.005) nP3 0.020 (6.193) 3

Therefore, the maximum flow rate for which the lining will be stable is 7.14 m3 /s . 5.6. The maximum shear stress on the channel boundary, τmax , and the average flow velocity, vave , in the channel are given by τmax = γyS0 1 2 1 vave = R 3 S02 n

(1) (2)

Since vave = vp when τmax = τp , and eliminating S0 from Equations 1 and 2 yields 1 2 vp = R 3 n

(

τp γy

)1

(

1

2

=

1

nγ 2

R4 y3

) 61

and taking γ = 9790 N/m2 gives vp =

(

1 n(9790)

125

1 2

R4 y3

) 16

1

τp2

1

τp2

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which simplifies to 0.010 vp = n

(

R4 y3

) 61

1

τp2

(3)

It is better to design a channel based on maximum permissible shear stress rather than maximum permissible velocity since the maximum permissible shear stress remains constant under all flow conditions, while the maximum permissible velocity varies with flow conditions, as evidenced by Equation 3. 5.7. From the given data: ds = 2.5 mm = 0.0025 m and SG = 2.65. At 20◦ C, γ = 9790 kN/m3 and ν = 1.00 × 10−6 m2 /s. Hence, ( ) √ [( ) ] ( ) √ ds 0.0025 γs − 1 gds = 0.1 0.1 [(2.65) − 1] (9.81)(0.0025) = 159 −6 ν γ 1.00 × 10 Using this value (159) in the the Sheilds diagram (Figure 5.5) yields an intersection point of around τ∗ = 0.050 and Re∗ = 120. Using the definition of τ∗ given by Equation 5.22, the critical shear stress, τc , is given by τc = τ∗ (γs − γ)ds = τ∗ (SG − 1)γds = (0.050)(2.65 − 1)(9790)(0.0025) = 2.0 N/m2 = 2.0 Pa Therefore, the (maximum) permissible stress on the bottom of the channel is 2.0 Pa . 5.8. Dividing Equation 5.28 by Equation 5.25 yields √ √ τps tan2 θ tan2 α − tan2 θ K= = cos θ 1 − = cos θ 2 τp tan α tan2 α Taking the cos θ under the square root yields √ cos2 θ tan2 α − sin2 θ K= tan2 α Multiplying the numerator and denominator by cos2 α yields √ cos2 θ sin2 α − sin2 θ cos2 α K= sin2 α Using the identities cos2 θ = 1 − sin2 θ and sin2 α = 1 − cos2 α in the numerator yields √ (1 − sin2 θ)(1 − cos2 α) − sin2 θ cos2 α K= sin2 α which simplifies to

√ K=

1 − sin2 θ − cos2 α sin2 α

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Using the identity cos2 α = 1 − sin2 α yields √ √ 1 − sin2 θ − (1 − sin2 α) sin2 α − sin2 θ = K= sin2 α sin2 α which simplifies to

√ 1−

K=

sin2 θ sin2 α

5.9. The tractive-force ratio, K, is given by Equation 5.29 as √ τps sin2 θ K= = 1− τp sin2 α Equations 5.16 and 5.17 indicate that the actual shear stress on the side of the channel, τs , is Ks times of the concurrent shear stress on the bottom of the channel. Therefore, as long as the tractive force ratio is greater than Ks , the side slope will be stable when the bottom is unstable. Hence, the condition for side-slope failure is √ sin2 θ 1− < Ks sin2 α which gives sin2 θ > 1 − Ks2 sin2 α or √

sin θ 1 − Ks2

> sin α

(1)

where it is noted that Ks is a function of θ as given by Equation 5.18 which can be expressed in the form   0.77 m ≤ 1.5    Ks = 0.066 + 0.67 1.5 < m < 5  tan θ    1.0 m≥5 and m = 1/ tan θ. When θ = 25◦ and α = 30◦ , then m=

1 = 2.14 tan θ

Ks = 0.066m + 0.67 = 0.811 √

sin θ 1 − Ks2

=√

sin 25◦ = 0.722 1 − 0.8112

sin α = sin 30◦ = 0.5 Therefore, Equation 1 is satisfied and so yes the side slope will become unstable before the bottom.

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5.10. From the given data: S0 = 0.01, b = 4 m, m = 2.5, d50 = 250 mm, d85 = 300 mm, and τp = 200 Pa. The side slope angle, θ, is given by ( ) ( ) 1 1 θ = tan−1 = tan−1 = 21.8◦ m 2.5 Since the stone lining is derived from an open gradation, the angle of repose, α, can be estimated from Table 5.2 (Equation 1) as ( ) ( ) d85 0.125 300 0.125 α = α0 = (37.1) = 38.0◦ d50 250 Therefore, the tractive force ratio, K, is given by Equation 5.29 as √ sin2 (21.8) K = 1− = 0.798 sin2 (38.0) The permissible shear stress on bottom of the channel, τp , is 200 Pa, and the permissible shear stress on the side of the channel, τps , is given by τps = Kτp = (0.798)(200) = 160 Pa The limiting condition for stability of the bottom lining is 200 = γyS0 = (9790)y(0.01) which yields y = 2.04 m. The tractive force ratio, K, is given by Equation 5.18 as K = 0.066m + 0.67 = 0.066(2.5) + 0.67 = 0.835 Therefore, the limiting condition for stability on the side lining is 160 = KγyS0 = (0.835)(9790)y(0.01) which yields y = 1.96 m. Therefore, the side lining controls the channel and the depth of flow should not be greater than 1.96 m . 5.11. From the given data: τp = 60 Pa, nb = 0.023, b = 2 m, m = 4, S0 = 0.005, and y = 0.80 m. Taking γ = 9790 N/m3 , the maximum shear stress in the straight segment of the channel, τb , is given by τb = γyS0 = (9790)(0.80)(0.005) = 39.2 Pa Therefore, the allowable bend factor, Kr , is Kr =

τr 60 = = 1.53 τb 39.2

According to Equation 5.31, Kr = 2.38 − 0.206

(r ) c

+ 0.0073

( r )2 c

T T ( r )2 (r ) c c 1.53 = 2.38 − 0.206 + 0.0073 T T

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which yields rc /T = 23.2 or 5.0. Since Equation 5.31 is valid for 2 < rc /T < 10, the only valid solution is rc /T = 5.0. Therefore, rc = 5.0T = 5.0[b + 2my] = 5.0[2 + 2(4)(0.80)] = 42 m So the radius of curvature of the channel should not be less than 42 m for the lining to remain stable. 5.12. Only rock can withstand vertical side slopes. A side slope of 3:1 is typical of roadside channels. 5.13. The recommended freeboard given by Equation 5.33 is F = 0.15 +

V2 2g

and so for F > 0.30 m the required velocity is given by 0.15 +

V2 > 0.30 2g

→

V >

√

2g(0.15)

→

V > 1.72 m/s

Also for channels on steep gradients with flow depths greater that 30 cm will require freeboards greater than 30 cm. 5.14. From the given data: b = 10 m, m = 2, S0 = 0.00053, n = 0.030, Q = 28 m3 /s, and rc = 100 m. Determine the normal depth using the Manning equation 1 A3 √ S0 n P 23 5

Q= which gives

1 (10y + 2y 2 ) 3 √ 28 = 0.00053 √ 0.030 (10 + 2y 5) 23 5

and solving for y gives y = 2.00 m and the corresponding velocity is given by V =

Q 28 = = 1.00 m/s A 10(2) + 2(2)2

Since y ≥ 0.30 m and V ≤ 1.72 m/s, the freeboard, F , estimated by Equation 5.34 is 0.30 m. The superelevation, hs , around a bend is given by hs =

V 2T V 2 [b + 4y] 1.002 [10 + 4(2)] = = = 0.018 m grc grc (9.81)(100)

The required additional freeboard around the bend is hs /2 = 0.009 m = 0.9 cm. Since this value is small and a conservative freeboard of 30 cm is already being used in the straight section of the channel, then maintaining a freeboard of 30 cm in the channel bend is appropriate. Hence the design flow depth is 2.00 m and the minimum required freeboard is 30 cm .

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5.15. From the given data: Q = 30 m3 /s, S0 = 0.002. For float-finished concrete, take n = 0.015. Use the best hydraulic section: b = 1.15y, m = 0.58, A = 1.73y 2 , P = 3.46y, and R = 0.5y. According to the Manning equation 1 A3 √ S0 n P 23 5

Q=

1 (1.73y 2 ) 3 √ 30 = 0.002 0.015 (3.46y) 23 5

which gives y = 2.30 m The flow velocity, V , is given by V =

Q 30 30 = = = 3.27 m/s 2 A 1.73y 1.73(2.30)2

Since y ≥ 0.30 m and V > 1.72 m/s, the freeboard, F , estimated by Equation 5.34 is F = 0.15 +

V2 3.272 = 0.15 + = 0.70 m 2g 2(9.81)

Therefore, for the most efficient channel, the minimum required channel depth is 2.30 m + 0.70 m = 3.00 m , the bottom width is 1.15(2.30) = 2.65 m and the side slope is 0.58:1 (H:V). 5.16. From the given data: S0 = 0.00042, Q = 4.5 m3 /s, and rc = 150 m. For unfinished concrete, take n = 0.017, and for the most efficient (rectangular) hydraulic section A = 2y 2 P = 4y T = 2y The Manning’s equation gives 1 A3 √ S0 n P 23 5

Q=

1 (2y 2 ) 3 √ 0.00042 0.017 (4y) 23 5

4.5 = which gives

y = 1.50 m Calculate the superelevation, hs , hs =

Q2 T (4.5)2 (2 × 1.5) V 2T = 2 = = 0.002 grc A grc (2 × 1.52 )2 (9.81)(150)

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The velocity, V , in the channel is given by V =

4.5 Q = = 1.00 m/s A 2(1.5)2

Since y ≥ 0.30 m and V ≤ 1.72 m/s, the freeboard, F , estimated by Equation 5.34 is 0.30 m, therefore, use a (straight-section) freeboard of 0.30 m. The additional freeboard on the channel bend (hs /2 = 0.001 m) is negligible, so use a freeboard of 0.30 m in the bend also. Hence the minimum required channel depth (with lining) is 1.50 m + 0.30 m = 1.80 m , and the required channel width is 2y = 2(1.50) = 3.00 m . 5.17. From the given data: S0 = 0.00032, Q = 4.4 m3 /s, and for smooth asphalt n = 0.013. Properties of the most efficient (hydraulic) triangular section are: A = y2 P = 2.83y m=1 The Manning equation gives 1 A3 √ S0 Q= n P 23 5

1 (y 2 ) 3 √ 0.00032 0.013 (2.83y) 23 5

4.4 = which yields

y = 2.01 m The velocity, V , in the channel is V =

Q 4.4 4.4 = 2 = = 1.09 m/s A y 2.012

Since y ≥ 0.30 m and V ≤ 1.72 m/s, the freeboard, F , estimated by Equation 5.34 is 0.30 m. The minimum depth of the channel to be excavated is therefore given by 2.01 m + 0.30 m = 2.31 m , with side slopes of 1:1 . 5.18. According to Table 5.4, n = 0.013. Using the most efficient hydraulic section, Table 5.1 gives A = 1.89y 2 ,

P = 3.77y,

T = 2.83y,

R=

A = 0.501y P

Substituting the geometric characteristics of the channel into the Manning equation yields 6= or

2 1 8 1 (1.89y 2 )(0.501y) 3 (0.0005) 2 = 2.051y 3 0.013

8

y 3 = 2.926

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which leads to y = 1.50 m and the flow area, A, is given by A = 1.89y 2 = 1.89(1.50)2 = 4.25 m2 and the average velocity, V , is Q 6 = = 1.41 m/s A 4.25 Since y ≥ 0.30 m and V ≤ 1.72 m/s, the freeboard, F , estimated by Equation 5.34 is 0.30 m. Therefore, the minimum depth of the (parabolic) channel should be 1.50 m + 0.30 m = 1.80 m . V =

The equation of the surface of the parabolic channel can be taken as y = ax2 which is symmetric about the y axis and passes through (0,0). The top width, T , is equal to 2x, and √ y T = 2x = 2 a For the most efficient hydraulic section, T = 2.83y = 2.83(1.50) = 4.25 m when y = 1.50 m, therefore √ 1.50 4.25 = 2 a which gives a = 0.332 and the equation of the parabolic channel is y = 0.332x2 5.19. From the given data: Q = 0.42 m3 /s, b = 0.4 m, m = 3, and S0 = 0.008. The flow area, A, and the wetted perimeter, P , can be expressed in terms of the flow depth, y, by the relations A = by + my 2 = 0.4y + 3y 2 √ √ P = b + 2y 1 + m2 = 0.4 + 2y 1 + 32 = 0.4 + 6.325y Step 1: Try a gravel mulch lining with d50 = 25 mm. Step 2: For typical gravel-mulch linings, the Manning’s n depends on the flow depth as shown Table 5.6. This functional relationship can be expressed as n(y) and the Manning equation gives 5

1 A 3 12 Q= S n P 23 0 5

1 1 (0.4y + 3y 2 ) 3 2 0.42 = 2 (0.008) n(y) (0.4 + 6.325y) 3

132

(1)

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Solving Equation 1 simultaneously with the n versus y relationship in Table 5.6 yields y = 0.349 m. Step 3: The maximum shear stress on the bottom of the channel, τb , is given by (assuming γ = 9790 N/m3 ) τb = γyS0 = (9790)(0.349)(0.008) = 27.3 Pa The side-shear-stress factor, Ks , is given by Equation 5.18 as Ks = 0.066m + 0.67 = 0.066(3) + 0.67 = 0.868 and so the maximum shear stress exerted on the side of the channel, τs , is given by τs = Ks τb = (0.868)(27.3) = 23.7 Pa Step 4: The permissible shear stress, τp , on the gravel-mulch lining is estimated from Table 5.7 for d50 = 25 mm as 19 Pa. Since the lining is non-cohesive, the permissible shear stress on the side of the channel, τps (=Kτp ), will be less than 19 Pa. Step 5: The maximum shear stress on bottom of the channel (27.3 Pa) is greater than the permissible shear stress on the gravel-mulch lining with d50 = 25 mm (19 Pa), and so the gravel-mulch lining with d50 = 25 mm is inadequate. Try another a larger stone size. Step 1: Try gravel mulch lining with d50 = 50 mm. Step 2: Solving the Manning equation (Equation 1) simultaneously with the n versus y relationship in Table 5.6 yields y = 0.389 m. Step 3: The maximum shear stress on the bottom of the channel, τb , is given by τb = γyS0 = (9790)(0.389)(0.008) = 30.5 Pa and the maximum shear stress exerted on the side of the channel, τs , is given by τs = Ks τb = (0.868)(30.5) = 26.5 Pa Step 4: The permissible shear stress, τp , on the gravel-mulch lining can be estimated from Table 5.7 for d50 = 50 mm as 38 Pa. The angle of repose, α, of the gravel mulch can be estimated from Table 5.2 (Equation 2) for subangular-shaped stones as α = α0 (d50 )0.00778 = (34.3)(50)0.00778 = 35.4◦ The side-slope angle θ is given by θ = tan−1

(

1 m

)

= tan−1

( ) 1 = 18.4◦ 3

The tractive-force ratio, K, is given by Equation 5.29 as √ √ sin2 θ sin2 (18.4) K = 1− = 1 − = 0.839 sin2 α sin2 (35.4) and hence the permissible shear stress on the side of the channel, τps is given by τps = Kτp = (0.839)(38) = 31.9 Pa

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Step 5: Since the maximum shear stress on bottom of the channel (30.5 Pa) is less than the permissible shear stress on the bottom of the channel (38 Pa), and the maximum shear stress on the side of the channel (26.5 Pa) is less permissible shear stress on the side of the channel (31.9 Pa), the lining is adequate. Final specification: The recommended channel design has a gravel mulch lining with d50 = 50 mm. 5.20. From the given data: S0 = 0.05%, Q = 0.2 m3 /s. Concrete Channel, √ Best Hydraulic Section: For the best hydraulic section, m = 1 , A = y 2 , P = 2 2y, and for unfinished concrete n = 0.017. Using the Manning equation gives 5

1 A3 1 Q= S0 n P 23 2 5

(y 2 ) 3 1 1 0.2 = √ 2 (0.0005) 0.017 (2 2) 3 2 which yields y = 0.640 m. For this depth of flow, A = y 2 = (0.640)2 = 0.4096 m2 Q 0.2 V = = = 0.49 m/s A 0.4096 Using a freeboard of 0.30 m gives a total depth of 0.640 m + 0.30 m = 0.940 m and hence the excavated volume per km length is Vexcav = (0.940)2 (1000) = 884 m3 Since the charge is $100/m3 , the total cost per km is 884 m3 × $100/m3 = $88,400/km . Gravel Mulch, d50 = 50 mm: For serviceability, take m = 3 . The best hydraulic section (m = 1) cannot be used because of likely slope-stability problems. For a triangular section, A = my 2 = 3y 2 √ √ P = 2 1 + m2 y = 2 1 + 32 y = 6.325y Using these geometric relationships in the Manning equation gives 5

1 A3 1 Q= S0 n P 23 2 5

1 (3y 2 ) 3 1 1.2 = 2 (0.0005) n (6.325y) 3 2

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which yields 3

y = 1.815n 8

(1)

Assuming that 0.5 m < y < 1.0 m, the n-function is given by n = 0.046 − 0.008y

(2)

Solving Equations 1 and 2 simultaneously yields y = 0.551 m and n = 0.0416. The assumption associated with Equation 2 is validated. The maximum shear stress on the side of the channel, τs , is given by τs = Ks γRS0

(3)

From the given data: Ks = 0.066m + 0.67 = 0.066(3) + 0.67 = 0.868 R=

A 3(0.551)2 = = 0.261 m P 6.325(0.551)

Substituting these calculated parameters into Equation 3, with γ = 9790 N/m3 , gives τs = (0.868)(9790)(0.261)(0.0005) = 1.1 Pa The permissible shear stress on the side of the channel, τps , is given by τps = Kτb

(4)

where τb = 38 Pa. For round stone, α = 31.5◦ (50)0.00778 = 32.5◦ ( ) −1 1 θ = tan = 18.4◦ 3 √ √ 2 sin θ sin2 18.4◦ K = 1− = 1 − = 0.809 sin2 α sin2 32.5◦ A 3(0.551)2 R= = = 0.261 m P 6.325(0.551) Substituting the calculated parameters into Equation 4 gives τps = (0.809)(38) = 31 Pa Since τs ≪ τps (i.e., 1.1 Pa ≪ 38 Pa), the channel is more than adequate. For y = 0.551 m and m = 3: A = my 2 = 3(0.551)2 = 0.911 m2 Q 0.2 V = = = 0.22 m2 A 0.911 Using a freeboard of 0.30 m gives a total depth of 0.551 m + 0.30 m = 0.851 m and hence the excavated volume per km length is Vexcav = 3(0.851)2 (1000) = 2170 m3 Since the charge is $70/m3 , the total cost per km is 2170 m3 × $70/m3 = $151,900/km .

135

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Comparison: For the concrete-lined best section the cost is $88,400/km, and for the gravel mulch section the cost is $151,900/km. The concrete lined channel is the better alternative in terms of cost. The difference is $151,900/km − $88,400/km = $63,500/km . 5.21. From the given data: b = 0.90 m, m = 3, S0 = 0.03, soil classification SC, PI = 16, e = 0.5, Q = 0.5 m3 /s, sod-grass lining in good condition, and h = 0.075 m. From the given channel dimensions, the area, A, and wetted perimeter, P , and hydraulic radius, R, are given by A = by + my 2 = (0.9)y + (3)y 2 = 0.9y + 3y 2 √ √ P = b + 2y 1 + m2 = (0.9) + 2y 1 + 32 = 0.9 + 6.325y R=

0.9y + 3y 2 A = P 0.9 + 6.325y

Step 1: Determine the flow depth in the channel. Equation 5.41 gives Cs = 106, and Equation 5.42 gives Cn = 0.35Cs0.10 h0.528 = 0.35(106)0.10 (0.075)0.528 = 0.142 The average shear stress on the channel boundary, τ0 , is given by Equation 5.44 as (assuming γ = 9790 N/m3 ) ( ) 0.9y + 3y 2 293.7(0.9y + 3y 2 ) τ0 = γRS0 = (9790) (0.03) = 0.9 + 6.325y 0.9 + 6.325y and the Manning’s n is given by Equation 5.43 as ]−0.4 ( [ ) 0.9 + 6.325y 0.4 293.7(0.9y + 3y 2 ) = 0.0146 n = Cn τ0−0.4 = (0.142) 0.9 + 6.325y 0.9y + 3y 2

(1)

The Manning equation gives 5

1 A 3 12 Q= S n P 23 0 5

1 1 (0.9y + 3y 2 ) 3 2 0.5 = 2 (0.03) n (0.9 + 6.325y) 3

(2)

Solving Equations 1 and 2 simultaneously yields y = 0.213 m and n = 0.032. Step 2: Determine the effective stress on the underlying soil. The maximum stress on the bottom of the channel, τb , is given by τb = γyS0 = (9790)(0.213)(0.03) = 62.6 Pa For sod grass in good condition, Table 5.9 gives Cf = 0.90, and assuming that d75 < 1.3 mm Equation 5.45 gives ns = 0.016, and the effective shear stress on the soil underlying the grass lining is given by Equation 5.46 as ( ) ( n )2 0.016 2 s τe = τb (1 − Cf ) = (62.6)(1 − 0.9) = 1.6 Pa n 0.032

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Step 3: Determine the the permissible shear stress on the soil underlying the vegetative lining and assess the adequacy of the lining. Since PI > 10, the soil is cohesive and the parameters to estimate the permissible shear stress are given in Table 5.10 as: c1 = 1.07, c2 = 14.3, c3 = 47.7, c4 = 1.42, c5 = −0.61, and c6 = 4.8 × 10−3 . The permissible shear stress on the underlying soil, τp,c , is then given by Equation 5.48 as τp,c = (c1 PI2 + c2 PI + c3 )(c4 + c5 e)2 c6 = [1.07(16)2 + 14.3(16) + 47.7][1.42 + (−0.61)(0.5)]2 (4.8 × 10−3 ) = 3.3 Pa Since the shear stress on the underlying soil (1.6 Pa) is less than the permissible shear stress on the underlying soil (3.3 Pa), the proposed grass lining is adequate . 5.22. From the given data: Q = 35 m3 /s, S0 = 0.1%, d75 = 30 mm (moderately rounded), and T = 17 m. For moderately angular gravel, Figure 5.7 gives α = 38.5◦ . Assume that the tractive force on the side is limiting. The tractive force ratio, K, is given by √ √ √ √ sin2 θ sin2 θ 2.58 2 K = 1− = 1− = 1 − 2.58 sin θ = 1 − (1) 2 2 ◦ 1 + m2 sin α sin 38.5 where the side slopes are m:1 (H:V). Equation 5.47 gives the permissible shear stress on the soil (= permissible shear stress on the bottom of the channel) as τp = 0.75d75 = 0.75(30) = 22.5 Pa The permissible shear stress on the side of the channel is therefore given by √ 2.58 τps = Kτp = 22.5 1 − 1 + m2

(2)

Assuming that 1.5 < m < 5, the side-shear stress factor, Ks is estimated by Equation 5.18 as Ks = 0.066m + 0.67 and so the maximum shear stress exerted on the side of the channel, τs , is given by τs = Ks γyS0 = (0.066m + 0.67)(9790)y(0.001) = (0.646m + 6.56)y

(3)

The maximum depth of flow for stability of the channel boundary, y, occurs when τs = τps and so combining Equations 2 and 3 gives √ 2.58 (0.646m + 6.56)y = 22.5 1 − 1 + m2 or 22.5 y= 0.646m + 6.56

137

√ 1−

2.58 1 + m2

(4)

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The top-width constraint requires that b + 2my = 17 → b = 17 − 2my

(5)

For d75 = 30 mm, Equation 5.45 estimates Manning’s n as 1

1

6 n = ns = 0.015d75 = 0.015(10) 6 = 0.022

and the Manning equation requires that 5

1 A 3 12 Q= S n P 23 0 5

35 =

1 1 (by + my 2 ) 3 2 √ 2 (0.001) 0.022 (b + 2y 1 + m2 ) 3

(6)

Equations 4 to 6 are three equations in three unknowns which yield m = 3.34 , b = 1.70 m , and y = 2.29 m. The channel is practical since m ≥ 3 as would be required for practicality, and the side-slope angle (θ = tan−1 (1/3.34) = 16.7◦ ) is much less than the angle of repose (38.5◦ ). 5.23. From the given data: Q = 0.5 m3 /s, S0 = 0.002, PI = 25, e = 0.4, and the soil is classified as CH. For practicality and functionality, take the side slope as 3:1 (H:V), so m = 3. Step 1: Determine the maximum flow depth for stability. For a soil classification of CH and PI = 25, Table 5.10 gives c3 = 0.097, c4 = 1.38, c5 = −0.373, and c6 = 48. Therefore, the permissible shear stress on the channel boundary, τp , is given by Equation 5.48 as τp = τp,c = [c1 PI2 + c2 PI + c3 ][c4 + c5 e]2 c6 = [0.097][1.38 − 0.373(0.4)]2 (48) = 7.1 Pa The maximum shear stress on the bottom of the channel, τb , is given by (for γ = 9790 N/m3 ) τb = γyS0 = (9790)(y)(0.002) = 19.6y The maximum allowable flow depth causes τb = τp , which requires that 19.6y = 7.1 Pa which yields y = 0.362 m. Step 2: Determine the bottom width of the channel. The flow must be of sufficient size to accommodate the design flow of 0.5 m3 /s. Manning’s n can be estimated by Equation 5.45 as n = 0.016, and the Manning equation requires that 5

Q=

5

1 1 A 3 12 1 [by + my 2 ] 3 2 S = √ 2 2 S0 0 nP3 n [b + 2y 1 + m2 ] 3 5

1 1 [b(0.362) + (3)(0.362)2 ] 3 2 0.5 = √ 2 (0.002) 0.016 [b + 2(0.362) 1 + 32 ] 3

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which yields b = 0.368 m. The corresponding flow area and average velocity are 0.53 m2 and 0.94 m/s respectively. The required freeboard, F , according to Equation 5.34 is 0.30 m. Hence, the designed channel should have a bottom width of 0.368 m , side slopes of 3:1 , and a total (minimum) depth of 0.362 m + 0.30 m = 0.66 m . Step 3: Compare the channel with the most efficient section. The bottom width of the most efficient section is given by Equation 5.11 as √ √ b = 2y( 1 + m2 − m) = 2(0.291)( 1 + 32 − 3) = 0.094 m Since the minimum required channel width (0.362 m) is greater than the bottom width of the most efficient section, then the most efficient section cannot be used . 5.24. From the given data: S0 = 0.003, d75 = 30 mm (moderately angular), d = 3 m, b = 6 m, and m = 2.5 Step 1: Calculate the stable flow depth. The maximum allowable shear stress on the bottom of the channel, τp , can be estimated using Equation 5.47 as τp = 0.75d75 = 0.75(30) = 22.5 Pa Assuming a freeboard of F = 0.30 m, the maximum allowable flow depth is d − F = 3 m − 0.30 m = 2.70 m. Check that flows with this depth will not erode the channel boundary. The side-shear-stress factor, Ks , is given by Equation 5.18 as Ks = 0.066m + 0.67 = 0.066(2.5) + 0.67 = 0.835 For d75 = 30 mm and moderately rounded, Figure 5.7 gives the angle of repose as α = 38.1◦ , and for m = 2.5 the side slope angle is θ = 21.8◦ . The tractive force ratio, K, is therefore given by Equation 5.29 as √ √ sin2 θ sin2 21.8◦ = 1 − = 0.798 K = 1− sin2 α sin2 38.1◦ Since K < Ks , the side slope is limiting for channel erosion. The permissible shear stress on the side, τps , and the shear stress exerted by the flowing water on the side, τs , are given by τps = Kτp = (0.798)(22.5) = 18.0 Pa τs = Ks τb = Ks γyS0 = (0.835)(9790)y(0.003) = 24.5y At the limit of stability, τs = τps , which requires that 24.5y = 18.0 Pa which yields y = 0.735 m. This result shows that the physical maximum flow depth in the channel (2.70 m) cannot be safely achieved without erosion. The capacity of the channel must be calculated using y = 0.735 m.

139

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Step 2: Determine the capacity of the channel. Manning’s n in the channel can be estimated using Equation 5.45 which yields 1

1

6 n = 0.015d75 = 0.015(30) 6 = 0.0264

and the capacity of the channel is given by the Manning equation as 5

1 A 3 12 Q= S n P 32 0 5

1 1 [by + my 2 ] 3 2 Q= √ 2 S0 n [b + 2y 1 + m2 ] 3 5

1 1 [(6)(0.735) + (2.5)(0.735)2 ] 3 2 = 8.30 m3 /s Q= √ 2 (0.003) 0.0264 [(6) + 2(0.735) 1 + 2.52 ] 3

The safe capacity of the channel is 8.30 m3 /s . 5.33. From the given data: b = 0.90 m, m = 3, S0 = 0.03, soil classification SC, PI = 16, e = 0.5, Q = 0.5 m3 /s, and vegetation lining with class D retardance. From the given channel dimensions, the area, A, and wetted perimeter, P , and hydraulic radius, R, are given by A = by + my 2 = (0.9)y + (3)y 2 = 0.9y + 3y 2 √ √ P = b + 2y 1 + m2 = (0.9) + 2y 1 + 32 = 0.9 + 6.325y R=

A 0.9y + 3y 2 = P 0.9 + 6.325y

Step 1: Determine the flow depth in the channel. Table 5.11 gives Cn = 0.147. The average shear stress on the channel boundary, τ0 , is given by Equation 5.44 as (assuming γ = 9790 N/m3 ) ) ( 293.7(0.9y + 3y 2 ) 0.9y + 3y 2 τ0 = γRS0 = (9790) (0.03) = 0.9 + 6.325y 0.9 + 6.325y and the Manning’s n is given by Equation 5.43 as [ ]−0.4 ( ) 293.7(0.9y + 3y 2 ) 0.9 + 6.325y 0.4 −0.4 n = Cn τ0 = (0.147) = 0.0151 0.9 + 6.325y 0.9y + 3y 2

(1)

The Manning equation gives 5

1 A 3 12 Q= S n P 23 0 5

0.5 =

1 1 (0.9y + 3y 2 ) 3 2 2 (0.03) n (0.9 + 6.325y) 3

Solving Equations 1 and 2 simultaneously yields y = 0.216 m and n = 0.032.

140

(2)

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Step 2: Determine the effective stress on the underlying soil. The maximum stress on the bottom of the channel, τb , is given by τb = γyS0 = (9790)(0.216)(0.03) = 63.4 Pa For mixed vegetation in good condition, Table 5.9 gives Cf = 0.75, and assuming that d75 < 1.3 mm Equation 5.45 gives ns = 0.016, and the effective shear stress on the soil underlying the grass lining is given by Equation 5.46 as ) ( ( n )2 0.016 2 s τe = τb (1 − Cf ) = (63.4)(1 − 0.75) = 4.0 Pa n 0.032 Step 3: Determine the the permissible shear stress on the soil underlying the vegetative lining and assess the adequacy of the lining. Since PI > 10, the soil is cohesive and the parameters to estimate the permissible shear stress are given in Table 5.10 as: c1 = 1.07, c2 = 14.3, c3 = 47.7, c4 = 1.42, c5 = −0.61, and c6 = 4.8 × 10−3 . The permissible shear stress on the underlying soil, τp,c , is then given by Equation 5.48 as τp,c = (c1 PI2 + c2 PI + c3 )(c4 + c5 e)2 c6 = [1.07(16)2 + 14.3(16) + 47.7][1.42 + (−0.61)(0.5)]2 (4.8 × 10−3 ) = 3.3 Pa Since the shear stress on the underlying soil (4.0 Pa) is greater than the permissible shear stress on the underlying soil (3.3 Pa), the proposed grass lining is inadequate . 5.26. From the given data, the channel properties are: b = 5 m, m = 3, d = 1 m, and S0 = 0.001. (a) Find capacity of grass channel. For grass with retardance C, Table 5.11 gives Cn = 0.220, and for sodded grass in good condition Table 5.9 gives Cf = 0.90. For sandyclay soil assume d75 < 1.3 mm so Manning’s n of the bare soil is given by Equation 5.45 as ns = 0.016. Assuming a required freeboard of F = 0.30 m, the maximum flow depth in the channel is y = d−F = 1 m − 0.30 m = 0.70 m. Using these variables, the relevant channel and flow properties are calculated as follows: A = by + my 2 = (5)(0.70) + (3)(0.70)2 = 4.97 m2 √ √ P = b + 2y 1 + m2 = 5 + 2(0.7) 1 + 32 = 9.43 m A 4.97 R= = = 0.527 m P 9.43 τ0 = γRS0 = (9790)(.527)(0.001) = 5.2 Pa n = Cn τ0−0.4 = (0.220)(5.2)−0.4 = 0.114 5

5

1 1 A 3 12 1 (4.97) 3 2 = 0.90 m3 /s Q= 2 S0 = 2 (0.001) nP3 0.114 (9.43) 3

τb = γyS0 = (9790)(0.70)(0.001) = 6.9 Pa ( ) ( n )2 0.016 2 s τe = τb (1 − Cf ) = (6.9)(1 − 0.90) = 0.01 Pa n 0.114

141

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Hence the effective stress on the underlying soil is 0.01 Pa. The permissible stress on the underlying soil can be calculated using the parameters in Table 5.10 for CL soil with PI = 14 and e = 0.4, which correspond to the parameters: c1 = 1.07, c2 = 14.3, c3 = 47.7, c4 = 1.48, c5 = −0.57, and c6 = 4.8 × 10−3 . Substituting these parameters into Equation 5.48 gives the permissible shear stress on the underlying soil, τp,c , as τp,c = [c1 PI2 + c2 PI + c3 ][c4 + c5 e]2 c6 = [(1.07)(14)2 + (14.3)(14) + (47.7)][(1.48) + (−0.57)(0.4)]2 (4.8 × 10−3 ) = 2.8 Pa Since τe ≪ τp,c (i.e., 0.01 Pa ≪ 2.8 Pa), the lining is adequate when y = 0.70 m, which corresponds to Q = 0.90 m3 /s . (b) Find capacity of bare-soil channel. Since the maximum bottom stress (5.2 Pa) is greater than the allowable soil shear stress (2.8 Pa), the maximum allowable depth under bare-soil conditions is less than 0.70 m. The limiting depth occurs when τ0 = τp,c , which requires that γRS0 = 2.8 (9790)R(0.001) = 2.8 which yields R = 0.286 m. Using the definition of the R requires that by + my 2 √ = 0.286 m b + 2y 1 + m2 5y + 3y 2 √ = 0.286 m 5 + 2y 1 + 32 which yields y = 0.340 m. For this flow depth, P = 7.15 m, A = 2.05 m2 , and Manning’s equation gives 5

Q=

5

1 1 A 3 12 1 (2.05) 3 (0.001) 2 = 1.76 m3 /s 2 S0 = nP3 0.016 (7.15) 23

Therefore the maximum flow rate under bare-soil conditions is 1.76 m3 /s . (Almost double the capacity when there is no lining. The is caused by the much lower value of n for the case of no lining.) 5.27. From the given data: m = 3, S0 = 1.4% = 0.014, and Q = 1.1 m3 /s. For a Bermuda grass of height 4 cm, Table 5.8 indicates a retardance classification of E. From the given channel dimensions, the area, A, and wetted perimeter, P , and hydraulic radius, R, are given by A = 3y 2 √ P = 2 1 + 32 y = 6.325y A R= = 0.4743y P

142

(1)

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Step 1: Determine the flow depth in the channel. For a retardance of E, Equation 5.56 gives a = 52.1, and Equation 5.55 gives Manning’s n as 1

1

1.22R 6 1.22R 6 n= = 0.4 1.4 52.1 + 19.97 log[R1.4 (0.014)0.4 ] 52.1 + 19.97 log[R S0 ]

(2)

The Manning equation requires that 5

1 A 3 12 Q= S n P 32 0 5

0.8 =

1 1 (3y 2 ) 3 2 2 (0.014) n (6.325y) 3

(3)

Solving Equations 1, 2, and 3 simultaneously yields y = 0.578 m . Step 2: Determine the maximum shear stress on the bottom of the channel. The maximum shear stress on the bottom of the channel, τb , is given by (assuming γ = 9790 N/m3 ) τb = γyS0 = (9790)(0.578)(0.014) = 79.2 Pa Step 3: Determine the the permissible shear stress on the channel lining. The permissible shear stress on a (retardance) class D lining, τp , is given in Table 5.12 as 16.8 Pa. Since the maximum shear stress exerted on the bottom of the channel (79.2 Pa) is greater than the permissible shear stress on the grass lining (16.8 Pa) the proposed lining is inadequate . 5.28. From the given data: Q = 10 m3 /s, S0 = 0.001, b = 3 m, m = 2, and the total depth of the channel is 2 m. The channel is lined with Kentucky bluegrass, which has an estimated retardance of C (Table 5.8). For any flow depth, y, the flow area, A, wetter perimeter, P , and hydraulic radius, R, are given by A = by + my 2 = 3y + 2y 2 √ √ P = b + 2 1 + m2 y = 3 + 2 5y = 3 + 4.472y R=

A 3y + 2y 2 = P 3 + 4.472y

(1)

Step 1: Determine the flow depth in the channel. For a retardance of C, Equation 5.56 gives a = 44.6, and Equation 5.55 gives Manning n as 1

1

1.22R 6 1.22R 6 n= = 0.4 1.4 44.6 + 19.97 log[R1.4 (0.001)0.4 ] 44.6 + 19.97 log[R S0 ]

(2)

and the Manning equation gives 5

1 A 3 12 Q= S n P 23 0 5

1 1 (3y + 2y 2 ) 3 2 10 = 2 (0.001) n (3 + 4.472y) 3

143

(3)

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Simultaneous solution of Equations (1) to (2) yields y = 2.31 m. Since this flow depth exceeds the depth of the channel (= 2 m), the channel is inadequate . 5.29. The required freeboard, F , and Manning’s n (for retardance C) are given by F = 0.152 +

Q2 2gA2

(1) 1

1.22R 6 n= 44.6 + 19.97 log(R1.4 S00.4 )

(2)

√ For a depth of flow, y, the geometric properties of the channel are: A = 4y 2 , P = 2 17y = 8.246y, R = A/P = 0.485y, T = 8y, and D = A/T = 0.5y. Substituting into Equation 2 gives 1

1

1.081y 6 1.22(0.485y) 6 = n= 1.4 0.4 44.6 + 19.97 log((0.485y) (0.01 )) 44.6 + 19.97 log(0.0575y 1.4 ) The maximum permissible shear stress is 98.9 Pa, therefore, at the limit of stability, γyS0 = 98.9 Pa → (9790)(y)(0.01) = 98.9 Pa → y = 1.01 m At this flow depth (y = 1.01 m), 1

1.081(1.01) 6 n= = 0.0543 44.6 + 19.97 log(0.0575 × 1.011.4 ) R = 0.485(1.01) = 0.490 m A = 4(1.01)2 = 4.08 m2 1 2 2 1 1 1 (4.08)(0.490) 3 (0.01) 2 = 4.67 m3 /s Q = AR 3 S02 = n 0.0543 Q 4.67 = = 1.14 m/s V = A 4.08 D = 0.5(1.01) = 0.505 V2 1.142 = = 0.262 (subcritical, OK) gD (9.81)(0.505) 1.142 F = 0.152 + = 0.218 m → Use F = 0.30 m 2(9.91)

Fr2 =

Therefore the maximum channel depth without lining is 1.01 + 0.30 = 1.31 m, and the corresponding flow capacity is 4.67 m3 /s . 5.30. From the given data: Q = 4 m3 /s, S0 = 0.8% = 0.008, y = 1 m, h = 15 cm, e = 0.45, d75 = 0.8 mm, and PI = 8. Take side slope as m = 3 .

144

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(a) Following the conventional design process: Cs = 106

(good condition)

Cn = 0.35(106)0.10 (0.15)0.528 = 0.205 A = by + my 2 = b(1) + 3(1)2 = b + 3 √ √ P = b + 2y 1 + m2 = b + 2(1) 1 + 32 = b + 6.325 A b+3 R= = P b + 6.325 ( ) ( ) b+3 b+3 (0.008) = 78.32 τ0 = γRS0 = (9790) b + 6.325 b + 6.325 [ ( )]−0.4 ( )−0.4 b+3 b+3 −0.4 n = Cn τ 0 = 0.0358 = 0.205 78.32 b + 6.325 b + 6.325 Substituting into the Manning equation yields 5

1 A 3 12 Q= S n P 23 0 ( )0.4 5 1 b+3 (b + 3) 3 2 4 = 27.91 2 (0.008) b + 6.325 (b + 6.325) 3 (b + 3)2.067 1.602 = (b + 6.325)1.067 which yields b = 0.345 m and n = 0.0472. (b) The maximum stress on the bottom of the channel is τb = γyS0 = (9790)(1)(0.008) = 78.32 Pa For sod grass in good condition, Cf = 0.90, and for the underlying soil ns = 0.016. The effective stress on the underlying soil is given by τe = τb (1 − Cf )

( n )2 s

n

( = (78.32)(1 − 0.90)

0.016 0.0472

)2 = 0.900 Pa

Since the soil is cohesionless, the permissible stress is τp = 1 Pa. Since τe < τp the lining is adequate . (c) The maximum shear stress on the bottom of the channel is given by τb = γyS0 = (9790)(1)(0.008) = 78.32 Pa In the alternative retardance-based approach, for Class C lining the permissible shear stress, τp , is 47.9 Pa. Since τb > τp the lining is inadequate .

145

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5.31. From the given data: b = 0.9 m, m = 3, S0 = 0.03, Q = 0.3 m3 /s, soil classification SC, PI = 16, and e = 0.5. For the proposed RECP lining, τI = 100 Pa. The flow area, A, wetted perimeter, P , and hydraulic radius, R, are given by A = by + my 2 = 0.9y + 3y 2 √ √ P = b + 2y 1 + m2 = 0.9 + 2y 1 + 32 = 0.9 + 6.325y R=

A 0.9y + 3y 2 = P 0.9 + 6.325y

Step 1: Determine the flow depth in the channel. For the RECP lining, the parameters a and b are given by Equations 5.62 and 5.63 as √ ( √ ( ) ( ) ) ( ) nupper nmid 0.036 0.033 b = −1.44 ln ln = −1.44 ln ln = −0.138 nlower nmid 0.040 0.036 −b a = nmid · τmid = (0.036) · (100)0.138 = 0.0680

The average shear stress on the channel boundary, τ0 , is given by (assuming γ = 9790 N/m3 ) ( τ0 = γRS0 = (9790)

0.9y + 3y 2 0.9 + 6.325y

)

( (0.03) = 293.7

0.9y + 3y 2 0.9 + 6.325y

)

The Manning’s n is given by Equation 5.61 as [ n=

aτ0b

= (0.0680) 293.7

(

0.9y + 3y 2 0.9 + 6.325y

)]−0.138

( = 0.0310

0.9y + 3y 2 0.9 + 6.325y

)−0.138 (1)

The Manning equation requires that 5

Q=

1 A 3 12 S n P 23 0 5

1 1 (0.9y + 3y 2 ) 3 0.3 = (0.03) 2 n (0.9 + 6.325y) 23

(2)

Solving Equations 1 and 2 simultaneously gives y = 0.188 m and n = 0.041. Step 2: Determine the effective stress on the underlying soil. The shear stress on the bottom of the channel, τb , is given by τb = γyS0 = (9790)(0.188)(0.03) = 55.2 Pa and the effective shear stress on the underlying soil, τe , is given by Equation 5.64 as ( ) ( )( ) ( τI ) 6.5 100 6.5 τe = τb − = 55.2 − = 2.1 Pa 4.3 τI 4.3 100

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Step 3: Determine the the permissible shear stress on the soil underlying the RECP lining and assess the adequacy of the lining. Since PI > 10, the soil is cohesive and the parameters to estimate the permissible shear stress are given in Table 5.10 as: c1 = 1.07, c2 = 14.3, c3 = 47.7, c4 = 1.42, c5 = −0.61, and c6 = 4.8 × 10−3 . The permissible shear stress on the underlying soil, τp,c , is then given by Equation 5.48 as τp,c = (c1 PI2 + c2 PI + c3 )(c4 + c5 e)2 c6 = [1.07(16)2 + 14.3(16) + 47.7][1.42 + (−0.61)(0.5)]2 (4.8 × 10−3 ) = 3.3 Pa Since the shear stress on the underlying soil (2.1 Pa) is less than the permissible shear stress on the underlying soil (3.3 Pa), the proposed lining is adequate .

5.32. From the given data: d50 = 0.15 m, γs = 25.9 kN/m3 , b = 0.6 m, m = 3, S0 = 2%, and Q = 1.13 m3 /s. For any flow depth, y, the wetted perimeter, P , the flow area, A, the top width, T , and the average depth, y¯, are given by √ √ P = b + 2y 1 + m2 = 0.60 + 2y 1 + 32 = 0.6 + 6.325y A = by + my 2 = (0.60)y + 3y 2 = 0.6y + 3y 2 T = b + 2my = (0.60) + 2(3)y = 0.6 + 6y y¯ =

A 0.6y + 3y 2 = T 0.6 + 6y

(1)

Step 1: Determine the flow depth in the channel. Assume that 1.5 ≤ y¯/d50 ≤ 185, then Manning’s n is given by Equation 5.67 as 1

0.319¯ y6

n=

(

2.25 + 5.23 log

y¯ d50

)

(2)

and the Manning Equation requires that 5

Q=

1 A 3 21 S n P 23 0 5

1 1 (0.6y + 3y 2 ) 3 2 1.13 = 2 (0.02) n (0.6 + 6.325y) 3

(3)

Solving Equations 1 to 3 simultaneously yields y = 0.501 m, y¯ = 0.292 m, and y¯/d50 = 1.95. Since 1.5 ≤ y¯/d50 ≤ 185, the assumed expression for n (Equation 2) is validated. Step 2: Determine the maximum shear stresses exerted on the bottom and sides of the channel. The maximum shear stress on the bottom of the channel, τp , is given by Equation 5.73 as (assuming γ = 9790 N/m3 ) τb = SF · γyS0 = SF · (9790)(0.501)(0.02) = 98.1 · SF Pa

147

(4)

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The side-shear-stress factor, Ks , is given by Equation 5.18 as Ks = 0.066m + 0.67 = 0.066(3) + 0.67 = 0.868 and so the maximum shear stress on the sides of the channel, τs , is given by τs = Ks τp = (0.868)(98.1 · SF) = 85.2 · SF Pa

(5)

Step 3: Determine the permissible shear stress on the bottom and sides of the channel and assess the adequacy of the lining. The shear-velocity Reynolds number, Re, under design flow conditions is given by Equation 5.77 as (assuming ν = 10−6 m2 /s) √ √ (9.81)(0.292)(0.02)(0.15) g y¯S0 d50 Re = = = 3.6 × 10−4 ν 10−6 Using this value of Re, the corresponding values of τ∗ and SF given by Equation 5.76 as τ∗ = 0.047

and

SF = 1.0

and hence the permissible shear stress on the bottom of the channel, τp , is given by Equation 5.75 as τp = τ∗ (γs − γ)d50 = (0.047)(25900 − 9790)(0.15) = 114 Pa The side-slope angle, θ is given by −1

θ = tan

( ) 1 = 18.4◦ 3

and the angle of repose, α can be estimated from Figure 5.6, which yields { 41.3◦ , very angular α= 38.5◦ , very rounded Since the riprap is subangular, taking the average of α for “very angular” and “very rounded” stones gives α = 39.9◦ and hence the tractive force ratio, K, is given by Equation 5.82 as √ √ sin2 θ sin2 (18.4◦ ) K = 1− = 1 − = 0.871 sin2 α sin2 (39.9◦ ) The permissible shear stress on the side of the channel, τps is given by 5.81 as τps = Kτp = (0.871)(114) = 99.3 Pa Using the calculated safety factor of SF = 1.0, the maximum shear stress on the bottom of the channel is given by Equation 4 as τb = 98.1(1.0) = 98.1 Pa and the maximum shear stress on the side of the channel is given by Equation 5 as τs = 85.2(1.0) = 85.2 Pa. Since the maximum shear stress on the bottom of the channel (98.1 Pa) is less than the permissible shear stress on the bottom of the channel (114 Pa), and the maximum shear stress on the sides of the channel (85.2 Pa) is less than the permissible shear stress on the sides of the channel (99.3 Pa), the proposed lining is adequate .

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5.33. From the given data: S0 = The lining will function as typical recommended value side-shear-stress factor, Ks ,

0.002, d50 = 25 mm (moderately rounded), and Q = 0.2 m3 /s. a gravel-mulch lining. Assume γs = 25.9 kN/m3 . Specify the of 3:1 (H:V) side slope (m = 3), and using Equation 5.18 the can be estimated as

Ks = 0.066m + 0.67 = 0.066(3) + 0.67 = 0.868 For d50 = 25 mm Figure 5.6 gives { 38.5◦ , very angular α= 31.5◦ , very rounded Since the lining is moderately, taking the average of α for “very angular” and “very rounded” stones gives α = 35◦ and hence the tractive force ratio, K, is given by Equation 5.82 as √ √ sin2 θ sin2 (18.4◦ ) K = 1− 1 − = = 0.835 sin2 α sin2 (35◦ ) Assume a value for the bottom width, b, substitute the given parameters into the following equations, and solve for the flow depth, y: A = by + 3y 2 √ P = b + 2y 1 + 32 T = b + 2(3)y A y¯ = T 1 0.319¯ y6 n= 2.25 + 5.23 log(¯ y /d50 ) 5

1 A 3 12 Q= S n P 23 0

(1)

For the solved value of y, verify that 1.5 ≤ y¯/d50 ≤ 185 to validate the assumed functional form of n. Determine Re using Equation 5.77 as (with ν = 10−6 m2 /s) √ g y¯S0 d50 Re = ν Based on the calculated value of Re, determine τ∗ and SF from Equation 5.76. The calculate the following shear stresses: τb = SF · γyS0 τs = K s τb τp = τ∗ (γs − γ)d50 τps = Kτp = 0.835τp

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If τb < τp and τs < τps the assigned value of b is adequate. For an optimal design it is recommended that b be changed incrementally in intervals of 0.1 m until these criteria are just satisfied. In the present case this occurs at b = 3.1 m, which has a flow depth of 9.1 cm. In accordance with the freeboard guidelines us a freeboard of 15 cm, so the minimum depth of the lined channel is 9.1 cm + 15 cm ≈ 24 cm. So the designed channel should have a bottom width of 3.1 m , side slopes of 3:1 , and a minimum lined depth of 24 cm .

5.34. From the given data: MT = 0.23 m, d50 = 0.15 m, γs = 25.9 kN/m3 , b = 0.60 m, m = 3, S0 = 9%, and Q = 0.28 m3 /s. For any flow depth, y, the wetted perimeter, P , the flow area, A, the top width, T , and the average depth, y¯, are given by

P = b + 2y

√

1 + m2 = 0.60 + 2y

√

1 + 32 = 0.60 + 6.325y

A = by + my 2 = (0.60)y + 3y 2 = 0.6y + 3y 2 T = b + 2my = (0.60) + 2(3)y = 0.6 + 6y y¯ =

0.6y + 3y 2 A = T 0.6 + 6y

(1)

Step 1: Determine the flow depth in the channel. Assume that 1.5 ≤ y¯/d50 ≤ 185, then Manning n is given by Equation 5.67 as 1

0.319¯ y6

n=

(

2.25 + 5.23 log

y¯ d50

)

(2)

and the Manning Equation requires that 5

1 A 3 21 S Q= n P 23 0 5

0.28 =

1 1 (0.6y + 3y 2 ) 3 2 2 (0.09) n (0.6 + 6.325y) 3

(3)

Solving Equations 1 to 3 simultaneously yields y = 0.244 m, y¯ = 0.158 m, and y¯/d50 = 1.05. Since y¯/d50 < 1.5, the assumed expression for n (Equation 2) is not validated. Assume that 0.3 < y¯/d50 < 1.5, then Manning n is given by Equation 5.67 as 1

y¯ 6 n= √ gf1 f2 f3

150

(4)

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where 0.28 √ (0.6y + 3y 2 ) (9.81)¯ y ( )0.453 ( ) ( )0.453 ( d50 y¯ 0.814 0.15 y¯ )0.814 = 1.14 = 1.14 T d50 0.6 + 6y 0.15 ( )log(0.755/β) 0.28Fr = β ( ) 0.6 + 6y 0.492 1.025((0.6+6y)/0.15)0.118 = 13.434 β 0.15 ( ) 0.6 + 6y −β = y¯

Fr = β f1 f2 f3

(5) (6) (7) (8) (9)

Solving Equations 3 to 9 simultaneously yields y = 0.182 m, y¯ = 0.123 m, and y¯/d50 = 0.82. Since 0.3 < y¯/d50 < 1.5 the assumed expression for n (Equation 4) is validated. It is also noted that Equation 5 gives Fr = 1.22, indicating that the flow is supercritical. Step 2: Determine the maximum shear stress exerted on the bottom of the channel. The default safety factor, SF, for gabions is 1.25, which is based on an assumed τ∗ = 0.10. The safety factor can also be calculated based on the friction-velocity Reynolds number, Re, where (assuming ν = 10−6 m2 /s) √ √ (9.81)(0.123)(0.09)(0.15) g y¯S0 d50 = Re = = 4.9 × 104 ν 10−6 and interpolating with Equation 5.76 gives SF = 1.03. Use the more conservative SF = 1.25. The maximum shear stress on the bottom of the channel, τb is therefore given by (assuming γ = 9790 N/m3 ) τb = SF · γyS0 = (1.25) · (9790)(0.182)(0.09) = 200 Pa Step 3: Determine the permissible shear stress on the bottom of the channel and assess the adequacy of the lining. Taking τ∗ = 0.10, Equation 5.88 gives τp = τ∗ (γs − γ)d50 = (0.10)(25900 − 9790)(0.15) = 242 Pa and taking Equation 5.89 gives τp = 0.0091(γs − γ)(MT + 1.24) = 0.0091(25900 − 9790)(0.23 + 1.24) = 216 Pa Therefore, taking the maximum of the two estimations gives τp = 242 pa. Since the maximum shear stress on the bottom of the channel (200 Pa) is less than the permissible shear stress (242 Pa), the proposed lining is adequate . 5.35. From the given data: m = 3, b = 0.9 m, S0 = 2%, h = 0.20 m, soil classification SC, PI = 16, e = 0.5, Q = 0.28 m3 /s. For a concrete lining, use nL = 0.013. For the grass lining, Equation 5.41 gives Cs = 106 and Equation 5.42 gives Cn = 0.35Cs0.10 h0.528 = 0.35(106)0.10 (0.20)0.528 = 0.239

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For a depth of flow, y, expressions for the flow area, A, and wetted perimeter, P , and low-flow wetted perimeter, PL , are given by A = by + my 2 = (0.9)y + (3)y 2 = 0.9y + 3y 2 √ √ P = b + 2y 1 + m2 = 0.9 + 2y 1 + 32 = 0.9 + 6.325y PL = b = 0.9 m Step 1: Determine the flow depth in the channel. The Manning’s n is given by Equation 5.99 as ( ) ( ) 3 ] 32 PL ns 2 PL + 1− n= nL P P nL ( )( [ ]2 0.9 ns ) 32 3 0.9 + 1− = (0.013) 0.9 + 6.325y 0.9 + 6.325y 0.013 [

(1)

For grassed sides ns is given by Equation 5.43 as ns = Cn τ0−0.4 = Cn [γRS0 ]−0.4 [ ( ) ]−0.4 0.9y + 3y 2 = (0.239) 9790 (0.02) 0.9 + 6.325y [ ]−0.4 0.9 + 3y 2 = 0.0290 0.9y + 6.325y

(2)

From the Manning equation 5

1 A 3 21 Q= S n P 23 0 5

1 1 (0.9y + 3y 2 ) 3 2 0.28 = 2 (0.02) n (0.9 + 6.325y) 3

(3)

Solving Equations 1 to 3 simultaneously yields the flow depth, y = 0.214 m and n = 0.046. Step 2: Determine the effective stress on the soil underlying the side of the channel. There is no need to calculate the maximum shear stress on the bottom of the channel, since with a concrete bottom lining erosion will not be an issue. For a side slope of 3:1 (H:V), the side-shear-stress-factor, Ks , can be estimated by Equation 5.18 as Ks = 0.066m + 0.67 = 0.066(3) + 0.67 = 0.868 and so the maximum shear stress on the sides of the channel, τs , can be estimated as (assuming γ = 9790 N/m3 ) τs = Ks τp = Ks [γyS0 ] = Ks [γyS0 ] = (0.868)[(9790)(0.214)(0.02)] = 36.4 Pa

152

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For mixed grass in good condition, Table 5.9 gives Cf = 0.75, for d75 < 1.3 mm (assumed) Equation 5.45 gives ns = 0.016, and the maximum effective shear stress, τe , on the soil underlying the grass lining on the side of the channel is given by Equation 5.46 as ( ) ( n )2 0.016 2 s τe = τs (1 − Cf ) = (36.4)(1 − 0.75) = 1.1 Pa n 0.046 Step 3: Determine the permissible shear stress on the side of the channel and assess the adequacy of the lining. Since PI > 10, the soil is cohesive and the parameters to estimate the permissible shear stress are given in Table 5.10 (for SC soil) as: c1 = 1.07, c2 = 14.3, c3 = 47.7, c4 = 1.42, c5 = −0.61, and c6 = 4.8 × 10−3 . The permissible shear stress on the underlying soil, τp,c , is then given by Equation 5.48 as τp,c = (c1 PI2 + c2 PI + c3 )(c4 + c5 e)2 c6 = [1.07(16)2 + 14.3(16) + 47.7][1.42 + (−0.61)(0.5)]2 (4.8 × 10−3 ) = 3.3 Pa Since the shear stress on the underlying soil (1.1 Pa) is less than the permissible shear stress on the underlying soil (3.3 Pa), the proposed grass lining is adequate . 5.36. From the given data: m = 4, b = 1.2 m, S0 = 1.5%, soil classification ML, PI = 12, e = 0.3, and Q = 0.7 m3 /s. For a concrete lining, use nL = 0.013. For class B vegetation, Table 5.11 gives Cn = 0.418. For a depth of flow, y, expressions for the flow area, A, and wetted perimeter, P , and low-flow wetted perimeter, PL , are given by A= P =

b + 2y

√

by + my 2 = (1.2)y + (4)y 2 = 1.2y + 4y 2 √ 1 + m2 = 1.2 + 2y 1 + 42 = 1.2 + 8.246y

PL =

b = 1.2 m

Step 1: Determine the flow depth in the channel. The Manning’s n is given by Equation 5.99 as [

( ) ( ) 3 ] 32 PL PL ns 2 n= + 1− nL P P nL [ ]2 ( )( 1.2 1.2 ns ) 32 3 = + 1− (0.013) 1.2 + 8.246y 1.2 + 8.246y 0.013

(1)

For grassed sides ns is given by Equation 5.43 as ns = Cn τ0−0.4 = Cn [γRS0 ]−0.4 ) ]−0.4 [ ( 1.2y + 4y 2 (0.015) = (0.418) 9790 1.2 + 8.246y [ ]−0.4 1.2y + 4y 2 = 0.0568 1.2 + 8.246y

153

(2)

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From the Manning equation 5

1 A 3 12 Q= S n P 23 0 5

0.7 =

1 1 (1.2 + 4y 2 ) 3 2 2 (0.015) n (1.2 + 8.246y) 3

(3)

Solving Equations 1 to 3 simultaneously yields the flow depth, y = 0.409 m and n = 0.081. Step 2: Determine the effective stress on the soil underlying the side of the channel. For a side slope of 4:1 (H:V), the side-shear-stress-factor, Ks , can be estimated by Equation 5.18 as Ks = 0.066m + 0.67 = 0.066(4) + 0.67 = 0.934 and so the maximum shear stress on the sides of the channel, τs , can be estimated as (assuming γ = 9790 N/m3 ) τs = Ks τp = Ks [γyS0 ] = Ks [γyS0 ] = (0.934)[(9790)(0.409)(0.015)] = 56.1 Pa For mixed grass in good condition, Table 5.9 gives Cf = 0.75, for d75 < 1.3 mm (assumed) Equation 5.45 gives ns = 0.016, and the maximum effective shear stress, τe , on the soil underlying the grass lining on the side of the channel is given by Equation 5.46 as ( ) ( n )2 0.016 2 s = (56.1)(1 − 0.75) = 0.5 Pa τe = τs (1 − Cf ) n 0.081 Step 3: Determine the permissible shear stress on the side of the channel and assess the adequacy of the lining. Since PI > 10, the soil is cohesive and the parameters to estimate the permissible shear stress are given in Table 5.10 (for ML soil) as: c1 = 1.07, c2 = 7.15, c3 = 11.9, c4 = 1.48, c5 = −0.57, and c6 = 4.8 × 10−3 . The permissible shear stress on the underlying soil, τp,c , is then given by Equation 5.48 as τp,c = (c1 PI2 + c2 PI + c3 )(c4 + c5 e)2 c6 = [1.07(12)2 + 7.15(12) + 11.9][1.48 + (−0.57)(0.3)]2 (4.8 × 10−3 ) = 2.1 Pa Since the effective shear stress on the underlying soil (0.5 Pa) is less than the permissible shear stress on the underlying soil (2.1 Pa), the proposed grass lining is adequate . 5.37. The area of the of the triangular bottom section is ( ) Tb Tb A1 = 2 2m1 and the area of the trapezoidal section above the triangular bottom section is [ ( ) ]( ) Tb Tb A2 = Tb + y − m2 y − 2m1 2m1

154

(1)

(2)

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Adding Equations (1) and (2) gives the total area of the channel as Tb A= 2

(

Tb 2m1

)

[

(

Tb + Tb + y − 2m1

)

]( m2

Tb y− 2m1

)

The perimeter of the triangular bottom section is ( ) √ Tb P1 = 2 1 + m21 2m1 and the perimeter of the trapezoidal section above the triangular bottom section is ( ) √ Tb P2 = 2 1 + m22 y − 2m1 Adding Equations (3) and (4) gives the total wetted perimeter of the channel as ) ) ( ( √ √ Tb Tb + 2 1 + m22 y − P = 2 1 + m21 2m1 2m1

155

(3)

(4)

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Chapter 6

Design of Sanitary Sewers 6.1. From the given data: A = 65 ha, 10% large lots, 75% small single-family lots, and 15% small two-family lots. The population densities are given as 6/ha for large lots, 75/ha for single-family lots, and 125/ha for two-family lots. Hence, the estimated population, P , of the residential development is given by P = 65[0.1(6) + 0.75(75) + 0.15(125)] = 4914 people Taking the average per capita flow rate as 350 L/d = 0.35 m3 /d, then the average total flow rate at the end of the design period is 4914(0.35) = 1720 m3 /d = 0.0199 m3 /s. Since the average flow at the beginning of the design period is 30% of the flow at the end of the design period, then the average flow when the sewers are first installed is 0.30(0.0199 m3 /s) = 0.00597 m3 /s. Using Equation 6.5, the peaking factors can be estimated as PFmin = 1.88Q−0.095 = 1.88(0.00597)−0.095 = 3.06 avg1 PFmax = 1.88Q−0.095 = 1.88(0.0199)−0.095 = 2.73 avg2 The maximum and minimum flows are estimated by multiplying the corresponding average wastewater flows by these factors, thus Maximum flow = 2.73(0.0199) = 0.0543 m3 /s = 54.3 L/s Minimum flow = 3.06(0.00597) = 0.0183 m3 /s = 18.3 L/s 6.2. From the given data, the total area of the city is 45 km2 = 4500 ha, and the residential area is 65% of 4500 ha = 2925 ha. Taking the per-capita flow rate as 500 L/d/person (= 5.79 × 10−6 m3 /s/person) gives the wastewater flows in the following table Type Large lots Small single-family lots Multi-story apartments Total

Area (ha) 0.15(2925) = 438.8 0.75(2925) = 2193 0.10(2925) = 292.5

157

Density (persons/ha) 6 75 2500

Population 2633 164531 731250 898414

Flow (m3 /s) 0.015 0.953 4.234 5.20

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The commercial sector of the city covers 25% of 4500 ha = 1125 ha, with a flow rate per unit area of 50,000 L/d/ha = 5.79 × 10−4 m3 /s/ha. Hence the average flow from the commercial sector is (5.79 × 10−4 )(1125) = 0.65 m3 /s. The industrial sector of the city covers 10% of 4500 ha = 450 ha, with a flow rate per unit area of 90,000 L/d/ha = 1.04 × 10−3 m3 /s/ha. Hence the average flow from the commercial sector is (1.04 × 10−3 )(450) = 0.47 m3 /s. The infiltration and inflow from the entire area is 1500 L/d/ha × 4500 ha = 6.75 × 106 L/d = 0.08 m3 /s. On the basis of these calculations, the average daily wastewater flow (excluding I/I) is 5.20 + 0.65 + 0.47 = 6.32 m3 /s. Since the average flow at the beginning of the design period is 35% of the flow at the end of the design period, then the average flow when the main sewer is first installed is 0.35(6.32 m3 /s) = 2.21 m3 /s. Using Equation 6.5, the peaking factors can be estimated as PFmin = 1.88Q−0.095 = 1.88(2.21)−0.095 = 1.74 avg1 PFmax = 1.88Q−0.095 = 1.88(6.32)−0.095 = 1.58 avg2 The maximum and minimum flows are estimated by multiplying the corresponding average wastewater flows by these factors and adding the I/I. Thus Maximum flow = 1.58(6.32) + 0.08 = 10.1 m3 /s Minimum flow = 1.74(2.21) + 0.08 = 3.9 m3 /s 6.3. From the given data: D = 760 mm, h = (3/4)×760 mm = 570 mm, and Q = 260 L/s = 0.260 m3 /s. Use Equation 6.13 to calculate θ, such that [ ( )] θ D h= 1 − cos 2 2 [ ( )] 760 θ 570 = 1 − cos 2 2 which yields θ = 4.189 radians. Substitute for θ in Equation 6.14 to determine A yields [ ] [ ] θ − sin θ 4.189 − sin(4.189) 2 A= D = (760)2 = 3.65 × 105 mm2 = 0.365 m2 8 8 The average velocity, V , is given by V =

Q 0.260 = = 0.712 m/s A 0.365

Therefore, the average velocity is estimated as 0.71 m/s .

158

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6.4. For circular pipe: ( A=

θ − sin θ 8

) D2

(1)

1 P = Dθ 2 A (θ − sin θ)D R= = P 4θ Q Q 8Q V = = ( θ−sin θ ) = 2 A (θ − sin θ)D2 D 8

(2) (3) (4)

The Manning equation is given by 1 2 12 R 3 S0 n and substituting Equations 1 to 4 into Equation 5 gives V =

(5)

[ ]2 8Q 1 (θ − sin θ)D 3 21 S0 = (θ − sin θ)D2 n 4θ which simplifies to 5

20.16Q =

1 (θ − sin θ) 3 8 12 D 3 S0 2 n θ3

and can be re-arranged to give − 12

θ− 3 (θ − sin θ) 3 − 20.16nQD− 3 S0 2

5

8

=0

6.5. From the given data: Q = 7 m3 /s, S0 = 0.01, D = 1.6 m, and assume n = 0.015. For uniform flow, − 12

θ− 3 (θ − sin θ) 3 − 20.16nQD− 3 S0 2

θ

− 23

5

8

− 38

5 3

(θ − sin θ) − 20.16(0.015)(7)(1.6)

− 12

(0.01)

=0 =0

which gives θ = 4.35 radians The flow depth, h, is therefore given by ( ) ( ) D θ 1.6 4.35 h= 1 − cos = 1 − cos = 1.25 m 2 2 2 2 The flow area, A, is given by ( ) ( ) θ − sin θ 4.35 − sin 4.35 2 A= D = (1.6)2 = 1.69 m2 8 8 and hence the flow velocity, V , is given by V =

Q 7 = = 4.14 m/s A 1.69

159

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When the pipe flow three-quarters full, 3 D h= D= 4 2

(

θ 1 − cos 2

)

which gives cos

θ = −0.5 2

θ = 4.19 radians Substituting into the uniform-flow equation gives (4.19)− 3 (4.19 − sin 4.19) 3 − 20.16(0.015)(7)D− 3 (0.01)− 2 = 0 2

5

8

1

which simplifies to D = 1.63 m 6.6. From the given data: Q = 3 m3 /s, S0 = 0.005, V = 2 m/s, and assume n = 0.013. The flow area, A, is given by Q 3 A= = = 1.5 m2 V 2 Also, ( ) θ − sin θ A= D2 = 1.5 m2 8 and solving for D gives

( D=

12 θ − sin θ

)1 2

According to the uniform-flow equation, −1

θ− 3 (θ − sin θ) 3 − 20.16nQD− 3 S0 2 = 0 ( )− 4 3 5 1 12 − 32 3 θ (θ − sin θ) − 20.16(0.013)(3) (0.005)− 2 = 0 θ − sin θ 2

5

8

θ− 3 (θ − sin θ) 3 − 0.405 = 0 2

1

which gives θ = 0.40 radians Therefore

( D=

12 θ − sin θ

)1

(

2

=

12 0.40 − sin 0.40

)1 2

= 33.7 m

The depth of flow, h, is given by ( ) ( ) D θ 33.7 0.40 h= 1 − cos = 1 − cos = 0.34 m 2 2 2 2

160

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6.7. From the given data: Q = 3.5 m3 /s, D = 1.4 m, n = 0.015, h/D = 0.5. Equation 6.13 gives [ ( )] 1 θ h = 1 − cos D 2 2 [ ( )] 1 θ 0.5 = 1 − cos 2 2

or

which leads to θ=π From Equation 6.17, the value of S0 is given by [ S0 =

]2

20.16nQD− 3

8

θ− 3 (θ − sin θ) 3 2

5

[ =

20.16(0.015)(3.5)(1.4)− 3 8

π − 3 (π − sin π) 3 2

5

]2 = 0.019

6.8. From Equation 6.17 − 12

θ− 3 (θ − sin θ) 3 − 20.16nQD− 3 S0 2

5

8

=0

which can be put in the form θ− 3 (θ − sin θ) 3 2

Q=

5

1

20.16nD− 3 S02 8

Extreme values of Q occur when 5

2

− 2 θ− 3 (θ − sin θ) 3 + θ− 3 53 (θ − sin θ) 3 (1 − cos θ) dQ =0= 3 8 −1 dθ 20.16nD− 3 S 2 5

2

0

which gives θ = 5.278 radians Finding d2 Q/dθ2 and substituting θ = 5.278 radians gives d2 Q/dθ2 < 0, which shows that a local maximum value of Q occurs at θ = 5.278 radians. The corresponding depth of flow in the circular pipe is given by ( ) ( ) h 1 θ 1 5.278 = 1 − cos = 1 − cos = 0.94 D 2 2 2 2 Therefore the maximum flow in a circular pipe occurs when h/D ≈ 0.94. 6.9. According to the Manning equation, the average velocity, V , is given by V =

1 2 12 R 3 S0 n

For full-pipe flow, A=

πD2 , 4

P = πD,

161

R=

A D = P 4

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and for half-pipe flow, A=

πD2 , 8

P =

πD , 2

R=

A D = P 4

Hence, for both full-pipe and half-pipe flow, 1 V = n

(

D 4

)2

1

3

S02

6.10. For a pipe flowing half full, A = πD2 /8, P = πD/2, R = D/4, and the Manning’s equation gives Q=

1 2 1 AR 3 S02 n

1 πD2 0.030 = 0.015 8

(

D 4

)2 3

1

(0.005) 2

which gives D = 0.301 m = 301 mm. The nearest commercial size is a diameter of 305 mm . For the Darcy-Weisbach (DW) equation, ks = (n/0.039)6 = 0.00324 m, µ = 0.001 N·s/m2 , and ρ = 998.2 kg/m3 . The DW equation gives ) ks 0.625µ + 3√ 12R ρR 2 8gS0 ] [ ( ( ) )√ 0.625(0.001) πD2 D 0.00324 0.030 = −2 8(9.81) (0.005) log + 3√ 8 4 12(D/4) (998.2)(D/4) 2 8(9.81)(0.005) √ Q = −2A 8gRS0 log

(

which yields D = 0.302 m = 146 mm. The nearest commercial size in 305 mm . √ The same Manning’s n should be used if 3.6 < R/ks < 360 and ks RS0 > 2.2 × 10−5 . When the pipe is half full, R (0.305/4) = = 24 ks 0.00324 √( ) √ 0.305 ks RS0 = 0.00324 × 0.005 = 6.326 × 10−5 4 √ When the pipe flows full, ks , R, and S0 are the same. Since ks RS0 > 2.2 × 10−5 , the flow is fully turbulent and Manning’s n will be the same for both half-full and full flow conditions. 6.11. According to Equation 6.17, for any circular pipe flowing under partially full conditions, − 21

20.16nQD− 3 S0 8

= θ− 3 (θ − sin θ) 3 2

5

(1)

When this same pipe is flowing full with the same flow rate, Q, then θ = 2π and Equation 6.17 gives − 12

20.16nQD− 3 S0 8

= (2π)− 3 [2π − sin(2π)] 3 = 2π 2

162

5

(2)

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Combining Equations 1 and 2 yields θ− 3 (θ − sin θ) 3 = 2π 2

5

which gives θ = 4.529 radians and 6.283 (= 2π) radians. Hence the flow rate under partially full conditions is the same as the flow rate under full-flow conditions when [ ( )] [ ( )] 1 θ 1 4.529 h = 1 − cos = 1 − cos = 0.82 D 2 2 2 2 Hence, when h/D = 0.82 the flow rate is the same as the full-flow flow rate. 6.12. From the given data: D = 915 mm = 0.915 m, S0 = 0.4% = 0.004, ϵ = 1 mm, T = 20◦ C, and h/D = 0.10, 0.50, and 1.00. At 20◦ C, the kinematic viscosity of water is ν = 10−6 m2 /s. For any given values of D (= 0.915 m) and h/D, the central angle, θ and the hydraulic radius, R, are given by Equation 6.13 and 6.16 as ( ) h −1 θ = 2 cos 1−2 D ( ) ( ) ( ) θ − sin θ θ − sin θ θ − sin θ A= D2 = (0.915)2 = 0.8372 8 8 8 ) ( ) ( ) ( 0.915 sin θ sin θ sin θ = 1− = 0.2288 1 − R = f racD4 1 − θ 4 θ θ Taking the unknown variable as Q, the following system of equations must be satisfied by Q: Re =

(Q/A)R ν

f=[

( log

(1) 0.25

ϵ 1.65 + 14.8R Re0.9 1

)]2

(2)

1

n = 0.1129R 6 f 2 Q=

θ

− 23

(3)

(θ − sin θ)

5 3

(4)

− 21

20.16nQD− 3 S0 8

Substituting known quantities in Equations 1 to 4 and solving for Q yields the following results:

(1) h/D 0.10 0.50 1.00

(2) Q (m3 /s) 0.027 0.618 1.236

(3)

(4)

(5)

n 0.0121 0.0126 0.0126

Typical n 0.0139 0.0145 0.0145

n/nfull 96% 100% 100%

163

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The input values of h/D are shown in Column 1, and the calculated Q and n values are shown in Columns 2 and 3, respectively. The “Typical” n value is obtained by increasing the calculated n value in Column 3 by 15% and is shown in Column 4. These values are 0.0139 , 0.0145 , and 0.0145 . The ratio of Manning’s n for the any given h/D to the full-flow Manning’s n is shown in Column 5. Based on these results, it is anticipated that Manning’s n will not vary much with flow depth and hence Manning’s n can be assumed to be constant. A typical value of n = 0.0145 could be assumed. 6.13. From the given data: Qdesign = 7 L/s = 0.010 m3 /s, S0 = 0.5% = 0.005, n = 0.013, and τc = 1.5 Pa. The hydraulic radius, R, required for self cleansing is derived from Equation 6.27 as (taking γ = 9790 N/m3 ) R=

τc 1.5 = = 0.0306 m γS0 (9790)(0.005)

For any given commercial-size diameter, D, determine θ and Q from Equations 1 and 2: ( ) sin θ D 1− R= 4 θ ( ) D sin θ 0.0306 = 1− (1) 4 θ θ− 3 (θ − sin θ) 3 2

Q=

5

− 12

20.16nD− 3 S0 8

θ− 3 (θ − sin θ) 3 2

Q=

5

(2)

20.16(0.013)D− 3 (0.005)− 2 8

1

The results of these calculations are tabulated as follows: D (mm) 610 685 760 840 915 1065

θ (radians) 1.13 1.07 1.01 0.96 0.92 0.85

Q (L/s) 5.64 5.96 6.26 6.56 6.84 7.35

Therefore commercial-size pipe diameters less than 1065 mm (42 in) would assure self cleansing. 6.14. From the given data: D = 0.915 mm = 0.915 m, S0 = 0.027% = 0.00027, Qmin = 15 L/s = 0.015 m3 /s, ϵ = 0.03 mm, Vpm = 0.60 m/s, and τc = 0.9 Pa. Under full-flow conditions, the velocity of flow in the pipe, Vfull , is given by Manning’s equation as Vfull =

1 nfull

2 3

1 2

Rfull S0 =

(

1 nfull

164

0.915 4

)2 3

1

(0.00027) 2 =

0.00615 nfull

(1)

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For a roughness height of 0.03 mm, nfull can be calculated using the Darcy-Weisbach equation and Equation 6.12 which yields nfull = 0.0102, and hence Equation 6.29 yields

Vfull =

0.00615 = 0.602 m/s 0.0102

Since Vfull > Vpm (i.e., 0.602 m/s > 0.60 m/s) the pipe configuration is adequate to assure self cleansing, as per the regulatory requirement.

Under minimum-flow conditions (Qmin = 0.015 m3 /s) in this pipe solution of the Manning equation (with variable n) yields n = 0.0101 and a hydraulic radius of R = 0.0757 m. Hence the minimum shear stress on the pipe boundary, τmin , is given by (taking γ = 9790 N/m3 )

τmin = γRS0 = (9790)(0.0757)(0.00027) = 0.20 Pa

Since the minimum shear stress (0.20 Pa) is less that the critical shear stress (0.9 Pa), the sewer pipe will not be self cleansing .

6.15. From the given data: D = 915 mm = 0.915 m, Qmax = 0.60 m3 /s, Qmin = 0.030 m3 /s, ϵ = 1.5 mm = 0.0015 m, τc = 2.0 Pa, T = 20◦ C, (h/D)max = 0.75, and Vlim = 4.0 m/s.

Step 1: From the given information, any pipe slope greater than 0.1% meets the minimumcover requirement. Therefore, Sref1 = 0.001.

Step 2: The pipe diameter under consideration is D = 915 mm.

Step 3: When Q = Qmax = 0.60 m3 /s and h/D = (h/D)max = 0.75, the following calcula-

165

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tions are made:

θmax Rmax Amax Remax ϵ Rmax fmax

[ ( ) ] h = 2 cos 1−2 = 2 cos−1 [1 − 2(0.75)] = 4.189 radians D max [ ] [ ] D sin θmax 0.915 sin(4.189) = 1− = 1− = 0.276 m 4 θmax 4 4.189 [ ] [ ] θmax − sin θmax 4.189 − sin(4.189) = D2 = (0.915)2 = 0.529 m2 8 8 (Qmax /Amax )Rmax (0.60/0.529)(0.276) = = = 3.13 × 105 ν 10−6 0.0015 = = 0.00543 0.276 )]−2 [ ( ϵ 1.65 = 0.25 log + 14.8Rmax Re0.9 max [ ( )]−2 0.00543 1.65 = 0.25 log + = 0.0215 14.8 (3.13 × 105 )0.9 −1

1

1

1

1

6 2 fmax = 0.1129(0.276) 6 (0.0215) 2 = 0.0133 nmax = 0.1129Rmax  2 − 83 20.16n Q D max max  Smax =  2 5 −3 θmax [θmax − sin θmax ] 3 [ ]2 8 20.16(0.0133)(0.60)(0.915)− 3 = = 0.00127 2 5 (4.189)− 3 [4.189 − sin(4.189)] 3

Based on these results, the minimum slope required for the sewer to flow no more than 75% full is 0.00127. Hence, Sref2 = 0.00127.

Step 4: When Q = Qmin = 0.030 m3 /s, for any given value of the pipe slope, Sref3 , the corresponding central flow angle, θmin , is determined by simultaneous solution of the

166

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following relations:

Rmin Amin Remin ϵ Rmin fmin

[ ] [ ] sin θmin 0.915 sin θmin D 1− = 1− = 4 θmin 4 θmin [ ] [ ] θmin − sin θmin θ − sin θmin min 2 = D = (0.915)2 8 8 (Qmin /Amin )Rmin (Qmin /Amin )Rmin = = ν 10−6 0.0015 = Rmin [ ( )]−2 ϵ 1.65 = 0.25 log + 14.8Rmin Re0.9 min 1

1

6 2 fmin nmin = 0.1129Rmin  2  2 − 38 − 83 20.16n Q (0.915) 20.16n Q D min min min min  =  Sref3 =  2 5 5 −3 − 23 θmin [θmin − sin θmin ] 3 θmin [θmin − sin θmin ] 3

Based on the value of θmin calculated from these equations, the value of Rmin is calculated and and the boundary shear stress, τmin is calculated using the relation

τmin = γRmin Sref3 The calculations in Step 4 are repeated for incremental values of Sref3 until τmin ≈ τc = 2.0 Pa. In the present case, these calculations yield Sref3 = 0.00306. Step 5: To meet the physical constraints, the maximum flow depth limitation, and the minimum boundary shear stress requirements, the minimum required pipe slope, S0 , is given by

S0 = max(Sref1 , Sref2 , Sref3 ) = max(0.00100, 0.00127, 0.00306) = 0.00306

Step 6: Determine the average velocity when Q = Qmax and S0 = 0.00306. Application of the Manning equation yields Vmax = 1.62 m/s. Since the maximum velocity (1.62 m/s) is less than the given limit (4.0 m/s) the calculated slope of 0.00306 is acceptable.

6.16. From the given data: A = 1700 ha, Pmin = 10000, Pmax = 50000, q = 250 L/d/person, and

167

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I/I = 2 m3 /d/ha. First determine the minimum and maximum flows: ( −3 ) 10 Qavg1 = Pmin q = (10000)(250) = 0.0289 m3 /s 86400 PFmin = 1.88Q−0.095 = 1.88(0.0289)−0.095 = 2.63 min 2 × 1700 = 0.0394 m3 /s QI/I1 = 86400 Qmin = PFmin × Qavg1 + QI/I1 = 2.63 × 0.0289 + 0.0394 = 0.115 m3 /s ( −3 ) 10 Qavg2 = Pmax q = (50000)(250) = 0.145 m3 /s 86400 = 1.88(0.145)−0.095 = 2.26 PFmax = 1.88Q−0.095 min QI/I2 = QI/I1 = 0.0394 m3 /s Qmax = PFmax × Qavg2 + QI/I2 = 2.26 × 0.145 + 0.0394 = 0.367 m3 /s It is required that S0 = 1% = 0.01. For D = 150 mm, the Manning roughness is 0.0106 for concrete in typical condition. For Q = Qmax = 0.367 m3 /s and h/D = 0.75, ( ) h 1 θ = 1 − cos D 2 2 ( ) 1 θ 0.75 = 1 − cos 2 2 which yields θ = 4.189 radians. Using the Manning equation (with D as the subject of the formula)  D= θ [ =

−1 20.16nQS0 2 − 23

3 8

 5

(θ − sin θ) 3

20.16(0.0106)(0.367)(0.01)− 2 1

]3 8

= 0.474 m

(4.189)− 3 (4.189 sin 4.189) 3 2

5

Tentatively select the closest commercial size pipe of 525 mm, readjust the Manning roughness to n = 0.0114 which corresponds to D = 525 mm and recalculate as follows: [ D=

20.16(0.0114)(0.367)(0.01)− 2

1

(4.189)− 3 (4.189 sin 4.189) 3 2

5

]3 8

= 0.487 m

Therefore tentative pipe diameter of 525 mm is confirmed. Check that the pipe is selfcleansing at the minimum flow when Q = Qmin = 0.115 m3 /s. Under this condition, the Manning equation gives − 12

θ− 3 (θ − sin θ) 3 − 20.16nQD− 3 S0 2

θ

− 32

5

8

− 83

5 3

(θ − sin θ) − 20.16(0.0114)(0.115)(0.525)

168

− 12

(0.01)

=0 =0

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which yields θ = 2.445 radians and hence ( ) ( ) D sin θ 0.525 sin 2.445 R= 1− = 1− = 0.0968 m 4 θ 4 2.445 τ = γRS0 = (9790)(0.0968)(0.01) = 9.5 Pa Since the minimum shear stress (= 9.5 Pa) is much greater than the minimum required shear stress (= 0.97 Pa) corresponding to a particle size of 1.5 mm, then the sewer is self cleansing. Check the maximum velocity, θ = 4.189 ) ( ) ( 2 4.189 − sin 4.189 2 θ − sin θ = (0.525) = 0.174 m2 A=D 8 8 Qmax 0.367 Vmax = = = 2.11 m/s A 0.174 Since the maximum velocity is less than the maximum allowable velocity of 3 m/s, a 525 mm diameter pipe meets all of the design requirements. 6.17. From the given data: D = 455 mm = 0.455 m, S1 = 0.4% = 0.004, S2 = 0.1% = 0.001, ϵ = 1.5 mm = 0.0015 m, and Q = 84.2 L/s = 0.0842 m3 /s. Since the inflow and outflow pipes are located opposite each other, the head loss coefficient can be taken as K = 0.15. Assume a temperature of 20◦ C so that ν = 10−6 m2 /s. Step 1: Determine the upstream flow conditions. The following equations are solved for the value of the central flow angle, θ1 : [ ] [ ] sin θ1 0.455 sin θmin D 1− = 1− R1 = 4 θ1 4 θmin [ ] [ ] θ1 − sin θ1 θ1 − sin θ1 2 A1 = D = (0.455)2 8 8 (Q/A1 )R1 (0.0842/A1 )R1 = Re1 = ν 10−6 [ ( )]−2 [ ( )]−2 ϵ 1.65 0.0015 1.65 f1 = 0.25 log + = 0.25 log + 14.8R1 Re0.9 14.8R1 Re0.9 1 1 1

1

n1 = 0.1129R16 f12  2  2 − 83 − 83 20.16n1 QD  =  20.16n12(0.0842)(0.455)  = 0.004 S1 =  2 5 5 −3 − θ1 [θ1 − sin θ1 ] 3 θ1 3 [θ1 − sin θ1 ] 3

(1)

These equations yield θ1 = 4.189 radians which corresponds to a flow depth y1 = 0.214 m and an average velocity V1 = 1.12 m/s. Step 2: Determine the downstream flow conditions. Use the same equations as in Step 1 with the exception that the slope in Equation 1 is taken as 0.001 instead of 0.004. This yields a flow depth y2 = 0.341 m and an average velocity of 0.643 m/s.

169

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Step 3: Determine the head loss at the manhole. Since the flow is straight through the manhole, the head loss coefficient can be estimated as K = 0.15 and the head loss, hL , at the manhole is given by Equation 6.30 as hL = K

(0.643)2 V22 = (0.15) = 0.003 m 2g 2(9.81)

Step 4: Determine the required drop at the manhole. The required drop, ∆z, is given by Equation 6.32 as [ 2 ] [ ] V2 V12 0.6432 1.122 ∆z = (y2 −y1 )+ − +hL = (0.341−0.214)+ − +0.003 = 0.087 m 2g 2g 2(9.81) 2(9.81) Therefore a minimum drop of 8.7 cm should be provided at the manhole to avoid backwater conditions. 6.18. From the given data: Q = 0.5 m3 /s, T = 23◦ C, S0 = 0.009, D = 1220 mm = 1.22 m, and n = 0.013. Substituting the given data into the Manning equation (Equation 6.17) gives − 12

θ− 3 (θ − sin θ) 3 − 20.16nQD− 3 S0 2

θ

− 23

5

8

− 83

5 3

(θ − sin θ) − 20.16(0.013)(0.5)(1.22)

(0.009)

− 12

=0 =0

which gives θ = 2.061 radians. The wetted perimeter, P , and the top width, B, are given by Dθ (1.22)(2.061) = = 1.257 m 2 ( ) 2 ( ) 2.061 θ = 1.22 sin = 1.046 m B = D sin 2 2 P =

Hydrogen sulfide is not a problem when Z ≤ 5000, hence Z = 0.308

EBOD P 1 1 S 2 Q3 B 0

5000 = 0.308

EBOD

1.257 (0.009) (0.5) 1.046 1 2

1 3

which gives EBOD = 1017 mg/L Since

EBOD = BOD5 · 1.07T −20

then 1017 = BOD5 · 1.0723−20 which gives BOD5 = 830 mg/L

170

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6.19. From the given data: BOD5 = 300 mg/L at 20◦ C, Q = 10 L/s = 0.01 m3 /s, D = 205 mm = 0.205 m, S0 = 0.1% = 0.001, and n = 0.013. Substituting the given data into the Manning equation (Equation 6.17) gives − 12

θ− 3 (θ − sin θ) 3 − 20.16nQD− 3 S0 2

θ

− 23

5

8

− 38

5 3

(θ − sin θ) − 20.16(0.013)(0.01)(0.205)

− 12

(0.001)

=0 =0

which gives θ = 4.16 radians. The wetted perimeter, P , and the top width, B, are given by (0.205)(4.16) Dθ = = 0.426 m 2 ( ) 2 ( ) θ 4.16 B = D sin = 0.205 sin = 0.179 m 2 2 P =

The Z value is given by Z = 0.308

EBOD P 300 P = 0.308 = 32, 300 1 1 1 1 B (0.001) 2 (0.01) 3 B S 2 Q3 0

Since Z > 10, 000, H2 S generation is expected . 6.20. (a) From the given data: Qmin = 0.15 m3 /min = 150 L/s, D = 915 mm, and the design particle size is 1.5 mm. From Table 6.8, Smin = 0.00942Q−0.5791 = 0.00942(150)−0.5791 = 0.000517 min (b) From the given data, Qmax = 0.29 m3 /s, and Table 6.4 gives n = 0.0118. Substituting these values into the Manning equation (Equation 6.17) and varying S0 until h/D = 0.75 yields S0 = 0.000233. This corresponds to a maximum velocity of 0.55 m/s, so the depth of flow controls the derived slope. Since this slope (= 0.000233) is less than the slope required for self cleansing (= 0.000517), then the design slope is the larger of the two, which is 0.000517 . (c) From the given data: Qavg = 0.25 m3 /s, T = 25◦ C, and BOD5 = 250 mg/L. Substituting into the Manning equation (with D = 915 mm, n = 0.0118, and S0 = 0.000517) yields θ = 3.21 radians and hence: P θ 3.21 = 1.60 = = B 2 sin(θ/2) 2 sin(3.21/2) EBOD = BOD5 × 1.07T −20 = 250 × 1.0725−20 = 351 Z = 0.308

EBOD 1 2

S0 Q

1 3

×

P 351 = 0.308 1 1 × (1.60) = 12093 B (0.000517) 2 (0.25) 3

Based on this value of Z, Table 6.7 indicates that sulfide generation is likely to occur . 6.21. From the given data: D = 535 mm, Qmax = 0.30 m3 /s, Qmin = 0.07 m3 /s = 70 L/s, and the design particle size is 1.5 mm.

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(a) From Table 6.8, the minimum slope is Smin = 0.00830Q−0.5767 = 0.00830(70)−0.5767 = 0.000716 min For a 535-mm diameter pipe in Typical condition, from Table 6.4 it can be estimated that n = 0.0114. Substituting these values along with Qmax = 0.30 m3 /s into the Manning equation (Equation 6.17) and varying S0 until h/D = 0.75 yields S0 = 0.000402. This corresponds to a maximum velocity of 0.48 m/s, so the depth of flow controls the derived slope. Since this slope (= 0.000402) is less than the slope required for self cleansing (= 0.000716), then the design slope is the larger of the two, which is 0.000716. Using this slope, the cover at the downstream manhole is 4.50 m − (2.50 m −150 × 0.000716 m) = 2.11 m. Since this is greater than the minimum cover of 2.00 m, use a pipe slope of 0.000716 . Maintain a diameter of 535 mm since both self-cleansing and flow capacity requirements are met. (b) At the upstream end, the crown elevation is 2.50 m and the invert elevation is 2.50 m − 0.535 m = 1.965 m . At the downstream end, the crown elevation is 2.50 m − 150 × 0.000716 m = 2.393 m and the invert elevation is 2.393 m − 0.535 m = 1.858 m . These elevations assume that no invert drop is necessary at the manhole. 6.22. (a) The assumed conditions are: full flow, V = 0.61 m/s, ks = 0.03 mm, and T = 15.6◦ C. At 15.6◦ C, ν = 1.12 × 10−6 m2 /s and for V = 0.61 m/s, D = 610 mm, and R = D/4 = 152.5 mm, we obtain the following results VD (0.61)(0.61) = = 3.32 × 105 ν 1.12 × 10−6 0.25 0.25 f=[ ( )]2 = [ ( )]2 = 0.0147 ks 5.74 0.03 5.74 log 3.7D + log 3.7(610) + (1.12×10−6 )0.9 Re0.9 1 1 1 1 n = 0.1129R 6 f 2 = 0.1129(0.1525) 6 (0.0147) 2 = 0.0100

Re =

For “typical-care” pipe, the calculated n value of 0.0100 is increased by 15% which yields n = 1.15(0.0100) = 0.0115. Which is exactly the same as recommended by ASCE, so agree with the recommended value. (b) For a design particle of diameter 1 mm, the critical tractive force, τc , is given by τc = 0.867(1)0.277 = 0.867 Pa The corresponding minimum slope, Smin is given by Smin =

τc 0.867 8.86 × 10−5 = = γR (9790)R R

(1)

where R is given by D R= 4

(

sin θ 1− θ

)

0.61 = 4

172

(

sin θ 1− θ

)

(

sin θ = 0.1525 1 − θ

) (2)

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For Q = 5 L/s = 0.005 m3 /s and n = 0.0115 the Manning equation requires that −1

2 θ− 3 (θ − sin θ) 3 − 20.16nQD− 3 Smin =0 2

5

8

−1

2 θ− 3 (θ − sin θ) 3 − 20.16(0.0115)(0.005)(0.610)− 3 Smin =0 2

5

8

−1

2 θ− 3 (θ − sin θ) 3 − 0.004331Smin =0 2

5

(3)

Solving Equations 1 to 3 simultaneously yields θ = 1.145 radians Smin = 0.00284 Using the ASCE formula gives { 0.00738(Q)−0.5772 = 0.00738(5)−0.5772 = 0.00291, for D = 600 mm Smin = 0.00759(Q)−0.5778 = 0.00759(5)−0.5778 = 0.00299, for D = 675 mm Interpolating for D = 610 mm yields Smin = 0.00292, which is 3% different from the calculated value. 6.23. Preliminary calculations and specifications: From the given data, the average wastewater flow at the beginning and end of design period, Qavg2 and Qavg2 , respectively, and the I/I flow can be expressed in convenient units as follows: Qavg2 = 250 L/d/person × 600 persons/ha = 150000 L/d/ha = 1.736 L/s/ha Qavg1 = 0.20Qavg2 = 0.20(1.736) = 0.347 L/s/ha QI/I2 = 100 m3 /d/km = 0.00116 L/s/m QI/I1 = QI/I2 = 0.00116 L/s/m

Since the design particle diameter is 1.0 mm, the design boundary shear stress (i.e., tractive force), τc , is given by Equation 6.28 as τc = 0.867d0.277 = 0.867(1.0)0.277 = 0.87 Pa Therefore, sewers in which the boundary shear stress is greater than 0.87 Pa under minimum-flow conditions will be taken as self cleansing. In accordance with conventional practice, the sewer system will be designed to meet the following additional constraints: Vmax = 3.5 m/s h ≤ 0.75 D Hydraulics of existing sewer: The results of the design computations are shown in Figure 6.1. The computations begin with Line 0, which is the existing sewer main that must be extended to accommodate the sewer lines in the proposed residential development.

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The average flow in the sewer main at the end of the design period is 0.400 m3 /s = 400 L/s, and the average flow at the beginning of the design period is 0.2(0.400) = 0.080 m3 /s = 80 L/s. Using Equation 6.5, the peaking factors and corresponding flows are given by PFmax = 1.88Q−0.095 = 1.88(0.400)−0.095 = 2.05 avg2 PFmin = 1.88Q−0.095 = 1.88(0.08)−0.095 = 2.39 avg1 Qmax = PFmax Qavg2 = (2.05)(0.400) = 0.820 m3 /s = 820 L/s Qmin = PFmin Qavg1 = (2.39)(0.08) = 0.191 m3 /s = 191 L/s Hence, the maximum flow is 820 L/s (column 10) and the minimum flow is 191 L/s (column 13). The n value for a 1220-mm concrete pipe in typical condition is 0.0121. With a slope of 0.008 (column 14) and a diameter of 1220 mm (column 15), the depth of flow at the maximum flow rate is 379 mm (column 17) and the corresponding maximum velocity is 2.65 m/s (column 18). Under minimum-flow conditions, the depth of flow is 184 mm with a corresponding boundary shear stress of 8.91 Pa (column 16). The invert elevation of the main sewer at MH 5 is 55.00 m (column 22) and the ground-surface elevation at MH 5 is 60.04 m (column 24). Sewer Line 1: The design of the sewer system begins with Line 1 on A Street, which goes from MH 1 to MH 2 and is 55 m long. The area contributing wastewater flow is 0.47 ha (column 7). The maximum and minimum wastewater flows are calculated as follows: QI/I2 = 0.00116 L/s/m × 55 m = 0.0638 L/s = 0.000064 m3 /s QI/I1 = QI/I2 = 0.000064 m3 /s Qavg2 = 1.736 L/s/ha × 0.47 ha = 0.816 L/s = 0.000816 m3 /s Qavg1 = 0.2Qavg2 = 0.2(0.000816) = 0.000245 m3 /s −0.44 PFmax = 0.281Q−0.44 = 6.42 avg2 = 0.281(0.000816) −0.44 PFmin = 0.281Q−0.44 = 10.9 avg1 = 0.281(0.000245)

Qmax,sewage = PFmax Qavg2 = (6.42)(0.000816) = 0.00524 m3 /s = 5.24 L/s Qmin,sewage = PFmin Qavg1 = (10.9)(0.000245) = 0.00178 m3 /s = 1.78 L/s Qmax = Qmax,sewage + QI/I2 = 5.24 + 0.0638 = 5.30 L/s Qmin = Qmin,sewage + QI/I1 = 1.78 + 0.0638 = 1.84 L/s Hence, the maximum flow is 5.30 L/s (column 10) and the minimum flow is 1.84 L/s (column 13). The n value for a 150-mm concrete pipe in typical condition is 0.0106. The minimum slope for self-cleansing, Smin calculated from the appropriate equation in

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Table 6.8 along with the ground slope, Sground as follows: Smin = 0.00539Q−0.5692 = 0.00539(1.84)−0.5692 = 0.00381 min Sground =

65.00 − 65.80 = 0.0218 55

Hence, to maintain a minimum cover of 1 m, specify the pipe slope, S0 , to equal the ground slope (= 0.0218). With a pipe slope of 0.0218 (column 14) and a diameter of 150 mm (column 15), the depth of flow at the maximum flow rate is 34 mm (column 17) and the corresponding maximum velocity is 1.77 m/s (column 18). Under minimum-flow conditions, the depth of flow is 26 mm with a corresponding boundary shear stress of 3.42 Pa (column 16). The pipe has adequate capacity, is self cleansing, and meets all design constraints. The drop in the sewer, ∆z, is given by ∆z = LS0 = (55)(0.0218) = 1.20 m The sewer invert at the upper end to have a 1.00 m cover is 65.00 m − 1.00 m − 0.150 m = 63.85 m (column 21), and the sewer invert at the lower end is 63.85 m − 1.20 m = 62.65 m. Sewer Lines 2 to 4: The design of Lines 2 and 3 follows the same sequence as for Line 1, with the exception that the wastewater flows in each pipe are derived from the sum of the contributing areas of all upstream pipes plus the pipe being designed and that the I/I flow in each pipe is the sum of all upstream I/I flows plus the I/I contribution to the pipe being designed. Using this approach, the invert elevation at the end of Lines 3 and 4 is 58.89 m (column 22), where the sewer laterals join the main sewer. The invert elevation of the sewer main is 55.00 m, which is 58.89 m − 55.00 m = 3.89 m below the invert of the laterals. A special drop-manhole structure will be required at this intersection. Sewer Line 5: The main sewer leaving MH 5 (Line 5) is designed next. The tributary area to Line 5 is the sum of the contributing areas of all contributing sewer laterals (Lines 1 to 4, 1.41 + 0.9 = 2.31 ha) plus the equivalent area of the average flow in the main sewer upstream of MH 5 (0.40 m3 /s ÷ 1.736 L/s/ha = 230.3 ha) plus the area that contributes directly to Line 5 (0.17 ha). Hence the total contributing area is 2.31 + 230.3 + 0.17 = 232.78 ha (column 7). The I/I contribution to Line 5 is the sum of the I/I contributions to all upstream laterals (0.285 + 0.105 = 0.390 L/s) plus the I/I contribution directly to Line 5 (70 m × 0.00116 L/s/m = 0.0812 L/s) for a total I/I contribution of 0.390 + 0.0812 = 0.471 L/s (column 8). The maximum and minimum flows are calculated using the peaking flow factors as previously described. Since the the required minimum slope is less than the practical slope of 0.08%, a slope of 0.08% is used. A manhole drop at the end of the pipe of 0.03 m is used to account for energy losses at the manhole, where laterals intersect.

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A Street A Street A Street A Street Main Street

2 3 5 5 12

To (4) 5

Manhole no.

1 2 3 4 5

Line no. Locaon From (1) (2) (3) 0 Main Street 1 2 3 4 5

Total (L/s) (10) 820

I/I (L/s) (11) -

1.78 2.67 3.30 2.57 185

Sewage (L/s) (12) -

1.84 2.83 3.58 2.67 185

Total (L/s) (13) 191

0.0218 0.0156 0.0236 0.0204 0.0008

Slope of sewer (14) 0.008

150 150 150 150 1220

Min tracve Diam force (mm) (Pa) (15) (16) 1220 8.90

Minimum flow

Sewage (L/s) (9) -

0.064 0.168 0.285 0.105 0.471

Maximum flow

I/I (L/s) (8) 5.30 8.03 9.98 7.65 828

Area

Length Increment Total (m) (ha) (ha) (5) (6) (7) 5.24 7.87 9.70 7.53 828

3.42 3.18 4.87 3.83 1.46

0.064 0.168 0.285 0.105 0.471

0.47 0.50 0.44 0.90 0.17

0.47 0.97 1.41 0.90 232.78

55 90 100 90 70

1.77 1.20 1.48 1.30 1.14

Max Max depth velocity (mm) (m/s) (17) (18) 379 2.65 34 61 61 55 731

Figure 6.1: Sewer Design Calculations

1.20 1.40 2.36 1.84 0.06

63.85 62.65 61.25 60.73 54.97

62.65 61.25 58.89 58.89 54.91

65.00 63.80 62.40 61.88 60.04

63.80 62.40 60.04 60.04 60.04

Sewer invert Ground surface elevaon elevaon Manhole invert Fall in Upper Lower Upper Lower end end drop sewer end end (m) (m) (m) (m) (m) (m) (19) (20) (21) (22) (23) (24) 55.00 60.04 0.03

176

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Chapter 7

Design of Hydraulic Structures 7.1. From the given data: D = 1.220 m, Q = 2.5 m3 /s, and S0 = 0.005. Assume that n = 0.013 for concrete pipe. Substituting in Equation 7.1 gives 5

(θ − sin θ) 3 θ

2 3

= 20.16

nQ (0.013)(2.5) = 5.453 √ = 20.16 8√ D S0 (1.220) 3 0.005 8 3

which yields θ = 4.052 radians. Substituting θ into Equation 7.2 gives [ ( )] [ ( )] D θ 1.220 4.052 y= 1 − cos = 1 − cos = 0.878 m 2 2 2 2 Therefore, the normal depth of flow is 0.878 m . Substituting the given data into Equation 7.3 gives (θ − sin θ)3 Q2 (2.5)2 ( θ ) = 512 5 = 512 = 120.7 gD (9.81)(1.220)5 sin 2 which yields θ = 4.0153 radians. Substituting θ into Equation 7.2 gives [ ( )] [ ( )] D θ 1.220 4.0153 y= 1 − cos = 1 − cos = 0.8681 m 2 2 2 2 Therefore, the critical depth of flow is 0.868 m . An approximation of the critical depth can be obtained by using Equation 7.4, which in this case gives [ ] [ 2 ]0.25 [ ][ ]0.25 1.01 Q 1.01 (2.5)2 yc = = = 0.857 m D0.26 g (1.220)0.26 9.81 7.2. From the given data: D = 0.5 m, n = 0.013, ke = 0.05, Cd = 0.95, S0 = 0.02, and L = 20 m. The elevation difference, ∆h, between the headwater and the top of the culvert at the exit is given by ∆h = 0.2 + S0 L = 0.2 + (0.02)(20) = 0.6 m Assume that the exit is not submerged, flow is either Type 2 or Type 3. Assume the flow is Type 2: v v u u 2g∆h 2(9.81)(0.6) u 2u Q = At = π(0.25) = 0.464 m3 /s t L 2 2(9.81)(0.013)2 20 4 + 0.05 + 1 2gn 4 + ke + 1 (0.5/4) 3

R3

177

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Determine if the culvert flows full. The Manning equation gives 1 A3 √ 1 (π0.252 ) 3 √ S0 = 0.02 = 0.534 m3 /s 2 nP3 0.013 (π0.5) 32 5

Qfull =

5

and the culvert flows full when the discharge exceeds 1.07(0.534) = 0.571 m3 /s. Since Q < 0.571 m3 /s, the culvert does not flow full, and the assumption of Type 2 flow is not supported. Assuming Type 3 flow, √ √ Q = Cd A 2gh = 0.95π(0.25)2 2(9.81)(0.25 + 0.2) = 0.55 m3 /s Since Q < 0.571 m3 /s, the culvert does not flow full, Type 3 flow is confirmed, and the discharge through the culvert is 0.55 m3 /s . 7.3. From the given data: b = 3.5 m, m = 2, S0 = 0.005, n = 0.025, Q = 3 m3 /s, L = 10 m, and Hp = 3 m. Find the normal depth of flow in the channel using the Manning equation: 5

1 A 3 12 Q= S n P 23 0 5

Q=

1 1 [by + my 2 ] 3 2 √ 2 S0 n [b + 2y 1 + m2 ] 3 5

3=

1 [3.5yn + 2yn2 ] 3 1 2 √ 2 (0.005) 0.025 [3.5 + 2y 1 + m2 ] 3

which gives yn = 0.465 m. This can be taken as the tailwater depth for the culvert. Assume Type II flow through the culvert, the flow is given by v u 2g∆h u Q = At 2gn2 L + ke + 1 4

(1)

R3

Taking the diameter to be D, ke = 1, and ∆h = 3 − D, then Equation 1 gives v πD2 u 2(9.81)(3 − D) u 3= t 2(9.81)(0.013)2 (10) 4 + 0.1 + 1 4 (0.25D) 3

Which gives D = 2.12 m and confirms Type II flow. For a reinforced concrete pipe, the next-larger commercial size is D = 2135 mm . 7.4. From the given data: Q = 1 m3 /s and H = 1 m . Assume Type 3 flow to use NIST equation. For the submerged condition, c = 0.0398, Y = 0.67, and S = 0. The NIST equation gives H = c · gFr2 + Y − 0.5S D [ ]2 1 1 = (0.0398) · (9.81) π √ + 0.67 − 0.5(0) 2 9.81D D 4D

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which gives two solutions D = 1.47 m or 0.568 m. Since only the latter solution is admissible take D = 0.568 m = 568 mm. This corresponds to Fr = 1.67 which is within the range of applicability of the NIST equation which requires Fr ≥ 0.7. The next higher commercial size is D = 610 mm . If a grooved end is selected, c = 0.0292, Y = 0.74, and the NIST equation gives H = c · gFr2 + Y − 0.5S D [ ]2 1 1 = (0.0292) · (9.81) π √ + 0.74 − 0.5(0) 2 9.81D D 4D

which gives two solutions D = 1.33 m or 0.528 m. Since only the latter solution is admissible take D = 0.528 m = 528 mm. This corresponds to Fr = 2.01 which is within the range of applicability of the NIST equation which requires Fr ≥ 0.7. The next higher commercial size is D = 535 mm . The corresponding appropriate equation is the orifice equation given by √ Q = Cd A gh where conventional values for Cd are 0.62 for square-edged entrances and 1.0 for rounded entrances. For a square-edged entrance, √ ( ) (π ) D 2 1 = (0.62) D (9.81) 1 − 4 2 which gives two solutions D = 1.94 m or 0.952 m. Since only the latter solution is admissible take D = 0.952 m = 952 mm. The next higher commercial size is D = 1065 mm . For a rounded entrance, 1 = (1.0)

(π 4

D

2

)

√

(

D (9.81) 1 − 2

)

which gives two solutions D = 1.98 m or 0.715 m. Since only the latter solution is admissible take D = 0.715 m = 715 mm. The next higher commercial size is D = 760 mm . Based on these results, using the non-NIST equation will lead to an overdesign of the culvert. 7.5. From the given data: T W = 10.00 m, D = 380 mm = 0.38 m, and L = 8 m. Since the water depth at the outlet is 10.00 m − 9.50 m = 0.50 m, and the culvert diameter is 0.38 m, then Type 1 flow is expected. Accordingly, the discharge, Q, is given by Equation 7.12 as √ 2g∆h Q=A (1) 4 2gn2 L/R 3 + ke + 1

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For a corrugated metal pipe, Table 7.3 gives n = 0.028 (conservative value) and Table 7.4 gives ke = 0.7 for a mitered entrance. Substituting into Equation 1 gives √ π 2(9.81)(HW − 10.00) 2 Q = (0.38) 4 4 2(9.81)(0.028)2 (8)/(0.38/4) 3 + 0.7 + 1 which simplifies to the following culvert performance curve √ Q = 0.236 (HW − 10)

(2)

If the culvert is required to pass 0.30 m3 /s, then the required headwater elevation satisfies the relation √ 0.3 = 0.236 (HW − 10) which gives HW = 11.62 m . When the tailwater elevation is 9.75 m, the flow through the culvert is Type 2, and the flow rate is based on the crest elevation of the culvert exit, which is equal to 9.50 m + 0.38 m = 9.88 m. Hence the discharge relation is given by √ Q = 0.236 (HW − 9.88) For HW = 11.62 m,

√ Q = 0.236 (11.62 − 9.88) = 0.31 m3 /s

Therefore the culvert capacity increases by 0.01 m3 /s or 3% . Type 3 flow is not a possibility in this case since the culvert is hydraulically long (i.e. L > 10D). 7.6. From the given data: culvert dimensions = 1.5 m × 1.5 m, n = 0.013, Cd = 0.95, ke = 0.05, S0 = 0.007, and L = 40 m. (a) Free outlet conditions. Assume Type 2 flow, where ∆h = 0.5 m + 0.007(40) = 0.78 m and therefore v v u u 2g∆h 2(9.81)(0.78) u 2u Q = At = (1.5) = 7.08 m3 /s t 40 L 2 2 2(9.81)(0.013) + 0.05 + 1 2gn 4 + ke + 1 4 2 (1.5 /6) 3

R3

Determine if the culvert flows full. Calculate the normal depth, yn , using the Manning equation, 5

1 An3 12 Q= S n P 23 0 n

5

1 1 (1.5yn ) 3 2 7.08 = 2 (0.007) 0.013 (1.5 + 2yn ) 3

(1.5yn )5 (1.5 + 2yn )2 yn = 1.22 m

1.33 =

180

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Therefore, the assumption that the culvert flows full (Type 2 flow) is not supported. Assuming Type 3 flow, √ √ Q = Cd A 2gh = 0.95(2.25) 2(9.81)(0.75 + 0.5) = 10.6 m3 /s Determine if the culvert flows full. Calculate the normal depth, yn , using the Manning equation, 5

1 An3 12 S Q= n P 23 0 n

5

1 (1.5yn ) 3 1 (0.007) 2 10.6 = 0.013 (1.5 + 2yn ) 23

(1.5yn )5 (1.5 + 2yn )2 yn = 1.7 m

4.468 =

Therefore, the assumption that the culvert does not flow full (Type 3 flow) is not supported. The flow is somewhere between Type 2 and Type 3 flow, and the discharge is in the range 7.08 to 10.6 m3 /s . (b) For a submerged outlet with tailwater 0.5 m above the crown of the culvert at the exit, ∆h = S0 L = 0.007(40) = 0.28 m and

v u u Q = At

v u 2g∆h 2(9.81)(0.28) 2u = (1.5) t = 4.25 m3 /s L 2 2(9.81)(0.013)2 40 4 + 0.05 + 1 2gn 4 + ke + 1 2 (1.5 /6) 3

R3

Find ∆h for Q = 10.6 m2 /s, where ] [ [ ] 10.62 2gn2 L Q2 2(9.81)(0.013)2 (40) ∆h = + 1 + ke + 1 = + 0.05 = 1.74 m 4 4 2gA2 2(9.81)(2.25)2 R3 (2.25/6) 3 So the headwater must be 1.5 + 0.5 + 1.74 − 0.28 = 3.46 m above the inflow-channel invert. For Q = 7.08 m3 /s, ∆h = 0.78 m. Therefore the range of headwater elevations is from 2.50 to 3.46 m above the channel invert. 7.7. From the given data: Q = 0.80 m3 /s, n = 0.013, H = 1 m, S0 = 2% = 0.02, L = 10 m, ke = 0.5, and Cd = 1. The data indicate that the difference between the headwater elevation and the crown of the culvert exit is given by ∆h = LS0 + H − D = (10)(0.02) + 1 − D = 1.2 − D where D is the culvert diameter. Assume the flow is either Type 2 or Type 3 flow. For Type 2 flow, Equation 7.12 gives √ π 2 2g(1.2 − D) 0.80 = D 4 4 2(9.81)(0.013)2 (10)/(D/4) 3 + 0.5 + 1

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which simplifies to

√ 1.02 = D2

4

19.62D 3 (1.2 − D) 4

0.211 + 1.5D 3 Solving iteratively for D gives D = 0.65 m Since H > 1.2D, inlet submergence verified. It must also be verified that the normal depth of flow is greater than the culvert diameter. When the culvert just flows full, Q=

1 2 1 AR 3 S02 n

where n = 0.013, A = π(0.65)2 /4 = 0.332 m2 , R = D/4 = 0.65/4 = 0.163 m, S0 = 0.02, and 2 1 1 (0.332)(0.163) 3 (0.02) 2 = 1.08 m3 /s 0.013

Qfull =

Therefore, when the flow is 0.80 m3 /s and the diameter is 0.65 m, the culvert does not flow full. This implies Type 3 flow, where √ Q = Cd A 2gh where Cd = 1 and

π √ 0.80 = (1) D2 2(9.81)(1 − D/2) 4

which yields D = 0.52 m Since A = π(0.52)2 /4 = 0.212 m2 , R = D/4 = 0.52/4 = 0.13 m, then the full-flow discharge is given by the Manning equation as Qfull =

2 1 1 (0.212)(0.13) 3 (0.02) 2 = 0.59 m3 /s 0.013

and the maximum flow rate in the culvert is 1.07(0.59) = 0.63 m3 /s. Therefore the culvert flows full and the flow is not Type 3. Since the flow is between Type 2 and Type 3, select the larger culvert diameter of 65 cm . 7.8. From the given data: H = 2 m, TW = 1 m, Q = 1 m3 /s, L = 15 m, and S0 = 0.015. Assuming Type 1 flow, then Equation 7.12 gives √ 2g∆h (1) Q=A 4 2 2gn L/R 3 + ke + 1 where A = πD2 /4 = 0.785D2 , ∆h = H − TW + S0 L = 2 − 1 + (15)(0.015) = 1.225 m, n = 0.013, R = D/4 = 0.25D, and taking ke = 0.50, Equation 1 gives √ 2(9.81)(1.225) 2 1 = 0.785D 4 2(9.81)(0.013)2 (15)/(0.25D) 3 + 0.50 + 1 √ 1 1 = 3.85D2 − 43 0.316D + 1.50

182

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Solving gives D = 0.61 m. This indicates that the exit is submerged, confirming Type 1 flow. Use a culvert with D = 61 cm . 7.9. From the given data: b = 2 m, m = 3, d = 3 m, and Q = 8 m3 /s. The minimum culvert size causes ponding to within 30 cm of the top of the approach channel. For a culvert diameter D with Type 3 flow, √ Q = Cd A 2gh √ ( ) (π ) D 2 8 = 0.62 D 2(9.81) 3 − 0.3 − 4 2 √( ) D 2 2.7 − 8 = 2.157D 2 which yields D = 1.645 m. The next step is to confirm the Type 3 flow assumption by demonstrating that the pipe does not flow full under design conditions. Taking n = 0.013 and assuming full-flow conditions, ( ) ( )2 1 1 1 2 8 1 1 πD2 D 3 2 π 2 Q = AR 3 S0 = S0 = D 3 S02 n n 4 4 16n 8 1 π (1.645) 3 (0.02) 2 = 8.054 m3 /s = 16(0.013) Since the actual flow rate is 8 m3 /s (< 8.054 m3 /s), the culvert does not flow full under design conditions and Type 3 flow is confirmed. The next larger commercial size is 1675 mm and this is the size that is recommended. The culvert should be constructed of reinforced concrete pipe . 7.10. From the given data: b = 2 m, Q = 4 m3 /s, S0 = 0.1, L = 25 m, ke = 0.1, and for a concrete culvert n = 0.013. Find the normal depth of flow in the culvert using the Manning equation, 5

1 A 3 21 Q= S n P 23 0 5

1 1 (2yn ) 3 2 4= 2 (0.1) 0.013 (2 + 2yn ) 3

which gives yn = 0.24 m Find the critical depth, yc , where Q2 A3 = c g Tc 2 4 (2yc )3 = 9.81 2 which gives yc = 0.74 m

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Therefore the culvert slope is steep. Assume Type 5 flow. Applying Equation 7.23 (neglecting the upstream velocity) gives √ Q = Ac

2g(∆h + V12 /2g − hi ) √ (

4 = (0.74 × 2)

42 2(9.81) ∆h − 0.1 × 2(9.81)(2 × 0.74)2 )

)

which gives ∆h = 0.41 m Since yc occurs at about 1.4yc downstream of the culvert entrance, the headwater depth is yc + ∆h − S0 (1.4yc ) = 0.74 + 0.41 − 0.1(1.4 × 0.74) = 1.05 m . Since the entrance is unsubmerged, Type 5 flow is confirmed. 7.11. From the given data: D = 1.83 m, L = 6.1 m, h1 = 1.170 m, and h2 = 0.920 m. For RCP take n = 0.013 and for vertical headwall take ke = 0.5. Assume Type 6 flow with negligible upstream velocity, hence the flow is described by ∆h −

Q2 V2 = hi + hf = 0.5 + S¯f L 2g 2gA21

where

( Sf =

nQ

(1)

)2 (2)

2

AR 3

From the culvert geometry and using Equation 8.113 gives h (m) 0.920 1.170

h/D 0.5027 0.6393

θ (rad) 3.154 3.706

A (m2 ) 1.325 1.776

P (m) 2.886 3.391

R (m) 0.4591 0.5237

Sf 2.718Q2 × 10−4 1.269Q2 × 10−4

Substituting into Equation 8.112 gives Q2 Q2 1 (1.170 − 0.920) − = 0.5 + Q2 2 2 2(9.81)(1.325) 2(9.81)(1.776) 2

(

2.78 + 1.269 2

)

10−4 × 6.1

which gives Q = 2.574 m3 /s. Since there are two barrels, the total flow through the culvert is 2 × 2.574 = 5.15 m3 /s . To verify Type 6 flow, show that y > yc within the culvert. Under critical flow conditions,

184

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with D = 1.83 m, Q2 A3 = g T 2 2.574 A3 = 9.81 T [( θ−sin θ ) 8

0.6754 =

D sin

D2 (θ)

[( θ−sin θ ) 0.6754 =

8

D sin

2

D2 (θ)

]3 ]3

2

(θ − sin θ)3 16.85 = sin(θ/2) which gives θ = 2.846 radians, and

[ ( )] D θ yc = 1 − cos = 0.780 m 2 2

Since y > yc , at both the entrance and the exit, Type 6 flow in confirmed. 7.12. From the given data: D = 0.915 m, Q = 1.5 m3 /s, L = 15 m, S0 = 0, ke = 0.2, and TW = 0.40 m. For a concrete culvert it can be assumed that n = 0.013. (a) For Type 2 flow, the difference between the headwater elevation and the crown of the culvert exit, ∆h, is given by Equation 7.11. From the given data, π 2 π D = (0.915)2 = 0.6576 m2 4 4 Q 1.5 V = = = 2.281 m/s A 0.6576 D 0.915 R= = = 0.2288 m 4 4 A=

and substituting into Equation 7.11 gives V2 V2 + 4 2g 2g R3 2 2 (0.013) (2.281) (15) (2.281)2 (2.281)2 H −D = + 0.2 + = 0.412 m 4 2(9.81) 2(9.81) (0.2288) 3 ∆h =

n2 V 2 L

+ ke

The calculated result that H − D = 0.412 m gives H = D + 0.295 m = 0.915 m + 0.412 m = 1.327 m (b) For the Type 2 flow variation shown in Figure 7.4, calculate the critical flow depth, yc . Under critical flow conditions, Equation 7.3 gives (θ − sin θ)3 Q2 (1.5)2 ( θ ) = 512 5 = 512 = 183.1 gD (9.81)(0.915)5 sin 2

185

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which yields θ = 4.3686 radians and hence [ ( )] [ ( )] D θ 0.915 4.3685 yc = 1 − cos = 1 − cos = 0.721 m 2 2 2 2 Therefore, assume that the HGL at the exit is (yc + D)/2 = (0.721 + 0.915)/2 = 0.818 m. Using the previously calculated value of ∆h = 0.412 m, the headwater depth is H = 0.818 m + 0.412 m = 1.23 m 7.13. The relationship between the depth of flow, y1 , at the culvert entrance and the depth of flow, y2 at the culvert exit can be estimated using the standard-step method, where [ ]1 Q2 y + 2gA2 2 L= ¯ Sf − S0 which can be put in the form L Q2 Q2 − y − (Sf 1 + Sf 2 − 2S0 ) = y1 + 2 2 2gA21 2gA22

(1)

where 2 nQ  = 2 A1 R13 2  nQ  = 2 A2 R23 

Sf 1

Sf 2

(2)

(3)

In this case, 5 2 3

A1 R1 =

A13 2

5

(2y1 ) 3

=

5 2 3

A2 R2 =

A23 2

2

(4)

2

(5)

(2 + 2y1 ) 3

P13

5

(2y2 ) 3

=

(2 + 2y2 ) 3

P23

Combining Equations 2 and 4, and taking n = 0.013 and Q = 10 m3 /s gives [

2

Sf 1 = (0.013)(10)

(2 + 2y1 ) 3 5

]2

4

= 0.00168

(2y1 ) 3

(2 + 2y1 ) 3 10

(6)

y13

and similarly 4

Sf 2 = 0.00168

(2 + 2y2 ) 3 10

y23

186

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The culvert tailwater depth, y2 , satisfies the Manning equation in the downstream channel, hence 5 1 2 1 1 A 3 21 2 3 Q = AR S0 = (8) S n n P 23 0 where Q = 10 m3 /s, n = 0.022, S0 = 0.005, b = 5 m, m = 2, and A = by2 + my22 = 5y2 + 2y22 √ √ P = b + 2 1 + m2 y2 = 5 + 2 1 + 22 y2 = 5 + 4.47y2 Substituting into Equation 8 gives 5

1 1 (5y2 + 2y22 ) 3 2 10 = 2 (0.005) 0.022 (5 + 4.47y 3

2

which yields y2 = 0.713 m

(9)

From the given data, L = 10 m, and combining Equations 1, 6, 7, and 9 gives   4 4 10  (2 + 2y1 ) 3 (2 + 2 × 0.713) 3 0.00168 + 0.00168 − 2(0.005) 10 10 2 0.713 3 y3 1

= y1 +

102 102 − 0.713 − 2(9.81)(2y1 )2 2(9.81)(2 × 0.713)2

which yields y1 = 0.677 m Therefore, the depth of flow at the culvert entrance is 0.68 m and the depth of flow at the culvert exit is 0.71 m . Since the culvert is 2 m × 2 m, then the culvert has adequate capacity when the flow is 10 m3 /s. It is interesting to note that the critical flow depth in the culvert is 1.36 m, the normal flow depth is 1.33 m, and hence there is a S3 water surface profile in the culvert. 7.14. From the given data: Q = 0.20 m3 /s, Er = 50.17 m, E0 = 48.01 m, b = 2 m, m = 2, S0 = 0.1% = 0.001, and for riprap n = 0.028 (typical). For normal flow in the exit channel. 5

Q=

1 A 3 12 S n P 23 0 5

1 1 (2y + 2y 2 ) 3 0.20 = (0.001) 2 √ 0.028 (2 + 2y 5) 23

which yields y = 0.224 m. Allow upstream ponding to within 46 cm (= 18 in.) of roadway gives a headwater depth of HW = (50.17 − 0.46) − 48.01 = 1.70 m

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Assume Type 2 flow and take ke = 0.05 gives v u 2g∆h u Q = At 2gn2 L + ke + 1 4 R3 v π u 2(9.81)(1.70 − D) 0.20 = D2 u t 2(9.81)(0.013)2 (15) 4 + 0.05 + 1 4 (D/4) 3

which yields D = 0.283 m. Verify the assumption that the culvert flows full. The culvert capacity, Q0 , is given by 5

1 A03 12 Q0 = 1.07 S n P 23 0 0

where n = 0.013, A0 = πD2 /4 = 0.0629 m2 , P0 = πD = 0.8891 m, and hence 5

1 1 (0.0629) 3 2 = 0.028 m3 /s Q0 = 1.07 2 (0.001) 0.013 (0.8891) 3

Since Q > Q0 the culvert flows full and Type 2 flow is confirmed. Also since the minimum required diameter (283 mm) is smaller than the minimum regulatory size (455 mm), use a culvert diameter of 455 mm . 7.15. For the drainage channel upstream and downstream of the culvert: b = 2 m, m = 3, d = 2 m, S0 = 0.5% = 0.005, and n = 0.018. For the culvert: D = 455 mm = 0.455 m, and S0 = 0.005. For both the channel and the culvert: Q = 1.5 m3 /s. The normal flow depth in the channel, yn is given by the Manning equation, 1 2 1 AR 3 S02 n 5 1 1 [2yn + 3yn2 ] 3 2 1.5 = √ 2 (0.005) 0.018 [2 + 2 10yn ] 3

Q=

which yields yn = 0.330 m. The culvert outlet is therefore not submerged. (a) Taking the required freeboard in the channel as 30 cm, the flow is either Type 2 or Type 3, with Type 2 more likely since the culvert is hydraulically long. Assuming Type 2 flow for a N-barreled culvert, v u Q 2g∆h u = At 2gn2 L (1) N + ke + 1 4 R3

188

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where ∆h = (h1 − h2 ) + S0 L = (1.70 − 0.455) + (0.005)(15) = 1.32 m n = 0.013 L = 15 m π π A = D2 = (0.455)2 = 0.163 m2 4 4 D 0.455 R= = = 0.114 m 4 4 ke = 0.5 (headwall with square-edged entrance) Q = 1.5 m3 /s Substituting into Equation 1 gives v u 1.5 2(9.81)(1.32) = (0.163)u t 2(9.81)(0.013)2 (15) N + 0.5 + 1 4 (0.114) 3

which yields N = 2.80. Therefore consider using a 3-barrel culvert. To confirm the Type-2 flow assumption, verify that each culvert flows full when Q = 1.5/3 = 0.5 m3 /s. The capacity of the culvert, Qref , is given by 1 2 2 1 1 1 Qref = 1.07 AR 3 S02 = 1.07 (0.163)(0.114) 3 (0.005) 2 = 0.22 m3 /s n 0.013

Since 0.5 m3 /s > Qref , the culvert flows full, Type 2 flow is confirmed, and a 3-barrel culvert is selected. (b) To find the headwater elevation, take N = 3 in Equation 1, v u 1.5 2(9.81)∆h = (0.163)u t 2(9.81)(0.013)2 (15) 3 + 0.5 + 1 4 (0.114) 3

which yield ∆h = 1.15 m, and hence the headwater depth, h1 , is given by h1 = ∆h + h2 − S0 L = 1.15 + 0.455 − (0.005)(15) = 1.530 m Use the direct step method to estimate the distance upstream to where the depth is

189

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equal to 50% of the depth at the culvert: y1 = 0.5(1.530) = 0.765 m A1 = (2)(0.765) + 3(0.765)2 = 3.29 m2 √ P1 = 2 + 2 10(0.765) = 6.84 m 3.29 A1 = = 0.481 m R1 = P1 6.84 Q 1.5 V1 = = = 0.456 m/s A1 3.29 2 [  ]2 nQ (0.018)(1.5)  = Sf 1 =  = 1.787 × 10−4 2 2 3 3 (3.29)(0.481) A1 R 1

y2 = 1.530 m A2 = (2)(1.530) + 3(1.530)2 = 10.08 m2 √ P2 = 2 + 2 10(1.530) = 11.68 m A2 10.08 R2 = = = 0.863 m P2 11.68 Q 1.5 V2 = = = 0.149 m/s A2 10.08  2 [ ]2 nQ (0.018)(1.5)  = Sf 2 =  = 0.8732 × 10−5 2 2 3 3 (10.08)(0.863) A 2 R2 S¯f = 0.5(Sf 1 + Sf 2 ) = 0.5(1.787 × 10−4 + 0.8732 × 10−5 ) = 0.9372 × 10−4 [ ] [ ] [ ] 2 1 0.4562 0.1492 y + V2g 0.765 + 2(9.81) − 1.530 + 2(9.81) 2 ∆L = ¯ = = 154 m 0.9372 × 10−4 − 0.005 Sf − S0 Therefore the distance upstream to where the water depth is 50% of the water depth at the culvert entrance is 154 m . 7.16. A flow scenario in which the culvert entrance is not submerged and the exit is submerged is illustrated below: The flow capacity of the culvert could be calculated as follows: (1) assume a flow, Q; (2) use backwater calculations to find where the water surface intersects the top of the pipe; (3) using this intersection point and the tailwater elevation, calculate the flow through the pressurized portion of the pipe; (4) repeat steps 1 to 3 until the flow in step 1 equals the flow in step 3. This would yield the capacity of the culvert under the given headwater and tailwater conditions. 7.17. From the given data, D = 450 mm = 0.45 m, L = 4 m, and S = 3% = 0.03. For an unsubmerged inlet (Type 5 flow), the appropriate USFHWA equation is Equation 7.26, which is given by M H Ec = + Kg 2 FrM − 0.5S (1) D D

190

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Figure 7.1: Culvert Flow which is applicable for Fr ≤ 0.6. Under critical-flow conditions (rectangular section), 3 3 E c = yc = 2 2

(

q2 g

) 31

3 = 2

(

Q2 gD2

) 13 (2)

For a box culvert with 45◦ wingwalls, Table 7.1 gives K = 0.026 and M = 1. Combining Equations 1 and 2 and substituting known parameters gives ( 2 )1 3 Q 3 M 2 gD2 H = + Kg 2 FrM − 0.5S D D ( )1 ]1 [ 3 Q2 3 1 2 9.81×0.452 H Q/0.452 = + 0.026(9.81) 2 √ − 0.5(0.03) 1.48 0.45 (9.81)(0.45) which simplifies to 2

H = 1.19Q 3 + 0.155Q − 0.00457

(3)

where H is in m and Q is in m3 /s. Equation 3 defines the culvert performance curve provided that Fr ≤ 0.6 √

Q/0.452 (9.81)(0.45)

≤ 0.6

which yields Q ≤ 0.264 m3 /s Therefore, for unsubmerged conditions, when Q ≤ 0.264 m3 /s the culvert performance curve is given by Equation 3. In the case where the culvert entrance is submerged (Type 3 flow), the appropriate USFHWA equation is Equation 7.15, which is given by H = cgFr2 + Y − 0.5S D

191

(4)

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which is applicable for Fr ≥ 0.7. For a box culvert with 45◦ wingwalls, Table 7.1 gives c = 0.0347 and Y = 0.86. Substituting known parameters into Equation 4 gives [ ]2 H Q/0.452 = 0.0347(9.81) √ + 0.86 − 0.5(0.03) 0.45 (9.81)(0.45) which yields H = 2.75Q2 + 0.258

(5)

where H is in m and Q is in m3 /s. Equation 5 defines the culvert performance curve provided that Fr ≥ 0.7 √

Q/0.452 (9.81)(0.45)

≥ 0.7

which yields Q ≥ 0.302 m3 /s Therefore, for submerged conditions, when Q ≥ 0.302 m3 /s the culvert performance curve is given by Equation 5. Using the analytic expressions given by Equations 3 and 5, the culvert performance curve is tabulated as follows: Q (m3 /s) 0.10 0.20 0.25 0.30 0.40 0.50 0.60

H (m) 0.27 0.43 0.50 0.51 0.70 0.95 1.25

The assumptions made for Type 3 and Type 5 flow are: low tailwater, hydraulically short culvert (for Type 3 flow), and a steep culvert slope (for Type 5 flow). For D = 0.45 m = 1.48 ft, and Q = 0.6 m3 /s = 21.2 ft3 /s, B = 0.45 m = 1.48 ft, Q/B = 21.2/1.48 = 14.3 (ft3 /s)/ft, and the USFHWA nomogram gives (for 30◦ to 75◦ wingwall flare) H = 3.1 D which yields H = 3.1D = 3.1(0.45 m) = 1.4 m This estimate of H obtained from the USFHWA nomogram (1.4 m) differs by approximately 10% from the calculated value of 1.25 m. This difference could be attributed to nomogram distortion (after repeated copying) and the uncertainty in reading the nomogram.

192

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7.18. From the given data: D = 1.220 m, Q = 2.3 m3 /s, L = 12 m, S0 = 0, ke = 0.3, and TW = 0.60 m. For a concrete culvert it can be assumed that n = 0.013. Since the culvert is horizontal, yn = ∞ and since the exit is not submerged the only possible flow types are Types 2, 3, and 6. Type 2 Flow: For Type 2 flow, the difference between the headwater elevation and the crown of the culvert exit, ∆h, is given by Equation 7.11. From the given data, π 2 π D = (1.220)2 = 1.169 m2 4 4 Q 2.30 V = = = 1.967 m/s A 1.169 D 1.220 R= = = 0.3050 m 4 4 A=

and substituting into Equation 7.11 gives V2 V2 + 2g 2g R 2 2 (0.013) (1.967) (12) (1.967)2 (1.967)2 + = 0.295 m H −D = + 0.3 4 2(9.81) 2(9.81) (0.3050) 3 ∆h =

n2 V 2 L 4 3

+ ke

The calculated result that H − D = 0.295 m gives H = D + 0.295 m = 1.220 m + 0.295 m = 1.515 m The culvert is not hydraulically long, since L = 9.8D. Therefore Type 2 flow is not guaranteed. Consider the other possibilities. Type 3 Flow: Application of Equation 7.15 requires that Fr > 0.7, and in this case V 1.967 Fr = √ =√ = 0.569 gD (9.81)(1.220) Therefore, since Fr < 0.7 application of Equation 7.15 is not validated and the flow cannot be Type 3. For future reference (Problem 7.19), perform the calculation anyway. For a culvert entrance flush with the headwall and with a grooved end, Table 7.1 gives c = 0.0292 and Y = 0.74. Substituting into Equation 7.15 gives H = 32.2 c Fr2 + Y − 0.5S0 D H = 32.2(0.0292)(0.569)2 + 0.74 − 0.5(0) 1.220 which yields H = 1.044 m. This result indicates that Type 3 flow will require a headwater depth of 1.044 m, which is less than the diameter of the pipe and further indicates that Type 3 flow will not occur.

193

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Type 6 Flow: For Type 6 flow, the headwater depth is calculated using Equation 7.29. Neglecting the headwater velocity, Equation 7.29 can be expressed as Q2 Q2 (H − TW) + 0 − = k + e 2g A¯2 2g A¯2

(

nQ ¯ 23 A¯R

)2 L

Solving this equation numerically (using the Solver package in an Excel spreadsheet) gives H = 1.143 m. This validates the assumption of Type 6 flow. Collectively, the results presented here have demonstrated that Type 6 flow is the most likely regime, and this will require a headwater depth of 1.143 m when the culvert is passing the design flow. However, the culvert is very close to being hydraulically long, in which case Type 2 flow might occur, with a required headwater depth of 1.515 m. For design purposes, it would be prudent to provide for a maximum headwater of 1.515 m . 7.19. The possible flow regimes are Types 2, 3, and 6, and the calculated headwater depths for these cases are as follows:

Type

Headwater depth (m)

2 3 6

1.515 1.044 1.143

Based on these results, the minimum-performance method would require a headwater depth of 1.515 m corresponding to Type 2 flow. This is the same as found previously. 7.20. For the given design flow rate, the tailwater elevation can be derived from the normal-flow condition in the downstream channel. Characteristics of the downstream trapezoidal channel are given as: b = 1 m, m = 2, So = 0.02, and n = 0.040. Taking Q = 3.00 m3 /s, the Manning equation gives Q=

1 2 1 AR 3 S02 n

1 3.00 = (yn + 2yn2 ) 0.040

(

yn + 2yn2 √ 1 + 2 5yn

) 23

1

(0.02) 2

which yields a normal-flow depth, yn = 0.673 m. Since the invert elevation of the downstream channel at the culvert outlet is 11.71 m, the tailwater elevation, TW, under the design condition is given by TW = 11.71 m + 0.673 m = 12.38 m Since the diameter of the culvert is 0.76 m and the tailwater depth is 0.673 m, the culvert outlet is not submerged; and since the roadway elevation is 13.50 m and the tailwater elevation is 12.38 m, the tailwater is below the roadway. Assuming that roadway overtopping (by the

194

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headwater) occurs under the design condition, the design flow rate is equal to the sum of the flow rate through the culvert and the flow rate over the roadway such that v u 3 2g∆h u Q = At 2gn2 L + Cd LR Hr2 (1) + ke + 1 4 R3

where Type 2 flow through the culvert is assumed. From the given data: Q = 3.00 m3 /s, D = 0.76 m, A = πD2 /4 = 0.454 m2 , n = 0.012 (Table 7.3 for concrete pipe, good joints, smooth walls), L = 20 m, R = D/4 = 0.19 m, ke = 0.7 (Table 7.4), LR = 15.0 m, and ∆h = HW + S0 L − D = (13.50 − 12.11 + Hr ) + 0.02 × 20 − 0.76 = 1.03 + Hr

(2)

Combining Equations 1 and 2 with the given data yields v u 3 2(9.81)(1.03 + Hr ) u 3 = 0.454t 2(9.81)(0.012)2 (20) + Cd (15)Hr2 + 0.7 + 1 4 0.19 3

3 √ 3 = 1.35 1.03 + Hr + 15Cd Hr2

The simultaneous solution this equation with the graphical relations in Figure 7.6 yields Cd = 1.50 Hr = 0.17 m The flow rate over the roadway, Qr , is given by 3

3

Qr = Cd LR Hr2 = (1.50)(15)(0.17) 2 = 1.58 m3 /s The corresponding flow rate through the culvert is equal to 3.00 m3 /s − 1.58 m3 /s = 1.42 m3 /s. The flow capacity of the culvert is given by 1 2 1 Qcapacity = 1.07 AR 3 S02 = 1.9 m3 /s n

Since this capacity exceeds the computed flow rate in the culvert, the culvert does not flow full and Type 3 flow is indicated. For Type 3 flow, 3 √ Q = Cd A 2gh + Cd LR Hr2 √

3 D 3 = Cd (0.454) 2(9.81)(road elevation + Hr − inlet invert − ) + 15Cd Hr2 2 √ ( ) 3 0.76 3 = Cd (0.454) 2(9.81) 13.5 + Hr − 12.11 − + 15Cd Hr2 2 3 √ 3 = 2.01Cd 1.01 + Hr + 15Cd Hr2

195

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which yields Cd = 1.385 Hr = 0.037 m The flow rate over the roadway, Qr , is given by 3

3

Qr = Cd LR Hr2 = (1.385)(15)(0.037) 2 = 0.15 m3 /s and the corresponding flow rate through the culvert is equal to 3.00 m3 /s − 0.15 m3 /s = 2.85 m3 /s. Since the flow through the culvert (2.85 m3 /s) exceeds the full-flow capacity of the culvert (1.9 m3 /s), Type 3 flow is not confirmed. Since neither Type 2 nor Type 3 flow can be confirmed, the flow conditions are intermediate between these types of flow. Type 2 conditions are more severe and should be taken as design conditions, in which case the depth of flow over the roadway is 0.17 m , the flow rate over the roadway is 1.58 m3 /s , and the flow rate through the culvert is 1.42 m3 /s . 7.21. From the given data: D = 0.450 m, H = 0.60 m, S0 = 0.01, and L = 3.6 m. From the given culvert characteristics it can be assumed that n = 0.013 and ke = 0.5. (a) Assuming Type 2 flow yields the following results: π 2 π D = (0.45)2 = 0.159 m2 4 4 ∆h = 0.6 − 0.45 + 3.6(0.01) = 0.186 m D 0.45 R= = = 0.1125 m 4v 4 u 2g∆h u Q = At 2gn2 L + ke + 1 4 R3 v u 2(9.81)(0.186) = (0.159)u = 0.232 m3 /s t 2(9.81)(0.013)2 (3.6) + 0.5 + 1 4 A=

(0.1125) 3

The next step is to confirm that the culvert flows full when Q = 0.232 m3 /s. According to the Manning equation, the full-flow capacity of the culvert is given by Qfull =

2 1 2 1 1 1 AR 3 S 2 = (0.159)(0.1125) 3 (0.01) 2 = 0.285 m3 /s n 0.013

Since the calculated flow rate assuming Type 2 is less than Qfull , the culvert doest not flow full and the assumption of Type 2 flow in not validated. Assume Type 3 flow and that the flow is given by the orifice equation, Cd = 0.62 D 0.45 h=H− = 0.6 − = 0.375 m 2 √2 √ Q = Cd A 2gh = (0.62)(0.159) 2(9.81)(0.375) = 0.267 m3 /s

196

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Since Q < Qfull , Type 3 flow is confirmed and the capacity according to the orifice equation is 0.267 m2 /s. Assume Type 3 flow and that the flow is given by the NIST equation, where c = 0.0398, Y = 0.67, and H = 32cFr2 + Y − 0.5S D 0.6 = 32(0.0398)Fr2 + (0.67) − 0.5(0.01) 0.45 which yields Fr = 0.722, which indicates that the NIST equation is valid (Fr > 0.7), and √ √ Q = Fr · A gD = (0.722) · (0.159) (9.81)(0.45) = 0.241 m3 /s Since Q < Qfull , Type 3 flow is confirmed and the capacity according to the NIST equation is 0.241 m2 /s. Since the NIST equation gives the least capacity, the capacity of the culvert should be taken as 0.241 m3 /s . (b) For the outlet design, first design the stone size of the riprap. Taking Q = 0.241 m3 /s and D = 0.45 m, calculate the depth of flow in the culvert using the Manning equation, 5

(θ − sin θ) 3 θ

2 3

= 20.16

nQ (0.013)(0.241) = 20.16 √ 8√ D 0.01 (0.45) 3 0.01 8 3

which yields θ = 3.988 radians. Hence, ( ) ( ) h 1 θ 1 3.988 = 1 − cos = 1 − cos = 0.705 D 2 2 2 2 h TW = D = (0.705)(0.45) = 0.317 m D ( )4 ( )4 0.044 Q 3 0.044 0.241 3 d50 = = = 0.0604 m = 60.4 mm TW D 0.317 0.45 Hence the median stone size should be 60 mm and the thickness should be 3 × d50 = 180 mm and length =

5.43Q D

3 2

=

5.43(0.241) 3

= 4.34 m

(0.45) 2

width = 3D + 0.4La = 3(0.45) + 0.4(4.34) = 3.09 m Since the calculated width is greater than three times the culvert diameter, the calculated width is confirmed as the required width. 7.22. From the given data: b = 7 m, y1 = 3.5 m, yg = 0.5 m. The gate discharge, Q, is given by √ √ Q = Cd byg 2gy1 = Cd (3)(0.5) 2(9.81)(3.5) = 12.43Cd For vertical gates, Cc = 0.61 and Cd = √

Cc y

1 + Cc yg1

=√

197

0.61 0.5 1 + 0.61 3.5

= 0.585

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Therefore Q = 12.43(0.585) = 7.27 m3 /s 7.23. A vertical gate with a rounded edge directs the flow under the gate in a downward direction and then gradually turns the flow into a horizontal direction. This produces a greater contraction than a sharp-edged gate which rapidly turns the flow into a horizontal direction. Consequently, the contraction coefficient of a vertical gate with a rounded edge is greater than the contraction coefficient of a vertical gate with a sharp edge. 7.24. The variables η and Cc are defined as follows yg y1

(1)

y2 yg

(2)

η = Cc and Cc = Combining Equations 7.44 and 7.45 gives Q= √

Cc 1+

y Cc yg1

byg

√

2gy1

Cc2 2 2 y b yg (2gy1 ) 1 + Cc yg1 ( ) Cc yg 2 ( ) ( ) 16 2 y1 y1 3 8Q 16η 2 y1 3 16 ( ) = = = yg y 1 + η y η(1 + η) gb2 y23 2 2 1 + Cc y 1 Q2 =

Equation 7.48 is given by y2 y3 = 2

√

( −1 +

8Q2 1+ 2 3 gb y2

) (4)

Substituting Equation 3 into Equation 4 and noting that y2 = Cc yg yields Cc y g y3 = 2

[√

16 1+ −1 η(1 + η)

]

which is the same as Equation 7.49. 7.25. From the given data: b = 5 m, Q = 16 m3 /s, yg = 0.7 m. The gate discharge is given by √ Q = Cd byg 2gy1 where for vertical gates, Cc = 0.61 and Cd = √

198

Cc y

1 + Cc yg1

(3)

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Contracted depth, y2 , is given by y2 = Cc yg = 0.61(0.7) = 0.427 m Find the downstream depth, y3 , using the hydraulic jump equation √ ( ) y2 8q 2 y3 = −1 + 1 + 3 2 gy2 where q= and therefore 0.427 y3 = 2

Q 16 = = 3.2 m2 /s b 5 √

( −1 +

8(3.2)2 1+ (9.81)(0.427)3

) = 2.01 m

7.26. From the given data: Q = 4 m3 /s, y2 = 1.1 m, b = 3 m, and q = Q/b = 4/3 = 1.33 m2 /s. The hydraulic jump equation gives √ ( ) y1 8q 2 y2 = −1 + 1 + 3 2 gy1 √ ( ) y1 8(1.33)2 1.1 = −1 + 1 + 2 (9.81)y13 which simplifies to

√

( y1

−1 +

1.44 1+ 3 y1

) = 2.2

and solving iteratively gives y1 = 0.24 m. For a vertical sluice gate, the coefficient of contraction is 0.61, and therefore the minimum gate opening to prevent the formation of a hydraulic jump is 0.24/0.61 = 0.39 m . The energy loss in the hydraulic jump is given by ∆E =

(y2 − y1 )3 (1.1 − 0.24)3 = = 0.60 m 4y1 y2 4(1.1)(0.24)

and the corresponding power loss, ∆P , is ∆P = γQ∆E = (9.79)(4)(0.60) = 23.5 kW 7.27. From the given data: Q = 10 m3 /s, b = 2 m, yg = 50 cm = 0.50 m, and y3 = 3.5 m. Assuming a coefficient of contraction, Cc , of 0.61, the depth of flow immediately downstream of the gate, y1 , is given by y1 = Cc yg = (0.61)(0.50) = 0.305 m

199

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The flow per unit width, q, in the lined channel is given by q=

Q 10 = = 5 m2 /s b 2

If y2 is the depth immediately before the hydraulic jump, and y3 (= 3.5 m) is the depth immediately after the hydraulic jump, then y2 and y3 are related by √ ) ( y3 1 8q 2 = −1 + 1 + 3 y2 2 gy2 √ ( ) 3.5 1 8(5)2 = −1 + 1 + y2 2 9.81y23 which yields y2 = 0.376 m Since y2 > y1 a hydraulic jump occurs downstream of the gate. If L is the distance along the channel required for frictional losses to cause the depth to increase from y1 to y2 , then L can be calculated using the direct-step method. In this case, y1 = 0.305 m A1 = 2(0.305) = 0.610 m2 P1 = 2y1 + b = 2(0.305) + 2 = 2.61 m 0.610 A1 = = 0.234 m R1 = P1 2.61 Q 10 V1 = = = 16.4 m/s A1 0.610  2 [ ]2 (0.015)(10) nQ  = S1 =  = 0.4193 2 2 3 (0.610)(0.234) 3 A1 R1 y2 = 0.376 m A2 = 2(0.376) = 0.752 m2 P2 = 2y2 + b = 2(0.376) + 2 = 2.752 m A2 0.752 R2 = = = 0.273 m P2 2.752 Q 10 V2 = = = 13.3 m/s A2 0.752  2 [ ]2 nQ (0.015)(10)  = S2 =  = 0.2247 2 2 3 (0.752)(0.273) 3 A2 R2 ¯ between sections 1 and 2 is given by and the average energy slope, S, 1 1 S¯ = (S1 + S2 ) = (0.4193 + 2247) = 0.3220 2 2

200

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Applying the energy equation using the direct step method yields [

]1 V2 2g 2

y+ L=

S¯ − S0

[

16.42 2(9.81)

0.305 + =

]

[ − 0.376 +

13.32 2(9.81)

0.3220 − 0

] = 14.4 m

Therefore, the hydraulic jump occurs 14.4 m downstream of the gate. Installation of baffle blocks or other energy-dissipation structures to increase friction losses at the gate outlet would cause a drowned hydraulic jump to occur at the gate. 7.28. From the given data: yg = 0.20 m, y1 = 1.7 m, and b = 1.5 m. Assuming Cc = 0.61, Equation 7.50 gives yg 0.20 η = Cc = 0.61 = 0.0718 y1 1.7 and, for free-flow conditions, Equation 7.45 gives the discharge coefficient, Cd , as Cd = √

Cc 0.61 = 0.589 =√ 1+η 1 + 0.0718

The free-flow discharge through the gate is given by Equation 7.44 as Q = Cd byg

√

√ 2gy1 = 0.589(1.5)(0.20) 2(9.81)(1.7) = 1.02 m3 /s

The distinguishing condition for the tailwater depth is given by Equation 7.49 as ] ] [√ [√ Cc yg 16 0.61(0.20) 16 1+ −1 = 1+ − 1 = 0.82 m y3 = 2 η(1 + η) 2 0.0718(1 + 0.0718) Therefore, the flow depth through the gate will be equal to 1.02 m3 /s as long as the tailwater depth is less than or equal to 0.82 m. If the tailwater depth exceeds 0.0.82 m, then the flow through the gate will be submerged and the flow will be reduced. 7.29. The energy equation (Equation 7.51) can be put in the form y = y1 +

Q2 Q2 − 2yb2 y12 2yb2 y22

(1)

and the momentum equation (Equation 7.52) can be put in the form y 2 = y32 +

2Q2 2Q2 − 2 2 gb y3 gb y2

(2)

Combining Equations 1 and 2 gives y32

2Q2 + 2 gb

(

1 1 − y3 y2

)

[

Q2 = y1 + 2gb2

201

(

1 1 − 2 2 y1 y2

)]2

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which simplifies to Q4 4g 2 b4

(

1 1 − y12 y22

)2 +

Q2 gb2

(

1 y1 2 2 − + − y1 y22 y3 y2

) + (y12 − y32 ) = 0

(3)

Define the following relations ( ξ=

)2 1 − 1 + 2(λ − 1) η

(4)

y1 y3 y2 = Cc yg yg y2 η = Cc = y1 y1 λ=

(5) (6) (7)

Combining Equations 3 to 7 yields [ ( ( 2 ) ( ) )2 ] ( 2 )2 Cc2 yg2 η2 1 Q 1 Q −ξ + 2 1− 2 =0 −1 4Cc2 yg2 η 2 gb2 y1 gb2 y1 η λ

(8)

Equation 8 is a quadratic equation in Q2 /(gb2 y1 ), and applying the quadratic formula to Equation 8 gives √ ( )2 ( ) 1 2− 1 − λ12 ξ ± ξ − 1 2 2 η Q = ( )2 gb2 y1 η2 1 − 1 2 2 2 2C y η c g

which simplifies to [

√ ξ−

ξ2 −

Q = Cc

(

1 η2 1 η

)2 ( −1 1−

1 λ2

)

]1

−η

2

byg

√

2gy1

7.30. From the given data: y1 = 5 m, b = 8 m, yg = 1.40 m, and y3 = 4.0 m. It must first be determined whether free-flow or submerged flow conditions exist. The distinguishing condition is given by Equation 7.49, where the contraction coefficient, Cc , can be assumed equal to 0.61, and η is given by Equation 7.50 as η = Cc

yg 1.40 = 0.61 = 0.17 y1 5

Substituting into Equation 7.49 gives the distinguishing condition as Cc yg y3 = 2

[√

] [√ ] 16 0.61(1.40) 16 1+ −1 = 1+ − 1 = 3.43 m η(1 + η) 2 0.17(1 + 0.17)

202

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Since the tailwater elevation (= 4 m) exceeds 3.43 m, then the flow is submerged and Equation 7.53 must be used to calculate the flow through the gate. From the given data, Equations 7.54 and 7.55 give y1 5 = = 1.25 y3 4 )2 ( )2 ( 1 1 − 1 + 2(λ − 1) = − 1 + 2(1.25 − 1) = 24.34 ξ= η 0.17

λ=

Substituting into Equation 7.53 gives [

√ ξ−

ξ2 −

Q = Cc

(

1 η2 1 η

−1

)2 (

1−

1 λ2

)

]1 2

byg

−η

√

[ √ ( 1 )2 ( 24.34 − 24.342 − 0.17 1− 2 − 1 = 0.61 = 38.8

1 0.17

2gy1

1 1.252

)

]1

− 0.17

2

√ (8)(1.40) 2(9.81)(5)

m3 /s

Therefore, under the given headwater, tailwater, and gate-opening conditions, the (submerged) discharge through the gate is 38.8 m3 /s . 7.31. From the given data: y1 = 5 m, b = 8 m, and Q = 50 m3 /s under free-flow conditions. Taking Cc = 0.61, Equation 7.50 gives η = Cc

yg yg = 0.61 = 0.122yg y1 5

and Equation 7.45 gives Cd = √

Cc 0.61 =√ 1+η 1 + 1.22yg

and Equation 7.44 gives the (free) flow through the gate as √ Q = Cd byg 2gy1 √ 0.61 50 = √ (8)(yg ) 2(9.81)(5) 1 + 1.22yg which yields yg = 1.88 m For yg = 1.88 m, and η = 0.122(1.88) = 0.229, the distinguishing condition is given by Equation 7.49 as [√ ] [√ ] Cc y g 16 0.61(1.88) 16 y3 = 1+ −1 = 1+ − 1 = 3.79 m 2 η(1 + η) 2 0.229(1 + 0.229)

203

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Since the tailwater depth of 4 m exceeds the distinguishing condition (y3 = 3.79 m), then submerged flow conditions exist and the flow through the gate is given by Equation 7.53. In this case, Equations 7.54 and 7.55 give y1 5 = = 1.25 y3 4 ( )2 ( )2 1 1 ξ= − 1 + 2(λ − 1) = − 1 + 2(1.25 − 1) = 11.8 η 0.229

λ=

and Equation 7.53 gives [

√ ξ−

ξ2 −

(

Q = Cc [ 11.8 −

1 η2 1 η

√

)2 ( −1 1−

1 λ2

)

]1 2

byg

−η

11.82

= 0.61

−

(

1 0.2292

1 0.229

−1

)2 (

1−

√ 2gy1 1 1.252

− 0.229

)

]1 2

√ (8)(1.88) 2(9.81)(5)

= 58.7 m3 /s Therefore, when the tailwater depth is 4 m, the flow through the gate is submerged and equal to 58.7 m3 /s . 7.32. From the given data: y1 = 5 m, y3 = 2 m, b = 3 m, and for a vertical gate with a sharp edge Cc = 0.61. At the distinguishing condition, ] [√ Cc y g 16 −1 (1) y3 = 1+ 2 η(1 + η) where η = Cc

yg yg → η = 0.61 = 0.122yg y1 5

Substituting into Equation 1 gives 0.61yg 2= 2

[√

] 16 1+ −1 0.122yg (1 + 0.122yg )

which yields yg = 0.383 m. So for gate openings < 0.383 m a hydraulic jump will occur downstream of the gate. For gate openings ≤ 0.383 m, Q = Cd byg where Cd = √

Cc y

1 + Cc yg1

√

2gy1

0.61 0.61 =√ =√ y 1 + 0.122yg 1 + 0.61 5g

204

(2)

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Substituting into Equation 2 gives the following discharge equation for yg ≤ 0.383 m: √ 0.61 Q= √ (3)yg 2(9.81)(5) 1 + 0.122yg which simplifies to Q= √

18.1yg , 1 + 0.122yg

for yg ≤ 0.383 m

For gate openings > 0.383 m, [

√ ξ−

ξ2 −

(

)2 ( −1 1−

1 η2

Q = Cc

1 η

1 λ2

)

]1

2

byg

−η

√

2gy1

where y1 5 = = 2.5 y3 2 yg yg η = Cc = 0.61 = 0.122yg y1 5 ( )2 )2 ( )2 ( 1 8.2 1 ξ= − 1 + 2(λ − 1) = − 1 + 2(2.5 − 1) = −1 +3 η 0.122yg yg

λ=

Substituting into Equation 3 gives the following discharge equation for yg > 0.383 m: [

√ ξ−

Q = 18.1

ξ2 −

(

1 η2

1 η

]1 2 )2 − 1 (0.84) yg ,

−η

for yg > 0.383 m

where ( ξ=

8.20 −1 yg

)2 +3

η = 0.122yg 7.33. Since

( Cc = 1 − 0.75

θ 90

)

To find θ that minimizes Cc , dCc 0.75 =− + 0.72 dθ 90

( + 0.36 (

θ 90

)

θ 90

1 =0 90

which simplifies to −0.00833 + 0.0000889θ = 0

205

)2

(3)

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or

θ = 93.7◦

Since

d 2 Cc >0 dθ2 then θ = 93.7◦ is a minimum. Since θ must be between 0◦ and 90◦ , then, within this range, the minimum value of Cc occurs at θ = 90◦ . 7.34. From Equation 7.73, 3

Q = Cw bH 2

(1)

If Hw is the height of the weir, then H + Hw = 2 m or H = 2 − Hw

(2)

Combining Equations 1 and 2 gives 3

Q = Cw b(2 − Hw ) 2 and re-arranging gives

( Hw = 2 −

Equation 7.74 gives

Q Cw b

)2 3

(3)

2 √ Cw = Cd 2g 3

(4)

and Equation 7.70 gives Cd = 0.611 + 0.075

H Hw

(5)

Combining Equations 4 and 5 gives 2 Cw = 3

(

H 0.611 + 0.075 Hw H = 1.80 + 0.221 Hw

)

√

2g (6)

Combining Equations 3 and 6 with Q = 10 m3 /s and b = 5 m, gives [

2 Hw = 2 − 1.80 + 0.221H/Hw

]2 3

Since H = 2 − Hw , this equation can be written as [

2Hw Hw = 2 − 1.58Hw + 0.442

206

]2

3

(7)

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Solving Equation 7 by trial and error gives Hw = 1.01 m and H = 2 − 1.01 = 0.99 m. Verify the validity of Equation 5: H 0.99 = = 0.98 < 5 Hw 1.01 Therefore, Equation 5 is valid. 7.35. A sill differs from a weir in that the height of the crest (Hw ) is much less than the height of water above the crest (H) such that H/Hw > 15 . The discharge coefficient for a weir (H/Hw ≤ 10) can be estimated using Equation 7.70, and the discharge coefficient for a sill (H/Hw ≥ 15) can be estimated using Equation 7.71. The efficacy of using Equation 7.72 to estimate the discharge coefficients for both sills and weirs can be measured by the percentage difference between Equations 7.70 and 7.72 for weirs and Equations 7.70 and 7.72 for sills. These results are presented in the following table: H/Hw 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Cd (Eq.7.72) 1.01 1.03 1.04 1.04 1.05 1.06 1.06 1.07 1.07 1.07 – – – – 1.07 1.07 1.07 1.07 1.07 1.07

Eq.7.70 −32% −26% −19% −13% −6% 0% 7% 13% 20% 27% – – – – – – – – – –

Eq.7.71 – – – – – – – – – – – – – – 9% 9% 8% 8% 7% 7%

These results indicate Equation 7.72 is most accurate in estimating the discharge coefficient of weirs when 5 ≤ H/Hw ≤ 7 and Equation 7.72 appears satisfactory for estimating the discharge coefficient of sills.

207

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7.36. From the given data: ymax = 3.0 m, Qmax = 3 m3 /s, and b = 5 m. The height, H, of water above the weir is given by H = 3 − Hw and the discharge equation is 3

Q = 1.83bH 2

3

3 = 1.83(5)H 2 which gives H = 0.475 m and therefore the height of the weir, Hw , is given by Hw = 3 − H = 3 − 0.475 = 2.52 m Verify H/Hw = 0.475/2.52 = 0.19 < 0.4, therefore the formulation is okay. 7.37. From the given data: Q = 5 m3 /s, b = 10 m, and H + Hw = 1 m Assuming H/Hw < 0.4, then 3

Q = 1.83bH 2 or

( H=

Q 1.83b

)2

(

3

=

5 1.83 × 10

)2 3

= 0.421 m

Hence Hw = 1 − H = 1 − 0.421 = 0.579 m Check H/Hw = 0.421/0.579 = 0.727 > 0.4, hence the assumed equation is not valid, and the general equation must be used, where 3

Q = Cw bH 2 ( ) √ 3 2 H 5= 0.611 + 0.075 2(9.81)(10)H 2 3 1−H Solution by trial and error gives H = 0.404 m and therefore the crest elevation of the weir is equal to 20 − H = 19.60 m . 7.38. From the given data: y1 = 3 m, b = 4 m, and yg = 0.5 m. When the weir is at its minimum height the distinguishing condition exists, in which case the discharge is free and a hydraulic jump occurs just downstream of the gate. A free discharge, with Cc = 0.61, gives Cc 0.61 Cd = √ =√ = 0.581 yg 1 + Cc y1 1 + 0.61 0.5 3 √ √ Q = Cd byg 2gy1 = (0.581)(4)(0.5) 2(9.81)(3) = 8.915 m3 /s

208

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The hydraulic jump equation is given by

[ ] √ y3 1 2 = −1 + 1 + 8Fr2 y2 2

Using Q = 8.915 m3 /s in the hydraulic jump equation yields y2 = Cc yg = (0.61)(0.5) = 0.305 m A2 = by2 = (4)(0.305) = 1.22 m2 Q 8.915 V2 = = = 7.307 m/s A2 1.22 V2 7.307 Fr2 = √ =√ = 4.224 gy2 (9.81)(0.305) [ ] √ 1 y3 2 = −1 + 1 + 8Fr2 y2 2 ] √ y3 1[ = −1 + 1 + 8(4.224)2 0.305 2 which yields y3 = 1.676 m. This is the depth upstream of the weir, so H + Hw = 1.676 m

(1)

The weir discharge is governed by the equation 3 2 √ Q = Cd 2gbH 2 3( ) 3 H √ 2 0.611 + 0.075 2(9.81)(4)H 2 8.915 = 3 Hw Combining Equations 1 and 2 gives ( 0.611 + 0.075

H 1.676 − H

)

(2)

3

H 2 = 0.7549

which yields H = 1.024 m and hence Hw = 1.676 − 1.024 = 0.652 m. Therefore, the minimum weir height to ensure submergence of the gate discharge is approximately 0.65 m . 7.39. From the given data: b = 5 m, Q = 1 m3 /s, H ≥ 0.5 m, and y ≤ 1.7 m. The discharge coefficient, Cd , is given by H Cd = 0.611 + 0.075 Hw Requiring H = 0.5 m and y = 1.7 m when Q = 1 m3 /s, Hw = y − H = 1.7 − 0.5 = 1.2 m H 0.5 = = 0.417 Hw 1.2 H = 0.611 + 0.075(0.417) = 0.642 Cd = 0.611 + 0.075 Hw √ 2 2 √ Cw = Cd 2g = (0.642) 2(9.81) = 1.90 3 3

209

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and the discharge equation requires 3

Q = Cw [b − 0.1nH]H 2 3

1 = 1.90[b − 0.1(2)(0.5)](0.5) 2 which gives b = 1.59 m and since Hw = 1.2 m, then the weir crest is 2 − 1.2 = 0.8 m below the top of the channel. 7.40. From the given data: Q = 55 m3 /s and b = 10 m. (a) The weir equations require that 3

Q = Cw (b − 0.2H)H 2 H Cd = 0.611 + 0.075 Hw 2 √ Cw = Cd 2g 3 H + Hw = 5 Combining the above equations and substituting the given data yields ( ) 3 2 5 − Hw √ 55 = 0.611 + 0.075 2(9.81)[10 − 0.2(5 − Hw )](5 − Hw ) 2 3 Hw ) ( 3 5 − Hw [10 − 0.2(5 − Hw )](5 − Hw ) 2 18.63 = 0.611 + 0.075 Hw which yields Hw = 2.95 m . Verify the validity of the assumed relation for Cd : H/Hw = (5-2.95)/2.95 = 0.69, which is less than 5, therefore, the assumed expression for Cd is applicable. (b) At the downstream end of the lake (Section 2), y2 = 5 m A2 = b2 y2 = (15)(5) = 75 m2 P2 = b2 + 2y2 = 15 + 2(5) = 25 m A2 75 R2 = = =3m P2 25  2 [ ]2 nQ (0.018)(55)  = Sf 2 =  = 4.027 × 10−5 2 2 3 3 (75)(3) A2 R 2

55 Q v2 = = = 0.733 m/s A2 75

210

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At the upstream end of the lake, the depth is y1 and A1 = b1 y1 = 15y1 P1 = b1 + 2y1 = 15 + 2y1 15y1 A1 R1 = = P1 15 + 2y1  2 Sf 1

 =

v1 =

4

3 (0.018)(55)  −4 (15 + 2y1 ) 10 ( ) 2  = 1.178 × 10 3 15y1 y13 (15y1 ) 15+2y 1

Q 55 3.667 = = A1 15y1 y1

Substituting into the energy equation, ]1 v2 2g 2

[ y+

∆L = ¯ Sf − S0 [

y1 + 100 =

(3.667)2 2(9.81)y12

]

[ − 5+

4.027×10−5 +1.178×10−4

4 (15+2y1 ) 3 10 y13

2

y1 + 100 =

0.6853 y12

0.7332 2(9.81)

]

− 0.005

− 5.027 4 3

1) 5.888 × 10−5 (15+2y − 0.004980 10

y13

which yields y1 = 4.50 m. Therefore, the depth at the upstream end of the lake is 4.50 m . 7.41. For the upstream location, y1 = 3.50 m A1 = 6(3.5) + 2(3.5)2 = 45.5 m2 V1 = Q/A1 = 0.549 m/s V12 = 0.01536 m 2g √ P1 = 6 + 2(3.5) 5 = 21.65 m [ ]2 2 (0.05)(25)(21.65) 3 S1 = = 0.0002803 5 (45.5) 3

211

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Applying the energy equation between the upstream location and the weir gives [ ] 252 [3.50 + 0.01536] − y2 + 2(9.81)(6y 2 2 2 +2y2 ) 100 = [ ] √ 2 2 (0.05)(25)(6+2y2 5) 3 5 2) 3 (6y2 +2y2

+0.0002803

− 0.001

2

which gives y2 = 3.57 m. For the weir, the Rouse equation gives ( ) 3 2 H √ Q= 0.611 + 0.075 2gbH 2 3 Hw This equation is valid as long as H/Hw < 5. The weir will be taken as a Cipolletti weir. Taking Z as the elevation of the weir crest, b as the length of the weir crest, and noting that the elevation of the bottom of the river at the weir location is 2.00 m − 0.1 m = 1.90 m, then H = 3.57 − Hw Z = Hw + 1.90 Combining the Rouse equation with the expression for H in terms of Hw gives the following design equation, ( ) 3 0.268 8.466 = 0.536 + b(3.57 − Hw ) 2 Hw This equation can be solved for alternative values of b. A possible design is: b = 2 m, Z = 2.65 m , Hw = 0.75 m, H = 2.82 m, H/Hw = 3.77 (which validates using the Rouse equation for Cd ). 7.42. From the given data: Q = 5 m3 /s, b = 10 m, and yn = 5 m. For a 5-m high suppressed sharp-crested weir, 3

Q = 1.83bH 2 3

5 = 1.83(10)H 2 which gives H = 0.421 m Check H/Hw = 0.421/5 = 0.084 < 0.4, hence the weir discharge formula is validated, and the new flow depth is 5 + 0.421 = 5.42 m . Consider a rectangular weir with b = 7 m, and a height of 5 m. 1st iteration, H = 0.5 m: H 0.5 = 0.611 + 0.075 = 0.62 Hw 5 √ 2 √ 2 Cw = Cd 2g = (0.62) 2(9.81) = 1.83 3 3 Cd = 0.611 + 0.075

212

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and the discharge equation gives 3

Q = Cw [b − 0.1nH]H 2 3

5 = 1.83[7 − 0.1(2)H]H 2 which yields H = 0.54 m 2nd iteration, H = 0.54 m: H 0.54 = 0.611 + 0.075 = 0.619 Hw 5 √ 2 √ 2 Cw = Cd 2g = (0.619) 2(9.81) = 1.83 3 3 Cd = 0.611 + 0.075

and the discharge equation gives 3

Q = Cw [b − 0.1nH]H 2 3

5 = 1.83[7 − 0.1(2)H]H 2 which yields H = 0.54 m Hence, if the weir is contracted, the flow depth is 5 + 0.54 = 5.54 m . For a 7-m wide Cipolletti weir, 3

Q = Cw bH 2 and taking Cw = 1.83 gives

3

5 = 1.83(7)H 2 which yields H = 0.53 m, and hence a flow depth of 5 + 0.53 = 5.53 m . Since H/Hw = 0.53/5 = 0.106 < 0.4, the applied formula is correct. 7.43. From the given data: H + Hw = 2 m, and Q = 5 m3 /s. The discharge formula is given by 3

Q = Cw bH 2 where

( ) 2 √ 2 H √ Cw = Cd 2g = 2g 0.611 + 0.075 3 3 Hw H = 1.80 + 0.221 Hw

If the top-width of the weir is 10 m, then ( b+2

213

H 4

) = 10

(1)

(2)

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or

H 2 Combining Equations 1 to 3, and substituting the given data yields [ ][ ] 3 H H 5 = 1.80 + 0.221 10 − H2 2−H 2 b = 10 −

(3)

which gives H = 0.42 m and hence Hw = 2 − 0.42 = 1.58 m and b = 10 −

H 0.42 = 10 − = 9.79 m 2 2

7.44. From the given data: Hw = 0.8 m, b = 1 m, and H = 0.2 m. Check H/Hw = 0.2/0.8 = 0.25 < 0.4, therefore 3

3

Q = 1.83bH 2 = 1.83(1)(0.2) 2 = 0.164 m3 /s When the weir is submerged,

[ ( y ) 3 ]0.385 Qs d 2 = 1− Q H

which gives [ Qs = Q 1 −

( y ) 3 ]0.385 d

2

H

[ = (0.164) 1 −

(

0.05 0.2

) 3 ]0.385 2

= 0.156 m3 /s

7.45. The flow over the weir is submerged, and hence either Equation 7.78 or Equation 7.79 can be used to estimate the ratio of submerged flow, Qs , to free flow, Q. From the given data: H = 2.5 m − 2.0 m = 0.5 m, and yd = 2.3 m − 2.0 m = 0.3 m. Equation 7.78 gives [ ( ) 3 ]0.385 [ ( y ) 3 ]0.385 0.3 2 Qs d 2 = 1− = 1− = 0.786 Q H 0.5 and Equation 7.79 gives Qs ( yd ) = 1+ Q 2H

√

yd 1− = H

(

0.3 1+ 2 × 0.5

)√ 1−

0.3 = 0.822 0.5

These results indicate that Qs /Q is in the range 0.786–0.822. From the given data: Hw = 2 m, and H/Hw = 0.5/2 = 0.25. Since H/Hw < 5, then either Equation 7.70 or Equation 7.72 can be used to estimate the discharge coefficient, Cd . According to Equation 7.70, Cd = 0.611 + 0.075

H = 0.611 + 0.075(0.25) = 0.630 Hw

214

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and Equation 7.72 gives [(

)10 ( )15 ]−0.01 14.14 H/Hw Cd = 1.06 + 8.15 + H/Hw H/Hw + 1 [( )10 ( )15 ]−0.01 14.14 0.25 = 1.06 + 8.15 + 0.25 0.25 + 1 = 1.01 These results indicate that Cd is in the range 0.630–1.01. From the given data, b = 5 m and the free-flow discharge over the weir is given by Equation 7.67 3 3 2 √ 2 √ Q = Cd 2gbH 2 = Cd 2(9.81)(5)(0.5) 2 = 5.22Cd 3 3

For Cd = 0.630, Q = 5.22(0.630) = 3.29 m3 /s and for Cd = 1.01, Q = 5.22(1.01) = 5.27 m3 /s. The minimum submerged flow, Qs , corresponds to Q/Qs = 0.786, which yields Qs = 0.786(3.29) = 2.59 m3 /s, and the maximum estimate of Qs corresponds to Q/Qs = 0.822, which yields Qs = 0.822(5.27) = 4.33 m3 /s. Available empirical formulae yield flow rates over the weir in the range 2.59– 4.33 m3 /s . For a more precise estimate of the flow rate, field calibration of this particular weir would be necessary. 7.46. From the given data: z1 = 101.40 m, z2 = 96.00 m, Q = 4.00 m3 /s, and b = 6 m. Also, H + Hw = 101.40 − 96.00 = 5.40 m. The weir equation gives: 3 2 √ Q = Cd 2gbH 2 3( ) √ 3 2 H 0.611 + 0.075 2(9.81)(6)H 2 4= 3 5.40 + H

which yields H = 0.51 m. Therefore, Hw = 5.40 − H = 4.89 m, and H/Hw = 0.014. Since H/Hw < 5, the equation used for Cd is valid. Based on these derived data, the crest elevation of the weir is 101.40 m − 0.51 m = 100.89 m . Using Villemonte’s formula, for a 90% flow reduction ( y )]0.385 Qs [ d = 1− Q H ( y )]0.385 Qs [ d = 1− Q H which yields yd /H = 0.386, and hence yd = 0.386(0.51) = 0.20 m. The limiting tailwater elevation is 100.89 m + 0.20 m = 101.09 m . 7.47. From the given data: S0 = 0.1% = 0.001, b = 2 m, m = 2, yT = 2 m, Q = 3 m3 /s, and y = 1.5 m.

215

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(a) For a contracted rectangular weir with the Rouse equation (Equation 7.70) used to estimate the discharge coefficient, the flow equation is given by 3

Q = Cw (b − 0.1nH)H 2 [ ] 3 2 √ = Cd 2g (b − 0.1nH)H 2 3 [ ( ) ] 3 2 H √ = 0.611 + 0.075 2g (b − 0.1nH)H 2 3 Hw Substituting the given parameters yields ) ] [ ( 3 H √ 2 3= 0.611 + 0.075 2g (2 − 0.2H)H 2 3 Hw

(1)

(2)

Also, since the maximum flow depth is 1.5 m, then H + Hw = 1.5 m

(3)

Solving Equations 2 and 3 gives H = 0.85 m Hw = 0.65 m These results were derived using the Rouse equation (Equation 7.70) to estimate the discharge coefficient, and the Rouse equation is strictly valid only when H/Hw < 5. In this case, H/Hw = 1.31 so the Rouse approximation is valid. However, the contraction formula given by Equation 1 is only valid when b > 3H, and in this case b = 2.35H and so the results do not validate using the contraction discharge formula given by Equation 1. Alternative estimates of Cd can be derived from either the Chaudhry equation (Equation 7.71) or the Swamee equation (Equation 7.72). The Chaudhry equation is for “sill-like” conditions which are unlikely to exist here, and so the general equation for estimating the discharge coefficient proposed by Swamee will also be considered. For a contracted rectangular weir with the Chaudhry equation (Equation 7.71) used to estimate the discharge coefficient, the flow equation is given by   [( )10 ( )15 ]−0.01 √ 3 14.14 H/H 2 w + 2g  (b − 0.1nH)H 2 (4) Q =  1.06 3 8.15 + H/Hw H/Hw + 1 Substituting the given parameters into Equation 4 and solving simultaneously with Equation 3 yields H = 0.66 m Hw = 0.84 m In this case b = 3.03H and so the results validate using the contraction discharge formula given by Equation 1.

216

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Analysis: It is apparent that the uncertainty in Cd (Rouse vs. Swamee) is such that the contraction discharge formula (Equation 1) might be valid in some cases. The results presented here collectively indicate that the required height of the weir is in the range of 0.65–0.84 m and that Cd estimated by the Chaudhry equation is higher than Cd estimated by the Rouse equation. To be conservative, the lower weir height should be selected along with the lower Cd (i.e., Rouse solution) so as to ensure that the total depth of 1.5 m is not exceeded. Therefore, the required weir can be taken as 0.65 m high , and 2 m wide . (b) When the weir is submerged, Villemonte’s formula gives [ ( y ) 3 ]0.385 Qs d 2 = 1− Q H where yd = (1.5 − 0.20) − 0.65 = 0.65 m Therefore,

[ ( ) 3 ]0.385 0.65 2 Qs = 1− 3 0.85

which yields Qs = 1.96 m3 /s Since the free-flow weir capacity is 3 m3 /s, then submergence reduces the weir capacity by 3 − 1.96 = 1.04 m3 /s . (c) For Qs /Q = 0.9, the submerged weir-flow equation gives [ ( y ) 3 ]0.385 2 d 0.9 = 1 − 0.85 which yields yd = 0.33 m The corresponding tailwater elevation is 0.65 m + 0.33 m = 0.98 m, and hence the difference between the headwater and tailwater elevations is 1.5 m − 0.98 m = 0.52 m. As long as the elevations of the headwater and tailwater differ by less than 52 cm the flow will deviate by less than 10% from the free-flow weir discharge. 7.48. The relative (percentage) difference, E, in using Equation 7.79 instead of Equation 7.78 to estimate Qs /Q is given by ] √ [( ] [[ ( yd ) 3 ]0.385 yd ) yd 2 1 + 2H 1− H − 1− H E= × 100% [ ( yd ) 3 ]0.385 2 1− H Which yields the following tabular relation:

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yd /H 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

E (%) 0.0 0.9 2.0 3.1 4.0 4.6 4.6 3.8 1.6

The maximum difference between Equations 7.78 and 7.79 is approximately 4.6% , which is within the uncertainty of Equations 7.78 and 7.79 in predicting the the submerged flow rates. On the basis of these results, neither equation is preferable . 7.49. From the given data: Q = 12 L/s = 0.012 m3 /s, and θ = 30◦ . The discharge equation requires that ( ) √ 5 8 θ Q = Cd 2g tan H2 15 2 and assuming that Cd = 0.58 gives √ 8 0.012 = (0.58) 2(9.81) tan 15

(

30◦ 2

)

5

H2

which simplifies to H = 0.25 m Since H > 50 mm, Cd = 0.58 is validated. 7.50. From the given data: H = 0.5 m, and θ = 50◦ . The discharge equation gives √ 8 Q = Cd 2g tan 15

( ) ( ◦) √ 5 5 θ 8 50 H 2 = (0.58) 2(9.81) tan (0.5) 2 = 0.11 m3 /s 2 15 2

7.51. From the given data: θ = 90◦ , and Q = 1 L/min = 1.667 × 10−5 m3 /s. The properties of water at 20◦ C are: σ = 0.0734 N/m, ρ = 998.2 kg/m3 , and µ = 0.001 N·s/m2 . Therefore 1

1

1

1

1 ρg 2 H 2 (998.2)(9.81) 2 H 2 Re = = = 3.13 × 106 H 2 µ 0.001 2 ρgH (998.2)(9.81)H 2 We = = = 1.33 × 105 H 2 σ 0.0734 1.19 1.19 0.0138 Cd = 0.583 + 1 = 0.583 + 1 1 = 0.583 + 5 (ReWe) 6 [(3.13 × 106 H 2 )(1.33 × 105 H 2 )] 6 H 12

218

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Substituting into the discharge equation gives ( ) √ 5 8 θ Q = Cd 2g tan H2 15 2 ( ) ( ◦) √ 5 8 0.0138 90 −5 1.67 × 10 = 0.583 + 2(9.81) tan H2 5 15 2 H 12 which gives H = 0.010 m 7.52. From the given data: H = 40 mm, and θ = 70◦ , therefore ( ) √ 5 8 θ Q = Cd 2g tan H2 15 2

(1)

Since H < 50 mm, Cd = 0.583 + 1

1.19 1

(ReWe) 6

1

1

1

(9.81) 2 (0.04) 2 g2H 2 = = 6.26 × 105 Re = ν 1.00 × 10−6 ρgH 2 (998.2)(9.81)(0.04)2 We = = = 215.2 σ 72.8 × 10−3 where a temperature of 20◦ C has been assumed. Substituting into Equation 1 gives ] [ ( ◦) √ 5 8 70 1.19 Q= 2(9.81) tan (0.04) 2 0.583 + 1 15 2 (6.26 × 105 × 215.2) 6 = 3.36 × 10−4 m3 /s = 0.336 L/s 7.53. From the given data: Q = 0.30 m3 /s, H + Hw = 1 m, and T = 1 m. By geometry, ( ) θ T = 2H tan 2

(1)

and the discharge relation is given by √ 8 Q = Cd 2g tan 15

( ) 5 θ H2 2

Combining Equations 1 and 2, and taking Cd = 0.58 gives ( ) √ 3 T 8 H2 Q = (0.58) 2(9.81) 15 2 ( ) √ 3 1 8 0.30 = (0.58) 2(9.81) H2 15 2

219

(2)

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which gives H = 0.577 m ( ) ( ) T 1 −1 −1 θ = 2 tan = 2 tan = 82◦ 2H 2 × 0.577 Hw = 1 − H = 1 − 0.577 = 0.423 m 7.54. (a) The Kindsvater and Shen equation can be expressed in the form ] [ ( ) 5 √ 2 5 8 θ ′ (H + k) Q= Cd 2g tan H2 5 15 2 H2 or

[ ( )5 ] ( ) 5 k 2 √ θ 8 ′ Cd 1 + 2g tan H2 Q= 15 H 2

Using the Kindsvater and Shen expression for Cd′ and k yields ( ) 4.42 − 0.1035θ + 1.005 × 10−3 θ2 − 3.24 × 10−6 θ3 Cd = (0.6072 − 0.000874θ + 6.1 × 10−6 θ2 ) 1 + 1000H

where θ is in degrees and H is in meters. (b) For θ = 90◦ , the Kindsvater and Shen parameters are Cd′ = 0.578

and k = 0.8835 mm

and so the Kindsvater and Shen equation can be put in the form ( ) 5 θ (H + k) 2 2 ( π ) ( H + 0.8835 ) 52 √ 8 = (0.5780) 2(9.81) tan 15 4 1000

√ 8 Q = Cd′ 2g tan 15

where H is in millimeters. The above equation simplifies to 2

H = 882.9Q 5 − 0.8835

(1)

To release 20,000 m3 in 24 hours requires an average flow rate of 20000/(24 × 60 × 60) = 0.2315 m3 /s. Using this flow rate in Equation 1 yields H = 490 mm = 0.49 m Therefore using a weir height of 0.49 m will release the stored water from the detention pond in no less than 24 hours.

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7.55. From the given data: θ = 90◦ , H2 = 20 cm = 0.20 m, and H1 = 1.20 − H2 = 1.20 − 0.20 = 1.00 m. The discharge, Q, over the weir is given by Equation 7.92 as √ ( ) 3 2 √ H2 2 Q = Cd 2gbH1 + Cvd A 2g H1 + 3 3

(1)

The discharge coefficient, Cd , is given by Equation 7.70 as Cd = 0.611 + 0.075

H1 H2

(2)

which yields Cd = 0.611 + 0.075

1.00 = 0.986 0.20

Since H1 /H2 = 5, Equation 2 is valid, and Cd can be taken as 0.986 in the weir design. The area, A, of the V-notch is given by ( ) ( ◦) θ 90 2 2 A = H2 tan = 0.20 tan = 0.04 m2 2 2 Taking Cvd = 0.6 and Q = 1.5 m3 /s, substituting into Equation 1 gives √ ( ) √ 3 0.20 2 1.5 = (0.986) 2(9.81)b(1.00) 2 + 0.60(0.04) 2(9.81) 1.00 + 3 3 which yields b = 0.48 m Therefore, a compound weir with a crest length of 48 cm will yield a discharge of 3 m3 /s when the water depth in the detention area is 1 m. 7.56. From the given data: Hw = 0.25 m, b = 1.5 m, y = 0.75 m, and the height of water, h1 , above the weir is h1 = y − Hw = 0.75 − 0.25 = 0.50 m. Q2 Q2 Q2 = h + = 0.50 + = 0.50 + 0.0403Q2 1 2gA2 2g(by)2 2(9.81)(1.5 × 0.75)2 0.65 0.65 0.65 Cd = ( 1 = 1 )1 = ( ) 2) 2 2 (3 + 0.161Q 0.50+0.0403Q2 2 H 1 + Hw 1+ 0.25 ]3 [ ]3 [ 2 2 1 1 2 0.65 2 2 2 Q = Cd bg 2 H = (0.50 + 0.0403Q ) 1 (1.5)(9.81) 3 3 (3 + 0.161Q2 ) 2 H = h1 +

which can be solved iteratively for Q to give Q = 0.343 m3 /s

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7.57. From the given data: L = 1 m, b = 1 m, and Hw = 0.3 m. The valid operating range is 0.08 < h1 /L < 0.33, so 0.08 m < h1 < 0.33 m. Q2 V2 = h1 + 2g 2gb2 (h1 + Hw )2 Q2 Hmin = 0.08 + = 0.08 + 0.353Q2 2(9.81)(1)2 (0.08 + 0.3)2 Q2 Hmax = 0.33 + = 0.33 + 0.0796Q2 2(9.81)(1)2 (0.5 + 0.3)2 0.65 Cd = ( )1 2 1 + HHw H = h1 +

Cd,min = ( 1+ Cd,max = ( 1+ Q = Cd bg Qmin =

0.65 0.08+0.353Q 0.3

)1 = 2

0.65 0.33+0.0796Q2

1 2

[

2

0.65

)1 = 2

1

(1.267 + 1.177Q2 ) 2 0.65 1

(2.10 + 0.265Q2 ) 2

0.3

2 H 3

]3 2

0.65 (1.267 + 1.177Q2min )

1 2

(1)(9.81)

1 2

[

2 (0.08 + 0.353Q2min ) 3

]3

2

which yields Qmin = 0.0223 m3 /s and Qmax =

0.65 1

(2.10 + 0.265Q2max ) 2

(1)(9.81)

1 2

[

]3 2 2 2 (0.33 + 0.0796Qmax ) 3

which yields Qmax = 0.146 m3 /s Therefore, the operating range for the weir is given by 0.0223 m3 /s < Q < 0.146 m3 /s 7.58. From the given data: b = 3 m, h1 + Hw = 4 m, Q = 5 m3 /s, and V12 (Q/A)2 [5/(3 × 4)]2 = = = 0.00885 m 2g 2g 2(9.81) which gives H = h1 +

V12 = h1 + 0.00885 2g

222

(1)

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The discharge rate is given by

(

√

Q = Cd gb

2 H 3

)3 2

(2)

and 0.65

Cd =

(3)

1

(1 + H/Hw ) 2

Combining Equations 1 to 3 and substituting the given data yields √

0.65

Q=

(1 + H/Hw )

1 2

0.65

5= [ 1+

h1 +0.00885 4−h1

( gb

]1

√

2

2 H 3

)3

2

[

2 9.81(3) (h1 + 0.00885) 3

]3

2

which gives h1 = 1.53 m

or

3.84 m

Hw = 4 − h1 = 0.16 m

or

2.47 m

This indicates two possible flow conditions:

Condition 1 2

h1 (m) 1.53 3.84

H (m) 1.54 3.85

Hw (m) 2.47 0.16

H/Hw

Cd

0.62 24.06

0.51 0.13

√

( )3 gb 32 H 2 (m3 /s) 9.77 38.64

Only Condition 1 gives a Cd that is typical of properly-operated broad-crested weirs (typically 0.527 < Cd < 0.540), so take Hw = 2.47 m . For proper operation of the weir, take h1 /L = 0.25, so L=

h1 1.53 = = 6.12 m 0.25 0.25

Therefore the weir should be 2.47 m high and 6.12 m long. 7.59. From the given data: b = 3 m, Q = 1.5 m3 /s, y = 2 m, and therefore h 1 = 2 − Hw

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and Q2 1.52 = (2 − H ) + = 2.003 − Hw w 2gb2 (h1 + Hw )2 2(9.81)(3)2 (2 − Hw + Hw )2 1 0.65 0.65 2 Cd = ( )1 = ( ) 1 = 0.459Hw 2 2 w 1 + HHw 1 + 2.003−H Hw ( )3 2 1 2 2 Q = Cd bg H 3 [ ]3 1 2 1 2 2 1.5 = 0.459Hw (3)(9.81) 2 (2.003 − Hw ) 3 H = h1 +

which gives Hw = 0.056 m, 1.33 m To achieve critical flow conditions over the weir use Hw = 1.33 m , which corresponds to h1 = 2 − Hw = 2 − 1.33 = 0.67 m The weir length, L, is then given by the relation 0.08 < h1 /L < 0.33 12.5 > L/h1 > 3.03 12.5h1 >

L

> 3.03h1

12.5(0.67) >

L

> 3.03(0.67)

8.38 m >

L

> 2.03 m

Therefore a weir length in the range 2.0 m < L < 8.4 m would be satisfactory. 7.60. From the given data: H = 11.70 m − 11.20 m = 0.50 m, Q = 0.75 m3 /s, and Hw = 11.20 − 9.10 m = 2.10 m. Using the Chow (1959) equation, Cd =

0.65 (1 + H/Hw )

1 2

0.65

=

1

= 0.584

(1 + 0.50/2.10) 2

Applying the weir equation, (

√

Q = Cd gb √

2 H 3

0.75 = 0.584 9.81b

)3

2

(

)3 2 2 0.50 3

which yields b = 2.13 m. For broad-crested weirs to function properly requires that 0.08 < h1 /L < 0.33. Using the Swamee (1988) equation with h1 /L = 0.20 yields [

(0.20)5 + 1500(0.20)1 3 Cd = 0.5 + 0.1 1 + 1000(0.20)3

224

]0.1 = 0.536

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Applying the weir equation, (

√

0.75 = 0.536 9.81b

)3 2 2 0.50 3

which gives b = 2.32 m. Therefore, the Swamee (1988) equation gives the more conservative design with b = 2.32 m and L = h1 /0.20 = 0.50/0.20 = 2.50 m . 7.61. When the depth is 0.50 m: Q = 0.08 m3 /s, h1 = 0.25 m, and Hw = 0.25 m. The discharge is given by [ ]3 2 2 ′√ (1) Q = Cd gb h1 3 where Cd′

(

) h1 = 0.95 + 0.89 − 0.38 h 1 + Hw ( )2 ( ) 0.25 0.25 = 0.95 + 0.89 = 0.938 − 0.38 0.25 + 0.25 0.25 + 0.25 h1 h1 + Hw

)2

(

Substituting known quantities into Equation 1 yields Q=

√ Cd′ gb

[

2 h1 3

]3

√

2

[

2 0.08 = (0.938) 9.81b (0.25) 3

]3

2

which gives b = 0.400 m . When the depth is 0.75 m: Q = 0.50 m3 /s. The total flow can be taken as the flow in the center section, Qc , plus the flow in the side sections, Qs . Take the side sections to each have the same width, bs . For flow in the center section: h1c = 0.50 m, Hwc = 0.25 m, and the discharge coefficient, Cdc , is given by ) h1c − 0.38 + 0.89 = 0.95 h1c + Hwc ( ( )2 ) 0.50 0.50 = 0.95 − 0.38 + 0.89 = 1.06 0.50 + 0.25 0.50 + 0.25 (

Cdc

h1c h1c + Hwc

)2

(

For flow in the side section: h1s = 0.25 m, Hws = 0.50 m, and the discharge coefficient, Cds , is given by (

Cds

) h1s = 0.95 − 0.38 + 0.89 h1s + Hws ( )2 ( ) 0.25 0.25 = 0.95 + 0.89 = 0.869 − 0.38 0.25 + 0.50 0.25 + 0.50 h1s h1s + Hws

)2

(

225

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The total flow, Q, is therefore given by Q = Cdc

√

[

2 gb h1c 3

]3 2

√

+ 2Cds g

(

B−b 2

)[

2 h1s 3

]3

2

[ ]3 ( )[ ]3 √ √ 2 2 2 B − 0.400 2 0.50 = (1.06) 9.81(0.400) (0.50) + 2(0.869) 9.81 (0.25) 3 2 3 which yields B = 1.720 m . For proper operation of the weir it is required that: h1 > 0.08 → L < L h1 < 0.33 → L > L

h1 →L< 0.08 h1 →L> 0.33

0.25 → L < 3.125 m 0.08 0.50 → L > 1.515 m 0.33

Based on these results, it would be appropriate to take L in the range of 1.52 m < L < 3.13 m . 7.62. The surface of the spillway in the quadrant downstream of the crest is given by Equation 7.116 as ( )n y 1 x = Hd K Hd The slope of the spillway is given by dy/dx where ( )n−1 ( )n−1 dy Hd x 1 n x = n = dx K Hd Hd K Hd If x = XDT when dy/dx = α (= downstream slope) then ( )n−1 n x α= K Hd which yields

Taking n = 1.85,

[ ] 1 n−1 XDT 1 = (Kα) Hd n [ ] 1 1.85−1 XDT 1 = (Kα) Hd 1.85

which simplifies to XDT = 0.485(Kα)1.176 Hd 7.63. Assuming that the spillway is sufficiently high that the ratio of the spillway height, P , to the design head, Hd , is greater than 3, Figure 7.28(a) gives the basic discharge coefficient as C0 = 2.18. Assuming that the effective head, He , is less than 9.1 m, then for the limiting case where the water pressure on the spillway is equal to −4.6 m, Equation 7.115 requires that He 1 = = 1.43 Hd 0.7

226

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and Figure 7.28(b) gives Cincl /Cvert = 0.99, and Figure 7.28(c) gives C/C0 = 1.05. The discharge coefficient, C, under maximum headwater conditions can now be calculated using the relation ( )( ) Cincl C C= C0 = (0.99)(1.05)(2.18) = 2.27 Cvert C0 From the given data, Q = 1600 m3 /s, Le = 40 m, and therefore the effective head, He , on the spillway is given by Equation 7.114 as [

Q He = CLe

]2 3

[

1600 = (2.27)(40)

]2

3

= 6.78 m

Since this calculated value of He is less than 9.1 m, the initial assumption that He < 9.1 m used in estimating the discharge coefficient is validated. The design head, Hd , corresponding to He = 6.78 m is given by Equation 7.115 as Hd = 0.7He = 0.7(6.78) = 4.75 m The depth of water, d, upstream of the spillway is given by d = 200 m − 170 m = 30 m and the approach velocity, vo , can be estimated by v0 =

Q 1600 = = 1.33 m/s Le d (40)(30)

with a velocity head, hv , given by hv =

1.332 v02 = = 0.09 m 2g 2(9.81)

Therefore, at the maximum pool elevation, the height of water above the spillway is He − hv = 6.78 m − 0.09 m = 6.69 m, and the required crest elevation of the spillway is 200 m − 6.69 m = 193.31 m . Having determined the required crest elevation of the spillway (= 193.31 m) to pass the design flow rate (= 1600 m3 /s), the next step is to determine the required shape of the spillway in the vicinity of the crest. Since the maximum pool elevation is 200 m and the bottom elevation behind the spillway is 170 m, the height of the spillway crest, P , is given by P = 200 m − 170 m − 6.69 m = 23.31 m and

P 23.31 = = 4.91 Hd 4.75

which indicates a “high spillway” with negligible approach velocity. Figure 7.27(a) gives K = 2.0 for the spillway coordinate coefficient, and the profile of the downstream quadrant of

227

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the spillway, taking n = 1.85, is given by ( )n y 1 x = Hd K Hd y 1 ( x )1.85 = 4.75 2.0 4.75 which simplifies to y = 0.133x1.85 Since the downstream slope of the spillway is 1:1.5 (H:V), then α = 1.5 and the horizontal distance from the apex to the downstream tangent point, XDT , is given by Equation 7.117 as XDT = 0.485(Kα)1.176 Hd XDT = 0.485(2.0 × 1.5)1.176 4.75 which gives XDT = 8.39 m Therefore, at a distance of 8.39 m downstream of the crest, the curved spillway profile merges into the linear profile of the spillway chute, which has a slope of 1:1.5 (H:V). The correspond1.85 = 0.133(8.39)1.85 = 6.80 m. ing y value is y = 0.133XDT Taking P/Hd = 4.91 in Figure 7.27(b) and (c) gives A/Hd = 0.28 and B/Hd = 0.165, which yields A = 0.28(4.75) = 1.33 m, B = 0.165(4.75) = 0.784 m, and the profile of the upstream quadrant of the spillway is given by Equation 7.118 as x2 (B − y)2 + =1 2 A B2 x2 (0.784 − y)2 + =1 1.332 0.7842 which simplifies to x2 + 2.878(0.784 − y)2 = 1.769

(1)

Since the upstream face of the spillway has a 1:1 slope, then Fs = 1 and the horizontal distance from the apex to the upstream tangent point is given by XU T =

A2 Fs [A2 Fs2 + B 2 ]

1 2

=

(1.33)2 (1) 1

[(1.33)2 (1)2 + (0.784)2 ] 2

= 1.146 m

Substituting x =1.146 m into Equation 1 gives two solutions: y = 0.386 m and y = 1.182 m. Referring to Figure 7.2, it is apparent why we have two solutions (y1 and y2 ), and that the value of y to be selected is the lower value of y = 0.386 m. Therefore, the shape of the spillway upstream of the crest is given by Equation 1 up to the point (1.146 m, 0.386 m) where it merges with the planar upstream face.

228

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Spillway Surface (0,0) (XUT,y1)

x

B (0,B)

A

(XUT,y2)

x2/A2 + (B-y)2/B2 = 1 y

Figure 7.2: Upstream Spillway Shape [As an aside, Equation 1 can be differentiated to give dy x = dx 2.878(0.784 − y)

(2)

At (−1.146 m, 0.386 m) Equation 2 gives dy/dy = 1, and at (−1.146 m, 1.182 m) Equation 2 gives dy/dy = −1. Since dy/dy = 1 at the upstream face, our result is confirmed.] 7.64. The discharge over the spillway is given by Equation 7.120 as 3

Q = CLe He2

(1)

Let X be the required crest elevation, then taking the pool elevation as 200 m, the effective head, He , is given by He = 200 − X (2) Combining Equations 1 and 2 yields 3

Q = CLe (200 − X) 2

(3)

The effective length of the spillway, Le , is given by Equation 7.121 as Le = L − wN − 2(N Kp + Ka )He

(4)

From the given data: w = 1 m, N = 2, for round-nosed piers Kp = 0.01, and for square abutments Ka = 0.2. Substituting these values, along with Equation 2 into Equation 4 yields Le = L − (1)(2) − 2(2 × 0.01 + 0.2)(200 − X) = L + 0.44X − 90

229

(5)

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Combining Equations 3 and 5 and putting Q = 500 m3 /s gives the following relationship between the crest length and the crest elevation, 3

500 = C(L + 0.44X − 90)(200 − X) 2 which can be put in the convenient form L=

500 3

C(200 − X) 2

− 0.44X + 90

(6)

Determination of the discharge coefficient requires specification of the crest height, P , which is given by P = X − 150 (7) where the bottom elevation behind the spillway is taken as 150 m. Also needed is the design head, Hd , given by Equation 7.114 for He > 9.1 m, and Equation 7.115 for He < 9.1 m. Using the above-derived equations, the relationship between the crest elevation, X, and the spillway length, L, is given in the following table for selected crest elevations between 195 m and 180 m. X (m) 195 190 185 180

P (m) 45 40 35 30

He (m) 5 10 15 20

Hd (m) 3.7 7.5 12.4 17.9

P/Hd

He /Hd

C0

Ci /Cv

C/C0

C

12.1 5.3 2.8 1.7

1.4 1.3 1.2 1.1

2.18 2.18 2.18 2.18

0.99 0.99 0.99 0.99

1.05 1.04 1.02 1.00

2.27 2.24 2.20 2.16

L (m) 23.9 13.5 12.5 13.4

7.65. From the given data: He = 0.50 m, Q = 0.75 m3 /s, Ka = 0 (abutments will be well-rounded), and P = 2.10 m. The design head, Hd , is given by Hd = 0.7He = 0.7(0.50) = 0.35 m and hence P/Hd = 2.10/0.35 = 6.00, and He /Hd = 0.50/0.35 = 1.43. The discharge coefficient is determined as: C0 = 2.18, C/C0 = 1.06 (using a vertical upstream face) and hence C = C0 ×

C = (2.18)(1.06) = 2.31 C0

Using the spillway equation, 3

Q = CLe He2 3

0.75 = (2.31)Le (0.50) 2 which yields Le = 0.918 m . Therefore, by using a spillway instead of a broad-crested weir a smaller crest length is required, hence a smaller “footprint” associated with the structure.

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For P/Hd = 6.00, K = 2.00, A/Hd = 0.282, B/Hd = 0.167, and n can be taken as 1,85. Therefore, A = 0.282(0.35) = 0.0987 m, B = 0.167(0.35) = 0.0585 m, and the elevation of the upstream spillway surface is given by ( )n y 1 x = Hd K Hd 1 ( x )1.85 y = 0.35 2.00 0.35 which yields y = 1.22x1.85 For the downstream surface of the spillway, x2 (B − y)2 + =1 A2 B2 x2 (0.0585 − y)2 + =1 0.09872 0.05852 which yields x2 (0.0585 − y)2 + = 10−3 9.74 3.42 7.66. From the given data: Le = 5.0 m, P = 4.2 m, Hd = 2.0 m, and hd = 2.0 m − 0.7 m = 1.3 m. From these data, P/Hd = 4.2/2.0 = 2.1, and Figure 7.28(a) gives the basic discharge coefficient, C0 , as 2.17. Since the upstream face of the spillway is vertical, and assuming the upstream velocity head is negligible, then He ≈ Hd and Figure 7.28(b) and (c) indicate that the free-flow discharge coefficient, C, is equal to C0 , and hence C = 2.17. The free-flow discharge, Q0 , is given by Equation 7.120 as [

Q20 Q0 = CLe He = (2.17)(5.0) 2.0 + 2gA2 3 2

] 32

[

Q20 = (2.17)(5.0) 2.0 + 2(9.81)(6.2 × 5)2

] 32

which yields Q0 = 31.9 m3 /s. This flow corresponds to an upstream velocity, va , and velocity head, va2 /2g, given by va =

Q0 31.9 v2 = = 1.03 m/s → a = 0.05 m A 6.2 × 5 2g

Therefore He = 2.0 m + 0.05 m = 2.05 m, and hd 1.3 = = 0.63 He 2.05 hd + d 1.3 + 4.8 = = 2.98 He 2.05 Using these parameter values in Figure 7.30 gives a discharge-coefficient reduction of approximately 0.75% caused by submergence. Hence, the adjusted discharge coefficient is C = (1 − 0.0075)(2.17) = 2.15

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The corresponding discharge, Q, under the given submerged condition is [

Q2 Q = CLe He = (2.15)(5.0) 2.0 + 2gA2 3 2

] 32

[

Q2 = (2.15)(5.0) 2.0 + 2(9.81)(6.2 × 5)2

] 32

which yields Q = 31.6 m3 /s. This flow corresponds to an upstream velocity, va , and velocity head, va2 /2g, given by va =

Q 31.6 v2 = = 1.02 m/s → a = 0.05 m A 6.2 × 5 2g

Since this approach velocity is approximately the same as calculated for the free-flow condition, no further iteration is necessary. The submerged discharge is 31.6 m3 /s . 7.67. From the given data: z0 = 0.50 m, z1 = 2.5 m, L = 10 m, and Hd = 0.70 m. The maximum allowable head on the spillway, He , is given by He =

Hd 0.7 = = 1.0 m 0.7 0.7

and

2.5 − 2 P = = 2.85 Hd 0.7

(a) From Figure 7.28, the discharge coefficient is given by C0 = 2.18. Under design conditions, H = 0.7 m, and the spillway discharge, Q, is given by 3

3

Q = C0 LH 2 = 2.18(10)(0.7) 2 = 12.8 m3 /s (b) Under maximum-flow conditions, H = He = 1.0 m, which corresponds to a maximum stage upstream of the spillway of 2.5 m + 1.0 m = 3.5 m , and using Figure 7.28 gives the corresponding discharge coefficient: H 1.0 C = = 1.43 → = 1.05 → C = (1.05)(2.18) = 2.29 Hd 0.7 C0 and hence the discharge capacity of the spillway under this condition is given by 3

3

Q = CLHe2 = (2.29)(10)(1.0) 2 = 22.9 m3 /s (c) If the maximum stage is exceeded, then the pressure on the downstream side of the spillway will likely lead to cavitation and possible pitting of the concrete structure. 7.68. For free discharge (when the gate is completely out of the water) the discharge, Q0 , is given by the relation 3 Q0 = Cd bH 2 (1) where Cd is the discharge coefficient, b is the width of the spillway, and H is the height of the water above the spillway crest. Under critical conditions (when the gate is above the water), the critical flow depth, yc , is calculated using the relation Q20 A3 Q2 = → 0 = (byc )3 b = b2 yc3 → yc = g T g

232

[(

Q20 b2

) ] 13 1 g

(2)

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When the gate is lowered such that the gate opening, yg , is less than yc , then the discharge is given by √ (3) Q = Cd byg 2gH From the given data, Cd = 2.2, b = 10 m, and H = 2 m, Equations 1 and 2 give 3

Q0 = (2.2)(10)(2) 2 = 62.2 m3 /s [( ) ]1 62.22 1 3 yc = = 1.58 m 102 9.81 Therefore, when the gate opening, yg , is greater than 1.58 m, the discharge is equal to 62.2 m3 /s. When the gate opening is less than yc , we have free orifice flow. For free orifice flow, take Cc = 0.61, which gives Cd = √

Cc 1+

y Cc Hg

0.61 0.61 =√ =√ yg 1 + 0.305yg 1 + 0.61 2

and hence Equation 3 gives Q= √

√ 38.2yg 0.61 (10)yg 2(9.81)(2) = √ m3 /s 1 + 0.305yg 1 + 0.305

(4)

Using Equation 4 gives the following results: Gate Opening, yg (m) > 1.58 1.58 1.40 1.20 1.00 0.80 0.60 0.40 0.20 0.00

Flow, Q (m3 /s) 62.2 49.6 44.8 39.2 33.4 27.4 21.1 14.4 7.4 0.0

7.69. From the given data: H = 2.00 m, ht = 0.70 m, h = 0.40 m, and L = 5 m. These data indicate that the flow through the gate is likely to be controlled-submerged flow. Assuming that C2 = 0.71, Equation 7.124 gives √ √ Q = C2 hL 2g(H − ht ) = (0.71)(0.40)(5) 2(9.81)(2.00 − 0.70) = 7.17 m3 /s Validation that the gate is both controlled and submerged can be determined by comparing the gate opening and tailwater depth to the critical dept, yc , given by ]1 [ 2 ] 13 [ (7.17)2 3 Q = = 0.59 m yc = gL2 (9.81)(5)2

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Since h < yc (0.40 m < 0.59 m) the flow is controlled, and since Hd > yc (0.70 m > 0.59 m) the flow is submerged. Therefore the assumed discharge equation is confirmed, and the estimated discharge is 7.17 m3 /s . 7.70. From the given data: X = 2.00 m, Hd = 7.00 m, a = 2.10 m, h = 0.50 m, R = 2.75 m, and L = 5.00 m. The gate angle, θ, is given by Equation 7.127 as ( ) ( ) −1 a − h −1 2.10 − 0.50 θ = cos = cos = 54◦ R 2.75 From the given data, X/Hd = 2.00/7.00 = 0.29 and θ = 54◦ , and Figure 7.34 gives the discharge coefficient as Cd = 0.67. When the ponded water is 1.00 m above the crest, the head on the gate opening, H, can be approximated by H = 0.87 m + 1.00 m −

h 0.50 cos(54◦ ) = 0.87 m + 1.00 m − cos(54◦ ) = 1.72 m 2 2

and the corresponding flow rate under gate is given by Equation 7.126 as √ √ Q = Cd Lh 2gH = (0.67)(5.00)(0.50) 2(9.81)(1.72) = 9.73 m3 /s This discharge is approximately 58% of the flow of 16.9 m3 /s when the gate opening is 1.00 m. 7.71. From the given data: Q0 = 10 m3 /s, H1 = 0.90 m, and h = 1.00 m. Under the given operating conditions the reservoir head on the gate seat, H2 , is H2 = H1 + h = 0.90 m + 1.00 m = 1.90 m and the head on the crest of the spillway is 1.00 m + 0.90 m − 0.25 m = 1.65 m. Assuming that the pool elevation remains as shown in Figure 7.37 when the gate is fully open, then H0 = 1.65 m. Substituting the given and derived parameters into Equation 7.128 gives 3

3

Q H2 −H2 = 2 3 1 Q0 H02 3

3

Q (1.90) 2 − (0.90) 2 = 3 10 (1.65) 2 which yields Q = 8.3 m3 /s . This discharge is approximately 84% higher than the flow of 4.5 m3 /s when the gate opening is 0.50 m. 7.72. From the given data: b = 10 m, and Q = 120 m3 /s. Since the water loses 15% of its energy flowing down the spillway, the elevation behind the spillway is 100 m, and the surface elevation of the downstream water is 80.00 m, then taking the elevation of the bottom of the stilling basin as X; the specific energy, E1 , at entrance to the stilling basin (i.e. base of the spillway) is given by E1 = (100 − X) − 0.15(100 − 80) = 97 − X

234

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Taking y1 as the depth of flow at the entrance to the stilling basin, Q2 2g(by1 )2 1202 97 − X = y1 + 2(9.81)(10y1 )2 E 1 = y1 +

which simplifies to X = 97 − y1 −

7.34 y12

(1)

Since the elevation of the water surface downstream of the stilling basin is 80.00 m, assume that the conjugate depth of the hydraulic jump, y2 , is given by y2 = 80 − X

(2)

and the relationship between y1 and y2 is given by the hydraulic jump equation √ ( ) y1 8q 2 y2 = −1 + 1 + 3 2 gy1 √ ( ) y1 8(120/10)2 y2 = −1 + 1 + 2 9.81y13 √ ) ( y1 117.4 y2 = −1 + 1 + 2 y13 Combining Equations 1 to 3 yields 7.34 y1 80 − 97 + y1 + 2 = 2 y1 which simplifies to 7.34 y1 1.5y1 + 2 − 2 y1

√

( −1 +

√ 1+

117.4 1+ y13

(3)

)

117.4 − 17 = 0 y13

The solutions to this equation are y1 = 17.1 m and y1 = 0.560 m. These depths correspond to subcritical and supercritical flows respectively. Since the upstream depth of flow must be supercritical, take y1 = 0.560 m (Fr = 9.1) and Equation 1 gives X = 97 − 0.560 −

7.34 = 73.0 m 0.5602

Therefore the stilling basin must have an invert elevation of 73.0 m for the hydraulic jump to occur within the stilling basin. Since the Froude number of the flow entering the stilling basin is 9.1 and the entrance velocity is 120/(0.560×10) = 21.4 m/s, then a Type II basin is required. This basin requires that the tailwater depth be 5% greater that the conjugate depth so 1.05y2 = 80 − X → y2 = 76.2 − 0.952X

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Repeating the previous calculations gives X = 72.7 m , y1 = 0.556 m, Fr = 9.2, v1 = 21.6 m/s, and y2 = 6.99 m. These results confirm that a Type II stilling basin is required. The dimensions of the chute blocks and dentated sill in the Type II stilling basin are given in Figure 7.41 in terms of the entering depth (y1 ) of 0.556 m and sequent depth (y2 ) of 6.99 m. The variables y1 and y2 used here correspond to the variables d1 and d2 used in Figure 7.41, hence d1 = 0.556 m and d2 = 6.99 m. The length, L, of the stilling basin is derived from Figure 7.41(c), where Fr = 9.2, L/d2 = 4.3, and hence L = 4.3(6.99) = 30.1 m . 7.73. The first step is to calculate the flow rate over the spillway under maximum-pool conditions. This is given by Equation 7.120 as 3

Q = CLe He2

(1)

When the pool elevation behind the spillway is 6.00 m, the effective head, He , is given by He = 6.00 m − 5.00 m = 1.00 m and Equation 7.115 gives the design head, Hd , (without piers) Hd = 0.7He = 0.7(1.00) = 0.70 m The effective length of the spillway crest, Le , is given by Equation 7.121 as Le = L − wN − 2(N Kp + Ka )He

(2)

Since there are no piers, N = 0, Ka = 0 (well rounded abutments), and L = 10 m, therefore Equation 2 gives Le = L = 10 m The height of the spillway crest, P , is 5 m, hence P 5 = =5 Hd 1 and

He 1 = = 1.43 Hd 0.7

Using these values in Figure 7.28 gives C0 = 3.95, C/C0 = 1.05, and hence ( ) C C= (C0 ) = (1.05)(3.95) = 4.15 C0 Converting from U.S. Customary units to SI units yields C = 0.552(4.15) = 2.29 Substituting the calculated parameters into Equation 1 yields 3

3

Q = CLe He2 = (2.29)(10)(1.00) 2 = 22.9 m3 /s Hence the flow rate down the spillway is 22.9 m3 /s.

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Neglecting energy loss down the spillway, at the entrance to the stilling basin the energy equation gives Q2 2gA2 22.92 6 = y2 + 2(9.81)(10y2 )2

E1 = y2 +

which yields y2 = 5.99 m

or

0.21 m

Since the flow at the base of the spillway will be supercritical, y2 = 0.21 m, and the flow velocity, V , and the Froude number, Fr, are given by 22.9 Q = = 10.9 m/s by (10)(0.21) 10.9 V Fr = √ = √ = 7.6 gy (9.81)(0.21) V =

Since Fr > 4.5 and V < 18 m/s, use a Type III stilling basin . 7.74. From the given data: Q = 100 m3 /s, wg = 3 m, w = 1 m, N = 3, ztr = 80.00 m, zbr = 67.00 m, and ztw = 70.00 m. The flow, Q, over the spillway crest is given by 3

Q = CLe He2

(1)

The equivalent length of the spillway crest, Le , is given by Le = L − wN − 2(N Kp + Ka )He Specify pointed-nose piers and rounded abutments, so Kp =0 and Ka = 0. Three gates will require two piers, so the spillway length, L, is given by L = 3 × gate width + 2 × pier width = 3(3) + 2(1) = 11 m The equivalent crest length is therefore given by Le = 11 m − (2)(1 m) − 0 = 9 m Assuming P/Hd > 3 and He < 9.1 m, then the discharge coefficient is C = 2.18. With piers, He /Hd = 1/0.74 = 1.35 which gives C/C0 = 1.04 and hence C = (1.04)(2.18) = 2.27. Substituting known quantities into Equation 1 gives 3

100 = (2.27)(9)He2 which yields He = 2.88 m. This validates the assumption that He < 9.1 m, and also gives Hd = 0.74He = 2.13 m. Estimate the velocity head upstream of the reservoir as follows: d = 80.00 m − 67.00 m = 13.00 m Q 100 v0 = = = 0.855 m/s Le d (9)(13) v02 0.8552 = = 0.04 m 2g 2(9.81)

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If h is the water level above the crest, then v02 = He 2g h + 0.04 = 2.88 h+

which yields h = 2.84 m and elevation of spillway crest = 80.00 m − 2.84 m = 77.16 m height of spillway, P = 77.16 m − 67.00 m = 10.16 m Since P/Hd = 10.17/2.13 = 4.77, the initial assumption that P/Hd > 3 is validated. The width of the spillway is the same as the crest length of the spillway and is therefore equal to 11 m. Since there is a 10% energy loss down the spillway, the specific energy at the entrance to the stilling basin, E1 , is estimated by (neglecting velocity heads) E1 = (80 − X) − 0.1(80 − 70) = 79 − X which requires that 79 − X = y1 +

1002 2(9.81)(11y1 )2

which simplifies to X = 79 − y1 −

4.21 y12

(2)

The hydraulic jump equation requires that √ ( ) y2 8(100/11)2 y2 = −1 + 1 + 2 (9.81)y13 which simplifies to y2 y2 = 2

√

( −1 +

67.40 1+ y13

) (3)

Consistency of y2 with the given tailwater elevation requires that y2 = 70 − X

(4)

Combining Equations 2 to 4 yields y1 = 0.555 m X = 64.76 m y2 = 5.24 m v1 = 16.4 m/s Fr1 = 7.02 Therefore the invert elevation of the stilling basin is 64.76 m . Since Fr1 > 4.5 and v1 < 18 m/s, a Type III stilling basin is required.

238

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7.75. From the given data, the average flow rate into the reservoir is 2500 m3 /s = 7.88 × 1010 m3 /year, and the average suspended-sediment concentration is 250 mg/L = 0.250 kg/m3 , therefore the average sediment load entering the reservoir is given by sediment load = inflow rate × suspended-sediment concentration = 7.83 × 1010

m3 kg × 0.250 3 = 1.96 × 1010 kg/year year m

The area of the reservoir is 850 km2 = 8.5 × 108 m2 , and the average depth of the reservoir is 18.7 m, therefore the reservoir storage capacity is given by storage capacity = area of reservoir × average depth = 8.5 × 108 m2 × 18.7 m = 1.59 × 1010 m3 and storage capacity 1.59 × 1010 m3 = = 0.20 year annual inflow 7.88 × 1010 m3 /year Based on this ratio of storage capacity to annual inflow (= 0.20 year), the percent of sediment trapped in the reservoir is estimated from Figure 7.48 as 93%. Since the average sediment load delivered by the river to the reservoir is 1.96 × 1010 kg/year, the rate at which sediment is accumulating in the reservoir is 0.93 × 1.96 × 1010 kg/year = 1.78 × 1010 kg/year. Since the bulk density of the sediment accumulating at the bottom of the reservoir is 1,600 kg/m3 , and the area of the reservoir is 850 km2 = 8.5 × 108 m2 , the rate at which sediment volume is accumulating is given by sediment trap rate sediment bulk density 1.78 × 1010 kg/year = = 1.11 × 107 m3 /year 1600 kg/m3

sediment volume accumulation rate =

Since the plan area of the reservoir is 850 km2 = 8.5 × 108 m2 , the rate of sediment accumulation on the bottom of the reservoir is given by sediment volume accumulation rate reservoir area 1.11 × 107 m3 /year = 0.013 m/year = 1.3 cm/year = 8.5 × 108 m2

rate of sediment accumulation =

At this rate, it will take approximately 1400 years for the reservoir capacity to decrease by 10% due to sediment accumulation. 7.76. From the given data, the cumulative inflow, demand, and difference between inflow and demand is tabulated below:

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Month

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Cumulative Inflow (×1010 m3 ) 0.23 0.50 0.85 1.11 1.42 1.75 1.99 2.24 2.64 3.07 3.42 3.74 4.00 4.20 4.47 4.81 5.17 5.50 5.80 6.13 6.47 6.81 7.13 7.42

Cumulative Demand (×1010 m3 ) 0.27 0.53 0.86 1.19 1.57 1.93 2.31 2.68 3.02 3.34 3.62 3.89 4.16 4.43 4.75 5.08 5.46 5.82 6.20 6.57 6.91 7.23 7.52 7.78

Cumulative Inflow − Demand (×1010 m3 ) −0.04 −0.03 −0.01 −0.08 −0.15 −0.18 −0.32 −0.44 −0.38 −0.27 −0.20 −0.15 −0.16 −0.23 −0.28 −0.27 −0.29 −0.32 −0.40 −0.44 −0.44 −0.42 −0.39 −0.36

These data indicate that the reservoir storage must be sufficient accommodate the cumulative deficit between demand and reservoir inflow that occurs between Month 3 and Month 21, and this deficit is equal to (−0.01 + 0.44) × 1010 m3 = 0.43 × 1010 m3 . Assuming that the reservoir is full at the beginning of this critical interval, then an active reservoir storage of 0.43 × 1010 m3 will be sufficient. 7.77. From the given data: H = 85 m, D = 0.60 m, L = 300 m, ks = 8 mm, Dj = 50 mm, kj = 0.8, k = 0.5, v2 = 2 m/s, and V2 = 6 m/s. The flow rate, Q, is determined by application of the energy equation between the upstream reservoir and the exit of the nozzle, which requires that H−

f L Q2 Q2 Q2 − k = j D 2gA2 2gA2j 2gA2j

Using the Swamee-Jain equation yields Q = 0.0597 m3 /s. The corresponding velocities in the delivery pipeline and nozzle jet are V = 0.21 m/s and Vj = 30.4 m/s, and the friction factor

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is f = 0.0425. Using these derived data yields the following results: he = H − hL = Y −

fL V 2 (0.0425)(300) 0.212 = 85 − = 84.95 m D 2g (0.60) 2(9.81)

Vj2 v2 V2 22 62 30.42 − k 2 − 2 = 84.95 − 0.8 − 0.5 − = 45.3 m 2g 2g 2g 2(9.81) 2(9.81) 2(9.81) PT = γQhT = (9.79)(0.0597)(45.3) = 26.5 kW hT 45.3 ηT = = = 0.53 he 84.95 hT = he − kj

For the given configuration, the expected power from the system is 26.5 kW with a hydraulic efficiency of 53% . 7.78. From the given data: H = 100 m, D = 2.0 m, L = 500 m, ks = 15 mm, Q = 20 m3 /s, ∆hDT = 5.0 m, and V = 0.8 m/s. The velocity (= Q/A) in the penstock can be calculated as Vp = 6.37 m/s. Using the given values of Q, D, and ks , and assuming that the kinematic viscosity of water, ν, is 10−6 m2 /s, (at 20◦ C), the friction factor, f , of the penstock can be calculated using the Swamee-Jain equation which yields f = 0.0345. The head, hT , extracted by the turbine is given by Equation 7.140 as hT = H −

V2 (0.0345)(500) 6.372 0.82 f L Vp2 − ∆hDT − = 100 − − 5.0 − = 77.16 m D 2g 2g 2.0 2(9.81) 2(9.81)

which gives PT = γQhT = (9.79)(20)(77.16) = 15100 kW = 15.1 MW Therefore the system will extract 15.1 MW of power from the water flowing through the turbine. 7.79. Assuming that the capacity of the turbines will be sufficient to pass a flow, Q of 1630 m3 /s at a head of 15 m, and taking the turbine efficiency, η, as 0.85, and γ = 9.79 kN/m3 , yields P = γQHη = (9.79)(1630)(15)(0.85) = 2.03 × 105 kW = 203 MW Therefore, to fully utilize the available head and anticipated flow rates, an installed capacity of 203 MW is required. 7.80. From the given data, it can be approximated that b = 10 m and Heff = 5 m. These approximations assume that the downstream channel is rectangular and the velocity head behind the dam is negligible. Taking ρ = 998 kg/m3 and using Equation 7.147 gives the maximum power, Pmax as Pmax = bpmax

( )5 ( )5 3 5 2 2 3 52 2 2 =b ρg 2 Heff = (10) (998)(9.81) 2 (5) 2 = 1.73 × 106 W = 1.73 MW 5 5

Therefore the theoretical maximum power that could be extracted at this location is 1.73 MW .

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Chapter 8

Probability and Statistics in Water-Resources Engineering 8.1. µx = σx2 =

n ∑ i=1 n ∑

xi f (xi ) = 3.06 (xi − µx )2 f (xi ) = 2.75

i=1 n 1 ∑ gx = 3 (xi − µx )3 f (xi ) = 0.425 σx i=1

8.2. ∫

∫

∞

∞

1 te−0.143t dt = 0.143 = 7.0 days 2 0.143 0 ∫0 ∞ ∫ ∞ σt2 = (t − µt )2 f (t)dt = 0.143 (t − 7.0)2 e−.143t dt 0 0 ∫ ∞ = 0.143 (t2 − 14t + 49)e−.143t dt [0 ] 2 1 49 = 0.143 − 14 + = 48.90 0.1433 0.1432 0.143 √ σt = 48.90 = 6.99 days ∫ ∞ ∫ 1 0.143 ∞ 3 gt = 3 (t − µt ) f (t)dt = (t − 7.0)3 e−0.143t dt 6.993 0 σt 0 ∫ 0.143 ∞ 3 = (t − 21t2 + 147t − 343)e−0.143t dt 6.993 0 [ ] 6 2 1 1 0.143 − 21 + 147 − 343 = 2.0 = 6.993 0.1434 0.1432 0.1432 0.143 µt =

tf (t)dt = 0.143

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8.3. (a) The area under the probability distribution curve must equal 1, therefore ∫

10

α dx = 1 x2 4 α 10 − =1 x 4] [ 1 1 −α − =1 10 4 which leads to α=

20 3

(b) If f (x) = ∫

then F (x) =

4

When x = 7

m3 /s,

x

20 (20/3) = 2 x2 3x

[ ] 20 20 x 20 1 1 ds = − = − 3s2 3s 4 3 4 x

[ ] 20 1 1 − = 0.715 F (7) = 3 4 7

and Prob(X > 7) = 1 − 0.715 = 0.285. The return period, T , is therefore given by T =

1 1 = = 3.5 years Prob(X > 7) 0.285

(c) ∫

10

∫

10

1 20 3 dx = ln x|10 4 = 6.11 m /s x 3 4 4 ) ∫ 10 ∫ ∫ ( 20 10 (x − 6.11)2 20 10 12.2 37.3 2 σ = (x − µ)f (x) dx = + 2 dx = 1− dx 3 4 x2 3 4 x x 4 [ ] 37.3 10 20 = x − 12.2 ln x − = 2.67 (m/s)2 3 x 4 µ=

20 xf (x) dx = 3

Therefore σ = 1.63 m/s 8.4. Let Q be the river flow, then since f (Q) = constant = c for Q ∈ [1, 2], ∫

2

c dQ = 1 1

cQ|21 = 1 c(2 − 1) = 1

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which gives c = f (Q) =

1 =1 2−1

Hence the pdf of Q is given by   1 f (Q) =



0

Q ∈ [1, 2] Q∈ / [1, 2]

2 , and skewness, g , are The mean, µQ , variance, σQ Q

2 Q2 = 1.5 µQ = Qf (Q)dQ = Q(1)dQ = 2 1 1 1 2 ∫ 2 ∫ 2 1 2 2 2 3 (Q − 1.5) (1)dQ = (Q − 1.5) = 0.083 (Q − 1.5) f (Q)dQ = σQ = 3 1 1 1 ∫

2

∫

2

gQ = 0 where gQ = 0 since the distribution is symmetrical about the mean. The flow with a return period of 50 years has an exceedance probability of 1/50 = 0.02, and a cumulative probability of 1 − 0.02 = 0.98. Hence, ∫ Q50 f (Q)dQ = 0.98 1 ∫ Q50 (1)dQ = 0.98 1 50 Q|Q = 0.98 1

Q50 − 1 = 0.98 which gives Q50 = 1.98 m3 /s . 8.5. (a) The probability of 1 wet year and 19 dry years is given by 20! (0.5)1 (0.5)19 = 1.91 × 10−5 1!19! (b) The probability of 10 wet years and 10 dry years is given by 20! (0.5)10 (0.5)10 = 0.176 10!10! 8.6. Probability of n successes in N trials for any single combination is pn (1 − p)N −n , and the number of combinations of n successes in N trials is N !/n!(N − n)!. Hence, the probability of n successes in N trials is N! pn (1 − p)N −n n!(N − n)!

245

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8.7. The probability of the design event being exceeded 5 times in 10 years is ( )5 ( )5 10! 1 24 = 2.1 × 10−5 5!(10 − 5)! 25 25 The probability of the design event being exceeded at least once in 10 years is ( )0 ( )10 10! 1 24 1− = 0.34 0!(10 − 0)! 25 25 8.8. The probability, P1 , that the design event is exceeded once in 5 years is ( )1 ( )4 5! 1 24 P1 = = 0.17 1!(4)! 25 25 and the probability, P0 , that the design event is not exceeded in 5 years is ( )0 ( )5 5! 1 24 P0 = = 0.815 0!(5)! 25 25 Therefore, the probability, P , that the design event is exceeded more than once is P = 1 − P0 − P1 = 1 − 0.815 − 0.17 = 0.015 8.9. Let p = exceedance probability of the design event. Therefore, the exceedance probability during the design life (10 years) is 1 − f (0), where f (0) =

10! 0 p (1 − p)10 = (1 − p)10 0!10!

Therefore, the exceedance probability, Pe , is given by Pe = 1 − (1 − p)10 Since the exceedance probability is given as 1%, then 0.01 = 1 − (1 − p)10 which reduces to p = 0.00100 The return period, T is therefore given by T = 1/p = 1/0.001 = 1000 years The cumulative probability of the design flow, Q, is F (Q), hence F (Q) = exp[− exp(−Q)] = 1 − p = 1 − 0.001 = 0.999 which yields Q = 6.9 m3 /s

246

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8.10. Probability of exceedance 1 year after construction 1! 1!1!

(

1 50

)1 (

49 50

)0 = 0.02

Probability that the 50-year storm will occur within the first 6 years is 6! 1− 0!6!

(

1 50

)0 (

49 50

)6 = 0.11

8.11. 184 = 5.26 days 35 1 λ= = 0.19 d−1 µt

µt =

Probability, P , of less than 1 week (= 7 days) between events given by P = 1 − e−λt = 1 − e−(0.19)(7) = 0.74 8.12. The average interval between floods is given as 10 years . Assume that the occurrence of floods is a Poisson process with an average rate of 1 per 10 years, which gives λ = 1/10 = 0.1/year. Therefore, the probability that there will be less than 6 months (= 0.5 years) between floods is given by the exponential distribution as F (0.5) = 1 − e−λt = 1 − e−(0.1)(0.5) = 0.0488 and therefore the probability of more than 6 months between floods is 1 − 0.0488 = 0.951 or 95.1% . 8.13. The gamma distribution is given by f (t) = With n = 1, f (t) =

λn tn−1 e−λt (n − 1)!

λ1 t0 e−λt = λe−λt (0)!

which is the same as the exponential distribution. 8.14. The probability density function of the time for three flood events is given by the gamma distribution, Equation 8.49. From the given data: n = 3, µt = 365 days, and λ=

n 3 = = 0.008219 d−1 µt 365

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The probability of up to t0 days elapsing before the third flood event is given by the cumulative distribution function ∫ ∫ t0 ∫ t0 n n−1 −λt (0.008219)3 t0 3−1 −0.008219t λ t e dt = t e dt F (t0 ) = f (t)dt = (n − 1)! (3 − 1)! 0 0 0 ∫ t0 −7 = 2.776 × 10 t2 e−0.008219t dt 0

According to Dwight (1961) (

∫ 2 at

at

t e dt = e

t2 2t 2 − 2+ 3 a a a

)

Using this relation to evaluate F (t0 ) gives −7

F (t0 ) = −2.776 × 10

[

−0.008219t

e

(

t2 2t 2 + + 2 0.008219 0.008219 0.0082193

)]t0 0

Taking t0 = 365 d (= 1 year) gives [

(

)]365 t2 2t 2 F (730) = −2.776 × 10 e + + 0.008219 0.0082192 0.0082193 0 )] [ ( )]} {[ ( = −2.776 × 10−7 e−3 1.621 × 107 + 1.081 × 107 + 0.360 × 107 − 1 3.602 × 106 −7

−0.008219t

= 0.577 Hence, the probability that 3 flood events will occur in less than 1 year is 0.577, and the probability that it will take more than 1 year to have 3 flood events is 1 − 0.577 = 0.423 or 42.3% . 8.15. From the given data: µx = 141.2 cm, σx = 28.2 cm, and the rainfall has a normal distribution. (a) For x = 127.0 cm, z=

x − µx 127.0 − 141.2 = = −0.504 σx 28.2

From Equation 8.67, F (z) = B where ]−4 1[ 1 + 0.196854|z| + 0.115194|z|2 + 0.000344|z|3 + 0.019527|z|4 2 ]−4 1[ 1 + 0.196854| − 0.504| + 0.115194| − 0.504|2 + 0.000344| − 0.504|3 + 0.019527| − 0.504|4 = 2 = 0.307

B=

(b) For x = 152.4 cm, z=

x − µx 152.4 − 141.2 = = 0.397 σx 28.2

From Equation 8.67, F (z) = 1 − B

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where ]−4 1[ 1 + 0.196854|z| + 0.115194|z|2 + 0.000344|z|3 + 0.019527|z|4 2 ]−4 1[ = 1 + 0.196854|0.397| + 0.115194|0.397|2 + 0.000344|0.397|3 + 0.019527|0.397|4 2 = 0.345

B=

therefore F (0.397) = 1 − 0.345 = 0.655 (c) The probability that the rainfall is between 127.0 cm and 152.4 cm is equal to 0.655 − 0.307 = 0.348 . 8.16. From the given data: µx = 620 m3 /s, σx = 311 m3 /s, and river capacity = 780 m3 /s. For log-normal distribution, ( ) σy2 exp µy + = 620 (1) 2 (620)2 [exp(σy2 ) − 1] = 3112

(2)

Solving Equations 1 and 2 yields µy = 6.32

and

σy = 0.473

The standard normal deviate, z, for y = ln 780 = 6.66 is z=

y − µy 6.66 − 6.32 = = 0.717 σy 0.473

and F (z) = 1 − B where ]−4 1[ 1 + 0.196854|z| + 0.115194|z|2 + 0.000344|z|3 + 0.019527|z|4 2 ]−4 1[ = 1 + 0.196854|0.717| + 0.115194|0.717|2 + 0.000344|0.717|3 + 0.019527|0.717|4 2 = 0.235

B=

therefore F (0.717) = 1 − 0.235 = 0.765 and the exceedance probability is 1 − 0.765 = 0.235 . 8.17. From the given data: µx = 114 cm, σx = 22 cm, and therefore ( ) σy2 exp µy + = 114 2 (114)2 [exp(σy2 − 1] = 222

249

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Solving Equations 1 and 2 yields µy = 4.72

and

σy = 0.191

For T = 100 years, the cumulative probability, F (z), of the standard normal deviate, z, is given by 1 F (z) = 1 − = 0.99 100 From the normal probability distribution tables, z = 2.33 and therefore 2.33 =

y100 − µy y100 − 4.72 = σy 0.191

which leads to y100 = 5.16 and x100 = ey100 = e5.16 = 174 cm If the rainfall is normally distributed, 2.33 =

x100 − µx x100 − 114 = σx 22

which yields x100 = 165 cm 8.18. From the given data: a = 95 m3 /s, b = 115 m3 /s, and f (x) =

1 1 = = 0.05 b−a 115 − 95

Therefore the probability, p, that x ≥ 100 m3 /s is p = 0.05(115 − 100) = 0.75 8.19. If y is the minimum value in sample set of n realizations, then F (y) = [1 − PX (y)]n and f (y) =

d F (y) = −n[1 − PX (y)]n−1 pX (y) dy

8.20. From the given data (on maximum annual values): µx = 480 m3 /s, and σx = 320 m3 /s. Therefore σx 320 =√ = 249.5 m3 /s 1.645 1.645 b = µx − 0.577a = 480 − 0.577(249.5) = 336.0 m3 /s X −b X − 336 Y = = a 250 a= √

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For a return period of 50 years, F (y) = 0.98 and 0.98 = exp[− exp(−y50 )] y50 = − ln[− ln(0.98)] = 3.90 x50 − 336 y50 = 250 x50 = 336 + 250y50 = 336 + 250(3.90) = 1311 m3 /s Therefore the annual maximum flow with a return period of 50 years is 1311 m3 /s For a return period of 100 years, F (y) = 0.99 and 0.99 = exp[− exp(−y100 )] y100 = − ln[− ln(0.99)] = 4.60 x100 − 336 y100 = 250 x100 = 336 + 250y100 = 336 + 250(4.60) = 1486 m3 /s Therefore the annual maximum flow with a return period of 100 years is 1486 m3 /s .

8.21. From the given data: µx = 710 mm, and σx = 112 mm. According to Equation 8.85, the scale and location parameters, a and b, are derived from µx and σx by σx 112 =√ = 87.32 mm 1.645 1.645 b = µx + 0.577a = 710 + 0.577(87.2) = 760.3 mm

a= √

Use these parameters to normalize the minima, where Y =

X −b a

Therefore, X = 600 mm corresponds to y=

600 − 760.3 = −1.836 87.32

The probability that y < −1.836 is given by Equation 8.88, where F (−1.836) = 1 − exp[− exp(−1.836)] = 0.147 and the probability that y > −1.836 is 1 − 0.147 = 0.853. The average interval between years where the rainfall exceeds 600 mm (y > −1.836) is therefore given by the return period, T , where 1 T = = 1.2 years 0.853

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8.22. From the given data: µx = 43 m3 /s, σx = 12 m3 /s, and µx = 43 = bΓ(1 + 1/a)

(1)

σx2 = 122 = b2 [Γ(1 + 2/a) − Γ2 (1 + 1/a)]

(2)

Eliminating b from Equations 1 and 2 yields the relation Γ(1 + 2/a) = 1.078 Γ2 (1 + 1/a) and solving for a gives a = 4.03 Substituting into Equation 1 gives b=

43 43 = = 47.5 Γ(1 + 1/a) Γ(1 + 1/4.03)

The extreme-value Type III distribution is therefore given by [ ( ] [ ( x )a ] x )4.03 F (x) = 1 − exp − = 1 − exp − b 47.5 and the probability that the annual low flow is less than 10 m3 /s is [ ( ) ] 10 4.03 F (10) = 1 − exp − = 0.0019 47.5 8.23. The cumulative distribution function (CDF) of an extreme-value Type III distribution is given by [ ( ) ] x−c a F (x) = 1 − exp − b−c where c = 9 m3 /s, and the corresponding return period, T , is given by [( ) ] 1 x−9 a T = = exp 1 − F (x) b−9

(1)

Since T = 10 years corresponds to x = 34 m3 /s and T = 50 years corresponds to x = 76 m3 /s, then [( ) ] [( ) ] 34 − 9 a 25 a 10 = exp = exp (2) b−9 b−9 [(

and 50 = exp which give

( ln(10) =

25 b−9

76 − 9 b−9

)a ]

[( = exp

67 b−9

)a

)a ]

( and

252

ln(50) =

(3)

67 b−9

)a

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Combining these equations yields ln 50 = ln 10

(

67 25

)a

which simplifies to a = 0.538 Substituting into Equation 2 gives [( 10 = exp ( ln 10 =

25 b−9

] ) 25 0.538 b−9 )0.258

which simplifies to b = 14.3 Hence, for T = 100 years, Equation 1 gives [( ] [( ] ) ) x100 − 9 0.538 x100 − 9 0.538 100 = exp = exp 14.3 − 9 5.3 ( )0.538 x100 − 9 ln 100 = 5.3 which simplifies to x100 = 99.6 m3 /s 8.24. From the given data: µx = 63.2 m3 /s, σx = 13.7 m3 /s, and gx = 1.86. In order to find the GEV cumulative distribution function, the parameters a, b, and c must first be determined. The shape parameter, c, can be determined directly from the skewness coefficient, gx , using Equation 8.100, where gx = 1.86 = sign(c)

−Γ(1 + 3c) + 3Γ(1 + c)Γ(1 + 2c) − 2Γ3 (1 + c) 3

[Γ(1 + 2c) − Γ2 (1 + c)] 2

which can be solved to give c = −0.0950. Rearranging Equation 8.99 gives b as √ √ c2 σx2 (−0.0950)2 (13.7)2 b= = = 9.26 m3 /s Γ(1 + 2c) − Γ2 (1 + c) 1.153 − 1.0652 and rearranging Equation 8.98 gives a as 9.26 b [1 − 1.065] = 56.9 m3 /s a = µx − [1 − Γ(1 + c)] = 63.2 − c −0.0950 Hence the cumulative distribution function, F (x), is given by Equation 8.97 as { [ } { [ ] } ] −0.0950(x − 56.9) 1/−0.0950 c(x − a) 1/c = exp − 1 − F (x) = exp − 1 − b 9.26

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which yields

{ } F (x) = exp − [1 + 0.0103(x − 56.9)]−10.5

For a return period of 50 years, F (x50 ) = 1 − 1/50 = 0.98 and { } F (x50 ) = 0.98 = exp − [1 + 0.0103(x50 − 56.9)]−10.5 which yields x50 = 101 m3 /s Therefore, an annual-maximum streamflow of 101 m3 /s has a return period of 50 years. 8.25. From the given data: ν = 15, and χ2 = 25, Appendix C.3 gives α = 0.05 Therefore the probability that the sum is greater than 25 is 0.05 . The expected value of the sum is ν = 15 . 8.26. According to the Gringorten formula PX (X > xm ) =

m−a N + 1 − 2a

and hence the return period, T , is given by 1 N + 1 − 2a = PX (X > xm ) m−a

T =

(1)

Taking N = 50 and a = 0.375, Equation 1 gives the return period of xm as T1 =

50 + 1 − 2(0.375) 50.25 = m − 0.375 m − 0.375

(2)

and taking N = 50 and a = 0.44, Equation 1 gives the return period of xm as T2 =

50 + 1 − 2(0.44) 50.12 = m − 0.44 m − 0.44

(3)

The variability in the return period, T , resulting from the uncertainty in a is given by T2 − T1 =

50.12 50.25 − m − 0.44 m − 0.375

(4)

The relation between T2 − T1 and m can be easily calculated using Equation 4, and the maximum (absolute) value of T2 − T1 occurs when m = 1 and is equal to 9.1 . The units of 9.1 depend on the nature of the measurements. If the interval between data points is 1 year, then the maximum error in return period is 9.1 years. 8.27. The ranking of flows between 1980 and 1996 are shown in Columns 1 and 2 of the following table:

254

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(1) Rank, m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

(2) Flow, qm (ft3 /s) 12,136 11,586 10,584 9,193 8,293 8,000 7,884 7,236 7,182 6,390 6,222 5,889 5,550 5,142 5,129 4,132 3,390

(3) Weibull P (Q > qm ) 0.06 0.11 0.17 0.22 0.28 0.33 0.39 0.44 0.50 0.56 0.61 0.67 0.72 0.78 0.83 0.89 0.94

(4) Gringorten P (Q > qm ) 0.04 0.09 0.15 0.21 0.27 0.33 0.38 0.44 0.50 0.56 0.62 0.67 0.73 0.79 0.85 0.91 0.96

In this case, N = 17, and the Weibull exceedance probabilities are given by Equation 8.119 as m m = P (Q > qm ) = 17 + 1 18 These probabilities are tabulated in Column 3. The Gringorten exceedance probabilities are given by (assuming a = 0.40)

P (Q > qm ) =

m−a m − 0.40 m − 0.40 = = N + 1 − 2a 17 + 1 − 2(0.40) 17.2

and these probabilities are tabulated in Column 4. 8.28. The first step in the analysis is to derive the theoretical frequency distribution. Appendix C.1 gives the cumulative probability distribution of the standard normal deviate, z, which is defined as z=

x − µx σx

where x is the annual rainfall. Converting the given rainfall amounts into standard normal deviates, z, yields:

255

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Rainfall (mm) 960 1064 1138 1201 1260 1318 1382 1455 1560

z

P (Z < z)

−1.28 −0.838 −0.521 −0.252 0.000 0.248 0.521 0.833 1.28

0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90

and therefore the theoretical frequencies are given by Range (mm) < 960 960 - 1064 1064 - 1138 1138 - 1201 1201 - 1260 1260 - 1318 1318 - 1382 1382 - 1455 1455 - 1560 > 1560

Theoretical Probability, pm 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10

Theoretical Outcomes, N pm 5 5 5 5 5 5 5 5 5 5

Based on the observed and theoretical frequency distributions, the chi-square statistic is given by Equation 8.114 as (4 − 5)2 (5 − 5)2 (6 − 5)2 (6 − 5)2 (5 − 5)2 (7 − 5)2 + + + + + + 5 5 5 5 5 5 (4 − 5)2 (3 − 5)2 (6 − 5)2 (4 − 5)2 + + + 5 5 5 5 = 2.80

χ2 =

Since both the mean and standard deviation were estimated from the measured data, the χ2 statistic has M − 1 − n degrees of freedom, where M = 10 (= number of intervals), and n = 2 (= number of population parameters estimated from measured data), hence 10 − 1 − 2 = 7 degrees of freedom. Using a 5% significance level, the hypothesis that the observations are drawn from a normal distribution is accepted if 0 ≤ 2.80 ≤ χ20.05 Appendix C.3 gives that, for 7 degrees of freedom, χ20.05 = 14.067, and since 0 ≤ 2.80 ≤ 14.067 the hypothesis that the annual rainfall is drawn from a normal distribution is accepted .

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8.29. Based on the given data, the sample and theoretical cumulative distribution functions of the annual rainfall are as follows: (1) Rainfall, x (mm) 960 1064 1138 1201 1260 1318 1382 1455 1560

(2) Normalized Rainfall, (x − µx )/σx −1.282 −0.838 −0.521 −0.252 0 0.248 0.521 0.833 1.282

(3) SN (x)

(4) PX (x)

(5) |PX (x) − SN (x)|

4/50 = 0.080 9/50 = 0.180 15/50 = 0.300 21/50 = 0.420 26/50 = 0.520 33/50 = 0.660 37/50 = 0.740 40/50 = 0.800 46/50 = 0.920

0.100 0.201 0.301 0.401 0.500 0.598 0.699 0.798 0.900

0.020 0.021 0.001 0.019 0.020 0.062 0.041 0.002 0.020

The normalized rainfall amounts in Column 2 are calculated using the rainfall amount, x, in Column 1, mean, µx , of 1260 mm, and standard deviation, σx , of 234 mm. The sample cumulative distribution function, SN (x), in Column 3 is calculated using the relation SN (x) =

k N

where k is the number of measurements less than the given rainfall amount, x, and N = 50 is the total number of measurements. The theoretical cumulative distribution function, PX (x), in Column 4 is calculated using the normalized rainfall (Column 2) and the standard normal distribution function given in Appendix C.1. The absolute value of the difference between the theoretical and sample distribution functions are given in Column 5. The maximum difference, D, between the theoretical and sample distribution functions is equal to 0.062, and occurs at an annual rainfall of 1318 mm. For a sample size of 9 and a significance level of 10%, Appendix C.4 gives the critical value of the Kolmogorov-Smirnov statistic as 0.388. Since 0.062 ≤ 0.388, the hypothesis that the measured data is from a normal distribution is accepted at the 10% significance level. 8.30. From the given data: µ = 1.6, therefore λ=

1 1 = = 0.625 µ 1.6

8.31. The mean and standard deviation of the outcomes (µx and σx ) are related to the mean and standard deviation of the log of the outcomes (µy and σy ) by σx2 = µ2x [exp(σy2 ) − 1]

µx = exp(µy + σy2 /2),

Application of the method of moments requires putting µx = 1.6, and σx = 1.2 which gives 1.6 = exp(µy + σy2 /2) 2

2

1.2 = 1.6

257

[exp(σy2 )

− 1]

(1) (2)

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Solving Equations 1 and 2 for µy and σy yields µy = 0.247 and σy = 0.668 ˆ = 35 m3 /s, σ ˆ = 15 m3 /s, and gˆ = 0.4. 8.32. From the given data: N = 41 (1940 to 1980), µ Substituting these values into Equations 8.122 to 8.124 yields σ ˆx 15 Sµˆx = √ = √ = 2.3 m3 /s 41 N √ √ 1 + 0.75ˆ gx 1 + 0.75(0.4) Sσˆx = σ ˆx = 15 = 1.9 m3 /s 2N 2(41) [ ]0.5 Sgˆx = 10A−B log10 (N/10) where Equations 8.125 and 8.126 give A = −0.33 + 0.08|0.4| = −0.30 B = 0.94 − 0.26|0.8| = 0.84 and therefore

[ ]0.5 Sgˆx = 10−0.30−0.84 log10 (41/10) = 0.39

Hence, the standard errors of the mean, standard deviation, and skewness are 2.3 m3 /s , 1.9 m3 /s , and 0.39 respectively. Using the ratio of the standard error to the estimated value as a measure of the accuracy of the estimated value, then Sµˆ 2.3 × 100 = × 100 = 6.6% µ ˆ 35 1.9 Sσˆ accuracy of standard deviation = × 100 = × 100 = 13% σ ˆ 15 Sgˆ 0.39 accuracy of skewness = × 100 = × 100 = 98% gˆ 0.4 accuracy of mean =

Based on these results, it is clear that the accuracies of the estimated parameters deteriorates as the order of the moments associated with the parameters increase. 8.33. The exponential distribution is given by f (t) = λe−λt

258

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Let θ = λ, then ′

L =

N ∏

f (ti , θ) =

i=1

N ∏

θe−θti

i=1

L = ln L′ =

N ∑

ln(θe−θti ) =

i=1 N ∑

∂L N =0+ − ∂θ θ

N N N N ∑ ∑ ∑ ∑ (ln θ − θti ) = ln θ − θ ti = N ln θ − θ ti i=1

i=1

i=1

i=1

ti = 0

i=1

Solving for θ gives N θ = ∑N

i=1 ti

Hence N λ = ∑N

i=1 ti

Using the moment method, λ=

1 = µ

1

∑N

i=1 ti

N = ∑N

i=1 ti

N

Hence the moment and maximum-likelihood methods yield the same result. 8.34. The normal distribution is given by f (x; µx , σx ) =

σx

1 √

[

1 exp − 2 2π

(

x − µx σx

)2 ]

where µx and σx are the parameters of the distribution. Use the log-likelihood function, L, where { } N ( √ ) 1 ( x − µ )2 ∑ i x L= ln f (xi ; µx , σx ) = − ln σx 2π − 2 σx i=1

Determine the values of µx and σx that maximize L, )( )} N { ( ∑ ∂L xi − µx 1 =0= − − ∂µx σx σx i=1 N ∑

xi − µ x σx2 i=1 ( )( )} N { ∑ ∂L 1 √ xi − µ x xi − µx =0= − √ 2π − − ∂σx σx σx2 σ 2π x i=1 } N { ∑ 1 (xi − µx )2 = − + σx σx3 =

i=1

259

(1)

(2)

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Equation 1 requires that N ∑

(xi − µx ) = 0

i=1

or N ∑

xi = µx

i=1

N ∑

1

(= µx N )

i=1

therefore

µx =

N 1 ∑ xi N i=1

Equation 2 requires that ( ) N N 1 ∑ 1 1 ∑ 1 = N = 3 (xi − µx )2 σx σx σx i=1

i=1

Multiplying by σx3 /N gives

σx2

N 1 ∑ = (xi − µx )2 N i=1

or v u N u1 ∑ σx = t (xi − µx )2 N i=1

8.35. For the N = 27 years of data, the rank (i) and corresponding flow (Qi ) are given in Columns 1 and 2 respectively in the following table:

260

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(1)

(2)

i

Qi (m3 /s)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 Sum

3 4 5 17 27 53 53 60 95 107 168 189 189 212 237 267 320 377 423 550 597 752 1062 1337 1683 1889 3359 14035

(3)

(4)

(5)

(1 − i)Qi

ln Qi

(ln Qi − µln Qi )2

4 10 51 108 265 318 420 760 963 1680 2079 2268 2756 3318 4005 5120 6409 7614 10450 11940 15792 23364 30751 40392 47225 87334 305396

1.099 1.386 1.609 2.833 3.296 3.970 3.970 4.094 4.554 4.673 5.124 5.242 5.242 5.357 5.468 5.587 5.768 5.932 6.047 6.310 6.392 6.623 6.968 7.198 7.428 7.544 8.119 137.835

16.051 13.829 12.219 5.161 3.273 1.288 1.288 1.021 0.304 0.187 0.000 0.019 0.019 0.063 0.132 0.233 0.440 0.684 0.888 1.452 1.656 2.304 3.470 4.381 5.398 5.948 9.087 90.793

The first probability-weighted moment, b0 , is given by Equation 8.134 as N 1 ∑ 1 b0 = Qi = (14035) = 519.8 N 27 i=1

The product (1 − i)Qi is shown in Column 3, and the second probability-weighted moment, b1 is given by Equation 8.135 as ∑ 1 1 b1 = (i − 1)Qi = (305396) = 435.0 N (N − 1) 27(27 − 1) N

i=2

The first L-moment, λ1 , is given by Equation 8.130 as λ1 = b0 = 519.8

261

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and the second L-moment, λ2 , is given by Equation 8.131 as λ2 = 2b1 − b0 = 2(435.0) − 519.8 = 350.3 The relationship between the L-moments, λ1 and λ2 , and the parameters of the lognormal distribution, µY and σY , are given in Table 8.2, which yields ( ) σY2 λ1 = 519.8 = exp µY + 2 ( ) ( ) σY2 σY λ2 = 350.3 = exp µY + erf 2 2 and solving for µY and σY gives µY = 5.29 σY = 1.39 Hence, using the L-moment method, the estimated mean and standard deviation of the lognormal distribution are 5.29 and 1.39 respectively. Column 4 in the above table gives ln Qi , Column 5 gives (ln Qi − µY )2 , and the the method of moments gives the mean and standard deviation (µY ,σY ) of the lognormal distribution as 1 ∑ 1 ln Qi = (137.835) = 5.11 N 27 i=1 v u √ N u 1 ∑ 1 2 t σY = (ln Qi − µY ) = 90.793 = 1.87 N −1 27 − 1 N

µY =

i=1

Hence, using the method of moments, the estimated mean and standard deviation of the lognormal distribution are 5.11 and 1.87 respectively. The difference between the mean and standard deviation (µY ,σY ) of the lognormal distribution estimated using the moment method (5.11,1.87) compared with the mean and standard deviation estimated using the L-moment method (5.29,1.39) is accounted for by the few outliers in the short period of record, which tend to make the L-moment parameter estimates more reliable than the moment parameter estimates. 8.36. The 10-year rainfall, x10 , can be written in terms of the frequency factor, K10 , as x10 = µx + K10 σx From the given data, µx = 152 cm, σx = 30 cm, and the exceedance probability, p, is given by 1 1 p= = = 0.10 T 10

262

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The intermediate variable, w, is given by Equation 8.144 as [ ( )] 1 [ ( )] 1 2 2 1 1 w = ln = ln = 2.146 p2 0.102 and Equation 8.143 gives the frequency factor, K10 , as 2.515517 + 0.802853w + 0.010328w2 1 + 1.432788w + 0.189269w2 + 0.001308w3 2.515517 + 0.802853(2.146) + 0.010328(2.146)2 = 2.146 − 1 + 1.432788(2.146) + 0.189269(2.146)2 + 0.001308(2.146)3 = 1.28

K10 = w −

The 10-year rainfall is therefore given by x10 = µx + K10 σx = 152 + 1.28(30) = 190 cm The 100-year rainfall, x100 , can be written in terms of the frequency factor, K100 , as x100 = µx + K100 σx The exceedance probability, p, is given by p=

1 1 = = 0.01 T 100

The intermediate variable, w, is given by Equation 8.144 as [ w = ln

(

1 p2

)] 1

[

2

= ln

(

1 0.012

)] 1 2

= 3.035

and Equation 8.143 gives the frequency factor, K100 , as 2.515517 + 0.802853w + 0.010328w2 1 + 1.432788w + 0.189269w2 + 0.001308w2 2.515517 + 0.802853(3.035) + 0.010328(3.035)2 = 3.035 − 1 + 1.432788(3.035) + 0.189269(3.035)2 + 0.001308(3.035)2 = 2.33

K100 = w −

The 100-year rainfall is therefore given by x100 = µx + K100 σx = 152 + 2.33(30) = 222 cm 8.37. From the given data, µx = 152 cm, σx = 30 cm, and the mean, µy , and standard deviation, σy , of the log-transformed variable, Y = ln X, are related to µx and σx by √ ( ) 1 µ4x σx2 + µ2x µy = ln , σ = ln y 2 µ2x + σx2 µ2x

263

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which lead to 1 µy = ln 2

(

1524 1522 + 302

) = 5.0

√

and

302 + 1522 = 0.195 1522 Since the rainfall, X, is lognormally distributed, Y (= ln x) is normally distributed, and the value of Y corresponding to a 10-year return period, y10 , is given by σy =

ln

y10 = µy + K10 σy where K10 is the frequency factor for a normal distribution and a return period of 10 years. Using Equation 8.143 gives K10 = 1.28, therefore y10 = 5.0 + 1.28(0.195) = 5.25 x10 = ey10 = e5.25 = 191 cm Similarly, using Equation 8.143 gives K100 = 2.33, therefore y100 = 5.0 + 2.33(0.195) = 5.45 x100 = ey100 = e5.45 = 233 cm If the rainfall is normally distributed, x10 = µx + σx K10 = 152 + 30(1.28) = 190 cm x100 = µx + σx K100 = 152 + 30(2.33) = 222 cm 8.38. From the given data: µx = 480 m3 /s, σx = 320 m3 /s, and gx = 0.87. The standard normal deviate corresponding to a 50-year return period, x′50 , is 2.06 (Appendix C.1), k = gx /6 = 0.87/6 = 0.145, and the frequency factor K50 is given by Equation 8.148 as } 1 { ′ K50 = [(x100 − k)k + 1]3 − 1 3k { } 1 = [(2.06 − 0.145)(0.145) + 1]3 − 1 3(0.145) = 2.50 The 50-year discharge, x50 , is therefore given by x50 = µx + K50 σx = 480 + (2.50)(320) = 1280 m3 /s The standard normal deviate corresponding to a 100-year return period, x′100 , is 2.33 (Appendix C.1), and the frequency factor K100 is given by Equation 8.148 as } 1 { ′ K100 = [(x100 − k)k + 1]3 − 1 3k { } 1 = [(2.33 − 0.145)(0.145) + 1]3 − 1 3(0.145) = 2.95

264

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The 100-year discharge, x100 , is therefore given by x100 = µx + K100 σx = 480 + (2.95)(320) = 1424 m3 /s 8.39. From the given data: gx = 3.0, 2.0, 1.0, 0.5, and 0.0, k = gx /6, x′100 , is 2.33 (Appendix C.1), and the frequency factor K100 is given by Equation 8.148 as K100 =

} 1 { ′ [(x100 − k)k + 1]3 − 1 3k

The values of K100 predicted by this equation are compared with those of Appendix C.2 in the following table: gx 3.0 2.0 1.0 0.5 0.0

K100 (Equation 8.148) 4.02 3.62 3.04 2.69 2.33

K100 (Appendix C.2) 4.05 3.61 3.02 2.69 2.33

The results shown in the above table demonstrate that the difference between the frequency factor given by Equation 8.148 and Appendix C.2 is very small for skewness coefficients in the range of [0,3]. These differences are all less than 1%. Based on these results, it appears that Equation 8.148 can be applied with reasonable accuracy outside of the [-1,+1] range of skewness coefficients. 8.40. From the given data, µx = 480 m3 /s, σx = 320 m3 /s,and the mean, µy , and standard deviation, σy , of the log-transformed variable, Y = ln X, are related to µx and σx by Equation 8.75, where ( ) σy2 , σx2 = µ2x [exp(σy2 ) − 1] µx = exp µy + 2 These equations can be put in the form 1 µy = ln 2

(

µ4x µ2x + σx2

which lead to µy =

1 ln 2

and

(

√ σy =

ln

√

) ,

σy =

4804 2 480 + 3202

ln

σx2 + µ2x µ2x

) = 5.99

3202 + 4802 = 0.606 4802

265

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From the given data, gy = 0.1, which yields k = gy /6 = 0.1/6 = 0.0167. The standard normal deviate corresponding to a 100-year return period, x′100 , is 2.33 (Appendix C.1), and the frequency factor for a 100-year return period, K100 , is given by Equation 8.148 as } 1 { ′ [(x100 − k)k + 1]3 − 1 3k { } 1 = [(2.33 + 0.0167)(0.0167) + 1]3 − 1 3(0.0167) = 2.40

K100 =

The 100-year log-discharge, y100 , is therefore given by y100 = µy + K100 σy = 5.99 + (2.40)(0.606) = 7.44 and the 100-year discharge, x100 , is x100 = ey100 = e7.44 = 1703 m3 /s 8.41. In Baton Rouge, Louisiana, Figure 8.12 gives a generalized skew coefficient GR of 0.0, and M SER is equal to 0.302. From the observed annual-maximum flood flows, G = −0.210, and for the 16-year period of record (1985-2000), N = 16. By definition A = −0.33 + 0.08|G| = −0.33 + 0.08| − 0.210| = −0.313 B = 0.94 − 0.26|G| = 0.94 − 0.26| − 0.210| = 0.885 M SES = 10[A−B(log10 (N/10))] = 10[−0.313−0.885(log10 (16/10))] = 0.321 and the weighted skew coefficient is given by GW =

(M SER )(G) + (M SES )(GR ) (0.302)(−0.210) + (0.321)(0.0) = = −0.102 M SER + M SES 0.302 + 0.321

Therefore, a skew coefficient of −0.102 is more appropriate than the measured skew coefficient of −0.210 for frequency analyses of flood flows in the Mississippi River at Baton Rouge, Louisiana. 8.42. From the given data: µx = 480 m3 /s, σx = 320 m3 /s, and the 100-year discharge, x100 , is given by x100 = µx + K100 σx = 480 + K100 320 For 25 years of data, the 100-year frequency factor is given by Table 8.3 as K100 = 3.729, and the 100-year discharge is given by x100 = 480 + (3.729)320 = 1670 m3 /s For ∞ years of data, the 100-year frequency factor is given by Equation 8.155 as √ { [ ( )]} 6 100 K100 = − 0.5772 + ln ln = 3.137 π 100 − 1

266

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which gives x100 = 480 + (3.137)320 = 1480 m3 /s Problem 8.20 estimates the 100-year discharge for an infinite number of measurements as 1486 m3 /s, which is almost identical to the estimate of 1480 m3 /s obtained here using equation 8.155. This difference can be attributed to roundoff. 8.43. From the given data: N = 15, and the ranked data are given in the following table: (1) Rank, m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

(2) Flow (m3 /s) 40 35 31 28 25 23 22 20 19 19 17 14 13 10 9

(3) m/(N + 1)

(4) (m − 0.4)/(N + 0.2)

0.0625 0.125 0.188 0.250 0.313 0.375 0.438 0.500 0.563 0.563 0.688 0.750 0.813 0.875 0.938

0.0395 0.105 0.171 0.237 0.302 0.368 0.434 0.500 0.566 0.566 0.697 0.763 0.829 0.895 0.961

The Weibull cumulative distribution function (CDF) is given in Column 3 by P (Q > qm ) =

m N +1

and the Gringorten CDF, with a = 0.4, is given in Column 4 by P (Q > qm ) =

m−a m − 0.4 = N + 1 − 2a N + 0.2

(a) Interpolating from the Weibull distribution gives P (Q > 30 m3 /s) = 0.209, and hence the return period, T , of 30 m3 /s is given by T =

1 1 = = 4.8 years 3 0.209 P (Q > 30 m /s)

(b) Interpolating from the Gringorten distribution gives P (Q > 30 m3 /s) = 0.193, and hence the return period, T , of 30 m3 /s is given by T =

1 1 = = 5.2 years 3 0.193 P (Q > 30 m /s)

267

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(c) From the given data, the mean, µq , is estimated by 1 ∑ qi = 21.7 m3 /s 15 15

µq =

i=1

and the standard deviation can be estimated by (the admittedly biased expression) v u 15 u1 ∑ σq = t (qi − 21.7)2 = 8.65 m3 /s 15 i=1

The Type I distribution requires that µq = b + 0.577a

and

σq2 = 1.645a2

21.7 = b + 0.577a

and

8.652 = 1.645a2

hence which gives and

a = 6.74

b = 17.8

(d) Using 6 classes to separate the data: (1) Range (m3 /s) 40

(2) Measured Outcomes 0 4 6 3 2 0

(3) Theoretical Probability, pm 0.025 0.286 0.380 0.198 0.075 0.036

(4) Theoretical Outcomes, N pm 0.375 4.29 5.70 2.97 1.13 0.54

The 6 classes in Column 1 consist of 4 interior classes of width (40−9)/4 = 7.75 m3 /s, and two extreme classes. The corresponding number of measured outcomes are shown in Column 2. The theoretical probability in Column 3 is derived from the Type I CDF F (y) = exp[− exp(−y)] where y=

q−b q − 17.8 = a 6.74

For q = 9 m3 /s, y = (9−17.8)/6.74 = −1.31, and F (y) = exp[− exp(1.31)] = 0.0246. For q = 16.75 m3 /s, y = (16.75−17.8)/6.74 = −0.156, and F (y) = exp[− exp(0.156)] = 0.311. Hence for 9 m3 /s < q < 16.75 m3 /s, pm = 0.311 − 0.0246 = 0.286. The remainder of the interval probabilities in Column 3 are calculated using this procedure. The theoretical number of outcomes in Column 4 are calculated by multiplying the number of outcomes

268

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(N = 15) by the theoretical probability in each interval. The chi-square statistic is given by χ2 =

6 ∑ (Xm − N pm )2 N pm

m=1

(0 − 0.375)2 (4 − 4.29)2 (6 − 5.70)2 (3 − 2.97)2 (2 − 1.13)2 (0 − 0.54)2 + + + + + 0.375 4.29 5.70 2.97 1.13 0.54 = 1.62

=

Since both the mean and standard deviation were estimated from the measured data, the χ2 statistic has M = 6 − 1 − 2 = 3 degrees of freedom. Using the 5% significance level, χ20.05 = 7.815, and since 1.62 < χ20.05 then the hypothesis that the data fits a Type I distribution is accepted . (e) For a return period, T , of 100 years, the frequency factor is given by Equation 8.155 as √ { √ { [ ( )]} [ ( )]} 6 T 6 100 KT = − 0.5772 + ln ln =− 0.5772 + ln ln = 3.14 π T −1 π 100 − 1 which gives x100 = µq + KT σq = 21.7 + 3.14(8.65) = 48.9 m3 /s 8.44. From the given data: T = 100 years, a = 55.2 m3 /s, b = 8.68 m3 /s, and c = −0.163. The annual-maximum streamflow with a return period of 100 years is given by Equation 8.156 as [ [ ( )c ] ( )−0.163 ] b T 8.68 100 x100 = a + 1 − ln = 55.2 + 1 − ln = 115 m3 /s c T −1 −0.163 100 − 1 Therefore, an annual-maximum streamflow of 115 m3 /s has a return period of 100 years. 8.45. From the given data: ⟨b⟩ = 2 m, ⟨m⟩ = 3, and ⟨y⟩ = 1 m. Denoting the estimated crosssectional area as ⟨A⟩, then ⟨A⟩ = ⟨b⟩⟨y⟩ + ⟨m⟩⟨y⟩2 = (2)(1) + (3)(1)2 = 5 m2 2 , can be estimated as The variance of A, σA )2 )2 )2 ( ( ( ∂A ∂A ∂A 2 2 2 σA = σb + σm + σy2 ∂b X=⟨X⟩ ∂m X=⟨X⟩ ∂y X=⟨X⟩

where

∂A = y|X=⟨X⟩ = ⟨y⟩ = 1 m ∂b X=⟨X⟩ ∂A = y 2 X=⟨X⟩ = ⟨y⟩2 = (1)2 = 1 m2 ∂m X=⟨X⟩ ∂A = b + 2my|X=⟨X⟩ = ⟨b⟩ + 2⟨m⟩⟨y⟩ = 2 + 2(3)(1) = 8 m ∂y X=⟨X⟩

269

(1)

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Substituting these derivatives and the given expected values into Equation 1 yields 2 σA = (8)2 (0.1)2 + (1)2 (0.3)2 + (1)2 (0.2)2 = 0.77 m4

The coefficient of variation of A is therefore estimated as √ σA 0.77 COV = = = 0.175 ≈ 18% ⟨A⟩ 5

270

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Chapter 9

Fundamentals of Surface-Water Hydrology I: Rainfall and Abstractions 9.1. The Weibull formula gives the return period, T , in terms of the rank, m, and total number of data points, n, as n+1 T = (1) m For the highest rainfall measurement of record, m = 1, and Equation 1 gives the corresponding return period as n+1 T = = n+1 1 9.2. For each time interval there are n = 50 ranked rainfall amounts of annual maxima. The relationship between the rank, m, and the return period, T , is given by T =

n+1 50 + 1 51 = = m m m

The return period can therefore be used in lieu of the rankings, and the given data can be put in the form: Return Period, T (years) 51 25.5 17

5 26.2 25.3 24.2

10 45.8 44.0 42.2

∆t in minutes 15 20 60.5 72.4 58.1 69.6 55.8 66.8

25 81.8 78.6 75.5

30 89.7 86.3 82.8

The rainfall increments with a return period, T , of 40 years can be linearly interpolated between the rainfall increments corresponding to T = 51 years and T = 25.5 years to yield: Duration (min) Rainfall (mm)

5 25.8

10 45.0

271

15 59.5

20 71.2

25 80.4

30 88.2

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and the average intensities for each duration are obtained by dividing the rainfall amounts by the corresponding duration to yield: Duration (min) Intensity (mm/h)

5 310

10 270

15 238

20 214

25 193

30 176

These points define the IDF curve for a return period of 40 years. 9.3. The ranked data were derived from a 8-year annual series (n = 8), where the return period, T , is given by 8+1 9 n+1 = = T = m m m Using this (Weibull) relation, the frequency distribution of the 10-min rainfall increments is given by Return period (y) 10-min rainfall (mm)

9 39.6

4.5 39.4

3.0 38.5

2.25 37.3

1.8 36.5

9.4. The calculated frequency distribution for the partial-duration series can be converted to the frequency distribution for the annual series by using the factors in Table 9.1. Applying linear interpolation in Table 9.1 leads to the following conversion factors: Return period (y) Factor

51 1.00

25.5 1.00

17 0.99

Interpolating from these data gives a conversion factor of 1.00 for the rainfall amounts corresponding to a 40-year return period. Hence, the IDF curve calculated from the partialduration series is the same as for the annual series, and (from Problem 9.2) is given by: Duration (min) Intensity (mm/h)

5 310

10 270

15 238

20 214

25 193

30 176

9.5. After rank-ordering the given data and dividing by the corresponding rainfall durations, the following results are obtained: Return Period (y) 5 10 25

1 2.2 2.7 3.2

2 1.6 2.0 3.4

Duration 4 6 1.0 0.7 1.1 0.8 1.8 1.2

The IDF equation to be matched has the form i=

a t+b

which can be put in the more convenient linear form 1 b 1 = t+ i a a

272

(h) 10 0.5 0.6 0.8

12 0.4 0.5 0.7

24 0.2 0.3 0.4

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Plotting 1/i versus t for a 5-year return period gives 1 = 0.174t + 0.342 i which yields a = 5.75 and b = 1.97 . 10 = 5.7 in., and R100 = 3.75 in.. For a return period, T , equal to 10 years, 9.6. R110 = 2.7 in., R24 1 then

1 1 =− = 9.49 years ln(1 − 1/T ) ln(1 − 1/10) 3.75 R100 x = 110 = = 1.39 2.7 R1

Tp = −

10 = 2.7/5.7 = 0.47 = 47%, then a = 30, b = 9.5, c = 0.81, and Equation 9.4 Since R110 /R24 1 1 1 gives

a = a1 R110 [(x − 1) log(Tp /10) + 1] = (30)(2.7)[(1.39 − 1) log(9.49/10) + 1] = 80.3 and therefore the 10-year IDF curve for Atlanta is given by i=

a 80.3 = c 1 (t + b1 ) (t + 9.5)0.81

where i is the rainfall intensity in inches/hour and t is the time in minutes. 10 = 5 in., and R100 = 3 inches. For a return period, 9.7. (a) New York City: R110 = 2.1 in., R24 1 T , equal to 10 years, then

1 1 =− = 9.49 years ln(1 − 1/T ) ln(1 − 1/10) R100 3 x = 110 = = 1.43 2.1 R1

Tp = −

10 = 2.1/5 = 0.42 = 42%, then a = 23.5, b = 7.7, c = 0.75, and Equation Since R110 /R24 1 1 1 9.4 gives

a = a1 R110 [(x − 1) log(Tp /10) + 1] = (23.5)(2.1)[(1.43 − 1) log(9.49/10) + 1] = 48.9 and therefore the 10-year IDF curve for New York is given by i=

a 48.9 = (t + b1 )c1 (t + 7.7)0.75

Wenzel (1982) gives the 10-year IDF curve for New York as (Table 9.2): i=

51.4 (t + 7.85)0.75

This is fairly close to the calculated IDF curve.

273

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10 = 6 in., and R100 = 3 inches. For a return period, T , (b) Los Angeles: R110 = 1.2 in., R24 1 equal to 10 years, then

1 1 =− = 9.49 years ln(1 − 1/T ) ln(1 − 1/10) 3 R100 x = 110 = = 2.5 1.2 R1

Tp = −

10 = 1.2/6 = 0.20 = 20%, then a = 9.0, b = 1.0, c = 0.50, and Equation Since R110 /R24 1 1 1 9.4 gives

a = a1 R110 [(x − 1) log(Tp /10) + 1] = (9.0)(1.2)[(2.5 − 1) log(9.49/10) + 1] = 10.4 and therefore the 10-year IDF curve for Los Angeles is given by i=

a 10.4 = c 1 (t + b1 ) (t + 1.0)0.50

Wenzel (1982) gives the 10-year IDF curve for Los Angeles as (Table 9.2): i=

10.9 (t + 1.15)0.51

This is fairly close to the calculated IDF curve. 9.8. Using the inverse-distance-squared method the weights, λi , are assigned to each cell in Figure 9.12 using the relation λi =

1 d2i ∑5 1 i=1 d2i

(1)

where di is the distance from the center of the grid cell to station i. The station weights assigned to each cell, calculated using Equation 1, are given Columns 3 to 7 in Table 9.1, and area, Ai of the catchment contained in each cell, i, is shown in Column 8. The rainfall amount, Pi , assigned to each cell is equal to the weighted average of the rainfall measurements, and this weighted average is given in Column 9. The total area, A, of the catchment is obtained by summing the values in Column 8 and is equal to 28.91 km2 . The average rainfall over the catchment, P , is given by 1 1 ∑ Pi Ai = 1691 = 58 mm P = A 28.91 The average rainfall over the entire catchment is therefore equal to 58 mm . 9.9. When equal weights are assigned to each rain gage, the estimated rainfall over the catchment area is simply equal to the arithmetic average of the measured rainfall at the gages. The average rainfall, P , over the catchment is therefore given by P =

60 + 90 + 65 + 35 + 20 = 54 mm 5

The rainfall calculated using the inverse-distance-squared method is probably more accurate, since it accounts for the spatial distribution of the rain gages in the catchment.

274

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Table 9.1: Calculation of Catchment-Averaged Rainfall (1) Row

(2) Column

1

1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6

2

3

4

5

6

7

(3) A 0.05 0.03 0.02 0.01 0.02 0.04 0.07 0.05 0.02 0.00 0.02 0.05 0.06 0.05 0.05 0.03 0.04 0.06 0.02 0.02 0.08 0.07 0.05 0.07 0.10 0.10 0.14 0.05 0.01 0.06 0.45 0.47 0.32 0.09 0.03 0.08 0.83 0.93 0.64 0.30 0.17 0.15

(5) (4) Weights, B C 0.21 0.07 0.13 0.05 0.07 0.04 0.04 0.03 0.06 0.06 0.09 0.13 0.40 0.08 0.29 0.08 0.10 0.05 0.00 0.00 0.05 0.07 0.09 0.17 0.69 0.06 0.60 0.08 0.28 0.12 0.11 0.13 0.09 0.22 0.10 0.30 0.93 0.02 0.91 0.03 0.46 0.22 0.18 0.45 0.10 0.58 0.10 0.54 0.79 0.05 0.74 0.09 0.37 0.35 0.08 0.81 0.02 0.94 0.07 0.74 0.37 0.09 0.31 0.14 0.22 0.37 0.07 0.78 0.03 0.91 0.07 0.75 0.09 0.05 0.03 0.03 0.11 0.20 0.11 0.50 0.09 0.66 0.09 0.63

(6) λi D 0.22 0.28 0.50 0.77 0.70 0.55 0.23 0.37 0.71 1.00 0.80 0.56 0.11 0.19 0.47 0.67 0.58 0.44 0.02 0.03 0.19 0.25 0.22 0.23 0.04 0.05 0.10 0.05 0.02 0.10 0.05 0.05 0.07 0.04 0.02 0.07 0.02 0.01 0.04 0.06 0.06 0.09

(7) E 0.45 0.51 0.37 0.15 0.16 0.19 0.23 0.22 0.11 0.00 0.07 0.13 0.08 0.08 0.08 0.06 0.07 0.10 0.01 0.01 0.05 0.05 0.05 0.06 0.02 0.02 0.04 0.02 0.01 0.03 0.03 0.03 0.03 0.02 0.01 0.03 0.01 0.00 0.02 0.03 0.03 0.04

Total

(8) Area, Ai (km2 ) 0.01 0.7 0.95 0.9 0.8 0.1 0.15 1 1 1 1 0.3 0.7 1 1 1 1 0.2 0.95 1 1 1 1 0.4 0.6 1 1 1 1 0.75 0 0.7 1 1 1 0.35 0 0.15 0.55 0.45 0.2 0 28.91

275

(9) Rainfall, Pi (mm) 43 37 35 36 38 42 58 51 41 35 39 44 75 70 54 45 47 50 87 86 68 59 58 57 82 81 69 64 64 62 69 68 66 64 64 63 62 61 62 63 63 62

(10) Pi × Ai (km2 ·mm) 0 26 33 32 30 4 9 51 41 35 39 13 53 70 54 45 47 10 83 86 68 59 58 23 49 81 69 64 64 47 0 48 66 64 64 22 0 9 34 28 13 0 1691

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9.10. For a 40-min storm, td = 40 min (= 0.667 h) and the average intensity, iave , is given by iave =

1628 1628 = = 85.7 mm/h (td + 8.16)0.76 (40 + 8.16)0.76

The peak rainfall intensity, ipeak , is therefore given by ipeak = 2.0iave = 2.0(85.7) = 171 mm/h From the given information, this peak rainfall intensity occurs at tp = 0.38td = 0.38(40) = 15 min. Therefore, the triangular hyetograph is described by td = 40 min,

tp = 15 min,

ipeak = 171 mm/h

9.11. For a 50-min storm, td = 50 min (= 0.833 h) and the average intensity, iave , is given by iave =

818 818 = = 37.1 mm/h 0.76 (td + 8.54) (50 + 8.54)0.76

The peak rainfall intensity, ipeak , is therefore given by ipeak = 2.0iave = 2.0(37.1) = 74.2 mm/h From the given information, this peak rainfall intensity occurs at tp = 0.44td = 0.44(50) = 22 min. Therefore, the triangular hyetograph is described by td = 50 min,

tp = 22 min,

ipeak = 74.2 mm/h

9.12. For a 60-minute storm with 11 time intervals, the time increment, ∆t, is given by 60 = 5.45 min 11 The average intensities for storm durations equal to multiples of ∆t are derived from the IDF curve using t = ∆t, 2∆t, . . . , 11∆t and the results are given in Column 3 of the following table: ∆t =

(1) Increment

1 2 3 4 5 6 7 8 9 10 11

(2) t

(3) i

(4) it

(min) 5.45 10.91 16.36 21.82 27.27 32.73 38.18 43.64 49.09 54.55 60.00

(mm/h) 110.1 85.74 71.06 61.12 53.92 48.40 44.05 40.50 37.55 35.06 32.92

(mm) 10.01 15.59 19.38 22.23 24.50 26.40 28.03 29.46 30.73 31.87 32.92

276

(5) Rainfall Amount (mm) 10.01 5.58 3.79 2.85 2.27 1.90 1.63 1.43 1.27 1.14 1.05

(6) Intensity (mm/h) 110 61.4 41.7 31.4 25.0 20.9 17.9 15.7 14.0 12.6 11.6

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The precipitation for each rainfall duration, t, is given in Column 4 (= Col.2 × Col.3), the rainfall amounts corresponding to the duration increments are given in Column 5, and the corresponding intensities are given in Column 6. In accordance with the alternating-block method, the maximum intensity (= 110 mm/h) is placed at the center of the storm, and the other intensities are arranged in descending order alternately to the right and left of the center block. The alternating-block hyetograph is therefore given by: Time (min) 0–5.45 5.45–10.91 10.91–16.36 16.36–21.82 21.82–27.27 27.27–32.73 32.73–38.18 38.18–43.64 43.64–49.09 49.09–54.55 54.55–60.00

Average Intensity (mm/h) 11.6 14.0 17.9 25.0 41.7 110 61.4 31.4 20.9 15.7 12.6

10 = 5.0 in., and R100 = 2.8 inches. For a return 9.13. For Boston, Massachusetts, R110 = 1.9 in., R24 1 period, T , equal to 10 years, then

1 1 =− = 9.49 years ln(1 − 1/T ) ln(1 − 1/10) R100 2.8 x = 110 = = 1.47 1.9 R1

Tp = −

10 = 1.9/5.0 = 0.38 = 38%, then a = 21, b = 7.0, c = 0.73, and Equation 9.4 Since R110 /R24 1 1 1 gives

a = a1 R110 [(x − 1) log(Tp /10) + 1] = (21)(1.9)[(1.47 − 1) log(9.49/10) + 1] = 39.5 and therefore the 10-year IDF curve for Boston is given by i=

1003 a 39.5 in./h = mm/h = c 0.73 1 (t + b1 ) (t + 7.0) (t + 7.0)0.73

For a 40-minute storm with 9 time intervals, the time increment, ∆t, is given by ∆t =

40 = 4.44 min 9

The average intensities for storm durations equal to multiples of ∆t are derived from the IDF curve using t = ∆t, 2∆t, . . . , 9∆t and the results are given in Column 3 of the following table:

277

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(1) Increment

(2) t

(3) i

(4) it

1 2 3 4 5 6 7 8 9

(min) 4.44 8.89 13.33 17.78 22.22 26.67 31.11 35.56 40.00

(mm/h) 169.30 133.19 111.27 96.30 85.38 76.99 70.33 64.88 60.35

(mm) 12.53 19.73 24.72 28.54 31.62 34.22 36.47 38.45 40.23

(5) Rainfall Amount (mm) 12.53 7.20 4.99 3.82 3.08 2.60 2.25 1.98 1.78

(6) Intensity (mm/h) 169 97.3 67.4 51.6 41.6 36.2 30.4 26.8 24.1

The precipitation for each rainfall duration, t, is given in Column 4 (= Col.2 × Col.3), the rainfall amounts corresponding to the duration increments are given in Column 5, and the corresponding intensities are given in Column 6. In accordance with the alternating-block method, the maximum intensity (= 169 mm/h) is placed at the center of the storm, and the other intensities are arranged in descending order alternately to the right and left of the center block. The alternating-block hyetograph is therefore given by: Time (min) 0–4.44 4.44–8.89 8.89–13.33 13.33–17.78 17.78–22.22 22.22–26.67 26.67–31.11 31.11–35.56 35.56–40.00

Average Intensity (mm/h) 24.1 30.4 41.6 67.4 169 97.3 51.6 36.2 26.8

9.14. From Wenzel (1982) for 10-year storms in Atlanta: i=

1628 64.1 in./h = mm/h (t + 8.16)0.76 (t + 8.16)0.76

For a 50-minute storm with 7 time intervals, the time increment, ∆t, is given by ∆t =

50 = 7.143 min = 0.119 h 7

The average intensities for storm durations equal to multiples of ∆t are derived from the IDF curve using t = ∆t, 2∆t, . . . , 7∆t and the results are given in Column 3 of the following table:

278

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(1) Increment

(2) t

(3) i

(4) it

1 2 3 4 5 6 7

(min) 7.143 14.29 21.43 28.57 33.71 42.86 50.00

(mm/h) 204.8 153.0 124.0 105.3 92.0 82.0 74.22

(mm) 24.37 36.44 44.31 50.12 54.73 58.57 61.85

(5) Rainfall Amount (mm) 24.37 12.07 7.87 5.81 4.61 3.84 3.28

(6) Intensity (mm/h) 204.8 101.4 66.1 48.8 38.7 32.3 27.6

The precipitation for each rainfall duration, t, is given in Column 4 (= Col.2 × Col.3), the rainfall amounts corresponding to the duration increments are given in Column 5, and the corresponding intensities are given in Column 6. In accordance with the alternating-block method, the maximum intensity (= 204.8 mm/h) is placed at the center of the storm, and the other intensities are arranged in descending order alternately to the right and left of the center block. The alternating-block hyetograph is therefore given by:

Time (min) 0–7.14 7.14–14.29 14.29–21.43 21.43–28.57 28.57–35.71 35.71–42.86 42.86–50.00

Average Intensity (mm/h) 27.6 38.7 66.1 204.8 101.4 48.8 32.3

9.15. For the rainfall in all intervals less than the storm duration to have the same return period, the maximum rainfalls for these time intervals must all occur in the same storm. This is very unlikely, and is therefore a conservative assumption.

9.16. In Miami, Florida, 24-hour storms are characterized by Type III rainfall, and the hyetograph is determined by multiplying the ordinates of the Type III hyetograph in Table 9.4 by PT = 260 mm to yield the following select points:

279

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Time

P/PT

(h) 0 3 6 9 12 15 18 21 24

0 0.032 0.072 0.148 0.500 0.848 0.922 0.968 1.000

Cumulative Precipitation, P (mm) 0 8.32 18.7 38.5 130 220 240 252 260

9.17. In Atlanta, Georgia, 24-hour storms are characterized by Type II rainfall, and the hyetograph is determined by multiplying the ordinates of the Type II hyetograph in Table 9.4 by PT = 175 mm. This hyetograph is compared with the Miami hyetograph (Problem 9.16) in the following table: Time (h) 0 3 6 9 12 15 18 21 24

P/PT (Type II) 0 0.035 0.080 0.147 0.663 0.856 0.922 0.965 1.000

Cumulative Precipitation (mm) Atlanta Miami 0 0 6.13 8.32 14.0 18.7 25.7 38.5 116 130 150 220 161 240 169 252 175 260

9.18. The rainfall intensity, i, is given by dP d i= = P24 dt dt

(

P P24

)

Using the given function yields

i = P24 ∗

 0.110e−0.504(12−t) + 1.007e−6.666(12−t) + 0.0112 t ≤ 12 h 0.110e−0.504(t−12) + 1.007e−6.666(t−12) + 0.0112 12 < t ≤ 24 h

Using the Wenzel (1982) 10-year IDF curve for Miami with t = 24 h = 1440 min yields 79.9 79.9 = = 0.393 in./h = 9.98 mm/h (t + 7.24)0.73 (1440 + 7.24)0 .73 = iave × 24 h = (9.98)(24) = 240 mm

iave = P24

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From the intensity function, the maximum rainfall intensity occurs when t = 12 h, which yields imax = P24 (0.110 + 1.007 + 0.0112) = 240(1.128) = 271 mm/h = 10.7 in./h 9.19. From the given data: td = 10 h, A = 200 km2 = 77 mi2 , and the areal-reduction factor is given by Equation 9.17 as 1

1

F = 1 − exp[−1.1td4 ] + exp[−1.1td4 − 0.01A] 1

1

= 1 − exp[−1.1(10) 4 ] + exp[−1.1(10) 4 − 0.01(77)] = 0.924 Therefore, for a local average rainfall of 193 mm, the average precipitation over a 200 km2 catchment is expected to be 0.924 × 193 mm = 178 mm . 9.20. Equation 9.17 can be written as 1 4

1 4

F = 1 − e−1.1td + e−1.1td e−0.01A

(1)

As td → ∞ the exponential terms involving td approach zero, and Equation 1 becomes F = 1 − 0 + 0 · e−0.01A = 1 Therefore, for long duration storms (td → ∞) the areal-reduction factor, F , is independent of the storm duration, td . Under these asymptotic conditions, F = 1 and therefore the areal-reduction factor, F , is also independent of the catchment area. 9.21. From the given data: P¯ = 202 mm, and σP = 65 mm. Taking km = 20 gives the probable maximum precipitation, Pm , as Pm = P¯ + km σP = 202 + (20)(65) = 1502 mm The PMP at this location is considerably less than the world’s largest observed 24-hour rainfall, which is given in Table 6.6 as 1870 mm. 9.22. The Chen method gives the IDF curve as i=

a (t + b1 )c1

(1)

where a is a constant given by a = a1 R110 [(x − 1) log(Tp /10) + 1]

(2)

10 derived from Figure 6.5, x is defined by a1 , b1 , and c1 are empirical functions of R110 /R24

x=

281

R1100 R110

(3)

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and Tp is the return period for the partial duration series, which is assumed to be related to return period, T , for the annual maximum series by the relation Tp = −

1 ln(1 − 1/T )

(4)

10 , and R100 are all constant (for a given location) and independent of Noting that R110 , R24 1 the return period, T , then Equations 2 to 4 collectively indicate that the constant a in the numerator of Equation 1 depends only on the return period, T . In other words, for a given location, a = f (T ) (5) 10 derived from Figure 9.10, then Similarly, since b1 and c1 are empirical functions of R110 /R24 b1 and c1 are constants that depend only on location and hence (t + b1 )c1 depends only on the duration, t, of the storm. In other words, for a given location,

(t + b1 )c1 = g(t)

(6)

Combining Equations 1, 5 and 6 gives that the Chen method always leads to an IDF curve of the form f (T ) i= g(t) 9.23. From the given data: P¯ = 320 mm, and σP = 156 mm. For an extreme storm with a return period, T , of 50,000 years, Equation 9.21 gives { [ } ] 1 0.13(km − 0.44) −7.69 F (km ) = 1 − = exp − 1 + (1) 50000 0.60 where km is the frequency factor with an exceedance probability of 50,000 years. Solving Equation 1 gives km = 14.67, and therefore the 24-hour rainfall extreme, P24 , with a return period of 50,000 years is given by Equation 9.20 as P24 = P¯ + km σP = 320 + 14.67(156) = 2609 mm

(2)

The IDF curve for the island is given by the Chen method in the form i=

f (T ) f (T ) = g(t) (t + 6.87)0.65

where g(t) is given as g(t) = (60t + 6.87)0.65

(3)

where t is in hours. Combining Equations 9.24, 2, and 3 gives the 8-hour rainfall as P8 =

8 g(24) 8 (60 × 24 + 6.87)0.65 P24 = (2609) = 1765 mm 24 g(8) 24 (60 × 8 + 6.87)0.65

Therefore, the 8-hour annual-maximum rainfall with a return period of 50,000 years is 1765 mm .

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9.24. According to Equation 9.28, for a 1-hour storm (td = 1 h) the maximum observed rainfall amount, Pm,obs , is given by Pm,obs = 422t0.475 = 422(1)0.475 = 422 mm d 9.25. Equation 9.28 gives the probable maximum precipitation, Pm,obs (in cm), as Pm,obs = 42.2t0.475 d where td is the storm duration in hours. Comparing the predictions given by Equation 9.28 with the observations in Table 9.5 gives: Duration 1 min 20 min 1 day 1 week 1 month 1 year

Observed (cm) 3.8 20.6 187 411 930 2650

Equation 9.28 (cm) 6.0 25.0 191 481 961 3148

Clearly, rainfall amounts given by Equation 9.28 consistently exceed observed maximum rainfall amounts. 9.26. For a 30-min storm, the average intensity, i, is given by the IDF equation as i=

2819 2819 = = 61 mm/h t + 16 30 + 16

and the precipitation amount, P , is given by P = it = (61)(

30 ) = 31 mm 60

The interception, I, of the wooded area can be estimated by Equation 9.31, where S = 5 mm, P = 31 mm, K = 6, E = 0.3 mm/h, and t = 30/60 h = 0.50 h, hence I = S(1 − e−P/S ) + KEt = 5(1 − e−31/5 ) + (6)(0.3)(0.50) = 5.0 + 0.9 = 5.9 mm The wooded area intercepts approximately 5.9 mm of the 31 mm that falls on the wooded area. Prior to clearing the wooded area, the rainfall reaching the ground in a 30-min storm is 31 − 5.9 = 25.1 mm. After clearing the wooded area, the rainfall reaching the ground is expected to be 31 mm, an increase of 24% over the incident rainfall prior to clearing the wooded area. 9.27. The hyetograph calculated in Problem 9.14 is:

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Time (min) 0–7.14 7.14–14.29 14.29–21.43 21.43–28.57 28.57–35.71 35.71–42.86 42.86–50.00

Average Intensity (mm/h) 27.6 38.7 66.1 204.8 101.4 48.8 32.3

The interception equation is given by: I = S(1 − eP/S ) + KEt where S = 9 mm, K = 8, E = 0.5 mm/h. Substituting data gives ( ) 1 −P/9 I = 9(1 − e ) + 8(0.5)t = 9(1 − e−P/9 ) + 0.0667t 60 where t is in minutes. Using the rainfall hyetograph from Problem 9.14: t (min) 0

∆P (mm)

P (mm) 0

I (mm) 0

∆I (mm)

∆P − ∆I (mm)

Incident Rainfall (mm/h)

3.2

0.1

0.8

3.0

1.6

13.4

2.7

5.2

43.7

1.9

22.5

0.6

11.5

96.6

0.4

5.4

45.4

0.5

3.3

27.7

3.3 7.14

3.3

3.2

4.6 14.29

7.9

6.2

15.8

8.9

40.2

10.8

52.3

11.4

7.9 21.43 24.4 28.57 12.1 35.71 5.8 42.86

58.1

11.8

3.8 50.00

61.9

189

12.3

9.28. Using the interception formula I = a + bP n for pine woods, it can be assumed that n = 0.5, a = 0 mm, and b = 1.01, hence (for P = 31 mm) I = a + bP n = 0 + 1.01(31)0.5 = 6 mm 9.29. The runoff fraction is given by the relation: Runoff fraction = =

Rainfall − Depression Storage Rainfall 30 mm − Depression Storage 30 mm

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These calculations are summarized in the following table: Surface Type Pavement: steep flat Impervious Lawns Pasture Forest Litter

Depression Storage (mm)

Runoff Fraction

0.5 2.5 1.9 3.8 5.1 7.6

0.98 0.92 0.94 0.87 0.83 0.75

Depression storage is most significant on forest litter and least significant on steep pavement . 9.30. Since the rainfall rate always exceeds the maximum infiltration rate, then ponding begins from t = 0, and the infiltration rate, f , as a function of time is given by f = fc + (f0 − fc )e−kt = 60 + (200 − 60)e−4t which leads to

f = 60 + 140e−4t

0 ≤ t ≤ 50 min

At the end of 50 min, the infiltration rate is given by f = 60 + 140e−4(50) = 60 mm/h If the rainfall rate were less than 200 mm/h, then ponding would not begin at the start of the storm and the infiltration rate at the end of 50 min would be greater than that calculated for the 200 mm/h storm. 9.31. Substituting the Horton infiltration model for fp requires that d [fc + (f0 − fc )e−kt ] = −k[fc + (f0 − fc )e−kt − fc ] = −k(f0 − fc )e−kt dt Evaluating the derivative on the left-hand side of this equation yields −k(f0 − fc )e−kt = −k(f0 − fc )e−kt which verifies that the Horton infiltration model satisfies the given differential equation. 9.32. From the given data: f0 = 150 mm/h, fc = 50 mm/h, and k = 3 min−1 = 180 h−1 . According to Equation 9.49, the cumulative infiltration, F , is related to the infiltration capacity, fp , by the relation [ ] fp fc f0 fc F = ln(f0 − fc ) + − ln(fp − fc ) − k k k k [ ] fp 50 150 50 = ln(150 − 50) + − ln(fp − 50) − 180 180 180 180

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which yields F = 2.113 − 0.2778 ln(fp − 50) − 0.005555fp

(1)

where fp is in mm/h. Each 10-minute increment of the storm will now be taken sequentially. t = 0–10 min: During this period the rainfall intensity, i (= 20 mm/h), is less than the minimum infiltration rate, fc (= 50 mm/h), and no ponding occurs. The cumulative infiltration, F , after 10 min is given by ( ) 10 = 3.33 mm F = i∆t = 20 60 t = 10–20 min: During this period, the rainfall intensity, i (= 40 mm/h), is less than the minimum infiltration rate, fc (= 50 mm/h), and no ponding occurs. The cumulative infiltration, F , after 20 min is given by ( ) 10 + 3.33 = 10.00 mm F = i∆t + 3.33 = 40 60 t = 20–30 min: During this period, the rainfall intensity, i (= 80 mm/h), exceeds the minimum infiltration rate, fc (= 50 mm/h), and therefore ponding is possible. At t = 20 min, F = 10.00 mm and Equation 1 gives 10.00 = 2.113 − 0.2778 ln(fp − 50) − 0.005555fp which yields an infiltration capacity, fp , of fp = 50 mm/h Since the rainfall rate (80 mm/h) exceeds the infiltration capacity (50 mm/h) from the beginning of the time interval, ponding begins at the beginning the time interval, at t = 20 min . 9.33. From the given data: f0 = 600 mm/h, fc = 30 mm/h, k = 0.5 min−1 = 30 h−1 , and d = 4 mm. The Horton cumulative infiltration function is given by 30 600 − 30 f0 − fc (1 − e−kt ) = t + (1 − e−0.5t ) k 60 30 = 0.5t + 19(1 − e−0.5t ) mm/h

F (t) = fc t +

(1)

where t is in minutes, and the infiltration capacity is given by f (t) = fc + (f0 − fc )e−kt = 30 + 570e−0.5t mm/h t = 0–10 min: Since i = 7 mm/h < fc , all the rain infiltrates and F (10) = 7 ×

286

10 = 1.17 mm 60

(2)

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t = 10–20 min: Since i = 25 mm/h < fc , all the rain infiltrates and ( ) 10 F (20) = F (10) + i∆t = 1.17 + 25 = 5.34 mm 60 t = 20–30 min: Given i = 45 mm/h, if all the rain infiltrates ( ) 10 F (30) = F (20) + i∆t = 5.34 + 45 = 12.8 mm 60 Using Equation 1, this corresponds to t = 1.95 min, and Equation 2 gives f = 245 mm/h. Therefore all the rain infiltrates. t = 30–40 min: Given i = 189 mm/h, if all the rain infiltrates ( ) 10 F (40) = F (30) + i∆t = 12.8 + 189 = 44.3 mm 60 Using Equation 1, this corresponds to t = 50.6 min, and Equation 2 gives f = 30 mm/h. Therefore ponding occurs during this time interval. Ponding occurs when t = t′ and f = 189 mm/h, or Equation 2 requires that ′

189 = 30 + 570e−0.5t which gives t′ = 2.55 min and Equation 1 gives ′

F (t′ ) = 0.5t′ + 19(1 − e−0.5t ) = 0.5(2.55) + 19(1 − e−0.5×2.55 ) = 15.0 mm Therefore, ponding occurs at time t′′ , where 189

(t′′ − 30) = 15 − 12.8 60

which yields t′′ = 30.7 min. Ponding begins 30.7 min after the start of the storm. It is important to note that t′ = 2.55 min corresponds to t′′ = 30.7 min, and therefore at the end of the time interval, t = 40 min, and t′ = (40 − 30.7) + 2.55 = 11.9 min, which gives F (40) = 0.5(11.9) + 19(1 − e−0.5×11.9 ) = 24.9 mm and

f (40) = 30 + 570e−0.5×11.9 = 31.5 mm/h

The rainfall during this time interval is 189(10/60) = 31.5 mm. Infiltration during this period is F (40) − F (30) = 24.9 − 12.8 = 12.1 mm, Since the depression storage is 4 mm, the runoff is 31.5 − (12.1 + 4) = 15.4 mm. t = 40–50 min: At t = 50 min, t′ = 11.9 + 10 = 21.9 min, and F (50) = 0.5(21.9) + 19(1 − e−0.5×21.9 ) = 29.9 mm

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and f (50) = 30 + 570e−0.5×21.9 = 30 mm/h The rainfall during this time interval is 97(10/60) = 16.2 mm. Infiltration during this period is F (50) − F (40) = 29.9 − 24.9 = 5 mm, Since the depression storage is full, the runoff is 16.2 − 5 = 11.2 mm. t = 50–60 min: At t = 60 min, t′ = 21.9 + 10 = 31.9 min, and F (60) = 0.5(31.9) + 19(1 − e−0.5×31.9 ) = 34.9 mm The rainfall during this time interval is 44(10/60) = 7.3 mm. Infiltration during this period is F (60) − F (50) = 34.9 − 29.9 = 5 mm, Since the depression storage is full, the runoff is 7.3 − 5 = 2.3 mm. t = 60–70 min: The rainfall rate (18 mm/h) is less than the infiltration capacity (30 mm/h) and therefore all of the rainfall will infiltrate, plus what is in depression storage. The rainfall during this time interval is 18(10/60) = 3 mm, and 4 mm is in depression storage. At 30 mm/h, the infiltration capacity during this interval is 30(10/60) = 5 mm. Hence the 3 mm of rainfall plus 2 mm from depression storage will infiltrate. A depth of 2 mm will remain in depression storage. The results of this analysis are summarized in the following table: Time

Rainfall

Infiltration

Depression Storage (mm)

Runoff

(min) 0

(mm)

(mm)

1.17

1.17

0

0

4.17

4.17

0

0

7.50

7.50

0

0

(mm)

10 20 30 31.5

12.1

4

15.4

16.2

5.0

0

11.2

7.3

5.0

0

2.3

3.0

5.0

−2.0

70.8

39.9

2.0

40 50 60 70 Total

0 28.9

The total runoff from the storm is 28.9 mm , which is (28.9/70.8)×100 = 40.8% of the rainfall.

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9.34. The basic equations of the Green-Ampt model are Equations 9.66 and 9.67. In the present case: Ks = 11 mm/h; θi = 21 (0.190+0.085) = 0.138; n = 0.45; and Φf = 110 mm. The infiltration capacity, fp , as a function of the cumulative infiltration, F , is Ks (n − θi )Φf F 11(0.45 − 0.138)(110) = 11 + F 378 = 11 + F

fp = Ks +

The cumulative infiltration as a function of time is given by ] [ F ′ Ks (t − tp + t ) = F − (n − θi )Φf ln 1 + (n − θi )Φf [ ] F 11(t − tp + t′ ) = F − (0.45 − 0.138)(110) ln 1 + (0.45 − 0.138)(110) = F − 34.3 ln(1 + 0.0291F )

(1)

(2)

If ponding occurs from t = 0, then 11t = F − 34.3 ln(1 + 0.0291F )

(3)

Each 10-minute increment in the storm will now be taken sequentially, and the computation of the runoff is summarized at the end of the problem. t = 0 - 10 min: During this period the rainfall intensity, i (= 20 mm/h) is greater than the saturated hydraulic conductivity (= 11 mm/h) so ponding is possible. If all the rainfall infiltrates, then the cumulative infiltration, F , at the end of the time interval is given by F = i∆t = 20(0.167) = 3.34 mm The corresponding infiltration capacity is given by Equation 1 as fp = 11 +

378 = 124 mm/h 3.34

Since the infiltration capacity at the end of the interval exceeds the rainfall rate during the interval, then no ponding occurs during this time interval. t = 10 - 20 min: During this period the rainfall intensity, i (= 40 mm/h) is greater than the saturated hydraulic conductivity (= 11 mm/h) so ponding is possible. If all the rainfall infiltrates, then the cumulative infiltration, F , at the end of the time interval is given by F = 3.34 + i∆t = 3.34 + 40(0.167) = 10.0 mm The corresponding infiltration capacity is given by Equation 1 as fp = 11 +

378 = 48.8 mm/h 10.0

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Since the infiltration capacity at the end of the interval exceeds the rainfall rate during the interval, then no ponding occurs during this time interval. t = 20 - 30 min: During this period the rainfall intensity, i (= 60 mm/h) is greater than the infiltration capacity at the end of the previous time interval (= 48.8 mm/h), so ponding begins at tp = 20 min = 0.333 h. The next step is to find the time, t′ , that it would take for 10 mm to infiltrate, if infiltration occurs at the potential rate from t = 0. Infiltration as a function of time is given by Equation 3, therefore 11t′ = 10.0 − 34.3 ln[1 + (0.0291)(10.0)] which leads to

t′ = 0.113 h

Therefore, the equation for the cumulative infiltration as a function of time after t = 0.333 h is given by Equation 2, which can be written as 11(t − 0.333 + 0.113) = F − 34.3 ln(1 + 0.0291F ) or 11(t − 0.220) = F − 34.3 ln(1 + 0.0291F )

(4)

At the end of the current time period, t = 30 min = 0.500 h, and substituting this value into Equation 4 leads to F = 16.6 mm Since the rainfall during this period is 10 mm, and the cumulative infiltration up to the beginning of this period is 10 mm, then the amount of rainfall that does not infiltrate is equal to 10 + 10 − 16.6 = 3.4 mm. This excess amount goes towards filling up the depression storage, which has a maximum capacity of 4 mm. Since the depression storage is not completely filled, there is no runoff. t = 30 - 40 min: Since the rainfall rate (= 110 mm/h) is now higher than during the previous time interval, ponding continues to occur. The time at the end of this period is 0.667 h, and substituting this value for t into Equation 4 gives the cumulative infiltration at the end of the time period as F = 21.7 mm Since the cumulative infiltration up to the beginning of this period is 16.6 mm, then the infiltrated amount during this period is 21.7 − 16.6 = 5.1 mm. The rainfall during this period is 18.3 mm, therefore the amount of rain that does not infiltrate is 18.3 − 5.1 = 13.2 mm. Since there is 0.6 mm in available depression storage, then the amount of runoff is 13.2 − 0.6 = 12.6 mm. t = 40 - 50 min: The rainfall rate (60 mm/h) is still higher than the infiltration capacity and ponding continues. The time at the end of this period is 0.833 h, and Equation 4 gives the cumulative infiltration at the end of this time period as F = 26.1 mm

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The cumulative infiltration up to the beginning of this period is 21.7 mm, therefore the infiltration during this time interval is 26.1 - 21.7 = 4.4 mm. The rainfall during this period is 10.0 mm, therefore the amount of rain that does not infiltrate is 10.0 − 4.4 = 5.6 mm. Since the depression storage is full, then all 5.6 mm is contributed to runoff. t = 50 - 60 min: The infiltration capacity of the soil at t = 50 min is given by Equation 1 as 378 fp = 11 + = 25.5 mm/h 26.1 Since the infiltration capacity at the beginning of the time interval (= 25.5 mm/h) is greater than the rainfall rate (= 20 mm/h), then infiltration of depression storage is occurring at the beginning of the time interval. If infiltration occurs at the potential rate, then the cumulative infiltration at the end of the time interval (t = 1 h) is given by Equation 4 as F = 30.2 mm The cumulative infiltration up to the beginning of this period is 26.1 mm, therefore the infiltration during this time interval is 30.2 − 26.1 = 4.1 mm. The rainfall during this period is 3.3 mm, therefore all 3.3 mm of the rain infiltrates, plus 0.8 mm from depression storage. There is no runoff during this time interval. The total rainfall during this storm is 51.6 mm, and the total runoff is 18.2 mm. The runoff is therefore equal to 35% of the rainfall. These calculations are summarized in the following table: Time

F

∆t

∆F

i

i∆t

(min) 0

(mm) 0

(h)

(mm)

(mm/h)

10 20

(mm)

0.167

3.34

20

3.3

0.0

0

0

0.167

6.66

40

6.7

0.0

0

0

0.167

6.6

60

10.0

3.4

3.4

0

0.167

5.1

110

18.3

0.6

4.0

12.6

0.167

4.4

60

10.0

0.0

4.0

5.6

0.167

4.1

20

3.3

−0.8

3.2

0

3.34 10.0

30

16.6

40

21.7

50

26.1

60 Total:

Total Storage (mm)

Runoff

(mm)

Depression Storage, ∆S (mm)

30.2 51.6

18.2

9.35. In this case: Ks = 1 mm/h; θi = 12 (0.321+0.221) = 0.271; n = 0.43; and Φf = 240 mm. The

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infiltration capacity, fp , as a function of the cumulative infiltration, F , is Ks (n − θi )Φf F 1(0.43 − 0.271)(240) =1+ F 38.2 =1+ F

fp = Ks +

The cumulative infiltration as a function of time is given by [ ] F ′ Ks (t − tp + t ) = F − (n − θi )Φf ln 1 + (n − θi )Φf [ ] F ′ 1(t − tp + t ) = F − (0.43 − 0.271)(240) ln 1 + (0.43 − 0.271)(240) = F − 38.2 ln(1 + 0.0262F )

(1)

(2)

If ponding occurs from t = 0, then t = F − 38.2 ln(1 + 0.0262F )

(3)

Each 10-minute increment in the storm will now be taken sequentially, and the computation of the runoff is summarized at the end of the problem. t = 0 - 10 min: During this period the rainfall intensity, i (= 20 mm/h) is greater than the saturated hydraulic conductivity (= 1 mm/h) so ponding is possible. If all the rainfall infiltrates, then the cumulative infiltration, F , at the end of the time interval is given by F = i∆t = 20(0.167) = 3.34 mm The corresponding infiltration capacity is given by Equation 1 as fp = 1 +

38.2 = 12.4 mm/h 3.34

Since the infiltration capacity at the end of the interval is less than the rainfall rate, ponding begins during this time interval. Determine the cumulative infiltration rate, F20 , corresponding to a potential infiltration rate of 20 mm/h: 20 = 1 +

38.2 F20

which leads to F20 = 2.01 mm Let tp be the time when ponding begins, then 20tp = 2.01 which leads to tp = 0.101 h

292

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So ponding begins at tp = 0.101 h. The next step is to find the time, t′ , that it would take for 2.01 mm to infiltrate, if infiltration occurs at the potential rate from t = 0. Infiltration as a function of time is given by Equation 3, therefore t′ = 2.01 − 38.2 ln[1 + (0.0262)(2.01)] which leads to

t′ = 0.049 h

Therefore, the equation for the cumulative infiltration as a function of time after t = 0.101 h is given by Equation 2, which can be written as (t − 0.101 + 0.049) = F − 38.2 ln(1 + 0.0262F ) or t − 0.052 = F − 38.2 ln(1 + 0.0262F )

(4)

At the end of the current time period, t = 10 min = 0.167 h, and substituting this value into Equation 4 leads to F = 3.1 mm Since the rainfall during this period is 3.3 mm, then the amount of rainfall that does not infiltrate is equal to 3.3 − 3.1 = 0.2 mm. This excess amount goes towards filling up the depression storage, which has a capacity of 9 mm. Since the depression storage is not full, there is no runoff. t = 10 - 20 min: Since the rainfall rate (= 40 mm/h) is now higher than during the previous time interval, ponding continues to occur. The time at the end of this period is 0.333 h, and substituting this value for t into Equation 4 gives the cumulative infiltration at the end of the time period as F = 4.9 mm Since the cumulative infiltration up to the beginning of this period is 3.1 mm, then the infiltrated amount during this period is 4.9 − 3.1 = 1.8 mm. The rainfall during this period is 6.7 mm, therefore the amount of rain that does not infiltrate is 6.7 − 1.8 = 4.9 mm. Since there is 8.8 mm in available depression storage, then there is no runoff. t = 20 - 30 min: Since the rainfall rate (= 60 mm/h) is higher than during the previous time interval, ponding continues to occur. The time at the end of this period is 0.500 h, and substituting this value for t into Equation 4 gives the cumulative infiltration at the end of the time period as F = 6.2 mm Since the cumulative infiltration up to the beginning of this period is 4.9 mm, then the infiltrated amount during this period is 6.2 − 4.9 = 1.3 mm. The rainfall during this period is 10 mm, therefore the amount of rain that does not infiltrate is 10 − 1.3 = 8.7 mm. Since there is 3.9 mm in available depression storage, then the runoff is 8.7 − 3.9 = 4.8 mm.

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t = 30 - 40 min: Since the rainfall rate (= 110 mm/h) is higher than during the previous time interval, ponding continues to occur. The time at the end of this period is 0.667 h, and substituting this value for t into Equation 4 gives the cumulative infiltration at the end of the time period as F = 7.3 mm

Since the cumulative infiltration up to the beginning of this period is 6.2 mm, then the infiltrated amount during this period is 7.3 − 6.2 = 1.1 mm. The rainfall during this period is 18.3 mm, therefore the amount of rain that does not infiltrate is 18.3 − 1.1 = 17.2 mm. Since there is no available depression storage, then the amount of runoff is 17.2 mm.

t = 40 - 50 min: Since the rainfall rate (= 60 mm/h) exceeds the infiltration capacity ponding continues to occur. The time at the end of this period is 0.833 h, and substituting this value for t into Equation 4 gives the cumulative infiltration at the end of the time period as

F = 8.3 mm

Since the cumulative infiltration up to the beginning of this period is 7.3 mm, then the infiltrated amount during this period is 8.3 − 7.3 = 1.0 mm. The rainfall during this period is 10.0 mm, therefore the amount of rain that does not infiltrate is 10.0 − 1.0 = 9.0 mm. Since there is no available depression storage, the amount of runoff is 9.0 mm.

t = 50 - 60 min: Since the rainfall rate (= 20 mm/h) exceeds the infiltration capacity ponding continues to occur. The time at the end of this period is 1.000 h, and substituting this value for t into Equation 4 gives the cumulative infiltration at the end of the time period as

F = 9.2 mm

Since the cumulative infiltration up to the beginning of this period is 8.3 mm, then the infiltrated amount during this period is 9.2 − 8.3 = 0.9 mm. The rainfall during this period is 3.3 mm, therefore the amount of rain that does not infiltrate is 3.3 − 0.9 = 2.4 mm. Since there is no available depression storage, then the amount of runoff is 2.4 mm.

The total rainfall during this storm is 51.6 mm, and the total runoff is 33.4 mm. The runoff is therefore equal to 65% of the rainfall. These calculations are summarized in the following table:

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Time

F

∆t

∆F

i

i∆t

(min) 0

(mm) 0

(h)

(mm)

(mm/h)

0.167

3.1

0.167

10

20

3.3

0.2

0.2

0

1.8

40

6.7

4.9

5.1

0

0.167

1.3

60

10.0

3.9

9.0

4.8

0.167

1.1

110

18.3

0.0

9.0

17.2

0.167

1.0

60

10.0

0.0

9.0

9.0

0.167

0.9

20

3.3

0.0

9.0

2.4

(mm)

3.1

20

4.9

30

6.2

40

7.3

50

8.3

60 Total:

Total Storage (mm)

Runoff

(mm)

Depression Storage, ∆S (mm)

9.2 51.6

33.4

9.36. Differentiating Equation 9.84 with respect to t, taking S as a constant, gives the infiltration rate, f , as dP (P + 0.8S)S dP dF dt − (P − 0.2S)S dt f= (1) = dt (P + 0.8S)2 Since the rainfall intensity, i, is related to the cumulative precipitation, P , by i=

dP dt

(2)

combining Equations 1 and 2 gives (P + 0.8S)Si − (P − 0.2S)Si (P + 0.8S)2 P Si + 0.8S 2 i − P Si + 0.2S 2 i = (P + 0.8S)2 S2i = (P + 0.8S)2

f=

This is the NRCS Curve Number model for infiltration rate (Equation 9.85). This model is unrealistic since it requires the infiltration rate to depend on the rainfall intensity. 9.37. The amount of rainfall can be estimated using the IDF curve. For a 1-hour 10-year storm, the average intensity is given by i=

203 = 9.40 cm/h (60 + 7.24)0.73

and the total amount of rainfall, P , in the storm is equal to (9.40)(1) = 9.40 cm. Since the minimum infiltration rate of the soil is 10 mm/h, then according to Table 9.15 it can be

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inferred that the soil is in Group A. The description of the area, residential with lot sizes on the order of 0.5 ac, and Group A soil is cited in Table 9.14 to have a curve number, CN, equal to 54. The curve number, CN, and soil storage, S (in mm), are related by Equation 9.81 as CN =

1000 10 + 0.0394S

and therefore the the storage, S, in this case is given as ( ) ( ) 1 1000 1 1000 S= − 10 = − 10 = 216 mm 0.0394 CN 0.0394 54 The runoff amount, Q, can be calculated from the rainfall amount, P (= 9.40 cm), and the maximum storage, S (= 21.6 cm) using Equation 9.80 where Q=

(P − 0.2S)2 , P + 0.8S

P > 0.2S

Since P (= 9.40 cm) > 0.2S (= 4.32 cm), then this equation is valid and Q=

[9.40 − 0.2(21.6)]2 = 0.97 cm 9.40 + 0.8(21.6)

Hence, there is 0.97 cm of runoff from the storm with a rainfall amount of 9.40 cm. 9.38. The amount of rainfall can be estimated using the IDF curve. For a 1-hour 10-year storm, the average intensity is given by i=

203 = 9.40 cm/h (60 + 7.24)0.73

and the total amount of rainfall, P , in the storm is equal to (9.40)(1) = 9.40 cm. Since the minimum infiltration rate of the soil is 10 mm/h, then according to Table 9.15 it can be inferred that the soil is in Group A. The description of the area, residential with lot sizes on the order of 0.5 ac, and Group A soil is cited in Table 9.14 to have an average curve number, CN, equal to 54. According to Table 9.16, for wet antecedent conditions the corresponding curve number can be taken as 74. The curve number, CN, and soil storage, S (in mm), are related by Equation 9.81 as 1000 CN = 10 + 0.0394S and therefore the the storage, S, in this case is given as ( ) ( ) 1 1000 1 1000 S= − 10 = − 10 = 89.2 mm 0.0394 CN 0.0394 74 The runoff amount, Q, can be calculated from the rainfall amount, P (= 9.40 cm), and the maximum storage, S (= 8.92 cm) using Equation 9.80 where Q=

(P − 0.2S)2 , P + 0.8S

296

P > 0.2S

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Since P (= 9.40 cm) > 0.2S (= 1.78 cm), then this equation is valid and

Q=

[9.40 − 0.2(8.92)]2 = 3.51 cm 9.40 + 0.8(8.92)

Hence, there is 3.51 cm of runoff from the storm with a rainfall amount of 9.40 cm. This runoff significantly exceeds the runoff of 0.97 cm under average conditions. 9.39. From the given data: td = 4.5 h, T = 10 y, d = 5 mm, f0 = 80 mm/h, fc = 8 mm/h, and k = 0.5 h−1 . The IDF curve for 10-y storms in Atlanta is

i=

64.1 in./h (t + 8.16)0.76

The formula for cumulative infiltration, F , is given by [

] fp fc f0 fc F = ln(f0 − fc ) + − ln(fp − fc ) − k k k k ] [ fp 80 8 8 ln(80 − 8) + − ln(fp − 8) − = 0.5 0.5 0.5 0.5 F = 228.4 − 16 ln(fp − 8) − 2fp The rainfall-intensities derived from the IDF curve at 30-min intervals are summarized in the following table:

Interval 1 2 3 4 5 6 7 8 9

time (min) 30 60 90 120 150 180 210 240 270

¯i (mm/h) 102.3 65.8 49.9 40.7 34.7 30.4 27.2 24.6 22.6

¯it (mm) 51.1 65.8 74.8 81.4 86.8 91.2 95.1 98.6 101.7

∆P (mm) 51.1 14.7 9.0 6.6 5.3 4.5 3.9 3.4 3.1

i (mm/h) 102.3 29.3 18.0 13.3 10.6 9.0 7.8 6.9 6.2

Based on the above rainfall intensities, the alternating-block hyetograph and runoff calculations are given in the following table:

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t (min) 0

∆t (h)

i∆t (mm)

∆F (mm)

F (mm)

fp (mm/h)

∆S (mm)

S (mm)

Q (mm)

6.2

0.5

3.1

3.1

3.1

78.6

0

0

0

7.8

0.5

3.9

3.9

7.0

76.9

0

0

0

10.6

0.5

5.3

5.3

12.3

74.5

0

0

0

18.0

0.5

9.0

9.0

21.3

70.5

0

0

0

102.3

0.5

51.1

31.6

53.0

56.7

5

5

14.5

29.3

0.5

14.7

19.7

72.6

48.3

0

0

0

13.3

0.5

6.6

6.6

79.3

45.6

0

0

0

9.0

0.5

4.5

4.5

83.7

43.7

0

0

0

6.9

0.5

3.4

3.4

87.2

42.3

0

0

0

i (mm/h)

30 60 90 120 150 180 210 240 270 Total

101.7

14.5

From the above table, ponding occurs within the 120–150 min time interval. @t = 120 min: F = 21.3 mm, i = 102.3 mm/h, fp = 70.5 mm/h and therefore ponding begins at t = 120 min. Using the Horton equation for infiltration at capacity,

F = fc t +

f0 − fc 72 (1 − e−kt ) = 8t + (1 − e−0.5t ) = 8t + 144(1 − e−0.5t ) k 0.5

Using this equation to find t′ , where ′

21.3 = 8t′ + 144(1 − e−0.5t ) which yields t′ = 0.2838 h, and so ′′ +0.2838)

F = 8(t′′ + 0.2838) + 144(1 − e−0.5(t

)

@t = 150 min: t′′ = 0.5 h, F = 52.96 mm, and fp = 56.66 mm/h. @t = 150-180 min: In this period i < fp . Assume that all water infiltrates under ponded

298

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conditions, taking t = 1 h gives F = 8(1.2838) + 144(1 − e−0.5(1.2838) ) = 78.48 mm infiltration during this period = 78.48 − 52.96 = 25.52 mm rainfall amount = 29.3 mm/h × 0.5 h = 14.65 mm depression storage infiltrated = 5 mm actual infiltration = 14.65 + 5 = 19.65 mm cumulative infiltration = 52.96 + 19.65 = 72.61 mm fp = 48.33 mm/h For all subsequent time intervals, all of the rain infiltrates. The total runoff is 14.49 mm (approximately 14.5 mm ) and the total rainfall is 101.68 mm, so using the CN equation gives (P − 0.2S)2 P + 0.8S (101.68 − 0.2S)2 14.49 = 101.68 + 0.8S Q=

which yields S = 1106 mm or 200 mm. Since P must be greater than 0.2S, take S = 200 mm = 7.87 in. and so 1000 1000 CN = = = 55.95 10 + S 10 + 7.87 Hence the curve number can be taken as 56 . 9.40. The average infiltration rate, f , is given by Equation 9.85 as f=

S2i (P + 0.8S)2

(1)

From the calculations in Problem 9.37: S = 216 mm, i = 94.0 mm/h, and P = 94.0 mm. Substituting into Equation 1 gives f=

(216)2 (94) = 62 mm/h (94 + 0.8 × 216)2

Therefore, the average infiltration rate during the storm (= 62 mm/h) significantly exceeds the minimum infiltration rate of 10 mm/h. 9.41. From Figure 9.8, PT = 9 in. = 22.86 cm = 228.6 mm. From Table 9.4, Type III rainfall. The 24-h hyetograph is therefore given by

299

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Time 1 2 3 4 5 6 7 8 9 10 11 12

i (mm/h) 2.3 2.3 2.7 2.5 3.2 3.4 3.9 5.9 7.5 9.4 13.9 57.2

Time

i (mm/h) 57.4 13.7 8.5 8.7 4.1 4.1 3.9 4.1 2.5 2.5 2.3 2.5

13 14 15 16 17 18 19 20 21 22 23 24

(a) Calculate interception: I = S(1 − e−P/S ) + KEt = 5(1 − e−228.6/5 ) + (7)(5)(1) = 40 mm Therefore, Fraction of rain reaching the ground =

228.6 − 40 = 83% 228.6

(b) For an infiltration capacity of 60 mm/h, the runoff is zero since the rainfall intensity never exceeds this value. For an infiltration capacity of 25 mm/h, runoff occurs in time steps 12 and 13: Time step 12: runoff = 57.2 − 25 − 5 = 27.2 mm Time step 13: runoff = 57.4 − 25 = 32.4 mm Total runoff = 27.2 + 32.4 = 59.6 mm (c) For the CN method: (P − 0.2S)2 P + 0.8S

Q=

For loamy sand, Q = 0 → P ≤ 0.2S → S ≥ CN ≤

P 0.2

=

228.6 0.2

= 1143 mm. Hence

1000 → CN ≤ 18 10 + 0.0394(1143)

For sandy loam, the CN equation gives 59.6 =

(228.6 − 0.2S)2 228.6 + 0.8S

which yields S = 3174 mm and 304 mm. Taking S = 304 mm (P > 0.2S) gives CN =

1000 = 46 10 + 0.0394(304)

300

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9.42. From the given data: f0 = 250 mm/h, fc = 44 mm/h, and k = 0.13 min−1 . The cumulative infiltration, F (τ ), is therefore given by Equation 6.34 as F (τ ) = fc τ +

f0 − fc (1 − e−kτ ) k

Substituting in the given data, and taking τ in hours, gives F (τ ) = 44τ + which simplifies to

(250 − 44) (1 − e−(0.13)(60)τ ) (0.13)(60)

F (τ ) = 44τ + 26.4(1 − e−7.80τ )

(1)

The corresponding infiltration capacity is given by the Horton equation (Equation 6.32) as fp = fc + (f0 − fc )e−kτ = 44 + (250 − 44)e−(0.13)(60)τ which simplifies to

fp = 44 + 206e−7.80τ

(2)

The 10-year 24-hour rainfall is given as 229 mm. Using 1-hour time increments, the temporal (Type II) rainfall distribution is given in Columns 1 to 3 of the Table 9.2, the 1-h rainfall increments are given in Column 4, and the average rainfall intensity in each time interval is given in Column 5. Up to t = 11 h. During this interval, the average rainfall intensity is less than the minimum infiltration capacity (44 mm/h) an hence there is no runoff. At t = 11 h, the cumulative infiltration, F11 , is equal to the rainfall amount of 54.0 mm. Substituting F = F11 = 54.0 mm into Equation 1 gives 54.0 = 44τ + 26.4(1 − e−7.80τ ) which gives τ = 0.632 h Substituting τ = 0.632 h into Equation 2 gives the infiltration capacity, f11 , at t = 11 h as f11 = 44 + 206e−7.80(0.632) = 45.5 mm/h t = 11 – 12 h. The average rainfall rate of 97.8 mm/h exceeds the infiltration capacity (45.5 mm/h), and runoff begins at t = 11 h. Using Equation 1, the cumulative infiltration, F12 , at t = 12 h is given by F12 = 44(0.632 + 1) + 26.4(1 − e−7.80(0.632+1) ) = 98.2 mm Therefore the infiltration during this period is 98.2 mm − 54.0 mm = 44.2 mm and the rainfall excess is 97.8 mm − 44.2 mm = 53.6 mm. Since the depression storage is 6 mm, the runoff during this interval is 53.6 mm − 6 mm = 47.6 mm.

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Table 9.2: 10-year 24-hour Rainfall Distribution in Miami (1) Time (h) 0

(2) P/PT 0

(3) P (mm) 0.0

1

0.011

2.5

2

0.023

5.3

3

0.035

8.0

4

0.048

11.0

5

0.064

14.7

6

0.080

18.3

7

0.100

22.9

8

0.120

27.5

9

0.147

33.7

10

0.181

41.4

11

0.236

54.0

12

0.663

151.8

13

0.776

177.7

14

0.825

188.9

15

0.856

196.0

16

0.881

201.7

17

0.903

206.8

18

0.922

211.1

19

0.938

214.8

20

0.953

218.2

21

0.965

221.0

22

0.977

223.7

23

0.989

226.5

24

1.000

229.0

302

(4) ∆P (mm)

(5) i (mm/h)

2.5

2.5

2.8

2.8

2.7

2.7

3.0

3.0

3.7

3.7

3.6

3.6

4.6

4.6

4.6

4.6

6.2

6.2

7.7

7.7

12.6

12.6

97.8

97.8

25.9

25.9

11.2

11.2

7.1

7.1

5.7

5.7

5.1

5.1

4.3

4.3

3.7

3.7

3.4

3.4

2.8

2.8

2.7

2.7

2.8

2.8

2.5

2.5

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t = 12 – 24 h. During this interval, the average rainfall rate remains below the minimum infiltration capacity (44 mm/h) and no further runoff occurs. The above analysis indicates that a runoff, Q, of 47.6 mm will result from a rainfall, P , of 229 mm. The Curve Number model gives (P − 0.2S)2 P + 0.8S (229 − 0.2S)2 47.6 = 229 + 0.8S Q=

which yields S = 360 mm and CN =

1000 1000 = = 41 10 + 0.0394S 10 + 0.0394(360)

Therefore the curve number for the site is 41 . The calculated curve number would be different for a 20-year 24-hour rainfall since the precipitation and runoff amounts would have a different relationship than for the 10-year 24-hour rainfall. This is not physically reasonable , since the curve number measures the storage available in the catchment, which should not depend on the rainfall amount. 9.43. The computations for this problem are summarized in Tables 9.3 and 9.4 for the 3-day and 1-day storms respectively. For each storm, the computation procedure is started by assuming an infiltration capacity, fp , and determine the intervals in which the rainfall intensity exceeded the (assumed) infiltration capacity. The rainfall excesses in these intervals are then calculated, and the runoff amount, Q, is the sum of rainfall excesses for the storm. The assumed infiltration capacities and corresponding runoff amount are shown in columns 1 and 2 in the summary table. The total rainfall amount, P (= 31.1 cm), is shown in column 3. The storage, S related the rainfall, P , to the runoff, Q, by the relation Q=

(P − 0.2S)2 P + 0.8S

which can be rearranged to give S = 5[2Q + P −

√ Q(4Q + 5P )]

Using the calculated values of Q (column 2) and P (column 3) gives the values of S in column 4, and the curve number, CN, is derived from S using the relation CN =

1000 10 + 0.3937S

and this result is shown in column 5. A plot showing the relationship between curve number and infiltration capacity for 1-day and 3-day storms is shown in Figure 9.1. When there is 21

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Table 9.3: 3-day Storm Runoff (1) fp (cm/h) 10 9 8 7 6 5 4 3 2 1 0

(2) Q (cm) 0.0 0.1 1.1 2.1 3.1 4.1 5.1 6.1 7.6 9.8 31.1

(3) P (cm) 31.1 31.1 31.1 31.1 31.1 31.1 31.1 31.1 31.1 31.1 31.1

(4) S (cm) 155.5 136.8 100.2 83.7 72.4 63.8 56.7 50.9 43.6 35.1 0.0

(5) CN 14 16 20 23 26 28 31 33 37 42 100

cm of available storage (S = 21 cm) then CN = 55 and the corresponding infiltration capacity is 0.8 cm/h for the 3-day storm and 1 cm/h for the 1-day storm. If the actual infiltration capacity is 4 cm/h, then a curve number of 31 should be used for a 3-day storm and a curve number of 40 for a 1-day storm. 9.44. For the parking lot and building, CN = 98 and the available storage, S1 , can be derived from Equation 9.81 as ( ) ( ) 1000 1 1000 1 − 10 = − 10 = 5 mm S1 = 0.0394 CN 0.0394 98 For a precipitation, P , of 180 mm, the rainfall excess, Q1 , from the parking lot and building is given by Equation 9.80 as Q1 =

[180 − (0.2)(5)]2 (P − 0.2S1 )2 = = 174 mm P + 0.8S1 180 + (0.8)(5)

The total area of the parking lot and building is equal to 5 ha + 10 ha = 15 ha, and the grassed area is 5 ha. The effective rainfall, Peff , on the grassed area is therefore given by Peff = 180 + 174

15 = 702 mm 5

The grassed area contains Type A soil, is in good condition, and Table 9.14 gives CN = 39. The available storage, S2 is derived from Equation 9.81 as ( ) ( ) 1 1000 1 1000 S2 = − 10 = − 10 = 397 mm 0.0394 CN 0.0394 39 The rainfall excess from the 5-ha grassed area, Q2 , is given by Q2 =

[702 − (0.2)(397)]2 (Peff − 0.2S2 )2 = = 380 mm Peff + 0.8S2 702 + (0.8)(397)

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Table 9.4: 1-day Storm Runoff (1) fp (cm/h) 10 9 8 7 6 5 4 3 2 1 0

(2) Q (cm) 5.5 6.0 6.5 7.0 7.5 8.0 8.8 9.8 11.5 15.2 31.1

(3) P (cm) 31.1 31.1 31.1 31.1 31.1 31.1 31.1 31.1 31.1 31.1 31.1

(4) S (cm) 54.3 51.4 48.8 46.3 44.0 41.9 38.7 35.1 29.8 20.8 0.0

(5) CN 32 33 34 35 37 38 40 42 46 55 100

and the rainfall excess, Q, from the entire 20-ha composite catchment is Q = 380

5 = 95 mm 20

Therefore a rainfall of 180 mm on the composite catchment will result in 95 mm of runoff. 9.45. The calculation of runoff from the composite catchment is the same in Problem 9.44 whether the runoff from the buildings are routed to the parking lot or to the grassed area. However, the effective rainfall on the parking lot, and hence the potential for flooding, is considerably higher when the buildings drain directly onto the parking lot. 9.46. For the parking lots and buildings CN = 98, and for the grassed area CN = 39 (Type A soil in good condition). Since there are 15 ha of parking lots and buildings, and 5 ha of grassed ¯ is given by area, the area-weighted curve number, CN, ¯ = 15(98) + 5(39) = 83 CN 15 + 5 The available storage, S, on the site can be derived from Equation 9.81 as ( ) ( ) 1 1000 1 1000 S= − 10 = 52.0 mm ¯ − 10 = 0.0394 0.0394 CN 83 For a precipitation, P , of 180 mm, the rainfall excess, Q, from the site is given by Q=

[180 − (0.2)(52)]2 (P − 0.2S)2 = = 130 mm P + 0.8S 180 + (0.8)(52)

Therefore a rainfall of 180 mm on the composite catchment will result in 130 mm of runoff. This result is 37% higher than the site runoff of 95 mm calculated by routing the runoff within

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120 100 80 60 40

Curve Number, CN

3-day 1-day

20 0 0

1

2

3

4

5

6

7

8

9

10

Infiltration Capacity (cm/h)

Figure 9.1: Curve Number versus Infiltration Capacity the site. If the roof drains are directly connected to the parking lot, the runoff calculated using an area-weighted curve number would be the same as calculated here. Hence, there would be no change . 9.47. It is preferable to route the rainfall excess on composite areas, rather than use weightedaverage curve numbers, since this approach reflects a more realistic accounting of the rainfall abstractions. 9.48. For I percent imperviousness, the area-weighted composite curve number, CNc , is given by ( ) ( ) I I CNc = CNp 1 − + 98 100 100 which simplifies to CNc = CNp +

I (98 − CNp ) 100

9.49. For I percent imperviousness, if R is the fraction of impervious area that is unconnected, and 50% of the unconnected impervious area runs onto a pervious surface and 50% runs onto an impervious surface, then the effective imperviousness is (1 − 0.5R)I, and the composite curve number is derived directly from the previous example and is given by CNc = CNp +

I(1 − 0.5R) (98 − CNp ) 100

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9.50. (a) Find t that gives the maximum runoff:

6000 t 100t × = t + 20 60 t + 20 t infiltration = 25 × = 0.417t 60 100t runoff(R) = − 0.417t t + 20 rainfall = it =

Therefore the maximum runoff occurs when dR (t∗ + 20)(100) − (100t∗ )(1) =0→ − 0.417 = 0 dt (t∗ + 20)2

which yields t∗ = 49.25 minutes. So the maximum 10-year runoff is

(100)(49.25) − 0.417(49.25) = 50.6 mm 49.25 + 20

(b)

200t t + 20 t infiltration = 50 × = 0.833t 60 200t runoff = − 0.833t t + 20

effective rainfall on the pervious area = 2it =

For the maximum runoff, again take dR/dt = 0, which again gives t∗ = 49.25 minutes (since R is simply double what it was before). So the adjusted 10-year maximum runoff from the entire catchment is 2 × 50.6 × 1/2 = 50.6 mm . The implication of this is that in some cases the arrangement of infiltration capacity does not make a difference.

9.51. From the given data: A = 2000 km2 . It is apparent from the measured streamflows that the peak streamflow occurs at t = 4.0 days and the inflection point occurs at t = 5 days.

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Constant-Discharge

Constant-Slope

Concave

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

Time (d) 0

Flow (m3 /s) 74

∆ (m3 /s)

Baseflow (m3 /s) 74

Runoff (m3 /s) 0

Baseflow (m3 /s) 74

Runoff (m3 /s) 0

Baseflow (m3 /s) 74

Runoff (m3 /s) 0

1

67

67

0

67

0

67

0

56

0

56

0

56

0

56

181

240

0

47

190

56

1864

423

1497

38

1882

56

551

607

0

607

0

56

112

168

0

168

0

56

52

108

0

108

0

56

26

82

0

82

0

56

13

69

0

69

0

56

7

63

0

63

0

56

1

57

0

57

0

51

0

51

0

51

0

−7 −11 2

56 181

3

237 1683

4

1920 −1313

5

607 −439

6

168 −60

7

108 −26

8

82 −13

9

69 −6

10

63 −6

11

57 −6

12

51

(a) For the constant-baseflow separation method, the baseflow is equal to the streamflow at t = 2 d. The baseflow is therefore taken as 56 m3 /s, which is shown in Column 4, and the corresponding direct runoff is shown in Column 5. The maximum rate of direct runoff is 1864 m3 /s . The depth of runoff, h, is given by [ ] 1 1 h= (181 + 1) + (1864 + 551 + 112 + 52 + 26 + 13 + 7) (1.0 × 86400) 2000 × 106 2 = 0.117 m = 11.7 cm (b) For the constant-slope baseflow separation method, the baseflow is delineated by a straight line from the lowest pre-peak flow at t = 2 d to the inflection point at t = 5 d. Since the corresponding flows at these times are 56 m3 /s and 607 m3 /s, the baseflow, Qb , for 2 d ≤ t ≤ 5 d is given by 607 − 56 Qb = 56 + (t − 2) = 56 + 183.7(t − 2) 5−2 The baseflow estimated using this equation is shown Column 6, and the corresponding direct runoff is shown in Column 7. The maximum rate of direct runoff is 1497 m3 /s . The depth of runoff, h, is given by [ ] 1 1 h= (1497) (1.0 × 86400) = 0.032 m = 3.2 cm 2000 × 106 2

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(c) For the concave baseflow separation method, the baseflow is delineated by a straight line extrapolated from the pre-peak hydrograph recession to the time of the peak flow and connected to a straight line from the time of the peak flow to the inflection point. The pre-peak hydrograph recession has a slope of (−7 − 11)/2 = 9 (m3 /s)/d, and so the baseflow between the beginning of the surface runoff at t = 2 d and the peak runoff at t = 4 d is given by Qb = 56 − 9(t − 2) which yields a baseflow at the time of the peak runoff (at t = 4 d) of Qb = 56 − 9(4 − 2) = 38 m3 /s. The baseflow between the peak at t = 4 d and the inflection point at t = 5 d (where Q = 607 m3 /s) is estimated by linear interpolation as Qb = 38 +

607 − 38 (t − 4) = 38 + 569(t − 4) 5−4

The estimated baseflows are shown Column 8, and the corresponding direct runoffs are shown in Column 9. The maximum rate of direct runoff is 1882 m3 /s . The depth of runoff, h, is given by ] [ 1 1 h= (190 + 1882) (1.0 × 86400) = 0.045 m = 4.5 cm 2000 × 106 2

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Chapter 10

Fundamentals of Surface-Water Hydrology II: Runoff 10.1. From the given data: C = 0.82, S0 = 0.02, n = 0.03, L = 50 m, and P24 = 80 mm. Using the conventional NRCS formulation given by Equation 10.20 gives 5.5 tf = √ P24

[

nL 2√ C 3 S0

]4 5

[ ]4 5 5.5 (0.03)(50) =√ = 4.5 min √ 2 80 (0.82) 3 0.02

If the imperviousness of the catchment is neglected, and the formulation given by Equation 10.21 is used, then [ ]4 [ ]4 5.5 nL 5 5.5 (0.03)(50) 5 √ √ tf = √ =√ = 4.1 min P24 S0 80 0.02 The limitations in using these NRCS formulas (Type IA rainfall and tf < 15 min) are addressed by using Equation 10.22, where N=

nL (0.03)(50) = 12.1 m √ = 2√ C S0 (0.82) 3 0.02 2 3

and according to Table 10.2 for Type IA rainfall a = 1.54 − 0.0184 ln P24 − 0.0356(ln P24 )2 = 1.54 − 0.0184 ln(80) − 0.0356(ln 80)2 = 0.776 b = 0.573 c = 0.0094 Substituting the values of N , a, b, and c into Equation 10.22 gives ln tf = a + b ln N + c[ln N ]2 = 0.776 + (0.573) ln(12.1) + (0.0094)[ln(12.1)]2 = 2.26 which yields tf = e2.26 = 9.6 min. Based on these results, the most accurate estimate of the time of concentration in the sheet flow regime is 9.6 min .

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10.2. (a) Kinematic Wave Equation. If the overland flow is assumed to be fully turbulent, the Manning form of the kinematic wave equation (Equation 10.13) can be used. From the given data: L = 80 m, ie = 70 mm/h, S0 = 0.008, and for an grass surface Table 10.1 gives n = 0.15. According to Equation 10.13, tc =

6.99 2 5

ie

(

nL √ S0

)3

6.99

5

=

(70)

2 5

(

0.15 × 80 √ 0.008

)3 5

= 24 min

If the overland flow is assumed to be in the transition range (which is more probable), then the Darcy-Weisbach form of the kinematic wave equation must be used. Since the Darcy-Weisbach C factor for grass is not available, bypass this calculation. (b) NRCS Method. Use n = 0.15, L = 80 m, S0 = 0.008, and P24 = 85 mm. According to the NRCS approach, assuming the sheet-flow regime (L ≤ 100 m) yields the following estimate of the time of concentration ( )4 ( )4 5.5 nL 5 5.5 0.15 × 80 5 √ √ tc = √ =√ = 30 min P24 S0 85 0.008 (c) Kirpich Equation. Use L = 80 m, and S0 = 0.008. According to the Kirpich equation, with a factor of 2 to account for a grass surface tc = 2(0.019)

L0.77 (80)0.77 = 2(0.019) = 7 min (0.008)0.385 S00.385

(d) Izzard Equation. Use S0 = 0.008, L = 80 m, ie = 70 mm/h, and a retardance coefficient, cr , given by Table 10.4 as 0.053∗ . The constant, K, is given by K=

2.8 × 10−6 ie + cr 1 3

=

2.8 × 10−6 × 70 + 0.053 1

= 0.266

(0.008) 3

S0

The Izzard equation gives the time of concentration as 1

tc =

530KL 3 2 3

1

=

530(0.266)(80) 3 2

= 36 min

(70) 3

ie

In this case, ie L = (0.070)(80) = 5.6 m2 /h, and therefore since ie L > 3.9 m2 /h, the Izzard equation is not strictly applicable. (e) Kerby Equation. Use L = 80 m, S0 = 0.008, and a retardance coefficient, r, given by Table 10.5 as 0.40. According to the Kerby equation tc = 1.44[LrS0−0.5 ]0.467 = 1.44[(80)(0.40)(0.008)−0.5 ]0.467 = 22 min 10.3. The maximum flow distance that should be described by sheet flow is 100 m (as per NRCS guidelines). ∗

Midway between closely clipped sod (0.046) and dense bluegrass (0.060).

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10.4. α and m are defined by the depth-discharge relation q = αy m

(1)

(a) Manning Equation. The Manning equation for a unit width is given by ( ) 1 1 2 2 5 1 1 1 21 2 2 3 3 S0 y 3 q = AR S0 = (y)(y) S0 = n n n Comparing this equation with Equation 1 gives α=

1 12 S , n 0

m=

5 3

(b) Darcy-Weisbach Equation. The Darcy-Weisbach equation for a unit width is given by (√ ) √ √ 1 1 1 1 3 8g 8g 8gS 0 AR 2 S02 = (y)(y) 2 S02 = y2 q= f f f Comparing this equation with Equation 1 gives √ α=

8gS0 , f

m=

3 2

10.5. The flow consists of both overland flow and channel flow. Use the kinematic wave equation to estimate the time of concentration of overland flow, t1 , where L = 30 m, ie = 50 mm/h, S0 = 0.01, and for an asphalt surface Table 10.1 gives n = 0.011. The kinematic wave equation (Equation 10.13) gives t1 =

6.99 2 5

ie

(

nL √ S0

)3 5

=

6.99 (50)

(

2 5

0.011 × 30 √ 0.01

)3 5

= 3 min

The flow area, A, in the drainage channel can be calculated using the Manning equation 1 2 1 1 Q = AR 3 S02 = A n n

(

A P

)2 3

5

1 A 3 12 S0 = S n P 23 0 1 2

where Q = 0.02 m3 /s, n = 0.013 (for concrete), and S0 = 0.006. The area, A, and wetted perimeter, P , can be written in terms of the depth of flow, d, in the drainage channel as A = 0.2d and P = 2d + 0.20, therefore the Manning equation can be put in the form 5

0.02 =

1 1 (0.2d) 3 2 2 (0.006) 0.013 (2d + 0.2) 3

and solving iteratively for d yields d = 0.12 m = 12 cm

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The flow area, A, in the drainage channel is given by A = 0.2d = (0.2)(0.12) = 0.024 m2 and therefore the flow velocity, V0 , is given by V0 =

Q 0.02 = = 0.83 m/s A 0.024

Since the length, L, of the drainage channel is 60 m, then the flow time, t2 , in the channel is given by L 60 t2 = = = 72 s = 1.2 min V0 0.83 The time of concentration, tc , of the entire catchment area is equal to the overland flow time, t1 , plus the channel flow time, t2 , hence tc = t1 + t2 = 3 + 1.2 = 4.2 min Because of uncertainties is estimating tc , the time of concentration of the catchment can reasonably be rounded to 4 minutes . 10.6. From the given data: the area of the watershed is 121 ha. The first step is to separate the base flow using the concave base-flow separation method. The inflection point occurs at 1000 h (t = 63 min) and a base flow of 0.009 m3 /s is assumed up to the time of the peak discharge. Using these points, the base flow is shown in Column 6 of Figure 10.1, and the corresponding runoff is shown in Column 7, where the resulting runoff volume is 5362 m3 . The rainfall characteristics are shown in Figure 10.2, where the cumulative rainfall volume of 34 465 m3 is shown at the bottom of Column 4. (a) Since the rainfall duration is 38 min, this indicates an average infiltration capacity, f0 , given by (34465 − 5362) 1000 f0 = × × 60 = 38 mm/h 121 × 104 38 The infiltration capacity must be increased from 38 mm/h to account for the times when the actual infiltration is equal to the rainfall rate (in which cases the rainfall rate is less than the infiltration capacity). As shown in Figure 10.2, an adjusted infiltration capacity of 75 mm/h will yield the the calculated direct runoff (5362 m3 ). This indicates that the watershed has predominantly Class A soils. (b) From the above analysis, P =

34465 × 1000 = 25.4 mm 121 × 104

Q=

5362 × 1000 = 4.43 mm 121 × 104

and

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Figure 10.1: Hydrograph Separation Applying the NRCS curve number equation, (P − 0.2S)2 P + 0.8S (28.5 − 0.2S)2 4.43 = 28.5 + 0.8S Q=

which yields S = 53.6 mm = 2.11 inches. This corresponds to a curve number, CN, given by 1000 1000 CN = = = 83 10 + S 10 + 2.11 Hence CN = 83 . (c) The time of concentration, tc , can be estimated by the NRCS relation tl = 0.6tc

315

(1)

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Figure 10.2: Rainfall/Runoff Analysis where tl is the time lag between the centroid of the rainfall excess and the time to peak of the runoff hydrograph. From the rainfall analysis in Figure 10.2, the centroid of the rainfall excess, t¯r , is given by 7.5(1609) + 21(3759) t¯r = = 17.0 min 1609 + 3759 and since the peak runoff occurs at t = 49 min, Equation 1 gives (49 − 17.0) = 0.6tc which yields tc = 53 min . (d) Practical uses of these data include calculating the watershed runoff for a given design rainfall. 10.7. The time of concentration, tc , is estimated by the kinematic wave equation (Equation 10.13) as ( )3 6.99 nL 5 √ tc = 2 S0 i5 e

where n = 0.25, L = 60 m, and S0 = 0.005, and therefore ( )3 6.99 0.25 × 60 5 174 √ tc = 2 = 0.4 min ie 0.005 ie5

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where ie is in mm/h. Equating the storm duration to the time of concentration, then the effective rainfall rate, ie , for a 10-year storm is given by the IDF relation as ie = Ci =

1500C mm/h (tc + 6.19)0.78

Combining the latter two equations, with C = 0.5, yields ie = (

1500(0.5) 750 )0.78 = ( )0.78 174 174 + 8.96 + 8.96 0.4 0.4 i i e

e

Solving by trial and error yields ie = 35 mm/h, and a corresponding time of concentration of 42 min, which exceeds the minimum allowable time of concentration of 5 minutes. The peak runoff, Qp , from the residential development is given by the rational formula as Qp = CiA = ie A where ie = 35 mm/h = 9.72 ×10−6 m/s, and A = 2 ha = 2 × 104 m2 , therefore Qp = (9.72 × 10−6 )(2 × 104 ) = 0.19 m3 /s 10.8. From the given data: A = 20 ha = 20 × 104 m2 , C = 0.7, n = 0.25, L = 100 m, and S0 = 0.6% = 0.006. The (maximum) effective rainfall intensity, ie , and corresponding time of concentration, tc , are given by the simultaneous solution of the following equations, 1020C (tc + 8.7)0.75 ( )3 6.99 nL 5 √ tc = 2 S0 5 ie

ie =

(1) (2)

Substituting the given data into Equations 1 and 2 yields 1020(0.7) 714 = 0.75 (tc + 8.7) (tc + 8.7)0.75 ( )3 6.99 0.25 × 100 5 223.8 √ tc = 2 = 0.4 ie 0.006 ie5

ie =

(3) (4)

Solving Equations 3 and 4 gives ie = 31.1 mm/h = 8.64 × 10−6 m/s and tc = 56.6 minutes. Using the rational formula to calculate the peak runoff, Qp , yields Qp = ie A = (8.64 × 10−6 )(20 × 104 ) = 1.73 m3 /s Repeating the previous calculations and changing each parameter by 10% to increase the peak discharge yields the following results

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Parameter

Value

A C n L So

22 ha 0.77 0.225 90 m 0.0066

ie (mm/h) 31.1 35.3 32.8 32.8 31.9

Qp (m3 /s) 1.90 1.96 1.82 1.82 1.77

On the basis of these results, it is clear that the highest increase in the peak runoff is associated with a 10% error in the runoff coefficient, C, and lowest increase in the peak runoff is associated with a 10% error in the slope, S0 . Assuming a worst-case 10% error in all parameters, then A = 22 ha = 22 × 104 m2 , C = 0.77, n = 0.225, L = 90 m, and S0 = 0.66% = 0.0066. The (maximum) effective rainfall intensity, ie , and corresponding time of concentration, tc , are given by the simultaneous solution of the following equations, 1020(0.77) 785.4 = 0.75 (tc + 8.7) (tc + 8.7)0.75 )3 ( 6.99 0.225 × 90 5 191.6 √ tc = 2 = 0.4 ie 0.0066 ie5

ie =

(5) (6)

Solving Equations 5 and 6 gives ie = 40.4 mm/h = 1.12 × 10−5 m/s and tc = 43 minutes. Using the rational formula to calculate the peak runoff, Qp , yields Qp = ie A = (1.12 × 10−5 )(22 × 104 ) = 2.46 m3 /s 10.9. For a given duration, storms with longer return periods generally have higher average intensities. Since infiltration losses are relatively insensitive to the rainfall intensity (particularly when the rainfall rate exceeds the infiltration capacity of the soil), then storms with higher average intensities will result is a greater fraction of runoff (i.e. a higher runoff coefficient). Therefore, storms with longer return periods will have higher runoff coefficients. 10.10. From the given data: A = 2 ha = 2 × 104 m2 , 40% impervious, L = 100 m, P2 = 12 cm, and S0 = 0.7% = 0.007. The runoff coefficient of the area can be taken as 0.4. The first step is to calculate the time of concentration, tc , using all available methods. • The kinematic wave equation with n = 0.012 (for asphalt) gives tc =

6.99 2 5

ie

(

nL √ S0

)3

(

6.99

5

=

(0.4i)

2 5

0.012 × 100 √ 0.007

)3 5

=

308 i0.4

which gives i = 1.66 × 106 t−2.5 c

318

(1)

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Combining Equation 1 with the given IDF curve yields 1.66 × 106 t−2.5 = c

2029 (tc + 7.24)0.73

which gives tc = 4.9 min • The NRCS method with only overland flow gives 5.5 tc = √ P24

(

nL √ S0

)4 5

5.5 =√ 120

(

0.12 × 100 √ 0.007

)4 5

= 4.2 min

• The Kirpich equation gives tc = 0.019

(100)0.77 L0.77 = 0.019 = 4.45 min 0.385 (0.007)0.385 S0

• The Izzard equation with cr = 0.0070 (for smooth asphalt) gives ( ) ( ) 1 −6 1 2.8×10−6 ×0.4i+0.0070 530 2.8×10 1 ie +cr L 3 530 (100) 3 1 S03 (0.007) 3 tc = = 2 2 (0.4i) 3 ie3 ( ) 1.12 × 10−6 i + 0.0070 = 23700 2 i3

(2)

Combining Equation 2 with the IDF curve gives i= [

( 23700

2029 1.12×10−6 i+0.0070 2

)

]0.73 + 7.24

i3

which gives i = 359 mm/h

and tc = 3.5 min

• The Kerby equation with r = 0.10 (for asphalt) gives ) ( ) ( 100 × 0.10 0.467 Lr 0.467 √ tc = 1.44 √ = 1.44 = 13.4 min S0 0.007 Based on the above results, the minimum expected time of concentration is 3.5 min with a corresponding rainfall intensity of 359 mm/h = 9.97 × 10−5 m/s. Using the rational method, the peak runoff, Qp , is given by Qp = CiA = (0.4)(9.97 × 10−5 )(2 × 104 ) = 0.798 m3 /s Since the design rainfall return period is 10 years, there is a probability of 0.1 or 10% of flooding in any given year.

319

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10.11. The time of concentration, tc , of the impervious area is given by tc =

6.99 2 5

ie

(

nL √ S0

)3 5

where n = 0.035, L = 30 m, and S0 = 0.005, and therefore tc =

6.99 2 5

ie

(

0.035 × 30 √ 0.005

)3 5

=

35.3 min i0.4 e

where ie is in mm/h. Taking the storm duration as tc , the 10-year effective rainfall rate, ie , is given by the IDF curve as ie = Ci =

1500C mm/h (tc + 8.96)0.78

Combining the latter two equations, with C = 0.9, yields ie = (

1500(0.9) 1350 )0.78 = ( )0.78 35.3 35.3 + 8.96 + 8.96 0.4 0.4 i i e

e

Solving by trial and error yields ie = 178 mm/h, and a corresponding time of concentration of 4.4 min, which is less than the minimum allowable time of concentration of 5 minutes. Taking tc = 5 min, the IDF relation yields ie =

1500(0.9) = 173 mm/h (5 + 8.96)0.78

The peak runoff, Qp , from the impervious area is given by the rational formula as Qp = CiA = ie A where ie = 173 mm/h = 4.81 ×10−5 m/s, and A = 0.5 ha = 5000 m2 , therefore Qp = (4.81 × 10−5 )(5000) = 0.24 m3 /s Problem 10.7 indicated that the peak runoff from the entire composite catchment is 0.19 m3 /s, and the present example shows that the peak runoff from the directly-connected impervious area is 0.24 m3 /s. The peak runoff from this development is therefore 0.24 m3 /s . 10.12. (a) For a single-family residential area, assume C = 0.40, and from the given data: A = 0.5 ha = 5000 m2 , tc = 10 min, and i=

1209 1209 = = 119 mm/h = 3.30 × 10−5 m/s (t + 8.86)0.79 (10 + 8.86)0.79

The peak runoff rate is given by the rational formula as Qp = CiA = (0.40)(3.30 × 10−5 )(5000) = 0.066 m3 /s

320

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(b) For t = 24 h = 1440 min the total rainfall amount is given by P = it =

1209 24 · = 0.0923 m = 9.23 cm 0.79 (1440 + 8.86) 1000

From the given data, S = 8 cm and the runoff, Q, is given by Q=

(P − 0.2S)2 (9.23 − 0.2 × 8)2 = = 3.72 cm = 0.0372 m P + 0.8S P + 0.8(8)

For the 10-ha site, the runoff volume, V , is given by V = QA = (0.0372)(10 × 104 ) = 3720 m3 Since S = 8 cm = 80 mm, the curve number for the site is given by CN =

1000 1000 = = 76 10 + 0.0394S 10 + 0.0394(80)

10.13. For CN = 79, the storage, S, is given by ( ) ( ) 1000 1 1000 1 − 10 = − 10 = 67 mm S= 0.0394 CN 0.0394 79 and since P = 100 mm, the runoff, Q, is given by Q=

[P − 0.2S]2 [100 − 0.2(67)]2 = = 49 mm = 4.9 cm P + 0.8S 100 + 0.8(67)

From Table 10.8, Fp = 0.93 (via interpolation). By definition, Ia 0.2S (0.2)(67) = = = 0.134 P P 100 Table 10.9 gives (via interpolation) Co = 2.53829, C1 = −0.61639, C2 = −0.15596, and since tc = 3.0 h, Equation 10.46 gives log(qu ) = C0 + C1 log tc + C2 (log tc )2 − 2.366 = 2.53829 − 0.61639 log(3.0) − 0.15596(log 3.0)2 − 2.366 = −0.157 which leads to qu = 0.697 (m3 /s)/(cm·km2 ) Therefore, according to the TR-55 method, the peak discharge, qp , is given by Equation 10.45 as qp = qu AQFp = 0.697(4.2)(4.9)(0.93) = 13.3 m3 /s

321

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10.14. For CN = 70, the storage, S, is given by ( ) ( ) 1 1000 1 1000 S= − 10 = − 10 = 109 mm 0.0394 CN 0.0394 70 and since P = 130 mm, the runoff, Q, is given by Q=

[P − 0.2S]2 [130 − 0.2(109)]2 = = 54 mm = 5.4 cm P + 0.8S 130 + 0.8(109)

From Table 10.8, Fp = 0.75. By definition, 0.2S (0.2)(109) Ia = = = 0.168 P P 130 Table 10.9 gives (via interpolation) Co = 2.25781, C1 = −0.50720, C2 = −0.09832, and since tc = 1.7 h, Equation 10.46 gives log(qu ) = C0 + C1 log tc + C2 (log tc )2 − 2.366 = 2.25781 − 0.50720 log(1.7) − 0.09832(log 1.7)2 − 2.366 = −0.230 which leads to qu = 0.589 (m3 /s)/(cm·km2 ) Therefore, according to the TR-55 method, the peak discharge, qp , is given by Equation 10.45 as qp = qu AQFp = 0.589(1.0)(5.4)(0.75) = 2.39 m3 /s 10.15. (a) Take n = 0.25 (lawns). Using kinematic-wave equation, tc = 6.99

(nL)0.6 (0.25 × 100)0.6 236.3 = 6.99 = 0.4 0.3 0.4 0.3 0.4 ie (0.005) ie ie S0

(1)

The 10-year IDF curve for Atlanta (from Table 9.2) is given by i=

64.1 (t + 8.16)0.76

in./h =

1628 (t + 8.16)0.76

mm/h

For a suburban area, take C = 0.33 and therefore ie =

1628(0.33) 537.2 = (t + 8.16)0.76 (t + 8.16)0.76

Combining Equations 1 and 2 gives tc = 19.12(tc + 8.16)0.304 which gives tc = 72.7 min

322

mm/h

(2)

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For the NRCS equation, the 2-year IDF curve for Atlanta is required. Using the Chen 10 = 5.5 in., R100 = 3.6 in., T = 2 years → T = 1.443 years, method: R110 = 2.6 in., R24 p 1 10 10 x = 1.385, R1 /R24 = 47.3%, a1 = 26, b1 = 9.0, c1 = 0.80, and a = 45.72. Therefore the 2-y IDF curve is given by i=

45.2 (t + 9.0)0.80

in./h

For a 2-year 24-h storm, 45.2 × 24 × 25.4 = 81.5 mm = 8.15 cm (24 × 60 + 9.0)0.80

P2 =

Therefore the overland travel time (= time of concentration) is given by tf = 0.0288

(nL)0.8 (0.25 × 100)0.8 = 1.103 h = 66.18 min = 0.0288 0.5 0.4 (8.15)0.5 (0.005)0.4 P2 S0

For the Kirpich equation, tc = 0.019

L0.77 1000.77 = 0.019 = 5.07 min (0.005)0.385 S00.385

For the Izzard equation (taking cr = 0.036), [ 530

2.8×10−6 ×537.2(tc +8.16)−0.76 +0.036 1

]

1

(100) 3

0.005 3

tc =

2

[537.2(tc + 8.16)−0.76 ] 3

which gives tc = 72.9 minutes. Check ie L, 537.2 = 19.0 mm/h (72.9 + 8.16)0.76 ) ( 1 (100) = 1.9 m2 /h ie L = 19.0 × 1000 ie =

So the Izzard equation is valid. For the Kerby equation (taking r = 0.40), ( tc = 1.44

Lr √ S0

)

( = 1.44

100 × 0.40 √ 0.005

) = 19.30 min

From the above analyses, the expected range of tc is 19 min < tc 9.6 × 10−14 Therefore, Manning’s equation is valid. Check the flow velocity: V =

Q = A

0.434 π 2 4 (0.462)

= 2.58 m/s

This velocity is adequate to prevent sedimentation and scour. 11.18 From the given data: S0 = 0.009 and Q = 0.50 m3 /s. Assume n = 0.013 for concrete pipe to account for aging and construction defects. The Manning equation (Equation 11.44) gives [

3.21Qn √ D= S0

]3

8

[

3.21(0.50)(0.013) √ = 0.009

373

]3 8

= 0.57 m

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Use the limitation given by Equation 11.43 to check whether the Manning equation is valid: √ √ n6 RS0 = (0.013)6 (0.57/4)(0.009) = 1.7 × 10−13 ≥ 9.6 × 10−14 Therefore the Manning equation is valid. Using a 57-cm pipe, the flow velocity, V , is given by Q 0.50 V = = π = 1.96 m/s 2 A 4 (0.57) This velocity exceeds the minimum velocity to prevent sedimentation (0.60 to 0.90 m/s), and is less than the maximum velocity to prevent excess scour (3 to 4.5 m/s). According to the Manning equation, the pipe should have a diameter of 57 cm . 11.19 The Darcy-Weisbach equation (Equation 11.44) gives [

0.811f Q2 D= gS0

] 15

[

0.811f (0.50)2 = (9.81)(0.009)

] 15

1

= 1.181f 5

which can be put in the form f = 0.435D5

(1)

The friction factor, f , also depends on D via the Colebrook equation (Equation 2.35) which is given by ( ) 1 ks /D 2.51 √ = −2 log √ + (2) 3.7 f Re f The equivalent sand roughness, ks , of concrete is in the range 0.3–3.0 mm (Table 2.1) and can be taken as ks = 1.7 mm. Assuming that the temperature of the water is 20◦ C, the kinematic viscosity, ν, is equal to 1.00 × 10−6 m/s2 , and the Reynolds number, Re, is given by Re =

VD 4Q 4(0.50) 6.37 × 105 = = = ν πDν πD(1.00 × 10−6 ) D

(3)

Combining Equations 1 to 3 gives √

(

1 0.435D5

= −2 log

0.0017/D + 3.7

2.51 √ 6.37×105 0.435D5 D

)

which simplifies to 1.52D− 2 = −2 log(4.59 × 10−4 D−1 + 5.97 × 10−6 D− 2 ) 5

3

and yields D = 0.57 m = 57 cm Therefore the Darcy-Weisbach equation requires that the sewer pipe be at least 57 cm in diameter. 11.20 From the given data: L = 65 m, Q = 0.5 m3 /s, Vsc = 0.9 m/s, and Vmax = 4.0 m/s. For RCP it is assumed that n = 0.013. The allowable pipe diameter and slope combinations are determined as follows:

374

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Step 1: The slope to meet the minimum-cover requirement is given as Sref1 = 0.001. Step 2: Assume D = 610 mm, which is equal to the diameter of of upstream pipe segment. Step 3: Use the Manning equation as the basis for design. For full-flow conditions, the Manning equation gives [ Sref2 =

[

]2

nQ

(0.013)(0.5)

=

8

0.312D 3

]2 = 0.00606

8

0.312(0.610) 3

Step 4: To determine the slope, Sref3 , required to meet the minimum-velocity requirement under full-flow conditions, the Manning equation gives [ Sref3 = 6.35

nVsc 2

[

]2 = 6.35

(0.013)(0.9)

D3

]2

2

= 0.00168

(0.610) 3

Step 5: The minimum slope, S0 , for a pipe diameter of 610 mm is then given by S0 = max(Sref1 , Sref2 , Sref3 ) = max(0.001, 0.00606, 0.00168) = 0.00606 Step 6: At a slope of 0.00606, the pipe flows full and the velocity is given by Vfull =

Q = A

0.5 π 2 4 0.610

= 1.7 m/s

Therefore, under design conditions, the velocity (1.7 m/s) is less than the maximum allowable velocity of 4.0 m/s. These calculations show that using a pipe diameter of 610 mm and a slope of 0.00606 will meet the design requirements. The slope of 0.00606 is greater than the ground slope of 0.001 and so the minimum-cover requirement is exceeded at the downstream end of the pipe segment. The required slope can be decreased by increasing the pipe diameter, and these options can be determined by repeating the above calculations for increased (commercial) pipe sizes. The results are as follows: D (mm) 610 685 760

S0 (-) 0.00606 0.00327 0.00188

11.21 The service manholes placed along the pipe will each cause a head loss, hL , where hL = K

375

V2 2g

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For inflow and outflow pipes aligned opposite to each other, K is between 0.12 and 0.32, and can be assigned an average value of K = 0.22. Since V = 1.96 m/s, the head loss, hL , is therefore given by (1.96)2 hL = 0.22 = 0.043 m 2(9.81) 11.22 From the given data: D1 = 0.610 m, S1 = 0.008, D2 = 0.610 m, S2 = 0.005, K = 0.35, and Q = 0.4 m3 /s. Assume n = 0.013 for RCP. Solving for the normal depths of flow in the upstream and downstream pipes using the Manning equation yields y1 = 0.375 m V1 = 2.12 m/s y2 = 0.445 m V2 = 1.75 m/s hL = K

V12 2.122 = 0.35 = 0.0804 m 2g 2(9.81)

Substituting these values into Equation 11.49 gives the required drop, ∆z, as ) ( ) 1.752 2.122 V22 V12 − +hL = (0.445−0.375)+ − +0.0804 = 0.077 m ∆z = (y2 −y1 )+ 2g 2g 2(9.81) 2(9.81) (

Therefore, the minimum required drop in the sewer invert is 0.077 m . 11.23 The calculations are described below and are summarized in Figure 11.2. Line 1--2: The characteristics of the contributing area are A = 2.00 ha, C = 0.70, and tc = 8.0 min. Substituting t = tc into the IDF equation yields i = 97 mm/h. The design flow, Q, is therefore given by (with appropriate unit conversion) Q = CiA = 0.7 × 97 mm/h × 2.00 ha = 0.38 m3 /s Several combinations of diameter, D, and pipe slope, S0 , can be used. With D = 535 mm and S0 = 0.00801 the slope is equal to the ground slope, and the velocity criteria are met. Since the upstream (U/S) ground elevation is 30.064 m, with minimum cover (1.0 m) the U/S invert elevation is 30.064 m − 1.0 m − 0.535 m = 28.529 m. Since the slope of the pipe is 0.00801 and the length of the pipe is 100 m, the downstream (D/S) invert elevation is 28.529 m − (0.00801)(100) = 27.728 m. Line 3--2: For this line, A = 0.71 ha, C = 0.55, and tc = 6.5 min which yields i = 104 mm/h and Q = 0.11 m3 /s. With D = 455 mm and S0 = 0.00713 the slope is equal to the ground slope, the pipe diameter is the minimum, and the velocity criteria are met. Since the upstream (U/S) ground elevation is 29.655 m, with minimum cover (1.0 m) the U/S invert elevation is 29.655 m − 1.0 m − 0.455 m = 28.200 m. Since the length of the pipe is 55 m, the downstream (D/S) invert elevation is 28.200 m − (0.00713)(55) = 27.808 m.

376

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Line 4--2: For this line, A = 0.86 ha, C = 0.62, and tc = 6.8 min which yields i = 103 mm/h and Q = 0.15 m3 /s. With D = 455 mm and S0 = 0.00605 the slope is equal to the ground slope, the pipe diameter is the minimum, and the velocity criteria are met. Since the upstream (U/S) ground elevation is 29.705 m, with minimum cover (1.0 m) the U/S invert elevation is 29.705 m − 1.0 m − 0.455 m = 28.250 m. Since the length of the pipe is 73 m, the downstream (D/S) invert elevation is 28.250 m − (0.00605)(73) = 27.808 m. ∑ Line 2--5: For this line A = 0.86 ha, C = 0.62, and CA = 3.60 ha for this and all upstream catchment areas contribution to Line 2–5. For all paths leading to Line 2–5 the maximum travel time is from the catchment of MH 1 with a travel time of 8.0 min + 0.8 min = 8.8 min. Using t = tc = 8.8 min in the IDF equation yields i = 94 mm/h, and hence the peak runoff, Q, is given by (with appropriate unit conversion) ∑ CA = 94 mm/h × 3.60 ha = 0.93 m3 /s Q = CiA = i With D = 760 mm and S0 = 0.00850 the slope is equal to the ground slope, and the velocity criteria are met. Determination of the upstream invert elevation of Line 2–5 requires an analysis of the pipelines entering MH 2 as shown below: Pipeline 1–2 3–2 4–2 2–5

Type Inflow Inflow Inflow Outflow

Diameter (mm) 535 455 455 760

Elevation (m) Invert Crown 27.728 28.263 27.808 28.263 27.808 28.263 27.503 28.263

All pipes enter MH 2 at 28.263 m and so the crown elevation of the exiting pipe (Line 2–5) should be set at 28.263 m and so the invert elevation of the U/S end of Line 2–5 is 28.263 m − 0.760 m = 27.503 m. The minimum invert elevation of pipes entering MH 2 is 27.728 m and so the invert drop at MH 2 is 27.728 m − 27.503 m = 0.225 m. Since the length of the pipe is 110 m, the downstream (D/S) invert elevation is 27.503 m − (0.00850)(110) = 26.568 m. Line 6--5: For this line, A = 0.30 ha, C = 0.75, and tc = 5.0 min which yields i = 113 mm/h and Q = 0.07 m3 /s. With D = 455 mm and S0 = 0.00767 the slope is equal to the ground slope, and the velocity criteria are met. The pipe does not flow full. Since the upstream (U/S) ground elevation is 28.558 m, with minimum cover (1.0 m) the U/S invert elevation is 28.558 m − 1.0 m − 0.455 m = 27.103 m. Since the length of the pipe is 30 m, the downstream (D/S) invert elevation is 27.103 m − (0.00767)(30) = 26.873 m. ∑ Line 5--7: For this line A = 1.20 ha, C = 0.90, and CA = 4.90 ha. For all paths leading to Line 2–5 the maximum travel time is from Line 2–5 with a travel time of 8.8 min + 0.7 min = 9.5 min. Using t = tc = 9.5 min in the IDF equation yields i = 91 mm/h, and hence the peak runoff, Q, is given by ∑ Q = CiA = i CA = 91 mm/h × 4.90 ha = 1.24 m3 /s

377

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With D = 840 mm and S0 = 0.00750 the slope is equal to the ground slope, and the velocity criteria are met. Determination of the upstream invert elevation of Line 5–7 requires an analysis of the pipelines entering MH 5 as shown below: Pipeline 2–5 6–5 2–5

Type Inflow Inflow Outflow

Diameter (mm) 760 455 840

Elevation (m) Invert Crown 26.568 27.328 26.873 27.328 26.488 27.328

The crown elevation of all pipes entering MH 2 is 27.328 m and so the crown elevation of the exiting pipe (Line 5–7) should be set at 27.328 m and so the invert elevation of the U/S end of Line 5–7 is 27.328 m − 0.840 m = 26.488 m. The minimum invert elevation of pipes entering MH 5 is 26.568 m and so the invert drop at MH 5 is 26.568 m − 26.488 m = 0.08 m. Since the length of the pipe is 90 m, the downstream (D/S) invert elevation is 26.488 m − (0.00750)(90) = 25.813 m.

378

L

From To (m) (1) (2) (3) 1 2 100 3 2 55 4 2 73 2 5 110 6 5 30 5 7 90

Pipeline

Area, A Inc. Total (ha) (ha) (4) (5) 2.00 2.00 0.71 0.71 0.86 0.86 1.50 5.07 0.30 0.30 1.20 6.57

(-) (6) 0.70 0.55 0.62 0.85 0.75 0.90

C

379

(mm) (13) 535 455 455 760 455 840

D

QFULL Full (m3/s) (m/s) (14) (15) 0.38 1.80 0.25 1.52 0.23 1.41 1.06 2.34 0.26 1.58 1.30 2.35

Figure 11.2: Storm-Sewer Design Calculations

tc AxC i Q Inc. Total Inlet System 3 (ha) (ha) (min) (min) (mm/h) (m /s) (7) (8) (9) (10) (11) (12) 1.40 1.40 8.0 8.0 97 0.38 0.39 0.39 6.5 6.5 104 0.11 0.53 0.53 6.8 6.8 103 0.15 1.28 3.60 7.4 8.8 94 0.93 0.23 0.23 5.0 5.0 113 0.07 1.08 4.90 7.0 9.5 91 1.24 V Design (m/s) (16) 2.05 1.48 1.50 2.64 1.35 2.68

Flow Invert Elevaon Time U/S D/S (min) (m) (m) (17) (18) (19) 0.8 28.529 27.728 0.6 28.200 27.808 0.8 28.250 27.808 0.7 27.503 26.568 0.4 27.103 26.873 0.6 26.488 25.813

Slope (-) (21) 0.00801 0.00713 0.00605 0.00850 0.00767 0.00750

Invert Drop (m) (20) 0.000 0.000 0.000 0.225 0.000 0.080

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Chapter 12

Design of Stormwater Management Systems 12.1. From the given data: A = 7.5 ha = 7.5 × 104 m2 , L = 100 m, W = 25 m, and d = 2.0 m. The water-quality volume, WQV, area of the lake, Alake , and ponded height, H0 , are given by WQV = 0.020A = 0.020(7.5 × 104 ) = 1500 m3 Alake = L × W = (100)(25) = 2500 m2 1500 WQV = = 0.600 m H0 = Alake 2500 Assuming that the WQV is added instantaneously to the lake, the outflow hydrograph, Q(t), satisfies the continuity equation: dH Q = Alake (1) dt where H is the height of the water surface above the control elevation. Combining Equation 1 with the given weir discharge equation, and substituting Alake = 2500 m2 , gives the following equation describing the stage hydrograph, 5

0.080H 2 = 2500

dH dt

Solving this equation with the initial condition that H = H0 = 0.600 m when t = 0 sec yields ( H=

1667 3586 − 0.080t

)2 3

and combining this result with the given weir discharge equation gives the outflow hydrograph as 18749 Q(t) = (2) 5 (3586 − 0.080t) 3

381

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where Q is in m3 /s and t is in seconds. The evacuation time is calculated using Equation 12.1 which requires that ∫

Te

O(t)dt = WQV ∫ 0

0 Te

18749 (3586 − 0.080t)

5 3

dt = 1500 m3

which yields Te = 28975 s = 8.0 h. Therefore, the evacuation time of the WQV in the detention pond is 8.0 h. From the given data, the average annual rainfall, d¯rain , is 1.20 m and the runoff coefficient, C, is 0.75; the average runoff, Q, and the volume of the pond, Vpond , are given by d¯rain AC (1.20)(7.5 × 104 )(0.75) = = 185 m3 /d 365 365 = LW d = (100)(25)(2.0) = 5000 m3

Q= Vpond

Therefore the detention time, Td , is given by Equation 12.2 as Td =

Vpond 5000 = 27 days = 185 Q

In summary, the detention pond will have an evacuation time of approximately 8.0 h and a detention time of approximately 27 days . 12.2. From the given data: D = 1.50 m, Z0 = 1.00 m, Z1 = 2.00 m, Lw = 0.40 m (orifice) and Lw = πD = π(1.50 m) = 4.71 m (top of riser), A0 = 0.40 m × 0.40 m = 0.16 m2 , and it can be assumed that Cw = 1.83 and Cd = 0.6. Substituting these values into Equation 12.9 yields the following results: Z (m) 1.00 1.20 1.40 1.60 1.80 2.00 2.20 2.40

Q Formula – Cw Lw (Z − Z0 )1.5 Cw Lw√ (Z − Z0 )1.5 Cd A0 √2g(Z − Z0 ) Cd A0 √2g(Z − Z0 ) Cd A0 2g(Z − Z0 ) √ Cw Lw (Z − Z1 )1.5 + Cd A0 √2g(Z − Z1 ) Cw Lw (Z − Z1 )1.5 + Cd A0 2g(Z − Z1 )

(m3 /s) 0.0 0.065 0.185 0.329 0.380 0.425 0.961 2.45

12.3 The storage volume, S, is given by ( S=

A0 + Ah 2

382

) h

(1)

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where A0 is the area of the base of the reservoir given by A0 = LW and Ah is the area at a height h above the base of the reservoir, given by ( ) 2h Ah = W + L tan α

(2)

(3)

Combining Equations 1 to 3 gives S=

L 2 h + (LW )h tan α

12.4 From the given data: A = 5 ha = 50000 m2 , and Q0 = 1 m3 /s. For a development with 0.4-ha (= 1-ac) lots and sandy loam (Type B) soil: CN = 68. For 25 mm of runoff, the runoff volume, V25 , is V25 = (0.025)(50000) = 1250 m3 Determine the stage/storage curve for the retention area. If the base of the retention area is L × L and the ponded depth is h, then the stage/storage curve is given by S = (L + 3h)2 h The maximum allowable ponding depth is 0.3 m and the minimum infiltration rate in Type B soil is 3.8 mm/h. So the time to infiltrate 0.3 m (= 300 mm) is 300/3.8 = 79 h. Since this is greater than the maximum of 72 h, then the maximum height in the retention area has to be reduced to 72 h × 3.8 mm/h = 274 mm = 0.274 m For a ponding depth of 0.274 m, h = 0.274 m, S = 1250 m3 , and the stage/storage curve gives 1250 = (L + 3 × 0.274)2 (0.274) which gives L = 66.72 m. The area at the top of the retention basin is therefore given by A = (L + 6h)2 = (66.72 + 6 × 0.274)2 = 4674 m2 Therefore, reserve 4674/50000 = 9.3% of the site area for retention. For a 25-y 1-day storm, the rainfall depth, P , is given by P = it =

7836t 7836(24) = = 179.2 mm 34.11 + 0.7052t 34.11 + 0.7052(24 × 60)

and the available site storage, S0 , is given by 1000 10 + 0.0394S0 1000 68 = 10 + 0.0394S0

CN =

383

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which gives S0 = 119 mm. The runoff depth, Q, is given by Q=

(P − 0.2S0 )2 (179.2 − 0.2 × 119)2 = = 88.00 mm P + 0.8S0 179.2 + 0.8 × 119

and this yields a runoff volume, V0 , given by V0 = (0.088)(50000) = 4400 m3 The height of ponded water corresponding to this runoff volume is derived from the stage/storage curve, V0 = (L + 3h)2 h 4400 = (66.72 + 3h)2 h which yields h = 0.912 m. The elevation of the weir crest is to be 0.274 m above the bottom of the retention area, and the crest width, b, can be initially estimated using the weir equation, 3

Q0 = 1.83bhw2 3

1 = 1.83b(0.912 − 0.274) 2 which gives b = 1.07 m . The final crest length will be determined by routing the runoff hydrograph through the detention basin. The discharge structure will consist of a weir box in which the water from the detention basin enters the box by flowing over the weir and exits the box through a culvert pipe. 12.5 The required detention basin volume is first estimated by subtracting the pre-development runoff volume, V1 , from the post-development runoff volume, V2 . From the given hydrographs: V1 = (30)(60)[2.0 + 7.5 + 1.7 + 0.90 + 0.75 + 0.62 + 0.49 + 0.30 + 0.18 + 0.50] = 26892 m3 and V2 = (30)(60)[3.5+10.6+7.5+5.1+3.0+1.5+0.98+0.75+0.62+0.51+0.25+0.12] = 61974 m3 A preliminary estimate of the required volume, V , of the drainage basin is V = V2 − V1 = 61974 − 26892 = 35082 m3 From the storage-elevation function, the head h corresponding to a storage volume of 35,082 m3 is 1.31 m. The maximum pre-development runoff, Q, is 7.5 m3 /s, and the weir equation gives 3 Q = 1.83bh 2 or b=

Q 1.83h

3 2

7.5

=

3

1.83(1.31) 2

384

= 2.73 m

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Based on this preliminary estimate of the crest length, use a trial length of 2.5 m, and the corresponding weir discharge equation is 3

3

3

Q = 1.83bh 2 = 1.83(2.5)h 2 = 4.58h 2 The post-development hydrograph can be routed through the detention basin using ∆t = 30 min, and the storage and outflow characteristics can be put in in the following form Elevation (m) 0 0.5 1.0 1.5

Storage, S (m3 ) 0 11022 24683 41522

Outflow, O (m3 /s) 0 1.62 4.58 8.41

2S/∆t + O (m3 /s) 0 13.87 32.01 54.55

The routing computations (using the modified Puls method) are summarized in the following table: Time (min) 0 30 60 90 120 150

Inflow, I (m3 /s) 0 3.5 10.6 7.5 5.1 3.0

2S/∆t − O (m3 /s) 0 2.68 12.60 21.96 24.54 23.26

2S/∆t + O (m3 /s) 0 3.5 16.78 30.70 34.56 32.64

O (m3 /s) 0 0.41 2.09 4.37 5.01 4.69

From these results, the maximum outflow is less than the pre-development discharge, and therefore a crest length of 2.5 m is adequate. From an economic viewpoint, it is not reasonable to increase the size of the outlet structure in order to make the post-development peak runoff equal to the pre-development peak runoff. In fact, it makes more (economic) sense to reduce the crest length of the outlet structure until the maximum elevation in the storage reservoir is equal to 1.5 m, making maximum use of the reservoir and using the shortest crest length. This condition occurs when b = 0.645 m , in which case the storage-outflow characteristics are given by Elevation (m) 0 0.5 1.0 1.5

Storage, S (m3 ) 0 11022 24683 41522

Outflow, O (m3 /s) 0 0.42 1.18 2.17

2S/∆t + O (m3 /s) 0 12.66 28.61 48.30

The routing computations (using the modified Puls method) are summarized in the following table:

385

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Time (min) 0 30 60 90 120 150 180 210 240

Inflow, I (m3 /s) 0 3.5 10.6 7.5 5.1 3.0 1.5 1.0 0.8

2S/∆t − O (m3 /s) 0 3.27 16.08 31.26 39.97 43.76 43.93 42.26 40.16

2S/∆t + O (m3 /s) 0 3.5 17.37 34.18 43.86 48.07 48.26 46.41 43.99

O (m3 /s) 0 0.12 0.64 1.46 1.95 2.16 2.17 2.07 1.92

The maximum water elevation in the reservoir is therefore equal to 1.5 m . 12.6 From the given data: A = 5 km2 = 5 × 106 m2 , WQV = 0.025 × 5 × 106 = 125000 m3 , and retention area = 0.5 × 106 m2 . Therefore, height of ponding of WQV =

125000 = 0.25 m 0.5 × 106

So the height of the crest of the discharge (weir) structure is 0.25 m . Need to route the discharge through a weir structure. The discharge equation of a sharpcrested weir is O = 1.83bH 1.5 Taking ∆t = 20 min = 1200 s, the storage table is given by Elev (m) 0 0.25 z

S (m3 ) 0 125000 5z × 105

O (m3 /s) 0 0 1.83b(z − 0.25)1.5

2S/∆t + O (m3 /s) 0 208.3 833.3z + 1.83b(z − 0.25)

Try b = 1 m. Routing the runoff through the retention area gives: t (min) 0 20 40 60 80 100 120

I (m3 /s) 0 7.4 14.8 11.1 7.4 3.7 0

2S/∆t − O (m3 /s) 0 7.4 29.6 55.5 74.0 85.1 88.8

386

2S/∆t − O (m3 /s) 0 7.4 29.6 55.5 74.0 85.1 88.8

O (m3 /s) 0 0 0 0 0 0 0

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Therefore, retaining the water-quality volume will retain all the runoff on site. The maximum volume of water in the retention area is equal to the runoff volume, which is given by 1 runoff volume, V = (120 × 60)(14.8) = 53280 m3 2 The maximum depth of ponding is therefore given by ponding depth =

V 53280 = = 0.107 m Ar 0.5 × 106

12.7 Under pre-development conditions: I = 10%, d = 0.25 cm, P = 98.5 cm, and the (annual) runoff, R, is given by Equation 10.150 as [ ( )] I R = 0.15 + 0.75 P − 3.004d0.5957 100 = [0.15 + 0.75(0.10)](98.5) − 3.004(0.25)0.5957 = 20.8 cm Since there are 62 storms/year, the runoff per storm is 0.208/62 = 0.00335 m/storm. Under post-development conditions: I = 65%, d = 0.15 cm, P = 98.5 cm, and the (annual) runoff, R, is given by Equation 10.150 as )] [ ( I P − 3.004d0.5957 R = 0.15 + 0.75 100 = [0.15 + 0.75(0.65)](98.5) − 3.004(0.15)0.5957 = 61.8 cm Since there are 62 storms/year, the runoff per storm is 0.618/62 = 0.00997 m/storm. Therefore, the required volume, V , of the storage area is given by V = (0.00997 − 0.00335)(10 × 104 ) = 662 m3 From the storage-elevation function, h = 1.10 m. Taking Q = 3 m3 /s, the orifice equation gives 3 Q √ √ A= = = 0.993 m2 0.65 2gh 0.65 2(9.81)(1.10) Therefore, the orifice diameter, D, is given by √ D=

0.993 ×

4 = 1.12 m π

12.8. From the given data: A0 = 5 ha = 5 × 104 m2 , A1 = 2.55 ha = 2.55 × 104 m2 , e0 = 6.200 m, e1 = 6.300 m, P = 25.4 cm, and S = 7 cm. (a) The runoff depth is given by the NRCS equation as Q=

[25.4 − 0.2(7)]2 [P − 0.2S]2 = 18.6 cm = 0.186 m = P + 0.8S 25.4 + 0.8(7)

387

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Therefore, the runoff volume, V0 , is given by V0 = QA0 = (0.186)(5 × 104 ) = 9300 m3 Between elevations e0 = 6.200 m and e1 = 6.300 m, the site storage, S01 is given by A1 2.5 × 104 = (6.300 − 6.200) = 1250 m3 2 2 and the additional above-ground storage required is 9300 − 1250 = 8050 m3 . If h is the height of the berm above elevation 6.300 m, then 8050 8050 = = 0.322 m h= A1 2.5 × 104 S01 = (e1 − e0 )

Therefore the elevation of the top of the berm is 6.300 m + 0.322 m = 6.622 m . (b) For 1.3 cm of runoff, the storage volume required is 0.013 × 5 × 104 = 650 m3 . If e is the top elevation of the V notch, then A1 = 650 (e − e0 ) 2 2.5 × 104 (e − 6.200) = 650 2 which yields e= 6.252 m and the height of the V-notch weir is 0.052 = 52 mm. The maximum discharge rate from the weir is 650/(24 × 60 × 60) = 0.00752 m3 /s. According to the Kindsvater and Shen equation ( ) √ 5 8 θ (H + k) 2 Q = Cd 2g tan 15 2 ( ) √ 8 θ −6 2 0.00752 = (0.6072 − 0.000874θ + 6.1 × 10 θ ) 2(9.81) tan 15 2 ( )5 52 + 4.42 − 0.1035θ + 1.005 × 10−3 θ2 − 3.24 × 10−6 θ3 2 1000 which yields a notch angle of θ = 166◦ . 12.9. Since the infiltration capacity of the soil is 50 mm/h, the runoff versus time is given in the following table: Time (min) 0–5 5–10 10–15 15–20 20–25 25–30 30–35 35–40 40–45 Total

Rainfall Rate (mm/h) 24.1 30.4 41.6 67.4 169. 97.3 51.6 36.2 26.8 544.4

388

Runoff Rate (mm/h) 0.0 0.0 0.0 17.4 119. 47.3 1.6 0.0 0.0 185.3

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Since ∆t = 5 min = 0.0833 h, and the catchment area, A, is 5 ha = 50 000 m2 , the volume of runoff, V , is given by V = (185.3 × 10−3 ∆t)(A) = (185.3 × 10−3 × 0.0833)(50000) = 772 m3 Therefore the storage volume required in the retention area is 772 m3 . Since the retention area covers 700 m2 , the depth, d, of ponding expected in the retention area during the design storm is 772 d= = 1.10 m 700 12.10. From the given data: A = 4 ha = 4 × 104 m2 , L = 200 m, S0 = 0.5%, ADCIA = 0.8 ha = 8000 m2 , LDCIA = 80 m, and SDCIA = 0.7%. The IDF curve for Miami for T = 25 years is given by 7836 mm/h 48.6T −0.11 + t(0.5895 + T −0.67 ) 7836 = mm/h −0.11 48.6(25) + t(0.5895 + 25−0.67 ) 7836 mm/h = 34.1 + 0.705t

i=

(1)

First, consider the entire contributing area. According to Table 10.7, for single-family residential areas, C = 0.4 is a mid-range value. This value should be increased by 10% to account for a 25-year return period storm, which yields C = 0.44. Using the kinematic wave equation (Equation 10.13) with n = 0.25 (typical of lawns in Table 10.1) gives tc = 6.99

(nL)0.6 (0.25 × 200)0.6 358 = 6.99 = 0.4 min 0.3 0.4 0.3 0.4 ie (0.005) ie ie S0

(2)

Taking t = tc in Equation 1 and substituting Equation 2 gives [ ] 7836 ie = Ci = 0.44 mm/h 34.1 + 0.705 i358 0.4 e

which yields ie = 36.8 mm/h = 1.02 × 10−5 m/s, and tc = 89 min. The peak runoff from the entire catchment is therefore given by Qp = CiA = ie A = (1.02 × 10−5 )(4 × 104 ) = 0.41 m3 /s Considering the DCIA consisting of asphalt pavement, Table 10.1 gives a mid-range value of C = 0.83, which should be increased by 10% to account for the 25-year return period of the storm, which yields C = 1.1(0.83) = 0.91. Using the kinematic wave equation with nDCIA = 0.011 (Table 10.1 for asphalt) gives tc = 6.99

(nDCIA LDCIA )0.6 (0.011 × 80)0.6 28.7 = 6.99 = 0.4 min 0.3 0.4 0.3 0.4 ie (0.007) ie ie SDCIA

389

(3)

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Taking t = tc in Equation 1 and substituting Equation 3 gives [ ] 7836 ie = Ci = 0.91 mm/h 34.1 + 0.705 28.7 i0.4 e

10−5

m/s, and tc = 3.5 min. Since tc < 5 min, take tc which yields ie = 195 mm/h = 5.28 × = 5 min, which gives an effective rainfall of [ ] 7836 ie = 0.91 = 190 mm/h = 5.28 × 10−5 m/s 34.1 + 0.705(5) The peak runoff from the entire catchment is therefore given by Qp = CiADCIA = ie ADCIA = (5.28 × 10−5 )(8000) = 0.42 m3 /s Therefore the peak runoff is determined by the DCIA and is estimated as 0.42 m3 /s . For a 25-year 3-day storm, the average intensity can be determined by taking t = 3 days = 4320 min in the equation for the IDF curve, which gives i=

7836 = 2.54 mm/h 34.1 + 0.705(4320)

which gives a 3-day rainfall amount, P , of P = 2.54(24)(3) = 183 mm For a residential area with 1/2-acre lots on Group B soil, the curve number, CN, can be estimated from Table 9.14 as CN = 70, where CN =

1000 10 + S

which yields S = 4.29 in. = 109 mm, and hence the runoff, Q, is given by Q=

(P − 0.2S)2 (183 − 0.2 × 109)2 = = 96 mm P + 0.8S 183 + 0.8 × 109

The volume of runoff, V , to be stored in the lake can be estimated by multiplying the runoff depth, Q (= 96 mm = 0.096 m) by the site area, A (= 4 × 104 m2 ), hence V = QA = (0.096)(4 × 104 ) = 3840 m3 At the control elevation, the receiving lake is 15 m × 15 m and has 6:1 side slopes above the control elevation. For a height, h, above the control elevation, the volume stored in the lake, Vs , can be estimated by Vs =

1 [15 × 15 + (15 + 12h)(15 + 12h)] h = 9h(25 + 20h + 8h2 ) 2

When Vs = 3840 m2 , then 3840 = 9h(25 + 20h + 8h2 ) which yields a (real) solution of h = 2.87 m, which corresponds to a setback distance of 2.87 m × 6 = 17.2 m .

390

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12.11 The volume, V of runoff corresponding to a depth of 2.5 cm = 0.025 m on an area of 20 ha = 2 × 105 m2 is given by V = (0.025)(2 × 105 ) = 5000 m3 At a minimum infiltration rate of 100 mm/h = 0.10 m/h, the maximum area, A of infiltration basin required to infiltrate 5000 m3 in 36 h is given by A=

5000 = 1390 m2 (0.10)(36)

12.12. For T = 10 years, the average storm intensity for a storm of duration t (in minutes) is given by 7836 7836 i= = −0.11 −0.67 48.6(10) + t(0.5895 + 10 37.72 + 0.8034t ) The rainfall depth, D, for a storm of duration t (in minutes) is given by ( ) it 7836 t 130.6t D= = = 60 37.72 + 0.8034t 60 37.72 + 0.8034t

(1)

For the maximum value of D, dD (37.72 + 0.8034t)(130.6) − (130.6)(0.8034) = dt (37.72 + 0.8034t)2

(2)

When dD/dt = 0, Equation 2 gives 4926 + 104.9t − 104.9t = 0 which yields 492.6 = 0 which is clearly impossible. Therefore, there is no maximum rainfall amount, and the rainfall amount monotonically increases with the duration of the storm. As t → ∞, then dD/dt → 0 and Equation 1 gives 130.6t D→ = 163 mm 0.8034t For a site with a runoff coefficient of 0.4, the maximum amount of runoff is 0.4 (163 mm) = 65 mm. If the site is capable of retaining 50 mm of runoff, then 65 mm − 50 mm = 15 mm of runoff will be discharged from the site. Therefore, yes a portion of the 10-year runoff will be discharged from the site. 12.13 The length of the swale is determined by Q fP

(1)

2 1 1 AR 3 S 2 n

(2)

L= and the Manning equation requires Q=

391

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For a triangular channel, A = my 2 √ P = 2 1 + m2 y

(3) (4)

where y is the depth of flow in the channel. Combining Equations 2 to 4 and re-arranging yields 3 3 2 1 1 n 8 Q 8 (1 + m ) 8 4 y=2 (5) 3 5 m 8 S 16 and combining Equations 1 and 4 yields L=

Q 2f 1 + m2 y √

(6)

Finally, combining Equations 5 and 6 to eliminate y gives 5

L = 151400

5

3

Q 8 m 8 S 16 3

5

n 8 (1 + m2 ) 8 f

where the conversion factor has been included for f in cm/h. 12.14 From the Manning equation: Q=

2 1 1 AR 3 S 2 n

(1)

and for a trapezoidal section A = (b + my)y √ P = b + 2y 1 + m2

(2) (3)

Combining Equations 1 to 3 and simplifying gives 5

5

1 y 3 (b + my) 3 Q= √ n (b + 2y 1 + m2 ) 23

(4)

For the most efficient trapezoidal section √ b = 2( 1 + m2 − m) y

(5)

b 2( 1 + m2 − m)

(6)

or y=

√

Combining Equations 4 and 6 gives [ ]5 5 3 √ bm 3 b + y 1 1 2( 1+m2 −m) Q= S2 2 [ ] √ n 2 3 √ 1+m b + 2(2b 1+m2 −z)

392

(7)

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Solving for y gives

[ ]2 3 √ 3 5 1+m2 Q 5 b + (√b1+m n5 2 −m) [ ] y= 3 bm S 10 b + 2(√1+m 2 −m)

(8)

Combining Equations 8 and 5 to eliminate b gives ]3 [ 8 Qn y = 1.19 √ 1 (2 1 + m2 − m)S 2

(9)

Combining Equations 3 and 9 gives the following expression for P , ]3 [ 8 √ Qn P = b + 2.38 1 + m2 √ 1 2 2 (2 1 + m − m)S

(10)

The swale length, L, is given by Q fP If L is in m, Q is in m3 /s, and f in cm/h, then L=

L = 360000

Q fP

(11)

Combining Equations 10 and 11 gives L= {

360000Q

[

b + 2.38

]3 8

Qn

√ 1 (2 1+m2 −m)S 2

√

} 1 + m2

(12) f

12.15 From the given data: Q = 0.01 m3 /s, m = 5:1 (H:V) = 5, S = 0.015, n = 0.030, f = 200 mm/h = 20 cm/h, and Equation 12.11 gives the required swale length, L, as 5

L = 151400

5

3

5

n 8 (1 + m2 ) 8 f 5

= 151400

3

Q 8 m 8 S 16 5

3

(0.01) 8 (5) 8 (0.015) 16 3

5

(0.030) 8 (1 + 52 ) 8 20

= 257 m 12.16 With b = 1 m, Equation 12.12 gives L= {

[

Qn

b + 2.38 ={ 1 + 2.38

360000Q

]3 8

√ 1 (2 1+m2 −m)S 2

[

360000(0.01) (0.01)(0.03)

√ 1 (2 1+52 −5)(0.015) 2

= 107 m

393

√

} 1 + m2

]3 8

√

f }

1 + 52

20

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12.17 From the given data: Q = 0.002 m3 /s, So = 0.02, and the average height of the vegetation is 100 mm. This given data covers the specifications in steps 1 to 3 of the design procedure. Step 4: The design depth in the swale should be at least 50 mm below the height of the vegetation (100 mm − 50 mm = 50 mm), with a maximum height of 75 mm. Therefore, in this case, the design flow depth is taken as 50 mm. Step 5: Use a trapezoidal section for the swale. Step 6: Use side slopes of 4:1 (m = 4), a bottom width b, and a depth y = 50 mm (= 0.050 m). The flow area, A, wetted perimeter, P , and hydraulic radius, R, are given by A = by + my 2 = b(0.050) + (4)(0.050)2 = 0.050b + 0.01 √ √ P = b + 2 1 + m2 y = b + 2 1 + 42 (0.050) = b + 0.412 A 0.050b + 0.01 R= = P b + 0.412

(1)

where 0.6 m < b < 2.5 m. Step 7: The Manning equation requires that 5

1 2 1 1 A 3 21 Q = AR 3 S02 = S n n P 23 0

In this case, 5

1 1 (0.050b + 0.01) 3 (0.01) 2 0.002 = 2 n (b + 0.412) 3

or

1 (0.050b + 0.01)5 = 8 × 10−6 n3 (b + 0.412)2

(2)

For Class E retardance, Manning’s n is given by Equation 5.55 as 1

n=

1.22R 6 52.1 + 19.97 log(R1.4 S00.4 )

(3)

Simultaneous solution of Equations 1, 2 and 3 gives b = 2.0 m and the corresponding flow velocity (= Q/A) is 0.0182 m/s. Since the flow velocity is less than the maximum velocity of 0.3 m/s, a swale with a bottom width of 2.0 m should be used. Step 8: Using a detention time of 5 minutes with the design velocity of 0.0182 m/s gives the length, L, of the swale as L = V t = (0.0182)(5 × 60) = 5.46 m which is shorter than the minimum length of 30 m. Therefore use a length of 30 m.

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In summary, a swale with a a trapezoidal cross-section, a bottom width of 2.00 m , side slopes of 4:1, and 30 m long should be used. 12.18 From the given data: Q = 0.03175 m3 /s and Kt = 10 m/d = 1.157 × 10−4 m/s. The length,L, of the exfiltration trench is given by CiAt Qt = [nW + Kt t]H [nW + Kt t]H

L=

Taking W = 1 m, H = 2 m, and n = 0.4 gives L=

0.03175t (0.03175)t = [(0.4)(1) + (1.157 × 10−4 )(2)]H 0.8 + 2.314 × 10−4 t

In this case L increases monotonically with t, so take t → ∞, which yields L = 137 m. With a factor of safety of 2, the trench length should be 274 m. So the dimensions of the exfiltration trench are 1 m × 2 m × 274 m . 12.19 From the given data: C = 0.5, A = 3 ha = 30000 m2 , and Kt = 35/3 = 11.7 m/d = 0.486 m/h (using a safety factor of 3). According to ASCE (1996) guidelines, the porosity, n, of the gravel pack can be taken as 40% (n = 0.4), and a trench width, W , of 1 m and height, H, of 2 m can be expected to perform efficiently. Substituting these values into Equation 12.24 gives (0.5)i(30000)t 15000it CiAt = = (1) L= (nW + Kt t)H (0.4 × 1 + 0.486t)(2) 0.8 + 0.972t where i in m/h, and t in hours are related by i=

0.403 (60t + 8.16)0.69

m/h

(2)

Combining Equations 1 and 2 gives the required trench length, L, as a function of the storm duration, t, as 6045t L= (3) (0.8 + 0.972t)(60t + 8.16)0.69 Taking the derivative with respect to t gives dL dt

=

[(0.8 + 0.972t)(60t + 8.16)0.69 ]6045 − 6045t[(0.8 + 0.972t)0.69(60t + 8.16)−0.31 60 + (60t + 8.16)0.69 0.972] [(0.8 + 0.972t)(60t + 8.16)0.69 ]2 0.69

=

6045(0.8 + 0.972t)(60t + 8.16)

− 250260t(0.8 + 0.972t)(60t + 8.16)−0.31 − 5876t(60t + 8.16)0.69 [(0.8 + 0.972t)(60t + 8.16)0.69 ]2

and the maximum-value criterion, dL/dt = 0, yields 6045(0.8 + 0.972t)(60t + 8.16)

0.69

− 250260t(0.8 + 0.972t)(60t + 8.16)

−0.31

0.69

− 5876t(60t + 8.16)

=0

which gives t = 0.628 h, and substituting into Equation 3 yields L = 192 m. On the basis of these results, a trench length of 192 m gives the trench volume required to handle the design storm without causing ponding. Since the seasonal high water table is 4.6 m below the ground surface, and the minimum allowable spacing between the bottom of the trench and the water table is 1.2 m, a (maximum) trench height of 4.6 m − 1.2 m = 3.4 m would still be adequate. The trench is designed with dimensions 1 m × 2 m × 192 m .

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12.20. The required length of trench is given by L=

CiAt (nW + Kt t)H

(1)

From the given data: W = 1 m, Kt = 5/2 = 2.5 m/d = 0.104 m/h (assuming a factor of safety of 2), H = 3 m, and for the gravel pack it can be assumed that n = 0.4. Taking the entire contributing area, then the average runoff coefficient, C, is given by C = 0.35(0.9) + 0.65(0.4) = 0.58 where the contributing area, A, is 0.03 ha = 300 m2 . Substituting known values into Equation 1 gives 58it (0.58)i(300)t L= = (2) (0.4 × 1 + 0.104 × t)(3) 0.4 + 0.104t In this case, i is in m/h and t is in hours. Using these same units in the IDF equation gives i=

7836(10−3 ) 7.836 = 37.7 + 0.803(60)t 37.7 + 48.2t

(3)

Combining Equations 2 and 3 yields ( ) 58t 7.836 454.5t L= = 0.4 + 0.104t 37.7 + 48.2t 15.1 + 23.2t + 5.01t2 The maximum-value criterion requires that dL (15.1 + 23.2t + 5.01t2 )(454.5) − 454.5t(23.2 + 10.02t) = =0 dt (15.1 + 23.2t + 5.01t2 )2 which yields t = 1.74 h and hence L = 11.2 m. Considering only the DCIA, C = 0.9, A = 0.35(300) = 105 m2 and Equation 1 gives L=

31.5t (0.9)i(105)t = 0.4 × 1 + 0.104t 0.4 + 0.104t

Since this value of L is equal to a fraction of the value of L given previously by Equation 2, it is clear that the minimum value of L will still occur at t = 1.74 h, and the corresponding value of L will be less than 11.2 m. Therefore, the required trench length is 11.2 m . For t = 1.74 h (= 6264 s) the average intensity is given by Equation 3 which yields i=

7.836 7.836 = = 0.0645 m/h 37.7 + 48.2t 37.7 + 48.2(1.74)

Therefore the depth of runoff retained by the trench is equal to Cit = (0.58)(0.0645)(1.74) = 0.065 m = 6.5 cm . 12.21. The first step is to determine whether the exfiltration trench should be designed for a rainfall return period of 10 years or a return period of a storm in which 25 mm of runoff occurs in 1

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hour. Since the runoff coefficient is 0.6, a rainfall of 25/0.6 = 42 mm is expected to cause a runoff of 25 mm. Taking i = 42 mm/h and t = 60 min in the IDF curves yields 42 =

7836 48.6T −0.11 + 60(0.5895 + T −0.67 )

which gives T = 0.48 years. Therefore, designing the trench for a rainfall with a 10-year return period (T = 10 years) will result in an exfiltration trench capable of retaining/exfiltrating more than 25 mm of runoff in 1 hour. For T = 10 years, the design IDF curve is given by i=

48.6(10)−0.11

7836 7836 = −0.67 + t(0.5895 + 10 37.73 + 0.8033t )

(1)

From the given data: C = 0.6, A = 0.5 ha = 5000 m2 , and Kt = 8/2 = 4 m/d (using a safety factor of 2) = 0.167 m/h. According to ASCE (1998) guidelines, the porosity, n, of the gravel pack can be taken as 40% (n = 0.4), and a trench width, W , of 1 m and height, H, of 2 m can be expected to perform efficiently. Since the depth to the water table is 3.5 m, there is sufficient space to accommodate a trench depth of 2 m and a vertical separation of at least 1.2 m between the bottom of the trench and the water table. Substituting these values into Equation 12.24 gives L=

(0.6)i(5000)t 3000it CiAt = = (nW + Kt t)H (0.4 × 1 + 0.167t)(2) 0.8 + 0.333t

(2)

where Equation 1 gives i=

7.836 7.836 = m/h 37.73 + 0.8033(60t) 37.73 + 48.20t

(3)

Combining Equations 2 and 3 gives the required trench length, L, as a function of the storm duration, t, as L=

23510t 1465t = 2 (0.8 + 0.333t)(37.73 + 48.20t) t + 3.176t + 1.880

(4)

Taking the derivative with respect to t gives dL (t2 + 3.176t + 1.880)(1465) − (1465t)(2t + 3.176) 1465(t2 − 1.880) = = − dt (t2 + 3.176t + 1.880)2 (t2 + 3.176t + 1.880)2 and the maximum-value criterion, dL/dt = 0, yields t2 − 1.880 = 0 which gives t = 1.37 h, and substituting in Equation 4 yields L = 248 m. On the basis of these results, a 248 m × 1 m × 2 m (L × W × H) trench is capable of handling all 10-year storms without causing surface ponding. Alternative Solution: The above solution applies the factor of safety to the trench hydraulic conductivity. In many

397

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cases, the factor of safety is applied to the trench length instead of the trench hydraulic conductivity. In this case, the solution is as follows: The first step is to determine whether the exfiltration trench should be designed for a rainfall return period of 10 years or a return period of a storm in which 25 mm of runoff occurs in 1 hour. Since the runoff coefficient is 0.6, a rainfall of 25/0.6 = 42 mm is expected to cause a runoff of 25 mm. Taking i = 42 mm/h and t = 60 min in the IDF curves yields 42 =

48.6T −0.11

7836 + 60(0.5895 + T −0.67 )

which gives T = 0.48 years. Therefore, designing the trench for a rainfall with a 10-year return period (T = 10 years) will result in an exfiltration trench capable of retaining/exfiltrating more than 25 mm of runoff in 1 hour. For T = 10 years, the design IDF curve is given by i=

48.6(10)−0.11

7836 7836 = −0.67 + t(0.5895 + 10 ) 37.73 + 0.8033t

(5)

From the given data: C = 0.6, A = 0.5 ha = 5000 m2 , and Kt = 8 m/d = 0.333 m/h. According to ASCE (1998) guidelines, the porosity, n, of the gravel pack can be taken as 40% (n = 0.4), and a trench width, W , of 1 m and height, H, of 2 m can be expected to perform efficiently. Since the depth to the water table is 3.5 m, there is sufficient space to accommodate a trench depth of 2 m and a vertical separation of at least 1.2 m between the bottom of the trench and the water table. Substituting these values into Equation 12.24 gives L=

(0.6)i(5000)t 3000it CiAt = = (nW + Kt t)H (0.4 × 1 + 0.333t)(2) 0.8 + 0.0.667t

(6)

where Equation 5 gives i=

7.836 7.836 = m/h 37.73 + 0.8033(60t) 37.73 + 48.20t

(7)

Combining Equations 6 and 7 gives the required trench length, L, as a function of the storm duration, t, as L=

23510t 731.3t = 2 (0.8 + 0.0.667t)(37.73 + 48.20t) t + 1.982t + 0.939

(8)

Taking the derivative with respect to t gives 731.3(t2 − 0.939) dL (t2 + 1.982t + 0.939)(731.3) − (731.3t)(2t + 1.982) = − = dt (t2 + 1.982t + 0.939)2 (t2 + 1.982t + 0.939)2 and the maximum-value criterion, dL/dt = 0, yields t2 − 0.939 = 0 which gives t = 0.939 h, and substituting in Equation 8 yields L = 187 m. Applying a factor of safety of 2 yields a required trench length of 2 × 187 m = 374 m. On the basis of these results, a 374 m × 1 m × 2 m (L × W × H) trench is capable of handling all 10-year storms without causing surface ponding.

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12.22. (a) In accordance with the definition of the trench hydraulic conductivity, the maximum flow rate out of the side of the trench and above the water table is given by Qu = 2KLDu (H2 − 0.5Du ) and the maximum flow rate out of the side of the trench and below the water table is given by Qs = 2KLDs H2 Hence the maximum total flow is given by Q = Qs + Qu = 2KL[Du (H2 − 0.5Du ) + Ds H2 ] which gives L=

Q K(2H2 Du − Du2 + 2H2 Ds )

(1)

(b) For tc = 10 minutes and T = 5 years, the average rainfall intensity is 308.5 48.6T −0.11 + t(0.5895 + T −0.67 ) 308.5 = 6.17 in./h = 4.35 × 10−5 m/s = −0.11 48.6(5) + (10)(0.5895 + 5−0.67 )

i=

From the given data, H2 = 2.088 m − 1.250 m = 0.838 m Du = 2.088 m − 0.305 m − 1.250 m = 0.533 m Ds = 1.250 m − (−2.484 m) = 3.734 m K = 1.74 × 10−3 s−1 Q = CiA = (0.6)(4.35 × 10−5 )(0.8 × 10−4 ) = 0.209 m3 /s Substituting into Equation 1 gives Q K(2H2 Du − Du2 + 2H2 Ds ) 0.209 = = 17.5 m −3 1.74 × 10 [2(0.838)(0.533) − (0.533)2 + 2(0.838)(3.734)]

L=

(c) The IDF curve can be expressed in the form i=

a′ b+t

where i is the rainfall intensity in m/s, t is the duration in seconds, and a′ and b are constants given by 0.130 0.5895 + T −0.67 2916T −0.11 b= 0.5895 + T −0.67

a′ =

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The runoff volume, V , resulting from a storm event of duration t is given by V = CiAt The storage volume, Vs , in the trench is given by V0 = 0.5W Du L and the exfiltrated volume, VS , out of the sides of the trench in time t is given by VS = 2KL[Du (2 −0.5Du ) + Ds H2 ]t Equating the runoff volume to the stored and exfiltrated volume gives V = V0 + VS CAa′ t = 0.5W Du L + 2KL[Du (2 −0.5Du ) + Ds H2 ]t b+t which can be rearranged and put in the form L=

at (b + t)(c + dt)

(2)

where 0.130CA 0.5895 + T −0.67 2916T −0.11 b= 0.5895 + T −0.67 c = 0.5W Du

a=

d = 2K[Du (H2 − 0.5Du ) + Ds H2 ] The maximum length occurs at t = t∗ , where dL =0 dt t=t∗ Combining Equations 2 and 3 yields √ t∗ =

bc d

and the (maximum) required trench length is given by L=

at∗ (b + t∗ )(c + dt∗ )

400

(3)

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(d) From the given data, 0.130CA 0.130(0.6)(0.8 × 104 ) = = 671 0.5895 + T −0.67 0.5895 + 5−0.67 2916T −0.11 2916(5)−0.11 b= = = 2628 0.5895 + T −0.67 0.5895 + 5−0.67 c = 0.5W Du = 0.5(1)(0.533) = 0.267

a=

d = 2K[Du (H2 − 0.5Du ) + Ds H2 ] = 2(1.74 × 10−3 )[(0.533)(0.838 − 0.5 × 0.533) + (3.734)(0.838)] = 0.01195 √ √ bc (2628)(0.267) ∗ t = = = 242 s = 4.04 min d 0.01195 (671)(242) L= = 17.9 m (2628 + 242)(0.267 + 0.01195 × 242) Choose a trench length of 17.9 m (instead of 17.5 m) to be more conservative. The difference is about 2%. (e) For T = 5 years and t0 = 10 minutes, i0 = 4.35 × 10−5 m/s 0.130CA 0.130CA a= = = 0.141CA −0.67 0.5895 + T 0.5895 + 5−0.67 b = 2628 √ √ bc 2628c c ∗ t = = → = 3.8 × 10−4 t∗ d d d ( a ) −5 4.35 × 10 Ci0 A 0.141 = → a = 3230dL0 L0 = d d at∗ (3230dL0 )t∗ L= = (b + t∗ )(c + dt∗ ) (2628 + t∗ )(c + dt∗ ) L 3230t∗ ( ) = L0 (2628 + t∗ ) dc + t∗ Since c/d L0 when t∗ < 600 s, which means that c/d < 3.8 × 10−4 (600) = 0.228, which, in terms of trench parameters gives 0.5W Du < 0.228 2K[Du (H2 − 0.5Du ) + Ds H2 ] 12.23. From the given data: A = 0.5 ha = 5000 m3 , dWQV = 2.5 cm = 0.025 m, Kt = 10 m/d, W = 1 m, H = 1.5 m, and n = 0.4. The water-quality volume, WQV, is WQV = A × dWQV = (5000)(0.025) = 125 m3

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If L is the length of the trench required to store the WQV, then nLW H = WQV (0.4)L(1)(1.5) = 125 m3 which yields L = 208 m. The evacuation time, Te , of the runoff stored in the trench is given by nLW H nW (0.4)(1) = = = 0.04 days = 1 hour Te = Kt HL Kt (10) A trench length of 208 m will provide adequate retention storage. The evacuation time of 1 h is much less than the regulatory limit of 3 days.

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Chapter 13

Estimation of Evapotranspiration 13.1. According to Equation 13.6,

( ln u2 = ln

(

2−d zom z−d zom

) ) uz

(1)

Taking h = 0.12 m, Equation 13.3 gives 2 2 d = h = (0.12) = 0.08 m 3 3 zom = 0.123h = 0.123(0.12) = 0.0148 m

(2) (3)

Combining Equations 1 to 3 gives ) ( ln 2−0.08 4.87 0.0148 u2 = ( z−0.08 ) uz = u z = u2 ln(67.6z − 5.41) ln 0.0148 This equation is (very nearly) the same as Equation 13.6. 13.2. From the give data, the average temperatures in March, April, and May are 14.1◦ C, 16.1◦ C, and 18.8◦ C respectively. According to Equation 13.33, the average monthly heat flux in April can be estimated by Gmonth = 0.07(Ti+1 − Ti−1 ) where Ti+1 = 18.8◦ C, Ti−1 = 14.1◦ C, and therefore Gmonth = 0.07(18.8 − 14.1) = 0.329 MJ/(m2 ·d) The equivalent evaporation rate can be expressed as evaporation rate =

Gmonth ρλ

where ρ = 998 kg/m3 , λ = 2.45 MJ/kg, and hence evaporation rate =

0.329 = 1.34 × 10−4 m/d = 0.13 mm/d (998)(2.45)

403

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13.3 The evapotranspiration rate from any vegetated surface is given by the Penman-Monteith (PM) equation (Equation 13.1) as   es −ea ∆(R − G) + ρ c 1  n a p ra  ( ) ETc = ρw λ ∆ + γ 1 + rs ra

The calculations to determine the variables in the PM equation are as follows • Adjust wind speed to 2 m above the ground. For grass, 4.87 4.87 u2 = uz = u3 = 0.921u3 ln[67.8z − 5.42] ln[67.8(3) − 5.42] For alfalfa, h = 0.50 m 2 2 d = h = (0.50) = 0.333 m 3 3 zom = 0.123h = 0.123(0.50) = 0.615 m ( ) ( ) ln 2−d ln 2−0.333 zom 0.0615 ) uz = ( 3−0.333 ) u3 = 0.875u3 u2 = u 2 = ( ln 0.0615 ln z−d zom • Calculate resistance factors. For grass, 0.00241 208 s/m = d/m ra = u2 u2 rs = 70 s/m = 8.10 × 10−4 d/m and for alfalfa, 0.00127 110 s/m = d/m u2 u2 rs = 45 s/m = 5.21 × 10−4 d/m

ra =

• The net radiation, Rn , is given by Rn = (1 − α)Rs + Ln where α can be taken as 0.23 for grass and 0.17 for alfalfa. The extraterrestrial solar radiation, S0 , is given by S0 =

24(60) Gsc dr (ωs sin ϕ sin δ + cos ϕ cos δ sin ωs ) π

where Gsc = 0.0820 MJ/(m2 ·min ) ( 2π dr = 1 + 0.033 cos J = 1 + 0.033 cos(0.0172J) 365 ( ) 2π δ = 0.4093 sin J − 1.405 = 0.4093 sin(0.0172J − 1.405) 365 ωs = cos−1 (− tan ϕ tan δ)

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The elevation of Greeley, Colorado, is 1462.4 m (z = 1462.4 m), and hence the clear-sky radiation, Rso , can be estimated by Rso = [0.75 + 2 × 10−5 z] = [0.75 + 2 × 10−5 (1462.4)]S0 = 0.78S0 The net longwave radiation, Ln , can be estimated by ( ) ( ) 4 4 Tmax,K + Tmin,K √ Rs Ln = −σ (0.34 − 0.14 ea ) 1.35 − 0.35 2 Rs0 For daily time intervals, it can be assumed that G = 0 MJ/(m2 ·d • Take λ = 2.45 MJ/kg • The pressure, p, at elevation z = 1462.4 m is given by [ ] ] [ 293 − 0.0065z 5.26 293 − 0.0065(1462.4) 5.26 p = 101.3 = 101.3 = 85.2 kPa 293 293 • The psychrometric constant, γ, is given by γ = 0.0016286

p 85.2 = 0.0016286 = 0.0566 kPa/◦ C λ 2.45

• The saturation vapor pressure, es , can be estimated by [ ( ) ( )] 17.27Tmax 17.27Tmin es = 0.3054 exp + exp Tmax + 237.3 Tmin + 237.3 • The vapor pressure gradient, ∆, can be estimated by ∆=

4098es (T + 237.3)2

where T is the average of the maximum and minimum temperatures. • The air density, ρa , can be estimated using the realtion ρa = 3.450

p kg/m3 T + 273

• The specific heat of moist air, cp , can be taken as 1.013 × 10−3 MJ/(kg·◦ C). • The density of water, ρw , can be taken as 998 kg/m3 . The above-itemized relations are used in the Penman-Monteith equation to calculate the reference grass and alfalfa ET, and these calculations are tabulated in Figure 13.1. These results indicate that, over the 7-day interval, the grass-reference ET varies in the range of 5.8–7.8 mm/d , and the alfalfa-reference ET varies in the range 7.7–10.1 mm/d . These results show that over a 7-day interval there can be significant variation in the reference ET (up to around 50%).

405

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Figure 13.1: Daily ET Spreadsheet

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Jan

Feb

Mar

Ap r

May

Jun

Jul

Aug

Sep

Oct

Nov

Dec

Tmean (oC)

17.3

19.1

20.9

22.5

25.1

26.4

27.2

27.2

26.2

24.8

21.4

18.8

Tmax (oC)

23.2

23.4

25.0

25.8

27.3

28.3

28.8

28.9

28.4

27.4

25.3

23.4

Tmin (oC) RHmin(%) RHmax (%) u10 (m/s)

8.5 70.9 96.1 3.52

12.1 71.9 94.8 3.60

15.1 71.3 96.0 3.84

17.8 67.2 94.0 3.45

22.0 74.0 94.1 2.91

24.0 79.5 95.0 2.66

24.9 80.6 93.9 2.49

24.8 81.1 95.8 2.44

24.2 80.8 96.9 2.56

21.0 75.8 95.2 3.11

16.1 76.0 96.9 3.39

10.3 74.3 97.1 3.35

12.51 2.63

14.60 2.69

18.18 2.87

20.03 2.58

21.70 2.18

19.54 1.99

19.95 1.86

18.33 1.83

15.97 1.91

16.11 2.33

13.69 2.54

11.59 2.51

-2

-1

Rs (MJ m d ) u2 (m/s) -1

ra (d m )

9.14E-04 8.94E-04 8.38E-04 9.33E-04 1.11E-03 1.21E-03

1.29E-03 1.32E-03 1.26E-03 1.03E-03 9.49E-04

9.61E-04

rs (d m-1)

8.10E-04 8.10E-04 8.10E-04 8.10E-04 8.10E-04 8.10E-04 0 0 0 0.03 0.11 0.12 12.51 14.60 18.18 19.35 19.25 17.13 2.84 2.88 3.17 3.32 3.63 3.85 1.11 1.41 1.72 2.04 2.64 2.98 1.76 1.87 2.13 2.25 2.69 3.01 0.15 0.15 0.14 0.13 0.11 0.10 0.69 0.68 0.77 0.78 0.71 0.74 0.72 0.71 0.79 0.80 0.74 0.77 -3.82 -3.72 -3.90 -3.88 -3.15 -2.93

8.10E-04 8.10E-04 8.10E-04 8.10E-04 8.10E-04 0.14 0.13 0.11 0.08 0 17.16 15.95 14.21 14.90 13.69 3.96 3.98 3.87 3.65 3.22 3.15 3.13 3.02 2.49 1.83 3.13 3.18 3.09 2.68 2.26 0.09 0.09 0.09 0.11 0.13 0.76 0.75 0.72 0.72 0.67 0.78 0.78 0.75 0.75 0.70 -2.88 -2.79 -2.77 -3.18 -3.34

8.10E-04 0 11.59 2.88 1.25 1.86 0.15 0.65 0.69 -3.55

Sn (MJ m-2 d-1) es(Tmax) (kPa) es(Tmin) (kPa) ea (kPa) e’ n/N f -2 -1 Ln (MJ m d ) -2

-1

Rn (MJ m d ) -1

(MJ kg ) (kPa oC-1) es (kPa) o

T ( C) (kPa oC-1) 3 a (kg/m ) 3 w (kg/m ) -1 o -1

cp (MJ kg

C )

8.69

10.88

14.28

15.47

16.09

14.20

14.28

13.16

11.45

11.72

10.35

8.04

2.45 0.06752 1.98

2.45 0.06752 2.14

2.45 0.06752 2.44

2.45 0.06752 2.68

2.45 0.06752 3.14

2.45 0.06752 3.42

2.45 0.06752 3.55

2.45 0.06752 3.56

2.45 0.06752 3.44

2.45 0.06752 3.07

2.45 0.06752 2.53

2.45 0.06752 2.07

15.85 0.12 1.21

17.75 0.13 1.20

20.05 0.15 1.19

24.65 0.19 1.17

26.15 0.20 1.17

26.85 0.21 1.17

26.85 0.21 1.17

26.30 0.20 1.17

24.20 0.18 1.18

20.70 0.15 1.19

16.85 0.12 1.21

998

998

998

998

998

998

998

998

998

21.80 0.16 ) 1.19 998

1.01E-03 1.01E-03 1.01E-03 1.01E-03 1.01E-03 1.01E-03

998 1.01E-03

1.01E-03 1.01E-03 1.01E-03 1.01E-03

998 1.01E-03

ETPM (mm d-1)

2.18

2.81

3.71

4.33

4.69

4.23

4.32

3.97

3.45

3.49

2.82

ETAbtew (mm d-1)

2.71

3.16

3.93

4.33

4.69

4.23

4.32

3.97

3.45

3.49

2.96

2.51

Diff (mm d-1

0.53

0.35

0.23

0.00

0.00

0.00

0.00

0.00

0.00

0.00

0.14

0.45

ETPM1 (mm d-1) Error-PM1 (%)

3.23 48

3.91 39

5.07 37

5.56 28

5.77 23

5.24 24

5.29 23

4.92 24

4.36 26

4.44 27

3.86 37

3.05 48

Figure 13.2: ET Calculations I

407

2.05

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13.4. (a) The PM calculations and the monthly albedos (given in the row after Rs ) that give the best agreement between the PM and Abtew equations are shown in Figure 13.2. Optimal values of α vary between 0 (not realistic) and 0.14. It appears that albedos are relatively small in the fall and winter when the sky is more cloudy, compared to the summer when there is relatively more sunshine and albedos are higher. The corresponding agreement between the PM and Abtew equations are shown in Figure 13.3. 8.00 PM Abtew

7.00

6.00

ET (mm/d)

5.00

4.00

3.00

2.00

1.00

0.00 1

3

5

7

9

11

Month

Figure 13.3: Final ET Values (b) If the wet season is defined as when rainfall exceeds ET, then the wet season in June to October and the dry season is November to May . These results are shown in Figure 13.4. Jan 31 7

Feb 28 8

Mar 31 11

Apr 30 13

May 31 15

Jun 30 13

Jul 31 13

Aug 31 12

Sep 30 10

Oct 31 11

Nov 30 8

Dec 31 6

Total

days ETPM (cm) ETPM1 (cm)

10

11

16

17

18

16

16

15

13

14

12

9

166

ETAbtew (cm) Rain (cm)

8 3

9 5

12 9

13 6

15 8

13 19

13 16

12 18

10 18

11 14

9 8

8 6

133 130

128

Figure 13.4: ET Calculations II (c) Comparing the ET using the full PM equation (ETPM ) with the ET using only the solar radiation (ETPM1 ) shows that the greatest discrepancies occur during the wet season (∼ 50%) and the least discrepancies occur during the dry season (∼ 25%). These results indicate that wind speed and humidity have a greater effect on dry-season ET than wetseason ET. One would suspect that the Abtew equation is most accurate during the wet season and least accurate during the dry season. 13.5 From the given data: u2 = 4 m/s, FET = 300 m, RHmean = 0.80, Ep = 152 mm, and Table 13.4 gives kp = 0.80. The reference evapotranspiration, ETo , is therefore given by ETo = kp Ep = (0.80)(152) = 122 mm 13.6 Equations 13.62 and 13.63 are used to calculate the pan coefficients shown in Figure 13.5. Comparing these data with Table 13.4 indicates that Equation 13.62 provides reasonable

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predictions for pans surrounded by short green crops, while Equation 13.63 is not in close agreement with Table 13.4 for dry bare areas.

Figure 13.5: Predicted Pan Coefficients 13.7. From the given data: ETp = 125 cm, P = 150 cm, and the index of dryness is given by ETp 125 = = 0.83 P 150 Using Equation 13.73 with w = 2.55 (for grassed catchments) gives [ ( )w ] 1 w P P ET =1+ − 1+ ETp ETp ETp [ ( ) ] 1 1 1 2.55 2.55 =1+ = 0.748 − 1+ 0.83 0.83 index of dryness =

Therefore, the actual evapotranspiration, ET, in central Florida can be estimated by ET = 0.748 ETp = 0.748(125 cm) = 94 cm

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Chapter 14

Fundamentals of Ground-Water Hydrology I: Governing Equations 14.1. The hydraulic conductivity of an aquifer is a better measure of the productivity of an aquifer than the porosity. The hydraulic conductivity measures the flow for a given head gradient, while the porosity is simply a measure of the amount of void space. 14.2. From the given data: K = 10 m/d n = 0.2 hB = hA − 10 cm hC = hA − 8 cm The seepage velocity along AB, vAB , is given by vAB = −

10 −0.10 K JAB = − = 0.005 m/d n 0.2 1000

This is the component of the seepage velocity along AB and is therefore not equal to the actual seepage velocity. The specific discharge along AC, qAC , is given by qAC = −KJAC = −10

−0.08 = 0.0016 m/d 500

and the specific discharge along AB, qAB , is given by qAB = nvAB = 0.2(0.005) = 0.001 m/d The components of specific discharge (= Darcy velocity) are therefore given by∗ qAB = 0.001 = qx cos 45◦ + qy sin 45◦ = 0.7071qx + 0.7071qy qAC = 0.016 = qx cos 50◦ − qy cos 40◦ = 0.6428qx − 0.7660qy ∗

The x-direction is to the east, and the y direction is to the north.

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Figure 14.1: Problem 14.2. Solving gives qx = 0.0019 m/d and qy = -0.00049 m/d . The components of the seepage velocity are given by qx 0.0019 = = 0.0095 m/d n 0.2 qy −0.00049 vy = = = −0.0025 m/d n 0.2

vx =

14.3. The mean position of the cloud is advected at the seepage velocity , not at the Darcy velocity. The seepage velocity is the rate at which ground water moves through the pores. A contaminant cloud “disperses” because the seepage velocity varies in space. 14.4. From the given data: Kwater = 20 m/d at 20◦ C. By definition K=k

γ µ

and for water at 20◦ C, γ = ρg = 998.2 × 9.81 = 9792 N/m3 µ = 1.002 × 10−3 N·s/m2 Substituting into Equation 1 gives† 20 †

1 9792 =k 86400 1.002 × 10−3

86400 s = 1 day

412

(1)

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which yields k = 2.37 × 10−11 m2 For PERC at 20◦ C, Appendix B.2 gives γ = ρg = 1623 × 9.81 = 15922 N/m3 µ = 0.9 × 10−3 N·s/m2 Substituting into Equation 1 gives K=k

γ 15922 = 2.37 × 10−11 × × 86400 = 36.3 m/d µ 0.9 × 10−3

Since KPERC > Kwater (36.3 m/d > 20 m/d), under the same piezometric gradient PERC will move faster than water. 14.5. From the given data: d10 = 0.5 mm and Uc = 3.5. The porosity, n can be estimated using Equation 14.22 as n = 0.255(1 + 0.83Uc ) = n = 0.255(1 + 0.833.5 ) = 0.388 For water at 20◦ C, ν = 1.004 × 10−6 m2 /s, and the hydraulic conductivity, K, is given by Equation 14.18 as K = α · ϕ(n) · d2e ·

g 9.81 = α · ϕ(n) · d2e · = 9.77 × 106 · α · ϕ(n) · d2e ν 1.004 × 10−6

where K is in m/s if de is expressed in meters. If de is expressed in mm and K in m/d, then Equation 14.23 is given by K = 8.44 × 105 · α · ϕ(n) · d2e Using the given data, K can be calculated using the empirical equations of Hazen, Slichter, Terzaghi, and Beyer which yield the following results Name Hazen

α (-) 6 × 10−4

ϕ(n) (-) 2.28

de (mm) 0.5

K (m/d) 289

Restriction met Y

Slichter

1 × 10−2

0.0445

0.5

94

Y

Terzaghi

8.4 × 10−3

0.0923

0.5

164

Y

Beyer

1.04 × 10−3

1

0.5

219

Y

The range of hydraulic conductivities calculated of 94-289 m/d is higher than the typical range of 0.1–50 m/d for medium sand given in Table 14.2.

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14.6. From the given data n = 0.20 and the grain size distribution gives: d10 = 0.124 mm, d17 = 0.167 mm, d20 = 0.180 mm, d60 = 0.711 mm, and Uc = d60 /d10 = 0.711/0.124 = 5.73. For water at 20◦ C, ν = 1.004 × 10−6 m2 /s, and the hydraulic conductivity, K, is given by Equation 14.18 as

K = α · ϕ(n) · d2e ·

g 9.81 = α · ϕ(n) · d2e · = 9.77 × 106 · α · ϕ(n) · d2e ν 1.004 × 10−6

(1)

where K is in m/s if de is expressed in meters. If de is expressed in mm and K in m/d, then Equation 14.23 is given by K = 8.44 × 105 · α · ϕ(n) · d2e

(2)

To facilitate calculation of de by the empirical formulae in Table 14.3, the given grain size distribution is more conveniently expressed in the following tabular form

Size (mm)

Fraction finer

4.760

0.98

2.000

0.86

0.840

0.64

0.420

Interval, i

Fraction retained, ∆gi

Upper Size, dgi (mm)

Lower Size, ddi (mm)

7

0.12

4.760

2.000

6

0.22

2.000

0.840

5

0.13

0.840

0.420

4

0.15

0.420

0.250

3

0.23

0.250

0.149

2

0.09

0.149

0.074

0.51

0.250

0.36

0.149

0.13

0.074

0.04

From the above-tabulated grain-size distribution: ∆g1 = 0.04 and d1 = 0.074 mm. Using the formulas in Table 14.3 with the grain-size distribution data to calculate the parameters in Equation 14.24 gives the following results

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Name Hazen

α (-) 6 × 10−4

ϕ(n) (-) 0.4

de (mm) 0.124

K (m/d) 3.1

Restriction met N

Slichter

1 × 10−2

0.00504

0.124

0.7

Y

Terzaghi

8.4 × 10−3

0.00569

0.124

0.6

N

Beyer

1.16 × 10−3

1

0.124

0.1

Y

Sauerbrei

3.75 × 10−3

0.0125

0.167

1.1

Y

Kr¨ uger

4.35 × 10−5

0.313

0.357

1.5

Y

Kozeny

8.3 × 10−3

0.0125

0.244

5.2

N

Zunker

1.24 × 10−3

0.0625

0.250

4.1

Y

Zamarin

8.3 × 10−3

0.0119

0.629

33.0

Y

USBR

2.87 × 10−4

1

0.180

7.8

N

Therefore the hydraulic conductivity is in the range of 0.1–33 m/d .

14.7. From the given data n = 0.25 and the grain size distribution gives: d10 = 0.015 mm, d17 = 0.050 mm, d20 = 0.055 mm, d60 = 0.150 mm, and Uc = d60 /d10 = 0.150/0.015 = 10. For water at 20◦ C, ν = 1.004 × 10−6 m2 /s, and the hydraulic conductivity, K, is given by Equation 14.18 as

K = α · ϕ(n) · d2e ·

g 9.81 = α · ϕ(n) · d2e · = 9.77 × 106 · α · ϕ(n) · d2e ν 1.004 × 10−6

(1)

where K is in m/s if de is expressed in meters. If de is expressed in mm and K in m/d, then Equation 14.23 is given by K = 8.44 × 105 · α · ϕ(n) · d2e

(2)

To facilitate calculation of de by the empirical formulae in Table 14.3, the given grain size distribution is more conveniently expressed in the following tabular form

415

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Size (mm)

Fraction finer

6.8000

1.00

5.0000

0.96

0.6000

0.83

0.4200

0.81

0.3000

0.79

0.2000

0.76

0.1800

0.64

0.0900

0.32

0.0750

0.275

0.0380

0.10

0.0220

0.10

0.0150

0.10

0.0095

0.09

0.0068

0.09

0.0015

Fraction retained, ∆gi

Upper Size, dgi (mm)

Lower Size, ddi (mm)

17

0.04

6.8000

5.0000

16

0.13

5.0000

0.6000

15

0.02

0.6000

0.4200

14

0.02

0.4200

0.3000

13

0.03

0.3000

0.2000

12

0.05

0.2000

0.1800

11

0.05

0.1800

0.1600

10

0.32

0.1600

0.0900

9

0.045

0.0900

0.0750

8

0.175

0.0750

0.0380

7

0.0

0.0380

0.0220

6

0.0

0.0220

0.0150

5

0.01

0.0150

0.0095

4

0.0

0.0095

0.0068

3

0.01

0.0068

0.0035

2

0.005

0.0035

0.0015

0.71

0.1600

0.0035

Interval, i

0.08 0.075

From the above-tabulated grain-size distribution: ∆g1 = 0.075 and d1 = 0.0015 mm. Using the formulas in Table 14.3 with the grain-size distribution data to calculate the parameters in Equation 14.24 gives the following results

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Name Hazen

α (-) 6 × 10−4

ϕ(n) (-) 0.9

de (mm) 0.015

K (m/d) 0.10

Restriction met N

Slichter

1 × 10−2

0.0105

0.015

0.02

Y

Terzaghi

8.4 × 10−3

0.0174

0.015

0.03

N

Beyer

1.02 × 10−3

1

0.015

0.19

N

Sauerbrei

3.75 × 10−3

0.0278

0.050

0.22

Y

Kr¨ uger

4.35 × 10−5

0.444

0.085

0.12

N

Kozeny

8.3 × 10−3

0.0278

0.071

0.98

N

Zunker

1.24 × 10−3

0.111

0.073

0.62

Y

Zamarin

8.3 × 10−3

0.0225

0.441

30.65

N

USBR

2.01 × 10−4

1

0.055

0.51

N

Therefore, among all formulas the hydraulic conductivity is predicted to be in the range of 0.02–30.65 m/d , however, among the valid formulas the hydraulic conductivity is in the range of 0.02–0.62 m/d . 14.8. Considering Figure 14.2, continuity requires that the flow into a cylinder of radius r be given by Qw = 2πrHne v which re-arranges to v= 14.9. The given relationship is

Qw 2πrHne

′ ′ Kij = K11 αi1 αj1 + K22 αi2 αj2

(1)

From Figure 14.3, α11 = cos θ α22 = cos θ α21 = cos(90 + θ) = − sin θ α12 = cos(90 − θ) = sin θ and substituting into Equation 1 gives ′ ′ K11 = K11 cos2 θ + K22 sin2 θ

417

(2)

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Figure 14.2: Problem 14.8.

Noting the identities 1 cos2 θ = (1 + cos 2θ) 2 2 sin θ = 1 − cos2 θ

(4)

sin 2θ = 2 sin θ cos θ

(5)

and substituting Equations 3 and 4 into Equation 2 gives ′ ′ K11 = K11 cos2 θ + K22 (1 − cos2 θ) ′ ′ ′ = K22 + (K11 − K22 ) cos2 θ ′ = K22 +

′ − K′ ′ K11 K ′ − K22 22 + 11 cos 2θ 2 2

which gives K11 =

′ ′ + K′ K11 K ′ − K22 22 + 11 cos 2θ 2 2

K22 =

′ + K′ ′ K11 K ′ − K22 22 − 11 cos 2θ 2 2

Similarly,

K12 = K21 = −

418

′ − K′ K11 22 sin 2θ 2

(3)

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Figure 14.3: Problem 14.9. 14.10. The given equations are: ′ + K′ ′ K11 K ′ − K22 22 + 11 cos 2θ 2 2 ′ ′ K ′ + K22 K ′ − K22 = 11 − 11 cos 2θ 2 2 ′ K ′ − K22 = K21 = − 11 sin 2θ 2

K11 =

(1)

K22

(2)

K12

(3)

Equations 1 and 2 combine to give ′ ′ K11 + K22 = K11 + K22

K11 − K22 = which means

(

K11 − K22 2

from Equation 3

)2

( =

( 2 K12

′ (K11

=

−

(4)

′ K22 ) cos 2θ

′ − K′ K11 22 2

′ − K′ K11 22 2

(5)

)2 cos2 2θ

(6)

)2 sin2 2θ

(7)

Combining Equations 6 and 7 gives [( ±

K11 − K22 2

]1

)2

2

+

419

2 K12

=±

′ − K′ K11 22 2

(8)

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where the identity cos2 2θ + sin2 2θ = 1 has been used. From Equation 4: ′ − K′ K11 K11 + K22 ′ 22 = − K22 2 2 ′ K ′ − K22 K11 + K22 ′ − 11 = − K11 2 2

(9)

+

(10)

Combining Equations 10 and 8 gives

′ K11

K11 + K22 = + 2

[(

K11 − K22 2

]1

)2

2

+

2 K12

+

2 K12

and combining Equations 9 and 8 gives

′ K22

K11 + K22 − = 2

[(

K11 − K22 2

]1

)2

2

14.11. Adding Equations 14.30 and 14.31 gives ′ ′ K11 + K22 = K11 + K22

(1)

and Equation 14.32 can be put in the form ′ ′ K11 − K22 =−

K12 sin 2θ

(2)

Substituting Equation 1 and Equation 2 into Equation 14.30 gives 2K11 = K11 + K22 −

2 K12 cos 2θ sin 2θ

which can be rearranged to give tan 2θ =

−2K12 K11 − K22

and solving for θ yields 1 2K12 θ = − tan−1 2 K22 − K11 According to the definition of θ, this is the angle of rotation a principal axis to the corresponding coordinate axis, counterclockwise positive. Therefore, the angle of rotation needed to reach a principal axis from the corresponding coordinate axis is θ′ = −θ, and is given by θ′ =

1 2K12 tan−1 2 K22 − K11

420

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14.12. From the given data: K11 = 15 m/d, K22 = 5 m/d, and K12 = K21 = 0 m/d. From Problem 14.2, the aquifer has a porosity, n, equal to 0.2, and using the linear relation h(x, y) = ax + by + c with the head data in Problem 14.2 gives 0 = a(0) + b(0) + c

(1)

−0.1 = a(707) + b(707) + c

(2)

−0.08 = a(321) + b(383) + c

(3)

Solving Equations 1 to 3 gives a = −1.91 × 10−4 ,

b = 4.91 × 10−5 ,

c=0

which means that ∂h = a = −1.91 × 10−4 ∂x ∂h J2 = = b = 4.91 × 10−5 ∂y J1 =

and therefore 15 K11 J1 = − (−1.91 × 10−4 ) = 1.43 × 10−2 m/d n 0.2 K22 5 v2 = − J2 = − (4.91 × 10−5 ) = −1.23 × 10−3 m/d n 0.2 v1 = −

The seepage velocity relative to the x1 axis is therefore 0.014 m/d @ −4.9◦ . For a homogeneous medium with K = 10 m/d, 10 K J1 = − (−1.91 × 10−4 ) = 9.55 × 10−3 m/d n 0.2 K 10 v2 = − J2 = − (4.91 × 10−5 ) = −2.46 × 10−3 m/d n 0.2 v1 = −

The seepage velocity relative to the x1 axis is therefore 0.00986 m/d @ −14.4◦ . Therefore, the anisotropic case yields a significantly higher seepage velocity. 14.13. From the given data: Kxx = 100 m/d, Kyy = 10 m/d, and Kxy = 0 m/d. For head gradient in east direction: Jx = −0.01 Jy = 0 qx = −Kxx Jx = −(100)(−0.01) = 1 m/d

421

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For head gradient in southeast direction: Jx = (−0.01) cos 45◦ = −0.00707 Jy = −(−0.01) sin 45◦ = 0.00707 qx = −Kxx Jx − Kxy Jy = −(100)(−0.00707) − (0)(0.00707) = 0.707 m/d qy = −Kyx Jx − Kyy Jy = −(0)(−0.00707) − (10)(0.00707) = −0.0707 m/d √ Therefore, the specific discharge has a magnitude of 0.7072 + 0.07072 = 0.711 m/d at an angle of tan−1 (−0.0707/0.707) = −5.7◦ to the x direction. The component of the flow in the direction of the (southeast) gradient is given by qJ = 0.711 cos(45◦ − 5.7◦ ) = 0.55 m/d The ground-water flow is in the direction of the head gradient only when the gradient is in the x-direction. 14.14. (a) Field measurements should include: • Water table measurements at three or more locations in the aquifer. The measurement locations should be selected such that the slope of the water table at the spill location can be calculated. • Measure of the hydraulic conductivity. This can be done in a variety of ways, such as a slug test, permeameter test on an aquifer core sample, or a pump test. • Measure or estimate the porosity. Can be measured in the laboratory from a core sample, or can be inferred from measurements in the same aquifer reported by others. The contaminant is expected to move with the seepage velocity in the aquifer. For an aquifer with a (isotropic) hydraulic conductivity, K, and porosity, n, the seepage velocity, vi , can be calculated using Darcy’s law as vi = −

K Ji n

(1)

where Ji is the i-component of the hydraulic gradient. The assumptions in using Equation 8.61 to estimate the rate and direction the contaminant will move are: • Aquifer is isotropic and homogeneous. • Horizontal hydraulic gradient is spatially homogeneous and temporally steady. • Vertical hydraulic gradient is negligible (i.e. Dupuit-Forchheimer assumption). (b) The formulae for the intrinsic permeability, k, given in Table 7.3 can be used to estimate the hydraulic conductivity from the results of a sieve analysis on the bag of sand. The hydraulic conductivity, K, can be calculated using the relation K=k

γ µ

(2)

where γ is the specific weight of the ground water, and µ is the dynamic viscosity of the ground water.

422

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(c) The average (effective) hydraulic conductivity, ⟨K⟩ of the aquifer should be estimated by the geometric mean, KG , of the measurements for two-dimensional flow, and if the flow being simulated is three-dimensional the effective hydraulic conductivity should be taken as KG (1 + σY2 /6) where σY is the standard deviation of the measured log-hydraulic conductivities. The rationale for this approach is that the probability distribution of hydraulic conductivities within an aquifer typically follow a log-normal distribution. 14.15. Values of K measured at 12 locations are given. Let Y = ln K, then µY = 5.360, σY2 = 0.4152, and therefore KG = eµY = e5.360 = 213 m/d For 2-D flow, ⟨K⟩ = KG = 213 m/d and for 3-D flow. ⟨K⟩ = KG (1 + σY2 /6) = (213)(1 + 0.4152/6) = 228 m/d 14.16. From the given data, σY2 = 0.4152 1 ∑ (Yi − ⟨Y ⟩)(Yi+1 − ⟨Y ⟩) = −0.0493 11 11

i=1

and

1 ∑ (Yi − ⟨Y ⟩)(Yi+2 − ⟨Y ⟩) = −0.16551 10 10

i=1

Therefore the correlation between measurements 100 m apart is −0.0493/0.4152 = −0.12 , and the correlation between measurements 200 m apart is −0.16551/0.4152 = −0.40 . 14.17. The exponential correlation function along the x1 axis is given by ρY (r1 ) = e−r1 /λ1 The integral scale, I, is ∫ ∞ ∫ I= ρY (r)dr = 0

0

∞

∞ e−r/λ dr = −λe−r/λ = −λ(0 − 1) = λ 0

14.18. From the definition of the Laplacian operator ∇2 f =

∂2f ∂2f ∂2f + + ∂x2 ∂y 2 ∂z 2

and x = r cos θ,

1

y = r sin θ 7−→ r = (x2 + y 2 ) 2 ,

423

θ = tan−1 (y/x)

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By the chain rule, ∂f ∂f ∂r ∂f ∂θ = · + · ∂x ∂r ∂x ∂θ ∂x

(1)

Noting that 1 du d tan−1 u = dx 1 + u2 dx then 1 1 r cos θ ∂r x x = (x2 + y 2 )− 2 (2x) = √ = cos θ = = 2 2 ∂x 2 r r x +y ∂θ ∂ 1 −y y r sin θ sin θ = tan−1 (y/x) = · 2 =− 2 =− 2 =− 2 2 ∂x ∂x 1 + (y/x) x x +y r r

(2) (3)

Combining Equations 1 to 3 gives ∂f ∂f sin θ ∂f = cos θ − ∂x (∂r ) r ∂θ ∂2f ∂ ∂f = ∂x2 ∂x ∂x [ ] [ ] ∂ ∂f sin θ ∂f sin θ ∂ ∂f sin θ ∂f = cos θ cos θ − − cos θ − ∂r ∂r r ∂θ r ∂θ ∂r r ∂θ 2 2 2 ∂ f 2 sin θ cos θ ∂f sin θ ∂f sin2 θ ∂ 2 f 2 sin θ cos θ ∂ f = cos2 θ 2 − + + + ∂r r ∂r∂θ r2 ∂θ r ∂r r2 ∂θ2 By the chain rule, ∂f ∂f ∂r ∂f ∂θ = · + · ∂y ∂r ∂y ∂θ ∂y and ∂r 1 = (x2 + y 2 )−1/2 (2y) = sin θ ∂y 2 ∂θ ∂ 1 1 x r cos θ cos θ = tan−1 (y/x) = · = 2 = = ∂y ∂y 1 + (y/x)2 x x + y2 r2 r therefore ∂f ∂f cos θ ∂f = sin θ + ∂y ∂r r ∂θ ( ) 2 ∂ f ∂ ∂f = 2 ∂y ∂y ∂y [ ] [ ] ∂ ∂f cos θ ∂f cos θ ∂ ∂f cos θ ∂f = sin θ sin θ + + sin θ − ∂r ∂r r ∂θ r ∂θ ∂r r ∂θ 2 2 2 ∂ f 2 sin θ cos θ ∂ f 2 sin θ cos θ ∂f cos θ ∂f cos2 θ ∂ 2 f = sin2 θ 2 + − + + ∂r r ∂r∂θ r2 ∂θ r ∂r r2 ∂θ2 Combining Equations 4 and 5 (and adding ∂ 2 f /∂z 2 ) to both sides gives ∇2 f =

1 ∂f ∂2f ∂2f 1 ∂2f + + + ∂r2 r ∂r r2 ∂θ2 ∂z 2

Cylindrical coordinates are useful in cases where there is cylindrical symmetry .

424

(4)

(5)

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14.19. The head distribution in a three-dimensional homogeneous aquifer is given by Equations 14.66 and 14.68 as 1 ∂2ϕ ∂2ϕ Ss ∂ϕ ∂ 2 ϕ 1 ∂ϕ + + + 2 = (1) 2 2 2 ∂r r ∂r r ∂θ ∂z K ∂t Under steady-state conditions, ∂ϕ/∂t = 0 and Equation 1 becomes ∂ 2 ϕ 1 ∂ϕ 1 ∂2ϕ ∂2ϕ + + + 2 =0 ∂r2 r ∂r r2 ∂θ2 ∂z

(2)

In the case where the head distribution is radially symmetric, partial derivatives with respect to θ are equal to zero, and Equation 2 becomes ∂ 2 ϕ 1 ∂ϕ ∂ 2 f + + 2 =0 ∂r2 r ∂r ∂z 14.20. The hydraulic conductivity, K, in the equivalent isotropic domain is given by √ √ K = 3 Kxx Kyy Kzz = 3 (30)(40)(13) = 25 m/d Therefore the given isotropic domain is equivalent to the anisotropic domain. The relation between the coordinates in the equivalent isotropic domain (x′ , y ′ , z ′ ) and the anisotropic domain (x, y, z) is given by Equation 14.69 as √ √ K 25 ′ x = x= x = 0.913x Kxx 30 √ √ K 25 ′ y = y= y = 0.791y Kyy 40 √ √ K 25 ′ z = z= z = 1.39z Kzz 13 Hence, the coordinates in the anisotropic domain corresponding to x′ = 100 m, y ′ = 100 m, z ′ = 10 m are x = 110 m, y = 126 m, z = 7.19 m At this location,

p + z = 25 m γ

therefore, the pore pressure, p, is given by p = γ(25 − z) = (9790)(25 − 7.19) = 1.74 × 105 Pa = 174 kPa ¯ xx and K ¯ yy , are given by Equations 14.98 14.21. The effective hydraulic conductivity components, K and 14.99 as n n ∑ ∑ i i ¯ xx = 1 ¯ yy = 1 K Kxx ∆zi and K Kyy ∆zi h h i=1

i=1

425

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When the water table is 2 m below the ground surface, h = 20 − 2 = 18 m, and therefore ¯ xx = 1 (7 × 3 + 9 × 2 + 14 × 4 + 11 × 4 + 6 × 4 + 2 × 1) = 9.2 m/d K 18 1 ¯ Kyy = (20 × 3 + 21 × 2 + 12 × 4 + 17 × 4 + 9 × 4 + 5 × 1) = 14.4 m/d 18 When the water table is 3 m below the ground surface, h = 20 − 3 = 17 m, and therefore ¯ xx = 1 (7 × 2 + 9 × 2 + 14 × 4 + 11 × 4 + 6 × 4 + 2 × 1) = 9.3 m/d K 17 ¯ yy = 1 (20 × 2 + 21 × 2 + 12 × 4 + 17 × 4 + 9 × 4 + 5 × 1) = 14.1 m/d K 17 ¯ xx increases slightly and K ¯ yy decreases slightly when the water table falls from 2 m Hence, K to 3 m below the ground surface. 14.22. The distribution of saturated thickness is given by Equation 14.4.1 as ( ) ( ) ∂ ∂h ∂ ∂h ∂h Kxx h + Kyy h + N = Sy ∂x ∂x ∂y .∂y ∂t Under steady-state conditions, this equation becomes ∂ Kxx ∂x

(

∂h h ∂x

)

∂ + Kyy ∂y

(

∂h h ∂y

) +N =0

This could describe the steady-state distribution of saturated thickness surrounding a pumping well, where the pumpage is exactly balanced by the recharge. 14.23. The distribution of piezometric head is given by Equation 14.112 as K

∂2ϕ S ∂ϕ ∂2ϕ N + K + = ∂x2 ∂y 2 b b ∂t

Under steady-state conditions, this equation becomes K

∂2ϕ ∂2ϕ N + K + =0 ∂x2 ∂y 2 b

This could describe the steady-state distribution of piezometric head surrounding a pumping well, where the pumpage is exactly balanced by the recharge from a semi-confining layer. 14.24. The storage coefficient, S, can be expressed in terms of the elastic properties of water and the elastic properties of a porous medium using Equation 14.103 (Bear, 1979) where ( α) S = γnb Ew + n

426

(1)

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From the given data: n = 0.3, b = 20 m, and α = 8 × 10−9 m2 /N. At 15◦ C, γ = ρg = (999.1)(9.81) = 9801 N/m3 , and Ew = (2.14 × 109 )−1 = 4.67 × 10−10 m2 /N. Substituting into Equation 1 yields ( ) 8 × 10−9 S = (9801)(0.3)(20) 4.67 × 10−10 + = 0.0016 0.3 The value of S estimated using the Lohman (1972) equation is given by S = 3 × 10−6 (20) = 0.00006 Hence the calculated storage coefficient using the Lohman (1972) equation is much different from that calculated using the Bear (1979) equation, Equation 1. The reason for this is that Lohman (1972) used 3 × 10−6 instead of γn(Ew + α/n) as the specific storage, Ss . As shown in Table 14.5, the specific storage, Ss , can vary from 10−3 to 10−7 , and 3 × 10−6 as assumed by Lohman (1972) is not representative of all aquifer materials, and certainly not the aquifer material in the present case. 14.25. The storage coefficient measures the volume of water released from storage due to the elasticity of the water and solid matrix. Since the elasticities are quite small, the volume of water released from storage in a confined aquifer due to a 1 m fall in the pressure head is also quite small. In contrast, the specific yield measures the volume of water released from storage in the pores of an unconfined aquifer as the porous medium is drained. Therefore a 1 m drop in the water table yields much more water from an unconfined aquifer than a 1 m drop of the piezometric head in a confined aquifer. 14.26. From the given data: Sy = 0.2 and ∆hc = 2∆hu . By definition, for the same volume, ∆V extracted from both formations, ∆V = Sy Au ∆hu = SAc ∆hc which gives

( S = Sy

Au Ac

)(

∆hu ∆hc

)

( = (0.2)

1 1000

)( ) 1 = 10−4 2

14.27. In terms of Cartesian coordinates (x′ , y ′ ), the head distribution in the isotropic aquifer is given by Qw ϕ(x′ , y ′ ) = ϕ0 − ln 2πT

( √

)

R′ x′2 + y ′2

Qw = ϕ0 − ln 4πT

(

R′2 x′2 + y ′2

) ,

r ′ < R′

where r′2 = x′2 + y ′2 and R′2 = X ′2 + Y ′2 . The coordinates (X ′ , Y ′ ) define the boundary of the zero-drawdown circle surrounding the well. Application of the head distribution in an isotropic medium to an anisotropic medium requires a scaling of the spatial coordinates and

427

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the transmissivity in accordance with Equations 14.118 and 14.124. These equations yield √ √ √ Tyy T T Txx = , = T = Txx Tyy ⇒ Txx Txx Tyy Tyy √ Tyy 2 T 2 x′2 = x = x Txx Txx √ Txx 2 T 2 ′2 y = y = y Tyy Tyy √ Tyy 2 T ′2 2 X = X = X Txx Txx √ T 2 Txx 2 ′2 Y = Y = Y Tyy Tyy Substituting these relationships into Equation 14 gives the head distribution in a homogeneous anisotropic aquifer as Qw ln ϕ(x, y) = ϕ0 − √ 4π Txx Tyy

(√

) √ Tyy /Txx X 2 + Txx /Tyy Y 2 √ √ , Tyy /Txx x2 + Txx /Tyy y 2

r 0.036 0.0030 20 h > 0.081 d t > 0.036

which indicates that storage in the semiconfining layer can be neglected. Satisfaction of either criterion for neglecting storage in the semiconfining layer is sufficient to neglect this effect. Since one of the criteria for neglecting storage in the semiconfining layer is satisfied, there is sufficient justification for neglecting this effect. Drawdown in the overlying aquifer can be neglected if

500

Ts 2 m /d

> 100T > (100)(26.8) m2 /d

500 m2 /d > 2680 m2 /d which indicates that drawdown in the overlying aquifer cannot be neglected. The alternate criterion for neglecting drawdown in the overlying aquifer is S ′ (b′ )2 10bK ′ 0.0015(2)4.5 d 20 h < 10(5)(0.003) 20 h < 0.20 d t
30 1− T b [ ( ) ] 0.12 (0.00031) 10 × 0.1 2 20 h > 30 1− d 26.8 5 20 h > 3.3 × 10−6 d Since both criteria for neglecting storage in the pumping well are satisfied, this effect can be neglected. The results of these analyses support the assumptions that: storage in the semiconfining layer can be neglected , drawdown in the overlying aquifer should be taken into account , and storage in the well can be neglected . Based on these results, the analysis of pump-test data using the Hantush-Jacob method and associated assumptions is not justified. 15.41. From the given data: H = 20 m, K = 40 m/d, (x, y) = (100 m, 100 m), T = KH = (40)(20) = 800 m2 /d, Sy = 0.2, and Qw = 200 L/s = 17280 m3 /d. Point A B C

(x, y) (m,m) (0,0) (200,200) (200,-200)

r (m) 141.4 141.4 316.2

u=

r 2 Sy 4T t

1.243/t 1.243/t 6.249/t

where t is in days. Using the Theis equation s=

Qw 17280 [W (u1 ) + W (u2 ) + W (u3 )] = [2W (u1 ) + W (u2 )] = 1.719[2W (u1 ) + W (u2 )] 4πT 4π(800)

This equation gives the drawdown as a function of time. To illustrate the use of this equation, start with t = 1 day: t (days) 1

u1

W (u1 )

u2

W (u2 )

1.243

0.1780

6.249

0.00033

s (m) 0.613

15.42. From the given data: rw = 0.5 m, Qw = 400 L/s = 34560 m3 /d, H = 24 m, K = 300 m/d, and S = 0.012. The equation describing the drawdown adjacent to a constant-head boundary is ( )] ( ) [ ( ) ( ) ( ) Qw R Qw 34560 r2 R r2 r2 s= ln − ln = ln = ln = 0.7639 ln 2πT r1 r2 2πT r2 2π(300 × 24) r1 r1 Applying this equation at (100 m, 0 m) and (−100 m, 0 m) gives

463

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Point

r1 (m) 100 100

(100,0) (−100,0)

r2 (m) 900 1100

s (m) 1.678 1.832

15.43. For Well No.2 at (0 m,200 m), and Qw = 400 L/s = 34560 m3 /d, induced drawdown, s2 , is given by ( ′) ( ′) Qw r2 r s2 = ln ′ = 0.7639 ln 2′ 2πT r1 r1 Point (100,0) (−100,0)

r1′ (m) 223.6 223.6

r2′ (m) 922.0 1118.0

s2 (m) 1.082 1.229

s1 (m) 1.678 1.832

s 2.760 3.061

If Well No.2 is to be located at (0,y), then @ (100 m,0 m), ( s2 = 0.7639 ln

1

(9002 + y 2 ) 2 1

(1002 + y 2 ) 2

)

( = 0.3820 ln

9002 + y 2 1002 + y 2

)

and the percent drawdown contributed by Well No.2 is % of total =

s2 100s2 × 100 = s1 + s2 s1 + s2

The value of y that gives a 1% error is y = 4917 m, and hence the coordinates of Well No.2 are (0 m, 4917 m) . 15.44. From the given data: rw = 0.8 m, Qw = 500 L/s = 43200 m3 /d, b = 24 m, K = 250 m/d, and Sy = 0.2. Assuming that the drawdown is small relative to the saturated thickness of the aquifer, the drawdown distribution surrounding the well is given by [ ] Qw (x − x0 − 2L)2 + (y − y0 )2 s= ln 4πT (x − x0 )2 + (y − y0 )2 where (x0 , y0 ) are the coordinates of the well, and L is the distance from the well to the constant-head boundary. Taking (x0 , y0 ) = (0,0) and L = 500 m, the drawdown distribution is given by [ ] ] [ 43200 (x − 1000)2 + y 2 (x − 1000)2 + y 2 s= ln = 0.5730 ln 4π(250 × 24) x2 + y 2 x2 + y 2 The seepage out of the canal, q, is given by ∫

+∞

q= −∞

∂s −Kb dy ∂x x=500 m

464

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where 2(x2 + y 2 )(x − 1000) − [(x − 1000)2 + y 2 ]2x ∂s x2 + y 2 = 0.5730 ∂x x=500 m (x − 1000)2 + y 2 (x2 + y 2 )2 x=500 m 1146 =− 5002 + y 2 Substituting into Equation 1 gives ∫ +∞ ∫ +∞ 1146 1 6 q= (250)(24) dy = 6.876 × 10 dy 2 + y2 2 + y2 500 500 −∞ −∞ ( y ) +∞ 1 = 6.876 × 106 tan−1 500 500 −∞ [ π ( π )] = 13752[tan−1 (+∞) − tan−1 (−∞)] = 13752 − − 2 2 = 43200 m3 /d Since the withdrawal rate from the canal (= 43200 m3 /d) is equal to the pumping rate, then 100% of the pumped water originates from the canal. 15.45. The drawdown distribution, s(x, y), is the sum of the drawdowns induced by two pumping wells placed symmetrically about an the impermeable boundary. Hence, [ ( ) ( )] ( 2) R R R Qw Qw ln + ln ln s(x, y) = sp + si = = ′ 2πT r r 2πT rr′ where R is the radius of influence, r is the radial distance of (x, y) from the pumping well, and r′ is the radial distance of (x, y) from the image well. From the given data, Qw = 400 L/s = 34560 m3 /d, T = Kb = (300)(24) = 7200 m2 /d, and R = 1200 m. Substituting the given data into the drawdown equation yields ( ( ) ) 10002 12002 34560 s(x, y) = ln = 0.764 ln 2π(7200) rr′ rr′ The drawdowns at (100 m, 0 m) and (−100 m, 0 m) are given in the following table: Location (100 m,0 m) (−100 m,0 m)

r (m) 100 100

r′ (m) 900 1100

s (m) 2.12 1.96

15.46. Using the method of images, the drawdown distribution is the sum of the drawdowns induced by four pumping wells placed symmetrically about the impermeable boundary. The real wells are at (0 m, 0 m) and (0 m, 200 m), the image wells are at (1000 m, 0 m) and (1000 m, 200 m), and the impermeable boundary is along the line x = 500 m. The drawdown distribution is therefore given by ( 2 ) ( 2 ) [ ( )] Qw R Qw R Qw R4 s= ln + ln = ln 2πT r1 r1′ 2πT r2 r2′ 2πT r1 r1′ r2 r2′

465

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where r1 is the distance from the well at (0 m, 0 m), r2 is the distance from the well at (0 m, 200 m), and r1′ and r2′ are the distances from the corresponding image wells. From the given data: Qw = 400 L/s = 34560 m3 /d, T = Kb = (300)(24) = 7200 m/d, and R = 1200 m, therefore [ ( )] [ ( )] 34560 12004 12004 s= ln = 0.764 ln 2π(7200) r1 r1′ r2 r2′ r1 r1′ r2 r2′ The drawdown calculations are summarized in the following table: Point

r1 (m) 100 100

(100 m, 0 m) (−100 m, 0 m)

r1′ (m) 900 1100

r2 (m) 224 224

r2′ (m) 922 1118

s (m) 3.60 3.30

The drawdown, s1 , at any location caused by the well at (0 m, 0 m) is given by ( 2 ) R Qw s1 = ln 2πT r1 r1′ and the drawdown, s2 , at any location caused by the well at (0 m, 200 m) is given by ( 2 ) Qw R s2 = ln 2πT r2 r2′

(1)

(2)

If the second well contributes 1% to the drawdown, then s2 = 0.01(s1 + s2 )

(3)

Combining Equations 1 to 3 gives ln(R2 /r2 r2′ ) = 0.01 ln(R4 /r1 r1′ r2 r2′ ) which can be written as

ln(12002 /r2 r2′ ) = 0.01 ln(12004 /r1 r1′ r2 r2′ )

This equation is satisfied at location (100 m, 0 m) when the second well is at (0 m, 1023 m), and is satisfied at (−100 m, 0 m) when the second well is at (0 m, 957 m). Hence, when the second well is further than (0 m, 1023 m) from the first well, the drawdown contributed by the second well is less than 1% of the total drawdown. 15.47. From the given data: rw = 0.5 m, Qw = 400 L/s = 34560 m3 /d, H = 24 m, K = 300 m/d, S = 0.012, and R = 1200 m. Since the radius of influence is 1200 m, and the required image wells would be at least 1800 m from the locations of interest, then there is no need to consider any of the image wells. The drawdown, s, 200 m from the well is given by s=

Qw R 34560 1200 ln = ln = 1.37 m 2πT r 2π(300 × 24) 200

Hence the drawdown 200 m east of the well is 1.37 m and the drawdown 200 m west of the well is 1.37 m .

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15.48. The drawdown distribution caused by a single well in an infinite aquifer is given by ( ) ( 2) [ ] Qw R Qw R Qw R2 s= ln = ln = ln 2πT r 4πT r2 4πT (x − x0 )2 + y 2 Taking the coordinates of the pumping well as (x0 ,0), then to account for the streams on both sides of the well, image injection wells must be placed at (−x0 ,0) and (2d − x0 ,0). These image wells create imbalances at the opposite boundaries that must be balanced by image pumping wells at (2d + x0 ,0) and (−2d + x0 ,0) respectively. In turn, these image pumping wells produce imbalances that must be countered by image injection wells at (−2d−x0 ,0) and (4d − x0 ,0). This sequence of corrections continues indefinitely, giving the total drawdown as { ln

R2 R2 R2 − ln + ln − (x − x0 )2 + y 2 (x + x0 )2 + y 2 (x − x0 + 2d)2 + y 2 R2 R2 R2 + ln − ln + ln (x + x0 − 2d)2 + y 2 (x − x0 − 2d)2 + y 2 (x + x0 + 2d)2 + y 2 } R2 R2 ln − ln + ··· (x − x0 − 4d)2 + y 2 (x + x0 + 4d)2 + y 2

Qw s= 4πT

which can be compactly written as ∞ Qw ∑ (x + x0 − 2nd)2 + y 2 s= ln 4πT n=−∞ (x − x0 − 2nd)2 + y 2

15.49. See Figure 15.3. 15.50. This can be shown by deriving the drawdown equation for an injection well in an unconfined aquifer. The governing flow equation is given by d2 (h2 ) 1 d(h2 ) + =0 dr2 r dr which can also be written as d dr

(

dh2 r dr

) =0

(1)

The boundary conditions for an injection well are dh 2πrw hK = −Qw dr r=rw h(rw ) = hw Integrating Equation 1 twice with respect to r yields h2 = A ln r + B

467

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Figure 15.3: Problem 15.49. Applying the boundary conditions yield A=−

Qw , πK

B = h2w − A ln rw

(3)

Substituting Equation 3 into Equation 2 yields the following expression for the piezometric head distribution in an unconfined aquifer surrounding an injection well ( ) Qw r 2 2 h = hw − ln (4) πK rw The drawdown, s, is defined by s(r) = H − h(r) in which case Equation 4 can be written as Qw (H − s) = (H − sw ) − ln πK 2

2

(

r rw

)

where sw is the drawdown at the well. The drawdown equation for an injection well can be put in the form ( ) Qw r ′ ′ s = sw − ln (5) 2πKH rw where

s2 s2 , s′w = sw − w 2H 2H The drawdown equation for a pumping well is ( ) Qw r ′ ′ s = sw + ln 2πKH rw s′ = s −

468

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where s′w at a pumping well is equal and opposite to s′w at an injection well for the same flowrate, Qw . Therefore, comparing Equations 5 and 6 demonstrates that the buildup distribution caused by injecting water at a rate Qw into a fully penetrating well in an unconfined infinite aquifer is exactly the same as the drawdown distribution caused by withdrawing water at a rate Qw from the same aquifer. 15.51. From the given data: h = 1 m, and the depth of the saturated zone is 24 m. According to the Ghyben-Herzberg equation, the depth of the saltwater interface below sea level is expected to be equal to 40h = 40 × 1 = 40 m. Hence, saltwater intrusion is not expected to be a problem at this location. 15.52. From the given data: x = 3 km = 3000 m, K = 40 m/d, and z = 50 m. Using the Glover solution, Equation 15.256 requires that √ √ √ ( ) 2 2 2 Q2 z= (Qx − C) = (Qx + QW ) = Qx + (1) Kϵ Kϵ Kϵ 2ϵK For the given information and assuming ϵ = 0.025, Equation 15.257 is a quadratic equation in Q in which the positive root of Q is given by ) (√ ) (√ x2 + z 2 − x = (0.025)(40) 30002 + 502 − 3000 = 0.417 m2 /d Q = ϵK Therefore, the fresh-water discharge per kilometer of shoreline is 0.417×1000 = 417 (m3 /d)/km. 15.53. (a) The Ghyben-Herzberg equation (Equation 15.248) is used to calculate the depth of the saltwater interface corresponding to the given water-table contours. The intersection of the saltwater interface with the bottom of the surficial aquifer is equated to the extent of saltwater intrusion, and the extent of saltwater intrusion corresponding to both high and low water table conditions should be shown on an area map (not included in this solution). (b) The freshwater inflow is calculated using Equation 15.252, which is given by Q=

K K 2 h ≈ (40)h2L 2Lϵ L 2L

Estimating the hydraulic conductivity, K, from the transmissivity contours and using the water-table elevations, hL , and distances L from the coast yields a freshwater inflow to Florida Bay of approximately 1–2 m2 /d in the wet season and 0.1–1 m2 /d in the dry season. 15.54. Let the water surface in the canal upstream of the gate be a distance h above the water surface downstream of the gate, which can be taken as 30 cm above mean sea level. The water surface in the canal upstream of the gate should be sufficient to keep the toe of the salt water wedge at the bottom of the aquifer, which is a distance z below the elevation of the sea water downstream of the gate. Since the elevation of the bottom of the canal is 3 m below sea level, then z = 24 + 3 + 0.3 = 27.3 m

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and the Ghyben-Herzberg equation gives h≈

1 1 = = 0.68 m 40z 40(27.3)

Hence the elevation of the water surface upstream of the gate should be at least 0.68 m + 0.30 m = 0.98 m above mean sea level to prevent salt water intrusion. 15.55. The maximum pumpage, Qm , is related to the distance, d, below the well by Qm = 0.6πd2 Kϵ where (at 20◦ C) ϵ=

(1)

1.025 − 0.998 ∆ρ = = 0.0271 ρf 0.998

From the given data: Qm = 40 L/s = 3456 m3 /d, and K = 500 m/d, therefore Equation 1 gives 3456 Qm = = 135 m2 d2 = 0.6πKϵ 0.6π(500)(0.0271) which gives d = 11.6 m Since the well is 15 m above the bottom of the aquifer, then the well will become unusable when the thickness of the saltwater wedge is 15 m − 11.6 m = 3.4 m . 15.56. From the given data: L = 1000 m, K = 75 m/d, b = 100 m, αT = 2 m, and qf = 0.1 m/d. It will be assumed that ϵ = 0.025. Without Dispersion: From the given data: λ∗ =

Kbϵ (75)(100)(0.025) = = 1.88 Lqf (1000)(0.1)

Substituting λ∗ into Equation 15.260 requires that [ ( 1 ( ) 1 − 1− Q∗ 2 Q∗ 1.88 = 2 1 − + ln [ ( π π 1+ 1−

Q∗ π Q∗ π

)1 ] 2

)1 ] 2

which yields Q∗ = 0.0594, and hence the maximum allowable flow rate, Qmax , is given by Qmax = Q∗ bLqf = (0.0594)(100)(1000)(0.1) = 594 m3 /d = 6.88 L/s With Dispersion: The modified buoyancy factor, ϵ∗ , is given by Equation 15.262 as [ [ ( )1 ] (α )1 ] 2 6 T 6 ∗ ϵ =ϵ 1− = (0.025) 1 − = 0.0120 b 100

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and hence λ∗ is taken as λ∗ =

Kbϵ∗ (75)(100)(0.0120) = = 0.900 Lqf (1000)(0.1)

Substituting λ∗ into Equation 15.260 yields Q∗ = 1.00, and hence the maximum allowable flow rate, Qmax , is given by Qmax = Q∗ bLqf = (1.00)(100)(1000)(0.1) = 10, 000 m3 /d = 116 L/s

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Chapter 16

Design of Ground-Water Systems 16.1. From the given data: Qw = 4.44 m3 /s, b = 35 m, K = 500 m/d, Sy = 0.2, rw = 0.5 m, sm = 2 m. The transmissivity, T , is calculated as (

2 T = Kb = 500 35 − 2

) = 17000 m2 /d

For a single well, uw =

2S rw (0.5)2 (0.2) y = = 2.015 × 10−9 4T t 4(17000)(365)

Since uw < 0.004, W (uw ) = −0.5772 − ln uw = −0.5772 − ln(2.015 × 10−9 ) = 19.45 From the Theis equation (for sw = 1 m),

Qw =

4πT sw 4π(17000)(1) = = 10983 m3 /d = 0.127 m3 /s = 127 L/s W (uw ) 19.45

Therefore, the number of wells required is

No. of wells =

4.44 = 34.9 wells 0.127

say 35 wells , and the pumpage, Qw , from each well is

Qw =

4.44 = 0.127 m3 /s = 127 L/s 35

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In a straight line, the center well is interfered with by 17 wells on each side. If the spacing of the wells is x, then { [ ( 2 ) ( ) ( )]} x Sy (2x)2 Sy (17x)2 Sy Qw 2= W (uw ) + 2 × W +W + ... + W 4πT 4T t 4T t 4T t { [ ( ) ( ) 2 2 x (0.2) (2x) (0.2) 0.127 × 86400 19.45 + 2 × W +W + ... + = 4π(17000) 4(17000)(365) 4(17000)(365) ( )]} (17x)2 (0.2) W 4(17000)(365) = 0.999 + 0.1027{W [8.058 × 10−9 (x)2 ] + W [8.058 × 10−9 (2x)2 ] + . . . + W [8.058 × 10−9 (17x)2 ]} which gives a spacing x = 1500 m These results, while accurate, are unrealistic. Increasing the allowable drawdown should be investigated. 16.2. The drawdown, sw , at a single well in the center of the 500 m × 500 m block of land is given by [ ( 2 ) ( 2 )] rw S y ri S y Qw W − 4W (1) sw = 4πT 4T t 4T t where ri is the distance to the image wells, and rw is the radius of the water-supply well. From the given data: H = 20 m, K = 75 m/d, Sy = 0.26, and rw = 0.800/2 = 0.400 m. For a drawdown of 2 m at the well, the average transmissivity in the wellfield can be estimated by ) ( ( sw ) 2 T =K H− = 1425 m2 /d = 75 20 − 2 2 Taking t = 365 days, gives [ 2 ] r Sy W w =W 4T t [ 2 ] r Sy W i =W 4T t

[

] (0.4)2 (0.26) = W (2.00 × 10−8 ) = 17.15 4(1425)(365) [ ] (500)2 (0.26) = W (0.0312) = 2.93 4(1425)(365)

(2) (3)

Combining Equations (1) to (3), and taking sw = 2 m gives 2=

Qw [17.15 − 4(2.93)] 4π(1425)

which yields Qw = 6595 m3 /d = 76.3 L/s Therefore, only 1 well is necessary for the wellfield. However, to permit servicing of the well, two wells 50 m apart in the center of the 500 m × 500 m block of land is recommended.

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16.3. The drawdown of a well adjacent to a fully penetrating river is given by sw =

Qw [W (u) − W (u′ )] 4πT

(1)

where u and u′ are the Boltzman variables evaluated at the real and image wells respectively. ′ = 199.5 From the given data: Qw = 83.3 L/s = 7200 m3 /d, T = (60)(20) = 1200 m2 /d, rw m, rw = 0.5 m, and taking t = 365 days and Sy ≈ n = 0.15 gives 2S rw (0.5)2 (0.15) y = = 2.14 × 10−8 4T t 4(1200)(365) W (u) = −0.5772 − ln u = −0.5772 − ln(2.14 × 10−8 ) = 17.08

u=

r′ 2w Sy (199.5)2 (0.15) = = 3.408 × 10−3 4T t 4(1200)(365) W (u′ ) = −0.5772 − ln u′ = −0.5772 − ln(3.408 × 10−3 ) = 5.10 u′ =

Therefore, taking the drawdown, sw , at the well as 1 m, the Equation 1 gives Qw =

4πT sw 4π(1200)(1) = = 1259 m3 /d W (u) − W (u′ ) 17.08 − 5.10

and therefore the number of wells required is given by No. of wells =

7200 = 5.71 ≈ 6 wells 1259

with each well having a pumping rate of 7200/6 = 1200 m3 /d = 13.9 L/s . The maximum drawdown occurs at the well(s) having two wells on one side and three wells on the other side. If x is the spacing between the wells, then u1 = u′1 = u2 = u′2 = u3 = u′3 = u4 = u′4 =

r 2 Sy (0.5)2 (0.15) = = 2.14 × 10−8 4T t 4(1200)(365) (199.5)2 (0.15) = 3.408 × 10−3 4(1200)(365) x2 (0.15) = 8.562 × 10−8 x2 4(1200)(365) (x2 + 2002 )(0.15) = 8.562 × 10−8 (x2 + 2002 ) 4(1200)(365) (2x)2 (0.15) = 3.425 × 10−7 x2 4(1200)(365) [(2x)2 + 2002 ](0.15) = 3.425 × 10−7 (4x2 + 2002 ) 4(1200)(365) (3x)2 (0.15) = 7.705 × 10−7 x2 4(1200)(365) [(3x)2 + 2002 ](0.15) = 7.705 × 10−7 (9x2 + 2002 ) 4(1200)(365)

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Defining ′

(

′

∆W = W (u) − W (u ) = ln u − ln u = ln then

(

3.408 × 10−3 2.14 × 10−8 ( 2 ) x + 2002 ′ ∆W2 = W (u2 ) − W (u2 ) = ln x2 ( 2 ) 4x + 2002 ′ ∆W3 = W (u3 ) − W (u3 ) = ln x2 ) ( 2 9x + 2002 ′ ∆W4 = W (u4 ) − W (u4 ) = ln x2

∆W1 = W (u1 ) − W (u′1 ) = ln

u′ u

)

) = 11.98

and the drawdown at the well is given by s=

Qw [∆W1 + 2∆W2 + 2∆W3 + ∆W4 ] 4πT

where s = 2 m, Qw = 1200 m3 /d, and hence [ ( 2 ) ( 2 ) ( 2 )] 1200 x + 2002 4x + 2002 9x + 2002 2= 11.98 + 2 ln + 2 ln + ln 4π(1200) x2 x2 x2 [(

or ln which simplifies to (

x2 + 2002 x2

x2 + 2002 x2

)2 (

)2 (

4x2 + 2002 x2

4x2 + 2002 x2

)2 (

)2 (

9x2 + 2002 x2

9x2 + 2002 x2

)] = 13.15

) = 5.14 × 105

which gives x = 62.2 m 16.4. From the given data: L = 100 m, W = 75 m, Qw = 467 L/s = 40,320 m3 /d, Sy = 0.2, and s = 2 m. Under existing conditions, ( 2 ) r Sy Qw s=4 W 4πT 4T t √ r = 502 + 37.52 = 62.5 m ) ( 40320 62.52 × 0.2 2=4 W 4πT 4T (365) ( ) 12834 0.5351 2= W T T [ ( )] 12834 0.5351 2= −0.5772 − ln T T

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which yields T = 72,100 m2 /d. With a new well (which is 137.5 m from the center of the rectangular area), Qw Qw W (u1 ) + W (u2 ) 4πT [ ( 4πT ( )) ( ( ))] Qw 0.5351 137.52 × 0.2 2= 4 −0.5772 − ln + −0.5772 − ln 4π(72100) 72100 4(72100)(365)

2=4

2 = 1.104 × 10−6 Qw [44.94 + 9.657] which yields Qw = 33181 m3 /d = 384 L/s. Therefore the total wellfield capacity is 5× 384 = 1920 L/s. The previous wellfield capacity is 4 × 467 L/s = 1868 L/s. The increase in capacity is 1920 L/s − 1868 L/s = 52 L/s . 16.5. The least drawdown at the site should be greater than 1.5 m. The potential points of interest that might have the least drawdown are at the center of the site and at the midpoint on each side of the site. Consider first the midpoint of the site, √ r 402 + 402 = 56.6 m r2 S (56.6)2 (0.16) = = 0.00267 4T t 4(1600)(30) Q Q s=4× (−0.5772 − ln u) = 4 × (−0.5772 − ln 0.00267) = 0.001064Q 4πT 4π(1600)

u=

which gives Q = 1409 m3 /day when s = 1.5 m. At the midpoint of each side of the site, √ r = 802 + 402 = 89.4 m r12 S (40)2 (0.16) = = 0.001333 4T t 4(1600)(30) (89.4)2 (0.16) r2 S u2 = 2 = = 0.006666 4T t 4(1600)(30) Q s=2× [(−0.5772 − ln u1 ) + (−0.5772 − ln u1 )] = 0.001041Q 4πT

u1 =

which gives Q = 1441 m3 /day when s = 1.5 m. Therefore the critical drawdown will be on the side of the site, and the required pumping rate for dewatering is 1441 m3 /day = 16.7 L/s . 16.6. If the concentration decays as a first-order process, c = e−λt c0 Defining T50 as the time when c/c0 = 0.5, then 0.5 = e−λT50 ln 0.5 = −λT50

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which gives T50 = −

ln 0.5 λ

or T50 =

0.693 λ

16.7. From the given data: R = 10−4 , and R is related to the viral concentration, c (in #/L), by R = 1 − (1 + 4.76c)−94.9 Hence, the allowable viral concentration is given by (1 − 10−4 )− 94.9 − 1 (1 − R)− 94.9 − 1 = = 2.21 × 10−7 /L c= 4.76 4.76 1

1

The travel time, t, from the school to the well is given by t=

πr2 bn Qw

where r = 100 m, b = 20 m, n = 0.17, and Qw = 400 L/s = 34560 m3 /d, hence t=

π(100)2 (20)(0.17) = 3.09 days 34560

If the decay constant, λ for viruses is 0.3 d−1 , then the maximum allowable concentration, c0 , at the school is related to the allowable concentration, c, at the well (2.21 × 10−7 /L) by c = e−λt c0 or c0 = ceλt which gives c0 = (2.21 × 10−7 )e0.3×3.09 = 5.58 × 10−7 /L 16.8. From the given data: Qw = 50 L/s = 4320 m3 /d, H = 25 m, and n = 0.2. At a distance r from the pumping well, the induced seepage velocity, v, is given by v=

Qw 4320 137.5 = = m/d 2πrHn 2πr(25)(0.2) r

An identical image recharge well located on the other side of the river will account for the effect of the river, and induce a seepage velocity such that the combined seepage velocity towards the (real) pumping well along a line joining the two wells is given by v=

137.5 137.5 41250 + = m/d r 300 − r r(300 − r)

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The velocity towards the pumping well is equal to −dr/dt, hence dr 41250 41250 =− = dt r(300 − r) r(r − 300) If r = 150 m when t = 0, then ∫

∫ t r′ (r′ − 300)dr′ = 41250dt′ 150 0 r t r′3 r′2 − 300 = 41250t′ 0 3 2 150 ( 3 ) ( ) r − 150r2 − 1.125 × 106 − 3.375 × 106 = 41250(t − 0) 3 which gives

r

] [ 3 1 r 2 6 t= − 150r + 2.25 × 10 41250 3

The contaminant reaches the well when r = rw = 0.15 m, hence the minimum travel time, tr , from the river is given by [ ] 0.153 1 2 6 − 150(0.15) + 2.25 × 10 = 54.5 days tr = 41250 3 Assuming first-order decay, then c = e−λtr = e−0.01×54.5 = 0.580 c0 From the given data, c = 1 µg/L, hence c0 =

1 = 1.72 µg/L 0.580

Therefore, the maximum allowable concentration in the river is 1.72 µg/L . This is a conservative result since: (1) inflow to the well from other parts of the river will have longer travel times, and (2) not all of the water entering the well comes from the river. 16.9. Required flow rate from wellfield, Q, is given by Q = (50000)(0.58) = 29000 m3 /d Neglecting interference, the allowable drawdown is 6/2 = 3 m, and the average transmissivity, T , is given by ¯ = 85(35 − 6/2) = 2720 m2 /d T = KH For a single well, uw =

2S rw (0.5)2 (0.15) y = = 9.442 × 10−9 4T t 4(2720)(365)

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Therefore, with less than 0.1% error (since u < 0.004), W (u) = −0.5772 − ln uw = −0.5772 − ln(9.442 × 10−9 ) = 17.90 The required pumpage for sw = 3 m, is Qw =

4πT sw 4π(2720)(3) = = 5729 m3 /d W (uw ) 17.90

Since the required flow is 29000 m3 /d, then the number of wells required is given by No. of wells =

29000 = 5.06 5729

Therefore, try using 6 wells . This requires 29000/6 = 4833 m3 /d = 55.9 L/s per well. For maximum spacing, place 4 wells on the corners of the square parcel of land. The area of the land is 50 ha = 500000 m2 , and so the dimensions are 707.1 m × 707.1 m. See Figure 16.1, which requires that

Figure 16.1: Problem 16.9. (

707.1 − x 2

)2

( +

707.1 2

)2 = x2

and gives x = 388 m At center well, u = u1 =

2S rw (0.50)2 (0.15) y = = 17.90 4T t 4(2720)(365)

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Assuming other 5 wells are 388 m away (a conservative estimate), then for each of these wells, u=

R 2 Sy (388)2 (0.15) = = 0.00569 4T t 4(2720)(365)

and from Table 15.1, W (u) = 4.601 Hence, the drawdown, sb , induced by the 5 boundary wells is Qw 4833 W (u) = 5 × 4.601 = 3.25 m 4πT 4π2720 and the drawdown at the central well, sc , is sb = 5 ×

Qw 4833 W (uw ) = 17.90 = 2.53 m 4πT 4π2720 The maximum drawdown induced by the wells is therefore less than 3.25 m + 2.53 m = 5.78 m. Hence the 6 wells will meet the drawdown limitation. sc =

Calculated the screen entrance velocity, vs =

Qw cπds Ls P

Assume c = 0.5, then vs =

4833 = 616 m/d = 0.41 m/min (0.5)π(0.5)(20)(0.5)

Since the maximum allowable entrance velocity is within the range 1.8–2.1 m/min, the screen is okay . 16.10. From the given data: rw = 0.5 m, K = 300 m/d, and Table 16.3 gives the maximum allowable screen entrance velocity, vs , as 3.7 m/min . The pumping rate, Qw , corresponding to this entrance velocity is given by 3.7 (0.5)π(2 × 0.5)Ls P = 9.69 × 10−2 Ls P m3 /s 60 where Ls is the screen length, and P is the percentage of open area. Qw = vs cπds Ls P =

The specification for the gravel pack and the screen are as follows: (since Uc of the aquifer matrix is 3.2) Uc between 1 and 2.5, with the 50% size not greater than 9 times the 50% size of the aquifer material. Select the screen slot size to be less than or equal to the 10% passing size of the gravel pack. 16.11. From the given data: b = 40 m, dmin = 0.25 mm, dmax = 0.5 mm, K = 1803 m/d, and Qw = 384 L/s = 23.04 m3 /min. Since the aquifer particles are distributed uniformly between dmin and dmax , then d10 = 0.275 mm d60 = 0.400 mm d60 0.400 Uc = = = 1.45 d10 0.275

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Without gravel pack: Since d10 > 0.25 mm and Uc < 3, the required slot size is in the range of d40 –d60 , which is 0.35–0.40 mm = 0.0138–0.0157 in. Therefore use a 15-slot screen . The entrance velocity, vs , is given by vs =

Qw cπds Ls P

Taking vs = 3.7 m/min, c = 0.5, and P = 0.15, gives 3.7 =

23.04 (0.5)πds Ls (0.15)

which yields ds Ls = 26.43 m2 . For Ls = 30 m, ds = 26.43/30 = 0.881 m. Therefore, use a screen length of 30 m and a screen diameter of 900 mm . With gravel pack: Since Uc of aquifer is less than 2.5, then for the gravel pack use Uc = 1–2.5 and d50 ≤ 6 × d50 = 6 × 0.375 which gives d50 ≤ 2.25 mm. For the gravel pack, specify d60 =2.25 mm and Uc = 2 , then d60 =2 d10 2.25 d10 = = 1.13 mm 2 The required slot size is in the range of d5 –d10 of the gravel pack, which is 1.02–1.13 mm = 0.040–0.045 in. Therefore use 40-slot screen . (Note that d5 is estimated by extrapolation using d60 and d10 . The entrance velocity, vs , is given by vs =

Qw cπds Ls P

Taking vs = 3.7 m/min, c = 0.5, and P = 0.15, gives 3.7 =

23.04 (0.5)πds Ls (0.15)

which yields ds Ls = 26.43 m2 . For Ls = 30 m, ds = 26.43/30 = 0.881 m. Therefore, use a screen length of 30 m and a screen diameter of 900 mm . This is the same screen length and diameter as without the gravel pack 16.12. From the given data: b = 30 m, K = 50 m/d, n = 0.15, Sy = 0.2, and Qw = 13.3 L/s = 0.8 m3 /min. According to Table 16.2 the recommended casing diameter is 250 mm (ID) and the minimum casing diameter is 200 mm. Use steel casing for structural stability. According to Table 16.4, the minimum screen diameter is 150 mm, and the optimal screen entrance velocity is given by Table 16.3 as 1.35 m/min. Assume a 10% open area and 50% clogging, and taking the screen diameter as 250 mm , equal to the casing diameter, then Equation 16.8 gives the required screen length as Ls =

Qw 0.8 = = 15.1 m cπds vs P (0.5)π(0.25)(1.35)(0.10)

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Therefore, the screen will extend over approximately the bottom half of the aquifer. The screen slot size is determined from the grain size distribution in the gravel pack. Since the aquifer material is uniformly distributed between 0.04 mm and 2.2 mm, then d10 = 0.26 mm, d30 = 0.69 mm, d50 = 1.12 mm, d60 = 1.34 mm, and Uc = d60 /d10 = 1.34/0.26 = 5.15. For Uc > 5, Table 16.6 gives the gravel pack characteristics. Noting that 6d30 = 6 × 0.69 = 4.14 mm and 9d30 = 9 × 0.69 = 6.21 mm, then taking Uc = 2.5 and the particle size distribution passing through d30 = (4.14+6.21)/2 = 5.18 mm gives d0 = 2.3 mm, d10 = 3.2 mm, and d100 = 12 mm. Therefore, the gravel pack should be uniformly distributed 2.3 mm × 12 mm sand , installed with a thickness of 10 cm , and the slot size of the screen should be 3.2 mm . 16.13. The key properties of the gravel pack gradations are given in Table 16.1. In accordance with the U.S. Bureau of Reclamation (1995) guidelines for specifying gravel packs, a companyspecific guide for selecting gravel packs and screens from Ricci Bros. Sand Co. is given in Table 16.2. Table 16.1: Key Properties of Gravel-Pack Gradations Type OOO P40 OO OON O 1 2 3 4

d5 (mm) 0.15 0.25 0.30 0.42 0.60 0.85 1.00 1.70 1.99

d10 (mm) 0.162 0.300 0.317 0.450 0.628 0.891 1.180 1.766 2.360

d30 (mm) 0.210 0.396 0.386 0.570 0.739 1.056 1.411 2.030 2.855

d50 (mm) 0.258 0.528 0.453 0.659 0.850 1.232 1.642 2.294 3.350

d60 (mm) 0.275 0.600 0.485 0.698 0.917 1.336 1.783 2.473 3.636

Uc 1.70 2.00 1.53 1.55 1.46 1.50 1.51 1.40 1.54

d30 /6 (mm) 0.035 0.066 0.064 0.095 0.123 0.176 0.235 0.338 0.476

d30 /9 (mm) 0.023 0.044 0.043 0.063 0.082 0.117 0.157 0.226 0.317

d50 /6 (mm) 0.043 0.088 0.076 0.110 0.142 0.205 0.274 0.382 0.558

d50 /9 (mm) 0.0287 0.0587 0.0503 0.0732 0.0944 0.137 0.182 0.255 0.372

16.14. According to VonHofe and Helweg (1998), for optimum performance the pump diameter should be approximately 60% of the screen diameter. To provide flexibility in placing the pump within the screened intake, the same diameter will be used for the casing and the screen. Therefore, using the 60% rule, 300 mm pump diameter = = 500 mm 0.60 0.60 The recommended borehole diameter is given by casing diameter = screen diameter =

borehole diameter = casing diameter + 2(100 mm) = 500 mm + 200 mm = 700 mm The screen length must be such that the screen entrance velocity, vs , is 1.8 m/min, where vs =

Qw cπds Ls P

From the given data: Qw = 41.7 L/s = 2.5 m3 /min, and P = 0.15. Taking c = 0.5 and ds = 500 mm = 0.5 m gives 2.5 1.8 = (0.5)π(0.5)Ls (0.15)

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Table 16.2: Guide to Selecting Gravel Packs and Screens from Ricci Bros. Sand Co., Inc. Aquifer Properties Gravel Pack Screen d30 d50 Slot Size (mm) (mm) (mm) < 2.5 > 0.043 OOO * > 0.088 P40 0.25 or 0.30 > 0.076 OO 0.30 > 0.110 OON 0.42 > 0.142 O 0.60 > 0.205 1 0.85 > 0.274 2 1.18 > 0.382 3 1.70 > 0.558 4 * 2.5–5 > 0.0287 OOO * > 0.0587 P40 0.25 or 0.30 > 0.0503 OO 0.30 > 0.0732 OON 0.42 > 0.0944 O 0.60 > 0.137 1 0.85 > 0.182 2 1.18 > 0.255 3 1.70 > 0.372 4 * >5 0.023–0.025 OOO * 0.044–0.066 P40 0.25 or 0.30 0.043–0.064 OO 0.30 0.063–0.095 OON 0.42 0.082–0.123 O 0.60 0.117–0.176 1 0.85 0.157–0.235 2 1.18 0.226–0.338 3 1.70 0.317–0.476 4 * *Screen size not available from Ricci Bros. Sand Co., Inc. Uc

which yields Ls = 11.8 m This puts the screen within the lower one-half to one-third of the aquifer and is therefore acceptable. Since Uc = d60 /d10 = 2.1/0.5 = 4.2 > 3.0 and d10 > 0.25 mm, a gravel pack is not required . Since 3 < Uc < 5 and d10 > 0.25 mm, Table 16.5 indicates that a slot size in the range of d40 − −d70 is desirable. Taking a midrange size of d55 , since d10 = 0.5 mm and d60 = 2.1 mm, interpolation gives d55 = 1.9 mm. Since a unit slot is equal to 0.025 mm, required slot size =

d55 1.9 = = 76 0.025 0.025

Using the next-lower slot size indicates a 75-slot screen . It is recommended that the pump intake be set in the middle of the screen . 16.15. From the given data: b = 20 m, K = 30 m/d, T = Kb = (30)(20) = 600 m2 /d, rw = 0.15 m,

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S = 10−4 , and t = 1 year = 365 days. Substituting these data into Equation 16.11 gives specific capacity =

4πT 4π(600) = = 289 m2 /d 2 S/4T t) W (rw W (0.152 × 10−4 /4 × 600 × 365)

= 3.35 (L/s)/m Since the allowable drawdown of the potentiometric surface is 5 m, the well yield is given by well yield = 3.35 (L/s)/m × 5 m = 16.8 L/s and therefore the maximum allowable pumping rate is 16.8 L/s . According to Table 16.8, a well with a specific capacity of 3.35 (L/s)/m has a moderate productivity . 16.16. From the given data: T = 270Sc , where T is the transmissivity in ft2 /d and Sc is the specific capacity in gpm/ft. Since 1 gpm = 5.451 m3 /d and 1 ft = 0.3048 m, the Fish and Stewart (1991) relation can be expressed in the form ) ( T 0.3048 = 270Sc (0.3048)2 5.451 or T = 1.40Sc where T is the transmissivity in m2 /d and Sc is the specific capacity in m2 /d. In theory, the relationship between the transmissivity and specific capacity is given by the Theis equation, which can be expressed in the form  ( r2 S )  w y W 4T tp  Sc T = 4π where tp is the pumping time corresponding to the specific capacity estimation. Using the Cooper-Jacob approximation, the Theis equation can be further approximated by [ ( 2 ) ] r Sy 1 T = −0.5772 − ln w + ln T Sc (1) 4π 4tp 2 S /4T t ≤ 0.004. Using the given data for r , t , S = Q/s , and which is valid when uw = rw y p w p c w taking Sy = 0.23, the value of T is calculated (iteratively) using Equation 1. A summary of the input and calculated results are shown in Table 16.3. Using the calculated value of T shown in Column 8 of Table 16.3, the value of uw is calculated to verify the validity of the CooperJacob approximation. In all cases, uw ≤ 0.004 confirming the validity of the approximation. For the given specific capacities (Sc ), the transmissivities are calculated using the Fish and Stewart (1991) equation and these calculated results are shown in Column 10 of Table 16.3. A plot of the transmissivity versus specific capacity using the Theis equation is compared with the Fish and Stewart (1991) equation in Figure 16.2. This comparison demonstrates that the Fish and Stewart (1991) approximation significantly overestimates the transmissivity in the surficial aquifer system of Miami-Dade county. A better approximation is

T = 0.886Sc

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Table 16.3: Estimation of Transmissivity from Specific Capacity in Surficial Aquifer, Miami-Dade County, Florida (1) Site

(2) Well

1

S-3011 S-3012 S-3013 S-3014 S-3005 S-3006 S-3007 S-3008 S-3009 S-3010

2

3

4

5 6 7

S-981 S-983 S-3065 S-3066 S-3045

(3) rw (m) 0.550 0.550 0.550 0.550 0.550 0.550 0.550 0.550 0.550 0.550 0.100 0.100 0.100 0.255 0.255 0.255 0.150 0.150 0.150 0.150 0.100

(4) Q (m3 /d) 38189 38189 38189 38189 15120 15120 15120 15120 15120 15120 2851 2851 3715 21168 21168 21168 6394 5962 6566 6566 5184

16.17. The theoretical relationship is

(5) sw (m) 4.600 5.270 4.150 4.300 0.518 1.067 0.610 1.219 0.610 0.610 0.305 0.305 0.457 1.280 1.219 0.975 1.280 1.280 1.219 1.219 0.671

(6) tp (d) 0.333 0.333 0.333 0.333 0.083 0.083 0.083 0.083 0.083 0.083 0.083 0.083 0.083 0.010 0.021 0.021 0.094 0.083 0.333 0.333 0.333

(7) Sc (m2 /d) 8302 7246 9202 8881 29189 14171 24787 12404 24787 24787 9348 9348 8130 16538 17365 21711 4995 4658 5387 5387 7726

(8) T (m2 /d) 7960 6426 8352 8034 25903 11677 21642 10076 21642 21642 10133 10133 8716 13062 14850 18991 4846 4443 5851 5851 9166

(9) uw (-) 6.555 × 10−6 8.120 × 10−6 6.248 × 10−6 6.495 × 10−6 8.058 × 10−6 1.787 × 10−5 9.644 × 10−6 2.072 × 10−5 9.644 × 10−6 9.644 × 10−6 6.809 × 10−7 6.809 × 10−7 7.916 × 10−7 2.748 × 10−5 1.209 × 10−5 9.450 × 10−6 2.848 × 10−6 3.494 × 10−6 6.633 × 10−7 6.633 × 10−7 1.882 × 10−7

(10) TFS (m2 /d) 11623 10145 12883 12434 40865 19839 34702 17365 34702 34702 13087 13087 11381 23153 24311 30395 6993 6521 7541 7541 10816

sw = αQw + β Qw

and the given data can be put in the form Qw (m3 /s) 0 0.05 0.10 0.15 0.20 0.25 0.30

sw /Qw (s/m2 ) − 4 5.3 5.7 7.35 8.4 9.4

A least-squares analysis of these points gives sw = 21.7Qw + 2.89 Qw Therefore the well loss coefficient is 21.7 s2 /m5 (= 0.006 min2 /m5 ), and the formation loss coefficient is 2.89 s/m2 .

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Transmissivity (m2/d)

45000 40000 35000

Fish and Stewart (1991)

30000 25000 20000

Best fit: T = 0.886 Sc

15000 10000 Measurements

5000 0 0

5000

10000

15000

20000

25000 Specific capacity (m2/d)

30000

35000

Figure 16.2: Specific capacity function for surficial aquifer system, Miami-Dade county, Florida . On the basis of the calculated well loss coefficient, the well is in good condition , where α < 0.05 min2 /m5 . At a pumping rate of 57.9 L/s (= 0.0579 m3 /s), sw = 21.7(0.0579) + 2.89 = 4.14 s/m2 Qw which gives a specific capacity of 1/4.14 = 0.242 m2 /s = 242 (L/s)/m. Hence, according to Table 16.8, the productivity is high . 16.18. The step-drawdown data should match the equation ( ) sw log − β = log α + (n − 1) log Qw Qw where, from the given data, β = 55.2 s/m2 = 6.389 × 10−4 d/m2 , and n − 1 = 1.22, which gives n = 2.22. The well efficiency, ew , is defined by ew = which gives

βQw × 100 βQw + αQnw (

α=

100β −β ew

)

1 Qn−1 w

Since ew = 32% when Qw = 10.3 L/s = 893 m3 /d, then the well loss coefficient, α, is given by ( ) 100 × 6.389 × 10−4 1 −4 α= − 6.389 × 10 = 3.41 × 10−7 d2 /m5 2.22−1 32 893

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At Qw = 16.7 L/s = 1440 m3 /d, the productivity of the well is measured by the specific capacity as Qw Qw 1 = = n sw βQw + αQw β + αQn−1 w 1 = 6.389 × 10−4 + (3.41 × 10−7 )(1440)2.22−1 = 326 m2 /d = 3.77 (L/s)/m

specific capacity =

Comparing this result with the classification system in Table 16.8 indicates that the well productivity is moderate at a pumping rate of 16.7 L/s. 16.19. The Rorabaugh method assumes that the drawdown, sw , and pumping rate, Qw , are related by sw = βQw + αQnw Plotting log(sw /Qw ) versus log Qw yields α = 0.0129, β = 0.288, and n = 1.5, which gives sw = 0.288Qw + 0.0129Q1.5 w where sw is in cm and Qw is in L/min. When Qw = 5.33 L/s = 500 L/min, the efficiency, ew , is given by ew =

βQw 0.288(500) × 100 = × 100 = 50% n βQw + αQw 0.288(500) + 0.0129(500)1.5

and the specific capacity is given by Qw Qw 1 = = n sw βQw + αQw β + αQn−1 w 1 = = 1.73 (L/min)/cm = 2.88 (L/s)/m 0.288 + 0.0129(500)0.5

specific capacity =

16.20. The empirical relation between the pumping rate, Q, and the drawdown, sw , can be expressed in the form sw = BQ + CQn which can be put in the form

(

log

sw −B Q

) = (n − 1) log Q + log C

Plotting log(sw /Q − B) versus log Q gives a linear relation for B = 0.0045 d/m2 , the slope of this line gives n−1 = 1.6 or n = 2.6, and the intercept gives log C = −7.854 or C = 1.40×10−8 . The empirical relation between sw and Q is therefore given by sw = 0.0045Q + 1.40 × 10−8 Q2.6 The well loss as a percentage of total losses is given by well loss =

CQn 1.40 × 10−8 Q2.6 × 100 = × 100 BQ + CQn 0.0045Q + 1.40 × 10−8 Q2.6

For the pumping rates in the pump-drawdown test

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Pumping Rate, Q (m3 /d) 500 1000 1500 2000 2500 3000 3500

Well Loss (%) 6 16 27 37 46 53 59

16.21. Assuming that estimates of the hydraulic conductivity and storage coefficient are available, then estimates of the drawdowns at proposed monitoring well locations as a function of time can be made. This calculation is essential to the design of the pump test, since the design pumping rate must be sufficient to produce drawdowns at the monitoring wells (during the pump test) that are measurable with a reasonable degree of accuracy. 16.22. The drawdown, s0 , induced by a well that begins pumping at t = 0 is given by the Theis equation as Qw W (u) s0 = 4πT where r2 S u= (1) 4T t If this well were to begin pumping at time t1 , then the induced drawdown, s1 , would be given by Qw s1 = W (u1 ) 4πT where r2 S (2) u1 = 4T (t − t1 ) Applying the principle of superposition (in time), then the drawdown, s, induced by a well that begins pumping at t = 0 and stops at t = t1 is given by s = s0 − s1 which yields s=

Qw [W (u) − W (u1 )] 4πT

where u and u1 are given by Equations 1 and 2 respectively. 16.23. From Equation 16.21

] Qw [ W (u) − W (u′ ) 4πT and the Cooper-Jacob approximations are given by s′ =

W (u) = −0.5772 − ln u ′

W (u ) = −0.5772 − ln u

489

(1)

(2) ′

(3)

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Combining Equations 1 to 3 gives ] ] Qw [ Qw [ (−0.5772 − ln u) − (−0.5772 − ln u′ ) = ln u′ − ln u 4πT ( ) 4πT Qw u′ = ln 4πT u

s′ =

Since u=

r2 S 4T t

then

u′ =

and

(4)

r2 S ′ 4T t′

S′t u′ = ′ u St

(5)

and combining Equations 4 and 5 gives s′ = or s′ =

Qw 4πT

Qw S′t ln ′ 4πT St ( ln

t S′ + ln t′ S

)

16.24. The recovery measurements should match the theoretical relation given by Equation 16.24, which can be written as ) ) ( ( t S′ Qw Qw ′ ln ′ + ln (1) s = 4πT t 4πT S The recovery measurements expressed as s′ versus ln t/t′ are: s′ (m) 1.01 0.9 0.83 0.75 0.7 0.61 0.55 0.6 0.42 0.37 0.31 0.26 0.23 0.19 0.15

t′ (min) 1 2 3 5 7 10 15 20 30 40 60 80 100 140 180

490

t (min) 241 242 243 245 247 250 255 260 270 280 300 320 340 380 420

ln(t/t′ ) 5.48 4.80 4.39 3.89 3.56 3.22 2.83 2.56 2.20 1.95 1.61 1.39 1.22 1.00 0.85

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From the given data: Qw = 33.3 L/s = 2880 m3 /d, t = 4 h = 240 min, and S = 0.0001. The best-fit line to the s′ versus ln t/t′ measurements is ( ) t s = 0.1872 ln ′ + 0.0143 t ′

and matching this equation with Equation 1 gives Qw = 0.1872 4πT S′ Qw ln = 0.0143 4πT S

(2) (3)

Solving Equation 2 for T gives T =

Qw 2880 = = 1220 m2 /d 4π(0.1872) 4π(0.1872)

and solving Equation 3 for S ′ gives S ′ = S exp

(

0.0143 Qw /4πT

(

) = 0.0001 exp

0.0143 0.1872

) = 0.00011

where S ′ is not significantly different from S. 16.25. Using the Theis equation, s=

Q W (u), 4πT

Suppose Q = 0.90Q0 , then 0.9Q0 s= W 4πT

u= (

r2 S 4T t

r2 S 4T t

)

Suppose s at measured r2 /t is 0.9 times the expected s, then the drawdown curve (and match point) is shifted downward by ln 0.9, and β = β0 − ln 0.9 where β0 is the vertical shift corresponding to a pumpage of Q0 . Hence, the transmissivity estimated by taking the flow rate equal to Q0 (when it is in fact equal to 0.9Q0 ) is T =

Q0 β0 −ln 0.9 e = 1.11T0 4π

where T0 =

Q0 β0 e 4π

Therefore a 10% error in Q leads to approximately an 11% error in T .

491

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16.26. From the given data: rw = rc = 0.076 m, and Le = Lw = b = 98 m. Determine the following quantities: Le 98 = = 1290 rw 0.076 C = 12.8 [ ] 1.1 12.8 −1 Re = + = 6.12 ln rw ln 1290 1290 Plot the drawdown, s, versus time, t, and this gives the following points (t, s) on the straightline portion of the relation: (0 s, 0.46 m) and ( 33 s, 0.149 m). Note that the s vs. t line becomes nonlinear after t = 33 s. Using the points on the straight-line portion of the curve: K=

0.0762 (6.12) 1 0.49 ln = 6.5 × 10−6 m/s = 0.56 m/d 2(98) 33 0.149

The support volume for this hydraulic conductivity is about 1 m × 98 m surrounding the hole used for the slug test. 16.27. The measured data of ln yt versus t indicates two straight line segments as shown in Figure 16.3. Fitting a straight line to the second linear segment yields

Figure 16.3: Plot of slug test measurements.

ln y = −0.0805t + 3.069 which has a slope, m, of −0.0805. From the given dimensions of the well and aquifer, Lw = 5.5 m, H = 80 m, rc = 0.076 m, Le = 4.56 m, rw = 0.12 m, and Le /rw = 4.56/0.12

492

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= 38.0. Equation 16.28 is appropriate for calculating ln(Re /rw ) (since the well is partially penetrating), and the dimensionless parameters A and B are given by Figure 16.8 as A = 2.6,

B = 0.42

Analyses by Bouwer and Rice (1976) indicated that if ln[(H − Lw )/rw ] > 6 then a value of 6 should be used in Equation 16.28. In this case, ln[(H − Lw )/rw ] = ln[(80 − 5.5)/0.12] = 6.43. Therefore, using ln[(H − Lw )/rw ] = 6 in Equation 16.28 yields } 1.1 A + B ln[(H − Lw )/rw ] −1 + ln(Lw /rw ) (Le /rw ) } { 2.6 + 0.42(6) −1 1.1 + = 2.37 = ln(5.5/0.12) (4.56/0.12)

Re ln = rw

{

and putting this result into Equation 16.27 gives rc2 ln(Re /rw ) m 2Le (0.076)2 (2.37) =− (−0.0805) = 1.21 × 10−4 m/s = 10 m/d 2(4.56)

K=−

16.28. The first step is to plot ln yt versus t, where the given measurements can be put in the form: t (s) 0 3 6 9 12 15 18 21 24

yt (mm) 700 392 260 137 91 47 31 16 11

ln yt 6.55 5.97 5.56 4.92 4.51 3.85 3.43 2.77 2.40

The plotted relation between ln yt and t is very much linear (r2 = 0.998), and is matched by the regression equation ln yt = −0.1752t + 6.5431 which has a slope, m, of −0.1752. From the given dimensions of the well and aquifer, Lw = 4 m, H = 12 m, rc = 75 mm, Le = 2 m, rw = rc + 110 mm = 185 mm, and Le /rw = 2/0.185 = 10.8. Equation 16.28 is appropriate for calculating ln(Re /rw ) (since the well is partially penetrating), and the dimensionless parameters A and B are given by Figure 16.8 as A = 1.8,

493

B = 0.25

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Substituting these data into Equation 16.28 yields Re ln = rw =

{ {

1.1 A + B ln[(H − Lw )/rw ] + ln(Lw /rw ) (Le /rw )

}−1

1.8 + 0.25 ln[(12 − 4)/0.185] 1.1 + ln(4/0.185) (2/0.185)

}−1 = 1.63

and putting this result into Equation 16.32 gives K=−

rc2 ln(Re /rw ) (0.075)2 (1.63) m=− (−0.1752) = 4.02 × 10−4 m/s = 34.7 m/d 2Le 2(2)

This result indicates an average (horizontal) hydraulic conductivity of 34.7 m/d in the immediate vicinity of the well. The transmissivity, T , of the aquifer can be estimated by T = KH = (34.7)(12) = 416 m2 /d 16.29. The relation between ln yt and t is still matched by the regression equation ln yt = −0.1752t + 6.5431 which has a slope, m, of −0.1752. From the given dimensions of the well and aquifer, Lw = H = 12 m, rc = 75 mm, Le = 2 m, rw = rc + 110 mm = 185 mm, and Le /rw = 2/0.185 = 10.8. Equation 16.29 is appropriate for calculating ln(Re /rw ) (since the well is fully penetrating), and the dimensionless parameter C is given by Figure 16.8 as C = 1.25 Substituting this value of C into Equation 16.29 yields ln

Re = rw

{

1.1 C + ln(Lw /rw ) (Le /rw )

}−1

{ =

1.1 1.25 + ln(12/0.185) (2/0.185)

}−1 = 2.64

and putting this result into Equation 16.32 gives K=−

rc2 ln(Re /rw ) (0.075)2 (2.64) m=− (−0.1752) = 6.50 × 10−4 m/s = 56 m/d 2Le 2(2)

This result indicates an average (horizontal) hydraulic conductivity of 56 m/d in the immediate vicinity of the well. The transmissivity, T , of the aquifer can be estimated by T = KH = (56)(12) = 672 m2 /d 16.30. The length, L, of the trench is given by Equation 16.38 as L=

Q Kt (W + H)

494

(1)

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where the dimensions H and W are illustrated in Figure 16.10. Following ASCE (1996) guidelines, specify W = 1 m (a typical backhoe dimension), and H = 2 m (a typical depth for vertical slope stability). The design injection rate, Q, is 5.79 L/s = 500 m3 /d, and the design trench hydraulic conductivity, Kt , is 35/2.5 = 14 m/d. Substituting these values for Q, Kt , W , and H into Equation 1 yields L=

Q 500 = = 11.9 m Kt (W + H) 14(1 + 2)

A trench dimension of 12 m long by 1 m wide by 2 m deep , when filled with gravel, will be capable of transferring water into the aquifer at a rate of 5.79 L/s (= 500 m3 /d). Since the seasonal high water table is 5.22 m below the ground surface, there is sufficient room to install a 2-m deep trench and still maintain a distance of at least 1.2 m between the bottom of the trench and the seasonal high water table. The next question is whether the aquifer will be able to transport the effluent away from the trench as fast as it is supplied, without causing the water table to rise to within 1.2 m of the bottom of the trench. The height of the water table above the base of the aquifer as a function of time is given by h2m (t)

=

h2i

2N νtS ∗ + K

(

L W √ , √ 4 νt 4 νt

) (2)

where hi = 15 m, N is given by N=

Q 500 = = 41.7 m/d LW (12)(1)

The hydraulic conductivity of the aquifer, K, is 70 m/d, and the parameter ν in Equation 2 is given by Kb K(hi + hm )/2 70(15 + hm )/2 = = = 159.1(15 + hm ) ν= Sy Sy 0.22 Substituting the trench dimensions (L = 12 m, W = 1 m), and aquifer properties into Equation 2 yields h2m

( ) 2(41.7) 1 12 ∗ √ = 15 + [159.1(15 + hm )]tS , √ 70 4 [159.1(15 + hm )]t 4 [159.1(15 + hm )]t ) ( 0.238 0.0198 ∗ √ = 225 + 189.6(15 + hm )tS ,√ (3) (15 + hm )t (15 + hm )t 2

This equation relates the thickness of the saturated zone below the trench to the time since the trench began operation. Values of hm for several values of t are shown in the following table:

495

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t (days) 1 10 100 1000 10000

hm (m) 15.28 15.37 15.46 15.54 15.62

where S ∗ (α, β) has been estimated using Equation 16.45. The calculated values of hm indicate that the saturated thickness under the trench steadily increases, and after 10,000 days (27.4 years) the saturated thickness will be 15.62 m. This indicates that the water table below the exfiltration trench will rise (mound) by about 0.62 m (= 62 cm) and will remain at an acceptable depth (> 1.2 m) below the trench. 16.31. The trench design in Problem 16.30 yielded a trench that is 2-m deep, and is 5.22 − 2 − 0.62 = 2.60 m above the mounded water table. If a backfill of 1 m is placed above the trench, the design is unaltered, and the trench will be 2.60 − 1 = 1.60 m above the mounded water table. This exceeds the minimum distance of 1.2 m, and therefore the trench dimensions are the same as in Problem 16.30 , with 1 m of backfill above the trench. 16.32. According to Hantush (1967), the aquifer thickness as a function of time is given by ( ) W 2N L √ , √ νtS ∗ h2m (t) = h2i + K 4 νt 4 νt

(1)

where hi = 7 m, K = 15 m/d, Sy ≈ n = 0.15, hm = 7.4 m (when the water table is within 1.2 m of the bottom of the trench), b = (hi + hm )/2 = (7 + 7.4)/2 = 7.2 m, W = 1 m, t = 365 days, and (15)(7.2) Kb = = 720 m2 /d ν= Sy 0.15 Substituting parameters into Equation 1 gives ( ) 2N 1 10 √ 7.42 = 72 + (720)(365)S ∗ , √ 15 4 (720)(365) 4 (720)(365) 54.76 = 49 + 35040N S ∗ (4.877 × 10−4 , 4.877 × 10−3 ) 54.76 = 49 + 35040N (3.609 × 10−5 ) which yields N = 4.55 m/d Since Q = N LW then Q = (4.55)(10)(1) = 46 m3 /d = 0.532 L/s Check whether the flow is limited by the trench hydraulic conductivity, where Q = Kt L(W + H) = 10(10)(1 + 1.5) = 250 m3 /d = 2.89 L/s

496

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Hence, the maximum allowable flow through the trench is controlled by the mounding limit and is equal to 0.532 L/s . 16.33. The trench length, L, is given by L=

Q Kt (W + H)

(1)

Following ASCE guidelines, use W = 1 m and H = 2 m. From the given data, Q = 5.79 L/s = 500 m3 /d, Kt = 20/2 m/d = 10 m/d, and therefore Equation 1 gives L=

500 = 16.7 m 10(1 + 2)

For a trench depth, H, of 2 m and 30 cm of backfill, there is 4 m − (2 m + 0.3 m) = 1.7 m between the bottom of the trench and the water table. The height, hm , of the water table after 20 years is given by Equation 16.39 as ( ) W L 2N ∗ 2 2 √ , √ νtS (2) hm (t) = hi + K 4 νt 4 νt where hi = 17 m, K = 7 m/d, W = 1 m, L = 16.7 m, and t = 20 years = 20(365) d = 7300 d. Hence, with b = 17 m, and Sy = 0.14, 500 Q = = 29.9 m/d LW (16.7)(1) Kb (7)(17 + hm )/2 ν= = = 25.0(17 + hm ) Sy 0.14

(3)

N=

(4)

Combining Equations (2) to (4) with the given data yields h2m (t)

2(29.9) = 17 + [25.0(17 + hm )](7300)S ∗ 7

(

1

2

= 289 + 1.559 × 10 (17 + hm )S 6

∗

(

4

16.7

)

√ , √ [25.0(17 + hm )](7300) 4 [25.0(17 + hm )](7300)

0.000585 0.00977 √ ,√ 17 + hm 17 + hm

) (5)

which gives hm = 21.22 m This indicates that the mound will rise 21.22 m − 17 m = 4.22 m above the water table. Since the bottom of the trench is only 1.7 m above the water table, then the mounding is unacceptable and a longer trench is required. The minimum trench length required is such that h2m = (17 + 1.7)2 = 350 m2 and

7(17 + 1.7/2) = 892.5 m2 /d 0.14 ( ) 2N W L ∗ √ , √ νtS = 350 − 172 K 4 νt 4 νt ν=

in which case

497

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which can be put in the form ( 2

500 L(1)

7

) ( ) (892.5)(7300)S ∗ 9.794 × 10−5 , 9.794L × 10−5 = 61 9.308 × 108 S ∗ (9.794 × 10−5 , 9.794L × 10−5 ) = 61 L

(6)

which yields L = 1420 m Therefore, the required trench dimensions are 1 m × 2 m × 1420 m . A exfiltration trench with these dimensions is clearly not practical, and an alternative means of disposal should be considered. 16.34. From the given data: L = W = 100 m, N = Q/(W L) = 5 m/d, K = 90 m/d, Sy = 0.2, hi = 35 m, and t = 20 years = 7300 days. Taking b= then

35 + hm hi + hm = 2 2

) ( m (90) 35+h Kb 2 = = 7875 + 225hm ν= Sy 0.2

(1)

Mounding beneath the pond is given by Equation 16.39 as h2m (t)

=

h2i

2N + νtS ∗ K

(

W L √ , √ 4 νt 4 νt

) (2)

Combining Equations 1 and 2 and substituting the given data yields ( ) 2(5) 100 100 √ (7875 + 225hm )(7300)S ∗ , √ 90 4 (7875 + 225hm )(7300) 4 (7875 + 225hm )(7300) ( ) 0.2926 0.2926 = 1225 + 811.1(7855 + 225hm )S ∗ √ ,√ 7875 + 225hm 7875 + 225hm

h2m = 352 + h2m

(3)

which yields hm = 48.2 m Therefore, after 20 years hm = 48.2 m and the water table beneath the pond rises 48.2 m − 35 m = 13.2 m. Since the bottom of the pond is 20 m above the water table at the beginning of recharge operations, this means that the recharge pond will still be adequate in 20 years. 16.35. From the given data: W = 5 m, L = 100 m, N = 0.2 m/d, K = 2 m/d, Sy =0.15, hi = 5.333 m, and t = 6 days.

498

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(a) Using the Hantush function √ hm

hm

hm

(

) W L √ , √ = 4 νt 4 νt v ( ) u u (2)(0.2) 5 100 t √ ν(6)S ∗ , √ = 5.3332 + 2 4 ν(6) 4 ν(6) √ ( ) 0.5103 10.21 √ , √ = 28.44 + 1.2νS ∗ ν ν h2i

2N + νtS ∗ K

(1)

where ν=

K(hi + hm )/2 2(5.333 + hm )/2 Kb = = Sy Sy 0.15

ν = 35.55 + 6.667hm

(2)

Equations 1 and 2 can be solved iteratively as indicated in the calculations: √ ) √ ) hm ν α(= 0.5103 β(= 0.5103 S ∗ (α, β) ν ν (m) 0.1080 8.00 88.89 0.0540 1.080 6.32 77.69 0.0574 1.158 0.1169 6.27 77.35 0.0580 1.161 0.1172

following tabulated hm (m) 6.32 6.27 6.27

Therefore the mounding height is 6.27 m . (b) Using the Swamee and Ojha (1997) approximation results in the calculations: √ ) √ ) hm ν α(= 0.5103 β(= 0.5103 S ∗ (α, β) ν ν (m) 6.27 77.35 0.0580 1.161 0.1240 6.32 77.69 0.0579 1.158 0.1138

following tabulated hm (m) 6.32 6.32

Therefore the Swamee and Ojha (1997) mounding height is 6.32 m , which is 0.8% different than calculated using the Hantush function. 16.35. From the given data: N = 45.7 cm/d = 0.457 m/d, hi = 30.5 m, hm = 45.7 m, T = 929 m2 /d, K = 929/30.5 = 30.5 m/d, n ≈ Sy = 0.2, ν = T /Sy = 4645 m2 /d, W = 30.5 m, and L = 3220 m. Using the Hantush equation, ( ) 2N W L 2 2 ∗ √ , √ hm = hi + νtS K 4 νt 4 νt ( ) 2(0.457) 305 3220 2 2 ∗ √ 45.7 = 30.5 + (4645)tS , √ 30.5 4 4645t 4 4645t which gives tS

∗

(

1.119 11.81 √ , √ t t

499

) = 8.321

(1)

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Assume β ≥ 3.0 and

S ∗ (α, β) = 1 − 4i2 erfc(α) = 1 + 4erfc(α)

Therefore, Equation 1 becomes [

(

t 1 + 4erfc

1.119 √ t

)] = 8.321

√ which gives t = 3.3 days and β = 11.81/ 3.3 = 6.51, which confirms β ≥ 3.0 as originally assumed. If t = 1 year = 365 days and N = 0.457 × 305/W = 139.4/W , then ( ) ( 139.4 ) 2 W 3220 W √ (4645)(365)S ∗ , √ 45.72 = 30.52 + 30.5 4 (4645)(365) 4 (4645)(365) which gives 1 ∗ S W

(

) W , 0.6182 = 7.478 × 10−5 5208

None of the approximate solutions appear applicable, so use tabulated values with the iteration formula ( ) W ∗ W = 13380S , 0.6182 (2) 5208 which converges to W = 10680 m. Therefore, for waterlogging in one year, the lake would have to be approximately 10.7 km wide .

500

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Chapter 17

Water-Resources Planning 17.1. If A is invested at the end of year j, then its value, Fj , at the end of year n is given by Fj = A(1 + i)n−j Therefore, if A is invested at the end of each year, the total future value, F , is given by F =

n ∑

Fj =

j=1

n ∑

A(1 + i)n−j = A

j=1

n ∑ (1 + i)n−j

(1)

j=1

Note the following, n ∑ (1 + i)n−j = (1 + i)n−1 + (1 + i)n−2 + . . . + 1 j=1

then

∑n

j=1 (1

(1 + i)n−1 + (1 + i)n−1 + and solving for

∑n

j=1 (1

+ i)n−j − 1

(1 + i) ∑n n−j − 1 j=1 (1 + i) (1 + i) ∑n n−j − 1 j=1 (1 + i) (1 + i)

= (1 + i)n−2 + (1 + i)n−3 + . . . + 1 = (1 + i)n−1 + (1 + i)n−2 + . . . + 1 =

n ∑

(1 + i)n−j

j=1

+ i)n−j yields n ∑

(1 + i)n−j =

j=1

(1 + i)n − 1 i

Combining Equations 1 and 2 yields F =A

(1 + i)n − 1 i

501

or

A i = F (1 + i)n − 1

(2)

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17.2. The sinking-fund factor (Equation 17.4) is given by A i = F (1 + i)n − 1

(1)

and the present-worth factor (Equation 17.3) is given by F = (1 + i)n P

(2)

Combining Equations 1 and 2 gives the capital-recovery factor as A F i A = × = × (1 + i)n P F P (1 + i)n − 1 which yields i(1 + i)n A = P (1 + i)n − 1 17.3. The present worth at year 0 of only the gradient is equal to the sum of the present worths of the individual values, where each value is considered a future amount. Hence, ( ) ( ) ( ) ( ) P P P P P =G , i, 2 + 2G , i, 3 + . . . + (n − 2)G , i, n − 1 + (n − 1)G , i, n F F F F Factor out G and use the P/F formula, [ ] 1 n−2 2 3 n−1 P =G + + + ... + + (1 + i)2 (1 + i)3 (1 + i)4 (1 + i)n−1 (1 + i)n Multiplying both sides by (1 + i) yields [ ] 1 2 3 n−2 n−1 P (1 + i) = G + + + ... + + (1 + i)1 (1 + i)2 (1 + i)3 (1 + i)n−2 (1 + i)n−1

(1)

(2)

Subtract Equation 1 from Equation 2 and simplify, [ ] [ ] 1 1 1 1 n iP = G + + ... + + −G (1 + i)1 (1 + i)2 (1 + i)n−1 (1 + i)n (1 + i)n The left-bracketed expression can be evaluated using the following identity, [ ] (1 + i)n − 1 1 1 1 1 = + + ... + + i(1 + i)n (1 + i)1 (1 + i)2 (1 + i)n−1 (1 + i)n Combining Equations 2 and 3 yields

[ ] G (1 + i)n − 1 n P = − i i(1 + i)n (1 + i)n

which simplifies to P (1 + i)n − in − 1 = G i2 (1 + i)n

502

(3)

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17.4. The series starts in year 1 at an initial amount A0 . The relation to determine the total present worth, P for the entire cash flow series can be achieved by multiplying each geometricallyincreasing annual return by the P/F factor 1/(1 + i)n , which yields A0 (1 + g) A0 (1 + g)2 A0 A0 (1 + g)n−1 + + + . . . + (1 + i)1 (1 + i)2 (1 + i)3 (1 + i)n [ ] 2 1 (1 + g) (1 + g) (1 + g)n−1 P = A0 + + + ... + (1 + i) (1 + i)2 (1 + i)3 (1 + i)n P =

(1)

Multiply both sides by (1 + g)/(1 + i), subtract Equation 1 from the result, factor out P , and obtain ) [ ] ( (1 + g)n 1 1+g P − 1 = A0 − 1+i (1 + i)n+1 1 + i Solve for P and simplify

 P = A0 

1−

(

1+g 1+i

i−g

)n  ,

g ̸= i

(2)

The term in brackets is the geometric-gradient series present worth factor for g ̸= i. When g = i, substitute g for i in Equation 1 to obtain [ ] 1 1 1 1 P = A0 + + + ... + 1+i 1+i 1+i 1+i which simplifies to P =

nA0 (1 + i)

(3)

Combining Equations 2 and 3 yields  n 1−( 1+g 1+i )   i−g

P = A0  

n 1+i

g ̸= i g=i

17.5. The two alternatives are considered as follows: Alternative 1: This alternative produces annual returns of $50,000 per year for all 30 years of the project. The present worth of these returns can be determined using the uniform-series present-worth factor (Equation 17.7) which can be put in the form [ ] (1 + i)n − 1 P = A i(1 + i)n In this case, A = $50,000, i = 0.04, n = 30, and ] [ (1 + 0.04)30 − 1 ($50, 000) = $864, 602 P = 0.04(1 + 0.04)30

503

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Therefore the present worth of the uniform annual returns is $864,602 . Alternative 2: This alternative produces a return of $17,000 at the end of the first year and increases by 6% each year for the duration of the 30-year design life. The present worth of these returns can be determined using the geometric-gradient present-worth factor (Equation 17.11) which can be put in the form ( )n   1 − 1+g 1+i  A1 P = i−g In this case, A1 = $17,000, g = 0.06, i = 0.04, n = 30, and  ( )30  1+0.06  1 − 1+0.04  P =  ($17, 000) = $655, 206 0.04 − 0.06 Therefore the present worth of the geometric-gradient annual returns is $655,206 . Based on these results, and the fact that both proposed alternatives have approximately the same cost, Alternative 1 produces the greatest return on investment. 17.6. For each alternative, the present worth of each cost and revenue item is calculated individually, and the net present value (NPV) is calculated as the sum of the present values of the cost and revenue items as follows: Alternative 1: Item initial investment, I

Formula P = I[

base revenue, Ar

P =

revenue gradient, Gr

P =

base operating cost, Ac

P =

operating cost gradient, Gc salvage value, F

P = P =

Parameters I = −$200, 000

]

(1+i)n −1 Ar n [ i(1+i) ] (1+i)n −in−1 Gr 2 (1+i)n i [ ] (1+i)n −1 Ac i(1+i)n

[

(1+i)n −in−1 2 n [ i (1+i) ] 1 (1+i)n

]

Gc

F

Present Worth −$200,000

i = 4%, n = 20, Ar = $50,000

$679,516

i = 4%, n = 20, Gr = $2,000

$223,129

i = 4%, n = 20, Ac = −$6,000

−$81,542

i = 4%, n = 20, Gc = −$4,000

−$446,259

i = 4%, n = 20, F = $20,000

$9,128 $183,972

Net Present Value Alternative 2: Item initial investment, I revenue, Ar , g cost, Ac , g salvage value, F

Formula P =[ I ] n 1−( 1+g 1+i ) P = Ar i−g [ ] 1+g n 1−( 1+i ) P = Ac i−g [ ] 1 P = (1+i)n F

Parameters I = −$100, 000 i = 4%, n = 20, Ar = $20,000, g = 6% i = 4%, n = 20, Ac = −$11,000, g = 6% i = 4%, n = 20, F = $10,000

Present Worth −$100,000 $463,695 −$255,032 $4,564 $113,227

Net Present Value

504

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Based on these results, Alternative 1 has the greatest net present value ($183,972), and Alternative 3 is not economically feasible. 17.7. All proposed alternatives are described by an initial cost, I, first-year revenue, R0 , revenue growth rate, gr , first-year cost, C0 , cost growth rate, gc , and design life, n, equal to 15 years. The return rate, i∗ , for each alternative satisfies the relation ( ) ( ) P P −I + , gr , i∗ , n R0 − , gc , i∗ , n C0 = 0 (1) R0 C0 where (P/A0 , g, i, n) is the geometric-gradient present-worth factor given by Equation 17.11, and (P/F, i, n) is the single-payment present-worth factor given by Equation 17.3. Combining Equations 17.3, 1, and 17.12 gives ( )n  ( )n    r c 1 − 1+g 1 − 1+g 1+i∗ 1+i∗  R0 −   C0 = 0 −I +  (2) i∗ − gr i∗ − gc Substituting given values of I, R0 , gr , C0 , and gc for each of the proposed alternatives yields the following results: Alternative 1 2

I $130,000 $75,000

R0 $16,000 $12,000

gr 4.5% 3.5%

C0 $6,000 $3,200

gc 5% 3.5%

i∗ 4.5%, 5% 3.5%

Since the minimum attractive rate of return is 6%, neither alternative is economically feasible. 17.8. From the given data: the design life, n, is 25 years, and the interest rate, i to be used in the analysis is 5%. The present value of the cost, C, is $800,000, the annual benefit, A, is $80,000, and the present value of the benefits, B is given by ( ) [ ] [ ] P (1 + i)n − 1 (1 + 0.05)25 − 1 B= , i, n A = A= ($80, 000) = $1, 127, 516 A i(1 + i)n 0.05(1 + 0.05)25 The disbenefit is that annual spending on other projects, AD , is reduced by $55,000 for the first eight years of the project, so the present value of the disbenefit, D is given by ( ] [ ] ) [ P (1 + 0.05)8 − 1 (1 + i)n − 1 D= , i, n AD = AD = ($55, 000) = $355, 477 A i(1 + i)n 0.05(1 + 0.05)8 The benefit cost ratio of the proposal, taking the disbenefit into account, is given by B−D $1, 127, 516 − $355, 477 = = 0.97 C $800, 000

or

B $1, 127, 516 = = 0.98 C +D $800, 000 + $355, 477

Since the benefit-cost ratio (0.97,0.98) is less than one, this proposal is not economically feasible.

505

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Solutions to Selected Problems using Mathcad® By Dixie M. Griffin, Jr.

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Chin 2.1 2nd and 3rd edition Water at 20oC is flowing in a 100 mm diameter pipe at an average velocity of 2 m/sec. If the pipe diameter is suddenly expanded to 150 mm, what is the new velocity in the pipe ? What are the volumetric flow rate and mass flow rate ? dia1  100  mm m v1  2  sec

dia2  150  mm ρ 20  998.2 

m 2

Q 

v2 

kg

π  dia1 4

m

 0.889

m

 π  dia 2  2  4   

3 N 3

m

3

 v1  0.016

Q

3

γ20  ρ 20  g  9.789  10

s

s

 π  dia 2  kg 1 ρ'20  ρ 20    v1  15.68 s  4 

I:\Mathcad application areas\Fluids-open channels-hydrology\HYDRAULIC PROCEDURES\Bernoulli's, energy equation, headloss equations all flows\ 5/13/2012

Chin 2-1 velocity in pipe.xmcd

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Chin 2-12

L during a fire. If the length of sec the service pipe is 40 m and the head loss in the pipe is not to exceed hL  45  m , calculate the minimum pipe diameter that must be used. Use the Colebrook - White equation in your calculations. A galvanized iron service pipe from a water main is required to deliver Q  300 

k  .15  mm

L  40  m

Discussion : Note that because we are only computing the headloss in the pipeline we do not need to write the entire energy equation, just the equation for the headloss in the pipe, use the Darcy Weisbach Equation. In addition, use the Colebrook - White equation to compute the friction factor, f The unknowns are D, f, V, and hL.

Solution procedure: Use a Mathcad solve block to solve the D-W equation, the C-W equation, and the continuity equation. Including two constraints, one on the pipe diameter to prevent excessive flow velocities and one on head loss to prevent excessive headloss. By adjusting the pipe diameter and headloss the flow velocity can be controlled for the given flow rate.

assume a temperature of 20o C, obtain the density and absolute viscosity of water: kg  3 N s , and μ  1.002  10  ρ  998.2  3 2 m m

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDRAULIC PROCEDURES \Bernoulli's, energy equation, headloss equations all flows\ 3/7/2012

Chin 2-11 pipe headloss using C-W eqn.xmcd

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Unknowns D, f, V, hL f  .01

V  10 

ft sec

D  6  in

hL  35  m

The solve block starts with the word "given" in math mode Given

hL = f 

L  D 

Q

2

2   π  D 2    4   2  g   

V=

Q

 π D 2     4 

 k   D  1 2.51 = 2  log    3.7 ρ D  V f   f μ   The two constraints below allow the user control the headloss and pipe diameter hL  45  m

D  9  in

The statement below initiates one of several user chosen algorithms to find the solution set for the equations given

 Dsoln     Vsoln     Find  D V f hL  fsoln  h   Lsoln  Print out the values determined to be the solution set

Dsoln  0.2  m

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDRAULIC PROCEDURES \Bernoulli's, energy equation, headloss equations all flows\ 3/7/2012

Vsoln  7.309 

m sec

fsoln  0.018

Chin 2-11 pipe headloss using C-W eqn.xmcd

hLsoln  8.59 m

2 of 3

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NR 

ρ  Dsoln  Vsoln μ

 1.665  10

6

flow regime is turbulent

By using a 0.2 m diameter pipe we will generate a head loss of hLsoln  8.59 m and a pipe velocity of Vsoln  7.309

m s

Check the solution values by plugging them into the governing equations to see if they are satisfied hLsoln 2   L Q f    2  soln Dsoln    π  Dsoln 2        2  g  4    

1

1 fsoln

Vsoln

1

Q 2

 π  D soln    4  

     2  log       

k Dsoln

3.7

  2.51   ρ  Dsoln  Vsoln   fsoln   μ 

1

check the flow rate

Q 

π  Dsoln 4

2

3

 Vsoln  0.3

m s

Summary : This file is a template of sorts for flow in a single pipeline. By changing the value of k we can determine the D-W friction factor for a variety of pipe materials. We can also simulate the headloss for various temperatures by changing the density and viscosity of water.

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDRAULIC PROCEDURES \Bernoulli's, energy equation, headloss equations all flows\ 3/7/2012

Chin 2-11 pipe headloss using C-W eqn.xmcd

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Chin 2-17 3rd edition

In the problem below we will: compute the various energy inputs and outputs to a pump/motor system use Mathcad's currency function to compute the monthly cost to run the pump

Water enters and leaves a pump in pipelines of the same diameter and approximately the same elevation. If the pressure at the pump inlet is pinlet  30  kPa and a pressure of pdischarge  500  kPa is desired for water leaving the pump, what is the head added by the pump, and what is the power delivered to the fluid.

γ  62.4 

ft hinlet 

houtlet 

3

lbf

Q  1 

3

pinlet γ

m sec

 3.061 m

pdischarge γ

 51.009 m

Headadded  houtlet  hinlet  47.948 m

The power per pound second being imparted to the liquid is P  γ  Q  Headadded  630.28  hp 5

P  4.7  10 W P  630.28  hp P  445.474 

BTU sec

In order to compute the power that must be supplied to the pump or stated another way, the output power of the motor driving the pump we need the pump efficiency which can only be obtained from the pump manufacturer. In this problem the pmp efficiency is effpump  60  %

The input power to the pump is therefore: Pinput 

γ  Q  Headadded effpump

 783.333  kW

However, the power you PAY for is the input power to the MOTOR. Motor efficiency is usually around 80%

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Chin 2-16 pump power.xmcd

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effmotor  80  %

The power that must be supplied to the motor is therefore Pinput_motor 

Assuming that power cost are cost 

γ  Q  Headadded effpump  effmotor

 979.167  kW

.05  $ and the pump operates approximately 60 hours per month the monthly kW  hr

cost to operate this pump will be. mo  30  day

Pinput_motor  cost  60 

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDRAULIC PROCEDURES \pumping calculations\ 9/24/2012

hr 3 $  2.938  10  mo mo

Chin 2-16 pump power.xmcd

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Chin 2-20 3 rd edition The top floor of an office building is 40 m above street level and is to be supplied with water from a municipal pipeline buried 1.5 m below street level. The water pressure in the distribution main is 450 kPa, the sum of the local loss coefficients in the building pipes is 10 and the pipe flow to be delivered to the top floor is 20 L/sec through a 150 mm diameter PVC pipe. The length of the pipeline in the building is 60 m and the water temperature is 20oC, and the water pressure on the top floor must be at least 150 kPa. Will a booster pump be necessary for the building ? If so what water horsepower must be supplied by the pump. Always define all variables at the top of the file, in this way you only need to change a variable value in place and the change will take affect throughout the file. pressure at street level : pstreet  450kPa

liter water flow at the top floor: Qtop  20  sec

pressure at the top floor : ptop  150  kPa

pipe diameter : dia  150  mm

sum of the minor loss coefficients : ksum  10 kg

density of water : ρ  998.2 

m water velocity : V 

Qtop

 π  dia 2     4 

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDRAULIC PROCEDURES \Bernoulli's, energy equation, headloss equations all flows\ 9/24/2012

pipe length : L  60  m

absolute viscosity : μ  .001002  Pa  sec

3

 1.132

m s

Chin 2-18 sizing booster pump.xmcd

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Use the Darcy-Weisbach equation to compute the headloss and the Jain equation to compute f. We have three unknowns and three equations, use a solve block. The way I did the problem is to write the energy equation WITH a pump term in it. If a pump is not needed I will get zero or a negative number for hp. Initial guesses for the Reynolds number, f, and the head supplied by the pump are given below Re  10

6

f  .01

hp  5  m

Given Reynolds Number Re =

ρ  dia  V μ

0.25

The Jain Equation f =

 5.74  0.9   Re 

2

log 

The energy equation:

pstreet ρ g

 1.5  m  ksum 

2 2 ptop V L V  f   hp =  40  m 2 g dia 2  g ρ g

 Resoln     fsoln   Find  Re f hp    hpsoln 

The amount of energy per time in

N m that must added by the pump hpsoln  11.926 m N 3 ft  lbf

One horsepower is 550 ft*lbf per second, therefore Hp  ρ  g  Qtop  hpsoln  1.722  10  Hp  3.131  hp

sec

Hp  2.335  kW

Note that Mathcad does the unit conversions automatically

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDRAULIC PROCEDURES \Bernoulli's, energy equation, headloss equations all flows\ 9/24/2012

Chin 2-18 sizing booster pump.xmcd

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Chin 2-21 3rd edition Water Pressure and Pump Power Needed Water is pumped from a supply reservoir to a ductile iron water transmission line, as shown in the figure below. The high point of the transmission line is is A which is 1 km downstream of the supply reservoir. The low point in 3

m , the the transmission line is at B, 1 km downstream of A. If the flowrate through the pipeline is Q  1  sec diameter of the pipeline is 750 mm, and the pressure at A is to be 350 kPa, then: (a) estimate the energy that must be added by the pump (b) estimate the power supplied by the pump (c) calculate the water pressure at B.

Use the Darcy Weisbach equation to compute head loss and the Swamee - Jain equation to compute the friction factor f. The water temperature in the winter may be assumed to be 10oC

Known values : density of water ρ  999.7 

kg m

3

unit weight of water γ  ρ  g absolute viscosity of water μ  1.307  10

3

 Pa  sec

pipe diameter dia  750  mm absolute roughness of pipe ks  .26  mm elevation at reservoir surface zres  7  m elevation at point A zA  10  m elevation at point B zB  4  m elevation at point A pA  350  kPa pressure at reservoir surface pres  0  kPa water velocity at surface of the reservoir vres  0 

m sec

length of pipe from reservoir to point A Lres_A  1  km length of pipe from A to B LA_B  1  km

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDRAULIC PROCEDURES \Bernoulli's, energy equation, headloss equations all flows\ 9/24/2012

Chin 2-19 water transmission line.xmcd

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compute the pipe velocity using the continuity equation

compute the Reynolds number in the pipe NR 

v 

Q

 π  dia 2     4 

 2.264

m s

ρ  dia  v 6  1.299  10 μ

Write the energy equation between the water surface and point A, there are two unknowns, v and f. Therefore, a second equation is needed for f. I choose to use the Colebrook White equation. Use a solve block to get solution.

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDRAULIC PROCEDURES \Bernoulli's, energy equation, headloss equations all flows\ 9/24/2012

Chin 2-19 water transmission line.xmcd

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initial guess for pump energy imparted hp  10 

ft  lbf lbf

initial guess for Darcy Weisbach friction factor f  .02 Given

pres γ

 zres 

vres 2 g

2

 f

Lres_A



dia

2 2 pA v v  hp =  zA  2 g 2 g γ

  ks   dia 1 2.51  = 2  log    3.7 f NR  f     hpsoln     Find h f  p   fsoln    The amount of energy imparted to each unit weight of fluid is found to be: hpsoln  44.508 

N m N

The Darcy Weisbach friction factor is found to be fsoln  0.016 The output power of the pump is Power  γ  Q  hpsoln  436.341  kW Power  585.143  hp

Power  413.571 

BTU sec

note the different units that can be used

6 cal

Power  6.253  10 

min

Now compute the pressure at point B, write the energy equation between the reservoir and point B, solve for the pressure at point B

pres γ

 zres 

vres 2 g

2

 fsoln 

 Lres_A  LA_B 

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dia

2 2 pB v v  hpsoln =  zB  2 g 2 g γ

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2 2  2 pres vres fsoln  v   LA_B  Lres_A   v pB  γ  hpsoln  zB  zres       51.409  psi 2 g 2 g 2  dia  g γ  

pB  354.453  kPa

Given that there is friction loss between A and B, why is the pressure at B larger than at A ?

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Chin Problem 2-25 Water flows at a rate of 10 cubic meters per second in a reinforced concrete box culvert 2 m x 2 m square. If the pipe is laid on a .002 slope what is the change in pressure over a distance of 500 m. The conduit flows full. 3

H  2  m

W  2  m

V 

m sec

Q  10 

S  .002

ρ  998 

kg m

Q m  2.5 W H s

lbf

γ  62.4 

ft

3

μ  .001  N 

sec m

2

L  500  m

3

Solution : This is a standard problem except that we have a noncircular conduit. In such cases the conduit is characterized by the hydraulic radius rather than the diameter. The hydraulic radius for any shape is the area of flow divided by the wetted perimeter. The problem is worked using two values of absolute roughness. H W For this conduit the hydraulic radius becomes: Rh   0.5 m 2 H  2 W

The Reynolds number for a noncircular pipe becomes: NR 

 

ρ  V  4  Rh μ

 4.99  10

6

0.25

To obtain f use the Jain Equation : f ks 

   log    

    5.74    N 0.9  R 

ks

 4 Rh 3.7

f ( .18  mm)  0.012

 

2

f ( 1.8  mm)  0.019

 

2

L V hf ks  f ks   4  Rh 2  g hf ( .18  mm)  0.936 m Δz = z2  z1 p1 γ

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 z1 

hf ( 1.8  mm)  1.525 m Δz  .002  500  m  1 m

2 2 2 p2 V L V V  f  =  z2  2 g 2 g 4  Rh 2  g γ

Chin 2-22 pressure drop in noncircular pipe.xmcd

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Δp = p1  p2

 L V  Δp ks  Δz  f ks   γ  4  Rh 2  g 

 



 

2



Δp ( .18  mm)  0.632  kPa Δp ( 1.8  mm)  5.145  kPa

unit weight water absolute roughness pressure change 62.4 lbs/ft3 1.8 mm 5.14 kPa 62.4 lbs/ft3 .18 mm ‐0.632 kPa

f 0.019 0.012

For the smooth concrete box culvert the pressure increases from point 1 to point 2, for the rough concrete the pressure drops from point 1 to point 2.

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Chin 2.3 2nd and 3rd editions The velocity distribution in a pipe is given by the equation:

  r  2 v ( r) = V0  1       R  where v ( r) is the velocity at a distance r from the pipe centerline, V0 is the centerline velocity, and R is the radius of the pipe. Calculate the average velocity and flowrate in the pipe in terms of R The pipe velocity is a maximum at the centerline and drops to zero at the pipe wall, the average velocity can be computed from.

1  Vave =   A 

R

V dA

0

A = 2π r dA = 2  π  r  dr

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Chin 2.3 2nd and 3rd editions.xmcd

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   2 π R  

R

1



2  r    2  π  r dr  V0   2  R 

V0  1 



0

Now lets plot the pipe velocity as a function of the distance from the pipe centerline just to see what it looks like, chose a vallue for R and V0

R  1  m

V0  1 

m sec

r  1  m .99  m  1  m

  r  2 v ( r)  V0  1       R  centerline  0 

m sec

Pipe Velocity vs Distance from Pipe Centerline 1 0.8 0.6 0.4 0.2 r

centerline

0  0.2  0.4  0.6  0.8 1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

v( r)

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Chin 2.3 2nd and 3rd editions.xmcd

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 1  Vave   2  π R 

R

  r  2 m V0  1      2  π  r dr  2  sec   R 

0

The average velocity is 1/2 m/sec when the centerline (maximum velocity) is 1 m/sec

The flowrate is given by multiplying the average velocity by the pipe area:

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Chin 2.3 2nd and 3rd editions.xmcd

Q=

V0 2

2

π R

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Chin Problem 2-32 3rd edition

Water flows at a velocity of 2 m/sec in a new 300 mm ductile iron pipe. Estimate the head loss over 500 m using the (a) Hazen Williams Equation (b) the Manning equation and (c) the Darcy Weisbach equation. Compare your results. Calculate the Hazen Williams roughness coefficient and Manning coefficient that should be used to obtain the same head loss as the Darcy Weisbach equation. equivalent sand grain roughness for ductile iron pipe: k  .26  mm define a millipascal: mPa 

pipe length: L  500  m

ρ  998.2 

kg m

3

Pa 1000

μ  1.002  mPa  sec

pipe diameter: D  300  mm

pipe velocity: V  2  π D

2

ν 

μ 6 m  1.004  10 s ρ

m sec

2

4

R 

π D 2

Darcy Weisbach Equation: hL = f 

L V  D 2 g

The Darcy Weisbach equation appropriate for all flow regimes. The friction factor f is computed using the Colebrook White equation. 1 f

= 2  log 



2.51  k  3.7  D N  f  R



Use the root function to solve the Colebrook-White Equation for f:

V 5 Reynolds number: NR  D   5.977  10 ν initial guess for Reynolds number: f  .015

 1  2  log  k  2.51  f  3.7  D N  f   R  f   

fsoln  root 

fsoln  0.02 2

L V The headloss using the Darcy Weisbach equation is: hL_DW  fsoln    6.638 m D 2 g Hazen Williams Equation: V = .894  CH  R

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0.63

 Sf

0.54

is an empirical friction loss equation

Chin 2-29 headloss comparison.xmcd

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From Chin, 2nd edition: The Hazen Williams is applicable to the flow of water at 16oC in pipes between 50 mm (2 in) and 1850 mm (72 in), and flow velocities less than 3 m/sec. (Mott 1994). Outside of these conditions , use of the H-W equation is strongly discouraged. To further support these quantitative limitations, Street and colleagues (1996) and Liou (1998) have shown that the H-W coefficient has a strong dependence on the Reynolds number and is most applicable where the pipe surface is relatively smooth and flow is in the early part of its transition to to rough flow. Furthermore, Jain and colleagues (1978) have shown that an error up to 39% can be expected in the evaluation of velocity by the H-W equation over a wide range of diameters and slopes . In spite of the cautionary notes, the Hazen Williams formula is frequently used in the United States for the design of large water supply pipes without regard to its limitations. This practice can potentially lead to litigation (Bombardelli and Garcia, 2003)

The Hazen Williams equation is empirical, this means it is not dimensionally consistent. The CORRECT way to handle such equations with Mathcad is to divide out ALL units, then attach the appropriate units to the entire equation.

Hazen Williams coefficient, typical value for new unlined ductile iron pipe: CH  130

L

hL  6.82 

m

D    m

1.17

 V  m  sec   CH

    

1.85

 m  6.175 m

Manning's Equation:

Manning' s n, typical value for new unlined ductile iron pipe: n  .013 

2

sec 1

m

3

1.0 3 Manning Equation : V =  R  Sf n 2

hL  6.35 

n  L V

2

4

D

3

hL  10.687 m

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part C - compute the Hazen Williams coefficient and Manning's n corresponding to the value of headloss obtained using the Darcy Weisbach equation. Manning's equation has been shown to be appropriate for flows in the fully rough region, where there is little dependence on the Reynolds number.

L

hL_DW = 6.82 

m

D    m

1.17

    

 V  m  sec   CH

  L  m CHsoln  root hL_DW  6.82  1.17  D      m

1.85

m

 V  m  sec   CH

    

1.85

    m CH   

CHsoln  125.017



nsoln  root  hL_DW  6.35 

  

4

D

3

2



n

  

s

nsoln  0.01 m

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2

n  L V

0.333

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Chin 2-38 3rd edition - 3 reservoir problem

Reservoirs A, B and C are connected as shown in the figure below. The water elevations in reservoirs A, B, and C are 100 m, 80 m, and 60 m, respectively. The three pipes connecting the reservoirs meet at junction J, with pipe AJ being 900 m long, BJ 800 m long, CJ 700 m long, and the diameters of all pipes equal to 850 mm. If all pipes are made of ductile iron and the water temperature is 20o C, find the flow into or out of each reservoir.

Discussion : Unknowns are the 3 flowrates. Pressure and elevation at point 2 are unknown but fixed. We can write the energy equation between reservoirs A and B, A and C and B and C. This gives three equations and three unknowns. However we must also restrict our solution vector in such a way that mass is conserved at the common joint , J.

Known quantities:

pA  0 

lbf in

2

LAJ  900  m

pC  0 

lbf in

2

LBJ  800  m

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pB  0 

lbf in

zA  100  m

zB  80  m

zC  60  m

2

LCJ  700  m

Chin 2-30 three reservoir problem.xmcd

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fAJ  .020

fBJ  .015

DAJ  0.85  m

fCJ  .020

DBJ  0.85  m

lbf

γ  62.4 

ft

DCJ  0.85  m

μ  .001  Pa  sec

3

ks  .26  mm

ρ  998.2 

kg m

AAJ 

π  DAJ

2

ABJ 

4

gal QAJ  1000  min

π  DBJ

2

ACJ 

4

gal QBJ  1000  min

π  DCJ

3

2

4

gal QCJ  100  min

Important note : Notice how the squared velocity terms in the equations below are written below, as Q  Q 2

rather than as Q . This is necessary because it allows for negative values of Q to be carried through the computations. Recall that a negative Q means the assumed direction is incorrect. If the Q terms are 2

written simply as Q the governing equations will not converge to a solution unless the assumed directions are the correct ones. Given

 QAJ  QAJ  2 pA LAJ  AAJ   zA  fAJ   γ

DAJ

2 g

 QBJ  QBJ    2     f  LBJ   ABJ BJ DBJ

2 g

   p  = B z B γ

energy equation between reservoirs 1 and 3

 QBJ  QBJ  2 pB LBJ  ABJ   zB  fBJ   γ 2 g D

 QCJ  QCJ    2  LCJ  ACJ  f    CJ D 2 g CJ

BJ energy equation between reservoirs 3 and 4

   p  = C z C γ

QAJ  QBJ  QCJ = 0 mass balance at the common joint, this not an equation as such, it is a constraint

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 QAJ  QAJ  2 pA LAJ  AAJ   zA  fAJ   γ

DAJ

2 g

 QCJ  QCJ    2     f  LCJ   ACJ CJ DCJ

2 g

   p  = C z C γ

energy equation between reservoirs 1 and 4

 ks   DAJ  1 2.51 = 2  log    3.7 fAJ  ρ  DAJ  QAJ       fAJ   μ  AAJ      ks   DBJ  1 2.51 = 2  log    3.7 fBJ  ρ  DBJ  QBJ       fBJ   μ  ABJ    

 ks   DCJ  1 2.51 = 2  log    3.7 fCJ  ρ  DCJ  QCJ       fCJ   μ  ACJ    

 QAJsoln     QBJsoln     QCJsoln     Find  QAJ QBJ QCJ fAJ fBJ fCJ  fAJsoln  f   BJsoln  f   CJsoln 

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3

m QAJsoln  2.819 s

fAJsoln  0.015

3

m QBJsoln  0.343 s fBJsoln  0.013

3

m QCJsoln  3.162 s fCJsoln  0.015

The solution implies that flow is leaving reservoir A and B and entering reservoir C

VAJsoln 

QAJsoln AAJ

VAJsoln  16.299 

ft sec

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDRAULIC PROCEDURES \PIPES-PIPING SYSTEMS-PIPING NETWORKS\ 9/24/2012

VBJsoln 

QBJsoln ABJ

VBJsoln  1.986 

ft sec

VCJsoln 

QCJsoln ACJ

VCJsoln  18.284 

Chin 2-30 three reservoir problem.xmcd

ft sec

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Chin 2-4 3rd Edition

Calculate the momentum correction coefficient, β, for the velocity distribution given by:

  r 2 v ( r) = V0  1     R 

  

The equation for the momentum correction coefficient is: β =

1 A  Vave

2

2 2

.

 2   v ( r) dA 2  A 2 4

   r 2 V0  r 2 2  V0  r v ( r) = V0  1     expand  v ( r) = V0   2 4    R   R R

R

  1  β= 2  2  V0    π R   0  2 

2 2 2 4   V 2  2  V 0  r  V 0  r   2  π  r dr  β = 4  0 2 4  3 R R  

The momentum correction factor is used to account for the fact that the velocity across a pipe is not constant as is assumed in the continuity equation

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Chin 2-48 3rd edition A pump is required to deliver 150 L/sec ( ± 10%) through a 300 mm diameter PVC pipe from a well to a reservoir. The water level in the well is 1.5 m below the ground surface and the water surface in the reservoir is 2 m above the ground surface. The delivery pipe is 300 m long, and local losses can be neglected. A pump manufacturer suggests using the following equation for the performance of the specified pump.

hp ( Q)  6  6.67  10

5

 Q  liter     sec 

2



where : hp has units of meters and Q is in liters per second. Note that the equation is NOT dimensionally consistent.

Define all parameters at the top of the file p1  0  psi v2  0 

p2  0  psi m sec

z1  1.5  m

f1  .02

f2  .0046

γ  62.4 

z2  2  m L  300  m

v1  0 

m sec

D  .3  m

lbf ft

3

Write the energy equation, including a term for the energy which must be supplied by a pump, hreqd, solve for hreqd

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Q

2

 π  D2    p1 v1 L  4   z1   f  γ 2 g 2 g D

2

2

Q  0 

 hreqd =

p2 γ

 z2 

v2

2

2 g

l l l 5   200  sec sec sec

Summary : I class I told you to use a Darcy Weisbach friction factor of 0.02. In Chin they compute it using the Jain equation and get a value of .0046. Note the difference in flow rate

p1

p2

v1

2

v2

2

2

system head curve 1 ( f = 0.02 ): hreqd1 ( Q)  z2  z1      2 g 2 g γ γ

p1

p2

v1

2

v2

8  L  Q  f1 2

2

system head curve 2: ( f = 0.0046 ) hreqd2 ( Q)  z2  z1      2 g 2 g γ γ

pump curve : hp ( Q)  6  6.67  10

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5

Chin 2-37 pump operating point.xmcd

Q    liter     sec 

5

π D g 2

8  L  Q  f2 2

5

π D g

2

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12 11.1 10.2 hreqd1 ( Q) 9.3 m

8.4

hreqd2 ( Q) 7.5 m 6.6 hp ( Q) 5.7 4.8 3.9 3

0

20

40

60

80 Q l

sec



100 Q l sec



120 Q l

140

160

180

200

95.7

sec

system head curve 1 f = 0.02 system head curve 2 f = 0.0046 pump curve In both instances the pump operating point occurs at the intersection of the pump curve and system head curve. The operating point for f = 0.02 is about 96 l/sec against a head of 5.3 m. For f = 0.0046 the flow is 148 l/sec against a head of 4.5 m

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Chin 2-50 3rd edition

A pump is to be selected to deliver water from a well to a treatment plant through a 300 m long pipeline. The temperature of the water is 20o the average elevation of the water surface in the well is 5 m below the ground surface, the pump is 50 cm above the ground surface and the water in the receiving reservoir at the water treatment plant is 4 m above the ground surface. The delivery pipe is made of ductile iron ( ks  .26  mm) with a diameter of 3

m 2 800 mm. If the selected pump has a performance curve equation of hp ( Q)  12  0.1  Q where Q is in and h p sec is in meters, what is the flowrate through the system? Calculate the specific speed of pump, and state what type of pump will be required when the pump motor has a rotation speed of 1200 rpm, neglect local losses.

z1  5  m

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z2  4  m

Chin 2-38 sizing pump and determining type.xmcd

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  π D 2 2  V  2 p1 V1 4 L V    z1   f   12  m  0.1   3 γ 2 g D 2 g  m     sec

2     2    m = p2  z  V2 2 2 g  γ     

  π D 2  V  2 4 L V   z1  z2 = f    12  m  0.1   3 D 2 g  m     sec ρ  998.2 

kg m

μ  1.002  10

3

 3 N s  2

2        m      

k  .26  mm

ω  1200  rpm

m

Problem asks for the flow rate. The unknowns are Q, f, NR, hL. Available equations 2

L V 1. friction loss hL = f   D 2 g 2. hp ( Q) = 12  0.1  Q

2

3. Colebrook White Equation

4. Reynolds Number NR =

NR  10

6

f  .012

V  2 

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1

 k   D  2.51 = 2  log    ρ D  V 3.7 f   f μ  

ρ D  V μ

hL  12  m m sec

hp  12  m

D  800  mm

L  300  m

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Given

 π D 2  V 4  hp = 12  m  0.1   3 m   sec

2

 L  V   D  2 g

hL =  f 

2

   m   

 k   D  1 2.51 = 2  log    3.7 ρ V D f   f μ  

NR 

  π D 2  V  2 4 L V   z1  z2 = f    12  m  0.1   3 D 2 g  m     sec

ρ D  V μ

2        m      

 NRsoln     fsoln     hLsoln   Find  NR f hL hp V    hpsoln   V   soln  NRsoln  1.594  10

6

fsoln  0.015

hLsoln  2.764 m

hpsoln  11.764 m hpsoln  38.596  ft

Vsoln  3.055

m s

2

Q 

3

π D m  Vsoln  1.536 s 4 4

Q  2.434  10  gpm

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Calculate the specific speed for the pump to be used:

1

ω  Q   rpm  gal 

 

Ns 

min

 

3

 hpsoln     ft 

2

Ns  1.209  10

4

4

1 2

( 24340 ) 

ω rpm 3

38.596

 1.209  10

4

4

According to the Table 2-3 the specific speed would require an axial flow pump, see picture below

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Problem 2.52 3rd edition Checking the NPSH of a Pump A pump lifts water through a 100-mm diameter ductile iron pipe from a lower to an upper reservoir. If the difference in elevation between the reservoir surfaces is 10 m, and the performance curve of the 2400-rpm pump is given by hp = 15 - 0.1Q2, where hp is in meters and Q in L/S, then estimate the flowrate through the system. If the pump manufacturer gives the required net positive suction head under these operating conditions as 1.5-m, what is the maximum height above the lower reservoir that the pump can be placed and maintain the same operating conditions?

Define known terms for the schematic shown: d  100mm

L  100m  3m

z1  0  m

z2  10  m

γ  62.4 

lbf ft

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Chin 2-39 suction lift pump NPSH calculations.xmcd

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A 

π d 4

2

p1  0  psi

p2  0  psi

zp  4  m

m sec

V1  0 

V2  0 

m sec

Lsuction  4  m

Discussion : Net positive suction head (NPSHA) available is the amount of pressure head required at the eye of the impeller. For this problem it is computed as:

NPSH A =

po γ

 Δz  hL 

pv γ

po

is the pressure head at the surface of the reservoir, Δz is the difference in elevation between the γ reservoir surface and the eye of the impeller on the suction side of the pump, hLis the pipe friction loss between the reservoir surface and the eye of the impeller, and pv is the vapor pressure of the liquid being pumped. where

Assume minor loss coefficients for the piping system and determine the total "K." Three values are assumed from the Fluids Mechanic text. The three minor loss areas are the entrance (K lower), the 90o bend at the pump (Kbend), and the exit (Kupper).

Klower  0.5

Kupper  1.0

Kbend  0.9

KTotal  Klower  Kupper  Kbend

KTotal  2.4

Lastly, a friction factor for ductile iron is assumed from the Fluids text as f  0.022 From the given information, plot the performance curve for the given pump from the provided equation with a dummy flow variable Q. Q  0 1  100 hp ( Q)  15  ( 0.1)  

 

Q liter sec

   

2

Derived from the energy equation, the total dynamic head of the pumping system is determined through the equation below. This equation is plotted against a dummy flow "discharge."

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Q  0

liter liter liter 1  100 s s s

2

2

2 2 p2 V2 V L V  z1   KTotal   f  =  z2  γ 2 g 2 g 2 g D 2 g γ

p1

V1

2 2  Q Q  2  p 2 2 V 2   p1 V1 A 2 L A system ( Q)    z2    z   K   f    1 2 g Total 2  g 2  g   γ d 2 g  γ

2

    

Once both curves are plotted on the same axis, their intersection is the pump operating point. Ideally the system should be designed so that the operating point corresponds to the highest or nearly highest pump efficiency. In addition the operating point should be located near the midpoint of the pump characteristic.

curves on next page

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Characteristic Curves 20 18 16 14

Head (m)

system( Q) m

12

hp ( Q)

10

10.89

8 6 4 2 0

0

1

2

3

4 Q liter s

5 

Q liter

6

7

8

9

10

6.42

sec

Flow (liters/sec) System Head Curve Performance Curve Operating Point liter s

Head  11m

Flow  6.5

Head  36.089  ft

Flow  103.027  gpm

To determine the available net suction head in the system above, Equation 2.122 from Chin's text is used. This value will provide the available net suction head for given length (3-m) from the suction of the pump to the surface of the lower reservoir. If this value exceeds the required minimum (1.5-m), cavitation will not occur. Therefore, the optimum flow condition for pumping system of 6.5 liter/s is substituted into the hL equation along with the friction and minor losses for the suction side of the pump, the assumed friction factor, f, and the zs which is the vertical distance between the suction end of the pump and the lower reservoir surface.

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The following values are the data required from the lower section of the pumping system.

f  0.022

d  100  mm

Klower  0.5

Δzs  3m Assumed to be pipe length, L as well

The friction and minor loss equations are manipulated in order for flow to be directly inserted.

Q = V A

V=

Q A

Area 

π d 4

2

2 2  Flow   Flow      f   zp  z1   Klower       Area     Area  hL_suction  2d  g 2g

hL_suction  0.158  ft

As indicated in Chin's text, atmospheric pressure is assumed for po ,and pv represents the saturation vapor pressure at the pump's suction end. Each pressure is defined below. Also, the density of water is defined below. po  1atm

pv  2340Pa

ρ  1000

kg m

3

With all required data defined, the NPSHA for the given conditions is evaluated. NPSH A 

po ρ g





 hL_suction  zp  z1 

pv γ

NPSH A  6.045 m

equation 122 Chin

NPSH A  19.834  ft

Therefore, with the given conditions (zs = 4-m), there is approximately 6-m of suction head provided at the pump suction. In order to determine the maximum zs (distance from the pump suction to the water's surface), symbolics are used to determine an equation for the maximum zs. This value occurs when the available suction head equals the manufacturer's recommendation of 1.5-m.

NPSH A  1.5m f  Lsuction   hL =

Flow    Area 

2d  g

2



 Flow    Area 

2

Klower  

2g

The headloss equation, hL, is substituted into the NPSHA equation in order to make zs the only unknown in the new equation. C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDRAULIC PROCEDURES \pumping calculations\ 9/24/2012

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2 2  Flow  Flow     Klower    po  f  Lsuction   Area      Area    Δz  pv NPSH A =   s ρ g ρ g  2d  g 2g 

Δzs 

po ρ g

 NPSH A 

pv ρ g

Klower  Flow



2

2

2



2  Area  g

Lsuction  Flow  f 2

 8.545 m

2  Area  d  g

Use the symbolic function to solve the NPSHA equation for the unknown, maximum zs. Δzs  8.545 m

Δzs  28.036  ft

Therefore, in order to prevent pump cavitation, the maximum distance which the pump suction can be placed above the lower reservoir's surface is approximately 8.5-m. This is confirmed in the hL and NPSHA equations below. 2 2   Flow   Flow   K  f  Lsuction    lower  Area     Area     hL    2d  g 2g  

NPSH A 

po ρ g

NPSH A  1.5 m

 hL  Δzs 

pv ρ g

NPSH A  4.921  ft

Therefore, the available suction head exceeds substantially the 1.5-m required by the pump manufacturer.

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Chin 2-39 suction lift pump NPSH calculations.xmcd

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Chin 2 - 53 3rd edition Water is being pumped from reservoir A to reservoir F through a 30 m long PVC pipe 150 mm in diameter. There is an open gate valve located at C, 90o bends (threaded) located at B, D, and E; and the pump performance curve is given as: hp ( Q)  20  4713  Q

2

where hp is the head added by the pump in meters and Q is the flowrate Q in m3/sec. The specific speed of the pump (in U.S. customary units) is 3000. Assuming that the flow is turbulent (in the smooth, rough, or transition range) and the temperature of the water is 20o C then (a) write the energy equation between the upper and lower reservoirs, accounting for entrance, exit, and local losses between A and F; (b) calculate the flowrate and velocity in the pipe; (c) if the NPSH at the pump operating point is 3.0 m, assess the potential for cavitation in the pump (for this analysis you may assume that the head loss in the pipe between the intake and the pump is negligible; and (d) use the affinity laws to estimate the pump performance curve when the motor on the pump is changed from 800 to 1600 rpm.

L  30  m KA  1

KB  0.9

D  150  mm

KC  0.2

KD  0.9

KA  KB  KC  KD  KE  KF  4.9

KE  0.9

KF  1

ΣK  4.9

The pipe material is PVC thus: ks  0  mm

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Chin 2-40 pump analysis.xmcd

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PA  0  psi

μ  1.002  10

PB  0  psi

3

 Pa  sec

ρ20  998.2 

VA  0

m sec

kg m

3 N 3

γ  ρ20  g  9.789  10 

3

VB  0 

m sec

zA  0  m

m

zB  10  m

The specific speed of the pump is given although it is not used directly. A specific speed of 3000 implies the pump will be of a radial flow type. Radial flow pumps can pump smaller flows against high heads. They are used in most pumps pumping wastewater. Multistage radial pumps are often used in deep water wells. The equation used for specific speed is given below. Although specific speed is not unitless the dimensions are rarely used. In the English system of units ω has dimensions of rpm, Q is in gpm, and hp is in feet 1

Ns =

ωQ

2

3

hp

4

Initial guesses

V  2 

m sec

3

hp  5  m

f  .0015

Q  10 

m sec

Re  10

6

Given 2

2

2 2 PB VB L V V  zA   f   ΣK   hp =  zB  γ 2 g 2 g 2 g D 2 g γ

PA

VA

2

Q  hp = 20  m  4713    3  m  m   sec  1 f

Re =

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDRAULIC PROCEDURES \pumping calculations\ 9/24/2012

= 2  log 

2.51    Re  f 

ρ20  D  V μ

V=

Q

 π D 2     4 

Chin 2-40 pump analysis.xmcd

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 Vsoln     hpsoln     fsoln   Find  V hp f Q Re    Qsoln   Re   soln 

hpsoln  12.109 m

fsoln  0.014

Vsoln  2.315

3

m s

m Qsoln  0.041 s

Re soln  3.46  10

5

3

m The operating point of the pump is: Qsoln  0.041 s

at hpsoln  12.109 m

The net positive suction head refers to the absolute pressure at the eye of the impeller and is particularly important when designing suction lift pump applications. The available NPSH can be computed. It is then compared to the required NPSH which must be obtained from the pump manufacturer.

The available net positive suction head is given by: NPSH A =

po

 Δz  hL 

γ the eye of the impeller. pv is the vapor pressure (absolute units) of water. Δz  3  m

hL  0  m

po  14.7 

NPSH A 

lbf in

2

po γ

γ  9.789  10

3

pv γ

, this the absolute pressure at

kg 2

m s

2

pv  2.337  kPa

 Δz  hL 

pv γ

 7.115 m

The required NPSH, specified by the pump manufacturer is 3 m, the available NPSH, computed above is 7.1 m thus we would not expect cavitation to occur if the pump is operated at the design condition. The Affinity Laws: The affinity laws refer to dimensionless groups of variables, obtained using the Buckingham Pi theorem, think of them like the Reynolds number. They can be used to predict the properties of geometrically similarly pumps in the same way the Reynolds number can be used to predict the forces on pipes.

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Chin 2-40 pump analysis.xmcd

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The two dimensionless groups used for pump design are the head coefficient,

g  hp 2

ω D Q ωD

3

2

, and the flow coefficient

where D is the inlet or outlet diameter, Q is the flowrate, ω is the rotation speed, and h is the head on the p

pump. 2

Q1 ω1 hp1 ω1 and According to the affinity laws, for a fixed D = = 2 Q2 ω2 hp2 ω2

ω1 Q1 =  Q = 0.5  Q2 ω2 2

hp1 =

ω1 ω2

2 2

 hp2

In this problem ω 1  800  rpm and ω 2  1600  rpm

ω1 ω2

2 2

 0.25



.25  hp2 = 20  4713  0.5  Q2

hp2 ( Q)  80.0  m  4713.0  

  

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDRAULIC PROCEDURES \pumping calculations\ 9/24/2012

Q  3

m

sec

  

2

m

2

hp ( Q)  20  m  4713  

Chin 2-40 pump analysis.xmcd

  

Q  3

m

sec

2

 m  

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3

Q  0 

3

3

m m m .01   1  sec sec sec

Pump Curves at 800 and 1600 rpm 80 72 64 hp ( Q) 56 48 m 40 hp2 ( Q) 32 m 24 16 8 0

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

Q m

3

sec

800 rpm 1600 rpm Observations : By increasing the rotation speed of the impeller we increase the shutoff head of the pump from 20 m to about 80 m, and increase the pump flow considerably at the same TDH. Above a TDH of 20 m the pump operating at 800 rpm delivers nothing, a geometrically similar pump operating at 1600 rpm delivers 0.11 m3/sec. or about 1750 gpm.

Shutoff head : The TDH above which a centrifugal pump delivers zero flow, the impeller will simply spin. Eventually the pump will heat up and kick out.

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Chin 2-40 pump analysis.xmcd

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Chin 2-55 3rd edition

If the performance curve of a certain pump model is given by:

hp = 30  .05  Q

2

where hp is in meters and Q in L/sec, what is the performance curve of a pump system consisting of n pumps in series? What is the performance curve of a pump system containing n of these pumps in parallel.

Solution : For pumps in series we add the heads at a given flowrate

Assume we have n pumps in series, the total head imparted by this series is Hp. Then the head imparted by EACH Hp pump will be: . Thus the pump curve for the system is: n

Hp n

= 30  .05  Q

2

one pump

2

Hp = 30  n  .05  Q  n

system with n pumps

For pumps in parallel we add flows corresponding to a given head Assume we have n pumps in parallel the characteristic for each pump in the system will be. Note that Hp is the same for n pumps as it is for 1 pump. Hp = 30  0.05  

Q  n

2

For n pumps in parallel multiply the flow of a single pump by n: 2

Hp = 30  0.05 

Q David - you show Hp = 30  0.05    n

 

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDRAULIC PROCEDURES \pumping calculations\ 9/24/2012

Q n

2

what am I missing ?

Chin 2-41 n pumps in series and parallel.xmcd

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Chin 2-56 3rd edition

Pumps in Series and Parallel A pump is placed in a pipe system in which the energy equation (system curve) is given by : hsystem = 15  .03  Q

2

where hp is the energy added by the pump in meters and Q is the flowrate through the system in L/sec. The performance curve of the pump is: hpump = 20  0.08  Q

2

What is the flowrate through the system ? If the pump is replaced by two identical pumps in parallel, what would be the flowrate through the system? If the pump is replaced by two identical pumps in series, what would be the flowrate in the system

Q  1 



System curve hsystem ( Q)  15  .03  

 

 

gal gal gal 2   200  min min min

Q    m 2

L

sec

   



Pump characteristic for single pump: hpump ( Q)  20  0.08  

 

 

Q    m 2

l

sec

   

n  2



Single pump characteristic modified for two pumps in parallel: h2pump_parallel ( Q)  20  0.08   

 

  l n    sec  Q

2

m   

Single pump curve modified pump characteristic for 2 pumps in series:



2  Q    m l      sec  

h2pump_series ( Q)  20  n  0.08  n  

 

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Chin 2-42 Pumps in series and Parallel.xmcd

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Operating Point Single Pump 20 18.6 17.2 hsystem( Q) 15.8 m 14.4 hpump ( Q) 13 m 11.6 16.28 10.2 8.8 7.4 6

0

1.4

2.8

4.2

5.6 Q l sec



7

8.4

Q

6.81

l

9.8

11.2

12.6

14

sec

System Curve Pump Curve Operating Point

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Chin 2-42 Pumps in series and Parallel.xmcd

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Pumps in Parallel I plotted the pump characteristic for 2 identical pumps in parallel in two ways: 1. Simply plot the curve for 1 pump against 2*Q 2. I modified the single pump curve equation to account for 2 identical pumps in parallel and used it. Both methods gave the same result.

2 Identical Pumps in Parallel 30 hsystem( Q) m hpump ( Q)

27.5 25 22.5

m hpump ( Q) m h2pump_parallel ( Q)

20 17.5 15

16.426

12.5

18.019

10 7.5 5

0

3

6

9

12 Q l

sec



Q l sec

15 

2 Q l sec



18 Q l

21

24

27

30

6.82 10.031

sec

system head curve single pump 2 identical pumps in parallel -single pump curve vs 2Q 2 identical pumps in parallel - single pump equation modified operating point 1 pump operating point 2 pumps 1 pump will deliver 6.9 l/sec against a head of 16.4 m, 2 identical pumps in parallel will deliver 10 l/sec against a head of 18 m.

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Chin 2-42 Pumps in series and Parallel.xmcd

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flow_increase 

10  6.9  44.928  % 6.9

head_increase 

18  16.4  9.756  % 16.4

Discussion - pumps in parallel Note that the flowrate is not doubled by using two pumps in parallel (a common misconception). This would only be the case if the system curve were a horizontal line. However, because the head required varies as the square of the pipe velocity the system curve rises at an increasing rate as the flow increases. Summary : Pumps in parallel provide the greatest benefit when the system curve is relatively flat, as the system curve steepens the benefit of 2 pumps in parallel is lost because of the sharp increase in head with increased flow. A steep system curve can result from using small pipelines or trying to pump too much flow into a system.

Pumps in Series Two Identical Pumps in Series 40 hsystem( Q)

36.5

m hpump ( Q)

33 29.5

m 2

26

hpump ( Q) m

22.5 19

h2pump_series ( Q) 16.36

15.5

18.91

12 8.5 5

0

1.4

2.8

4.2

Q l sec



5.6 Q l

sec



7 Q l

sec



8.4 Q liter

9.8

11.2 12.6

14

6.75 11.41

sec

System curve one pump two pumps in series operating point one pump operating point two pumps in series operating point 2 pumps in series

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Chin 2-42 Pumps in series and Parallel.xmcd

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A single pump delivers 6.8 liters per second against a head of 16.4 meters, 2 identical pumps in series deliver 11.4 liters per second against a head of 18.9 m. Pumps in series provide the greatest benefit when the system curve is steep. If the system curve is flat 2 pumps in series will increase the flow.

increasehead 

18.9  m  16.4  m  15.244  % 16.4  m

11.4  increaseflow 

liter liter  6.8  sec sec  67.647  % l 6.8  sec

Two pumps in series are used to deliver a flow against higher heads than a single pump can accommodate. Note however that the effect of two pumps in series depends to some extent on the shape of the system curve. The benefit of 2 pumps in series is greatest when the system curve is steep (large increase in head per unit flow increase). In the problem here the system curve is relatively flat, as a result 2 pumps in series actually produce a larger flow rate (11.4 l/sec) than two pumps in parallel (10 l/sec).

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Chin 2-42 Pumps in series and Parallel.xmcd

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Chin 2-44 3rd edition Population Estimates The design life of a planned water distribution system is to end in the year 2030, and the population in the town has been measured every 10 years since 1920 by the U.S census bureau. The reported populations are listed below. Estimate the population of the town using (a) graphical extension, (b) arithmetic growth projection, (c) geometric growth projection (d) declining growth projection (assume a saturation population of 100,000 persons), and (e) logistic growth curve projection.

In this problem (part a only) I am going to use the regress and interp functions to draw a 2nd order polynomial through the data points and extend it so I can estimate the population in the year 2030.

 out1      out2 

year 1920 1930 1940 1950 1960 1970 1980 1990

population 25521 30208 30721 37253 38302 41983 56451 64109

i  0  7

yeari  out1i

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDRAULIC PROCEDURES \Demand and Fire flow computations\ 9/24/2012

populationi  out2i

Chin - 2-44 Population Estimates part a using regress.xmcd

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First, I simply plot the given data to get a feel for what it looks like

Population Data

populationi

1 10

5

9.2 10

4

8.4 10

4

7.6 10

4

6.8 10

4

6 10

4

5.2 10

4

4.4 10

4

3.6 10

4

2.8 10

4 4

2 10 3 1.9 10

1.928 10

3

1.956 10

3

1.984 10

3

2.012 10

3

2.04 10

3

yeari

Population Data Now I used regress and interp to develop a 2nd order polynomial, and plot it, the red solid line

vs  regress ( year population 2) yr  1920 1921  2030 pop ( yr)  interp ( vs year population yr)

Now, according to Mathcad help the last three numbers in the vs vector are the actual coefficients in the equation produced by the regress algorithm. 3     3     2   vs   2.590833  10 7     2.698621  10 4     7.034643 

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Chin - 2-44 Population Estimates part a using regress.xmcd

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So, the 2nd order polynomial using the coefficients produced by regress is given below and plotted on the same set of axis, as expected they plot on top of each other.

pp ( yr)  vs3  vs4  yr  vs5  yr

2

Data Extrapolated with 2nd Order Polynomial 1.2 10

5

1.1 10

5

1 10

5

pop ( yr)

9 10

4

populationi

8 10

4

7 10

4

115390

6 10

4

5 10

4

4 10

4

3 10

4

pp ( yr)

4

2 10 3 1.9 10

1.93 10

3

1.96 10

3

1.99 10

3

2.02 10

3

2.05 10

3

yr yeari 2030 yr

Interpolated 2nd order polynomial population data points population at 2030 polynomial with regress generated coefficients

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Chin - 2-44 Population Estimates part a using regress.xmcd

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Chin 3-2 3rd edition Population estimates The design life of a planned water distribution system is to end in the year 2030, and the population in the town has been measured every 10 years since 1920 by the U.S census bureau. The reported populations are listed below. Estimate the population of the town using (a) graphical extension, (b) arithmetic growth projection, (c) geometric growth projection (d) declining growth projection (assume a saturation population of 100,000 persons), and (e) logistic growth curve projection.

 out1      out2 

year 1920 1930 1940 1950 1960 1970 1980 1990

population 25521 30208 30721 37253 38302 41983 56451 64109

k

765.8

i  0  7

yeari  out1i

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populationi  out2i

Chin - 2-44 Population Estimates.xmcd

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(a) Graphical Extension - By extending the curve suggested by the data I estimate a 2030 population of approximately 92,000 persons

Graphical Extension Population Projection

populationi

1 10

5

9.2 10

4

8.4 10

4

7.6 10

4

6.8 10

4

6 10

4

5.2 10

4

4.4 10

4

3.6 10

4

2.8 10

4 4

2 10 3 1.9 10

1.926 10

3

1.952 10

3

1.978 10

3

2.004 10

3

2.03 10

3

yeari

Graphical Extension

(b) Arithmetic Growth Projection P ( t) = Po  karithmetic  t Po  population7

To obtain k I used the last two population estimates: karithmetic 

population7  population6 10

 765.8

2030  1990  40 P ( t)  Po  karithmetic  t P ( 40)  9.474  10

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( c ) Geometric Growth Projection P ( t) = Po  e

kgeometric  t

we have two unknowns Po and kgeometric

Use 1940 population 30,721 and the 1990 population 64,109 as the basis for estimating the constants we get P1970  41983

P1990  64109

t1970  0  yr

t1990  20  yr

Po  41983

P1990 = P1970  e

kgeometric  t1990

 P1990    P1970   0.021  1

ln  kgeometric 

P2030  Po  e

t1990

yr

kgeometric  40  yr





 9.79  10

( d ) Declining Growth projection P ( t) = Psat  Psat  Po  e

4

 kdeclining  t

Psat  100000 P1970  4.198  10

4

t1970  0

P1990  6.411  10

4

t1990  20  yr

kdeclining 

t2030  50  yr

.05 yr

Given





 kdeclining  t1970





 kdeclining  t1990

P1970 = Psat  Psat  Po  e P1990 = Psat  Psat  Po  e

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Posoln      Find P k  o declining  kdeclining_soln    Posoln  4.198  10

4

kdeclining_soln  0.024 





P2030  Psat  Psat  Posoln  e

 kdeclining_soln  t2030

P2030  8.254  10

( e ) Logistic Growth Curve P ( t) =

1 yr

4

Psat 1  a e

b t

P1920  25521

t1920  0  yr

P1940  30721

t1940  20  yr

P1990  64109

t1990  70  yr

a  1.65

b 

.045 yr

Psat  100000

Given

P1920 =

Psat 1  a e

P1990 =

Psat 1  a e

P1940 =

b  t1990

Psat 1  a e

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b  t1920

b  t1940

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 asoln     bsoln   Minerr  a b Psat    Psat_soln 

ERR  1.803  10

asoln  6.1  10

7

bsoln  0.014 

3

1 yr

Psat_soln  1.482  10

12

t  0  yr 1  yr  150  yr

P ( t) 

Psat_soln 1  asoln  e

2 10

5

1.6 10

5

1.2 10

5

8 10

4

4 10

4

bsoln  t

P ( t)

0

0

30

60

90

120

150

t yr

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Chin 2-45 3rd edition A city founded in 1950 had a population of 13,000 in 1960; 125,000 in 1975; and 300,000 in 1990. Assuming that the population growth follows a logistic curve, estimate the saturation population of the city. From Chin " Extrapolation beyond 10 years, called long term projections, involve fitting an S shaped curve to the historical population trends and then extrapolating using the fitted equation. The most commonly fitted S curve is the so called logistic curve, described by the following equation.

P ( t) =

Psat 1  a e

b t

In this equation a and b are constants to be fitted from the data, t is the time since data collection began, Psat is the saturation population. We have three population estimates, therefore we can write three logistic curve equations and solve them simultaneously for a, b, and Psat P1960  13000

t1960  0  yr

P1975  125000

t1975  15  yr

P1990  300000

t1990  30  yr

a  10

b 

1 yr

Psat  500000

Given P1960 =

Psat 1  a e

P1975 =

Psat 1  a e

P1990 =

b  t1960

b  t1975

Psat 1  a e

b  t1990

 asoln     bsoln   Find  a b Psat    Psatsoln  C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDRAULIC PROCEDURES \Demand and Fire flow computations\ 9/24/2012

Chin - 2-45 Logistic Growth Population Estimate.xmcd

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asoln  24.689

bsoln  0.18 

1 yr

Psatsoln  3.34  10

5

t  0  yr 1  yr  40yr

Psatsoln

P ( t) 

1  asoln  e

bsoln  t

Now that we have the various constants we can plot the Logistic Growth curve

Logistic Growth Curve 3.5 10

5

3.15 10

5

2.8 10

5

2.45 10

5

2.1 10

5

125000 1.75 10

5

1.4 10

5

1.05 10

5

7 10

4

3.5 10

4

P ( t) 13000

300000

0

0

4

8

12

16 t yr

20

24

28

32

36

40

0 15 30

Logistic Growth Curve 1960 population 1975 population 1990 population

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Chin - 2-45 Logistic Growth Population Estimate.xmcd

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Chin Problem 3-5 3rd edition Estimate the flowrate and volume of water required to provide adequate fire protection to a five story office building constructed of jointed masonry. The effective floor area of the building is 5000 m2.

Solution : The most popular method for computing fire flows is that developed by the Insurance Services Office (ISO) which is an organization representing the fire insurance underwriters. The required fire flow can be estimated using:





NFFi = Ci  Oi  Xi  Pi

NFF is the needed fire flow at location i. Ci is the construction factor based on the size of the building and its construction: Ci = 220  F 

Ai

Ai is the effective floor area, typically equal to the area of the largest floor plus 50% of the area of all other floors. F is a coefficient based on the class of construction

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Chin - 2-47 fire demand and volume.xmcd

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Effective area : Ai  5000  m

Ci  220  F 

2

F  1

Ai m



liter

Oi  0.85 4 liter

 1.556  10 

2 min

min

liter round C to the nearest 1000 liter/min, thus Ci  16000  i min

 Xi  Pi = 1.4 



NFFi = Ci  Oi  Xi  Pi

4 liter NFF  Ci  Oi  1.4  1.904  10  min

Round NFF to nearest 2000 liter/min 4

NFF  1.904  10 

l min

From Table 2-10 Chin the required fire flow duration is 5 hours, t  5  hr thus the needed volume is

3

Vol  t  NFF  5.712  10  m

3

6

Vol  5.712  10 L

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Chin - 2-47 fire demand and volume.xmcd

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Chin Problem 3-5 Estimate the flowrate and volume of water required to provide adequate fire protection to a five story office building constructed of jointed masonry. The effective floor area of the building is 5000 m2.

Solution : The most popular method for computing fire flows is that developed by the Insurance Services Office (ISO) which is an organization representing the fire insurance underwriters. The required fire flow can be estimated using:





NFFi = Ci  Oi  Xi  Pi

NFF is the needed fire flow at location i. Ci is the construction factor based on the size of the building and its construction: Ci = 220  F 

Ai

Ai is the effective floor area, typically equal to the area of the largest floor plus 50% of the area of all other floors. F is a coefficient based on the class of construction

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Chin - 2-47 fire demand and volume.xmcd

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Effective area : Ai  5000  m

Ci  220  F 

2

F  1

Ai m



liter

Oi  0.85 4 liter

 1.556  10 

2 min

min

liter round C to the nearest 1000 liter/min, thus Ci  16000  i min

 Xi  Pi = 1.4 



NFFi = Ci  Oi  Xi  Pi

4 liter NFF  Ci  Oi  1.4  1.904  10  min

Round NFF to nearest 2000 liter/min 4

NFF  1.904  10 

l min

From Table 2-10 Chin the required fire flow duration is 5 hours, t  5  hr thus the needed volume is

3

Vol  t  NFF  5.712  10  m

3

6

Vol  5.712  10 L

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Chin - 2-47 fire demand and volume.xmcd

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Chin problem 2-46

The average demand of a population served by a water system is 580 L/cap*day and the population at the end of the design life is expected to be 100,000 people. Estimate the maximum daily demand and maximum hourly demand at the end of the design life. From Chin - pg. 69 Variations in demand - Water demand generally fluctuates on a seasonal and daily basis, being below the average daily demand in the early morning hours and above the average daily demand during midday hours. Typical cycles in water demand are shown in Figure 2.24, pg. 60 of Chin. On a typical day in most communities, water use is lowest at night (11 p.m. to 5 a.m.) when most people are asleep. Water usage rises rapidly in the morning ( 5 am to 11 am ) followed by moderate use through midday ( 11 am to 6 pm ). Use then increases in the evening ( 6pm to 10 pm) and drops rather quickly around 10 pm. During a 24 hour period demand can range from 25% to 175% of average. To account for variation in demand we commonly use peaking factors, shown below.

person  1

L Qcap  580  person  day population  100000  person 3

m Average daily demand: Qcap  population  0.671 s





7 gal peaking factor for maximum daily demand, 1.80: Qmax_daily  1.8  Qcap  population  2.758  10  day 4 gal Qmax_daily  1.915  10  min 8 liter Qmax_daily  1.044  10  day C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDRAULIC PROCEDURES \Demand and Fire flow computations\ 9/24/2012

Chin 2-46.xmcd

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



7 gal peaking factor for maximum hourly demand, 3.25: Qmax_hourly  3.25  Qcap  population  4.98  10  day 4 gal Qmax_hourly  3.458  10  min 8 L Qmax_hourly  1.885  10  day

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Chin 2-46.xmcd

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Chin Problem 3-6 3rd edition What is the maximum fire flow and corresponding duration that can estimated for any building ? Solution : The solution comes directly from Table 2-10 Chen. The maximum fire flow is 45,000 L/min for a duration of 10 hours.

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Chin 2-48 fire flow.xmcd

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Problem 3-7 and 3-10 3rd edition Chin A water supply system is being designed to serve a population of 200,000 persons with an average per capita demand of 600 L/cap*day and a needed fire flow of 28,000 L/min. If the water supply is drawn from a river, what should be the design capacity of the supply pumps and treatment plant ? For what duration must the fire flow be sustained and what volume of water must be kept in the service reservoir to accommodate a fire ? What should be the design capacity of the distribution pipes ? Solution : Note that we have a service reservoir to hold the needed fire flow, it does not come from the system directly during a fire. Chin pg. 73 "Typically the delivery pipelines from the water source to the treatment plant as well as the treatment plant itself are designed based on the maximum daily demand." person  1 liter Qcap  600  person  day

population  200000  person

fire_flow  28000 

liter min

For this problem the maximum daily demand is obtained using a peaking factor of 1.8, PF daily  1.8, PF hour  3.25 So, the required capacity of the low service pump should be (according to Table 2.11 Chin) 3

m Qmax_day  Qcap  PF daily  population  2.5 s

AND we should have a reserve pump available.

Fire flow must be sustainable for 7 hours, Table 2-10 Chin Volfire_flow  7  hr  28000 

liter 4 3  1.176  10  m min

Chin - page 73 " The flowrates and pressures in a distribution system are analyzed under both (1) maximum daily demand plus fire flows and (2) maximum hourly demand. The largest value governs. for this problem the peaking factor for the maximum hourly demand is 3.25. 3

m Max daily plus fire: Qmax_day  fire_flow  2.967 s

3

3

m Max hourly flow: Qcap  population  PF hour  4.514 s

l

Qmax_day  fire_flow  2.967  10  sec 8 l Qmax_day  fire_flow  2.563  10  day

3 l Qcap  population  PF hour  4.514  10  sec 8 l Qcap  population  PF hour  3.9  10  day

Thus the maximum hourly flow governs. C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDRAULIC PROCEDURES \Demand and Fire flow computations\ 9/24/2012

chin 2-49 and 2-51 fire flow design.xmcd

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Problem 2-51 Now, compute the volume of storage required for the elevated storage reservoir in the problem above. Solution : From Chin, pg. 82 - (1) adequate volume to supply peak demands in excess of the maximum daily demand using no more than 50% of available storage (2) adequate volume to supply the critical fire demand in addition to the volume required for meeting maximum daily demand fluctuations. and (3) adequate volume to supply the average daily demand of the system for the estimated duration of a possible emergency. (1) Conventional design practice is to rely on pumping to meet the operational demands up to the maximum daily demand, where detailed demand data is not available. The storage needed to supply peak demands beyond this is generally no more than 20-25% of the maximum daily demand.

4

Vpeak  25%  Qcap  PF daily  population  1  day  5.4  10  m

3

(2) The critical fire duration is 7 hours, thus the required fire volume is: 4

Fire_Volume  7  hr  ( fire_flow)  1.176  10  m

3

6

Fire_Volume  3.107  10  gal

( 3 ) Emergency storage is generally necessary to provide water during power outages, breaks in water mains, problems at treatment plants and other unexpected shutdowns of the water supply system. Emergency volumes for most municipalities generally vary between one and two days at average demand.

8

Qave_day  Qcap  population  1  day  1.2  10 L 5

Therefore the total reservoir volume should be: Qave_day  Fire_Volume  Vpeak  1.858  10  m

3

7

Qave_day  Fire_Volume  Vpeak  4.907  10  gal 8

Qave_day  Fire_Volume  Vpeak  1.858  10 L This is 49 million gallons !! Alot of water !

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chin 2-49 and 2-51 fire flow design.xmcd

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Chin - problem 3-8 3rd edition What is the minimum acceptable water pressure in a distribution system under average daily demand conditions ?

Thus the minimum pressure would be pmin  240  kPa pmin  34.809  psi

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Chin 4-10 3rd edition Water flows at 20 m3/sec in a trapezoidal channel that has a bottom width of 2.8 m, side slopes of 2H:1V, longitudinal slope of 0.01, and a Manning's n of 0.015 (a) Use the Manning equation to find the normal depth of flow, and (b) determine the equivalent sand grain roughness of the channel. Assume the flow is fully turbulent. 3

Q  20 

m sec

SS  2

A ( d)  w  d  2 

1  SS  d  d 2

n  0.015 

w  2.8  m

P ( d)  w  2 

sec R ( d) 

1

m

3 2

S  0.01

2

d  ( SS  d)

2

A ( d) P ( d)

1

1 3 2 Q =  A ( d)  R ( d)  S n

d  2  m 2 1    1 3 2  soln  root  Q   A ( d)  R ( d)  S d n  

soln  0.905 m For fully turbulent flow the friction factor depends only on the relative roughness. If this is the case then the following equation can be used: 1

 ks    12  R 

= 2  log  f

This equation contains the equivalent sand grain height, ks, but also the Darcy Weisbach friction factor, f . In order to get f equate the values of C from Nikuradse, Colebrook, and Chezy and the value of C from Manning

C=

8 g f

1

Nikuradse, Colebrook, and Chezy

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Chin 3-11 Mannings equation.xmcd

6

R C= n

Manning

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1 6

8 g R which can be solved for f = f n

equating these two we get:

f=

8 g n

2

1 3

R

A ( soln) P ( soln)

R ( soln) 

R ( soln)  0.609 m

f 

8 g n

2 1

R ( soln)

3

   12  R ( soln)  

1

 0.021

= 2  log  f

ks

Solve for ks using the root function ks  .1  mm

 1

kssoln  root 

 f

   ks  12  R ( soln)   

 2  log 

ks

kssoln  2.506  mm

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Chin 4-11 3rd edition Estimating Manning's n

It has been shown that in fully rough turbulent flow Manning's n can be related to the height d of the roughness projections on the channel surface by the relationship 1

n = .039d

6

If the estimated roughness height, d in a channel is 30 mm, then determine the percentage error in Manning's n resulting from a 70% error in estimating d Compute the diameter resulting from a 70% error resulting from overestimation and underestimating d .7  ( 30  mm)  21  mm

30  mm  21  mm  9  mm

30  mm  21  mm  51  mm

1

n ( d)  .039  d

6

d  1  mm 2  mm  90  mm 30 18.571 n ( d)  n ( 30  mm) n ( 30  mm)

%

7.143 0

 4.286

0  18.81

 15.714

9.24

 27.143  38.571  50

0

18

36

54

d

30 9 51

mm

72

90

n vs d d = 30 mm d = 9 mm d = 51 mm Underestimating d by 70% results in a 19% error in n, overestimating d by 70% results in a 9.25% error in n

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Chin 3-12 estimating n.xmcd

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Chin 4-14 3rd edition Assessing Conditions for Fully Turbulent Flow

Show that the turbulence condition

u  ks

 70 can be put in the form: ks 

ν

R  So  2.2  10

5

A trapezoidal concrete channel with a bottom width of 3 m and side slopes of 2H:1V is estimated to have an equivalent sand grain roughness of 3 mm, and is laid on a slope of 0.1%. Determine the minimum flow depth for fully turbulent conditions to exist. Can Manning's equation be used at this flow depth ? If turbulent flow conditions do not not exist at a given flow depth, give the equation that should be used to relate the flowrate to the depth of flow. Now u =

g  R  S , substituting this into the shear velocity Reynolds number results in: g  R  S  ks

 70 ν known values given below ν  1  10

6 m



2

sec

b  3  m

SS  2

R  S  ks =

70 

ks  3  mm

So  .1  %

70  ν g

ν 2

m

divide out units and evaluate RHS

sec

 2.235  10

5

thus

g

R  S  ks = 2.235  10

5

m 2

sec

The area of a trapezoidal channel A = b  y  SS  y

2

The wetted perimeter of a trapezoidal channel : P = b  2y 

The hydraulic radius is R =

b  y  SS  y b  2y 

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1  SS

2

2

1  SS

2

Chin 3-14 evaluation turbulencs condition (2).xmcd

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Foe fully turbulent flow

R  S  ks  2.2  10

5

b  y  SS  y

or

b  2y 

2

1  SS

2

 S  ks  2.2  10

5

The smallest depth at which fully turbulent flow can occur is given by solving:

b  y  SS  y b  2y 

2

1  SS

2

 So  ks = 2.2  10

5

y  .1  m 3   2  b  y  SS  ( y) 2 5 ysoln  root   So  ks  2.2  10  m y  b  2y  1  SS 2   

ysoln  5.618  cm Thus the depth must be greater than 5.6 cm for flow to be fully turbulent, at this depth compute R

R 

Thus

b  ysoln  SS  ysoln

2

b  2ysoln 

2

R  17.926 which is in the range where ks

1  SS

n 1

ks

 0.054 m

can be considered constant: 4 

R  500, Yen(1991), ks

6

see Figure 3.4 in Chin, thus Manning's equation is considered appropriate

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Chin 4-15 3rd edition Water flows at a depth of 4 m in a trapezoidal, concrete-lined channel section with a bottom width of 4 m and side slopes 3H:1V. The slope of the channel is .0001 ft/ft and the water temperature is 20oC. Assess the validity of using sec Manning's Equation , assuming n  .013  1

m define a "millipascal" mPa 

3

1  Pa 1000

o

density of water at 20 C: ρ  998.2 

kg m

3

absolute roughness of concrete: .3 to 3 mm use a value of ks  1.3  mm from Table 2.1 side slope H:V SS  3 NOTE - It's easier to write equations for the area and wetted perimeter if side slope is defined as H:V depth of flow d  4  m bottom width bottom  4  m longitudinal slope of channel S  .0001 write equation for area of flow as a function of flow depth: A ( d)  bottom  d  2 

write equation for wetted perimeter as function of flow depth: P ( d)  bottom  2 

define the hydraulic radius as: RH ( d) 

A ( d) P ( d)

1 2  SS  d 2 2

( SS  d)  d

A ( d)  64 m

2

2

RH ( d)  2.184 m 2 3

1

1 A ( d)  2 Q   A ( d)    S n  P ( d)  3

Q  82.882

m s

Below I assess the applicability of using Manning's equation to solve this problem

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From Chin, pg 105 Chin describes the research and assumptions underlying the Manning, Chezy and Darcy Weisbach equations. He concludes that in order to make proper use of Manning's Equation the following condition must be met to insure hydraulically rough conditions (pg 107): 6

n 

RH  S  9.6  10

6

 n    sec   1   3   m 

R H ( d) m

 14

S  7.134  10

 14

Since the condition specified above is not met Manning's equation is not strictly applicable because conditions are not hydraulically rough. In reality Manning's equation is always used for open channel flow calculations

Alternate analysis by Sturm (pg 108)

From Chin: In summary the validity of Manning's equations relies on two assumptions (1)

n 1

ks the flow is fully turbulent. These assumptions require that 4 

s

The problem statement says that n  0.013 m

Since

R H ( d) ks

 1.68  10

3

0.333

RH ks

 500 and that ks 

and I have chosen ks  1.3  10

exceeds both Sturm's and Hager's criterion

n 1

ks

3

is constant and (2)

6

R  Sf  2.2  10

5

m

cannot be considered constant and

6

Manning's equation cannot be used

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This value above is NOT in the range given by Hager (1999) 3.6  4

RH ks

R  360 as well as that by Sturm ks

 500

ks 

The value for this problem is:

ks



m Manning's equation cannot be used.

R H ( d) m

R  Sf  2.2  10

 S  1.921  10

5

5

thus the flow is not fully turbulent and

Summary : Manning's equation not strictly applicable for this problem

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Chin 4.18 3rd edition Channel transition

Water flows at 18 cms in a trapezoidal channel with a bottom width of 5 m and side slopes of 2:1 (H:V) . The depth of flow in the channel is 2 m. If a bridge pier of width 50 cm is placed in the middle of the channel, what is the depth of flow adjacent to the pier. What is the maximum width of bridge pier that will not cause a rise in the water surface upstream.

width  5  m

y  2  m

3

Q  18 

widthpier  50  cm

m sec

2

T  5  m  8  m

Flow area without the bridge pier: A  width  y 

1 2  2  2  y  y  18 m 2

Compute the upstream Froude number

2

Fr 

Q T g A

3

 0.271 subcritical flow regime, thus we must chose the subcritical flow depth

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Chin 3-33 bridge pier place in channel.xmcd

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upstream specific energy SE  y 

Q

2

 2.051 m

2

A  2 g Now compute the area of flow with the pier in place





1 Apier ypier  width  ypier   2  2  ypier  ypier  widthpier 2 Now compute the downstream specific energy as a function of the downstream depth. Intersection(s) of the downstream specific energy curve with the upstream specific energy level denote possible flow depths





SEpier ypier  ypier 

Q

2 2

1  width  y   pier  2  2  2  ypier  ypier  widthpier  2  g   ypier  .5  m 0.55  m  3  m

As shown below there are two possible depths, one corresponding to subcritical flow (1.99') and one corresponding to supercritical flow (0.54'). Since the upstream flow regime is subcritical the chose the subcritical downstream depth.

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Chin 3-33 3.2

SE with pier f(yp) SE w/o pier subcritical depth supercritical depth

3 2.8



SEpier ypier m



2.6 2.4

SE m

2.2

2.051 2.051

2 1.8 1.6 1.4 1.2 0.5

0.8

1.1

1.4

1.7

2

2.3

2.6

2.9

3.2

3.5

ypier ypier  1.99 0.54 m m

The effect of the pier on depth of flow is negligible

A choke will occur if the width of the pier is such that the depth adjacent to the pier reaches critical depth and the Froude number equals 1.0. 3

2 Ac Q where Q is the At critical conditions in a channel the following relationship can be shown to be true: = g Tc flow, Tc is the top width of the channel at critical flow, and Ac is the area of flow at critical conditions.





2

At critical conditions the area of flow in a trapezoidal channel is: Ac = 5  m  w p  yc  2  yc , where w p is the width of the pier.





At critical conditions the top width in a trapezoidal channel is: Tc = 5  m  w p  4  yc

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3

2 Ac Q but two unknowns yc and w p we need a second equation. At critical So far we have one equation, = g Tc

conditions it is also true that: y1 

V1

2

2 g

= yc 

Vc

2

= yc 

2 g

Q

2

2

Ac  2  g

Using a solve block we can solve all 4 equations simultaneously for Ac Tc yc w p Tc  15  m

Ac  50  m

2

w p  10  m

yc  1.5  m

Given 3

2 Ac Q = g Tc









A c = 5  m  w p  yc  2  yc

2

T c = 5  m  w p  4  yc

y

Q 2

2

A  2 g

= yc 

Q

2

2

Ac  2  g

 ycsoln     Acsoln     Find  yc Ac wp Tc  wpsoln  T   csoln 

ycsoln  1.614 m

Acsoln  6.149 m

2

w psoln  4.419 m

Tcsoln  7.038 m

If the pier is no wider than 4.42 m the upstream flow will not be choked.

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Check the solution values by plugging them back into the governing equations, they work

Acsoln

Q

2  5 m  w psoln  ycsoln  2  ycsoln  

2

g 3

1

Tc

1

 5  m  wp  4  yc y

Q

2

2

Acsoln

A  2 g

Tcsoln

Q

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ycsoln 

 15

2

1

2

Acsoln  2  g

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Chin 4-22 3rd edition The sections shown in the floodplain in Figure 3.5 have the following values of Manning's n i  0  6

1 2  3 Data   4  5 6  7

0.04 

0.030 



0.015  0.013 



0.017  0.035 



0.060 

ni  Datai 1

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Below I compute the area and wetted perimeter of each piece of the cross section Area0 

1 2  5  m  4  5  m  50 m 2

Area1  5  m  100  m  500 m

Area2 

P2 

P4 

2

y2 

2

3  m  ( 2  3  m)  6.245 m

1 2  5  m  4  5  m  50 m 2

P6 

5 m  8 m  6.5 m 2

y3  8  m

2

2

3  m  ( 2  3  m)  6.245 m

y4 

P5  150  m

2

y0  2.5  m y1  5  m

P3  15  m

1 2  ( 5  m  8  m)  6  m  39 m 2

Area6 

2

P1  100  m

2

Area5  150  m  5  m  750 m

2

( 5  m)  ( 4  5  m)  20.616 m

2

1 2  ( 5  m  8  m)  6  m  39 m 2

Area3  15  m  8  m  120 m

Area4 

P0 

5 m  8 m  6.5 m 2

y5  5  m

2

2

( 5  m)  ( 4  5  m)  20.616 m

y6  2.5  m

6

Rhyd2 



Areai

 6   i 

 Pi  

i  0



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0

 4.857 m

Rhyd  i

Areai Pi

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Horton / Einstein

2

 6   i  ne1      

3     2  Pi   ni    0  6  Pi   i  0 

3



 0.034



Assumption : Total Cross section mean velocity equal to subarea mean velocity Einstein and Banks 1

 6   i  ne2      



Pi   ni   0  6   Pi  i  0 

2

2 



 0.035



Assumption : Total resistance force equal to the sum of the subarea resistance forces

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Lotter 5

6



ne 

 Pi   Rhyd2 3

i  0

 0.026

5

Pi   Rhyd 

6





i  0

3

i

ni

Assumption : Total discharge equal to the sum of the subarea discharges

Krishnamurthy and Christensen

3     2 Pi   yi  ln  ni

6



i  0 3   2 Pi   yi 

6



 3.528

i  0





ln ne4 = 3.55

ne4  0.0287

Assumption : Logarithmic velocity distribution over depth y for a wide channel

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From Chin, pg 111: "It is interesting to note that the equations for estimating composite roughness require both the hydraulic radius and wetted perimeter.

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Chin 4-23 3rd edition Compound n and Extent of Flood Flow onto Floodplain Consider the drainage channel and adjacent floodplains shown in the figure below. The Manning roughness coefficients are given by: Section

n nL  0.040 

Left floodplain

sec 1

m

3

nmain  0.016 

Main channel

right floodplain

nR  0.050 

sec

m sec

1 3

1

m

3

The longitudinal channel slope is 0.5%. Field tests have shown that the Horton (1933) formula for the composite channel roughness best describes this channel. Find the capacity of the main channel and the lateral extent of the flow onto the floodplains for the 100 year of 1590 m3/sec.

The Horton equation for the composite channel roughness is given as: 2

 N   i  ne =  



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1

3     P i  ni 2       P 

3

Chin 3-20 composite channel roughness (2).xmcd

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Channel parameters defined below (main channel slopes defined as run/rise, flood plain channels defined rise/run): width of main channel: w main  30  m water depth in main channel: depthmain  3  m 3 side slope (run/rise) for the left side of the main channel: SSleft_main  3 1 2 side slope (run/rise) for the right side of the main channel: SSright_main  2 1 side slope (rise/run) for left floodplain: SSL  1  % side slope (rise/run) for right floodplain: SSR  4  % longitudinal slope of main channel: S  0.5  %

Area of main channel:

1 2 1 2 2 Amain  w main  depthmain   SSleft_main  depthmain   SSright_main  depthmain  112.5 m 2 2

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Wetted perimeter of main channel



 SSleft_main depthmain 2  depthmain2   46.195 m 2 2   SSright_main  depthmain  depthmain 

Pmain  w main 

 

Hydraulic radius of main channel Rmain 

1

Amain Pmain

 2.435 m

2

1

3

2

capacity of main channel: Qmain  A R S nmain main main

3

m  899.958 s

Horizontal distance along the floodplain resulting from a rise of 1 foot above the top of the main ft channel on the left side: SSLFP  100  ft

Horizontal distance along the floodplain resulting from a rise of 1 foot above the top of the main ft channel on the right side: SSRFP  25  ft The capacity of the main channel once y > 3 meters is computed as the main channel top width times the quantity ( y  3  m)

top width of main channel: T  30  m  2  3  m  3  3  m  45 m capacity of main channel when y > 3 m : Amain  T  ( y  3  m)

Now, to determine the capacity of the entire channel we need to determine the effective n for the channel, however this requires that we come up with the wetted perimeter for the floodplains. We can't do this because we don't know the depth y corresponding to the 100 flood flow. However we can write the effective n as well as the wetted perimeter and area of the entire channel in terms of the unknown depth, y. Solve numerically for y.

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Use the Horton equation to determine the effective (compound n) for the channel.

AFP ( y) 

1 1  ( y  3  m)  SSLFP  ( y  3  m)   ( y  3  m)  SSRFP  ( y  3  m) 2 2

PFP ( y) 

2

( y  3  m)  SSLFP 

PLFP ( y) 

PRFP ( y) 

( y  3  m)  SSRFP

( y  3  m)  SSLFP

2

2

( y  3  m)  SSRFP

2

The Horton/Einstein equation for the effective n value written in terms of y 2 3 3 3   2 2 2  PLFP ( y)   nL  PRFP ( y)   nR  Pmain   nmain  ne ( y)     PLFP(y)  PRFP (y)  Pmain  

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Chin 3-20 composite channel roughness (2).xmcd

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3

m Q100  1590  sec

Rchannel ( y) 

AFP ( y)  Amain  T  ( y  3  m)





PLFP ( y)  PRFP ( y)  Pmain

Below is Manning's equation: 2

1

1 3 2 Q100 =  A ( y)  Amain  T  ( y  3  m)  Rchannel ( y)  S   ne ( y)  FP y  4  m

solve for the depth, y, numerically 2 1    1 3 2  soln  root Q100   AFP ( y)  Amain  T  ( y  3  m)  Rchannel ( y)  S y     n e ( y)  

soln  5.501 m

The depth in the channel corresponding to the 100 year flood flow is therefore soln  5.501 m Extent of water along left bank: ( soln  3  m)  100  250.115 m

Extent of water along right bank: ( soln  3  m)  25  62.529 m

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Chin 4-27 3rd edition 3

m in a trapezoidal channel with a bottom width of 2m and side slopes of 2H:1V. Over a sec distance of 100 m the bottom expands to 2.5 m, with the side slopes remaining constant at 2H:1V If the depth of flow at both of these sections is 1 m and the channel slope is .001, calculate the head loss between the sections. What is the power dissipated in kilowatts. Water flows at 8.4 

3

Q  8.4 

m sec

b1  2  m

SS  2

b2  2.5  m

y1  1  m

y2  1  m

So  .001

L  100  m SS1  2

SS2  2

γ  62.4 

lbf ft

V1

The specific energy at the upstream end of the channel is: y1  V2

3

2

2 g

while that at the downstream end of the

2





. The energy lost due to friction between the sections is given by: L  Sf  So where Sf is channel is y2  2 g the "friction slope" in ft/ft, energy lost per foot of channel. So is the bottom slope of the channel 2

A1  b1  y1  SS1  y1  4 m

2

2

A2  b2  y2  SS2  y2  4.5 m

V1 

Q m  2.1 A1 s

V2 

2

Q m  1.867 A2 s

The equation below says "the difference in specific energies between sections 1 and 2 is equal to the difference between the drop in channel elevation due to its slope (ft) and the loss due to friction (ft)

y1 

V1

2

2 g

V2

 z1  hL = y2   z2 2 g

let hL = L  Sf

y1 

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V1

2

2 g

2

 y2 

let L  So = z1  z2

V2

2

2 g



= L  Sf  So

Chin 3-23 friction loss in 100 m of channel.xmcd

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y1 

V1

2

2 g

 y2 

V2

2

2 g

 0.047 m



.047  m = L  Sf  So

The friction slope in ft/ft is given by Sf 

 0.047  m  L  So L

 1.47  10

3

The total energy lost is: hL  L  Sf  0.147 m The power dissipated is P  γ  Q  hL  12.104  kW

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Chin 4-29 3rd edition 3

m flowing in a rectangular channel with a width of w  5  m. If sec the depth of flow is equal to y  3  m, is the flow super or subcritical ? Determine the critical depth for a flow of Q  30 

Q

Fr =

V g y

A

=

g y

A  y  w Q

Fr 

A

g y

 0.369

Since the Fr number is less than 1 the flow is subcritical and the depth of flow is greater than the critical depth

Froude number less than 1 - flow subcritical - normal depth > critical depth Froude number greater than 1 - flow supercritical normal depth < critical depth

A Froude number equal to one corresponds to the minimum energy per kg of water needed to maintain a specific flow profile.

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Chin 3-25 Froude number calculation (2).xmcd

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Chin 4-30 3rd edition Critical Depth in a Trapezoidal Channel Determine the critical depth for a flow of 50 m3/sec in a trapezoidal channel with a bottom width of 4 m and side slopes of 1.5H:1V . If the depth of flow is 3 m, calculate the Froude number and determine whether the flow is subcritical or supercritical. 3

w  4  m

SS  1.5

d  3  m

Q  50 

m sec

To start with we are looking for critical depth. At critical depth the Froude number equals 1.0. Write an expression for the Froude number in terms of the critical depth.

A ( d)  w  d  2 

1  SS  d  d 2

Fr = 1 =

V ( d) =

2

P ( d)  w  2 

d  ( SS  d)

2

V g  Dc Q

 w  d  2  1  SS  d  d   2   2

It's a trapezoidal channel so we use the hydraulic depth

V A . Write the expression for hydraulic Dc = = g T

depth in terms of critical depth

w  dc  2  Dc =

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1  SS  dc  dc 2

w  2  dc  1.5

Chin 3-26 - critical depth in a channel.xmcd

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Now write the expression for the Froude number in terms of the critical depth and set it equal to 1.0, solve numerically

Q

 w  d  2   SS  d  d   c c c 2   1

1=

g

1  SS  dc  dc 2 w  2  dc  1.5

w  dc  2 

dc  1  m

   dc 1  w  dc  2   SS  dc  dc  2 g  w  2  dc  1.5 

   dcsoln  root 1     

Q

 w  d  2  1  SS  d  d   c c c 2  

critical depth dcsoln  1.96 m

Q   A d   csoln  Critical specific energy : Ec  dcsoln   2 g

2

 2.649 m

Now assume the channel depth is d  3  m, compute the Froude number

A ( d)  w  d  2 

V 

Q A ( 3  m)

1  SS  d  d 2

 1.961

m s

T ( d)  4  m  2  ( d)  1.5

D 

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A ( d) T ( d)

 1.962 m

Chin 3-26 - critical depth in a channel.xmcd

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Fr 

V g D

 0.447

flow is subcritical

d  0  m .1  m  15  m

 Q     A ( d)  E ( d)  d 

2

E ( 3  m)  3.196 m

2 g

Specific Energy Curve, Q = 50 cu. m./sec. 16 14.4 12.8 11.2 d m

9.6 8

3 1.96

6.4 4.8 3.2 1.6 0

0

5

10

15

20 E ( d) m

25

30

35

40

45

50

3.196 2.649

Specific Energy Curve depth of 3 m critical depth

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Chin 3-26 - critical depth in a channel.xmcd

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Chin 4-31 Flow Constriction in Channel A rectangular channel 2 m wide carries 3 cms of water at a depth of 1.2 m. If an obstruction 40 cm wide is placed in the channel, find the water water surface elevation at the constriction. What is the minimum width of the constriction that will not cause the upstream water surface elevation to rise.

Think of the obstruction as a bridge pier

depth of flow upstream of obstruction: y1  1.2  m 3

flow rate in channel: Q  3 

m sec

channel width upstream of obstruction: bup  2  m channel width at the obstruction: bobs  2  m  40  cm  1.6 m width of the obstruction: w obs  40  cm

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Chin 3-27 Obstruction in channel (2).xmcd

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Q m upstream velocity: Vup   1.25 bup  y1 s

Vup

upstream velocity head:

2

 0.08 m

2 g

2

Specific energy is defined to be: SE = y 

v 2 g

Q

 bup y1 2

Upstream of the obstruction the specific energy is: E1  y1 

 1.28 m

2 g

Q

The Froude number upstream of the obstruction is: Frup 

2

2

 bup y1 2

 0.133 thus the flow regime is

g  y1

subcritical. The depth of flow at the obstruction must also result in a subcritical flow regime

If we ignore energy losses between the upstream channel section and the channel section AT the obstruction then the two specific energy levels are the same E1 = E2 Q

E 1  y2 

2

 bobs  y2 2

= 0

2 g

Solve for y2 using the root function y2  1.0  m

   y2soln  root E1  y2  

Q

2

 bobs  y2 2 g

2

   y2 

The depth of flow AT the obstruction is: y2soln  1.142 m Check the Froude number at the obstruction to make sure its still subcritical

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Chin 3-27 Obstruction in channel (2).xmcd

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Q

Frobs 

Vobs 

 bobs  y2soln 2

 0.241

g  y2soln

Q m  1.641 y2soln  bup  40  cm s





Velocity head at obstruction :

Δvelocity_head 

Vobs 2 g

2



Vobs 2 g

Vup

2

 0.137 m

2

2 g

 0.058 m

Δy  1.2  m  1.142  m  0.058 m

Soooo we have traded 0.058 m of depth for 0.058 m of velocity head

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Part b. Find the minimum channel width that creates critical conditions at the constriction At critical conditions the flow contains just enough energy to avoid a choked condition, the Froude number = 1.0.

For a rectangular channel the critical depth equals 2/3 of the specific energy at that point.

2 Ignoring losses the critical depth AT the obstruction equals: yc   E1  0.853 m 3 Q

E 1 = yc 

2

 yc bc 2 2 g

solve for the minimum channel width necessary to prevent choking 2 Q    2  g  yc  E 1  yc bc   2 Q   2 g y  E  y c 1 c 

bc 

    1.216   m  1.216     

2 Q 2

g  yc 

E 1  yc

amount_of_constriction  2  m  bc  0.784 m If the channel becomes any narrower the upstream water surface elevation will increase, i.e. upstream depth will be greater than 1.2 m

Vc 

Q m  2.892 yc  b c s

Vc

2

2 g

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 0.427 m

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Fr 

Vc

g  yc

1

If the obstruction is larger than 78.4 cm in diameter then the water surface upstream of the obstruction will rise in order gain energy to get around the obstruction.

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Chin 4-34 3rd edition 3

m . In order to pass the sec flow under a roadway, the channel is constricted to a width of 6 m . Under design conditions, the depth of flow just upstream of the constriction is 1.00 m, and the contraction takes place over a distance of 7 m. A lined rectangular concrete drainage channel is 10 m wide and carries a flow of Q  8 

(a) If the energy loss in the contraction is equal to

V1

2

, where V1 is the average velocity upstream of the 2 g contraction, what is the depth of flow in the contraction ? (b) Does consideration of energy losses have a significant effect on the depth of flow in the contraction ? 3

(c) If the width of the constriction is reduced to 4.5 m and a flow of 8 

m is maintained, determine the depth of flow sec

in the constriction (include energy losses) (d) If reducing the width of the constriction to 4.5 m influences the upstream depth, determine the new upstream depth

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Chin 3-29 Flow Constriction.xmcd

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This is a pretty standard specific energy problem. We can compute the S.E. and the Froude number of the flow upstream of the constriction. We then account for the energy loss and compute the depth downstream based on the remaining S.E. 3

b1  10  m

y1  1.0  m

Q  8 

m sec

b2  6  m

L  7  m

Q m upstream velocity of flow : V1   0.8 b 1  y1 s Q

upstream Froude number :

y1  b1

 0.255

y1  g

According to the problem statement the energy loss between sections 1 and 2 is V1

V1

2

2 g

thus we can write:

2

E2 = E1  2 g

The specific energy in the constriction equals the upstream energy minus the loss,

V1

2

2 g

Now write the S.E. equation in terms of the flow, depth of flow, and channel geometry

Q

 b2 y2 2

y2 

Q

2

2 g

Q

= y1 

2

 y1 b1 2 2 g

Q



2

2 g

 b1 y1 2 2 g

Q

 y1 b1 2

 0.033 m

Q 2

y1 

2

 0.091  m

2

2

 y1 b1 2 2 g

 1.033 m

3

b 2  2g

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Chin 3-29 Flow Constriction.xmcd

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Note to 411 people: I used the polyroots function to solve for the downstream depth simply to become familiar with it. I would not suggest you use it or symbolics to solve for it. Use either a graphical approach or the "root" function.

y2 

.091  m y2

3

2

 

 1.033  m  .033  m = 0

p y2  .091  y2

2

 y2  1

Solve for y2 numerically using the polyroots function

 0.091    0   v  p  y2 coeffs   1     1   0.268  r  polyroots ( v)  m   0.385  m  0.883    Polyroots returns 2 positive roots one for supercritical flow, one for subcritical flow. To determine which we have determine the flow regime upstream of the contraction

The upstream Froude number is: Fr1 

V1 g  y1

 0.255 subcritical, therefore the flow regime in the

contraction must also be subcritical, use the larger value y2  r2  0.883 m

The upstream depth (1.0 m) drops to 0.883 m after the constriction

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Now perform the same calculation but ignore the energy loss: V1

2

E1  y1   1.033 m 2 g Q

y2 

2

 b2 y2

Q

2

 b2 2

 1.033  m = 0

2 g

2

2 g





p y22  y22  .091  y22

2

 0.091  m

3

 1.033

 0.091    0   v  p  y22 coeffs   1.033     1 

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 0.265  r  polyroots ( v)  m   0.371  m  0.927   

Since the upstream flow is subcritical the flow in the contraction is also subcritical y2b  r2  0.927 m

The difference in depth neglecting the energy loss is:

.927  m  .883  m  4.746  % .927  m 3

(c) If the width of the constriction is reduced to 4.5 m and a flow of 8 

m is maintained, determine the depth of flow sec

in the constriction (include energy losses)

What is really being said here is "by reducing the downstream channel width to 4.5 m do we create a hydraulic choke" 3

b1  10  m

y1  1.0  m

Q  8 

m sec

b2  4.5  m

V1

L  7  m

2

E2 = E1  2 g

Q

y2 



2

Q



b2  y2

2 g

2

= y1 



2

Q



y1  b1

2 g

2





Q

2



b1  y1

2

y1 

2 g

 y1 b1 2 2 g

Q

y2 

.161  m y2

2

3

2

2

 y1 b1 2

 1.033  m  .033  m = 0

2 g Q 2

2

 0.161  m

 1.033 m

 0.033 m

3

b 2  2g





p y2c  .161  y2c

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2

 y2c  1

Chin 3-29 Flow Constriction.xmcd

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Solve for y2 numerically using the polyroots function

 0.161    0   v  p  y2c coeffs   1     1 

 0.346  r  polyroots ( v)  m   0.673  0.112i  m  0.673  0.112i   

None of the roots are real and positive therefore the flow is choked and critical flow exists within the contraction. Write the equation for the Froude number, set it equal to 1.0 and solve for the critical depth.

2

2

Fr = 1 =

Q g  y2c Q

1=

2

 b2 y2c 2 g  y2c

2   3  Q  2 1  3 3 b2  g    2  Q 3   1  3  i     2   2 y2c   2 1  3 3  b2  g   2 3 i   3 1  Q   2  2   2 1  3 3  b2  g 

         0.686       0.343  0.594i  m   0.343  0.594i            

The depth of flow in the contraction is critical depth y2c  0.686 m 0 ycrit  y2c 0

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d) If reducing the width of the constriction to 4.5 m influences the upstream depth, determine the new upstream depth

Now at critical conditions in a rectangular channel we know that E2 

E1 

Q

y1d 

2

 b1 y1d 2 2 g

Q



V1

2

2 g

= E2

2

Q

 b1 y1d 2 2 g

3 y  1.028 m 2 crit

= 1.028  m

2

 b1 2 2 g

 0.033  m

3

y1d  1.028  m The depth in the contraction at critical conditions is 0.686 m. The upstream depth increases from 1.0 m to 1.028 m

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Chin 4-39 3rd edition 3

m with a bottom width of b  4.5  m and side slopes of SS  1.5 sec (H:V) . The depth of flow in the channel is yup  1.9  m. If a step of Δh  15  cm is placed in the channel, what is Water flows in a trapezoidal channel Q  15 

the depth of flow over the step ? What is the maximum height of the step that will not cause water surface upstream of the step ?

Area of a trapezoidal channel A ( y)  b  y  2 

1  y  SS  y 2

(a) What is the depth of flow over a 15 cm step placed in the channel ? Write the energy equation from some point upstream to the step, solve for the depth on the step.

Q

yup 

2



2

A yup

 Δh = ystep 

2 g

Q

2

1 b  y   step  2   ystep  SS  ystep 2  

ystep  1  m

    ysoln  root yup  

Q

2



2

A yup

2 g

2  Q  2  1   b y  2   y  SS  y   step step  2 step   Δh  ystep   2  g 

         ystep  

The depth on the step caused by a 15 cm step height ysoln  1.733 m

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Chin 3-34 step in trapzoidal channel (2).xmcd

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( b ) Find the height of the step that will cause critical depth on the step assume a value for critical depth yc  1.5  m assume a value for step height Δh  15  cm

Our unknowns are critical depth and the height of the step that will cause critical depth, therefore we need 2 equations containing these unknowns. We will use the specific energy equation and the equation for the Froude number, which in this case is set equal to 1.0

Given

Q

Specific Energy: yup 

Q

2





A yup

2 g

2

2

( b 2  Δh  SS)  y  2  1  y  SS  y   c c 2 c   Δh = yc   2 g

2

Q

Froude Number: 1 =

( b 2  Δh  SS)  y  2  1  y  SS  y   c c 2 c  

( b  2  Δh  SS)  y  2  1  y  SS  y   c c 2 c  g  b  2  Δh  SS  2  yc  SS

 ycsoln     Find y Δh c   Δhsoln    ycsoln  0.719 m Δhsoln  0.921 m Critical depth is ycsoln  0.719 m and the height of the step that will produce it without raising the upstream water surface elevation is Δhsoln  920.871  mm

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Chin 3-34 step in trapzoidal channel (2).xmcd

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Chin 3-34 step in trapzoidal channel (2).xmcd

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Chin 4-46 3rd edition Supercritical and Subcritical Flow in a Rectangular Channel 3

Water flows at Q  30  n  .035 

sec 1

m in a rectangular channel with a width of w  8  m. Manning's n for the channel is sec

. Determine the range of channel slopes that would be classified as steep and the range that

3

m would be classed as mild. Discussion : The dividing line between steep and mild slopes occurs where the Froude number equals 1.0. I wrote the equation for the Froude number and solved for the depth when it equaled 1.0. I then used this depth in Manning's equation to determine the critical slope. Slopes steeper than this value will produce supercritical flow, slopes less steep will produce subcritical flow.

Q wy

Fr =

g y 2

 Q g    Fr  w  y ( Fr)  

3

g

y ( 1)  1.128 m at this depth the Froude number equals 1.0 2 3

Q=

1

1 y ( 1)  w  2  y ( 1)  w    S n 2  y ( 1 )  w   2

Q n

Sc 

2 4

w  y ( 1)   2

2

w  y ( 1)    w  2  y ( 1) 

 0.016

3

Sc  1.607  % Slopes larger than .016 will produce supercritical flow, slopes less than this value will produce subcritical flow

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Chin 3-37 critical slope in channel (2).xmcd

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Chin 4-7 Applicability of Manning's Equation Water flows at a depth of 2.2 m in a trapezoidal, concrete lined channel section ( ks  2  mm ) with a bottom width of 3.6 m and side slopes of 2H:1V . The longitudinal slope of the channel is So  .0006 and the water temperature is 20o C. Assuming uniform flow conditions, estimate the average velocity and flowrate in the channel. Use both the Darcy-Weisbach and Manning equations and compare your answers. ks is the equivalent sand grain roughness of the channel SS  2

y  2.2  m

ρ  998.2 

kg m

The Darcy-Weisbach equation is: V =

γ  ρ  g

3

8 g  f

b  3.6  m

μ  1.002  10

3

 Pa  sec

R  Sf where f is the Darcy-Weisbach friction factor, R is the hydraulic

radius and Sf is the friction slope. 2

The area of a trapezoidal channel can be obtained from : A  b  y  SS  y  17.6 m The wetted perimeter of a trapezoidal channel can obtained from : P  b  2 

The hydraulic radius is therefore: R 

2 2

1  SS y  13.439 m

A  1.31 m P

In the VAST majority of real world flow situations the flow can be considered fully turbulent, thus we can estimate the friction factor from the equation below (ASCE, 1963)

 ks    12  R 

1

= 2  log  f f  .03

 1

fsoln  root 

 f

 ks    f  12  R  

 2  log 

fsoln  0.016

Now estimate the flow velocity: V 

8 g  fsoln

The flow rate is then Q  A  V  34.044

m s

R  So  1.934

m s

3

Now, in order to obtain this result we assumed the flow was fully turbulent, we can check this assumption by computing the Reynolds number

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Chin 3-8 D-W and Manning'e equation (2).xmcd

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ρ 4 R  V 7  1.009  10 The flow is, in fact fully turbulent μ

NR 

Now recompute the friction factor using equation 3-29, ASCE, 1963

 1

fcheck  root  

 ks 2.5   12  R NR  

 2  log 

 f

 f f   

fcheck  0.017 This is very close to the value obtained earlier and confirms the values of flow and velocity obtained 2

1

1 3 2 Manning's equation can be written as: Q =  A  R  So n Table 3-1 suggests a midrange roughness value for concrete of n  .015 

sec 1

m 2

1

3

3

1 m 3 2 Q   A  R  So  34.403 s n

very close to the value obtained with the Darcy-Weisbach equation

The following inequality must be satisfied for Manning's equation to be valid: 6

n 

R  So  9.6  10

6

n 

R  So 6

 14

 3.193  10

 13

s

1.5

m

computed value 3.193  10

 13

 9.6  10

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 14

suggests Manning's equation is valid

Chin 3-8 D-W and Manning'e equation (2).xmcd

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1

The relationship between the 90 percentile bed material and Manning's n is given as n = .038  d R as 2.5   250 d

 n  sec 1   m3 d    .038

6

and is valid as long

6

    3   m  3.783  10 m 

R  346.186 which is outside the range given, this suggests that this relationship is not satisfied d and STRICKLY speaking, Mannings equation may not be applicable. For this problem

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Chin 3-8 D-W and Manning'e equation (2).xmcd

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Chin 4-8 3rd edition Channel Flow using D-W equation and applicability of Manning's Equation Water flows in a trapezoidal channel that has a bottom width of b  5  m side slopes of 2H:1H and a longitudinal slope of So  .0001. The channel has an equivalent sand grain roughness of ks  1  mm. Calculate the uniform 3

m . Is the flow hydraulically rough, smooth or in transition. sec

flow depth in the channel when the flowrate is Q  18  Would Manning's equation be valid in this case?

μ  1.002  10

3

 Pa  sec

ρ  998.2 

kg m

3

We will use the Darcy Weisbach Equation

eqn 1

V=

8 g  f

R  So or Q = A 

8 g  f

R  So

The area of flow in a trapezoidal channel can be obtained from : A = 5  y  2  y The wetted perimeter can be obtained from : P = 5  2  The hydraulic radius is : R =

5  2

5 y

5  2

5 y

2

5 y

eqn 2

If we assume the flow is fully turbulent f can be obtained from:

eqn 3

1

 ks    12  R 

= 2  log  f

We can solve equations 1 through 3 for f, R, and y using a solve block guesses

f  .014

y  3.185  m

R  2.38  m

Given





Q = 5  m y  2  y

2

5  m y  2  y

2

R=

5 m  2



1

8 g  f

R  So

5 y

 ks     12  R  

=  2  log  f



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Chin 3-9 D-W and Manning'e equation.xmcd

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 fsoln     ysoln   Find ( f y R)    Rsoln  fsoln  0.014

ysoln  2.184 m

Rsoln  1.385 m

Using the solution values compute the D-W flow velocity and the Reynolds number 8 g

V 

fsoln





Rsoln  So  0.88

8 g

ρ  4  Rsoln  

 fsoln

NR 



m s



Rsoln  So

  4.858  10 6

μ

Hydraulically smooth turbulent flow occurs when the Reynolds number is less than 10 transitional or rough

5

thus the flow is either

The equation below was taken from Chin and can be used for rough turbulent flow 1

ks  2.5    12  Rsoln NR  fsoln   

= 2  log 

fsoln

 1

fsoln  root  

ks  2.5   12  Rsoln NR  

 2  log 

 f

 f f   

fsoln  0.014 Using the values obtained for R, and NR we recompute the friction factor getting back our original assumption and confirming our assumption of fully rough turbulent flow. Shear Velocity Reynolds Number Approach

Now compute the shear velocity Reynolds number

determine the shear velocity: u 

u  ks ν

g  Rsoln  So  0.037

m s

2

the kinematic viscosity ν 

μ 6 m  1.004  10 s ρ

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Chin 3-9 D-W and Manning'e equation.xmcd

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u  ks

 36.718 This parameter is in the transitional region of turbulent flow, between 4 and 100, this ν appears to contradict our earlier result.

Now examine the applicability of Manning's equation:

V=

8 g  f

2

R S

V=

1

1 3 2 R S n

Set the D-W equation equal to Manning's equation and simplify 2

8 g = f

1

1 3 2 R S n 1

1

2

2

R S

1 6

=

R n

1 6

8 g R = f n

Using the f value found previously, solve for Manning's n 1

n 

2  Rsoln 4

6

s

 0.014

g fsoln

m

0.333

6

Check the criteria for validity of Manning's equation: n 

6

 n    sec   1   3   m  Rsoln ks

Rsoln m

 So  9.741  10

 1.385  10

3

 14

 14

> 9.6  10 applicable

R  S  9.6  10

 14

Manning's equation appears

This value must be less than 500 for Manning's equation to be valid

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Chin 3-9 D-W and Manning'e equation.xmcd

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Chin 7-16 3rd edition A 450 mm by 450 mm concrete box culvert is being proposed to provide drainage across a rural highway. The culvert is to be placed in a vertical headwall with 45o wingwall flares, 4 m long, and is to have a slope of 3%. Use the U.S. Federal Highway Administration equations 3-231 to 3-238 to calculate the culvert performance curve for flowrates up to 6 cms. State your assumptions in using these equations, and confirm your results for a flowrate of 0.6 cms using the USFHWA nomograph shown in Figure 3-59.

Discussion : There are two basic types of culvert equations, those derived from energy considerations and those developed from empirical data. Only the empirical equations take into account the entrance details such as the wing walls thus we will concentrate on those to solve this problem. NOTE : The empirical equations are NOT dimensionally consistent and were developed for U.S. customary units. Keep this in mind when using Mathcad. First, determine the depth of flow and whether the slope is mild or steep.

w  450  mm

n  .013 

3

sec

S  .03

1

m

3 2

Q=

1 n

 w  y  

wy

m Q  0.6  sec

3

1

 S 2 

y  1  m

 w  2 b 

2  1  3  1 wy  2  ysoln  root Q   w  y     S y n   w  2 w  

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Chin 3-102 box culvert performance.xmcd

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ysoln  1.279 ft V 

Fr 

Q ft  11.217 w  ysoln s V g  ysoln

 1.748

The Froude number is greater than 1 for all flows, thus the slope is steep At low flows the entrance will not be surcharged. The water depth will pass through critical depth at the culvert entrance. Assuming type 5 flow the empirical equation is:

Ec H  =  K  D D

M

  0.5  S 0.5   A D  Q

D  450  mm  1.476  ft Q  Q

S  0.03

K  .026 A  D

M  1

2 1

3  Q  For a rectangular channel the critical energy equals Ec    2 2   g D  2

Ec H =  K   D D

3

M

  0.5  S 0.5   A D  Q

Because the equation is empirical it is not dimensionally consistent, thus I divided out units and attached the units required

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Chin 3-102 box culvert performance.xmcd

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         3   2         H ( Q)     

1

    2 D      ft    

 Q    ft 3     sec     g ft 2

sec

2

3

D ft

Q   3 ft  D sec   K  2 0.5 ft  D D           ft   ft 

       M         D  0.5  S  D   ft ft   ft    

This equation defines culvert performance provided that:

Q A D

0.5

 3.5

 A  D  0.5 ft 3     ft 2  ft   sec  

Q  3.5  

3  A  D  0.5 ft 3 ft     9.27   ft 2  ft   sec sec  

3.5  

Type 5 flow and this equation govern up to a flow of 9.27 cfs

3

3

3

ft ft ft Q  .1  .2   9  sec sec sec

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Type 5 Flow vs Depth

1.8 1.62 1.44 1.26 1.08 H ( Q)

0.9 0.72 0.54 0.36 0.18 0

0

0.9

1.8

2.7

3.6

4.5

5.4

6.3

7.2

8.1

9

Q

At 6 cms the upper end of the box culvert is surcharged and the culvert behaves like an orifice. The empirical equation governing Type III flow is

2

  Y  0.5  S 0.5   A D 

H = c   D

Q

c  .0347

Y  0.86 2

Q   3   ft   sec H2 ( Q)  c   0.5   A  D    2  ft    ft

     D D  D   ft  Y  ft  0.5  S  ft   ft       Q

This equation is valid as long as :

A D Q  4 A  D

4

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A D   2 ft ft  

0.5

 4

0.5

3



0.5

3

ft ft  10.594  sec sec

Chin 3-102 box culvert performance.xmcd

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3

Q  .1 

3

3

ft ft ft .2   20  sec sec sec

The user defined conditional function, below plots both functions on the same set of axis 3

elev ( Q) 

H ( Q) if Q  9.27 

ft sec 3

H2 ( Q) if Q  10.6 

ft sec

Culvert Performance 4.5 4.05 3.6 3.15 2.7 elev ( Q) ft

2.25 1.8 1.35 0.9 0.45 0

0

2

4

6

8

10

12

14 Type 16 3 18

Q ft

20

flow

3

sec

Type 5 flow

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Chin, Problem 7-6 3rd edition

A concrete pipe culvert n  .013 

sec 1

m

3

is to be sized to pass Q  .8

m when the headwater depth is 1 m. sec

3

HW  1  m Site conditions require that the pipe slope be S  2% and the length of the culvert be L  10  m. If the culvert entrance conditions are such that ke  0.5 and Cd  1, and the discharge is free, determine the diameter of the culvert.

The flow is not Type I according to Chin since the outlet is not submerged. It is possibly Type II or Type III. Type II occurs if the normal depth is greater than the barrel diameter. This implies water is being admitted fast enough to keep the barrel full. Type III occurs if the inlet cannot admit water fast enough to keep the barrel full.

Type I and II flows are said to be under outlet control since the barrel flows full and it is the barrel friction which determine the capacity.

Discussion : Inlet is surcharged, outlet is free (not surcharged). Type II flow occurs if the normal depth of flow is greater than the diameter of the culvert. Type III occurs if the entrance is submerged and the culvert cannot admit water fast enough to keep the pipe full. Assume Type II or III flow. For Type II flow we can write the energy equation between points 1 and 3, neglecting the losses between section 1 and the barrel and section 3 and the barrel. Δh = hi  hf  ho

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Chin 3-97.xmcd

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2

V inlet loss hi = ke  2 g 2

friction loss using Manning's equation: hf =

2

n V L 4

R

3

2  2 2 Vd  V V . In many cases this expression is simplified to , with Vd outlet loss ho = ko   α   αd  h = o 2  g  2 g  2 g

assumed equal to zero.

2

Combining equations we get : Δh =

2

n V L 4

R

2

 ke 

2

V V  2 g 2 g

3

For type I flow ∆h is the difference in water surface elevations, for type II flow ∆h is the difference between the headwater elevation and the elevation of the crown. The equation above can be solved for Q

Q = A

2  g  Δh

 2  g  n 2 L   k 1 4   e   3  R 

In the D-W equation is used for friction loss we get: Q = A 

2  g  Δh f  L      ke  1  4 R 

If we assume Type II flow then Δh = L  S  H  D = 1.2  D 2

Q=

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π D  4

2  g  ( 1.2  D)

 2  g  n 2 L   K 1 e 4     3  R 

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D  .5  m



2

π D  4

Dsoln  root Q 

     



2  g  ( 1.2  m  D)

 2  g  n 2 L     ke  1 4    D 3       4 

D

     

Dsoln  0.652  m

O.K. - So far so good. However now we must verify the assumption we made for type II flow, that the normal depth is greater than the barrel diameter 2 3

2

Q=

1

1 π D  D  2    S 4 n 4  

2   1 3 2   1 π D  D  2 NDsoln  root Q       S D 4 n  4 

NDsoln  0.582  m Uh - Oh the normal depth is less than the barrel diameter, the pipe is not flowing full. Try Type III flow. In type III flow the barrel inlet cannot admit water fast enough to keep the barrel full. The inlet acts like an orifice. The orifice equation is: Q = Cd  A 

2 g h

2

Q = Cd 



π D  4

DIII  root Q  Cd 



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2  g  ( 1.2  m  D) 2

π D  4

Chin 3-97.xmcd



2  g  ( 1.2  m  D) D



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DIII  0.53  m

A 

π  DIII 4

 0.221  m

2

The corresponding flow full discharge is: 2 3

1

3  DIII  1 m 2 Qfull   A     S  0.624  sec n  4 

Thus the culvert is flowing full at .624 cms and could NOT be flowing partially full at .80 cms. The flow regime is somewhere between type II and type III so we will use the larger diameter, 0.65 m

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Chin 7-7 3rd edition A circular concrete culvert is designed to pass water under a roadway. Under design conditions, the maximum 3

headwater depth is 2 m, the tailwater depth is 1 m, the flow through the culvert is Q  1 

m , the length is 15 m, and sec

the slope is 1.5% . Determine the size of the culvert. Discussion : At this point we do not know whether the inlet or outlet is submerged. The conservative thing thing to do in such cases is assume Type II flow (outlet control) where both the inlet and outlet are submerged, the pipe flows full, and the downstream velocity is zero. The equation for discharge through a culvert under Type I or II flow is:

ke  .5

HW  2  m

TW  1  m

L  15  m

S  1.5%

head loss through the pipe due to friction is: hf  S  L  0.225 m

n  .013 

sec 1

m

3

for Type I and II flow regimes Δh = HW  So  L  TW

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Chin 3-98 Type I culvert Flow and HY-8 output.xmcd

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2 2

Δh =

n V L 4

R

2

 kexit 

2

V V  2 g 2 g

3

2  g  Δh

Q = A

 2  g  n 2 L   k 1 4   e   3  R 

In this equation ∆h is the elevation difference between the HW elevation and the TW elevation: Δh  HW  S  L  TW  1.225 m

R=

D 4



Dsoln  root Q 

     

A=

2

π D  4

π D 4

2

D  1  m

2  g  Δh

 2  g  n 2 L     ke  1 4    D 3       4 



D

     

Dsoln  0.614 m

Thus, the required diameter of the pipe to carry the design flow under these conditions is about 61 cm. Since the TW is 1 m, the outlet is submerged and Type II flow is confirmed. Try using a 61 cm diameter culvert, assume type II flow, see what the discharge is, check to see if normal depth is > culvert diameter to confirm type II flow Use Manning's equation to determine flow velocity assuming pipe flows full. d  Dsoln C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \Culvert sizing\ 9/24/2012

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A ( d) 

π d 4

2

P ( d)  π  d 2 3

1

1  A ( d)  m 2 V      S  2.701 n  P ( d)  s

    π  d2 2   4 1 π d depthnormal  root Q    4 n   πd

2

   

3

  1   2  S d 

depthnormal  2.191  ft

The normal depth in the culvert is greater than the pipe diameter implying the flow is pressurized and we are dealing with type II flow

Fr 

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V A ( d) g d

 1.242

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HY - 8 output from Chin 3-98 In order to get the specified head and tailwater depths HY-8 used a 610 mm diameter culvert. Bottom width of downstream channel is 0.8 m.

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Chin 7-9 3rd edition 3

m . The culvert is to be laid on a slope of sec .1 m/m, is 25 m long, and have a low tailwater elevation. Assuming an entrance loss coefficient of ke  0.1, A 2m x 2m concrete box culvert is to handle a design flow of Q  4 

determine the headwater depth under design conditions.

Solution : culvert slope : S  .1 

m m

culvert width : w  2  m culvert height : H  2  m

culvert length : L  25  m entrance loss coefficient : ke  .1 Manning's n for concrete: n  .013 

sec 1

m

3

First find the normal depth in the culvert using Manning's equation: A ( d)  w  d

P ( d)  2  d  w

R ( d) 

Q=

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1 n

A ( d) P ( d) 2

1

3

2

 A R  S

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d  1  m 2 1    1 3 2  yn  root  Q   A ( d)  R ( d)  S d n  

yn  0.244 m This is the depth of flow IN the culvert NOT the HW depth Now find critical depth to determine if the slope of the box culvert is hydraulically steep or mild

1

  Q  2    w   yc     g 

3

 0.742 m

yc  yn Thus the flow regime is supercritical and the slope of the culvert is steep, critical depth occurs at about 1.4*yc downstream of the culvert entrance. Looking at page 187 in Chin it appears this may be type 5 flow. This will be confirmed if we find that the headwater depth does not submerge the culvert inlet.

In order to get an expression for type V flow through a culvert write the energy equation between the HW elevation and the culvert entrance.

HW 

VH

2

2 g

Vc

2

 hent = yc  2 g 2

V hent = ke  2 g

Q

 HW  yc 

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VH

2

2 g

 ke 

VH

2

2 g

2 2

=

Ac

2 g

Chin 3-99 Type 5 flow.xmcd

Δh = HW  yc

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Q

2

2 2  V 2 Ac VH  H Δh    ke  = 2  g  2 g  2 g

 A  V 2  V 2  k  2  g  Δh   c  H H e Q=    A  V 2  V 2  k  2  g  Δh  H H e  c  solve the expression for Q results in: 2   VH Q = Ac  2  g   Δh   hent 2 g  

1

Defining Cd 

1  ke

we can write Q = Cd  Ac 



2  g  HW  yc



The flow area corresponding to the critical depth is given by: Ac  yc  w

g  yc  Q = Cd  Ac 



2  g  HW  yc



HW 

Q

2

2

2  Ac  Cd g

2

 1.15 m

Thus the headwater does not submerge the inlet ( 2 m high) and type V flow is confirmed for a design flow 3

m of Q  4 s

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.

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Chin 8-1 3rd edition A flood control system is designed such that the probability that the system capacity is exceeded X times in 30 years is given by the following discrete probability distribution.

 out1      out2 

number of exceedances, x 0 1 2 3 4 5 6 7 8 9

probability of occurrence, f(x) 0.04 0.14 0.23 0.24 0.18 0.1 0.05 0.02 0.01 0

This a discrete distribution because exceedances come in discrete numbers 1,2,3....you can't have 1.35 exceedances. OK - each entry in the right hand column is the probability distribution that the system in question will be exceeded the number of times listed in the corresponding left column, for example, there is a 14% chance that the system capacity will be exceeded once in 30 years. The mean number of exceedances in 30 years is simply the probability weighted average of the number of exceedances in each year. i  0  9 read the number of exceedances and the corresponding probability in each year from the Table above no_exceedancesi  out1i Pri  out2i

Chin 4-1 moments of a distribution.xmcd 9/24/2012

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Discrete Probability Distribution

Pri Pri

0.25 0.225 0.2 0.175 0.15 0.125 0.1 0.075 0.05 0.025 0

0

1

2

3

4

5

6

7

8

9

no_exceedancesi

sum the product of the number of exceedances and the corresponding probability to obtain the probability mean number of exceedances in 30 years. Note that for the number of exceedances > 9 the probability is zero

9

μ 



 no_exceedancesi  Pri

i  0

The expected (most likely) number of exceedances in thirty years is: μ  3.06 or 3.0

The standard deviation is the average magnitude of the deviation of the random variable from its mean value. 9

var 



 no_exceedancesi  μ 2  Pri  

i  0

var  2.75 Std_dev 

var  1.658

The average number of exceedances in a 30 year period is μ  3.06 and the average variation about this number is 1.66 exceedances.

Chin 4-1 moments of a distribution.xmcd 9/24/2012

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The skew coefficient tells us how the distribution departs from a bell shaped curve. A positive value indicates the distribution is skewed, that is, has a longer tail to the right.

The skew is slightly positive, indicating a slight tail to the right.

skew 

9

1 3

var





 no_exceedancesi  μ 3  Pri  

2 i  0

skew  0.424 Note that the distribution DOES have a slightly longer tail to the right.

Chin 4-1 moments of a distribution.xmcd 9/24/2012

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Chin 8-12 3rd edition A drainage system is designed for a storm with a return period of 10 years. What is the average interval between such events ? What is the probability that there will be more than six months between floods ? 1 The average interval between occurrences of a 10 yr. event is 10 years, μt  10  yr, λ  μt

Solution : Part A - Based on the definition of return period, the average interval between events will be 10 years. Recall that a 10 year event occurs "on the average" once in ten years. Therefore the answer to question is "10 years"

Part B What is the probability that there will be more than six months between floods ? First what statistical distribution can we use to answer this question ? According to Chin "The exponential distribution describes the probability distribution of the time intervals between occurrences of an event in a continuous Poisson process.

n

Start with a Poisson Process f ( n) = e

 λt

( λ  t)  e n

 λt

. The probability of zero occurrences (n = 0) in a time interval t is

. Thus, the probability of one or more occurrence is a time interval t is 1  e

 λt

. NOTE THAT THIS IS THE

SAME THING AS SAYING THAT THE PROBABILITY OF THE TIME INTERVAL BEING LESS THAN t IS 1  e Thus 1  e

 λt

 λt

.

is the cumulative probability distribution of the time between events being less than or equal to t.

The probability density function of the time between events can thus be obtained by differentiating the cumulative distribution. λ  λ





 λt  λt d  λe 1e dt  λt

. The mean value for the time between The Exponential probability density function is therefore: f ( t) = λ  e 1 1 sooooo... λ = . If μt is the average time interval between events where the event has a events is μt = λ μt single year probability of occurrence of p then λ = p. p is often called the single year probability of occurrence.

Chin 4-12 Exponential distribution applied to hydrologic events.xmcd 9/24/2012

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The probability of a period of 6 months or less between storm events is simply the area under the exponential distribution from 0 to 180 days. 180day

  0

λe

 λt

dt  1  e

 180  λ  day

 0.048

Thus the probability of a period greater than 6 months is 1 minus the probability of a value of 6 months or less: 180day

 1 0

λe

 λt

dt  e

 180  λ  day

 0.952

We can describe the calculations above using a plot of the exponential distribution. In the plot below the probability of an interval less than 0.5 years is given by the area under the curve between 0 and 180 days, or 6 months. Now, if the total area under the curve is 1 (see below), the probability of an interval greater than 6 months is 1 minus the probability of an interval less than 6 months. We can demonstrate that the exponential distribution has the required probability that the area under the curve is 1 by simply integrating from zero to a big number. 100  yr

  0

λe

 λt

dt  1  e

 100  λ  yr

1

t  1  day 2  day  10  yr

Exponential probability density function f ( t λ )  λ  e

 λt

Plot the distribution below

Chin 4-12 Exponential distribution applied to hydrologic events.xmcd 9/24/2012

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Exponential Probability Distribution - Chin 4-12

f ( t λ)

7 10

9

6.3 10

9

5.6 10

9

4.9 10

9

4.2 10

9

 λ 9   2  3.5 10

f  t 

f ( t 2λ) 2.8 10 9 2.1 10

9

1.4 10

9

7 10

 10

0

0

1

2

3

4

5

6

7

8

9

10

t yr

lambda = .10 lambda = .05 lambda = .20

The plot above implies that there are higher probabilities of shorter interstorm intervals than longer ones. We can obtain the same results using the cumulative exponential distribution, the probability of 1 or more events within a 6 month period is, that is, the probability of an interval of less than 6 months is: t  .5  yr F ( t)  1  e

 λt

F ( .5  yr)  0.049



Therefore the probability of an interval greater than 6 months is: 1  1  e

 λt

  0.951

t  1  day 2  day  10  yr F ( t)  1  e Chin 4-12 Exponential distribution applied to hydrologic events.xmcd 9/24/2012

 λt

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Cumulative Exponential Distribution Chin 4-12 0.25 0.225 0.2 0.175 F ( t) .049

0.15 0.125 0.1 0.075 0.05 0.025 0

0

0.1

0.2

0.3

0.4

0.5 t yr

0.6

0.7

0.8

0.9

1

.5

Each point on the y axis of the curve above represents the cumulative probability of an interval equal to or less than the corresponding value on the x axis.

Chin 4-12 Exponential distribution applied to hydrologic events.xmcd 9/24/2012

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Chin 8-14 3rd edition - the Poisson and the Gamma Distribution A residential development in the southern U.S, floods whenever more than 10.2 cm of rain falls in 24 hours. In a typical year there are three such storms. Assuming the occurrence of these events constitute a Poisson process, what is the probability that it will take more than one year to have three such events.

Solution and discussion: The problem statement simply says what is the probability that more than 1 year will elapse before three such events occur. It does not say how many events occur within the year, it could be 0, 1, or 2. So it is equivalent to ask "What is the probability of less than 3 events in a year". This would be the sum of the probability of 1 event + probability of 2 events + probability of 3 events. This problem can be done two ways (1) using the Poisson distribution and (2) using the Gamma distribution

The Poisson Distribution

n

 λt

( λ  t)  e . Here t is an n interval of time,  is the average number of occurrences per unit of time and n is a specified number of occurrences within that interval of time. f(n) is the probability of a specified number of occurrences in the specified time interval, t. The probability distribution for a time continuous Poisson process is given by: f ( n) =

the average rate of occurrence based on the data is: λ 

1 3  365 day

the time interval in question is: t  1  yr n

f ( n) 

( λ  t)  e n

 λt

The probability that less than 3 events occur in one year equals the sum of the probabilities of 0, 1 or 2 events in 1 year. Thus, the probability a year will elapse before the third event IS THE SAME. Probability of less than 3 events in one year: f ( 0)  ( f ( 1)  f ( 2) )  0.423 . Thus the third event MUST occur after 1 year WITH the same probability. f ( 0)  ( f ( 1)  f ( 2) )  42.274  %

The question asks: what is the probability that it will take more than 1 year to have 3 such events

Chin 4-14 Poisson and Gamma Distributions.xmcd 9/24/2012

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The Gamma Distribution

The GAMMA distribution describes the probability of a time interval t until the nth occurrence of a Poisson process. IF the rainfall events described in the problem can be described as a Poisson process, we are asked "what is the probability that it will take longer than 1 year to have 3 such occurrences".

n n 1

The gamma distribution is: f ( t) =

 λt

e λ t ( n  1)

Where f(t) is the probability that a time interval t will elapse before the occurrence of n events n  3

The probability that a time interval of at least 365 days will elapse before 3 such occurrences is given by 365  day

  obtaining and evaluating the cumulative Gamma distribution: F ( t)   0

3 3 1

 λt

e λ t ( n  1)

dt

There is a 57% probability that 365 days or less will elapse before 3 such events occur: F ( 365)  0.577

Therefore there is a 1  F ( 365)  0.423 or 42% probability that more than 365 days will elapse before the occurrence of three such events

Chin 4-14 Poisson and Gamma Distributions.xmcd 9/24/2012

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Chin 8-15 3rd edition Application of the Normal Distribution The annual rainfall in Everglades National Park has been estimated to have a mean of μ  141.2  cm and a standard deviation of σ  28.2  cm . Assuming the annual rainfall data is normally distributed what is the probability of (a) an annual rainfall less than 127 cm, (b) an annual rainfall less than 152.4 cm , and (c) an annual rainfall between 127.0 and 152.4 cm

Solution : The equation for the normal probability distribution is: f ( x μ σ ) 

 1  x  μ  2  exp     σ  2 π 2  σ   1

The probability of having an annual rainfall less than 127 cm is the cumulative probability between zero and 127 cm.

Note that the normal distribution is defined from - ∞ to + ∞ . However, in this case we can integrate from zero to the specified values because, for all practical purposes the left tail of the distribution reaches the x axis before x gets to zero. x  0  cm 1  cm  250  cm

Probability Distribution Annual Rainfall 3 2.7 2.4 2.1 f ( x μ σ) 1.8 σ  1.5  f  x μ   2  1.2  0.9 0.6 0.3 0

0

25

50

75

100

125

150

175

200

225

250

x cm

std deviation = 28.2 cm std. deviation = 14.1 cm Question : Is the normal distribution a continuous or discrete distribution ?

Chin 4-15 Everglades National Park - Annual C:\Mathcad application areas\Mathcad Rainfall.xmcd application areas\Fluids-open channels9/24/2012 hydrology\HYDROLOGY PROCEDURES \probability and statistics in hydrology\ 12:25 PM

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The probability of an annual rainfall less than 127 cm equals the area under the curve from zero to 127 cm 127  cm

    0cm

1 σ

2 π

 1  x  μ  2    dx  0.307 2  σ  

 exp 

There is a 30.7% probability of having an annual rainfall less than 127 cm. Stated differently, 30.7% of all annual rainfall values in the future can be expected to be at or below 127 cm. pnorm ( 127 141.2 28.2)  0.307

The probability of an annual rainfall less than 152.4 cm equals the area under the curve from zero to 152.4 cm

152.4  cm

    0  cm

1 σ

2 π

pnorm  152.4 



152.4  cm

    127  cm

1 σ

pnorm  152.4 



μ cm

2 π



 1  x  μ  2    dx  0.654 2  σ  

 exp 

μ σ     0.654 cm cm 

 1  x  μ  2    dx  0.347 2  σ  

 exp 

σ  μ σ      pnorm  127    0.347 cm cm  

cm 

There is a 34.7% probability of having an annual rainfall between 127 and 152.4 cm.

Chin 4-15 Everglades National Park - Annual C:\Mathcad application areas\Mathcad Rainfall.xmcd application areas\Fluids-open channels9/24/2012 hydrology\HYDROLOGY PROCEDURES \probability and statistics in hydrology\ 12:25 PM

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8 - 16 3rd edition Chin

Annual maximum discharges in a river show a mean of μx  620 and a standard deviation σ x  311. If the capacity of the river is 780 m3/sec and the flow can be assumed to follow a log-normal distribution, what is the probability that the maximum discharge will exceed the channel capacity ? Discussion : In cases where the random variable X is equal to the product of n random variables X1 X2  XN we can write ln X1  ln X2  ln X3 thus ln ( X) will be normally distributed and X is said to have a log normal distribution.

 

 

 

Define the random variable Y, by the relation :

Y = ln ( X)

Thus, the probability distribution of Y can be written as:

  ln ( x)  μ 2   y  f ( x) =  exp  2   x  σy  2  π 2  σy   1

2  σ y  the relationship between the mean , μx and log-mean μy is given by: μx = exp  μy  2  

the relationship between the standard deviation of the data and the std deviation of the logs of the data is given as: σ x =

2 2 μx   exp  σ y   1









We are given μx and σ x. Use a solve block to get μy and σ y the mean and standard deviation of the logs of the data

σ y  2

μy  3

Given 2  σ y  μx = exp  μy   2  

σx =

Chin 4-16 Log Normal distribution - Annual River max flows.xmcd 9/24/2012

2 2 μx   exp  σ y   1









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 μysoln     Find μ σ  y y  σ ysoln     μysoln   6.318      σ ysoln   0.474   

Now, the probability that the river capacity will be exceeded equals 1 minus the probability that it will not be exceeded.

x

  1 F ( x)    x σ ysoln   0

2   ln ( x)  μ  ysoln   dx  exp 2   2 π 2  σ ysoln  

F ( 780)  0.765 1  F ( 780)  0.235 3

The probability that the river flow will be less than or equal to 780 

m equals : F ( 780)  76.469  % . sec

3

The probability that the river flow will exceed 780 

m equals 1  F ( 780)  23.531  % sec 3

We may say that approximately 23% of all future annual maximum discharges will be greater than 780 

m . That is, sec

about 1 in every 4 annual discharges.

Chin 4-16 Log Normal distribution - Annual River max flows.xmcd 9/24/2012

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Chin 8-17 3rd edition Log Normal Distribution The annual rainfall in a rural town is shown to be log-normally distributed with a mean of μx  114  cm and a standard deviation of σ x  22  cm. Estimate the rainfall amount having a return period of 100 years . How would your estimate differ if the rainfall were normally distributed ? In this problem the random variable is the annual rainfall, X. In practical terms a data set is said to be log-normally distributed when the logs (natural or common) appear to be normally distributed.

  ln ( x)  μ 2   y  for x  0 The log normal probability distribution is: f ( x) =  exp  2   x  σy  2  π 2  σy   1

The relationship between the mean of the data and the mean of the ln of the data is:

2  σ y  μx = exp  μy  2  

A similar relationship can be written for the standard deviation of the data 2 2 2 σ x = μx   exp  σ y   1    

First use a solve block to determine μy and σ y μy  2

σ y  3

σ x  22  cm

Given 2  σ y  = exp  μy   cm 2  

μx

2

2

 σx   μx   2   =    exp  σ y   1  cm   cm   μysoln     Find μ σ  y y  σ ysoln    C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \probability and statistics in hydrology\ 9/24/2012

Chin 4-17 Log normal distribution.xmcd

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The coefficient of variation is: Cv 

σx μx

 0.193

3

The skew coefficient is: gx  3  Cv  Cv  0.586 indicating a distribution slightly skewed to the right.

The mean and standard deviation of the natural logs of the data are μysoln  4.718 , and σ ysoln  0.191 The 100 year rainfall event has a single year probability of occurrence of 1%. Thus we are looking for the rainfall value such that 99% of future rainfall values will less than it, or conversely, 1% will be greater.

175  cm

      0

2    x      ln   μ  ysoln 1 cm    dx  0.99  exp    2   x  σ ysoln  2  π 2  σ ysoln  

Thus the 100 year rainfall is approximately 175 cm Now plot the log normal distribution based on the data statistics determined in the problem

x  0  cm 1  cm  300  cm

2    x      ln   μ  ysoln 1 cm    f ( x)   exp    2   x  σ ysoln  2  π 2  σ ysoln  

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Chin 4-17 Log normal distribution.xmcd

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Log Normal Distribution 2 1.8 1.6 1.4 1.2 f ( x)

1 0.8 0.6 0.4 0.2 0

0

30

60

90

120

150

180

210

240

270

300

x cm

Note that the distribution is slightly skewed to the right. This is consistent with the small positive skew coefficient, gx  0.586 . A symmetrical distribution has a skew coefficient of zero.

Normally Distributed Rainfall 165  cm

     0

  x  μ 2  x  dx  0.99  exp   2 σ 2  σx  2  π x   1

If the data were normally distributed the 100 year rainfall would be about 165 cm.

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Chin 4-17 Log normal distribution.xmcd

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Chin 8-2 3rd edition

The probability distribution of the time between floods in a residential area is given by:

NOTE : This is an exponential distribution



f ( t)  .143 e

 for t  0

 .143  t

f ( t)  0 for t  0

If we integrate the distribution from 0 to plus infinity we get 1, as we must for a proper statistical distribution



∞



  .143  t  .143 e dt  1 0

t is the interval between floods in days. Estimate the mean, standard deviation, and skew of t

∞





  .143  t  μt   t  .143 e  dt 0 μt  7.0 Thus, the expected value of t is 7 days

∞

 vart   0

 t  μt 2 .143 e  .143 t dt 2

vart  13.98  μt  1.0  μt  97.80

σ t 



vart  6.996



The expected value of x  μt is 6.99 days. Note that this variable can only take on positive values

Chin 4-2 moments of a distribution.xmcd 9/24/2012

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gt 

∞

  3 σ t 0 1

 t  μt 3 .143 e  .143 t dt

2

gt 

3

293.41  μt  20.97  μt  1.0  μt  2051.83 σt

gt  1.995

3



 3 is 1.99 days3

The expected value of x  μt

Note that this value can be positive or negative. If the skew coefficient is positive it implies that x > μt and the distribution has a long tail to the right.

t  .1 .2  20



f ( t)  .143 e



 .143  t

Flood Interval Distribution probability Intensity

0.16 0.144 0.128 0.112 0.096 f ( t)

0.08 0.064 0.048 0.032 0.016 0

0

2.2

4.4

6.6

8.8

11

13.2 15.4 17.6 19.8

22

t

The curve suggests that the flood intervals tend to be short, with fewer longer values

Chin 4-2 moments of a distribution.xmcd 9/24/2012

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If we integrate this probability density function we see that 86.5 % of future intervals will be within plus or minus 1 standard deviation (7 hr) of the mean (7 hr)

14

  0

Chin 4-2 moments of a distribution.xmcd 9/24/2012

f ( t) dt  0.865

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Chin 8-20

The Extreme value Type I Distribution, a.k.a. the Fisher-Tippett distribution Type I distribution, a.k.a. the double exponential distribution. 3

m and a standard deviation of The annual maximum discharges in a river show a mean of μx  480  sec 3

m . Assuming the annual maximum flows are described by an extreme value Type I distribution use sec equation 4- 85 to estimate the annual maximum flowrates corresponding to return periods of 50 and 100 years. σ x  320 

Discussion : The Extreme Value Type I (Gumbel) requires that the parent distribution be unbounded in the direction of the extreme value. Specifically, Type I distributions require that the parent distribution falls off in an exponential manner, such that the upper tail of the cumulative distribution can be expressed in the form:

P X ( x) = 1  e

 g ( x)

where g(x) is an increasing function of x

In using the Type I distribution with the normal distribution as the parent the probability density function for the Extreme value Type I distribution is given by:

f ( x) =

1 x  b   x  b   exp  ±    exp  ±   a   a    a 

where ∞  b  ∞ , ∞  x  ∞, and a  0 That is we assume that the distribution of flows within a year can be described by a normal distribution. This being the case, the distribution of the maximum flow for each year is given by the distribution above. a and b are referred to as "scale parameters" and b is actually the mode of the distribution.

The mean, standard deviation and skew can be computed from a and b and when seeking maximum values are: 2

μx = b  .577a

σ = 1.645  a

2

gx = 1.1396

The mean, standard deviation, and skew for minimum values are: μx = b  .577a

By using the transformation: y =

Chin extreme value Type I (Gumbel) 420.xmcd 9/24/2012

2

σ = 1.645  a

2

gx = 1.1396

xb we can write: f ( y) = exp ( ±y  exp ( ±y) ) a

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which yields the following cumulative distributions for the maximum yearly and minimum yearly flows. maxima : F ( y) = exp ( exp ( y) ) minima : F ( y) = 1  exp ( exp ( y) ) In order to solve the problem at hand we must first find the values of a and b.

3

μx  480

3

m s

σ x  320 3

a  5 

m s

3

m sec

m sec

b  4 

Given μx = b  .577a σx =

1.645  a

2

 asoln     Find ( a b)  bsoln     asoln   249.498  m 3     bsoln   336.04  s   These are the values of a and b for the cumulative distribution of annual maximum values

Y=

X  bsoln Xb = a asoln

for a 100 year event the cumulative probability is 0.99 thus we can write:





.99 = exp exp y100



 y100

0.99 = e

e

y100  4.60

Chin extreme value Type I (Gumbel) 420.xmcd 9/24/2012

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y100 =

x100  bsoln asoln

x100  y100  asoln  bsoln  1.484  10

3m

3

s

There is a 1% chance that the annual maximum flow in a given year will be greater than x100  1.484  10

3m

3

s

for a 50 year event the cumulative probability is 0.98, thus we can write:





.98 = exp exp y50



 y50

0.98 = e

e

y50  3.90 x50  bsoln

y50 =

asoln

x50  bsoln  asoln  y50  1.309  10

3m

3

s

There is a 2% chance in any given year that the maximum flow for that year will be greater than x50  1.309  10

3m

3

s

Plot the cumulative distribution for the annual maximum flows: y  0 .1  10 maxima : F ( y)  exp ( exp ( y) )





3

ft x ( y)  bsoln  asoln  y  sec

Chin extreme value Type I (Gumbel) 420.xmcd 9/24/2012

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Cumulative Extreme Value Type I Distribution Probability of flow less than or equal to

1 0.9 0.8 0.7 0.6 F ( y) 0.5 0.4 0.3 0.2 0.1 0 0

375

750

1.125 10

3

1.5 10

3

1.875 10

3

2.25 10

3

2.625 10

3

3 10

3

x ( y) ft

3

sec

Annual Maximum Flow (cfs)

Chin extreme value Type I (Gumbel) 420.xmcd 9/24/2012

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Chin extreme value Type I (Gumbel) 420.xmcd 9/24/2012

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Chin 4-27 3rd edition

Frequency Analysis Procedures In this document we will explore procedures for developing mathematical models that allow for prediction of events having a specified probability of occurrence. We will use problem 4-27 from Chin's Water Resources Engineering text as an example of how this is done. The procedure is basically the same for any distribution and any type of hydrologic event. Note that the procedure requires data. The data set is analyzed to see if it fits an assumed cumulative statistical distribution. If it does we can then use the assumed distribution as a model to compute events with a specified probability of occurrence. This document also illustrates the use of several Mathcad functions useful in this regard

Chin 4-27 Peak Annual Flow in a River Assume then verify that the data fits a cumulative normal distribution, make a frequency analysis of the peak annual flows for the period 1980-1996. Plot the data using the the Weibull plotting position formula as well as the Gringorten plotting equation. Based on the frequency curve estimate the following. (a) the 50 yr. peak annual flow (b) the probability that the flowrate in any single year will exceed 8500 cfs. i  0  15

DATA 

...\Chin 4-27.txt

This data set is a 2 column 16 row matrix. The first (zeroth in Mathcad) column contains the year and the second column contains the peak annual flow for each year. Note that the data values are yearly values. This allows the data values to be considered independent of each other.

0

DATA 

Chin 4-27 Frequency analysis.xmcd last save 9/24/2012 / 12:32 PM

1

0

1.98·10 3

8·10 3

1

1.981·10 3

5.55·10 3

2

1.982·10 3

3.39·10 3

3

1.983·10 3

6.39·10 3

4

1.984·10 3

5.889·10 3

5

1.985·10 3

7.182·10 3

6

1.986·10 3

1.058·10 4

7

1.987·10 3

1.159·10 4

8

1.988·10 3

8.293·10 3

9

1.989·10 3

9.193·10 3

10

1.99·10 3

5.142·10 3

11

1.991·10 3

7.884·10 3

12

1.992·10 3

4.132·10 3

13

1.993·10 3

...

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partial view of data set

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Rename the second column of the data set "flow" . Then compute the mean and standard deviation of the data. Note that Stdev is used rather than stdev, why ? 3

flowi  DATA i 1  μ  mean ( flow) σ  Stdev( flow)

ft sec

rows ( DATA )  17 3 ft

μ  7.357  10  3

σ  72.567 m  s

3

sec

1

Now we need to sort the evaporation data from smallest to largest.

Sorted_flow  sort ( flow)

This equation is similar to the equation used to compute dice probabilities except we use the ranks of the data rather than the data values themselves. We also divide by [max(rank) +1] to avoid a probability of zero.

The probability of a flow being smaller than the flow with rank 1 (smallest flow) is probability of a future flow being smaller than the largest flow is

1  0.063 . Conversely, the 16

18  0.947 19

The probability of occurrence of a ranked flow is obtained from the Weibull plotting equation, rank . max ( rank)  1 Rankdemo ( x) 

sorted  sort ( x) for i  0  length ( x)  1 m  match  xi sorted  1 ranki  mean ( m) rank

max ( Rankdemo ( flow) )  16

new_rank  sort ( Rankdemo ( flow) ) The Mathcad function "Rank" assign ranks to all data and computes the average rank for multiple values. YOU DO NOT HAVE THIS FUNCTION. Chin 4-27 Frequency analysis.xmcd last save 9/24/2012 / 12:32 PM

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a  .375

probi 

new_ranki max ( new_rank)  1

probG  i

Weibel plotting equation based on the Rank function

new_ranki  a max ( new_rank)  1  2  a

Below is a probability plot plotted on arithmetic axes. Thus, there is no expectation that the plot should approximate a straight line if the data are normally distributed. If the data were: 1. plotted on normal probability paper AND 2. they were normally distributed, they would approximate a straight line.

Probability of Peak Annual Flow Being Less Than

4

Flow - cfs

Sorted_flowi ft

3

sec

Sorted_flowi ft

3

sec

1.3 10 4 1.2 10 4 1.1 10 4 1 10 3 9 10 3 8 10 3 7 10 3 6 10 3 5 10 3 4 10 3 3 10

0

0.1

0.2

0.3

0.4

0.5 probi probG

0.6

0.7

0.8

0.9

1

i

Probability of Non-exceedance Probability based on Weibell sort Probabiliy based on Gringorten sort

(a.) Estimate the 50 yr. peak annual flow rate. Stated a different way, estimate the peak flow rate which has a probability of occurrence of .02 or a probability of non-occurrence of 0.98 in any given year. Solution : There are several possible ways to try to answer this question. However, it important to note one thing first and that is WE ARE EXTRAPOLATING BEYOND THE DATA USED TO CREATE THE PROBABILITY PLOT. Note that largest probability computed from the data is less than 98%. In classical statistics this a BIG no-no. However, in the real world it is done when there is no other choice. Presumably, we have no more peak annual flow data . Now, the ways to answer the question:

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1. We can attempt to read the value off of the curve above. However, extrapolation is difficult since, if the data were normally distributed, they would not plot as a straight line on an arithmetic scale. Thus, we don't have any rational way to extrapolate from the data.

2. Plot a cumulative probability curve on arithmetic axis using pnorm. On the same set of axes plot the probability occurrence based on the actual data. Decide based on visual examination, if the data based probability plot in approximately coincident with the cumulative normal probability distribution. If so, read the peak flow value corresponding to an exceedance value of 2% from the normal probability plot. I get about 10 using "trace". Note that the data based probability plot seems to match the cumulative normal plot in the region where we have data. An alternative procedure, once it has been assumed that the data are normally distributed is to find the value directly using qnorm.

3. Book Method - Manually rank and plot the data vs probability on normal probability paper. Draw a straight line on the paper using the mean and standard deviation of the data and the characteristics of a normal distribution. Visually determine whether or not the data falls along this straight line, if so the data may be assumed normally distributed and needed values estimated from a straight line through the data. I did not do this here because Mathcad does not have a probability scale, I don't have any normal probability paper handy, and I have long since become too lazy to do problems manually.

3

x  3000 

3

3

ft ft ft 4000   15000  sec sec sec

Chin 4-27 Frequency analysis.xmcd last save 9/24/2012 / 12:32 PM

plotting variable - cumulative normal distribution

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1.5 10

4

x

1.38 10

4

3

1.26 10

4

sec

1.14 10

4

Sorted_flowi

1.02 10

4

3

9 10

3

sec

7.8 10

3

6.6 10

3

5.4 10

3

4.2 10

3

ft

ft

Sorted_flowi ft

3

sec

Chin 4-27 Annual Peak Flow

.98

3

3 10 0.05

0.144 0.238 0.332 0.426

0.52

0.614 0.708 0.802 0.896

0.99

 x  μ  σ  prob prob i Gi 3 3 3  ft ft   ft  sec sec sec 

pnorm 

Cumulative normal distribution Weibell Sort Gringorten sort

At this point I would make the call as to whether or not the data set fits a normal distribution. If it does then the calculations below can be done defensibly

The value of evaporation corresponding to a non exceedance vale of 98% is about 10. It is noted above with lines.

qnorm  .98 

 

μ ft

3

sec



σ ft

3

sec

  1.262  10 4   

A second method is simply use "qnorm" to output the

value. Note this nearly the same value as read from the plot (it better be!)

(b) Estimate the probability that the peak annual flow rate in any single year will exceed 11,000 cfs.

Solution : This simply the same question stated in a different way. Once we believe the data to be normally distributed we can simply read the probability off the plot above or obtain it directly using pnorm. 1 - probability the peak annual flow is less than or equal to 11,000 cfs"

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

1  pnorm  11000 

 

μ

ft

3

sec



σ

ft

3

sec

  0.078   

The probability that the next peak annual flow will be greater than or equal to 11,000 cfs is about 7.8% . Stated another way if we had an infinite number of 16 year data sets for this river, the next annual flow would be less than 11,00 cfs in 7.8% of them.

Summary : The process requires a data set of independent values. These were ranked and the probability of occurrence of each was computed using the Weibull as well as the Gringorten plotting position equations. The probability of occurrence was plotted against the magnitude of the value. This is often called a reference distribution. On the same set of axis was plotted a cumulative normal distribution using the mean and standard deviation computed from the data. A decision is made whether or not the reference distribution fits the cumulative normal distribution. If it does then we can the equation for the cumulative normal as a model for predicting future values. NOTE : IN ALL CASES THE CALCULATIONS ARE BASED ON THE ASSUMPTION, PREVIOUSLY MADE, THAT FUTURE DATA WILL FOLLOW A NORMAL DISTRIBUTION WITH THE SAME STATISTICAL CHARACTERISTICS AS THE SAMPLE USED HERE. IF THIS ASSUMPTION IS NOT DEFENSIBLE THE COMPUTATIONS ARE USELESS, REGARDLESS OF HOW THEY ARE DONE. Note : In reality you can rank the data from largest to smallest or smallest to largest as long as you reverse the probabilities.

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Chin 8-3 3rd edition The annual maximum flowrates in a river are known to vary between 4 and 10 m3/sec. Furthermore the probability distribution of these flows is given by: f ( x) =

α x

2

Here x is an annual maximum flow. (a) Determine the value of α, (b) the return period of of a maximum flow of 7 m3/sec. and (c) the mean and standard deviation of the annual maximum flows. Solution part a:

x= 10

 If the equation given is a valid probability distribution then:   x= 4

α x

2

dx = 1 that is the sum of the probabilities

of all possible maximum annual flows must be 1 x= 10

  x= 4

αx

2

dx

α α 3 α  simplify  10 4 20 3 α = 1 20

α 

20 3

solution to part b: The probability distribution of annual maximum flows in a river is a continuous distribution, thus there is no such thing as the probability of a flowrate of exactly 7 cms. What we are really interested in is the probability of a flow equal to or greater than 7 cms.

3

The return period of a flow of 7 

3

m m is the probability of a flow equal to or greater than 7  in magnitude. This sec sec

 3 m  Q, the cumulative probability of a flow less than or equal to 7 m /sec is obtained  sec  from the cumulative probability distribution. 

can be obtained as 1  Pr  7 

3

Chin 4-3 - probability distribution from data.xmcd 9/24/2012

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The cumulative probability distribution is the probability of a flow equal to or less than a specific value. It is obtained by integrating the probability distribution from its lower endpoint to the value of interest. The probability of a flow greater than this value is equal to (1 - probability of a flow less than this value).

x

 20 20 5 dx ===========> F ( x)  F ( x) =   3 3 x  3 x2 4

3

Cumulative probability of a flow less than or equal to 7 m /sec is F ( x) 

20 5 , F ( 7)  71.429  %  3 3 x

Therefore the probability of a flow greater than 7 is: 1  F ( 7)  28.571  %

1 This corresponds to a return period of TR   yr  3.5  yr ( 1  F ( 7) )

solution part c

Assuming that xi is an outcome of a continuous random variable, X. f(xi) is the continuous probability distribution of X, and g(xi) is an arbitrary function of x, then the expected value of the function g, represented by: ∞

 =  g ( x )  f ( x ) dx  ∞

One such function g(x) is simply x. The expected value of x is termed the mean of the distribution and is commonly given the symbol , μ 10

 To determine the mean of this distribution we write : μ    4 3

x

20 3 x

2

dx

3

20  ln ( 5) 20  ln ( 2)  m m μ     6.109  3 s 3   sec Chin 4-3 - probability distribution from data.xmcd 9/24/2012

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A second such function is g ( x)  ( x  μ)

2

μ  6.109 10

 var =   4

20

2

( x  μ) 

3 x

2

dx

6

6

 40  μ  ln ( 4)  40  μ  ln ( 10)  μ 2  40  m  2.685 m  2 3 3   sec 2 s

var  

3

σ 

σ  1.639

var

m s

2

Thus the expected value of the quantity ( x  μ) is called the variance and the square root of the variance is called the standard deviation. 3

α  6.667

f ( x) 

x  4 

α F ( x) 

2

x

 m3     sec 

Chin 4-3 - probability distribution from data.xmcd 9/24/2012

3

3

m m m 4.1   10  sec sec sec

5  3

2

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20 3

x 3

m

sec

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Annual Maximum Flow Probability Distributions

f ( x)

0.417

1

0.382

0.9

0.347

0.8

0.312

0.7

0.277

0.6

0.242

0.5 F ( x)

0.207

0.4

0.172

0.3

0.137

0.2

0.102

0.1

0.067

4

4.6

5.2

5.8

6.4

7

7.6

8.2

8.8

9.4

0 10

x m

3

sec

Probability Distribution Cumulative Probability Distribution NOTES : Note that the distribution is not symmetrical, the probability of getting a "smaller" maximum flow is greater than the probability of getting a "larger" maximum flow. The maximum cumulative probability is 1.0 as it should be.

Chin 4-3 - probability distribution from data.xmcd 9/24/2012

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Chin 8-5 3rd edition The Binomial Distribution The average annual rainfall over over a certain area is estimated to be 1524 mm, and the rainfall is as likely to be above average in any year (wet year) as it is to be below average. For any 20 yr. interval, what is the probability of experiencing the following events: (a) one wet year with 19 dry years (b) 10 wet years and 10 dry years Discussion : This is a straightforward Binomial Distribution problem. Each year is a trial. Each trial is assumed independent of all other trials. Based on the problem statement, the probability of having a rainfall amount exceeding 1524 mm in any given year is 50%, thus p  0.5.

known values: p is the single year probability of occurrence (success) N is the number of independent trials n is the number of "successes" p  .5 N n  ( N  n)

p

n

N  20

is the number of ways of having n successes in N trials, called the binomial coefficient

is the probability of n successes, p*p*p*p n times

( 1  p)

N n

is the probability of N  n failures (1- p) (N-n) * (1-p) (N-n) -1 * (1-p) (N-n)-2 ...... N-n times.

The Binomial distribution given by: f ( n) 

N n N n  p  ( 1  p) n  ( N  n)

The probability of 1 success followed by 19 "failures" is given by : f ( 1)  1.907  10

3

%

The probability of 10 successes and 10 failures is given by: f ( 10)  17.62  %. This says that in successive 20 yr. periods there is nearly a 1 in 5 chance of having 10 wet years and 10 dry years. You may NOT assume that the successes occur in a single string.

for n  1 the number of ways of having 1 success in 20 trials is

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N  20 n  ( N  n)

Chin 4-5 Binomial Distribution.xmcd

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for n  10 the number of ways of having 10 success in 20 trials is

N 5  1.848  10 n  ( N  n)

1

the probability of 1 success is: p  0.5 the probability of 19 failures is: ( 1  p) the probability of 10 successes is p

10

the probability of 10 failures is ( 1  p)

20 1

 1.907  10

 9.766  10

20 10

6

4

 9.766  10

4

Putting the pieces together we get the values below which match the solutions above

20 1 20 1 5  p  ( 1  p)  1.907  10 1  ( 20  1)

20 10 20 10  p  ( 1  p)  0.176 10  ( 20  10)

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Chin 4-5 Binomial Distribution.xmcd

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e

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Chin 4-5 Binomial Distribution.xmcd

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Chin 8-7 3rd edition A flood control system is designed for a rainfall event with a return period of 25 years. What is the probability that the design storm will be exceeded 5 times in 10 years ? What is the probability the design storm will be equaled or exceeded at least once in 10 years. Solution

Consider each year a Bernoulli trial with a constant probability, p. The distribution of exceedances is then given by the Binomial distribution. The return period,TR  25, thus p  N  10

f ( n) 

1  0.04 TR

n  5

N n N n  p  ( 1  p) n  ( N  n)

The probability of 5 exceedances in 10 years is f ( 5)  2.104  10

5

We can accomplish the same thing using the "canned" distribution in Mathcad

dbinom ( n N p) dbinom  5 10 



1  5  2.104  10  TR



where : n is the number of "successes" N is the number of Bernoulli trials p is the single year probability of occurrence

The probability of at least one exceedance is equal to 1 minus the probability of no exceedances in the 10 year period. n  0 Pr  1 



N n  ( N  n)

n

 p  ( 1  p)

N n

 

Pr  0.335 The expected number of exceedances during a 10 year period is : N  p  0.4 The standard deviation of the mean is : σ 

Chin 4-7 Bernoulli distribution.xmcd 9/24/2012

N  p  ( 1  p), σ  0.62

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Chin 8-9 Stormwater System A stormwater-management system is to have a design life of 10 years, and a runoff event is to be used such that Design there is a 1% chance that the system capacity will be exceeded during the ten year design life. Determine the 3

m return period of the design runoff event. If the cumulative distribution function of the runoff in is the sec cumulative distribution function is given by a Gumbel distribution: F ( Q) = exp ( exp ( Q) )

F ( Q) is the cumulative probability of flow being equal to or less than Q Discussion : This problem can be broken down into two pieces, each involving a different statistical distribution. First we need to determine the single year probability of an event that will result in a total probability of 1% that the system will have one or more exceedances over its 10 year life span. Consider the outcome in any year independent of any other year. Thus we have 10 Bernoulli trials. The probability of occurrence in any given year of the design event is unknown, however the probability of system failure 1 or more times during the design life of 10 years is 1%. What we are looking for is the single year probability of such an event.

What we are really saying is that the probability that the system capacity will be exceeded 1 or more times in 10 years is 1%. This can be computed as [1 - the probability that the system capacity is exceeded 0 times in ten years]. Using the binomial distribution with N  10, n  0

The binomial distribution can be written as: f ( n) =

f ( n) =

N n N n  p  ( 1  p) n  ( N  n)

N n N n 10  p  ( 1  p) float 3  f ( 0) = ( p  1) n  ( N  n)

The probability of zero failures during the design life is : ( p  1)

10

So the probability of 1 or more failures is 1 minus the probability of zero failures .01 = 1  ( p  1) for the single probability p gives the following result.

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Chin 4-9 Drainage system failure Gumbel and Bernoulli Distributions.xmcd

10

solving this

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.01 = 1  ( p  1)

0.00100453     1.999    0.191796  0.587195i     0.691293  0.950101i   0.691293  0.950101i  solve   float 6  1.30871  0.950101i   1.30871  0.950101i     1.8082  0.587195i   1.8082  0.587195i     0.191796  0.587195i 

10

None of the solutions containing i make sense, in addition 2 gives a probability of 200% which makes no sense, thus the correct answer is: p  0.001. This is the single year probability of the storm event which results in a 1% chance of the drainage system capacity being exceeded 1 or more times in 10 years.

The return period of this event is therefore: T 

1 3  1  10 years p

Thus the probability of a flow equal to or less than is 1 minus the single year probability of such an event F ( Q)  .999 F ( Q) = exp ( exp ( Q) )

.999 = exp  exp  

  

  

Q  3

m

sec

  

solving for the flowrate we get: 3

Q  

3

1.0  m  ln ( 0.0010005003335835335001) m  6.907 s sec

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Chin 4-9 Drainage system failure Gumbel and Bernoulli Distributions.xmcd

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3

m has a single year probability of .001. Such a flow results in a 1% chance of the sec drainage system being overloaded 1 or more times during its 10 year design life. Summary : A flow of 6.9 

3

The computed flow rate is 6.9 

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m 8 gal , 157 million gallons per day !  1.575  10  sec day

Chin 4-9 Drainage system failure Gumbel and Bernoulli Distributions.xmcd

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Chin 9-12 3rd edition The "alternating block method" is a simple way of developing a design hyetograph with a specified return period, T, from an I-D_F curve. The design hyetograph produced by this method results is n successive rainfall intensity intervals of t over a total time period, Td = n  Δt . First, select the design return period and duration for the storm. Next, choose an interval, t. Next, from an IDF curve determine the intensity and rainfall depth for intervals t, 2t, 3t,.....Td, each with a return period of T . Determine the depth of precipitation that occurred in each time interval by subtracting the depths at successive times. For example, if t = 1 hr. the depth of precipitation in the first hour is the intensity in in/hr times 1 hr. The depth of precipitation in the second hour is the intensity corresponding to 2t minus the intensity at t, time 1 hour. This process is continued for all successive time intervals

Problem : Use the alternating block method and the IDF curve given below to compute the hyetograph for a 10 year, 1 hour using 11 time intervals

td 

Example Calculation using

60  min  5.455  min 11

818

i 

  td   8.54  min 

mm mm  110.109  0.76 hr hr 

td is the duration of the storm; for this procedure it varies from 0 minutes out to 60 minutes output1  1 2 3 4 5 6 7 8 9 10 11

time min 5.45 10.90 16.35 21.80 27.25 32.70 38.15 43.60 49.05 54.50 59.95

mm/hr 110.1 85.7 71.05 61.1 53.9 48.4 44.07 40.5 37.5 35.06 32.9

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \Design Storms-hyetographs-precipitation\ 9/24/2012

Depth, in mm 10.0191 15.5974 19.3967 22.2404 24.5245 25.8746 28.0726 29.484 30.7125 31.9046 32.9329

sequential depth mm 10.0191 5.5783 3.79925 2.84375 2.2841 1.35014 2.19795 1.41141 1.2285 1.1921 1.0283

mm per hr

hyetograph

110.2101 61.3613 41.7918 31.2813 25.1251 14.8515 24.1775 15.5255 13.5135 13.1131 11.3113

13.11 14.85 24.17 31.28 61.36 110.21 41.79 25.12 15.52 13.51 11.31

Chin - 5-12 alternating block method.xmcd

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i  0  10

precipi  output1i 

mm hr

Δt  5.45  min

10 year, 2 hour storm, 10 min. interval 120 108 96 84 precipi mm hr

72 60 48 36 24 12 0

0

6

12

18

24

30

36

42

48

54

60

i  Δt min

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Chin - 5-12 alternating block method.xmcd

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9 -11 3rd edition Chin The IDF curve for 10 year storms in Santa Fe, New Mexico, is given by:

818

i=

 td  8.54

0.76

where i is the average intensity during the event in mm/hr and td is the duration in minutes. Assuming that the mean of the rainfall distribution is equal to 44% of the rainfall distribution, estimate the triangular hyetograph for a 50 minute storm

td  50  min

tp tave = 3 1 td td tave = .44 td

tave  .44  td

Yen and Chow as well as El-Jabi and Sarraf studied a wide variety of storms and found that

tave was typically td

in the range of .32 to .51 (Chin, pg 355). Mcenroe found a much smaller time to peak for 1 hour storms with return periods equal to or greater than two years. tp  3  tave  td tp  16  min

i 

818

  td   8.54  min 

0.76



mm hr

i  37.11 

mm hr

vol  td  i  30.925  mm 1  i  t = td  i 2 p d mm ip  2  i  74.22  hr Note : It appears that the peak intensity will always equal twice the average intensity for this method

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Chin 5-11 rainfall hyetograph from IDF equation.xmcd

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Chin 9-13 3rd edition

Derive the IDF curve for Boston, Mass. using the Chen method. Use the derived IDF curve with the alternating block method to determine the the hyetograph for a storm with a return period of 10 years and a duration of 40 minutes. Use 9 time intervals to construct the hyetograph. T  10  yr min  4.444  min 9

Δt  40 

Background and Commentary: Chin notes that the Chen procedure can be used to develop IDF curves for regions where they do not exist. The method relies on data from TP - 40: the 10 yr, 1 hr rainfall, R10_1, the 10 yr. 24 hour rainfall R10_24, and the 100 yr, 1 hr rainfall R100_1 .

The IDF curve can then be expressed by an equation of the form: a

i=

 t  b 1

c1

  Tp   a = a1  R10_1  ( x  1)  log    1 10 

x=







R100_1 R10_1

Tp is the return period for the partial duration series which is assumed to be related to the return period for the annual maximum series by the relation

Tp  

R10_1  2  in

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1 1   ln  1  T    yr  

 yr  9.491  yr

R10_24  4.8  in

R100_1  3in

Chin 5-13 - Chen Method and alternating block comparison.xmcd

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  Tp   a = a1  R10_1  ( x  1)  log    1 10 





x 



R10_1 R10_24 a1  24

R100_1 R10_1

 1.5

 0.417

b1  8

c1  .74





Tp





 yr   R10_1    1 a  a1   ( x  1)  log  in   10   a  47.456

47.5

i=

( t  8)

 out1      out2 



in

.74 hr

1212

=

( t  8)

.74



mm hr

time, td minutes col. 1

intensity mm/hr col. 2

i*t mm col. 3

incremental rainfall (mm) col. 4

intensity  (mm/hr) col. 5

sorted intensity  mm/hr col 6

4.44 8.88 13.32 17.76 22.2 26.64 31.08 35.52 39.96

187.64 149.71 125.95 109.50 97.34 87.95 80.44 74.28 69.13

13.89 22.16 27.96 32.41 36.02 39.05 41.67 43.97 46.04

13.89 8.27 5.80 4.45 3.61 3.03 2.62 2.31 2.07

187.64 99.25 69.65 53.40 43.26 36.38 31.42 27.69 22.74

27.69 36.38 53.40 99.25 187.64 69.65 43.26 31.42 22.74

i  0  8 timei  out1i  min

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intensityi  out2i 

Chin 5-13 - Chen Method and alternating block comparison.xmcd

mm hr

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10 year - 40 minute Design Hyetograph - Boston Mass. 200 182 164 146 intensityi mm

128 110

hr

92 74 56 38 20

0

0.07

0.14

0.21

0.28

0.35

0.42

0.49

0.56

0.63

0.7

timei hr

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Chin 5-13 - Chen Method and alternating block comparison.xmcd

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Discussion : McCuen page 213

Soil Conservation Service 24-hour Storm Distributions The SCS developed four dimensionless rainfall distributions using the weather bureau's Rainfall Frequency Atlases (NWS 1961). The rainfall frequency data in these atlases are for areas less than 400 mi2, for durations up to 24 hr. and for frequencies from 1 to 100 years. Data analysis indicated for major storm distributions, each occurring in a different region of the country. These four distributions were given the types I, IA, II and III.

Chin 9-16 3rd edition

The precipitation resulting from a 25 year, 24 hour storm in Miami, Florida, is estimated to be 260 mm. Calculate and plot the NRCS hyetograph

DATA 

...\scs type I IA II III storms.txt

SCS dimensionless storm data from Table 4-11 McCuen

rows ( DATA )  50 i  0  48 houri  DATA i 0  hr type_Ii  DATA i 1

type_IAi  DATA i 2

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type_IIi  DATA i 3

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type_IIIi  DATA i 4

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SCS Dimensionless Storms

1 0.9 0.8 type_Ii 0.7 type_IAi type_IIi type_IIIi

0.6 0.5 0.4 0.3 0.2 0.1 0

0

2.5

5

7.5

10

12.5

15

17.5

20

22.5

25

houri hr Type I SCS storm type IA SCS storm type II SCS storm type III SCS storm

i  1  48

Time_inci  type_I_inci  type_Ii  type_Ii 1 type_II_inci  type_IIi  type_IIi 1

houri  houri 1 2 type_IA_inci  type_IAi  type_IAi 1 type_III_inci  type_IIIi  type_IIIi 1

P  260  mm

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Type III SCS Storm

60 54 48 42 type_III_inci  P mm

36 30 24 18 12 6 0

0

2.5

5

7.5

10

12.5

15

17.5

20

22.5

25

Time_inci hr

48



 type_III_inci  P  260  mm

i  0

Chin 5 - 17

The precipitation resulting from a 25 year, 24 hour storm in Atlanta, Georgia is estimated to be 175 mm. Calculate the NRCS 24 hr. hyetograph, and compare it with the 25 yr., 24 hr hyetograph for Miami calculated in problem 5-16

P  175  mm

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Type III SCS Storm

70 63 56 49 type_II_inci  P mm

42 35 28 21 14 7 0

0

2.5

5

7.5

10

12.5

15

17.5

20

22.5

25

Time_inci hr

48



 type_II_inci  P  175  mm

i  0 24



 type_II_inci  P  116.06  mm

i  0

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Problem 9-2 3rd edition Chin A rainfall record contains 50 years of measurements at 5 minute intervals. The annual maximum rainfall amounts for intervals of 5, 10, 15, 20, 25, and 30 minutes are ranked as follows where the rainfall amounts are in millimeters. Calculate the IDF curve for a return period of 40 years

 out1      out2 

time interval in minutes 15 20 25

5

10

Rank 1 2 3

30

26.2 25.3 24.2

45.8 44 42.2

60.5 58.1 55.8

72.4 69.6 66.8

81.8 78.6 75.5

89.7 86.3 82.8

Return Period 51.00 40.00 25.50 17.00

314.40 309.74 303.60 290.40

274.8 270.14 264 253.2

242 237.86 232.4 223.2

217.2 213.58 208.8 200.4

196.32 193.01 188.64 181.2

179.4 176.47 172.6 165.6

I interpolated the 40 year IDF within the Excel spreadsheet : T

duration  out1  min

Weibull Formula

T mm

intensity  out2 

hr

m allows computation of the probability of occurrence : where m is rank of data, N is number N1

of data points.

Return period is the reciprocal of the probability of occurrence

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40 Year IDF Curve 320 304 288 272 intensity 256 mm 240 hr

224 208 192 176 160

5

7.5

10

12.5

15

17.5

20

22.5

25

27.5

30

duration min

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9 - 26 Chin 3rd edition Effect of Interception on Precipitation reaching the Ground A pine forest is to be cleared for development in which all the on the site will be removed. The IDF curve for the area is given by: i=

mm 2819  hr t  16 min

where I is the rainfall intensity in mm/hr and t is the duration in minutes. The storage capacity of the trees in the forest is estimated as 5 mm, the leaf index is 6, and the evaporation rate during a storm is estimated as 0.3 mm/hr. Determine the increase in precipitation reaching the ground during a 30 minute storm that will result from clearing the site. This problem uses a rather sophisticated equation for estimating ( I ):

I = S   1  exp 

P     K  E  t  S 



where : I - interception in mm S - Available storage in mm, generally in the range of 3 - 4 mm for fully developed pine trees S  5  mm P - Amount of precipitation during storm, mm, from IDF curve K - Leaf Index, K  6 E - evaporation during storm, E  0.3 

mm hr

t - storm duration, t  30  min

i 

2819 t  16 min



mm mm  61.283  hr hr

P  i  t  30.641  mm i  t    I  S   1  exp     K  E  t  5.889  mm    S  

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precipafter  i  t  30.641  mm precipbefore  i  t  I  24.752  mm Increase  precipafter  precipbefore Increase  5.889  mm

Clearing the trees results in about 6 mm more precipitation hitting the ground

Chin 9 - 28 3rd edition For the pine forest described in problem 5-26, estimate the interception using a Horton-type empirical equation of n

the form I = a  b  P where a and b are constants and P is the precipitation amount. from Chin, pg 376 "typical values are n  1 (for most vegetative covers); a between .02 mm for shrubs and 0.05 mm for pine woods; and b between .18 and .20 for orchards and woods and 0.40 for shrubs." a  .05



b  .19

n  1

1 P      mm  5.872  mm  mm  

I  a  b  



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Chin 9 - 34 3rd edition The Green Apt Method

An area consists almost entirely of sandy loam, which typically has a saturated hydraulic conductivity of mm , an average suction head of -110 mm, a porosity of 0.45, field capacity of .190, wilting point of Ks  11  hr 0.085, and depression storage of 4 mm. The design rainfall is given as:

Interval (min) 0‐10 10‐20 20‐30 30‐40 40‐50 50‐60

Average Rainfall (mm/hr) 20 40 60 110 60 20

suction pressure: Φf  110  mm

water content : θ i 

.190  .085 2

porosity : n  .45

potential infiltration : fp = Ks 





K s  n  θ i  Φf F

The equation below relates the accumulated infiltration (occurring at the potential infiltration rate) to time F  Ks t  tp  t'p = F  n  θ i  Φf  ln 1  n  θ i  Φf   



t = 0  10  min

mm i0_10  20  hr







Δt  10  min





mm i0_10  20   Δt hr

mm mm i0_10  20  thus During this period the rainfall intensity is greater than the saturated hydraulic conductivity, Ks  11  hr hr ponding is possible. The maximum volume of infiltrated water, F, equals the volume of rain that falls in the 10 minute period. If all the rainfall infiltrates the accumulated infiltration at the end of the period will be: F10  i0_10  Δt  3.333  mm

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Using the volume of rain that fell in 10 minutes, the corresponding infiltration capacity at the end of 10 minutes is :



fp  Ks 







K s  n  θ i  Φf  mm   124.438  F10 hr



The infiltration capacity at the END of the interval exceeds the precipitation rate (20 mm/hr), thus no ponding occurs, the entire 3.33 mm infiltrates. t = 10  20  min mm i10_20  40  hr

i10_20  Δt  6.667  mm

During this period the rainfall intensity is greater than the saturated hydraulic conductivity so ponding is possible too. If all the rainfall infiltrates the cumulative infiltration, F, at the end of the period will equal what has accumulated prior to this time period plus what accumulates during this period: F20  F10  i10_20  Δt  10  mm Now use F20 to compute the infiltration rate at the end of 20 minutes



The corresponding infiltration capacity is : fp  Ks 







K s  n  θ i  Φf  mm   48.813  F20 hr



Since the infiltration capacity is greater than the precipitation rate there is no ponding during this time interval and the 6.667 mm infiltrates. The total infiltrated up to this point is: 6.67  mm  3.33  mm  10  mm

t = 20  30  min

mm i20_30  60  hr

i20_30  Δt  10  mm

mm which greater than the potential At the start of this time interval the rainfall rate increases to i20_30  60  hr infiltration at 20 minutes computed at the end of the previous time period. Thus, ponding begins immediately tp  20  min. During this interval i20_30  Δt  10  mm will fall. Compute the accumulated infiltration at the end of this time interval F30  F20  i20_30  Δt  20  mm The 10 mm that falls during this period infiltrates at the potential rate, we need to know how long this takes:

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F20   Ks  t'p = F20  n  θ i  Φf  ln 1   n  θ i  Φf





F20  Φf  ln 



F20

 Φf   n  θ i

t'p 















 1  n  θ i



 6.668  min

Ks

The 10 mm takes t'p  6.668  min to infiltrate. Therefore the equation for the cumulative infiltration at the end of t  tp  t'p = 16.7  min is F   Ks  t  tp  t'p = F  n  θ i  Φf  ln 1   n  θ i  Φf  













At the end of this period t  .5  hr, tp  20  min, t'p  6.668  min the accumulated infiltration is computed below using the root function: initial guess: F30  10  mm F30      soln30  root Ks  t  tp  t'p  F30  n  θ i  Φf  ln 1   F30 n  θ i  Φf























soln30  16.599  mm F30  soln30 F30  16.599  mm What this equation is telling us is that if during the first 30 minutes of rainfall, infiltration occurs at the potential rate for a time period equal to t  tp  t'p  16.668  min ., the volume in the ground at the end of 30 minutes will be F30  16.599  mm

The AVERAGE infiltration rate during this period was:

F30 30  min

 33.198 

mm hr

At the beginning of this interval (20-30 min) 10 mm of precipitation had fallen, 10 mm falls during this interval. However, the accumulated infiltration at 30 minutes is only F30  16.599  mm . Thus 20  mm  F30  3.401  mm did not infiltrate. Since there is 4 mm of depression storage this will go toward filling that. The depression storage remaining is 4  mm  3.401  mm  0.599  mm

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t = 30  40  min mm i30_40  110  hr

i30_40  Δt  18.333  mm

The rainfall rate is now even higher than the previous interval and ponding continues. The elapsed time at the end of this interval is t  40  min thus the accumulated infiltration at the end of the period is: F40  30  mm F40      soln40  root Ks  t  tp  t'p  F40  n  θ i  Φf  ln 1   F40 n  θ i  Φf























soln40  0.022 m F40  soln40  21.728  mm The increase in infiltrated amount is during the 30-40 minute interval F40  F30  5.129  mm . The remaining depression storage is 4  mm  3.401  mm  0.599  mm and gets used up here





Thus, the runoff during this period is: i30_40  Δt  F40  F30  .599  mm  12.605  mm

t = 40  50  min

mm i40_50  60  hr

i40_50  Δt  10  mm

The rainfall rate is still higher than the infiltration capacity so ponding continues. The elapsed time is t  50  min and the accumulated infiltration is: F50  30  mm F50      soln50  root Ks  t  tp  t'p  F50  n  θ i  Φf  ln 1   F50 n  θ i  Φf























F50  soln50  26.195  mm

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The accumulated infiltration up to this point is F40  21.728  mm thus the increase in infiltration is F50  F40  4.467  mm. The rainfall during this period is 10 mm so the difference 10  mm  F50  F40  5.533  mm





. This is all runoff, depression storage is full.

t = 50  60  min mm i50_60  20  hr

i50_60  Δt  3.333  mm

Now the rainfall has dropped to the point where it may be lower than the infiltration rate



fp  Ks 







K s  n  θ i  Φf  mm   25.435  F50 hr



Since the infiltration rate at the beginning of the period is greater than the precipitation rate the depression storage will begin to infiltrate. The elapsed time at this point is t  1  hr F60  30  mm F60      soln60  root Ks  t  tp  t'p  F60  n  θ i  Φf  ln 1   F60 n  θ i  Φf























F60  soln60 F60  30.262  mm

Now the cumulative infiltration up to this point is F50  26.195  mm thus the increase in infiltration is F60  F50  4.068  mm Thus, all the rainfall from this period plus F60  F50  i50_60  Δt  0.734  mm from





depression storage will infiltrate. There is no runoff during this time interval.

The amount remaining in depression storage at 60 minutes is 4  mm  .734  mm  3.266  mm

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Chin 5-34 Green Ampt Infiltration.xmcd

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 output1   output2    output3   

Del F

i

fp

i*Del t

Available Depression Storage

Total storage

Runoff mm

3.34

20

124.4

3.3

4

0

0

6.66

40

48.8

6.7

4

0

0

6.6

60

10

0.6

3.4

0

5.1

110

18.3

0

4

12.6

4.4

60

10

0

4

5.6

4.1

20

3.3

0

3.2

0

i  0 2  10

precipi  output1i 

mm hr

Δt  .167  hr

timei  output3i  min runoff i  output2i  mm

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Chin 5-34 - Sandy Loam Soil 15

100

10

precipi

runoffi

 mm   hr   

mm 50

0

5

0

6

12

18

24

30

36

42

48

54

0 60

timei min

precipitation runoff

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Chin 9 35 3rd edition

An area consists almost entirely of sandy clay soil, which typically has a saturated hydraulic conductivity of mm , an average suction head of 240 mm, a porosity of 0.43, field capacity of .321, wilting point of Ks  1  hr 0.221, and depression storage of 9 mm. The design rainfall is given as:

Interval (min) 0‐10 10‐20 20‐30 30‐40 40‐50 50‐60

Average Rainfall (mm/hr) 20 40 60 110 60 20

field_capacity  .321

suction pressure: Φf  240  mm

wilting_pt  .221

water content : θ i 

field_capacity  wilting_pt  0.271 2

dep stor  9  mm

porosity : n  .43

potential infiltration rate : fp = Ks 





K s  n  θ i  Φf F

NOTE : The potential infiltration rate is governed by how much water has accumulated in the soil (F). F can be produced during periods when the rainfall is less than the potential infiltration rate and/or during periods when the potential infiltration rate governs water movement into the soil. The equation below relates the accumulated infiltration (occurring at the potential infiltration rate) to time and indirectly to the potential infiltration rate F  Ks t  tp  t'p = F  n  θ i  Φf  ln 1   n  θ i  Φf

















If ponding begins immediately then we can write: Ks ( t) = F  n  θ i  Φf  ln 1 



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Chin 5-35 Sandy Clay - Green Ampt Infiltration.xmcd







F  n  θ i  Φf  



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This equation can be used to obtain the time it takes for F (infiltrating at fp) to get into the soil OR for a given time how much infiltration has accumulated.

t = 0  10  min

mm i0_10  20  hr

Δt  10  min

 i0_10  Δt  3.333  mm

mm thus During this period the rainfall intensity is greater than the saturated hydraulic conductivity, Ks  1  hr ponding is possible. At the start of the event fp is big because F is small, as F increases fp drops. If all the rainfall infiltrates, the accumulated infiltration at the end of the period will be i0_10  Δt  3.333  mm: mm i0_10  20  hr F10  i0_10  Δt  3.333  mm The corresponding infiltration capacity when F10  3.333  mm is given by:



fp  Ks 







K s  n  θ i  Φf  mm   12.448  F10 hr



The infiltration capacity at the END of the interval is less than the precipitation rate (20 mm/hr), thus ponding occurs somewhere in the 10 minute interval. As infiltration continues, F increases and fp continually decreases. Sometime during the first 10 minutes the infiltration rate was equal to the precipitation rate, this is the time when ponding started. To find the time till ponding began first find the accumulated infiltration when the potential infiltration rate was 20 mm/hr: fp = Ks 





K s  n  θ i  Φf F

mm fp  20  hr The quantity that has infiltrated into the ground when the infiltration rate equals the precipitation rate equals 20 mm/hr is: K s  Φf  n  θ i F0    2.008  mm Ks  fp



C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \infiltration and initial abstractions\Green Apt Procedure\ 9/24/2012



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The precipitation rate is constant at 20 mm/hr during the time interval, now find the time till ponding: 20 

tp 

mm  t = 2.008  mm hr p

2.008  mm

 6.024  min ponding begins to occur 6.024 minutes after precipitation begins, prior to this water mm 20  hr could infiltrate at a faster rate than it was hitting the ground surface Now find out how long it will take for this amount to infiltrate into the ground at the potential infiltration rate. The equation below is the result of integrating the infiltration rate w.r.t time. Thus, it gives the accumulated infiltration in the ground assuming infiltration started at time zero and ended when the volume in the ground was F0: F0   Ks  t'p = F0  n  θ i  Φf  ln 1   n  θ i  Φf  







F0   F0  n  θ i  Φf  ln 1   n  θ i  Φf  



t'p 









Ks

 3.064  min

It takes 3.064 min. for the 2.008 mm to infiltrate at the fp rate Now find how much water has infiltrated by the end of the first time interval.

t  10  min initial guess: F10  10  mm F10      soln10  root Ks  t  tp  t'p  F10  n  θ i  Φf  ln 1   F10 n  θ i  Φf













infiltration occurs at the potential infiltration rate for:











 t  tp  t'p  7.04  min

soln10  3.071  mm F10  soln10  3.071  mm O.K. - The amount of rain that fell was i0_10  Δt  3.333  mm and the amount infiltrated was F10  3.071  mm thus the amount that was unable to infiltrate and went to depression storage is i0_10  Δt  F10  0.262  mm

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The depression storage started at 9 mm. The amount remaining is at the end of this interval is:





dep stor_10  dep stor  i0_10  Δt  F10  8.738  mm

mm i10_20  40  hr

t = 10  20  min

i10_20  Δt  6.667  mm

During this period the rainfall intensity is greater than the infiltration rate so ponding continues. Compute the accumulated infiltration at the end of the period.

t  20  min initial guess: F20  10  mm

F10      soln20  root Ks  t  tp  t'p  F10  n  θ i  Φf  ln 1   F10 n  θ i  Φf























soln20  4.847  mm F20  soln20

The rainfall during the period is: i10_20  Δt  6.667  mm The amount infiltrated during this period is: F20  F10  1.776  mm





The amount that does not infiltrate is: i10_20  Δt  F20  F10  4.891  mm

The excess rainfall goes into depression storage, thus the depression storage remaining is:





dep stor_20  dep stor_10  i10_20  Δt  F20  F10   3.847  mm  

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mm i20_30  60  hr

t = 20  30  min

i20_30  Δt  10  mm

mm . At the start of this time interval the rainfall rate increases to i20_30  60  hr t  30  min F   Ks  t  tp  t'p = F  n  θ i  Φf  ln 1   n  θ i  Φf  













initial guess: F30  10  mm

F30      soln30  root Ks  t  tp  t'p  F30  n  θ i  Φf  ln 1   F30 n  θ i  Φf























At the end of this period the accumulated infiltration is: soln30  6.169  mm F30  soln30  6.169  mm

Quantity infiltrated: F30  F20  1.322  mm Rainfall i20_30  Δt  10  mm





Quantity unable to infiltrate i20_30  Δt  F30  F20  8.678  mm Depression storage remaining at 30 minutes :





dep stor_30  dep stor_20  i20_30  Δt  F30  F20   4.831  mm   Thus depression storage has been used up. Thus the runoff during this period is: runoff20_30  4.831  mm

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mm i30_40  110  hr

t = 30  40  min

i30_40  Δt  18.333  mm

The rainfall rate is now even higher than the previous interval and ponding continues. The elapsed time at the end of this interval is t  40  min thus the accumulated infiltration at the end of the period is: F40  30  mm F40      soln40  root Ks  t  tp  t'p  F40  n  θ i  Φf  ln 1   F40 n  θ i  Φf













soln40  7.282  mm











F40  soln40  7.282  mm

The increase in infiltrated amount is F40  F30  1.113  mm .





Thus the runoff during this period is: i30_40  Δt  F40  F30  17.221  mm

t = 40  50  min

mm i40_50  60  hr

i40_50  Δt  10  mm

The rainfall rate is still higher than the infiltration capacity so ponding continues. The elapsed time is t  50  min and the accumulated infiltration is: F50  30  mm F50      soln50  root Ks  t  tp  t'p  F50  n  θ i  Φf  ln 1   F50 n  θ i  Φf























F50  soln50  8.267  mm infiltration F50  F40  0.985  mm

rainfall : i40_50  Δt  0.01 m





runoff : runoff40_50  i40_50  Δt  F50  F40  9.015  mm

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The accumulated infiltration up to this point is F40  0.287  in thus the increase in infiltration is





F50  F40  0.985  mm. The rainfall during this period is 10 mm so the difference 10  mm  F50  F40  9.015  mm . This is all runoff.

t = 50  60  min

mm i50_60  20  hr

i50_60  Δt  3.333  mm

Now the rainfall has dropped to the point where it may be lower than the infiltration rate



fp  Ks 







K s  n  θ i  Φf  mm   5.616  F50 hr



Since the infiltration rate at the beginning of the period is greater than the precipitation rate the depression storage will begin to infiltrate. The elapsed time at this point is t  1  hr F60  30  mm F60      soln60  root Ks  t  tp  t'p  F60  n  θ i  Φf  ln 1   F60 n  θ i  Φf























F60  soln60  9.163  mm rainfall : i50_60  Δt  3.333  mm infiltration : F60  F50  0.897  mm





runoff50_60  i50_60  Δt  F60  F50  2.437  mm

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Summary of Results  output1   output2    output3   

Time rainfall min mm/hr 0

F time Del F i*Del t mm interval (hrs.) 0

20 10

Runoff mm

0.167

3.071

3.34

0

0.167

1.776

6.68

4.904

5.173

3.827

0

0.167

1.323 10.02

8.697

13.87

‐4.87

4.87

0.167

1.11

18.37

17.26

31.13

0

17.26

0.167

0.99

10.02

9.03

40.16

0

9.03

0.167

0.89

3.34

2.45

42.341

0

2.45

3.071 40

20

4.847 60

30

6.17 110

40

7.28 60

50

8.27 20

60

Excess Total available  Infiltration storage depression storage 9 0.269 0.269 8.731

9.16

i  0 2  10

precipi  output1i 

mm hr

runoff_volumei  output2i  mm

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \infiltration and initial abstractions\Green Apt Procedure\ 9/24/2012

Δt  .167  hr

timei  output3i  min

Chin 5-35 Sandy Clay - Green Ampt Infiltration.xmcd

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Chin 5-35 Sandy Clay 20

15 100 precipi

 mm   hr   

10

runoff_volumei mm

50 5

0

0

10

20

30

40

50

0 60

timei 5  min min

precipitation runoff Comparing this solution to 5-34 we see that the runoff begins sooner and lasts until the precipitation stops

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Chin 9-37 3rd edition Drainage facilities are to be designed for a rainfall return period of 10 years and a duration of 1 hour. The IDF curve is given by: t  60  min 203

i ( t) 

 t  7.24    min 

.73

i ( 60  min)  9.404 



cm hr

cm hr

P  i ( 60  min)  t  94.043  mm

where i is the rainfall intensity in cm/hr and time is the rainfall duration in minutes. The minimum infiltration rate is 10 mm/hr and the area to be drained is primarily residential lots with sizes on the order of .2 ha (0.5 ac). Use the NCRS method to estimate the total amount of runoff in cm, assuming the soil is in average condition at the beginning of the design storm. Notes : The minimum infiltration rate places the soil in the area in group A. The antecedent runoff condition is ARC II, normal conditions. From Table 5.14, pg 397 I chose a curve number of 54 CNII  54 corresponding to class A soils with 1/2 acre lot sizes. 1000 CNII = 10  .0394  S S  



25380.71   253.80  mm CNII



S  216.213  mm .2  S  43.243  mm Thus P > .2*S 2

Q 

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \SCS Procedures\ 9/24/2012

( P  .2  S)  9.665  mm P  0.8  S

Chin 5-37 and 5-38 Curve number method.xmcd

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Summary : Rainfall is P  94.043  mm and runoff is Q  9.665  mm

About

Q  10.277  % of the precipitation runs off from this catchment P

Problem 9-38 3rd edition Repeat problem 5-37 for the case in which a heavy rainfall occurs within the previous 5 days and the soil is saturated. In this case we have ARC III . We can convert the CNII to a CNIII as follows: CNII CNIII  .43  .0057  CNII CNIII  73.191

1000 CNIII = 10  .0394  S

S  



25380.71   253.80  mm CNIII



S  0.093 m

2

Q 

( P  .2  S)  33.798  mm P  0.8  S

Q  35.939  % that is, nearly 40% of incident precipitation runs off as compared to 10 % in the P previous problem In this case

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Chin 5-37 and 5-38 Curve number method.xmcd

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Chin 9-43 3rd edition A proposed 20 ha. development includes 5 ha of parking lots, 10 ha of buildings, and 5 ha of grassed areas. The runoff from the parking lots and buildings are both routed directly to the grassed areas. If the grassed area contains Type A soil in good condition , estimate the runoff from the site for a 180 mm rainfall event. Total_area  20  hectare

parking  5  hectare

buildings  10  hectare

grass  5  hectare

P  180  mm The curve number for the impervious areas is obtained from Table 5-14: CNimp  98 The curve number for Type A soils in good condition is: CNper  39 Estimate the rainfall excess from the buildings and parking lot by first estimating the storage for this area: Qimp =

( P  .2  Simp) P  .8  Simp

2

1000 CNimp = 10  .0394  Simp

25380.7  Simp    253.8  mm  5.187  mm  CNimp 

The total runoff from the impervious areas is Qimp 

 P  .2 Simp 2 P  .8  Simp

 173.922  mm

Now, the total rainfall excess crossing the grassed area is a area weighted average of the pervious and impervious areas. This is equivalent converting the depth on the impervious area to an equivalent depth over the grassed area: Volume of water onto grassy area In words - the volume of water onto the grass is made up of 1. runoff from the roofs and parking lot, 2. the direct precipitation onto the grass. The sum of these two is converted to an effective depth by dividing the volume by the area of the grass.

Peff 

Peff 

grass Total_area

= Qimp 

( parking  buildings) Total_area

 P

Qimp  buildings  P  grass  Qimp  parking grass

grass Total_area

 701.766  mm

Now, using the effective precipitation just computed

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Chin 5-42 and 5-43 Runoff from mixed area.xmcd

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Now determine the runoff from the grassed area:

Qper =

 Peff  .2 Sper 2 Peff  .8  Sper

1000 CNper = 10  .0394  Sper

 25380.7  253.8  mm Sper     CNper 

Qper 

 Peff  .2 Sper 2 Peff  .8  Sper

 379.988  mm

Now convert the depth of runoff from the pervious area to an equivalent depth over the entire catchment.

grass The runoff from the entire development is therefore: Peff  Qper   94.997  mm Total_area

The percentage of the precipitation that runs off is:

Peff P

 52.776  %

9-44 Repeat 9 - 43 for the case where the runoff from the buildings, specifically the roofs is directed directly onto the parking lots. Based on this result what can you infer about the importance of directing roof drains to pervious areas

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Chin 5-42 and 5-43 Runoff from mixed area.xmcd

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Now compute the runoff from the parking lot and roofs of buildings:

Qimp =

 P  .2 Simp 2

1000 CNimp = 10  .0394  Simp

P  .8  Simp

 25380.7  253.8  mm  5.187  mm Simp     CNimp 

Qimp 

 P  .2 Simp 2 P  .8  Simp

 173.922  mm

The runoff from the grassed area is:

Qper =

 P  .2 Sper

2

P  .8  Sper

1000 CNper = 10  .0394  Sper

25380.7  Sper    253.8  mm  396.987  mm  CNper 

Qper 

 P  .2 Sper P  .8  Sper

2

 20.34  mm

The depth of runoff from the entire catchment is : Qimp 

( buildings  parking) ( grass)  Qper   135.526  mm Total_area Total_area

The real significance of this problem is that you have 174 mm of water going onto the parking lot which may well create hazardous conditions unless proper drainage is provided. Note also that the depth of runoff over the entire catchment is larger than if the runoff from the impervious areas is conveyed over the grass and allowed to infiltrate.

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Chin 5-42 and 5-43 Runoff from mixed area.xmcd

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Qimp 

( buildings  parking) ( grass)  Qper  Total_area Total_area  75.292  % P

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Chin 5-42 and 5-43 Runoff from mixed area.xmcd

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Chin 9-5 3rd edition Development of IDF curves from data The rainfall data in the table below were compiled by determining the maximum amount of rainfall that occurred in a given time period during each given year. Use these data to develop IDF curves for 5, 10, and 25 year return periods. If these IDF curves are to be fitted to a function of the form :

i=

a tb

Estimate the values of a and b for a 5 year return period I created an Excel component and entered the data into it, below one_hr     two_hr    four_hr     six_hr   ten_hr    twelve_hr     twentyfour_hr 

1946 1947 1948 1949 1950 1951 1952 1953 1954 1955 1956 1957 1958 1959 1960 1961 1962 1963 1964 1965 1966

1 ‐hr 2‐hr 4‐hr 6‐hr 10‐hr 12 hr Maximum rainfal amounts for each time period in each year 1.7 2.9 4.6 4.8 4.8 4.8 1.9 2.6 2.6 3 3.7 3.7 1.9 2.3 2.7 2.9 3.1 3.1 1.9 2.4 2.8 4 4.7 4.9 1.9 1.9 2.2 2.5 2.5 2.5 2.1 2.4 3 3.5 4 4.2 1.7 2.5 2.7 2.8 3 3 1.7 3.6 3.6 3.6 4.3 4.3 1.9 2.1 2.8 3 3.8 3.8 2.4 3.7 4.5 4.5 4.6 4.6 1 1.3 1.6 1.7 1.7 1.7 2.7 2.8 3.2 3.7 3.7 3.8 1.3 1.5 2 2.6 3.5 3.8 2.8 3.2 3.4 4.1 5.2 5.7 2 2.1 2.8 4.1 5.6 6 1.8 2.7 2.8 2.8 2.8 2.8 1.5 2.6 2.8 3.1 3.8 4.5 1.4 2.5 3.1 3.7 4.2 4.3 1.6 1.8 2.2 2.6 3.6 3.6 1.2 1.7 2.4 2.7 3 3.1 1.4 1.5 1.5 2.3 2.5 2.7

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \Developing IDF curves\ 9/24/2012

Chin 5-5 model of IDF curves - Splines.xmcd

24 hr 5.1 4.8 3.2 5.7 3 5.1 4 4.7 3.8 5.4 1.8 4.4 4.1 6 7.9 2.8 6.4 4.5 4.1 3.3 2.8

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Commentary :

In order to develop IDF curves we must have an annual maximum rainfall series for a number of years. In this case we have the maximum rainfall amounts for 1 hr, 2 hr, 4 hr, 6 hr, 10 hr, 12 hr, and 24 hr beginning in 1946, running to 1990. First we need to compute the rainfall intensity corresponding to each amount by dividing by the corresponding time then rank the intensities from largest to smallest.

First compute the rainfall intensities In order to proceed from this point I first read out each column of data to Mathcad and gave it units. Then I divided by the duration in hours. i  0  44 rows ( one_hr)  45

one_hr_raini 

six_hr_raini 

one_hri cm two_hr_raini 

1  hr

six_hri  cm 6  hr

ten_hr_raini 

two_hr i  cm

four_hr_raini 

2  hr

ten_hri  cm twelve_hr_raini 

10  hr

twentyfour_hr_raini 

four_hri  cm 4  hr

twelve_hri  cm 12  hr

twentyfour_hri  cm 24  hr

Sort data from largest to smallest, assign the largest value the lowest rank We need to compute the corresponding rank of each intensity with the largest one getting the lowest rank. I did this using the Mathcad program below. First I sorted the data. The sort function sorts from smallest to largest . I then reversed the sort so I have the data sorted from largest to smallest. Finally, I compute the ranks of the data, sorted in this way. The algorithm below accounts for multiple occurrences of the same value.

Rankdemo ( x) 

sorted  reverse ( sort ( x) ) for i  0  length ( x)  1 m  match  xi sorted  1 ranki  mean ( m) rank

match(z, A) Looks in the reverse sorted vector, for a given value, xi, and returns the index of its position. The index is placed in the location corresponding to the values location in the unsorted vector. If there is more than one of the same value it returns the mean of the indices.

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Chin 5-5 model of IDF curves - Splines.xmcd

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Example of Rankdemo usage: Rankdemo ( x) 

sorted  reverse ( sort ( x) ) for i  0  length ( x)  1 m  match  xi sorted  1 ranki  mean ( m) rank

 0.5     0.3   0.2    0.3  x    0.1     0.3   0.4     0.2 

 1     4   6.5    4  Rankdemo ( x)    8     4   2     6.5 

Below are two vectors, the left contains the ranks of the data in the vector on the right

0 27

0

1.7

1

17.5

1

1.9

2

17.5

2

1.9

3

17.5

3

1.9

4

17.5

4

1.9

5

10.5

5

2.1

6

27

6

1.7

Rankdemo ( one_hr_rain)  7

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \Developing IDF curves\ 9/24/2012

0

0

27

one_hr_rain  7

8

17.5

8

1.9

9

8

9

2.4

10

45

10

1

11

5.5

11

2.7

12

40.5

12

1.3

13

4

13

2.8

14

12.5

14

2

15

...

15

...

Chin 5-5 model of IDF curves - Splines.xmcd

1.7



cm hr

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Note that the largest value in the data set one_hr_rain has the lowest rank and the smallest value has the highest rank ranks_1_hr  Rankdemo ( one_hr_rain)

ranks_2_hr  Rankdemo ( two_hr_rain)

ranks_4_hr  Rankdemo ( four_hr_rain)

ranks_6_hr  Rankdemo ( six_hr_rain)

ranks_10_hr  Rankdemo ( ten_hr_rain)

ranks_12_hr  Rankdemo ( twelve_hr_rain)

ranks_24_hr  Rankdemo ( twentyfour_hr_rain)

Compute return period using the Weibull formula return_period_1_hr_raini 

46  yr  rnd ( .001)  yr ranks_1_hri

return_period_4_hr_rain 

46  yr ranks_4_hr

return_period_10_hr_rain 

46  yr ranks_10_hr

return_period_2_hr_rain 

return_period_6_hr_rain 

46  yr ranks_6_hr

return_period_12_hr_rain 

return_period_24_hr_rain 

46  yr ranks_2_hr

46  yr ranks_12_hr

45  yr ranks_24_hr

one_hr_rain  sort ( one_hr_rain)

return_period_1_hr_rain  sort ( return_period_1_hr_rain)

two_hr_rain  sort ( two_hr_rain)

return_period_2_hr_rain  sort ( return_period_2_hr_rain)

four_hr_rain  sort ( four_hr_rain)

six_hr_rain  sort ( six_hr_rain)

ten_hr_rain  sort ( ten_hr_rain)

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \Developing IDF curves\ 9/24/2012

return_period_4_hr_rain  sort ( return_period_4_hr_rain)

return_period_6_hr_rain  sort ( return_period_6_hr_rain)

return_period_10_hr_rain  sort ( return_period_10_hr_rain)

Chin 5-5 model of IDF curves - Splines.xmcd

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twelve_hr_rain  sort ( twelve_hr_rain)

return_period_12_hr_rain  sort ( return_period_12_hr_rain)

twentyfour_hr_rain  sort ( twentyfour_hr_rain)

return_period_24_hr_rain  sort ( return_period_24_hr_rain)

t  1  yr 2  yr  50  yr

intensity1hr ( t)  interp ( cspline ( return_period_1_hr_rain one_hr_rain) return_period_1_hr_rain one_hr_rain t) cm intensity1hr ( 5  yr)  2.171  hr

One Hour Rainfall Intensity Curve 10

intensity1hr ( t) cm hr

1 one_hr_raini cm hr

0.1

2

6.8

11.6

16.4 t yr

21.2 

26

30.8

35.6

40.4

45.2

50

return_period_1_hr_raini yr

splined data actual data

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Chin 5-5 model of IDF curves - Splines.xmcd

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Plot rainfall amount vs return period for each duration 4 3.6

Rainfall Intensity

3.2 2.8 2.4 2 1.6

1 hour rainfall 2 hour rainfall 4 hour rainfall

1.2 0.8 0.4 0

0

5

10

15

20

25

30

35

40

45

50

Return Period (years)

1.8 1.62

Rainfall Amount)

1.44 1.26 1.08 0.9 0.72

six hour rainfall ten hour rainfall twelve hour rainfall

0.54 0.36 0.18 0

0

5

10

15

20

25

30

35

40

45

50

Return Period (years)

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Chin 5-5 model of IDF curves - Splines.xmcd

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0.5 0.455

Rainfall Amount)

0.41 0.365 0.32 0.275 0.23 0.185 0.14

twentyfour hour rainfall

0.095 0.05

0

4.5

9

13.5

18

22.5

27

31.5

36

40.5

45

Return Period (years)

Read the intensity corresponding to each return period for each duration and place value in matrix below. I used the trace function to assist in doing this. i  0  6

 1 2.2 2.73  2 1.6 1.98   4 1.01 1.12 DATA   6 .696 .83   10 .478 .590  12 .41 .499   24 .23 .297

3.2 

3.35 



1.85  1.24 



.81 

.689 



.431 

Extract each column and give it a name and units, then plot the rainfall amounts vs their return periods durationi  DATA i 0  hr C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \Developing IDF curves\ 9/24/2012

five_yri  DATA i 1 

Chin 5-5 model of IDF curves - Splines.xmcd

cm hr 7 of 11

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ten_yri  DATA i 2 

cm hr

twentyfive_yri  DATA i 3 

cm hr

3.5

5-year IDF curve 10-year IDF curve 25-year IDF curve

3.15

five_yri cm

2.8

hr

2.45

ten_yri

2.1

 cm   hr   

1.75 1.4

twentyfive_yri cm

1.05 0.7

hr

0.35 0

0

2.5

5

7.5

10

12.5

15

17.5

20

22.5

25

durationi hr

Now we are asked to evaluate the constants a and b in the equation for the IDF curve To do this I rewrote the equation in a linear form, then using the data from the 5 year IDF curve evaluated the slope and intercept of the line of best fit. Now, rewrite the equation for the IDF curve in a linear form where

1 1 is the dependent variable, is the slope and i a

b is the intercept. a 1 tb b t = =  i a a a Use slope and intercept function to get slope and intercept, then get a and b

slope_IDF  slope  duration 



slope_IDF  16.968

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \Developing IDF curves\ 9/24/2012

1 m

a 

  five_yr  1

1 slope_IDF

Chin 5-5 model of IDF curves - Splines.xmcd

 0.059 m

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 

b_over_a  intercept  duration 

b_over_a  1.221  10

1   five_yr 

5 s

3

b  b_over_a  a  7.195  10 s

m

Create a dummy plotting variable, plot the data points and the linear model on the same set of axis

x  0  hr 1  hr  25  hr pred ( x)  slope_IDF  x 

b a

The plot below shows that the data fits a linear model

Linear Model for Verifying Fit 5 4.5 4 1

3.5

five_yr hr cm

pred ( x) hr cm

3 2.5 2 1.5

Data points Linear Model

1 0.5 0

0

2.5

5

7.5

10

12.5

duration hr

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \Developing IDF curves\ 9/24/2012

15 

17.5

20

22.5

25

x hr

Chin 5-5 model of IDF curves - Splines.xmcd

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Finally, using the values of a and b plot the 5-yr IDF curve using i =

t  0  hr 1  hr  24  hr

intensity( t) 

a tb

a tb

Computed 5-yr IDF Curve with Data 3

computed IDF curve Actual Data 5 yr , 10 hour rainfall intensity, cm/hr

2.7 intensity ( t) 2.4 cm hr

five_yri cm hr

.478

2.1 1.8 1.5 1.2 0.9 0.6 0.3 0

0

2.5

5

7.5

10 t hr

12.5 

durationi hr

15

17.5

20

22.5

25

10

And finally...the interpretation of the curve. The rainfall amount expected ON THE AVERAGE once every 5 years, lasting 10 hours, will have an average intensity of .478 cm/hr

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Chin 5-5 model of IDF curves - Splines.xmcd

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0 0

1.704

1

2.629

2

2.629

3

2.629

4

2.629

5

4.381

6

1.704

return_period_1_hr_rain  7 8

1.704

9

5.75

10

1.023

11

8.364

12

1.136

13

11.501

14

3.681

15

...

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \Developing IDF curves\ 9/24/2012

 yr

2.629

Chin 5-5 model of IDF curves - Splines.xmcd

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Chin 10- 2 3rd edition

A catchment with a grass surface has an average slope of 0.8% and the distance from the catchment boundary to the outlet is 80 m. For a 30 minute storm with an effective rainfall rate equal to 70 mm/hr estimate the time of concentration using (a) the kinematic wave equation, (b) the NRCS method, (c) the Kirpich method, (d) the Izzard equation, and (e) the the Kerby Equation.

The various constants required for the methods used are listed below: mm ie  70  hr

L  80  m

So  .8%

r  .4

n  .15

Lsheet  80  m

P2_yr  30  min  70 

mm hr

P2_yr  3.5  cm

The kinematic wave equation

tc  6.99 

n  L     m  ie   mm  hr

   

0.6

 min

0.4

So

0.3

tc  24.156  min

The NRCS Method - Here I assumed sheet flow entirely. According to Chin and other authors the maximum distance over which sheet flow should be assumed is 100 m or 300 ft

sheet flow 0.8

 Lsheet  n   m   tsheet  .0288   hr  46.518  min 0.5  P2_yr  0.4    So cm  

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Chin 5-48 time of concentration.xmcd

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This is an overestimate of tc, since P2 should be the 24-h rainfall rather than the 30-min rainfall.

In Louisiana, the 2-yr, 24 hr. rainfall is about 4.6 inches. Therefore, the time of concentration for a 24 hr event is: P2_yr_La  4.6  in

sheet flow tsheet  .0288 

 Lsheet  n   m    P2_yr_La   

 hr  25.46  min

0.5

 

cm

0.8

 So

0.4

The Kirpich Equation .77

L   m tc  .019     min 0.385 So

tc  3.56  min

The Izzard Equation cr  .060

tc  10  min

K  10

Given

1

530  K   tc =

L   m 2

 ie   mm  hr

   

3

2.8  10

6



 min

mm

 cr

hr

K=

1

3

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \Rational Equation and Procedure\ 9/24/2012

ie

So

3

Chin 5-48 time of concentration.xmcd

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 Ksoln     Find K t  c  tcsoln   

Ksoln  0.301

tcsoln  40.468  min

2

2

m m which is greater than the allowable maximum of 3.9  so the Izzard However note that ie  L  5.6  hr hr equation is not strictly applicable.

The Kirby Equation  L  r  m  tc  1.44    So   

.467

 min

tc  22.433  min

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Chin 5-48 time of concentration.xmcd

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Chin 10-21

2

In a watershed with a area of A  315  km the measured rainfall and runoff in a small river leaving the catchment are given below.

 out1      out2  Time (hr) Rainfall (cm) Effective Rainfall River Flo w  (m3/sec) 0 100 1 0.5 0 100 2 300 3 2.5 2 700 4 1000 5 2.5 2 800 6 600 7 0.5 0 400 8 300 9 200 10 100 11 100

i  0  10

3

timei  out1i  hr

DROi  out2i 

m sec

Next we can pass smooth curve through the data points using a linear spline with interpolation vs  lspline ( time DRO)

interp ( vs time DRO t)

t  0  hr .1  hr  10  hr interpolation value C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \Unit Hydrographs and Convolution\Synthetic Unit Hydrographs\Unit Hydrographs and Convolution\

Chin 5-62 base flow separation, UH, convolution.xmcd

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900 800 interp ( vs time DRO t) 700 3 600 m 500 sec 400 DROi 300 3 m 200 sec 100 0  100

0

1

2

3

4

5 t hr



6

8

9

10

hr

10  hr

 The volume obtained by integrating the spline is:  0hr

7

timei

9

interp ( vs time DRO t) dt  3.333  10  gal

10  hr

 A  runoff =  0hr

interp ( vs time DRO t) dt

10  hr

runoff 

  0

interp ( vs time DRO t) dt A

 4.006  cm

The total rainfall is 6 cm, the effective rainfall (runoff) is 4 cm, thus we have 2 cm of losses. Assume the losses are evenly distributed over the entire rainfall event we get the effective rainfall from hour 1 to hour 3. Thus the duration of effective rainfall is 2 hours, which is also the duration of the UH hydrograph.

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \Unit Hydrographs and Convolution\Synthetic Unit Hydrographs\Unit Hydrographs and Convolution\

Chin 5-62 base flow separation, UH, convolution.xmcd

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Since the total runoff is 4 cm over the catchment area we can produce a Unit Hydrograph by dividing all ordinates by four

 

vs  lspline  time 

 

DRO   4 

interp  vs time 

DRO  t 4 

t  0  hr .1  hr  10  hr interpolation value

250

 

interp  vs time 

m

3

sec

4

4

220  t 190



3

160

sec

130

m

DRO

DRO

100 i

70 40 10  20  50

0

1

2

3

4

5 t



hr 10  hr

 The volume obtained by integrating the spline is:   0hr

10  hr

 A  runoff =   0hr

10  hr

runoff 

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \Unit Hydrographs and Convolution\Synthetic Unit Hydrographs\Unit Hydrographs and Convolution\

   0

6

7

interp  vs time 

 

10

hr



interp  vs time 

9

timei

interp  vs time 



8

DRO  8 t dt  8.333  10  gal 4 

DRO 4

t dt



DRO  t dt 4 

A

Chin 5-62 base flow separation, UH, convolution.xmcd

 1.001  cm

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 0     50   150     225   175  m 3 UH1hr     125  sec  cm  75     50   25     0 





rows UH1hr  10

 0     50   150   225     175   125    m3 UH1hr_mod   75    50  sec  cm    25   0     0   0     0 

1    3 P     cm 4    2 



rows ( P)  4



rows UH1hr  rows ( P)  1  13

1    3  4  2    0  0    P   0   cm 0    0  0    0  0    0 

1    2  3  4    5  6    time   7   hr 8    9   10     11   12     13 

n  0  12 n

Qn 



i  0 C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \Unit Hydrographs and Convolution\Synthetic Unit Hydrographs\Unit Hydrographs and Convolution\

 Pi  UH1hr_mod  n i 

Chin 5-62 base flow separation, UH, convolution.xmcd

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rows ( Q)  13

vs  lspline ( time Q)

interp ( vs time Q t)

t  0  hr .1  hr  16  hr interpolation value

interp ( vs time Q t) m

3

sec

Q m

3

sec

2 10

3

1.8 10

3

1.6 10

3

1.4 10

3

1.2 10

3

1 10

3

800 600 400 200 0

0

1.6

3.2

4.8

6.4

8 t hr



9.6

12.8

14.4

16

hr

13  hr

  0hr

11.2

time

9

interp ( vs time Q t) dt  8.309  10  gal

13  hr

  0hr

interp ( vs time Q t) dt A

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \Unit Hydrographs and Convolution\Synthetic Unit Hydrographs\Unit Hydrographs and Convolution\

Chin 5-62 base flow separation, UH, convolution.xmcd

 9.985  cm

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Chin 1-22 Snyder Unit hydrograph

Develop a 3-hour Snyder Unit Hydrograph for a 100 km2 watershed where the main stream is 15 km long and the distance from the watershed outlet to the point on the the stream nearest the centroid of the watershed is 7 km. Assume Cp  0.7 and Ct  1.8 A  100  km

2

L  15  km

Lc  7  km

The watershed area is: A  100  km

Cp  0.7

Ct  1.8

tR  3  hr

2

The length of the main stream is : L  15  km The length of the stream from outlet to a point on the stream nearest the centroid is: Lc  7  km The Snyder peaking factor is Cp  0.7 and Ct  1.8 The lag time for the watershed, defined as the time interval from the centroid of the rainfall hyetograph to the peak

 L Lc  flow value on the hydrograph is: tl  .75  Ct      km km 

0.3

 hr  5.454  hr

tl This lag time corresponds to an excess rainfall duration of: tr   0.992  hr , however we want an excess rainfall 5.5 duration of tR  3  hr





Therefore the adjusted lag time is: tlR  tl  .25  tR  tr  5.956  hr

  tR  hr tlR    hr  7.456  hr The time until the peak flow rate is therefore: tp    hr   2 A

Cp  The peak flow rate is: QpR  2.75 

km tlR

2

3

3

m 3 ft   1.141  10  sec sec

hr

tlR   The initial time base of the hydrograph is : tb   72  3    hr  89.867  hr hr  

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \Unit Hydrographs and Convolution\Synthetic Unit Hydrographs\Snyder Unit Hydrograph\ 5/9/2012

Chin 5-65 3 -hr Synder UH.xmcd

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tb tb should be on the order of 5, in this case  15.089 . When the value is >> 5 it is tlR tlR

According to Chin values of

suggested to take

tb = 5 . This results in a time base of tb  5  tlR  29.779  hr tlR

Now compute the width of the hydrograph at 50% and 75% of the computed peak flow

W50 

2.14

      

QpR m

3

sec

A km

2

      

1.08

 hr  7.247  hr

W75 

1.22

      

      

QpR m

3

sec

A km

2

3

tinit  0  hr

qinit  0 

ft sec  in

1.08

hr  4.132  hr

3

qfinal  0 

ft sec  in

EXAMPLE :

Create a point that has a time value of : t1  tp 

3 W50 ft  5.04  hr and a flow value of 50%  QpR  570.711  sec 3

2 Another point with the same flow is plotted at: t2  tp  W50   12.287  hr 3.

The 75% points are done the same way:



W75 



3 

t3   tp 



t3  6.079  hr

2 t4  tp  W75  3

t4  10.21  hr

3

75%  QpR  856.066 

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \Unit Hydrographs and Convolution\Synthetic Unit Hydrographs\Snyder Unit Hydrograph\ 5/9/2012

ft sec

Chin 5-65 3 -hr Synder UH.xmcd

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 tinit   hr    t1  hr    t  3  hr    tp DATA   hr    t  4  hr    t2   hr    tb  hr  

  ft   sec  .5  QpR   3 ft  sec  .75  QpR   3  ft  sec  QpR   3 ft  sec  .75  QpR   3  ft  sec  .5QpR   40  3 ft   sec  qfinal   3 ft  sec  qinit 3

i  0  6

3

timei  DATA i 0  hr

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \Unit Hydrographs and Convolution\Synthetic Unit Hydrographs\Snyder Unit Hydrograph\ 5/9/2012

qi  DATA i 1 

ft sec

extract and rename hydrograph points

Chin 5-65 3 -hr Synder UH.xmcd

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vs  lspline ( time q)

interp ( vs time q t)

create spline and interpolate points

t  0  hr .01  hr  tb

Three Hour Snyder Unit Hydrograph 1.08 10

3

960

qi ft

1.2 10

3

840

3

720

sec

interp ( vs time q t) ft

3

sec

600 480 360 240 120 0

0

3

6

9

12

15 timei hr



18

21

24

27

30

t hr

Basic Snyder Unit Hydrograph - straight line segments Snyder Unit Hydrograph - smooth curve

Now we need to check the runoff volume (area under smooth curve) to be sure it is approximately 1 area-inch Reqd_Vol  A  1  cm 6

Reqd_Vol  1  10  m

3

This is the needed volume t

b 6 3 The volume obtained by integrating the spline is:  interp ( vs time q t) dt  1.093  10  m 0  hr t

b  interp ( vs time q t) dt 0  hr A

 1.093  cm

The two values are very close

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Chin 5-65 3 -hr Synder UH.xmcd

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Chin 10-34

The Storage-Indication Procedure for Level Pool Reservoir Routing This is a semi-graphical procedure for developing an outflow hydrograph from a level pool reservoir. It is based on conservation of mass. What we have here is water entering a level pool reservoir and leaving at the same time I1 and I2 are sequential inflow values. O is the reservoir outflow and S(O) is volume stored at time t mass in - mass out = time rate of change of storage IO=

ΔS Δt

This equation can be rewritten in the form below where the inflow and outflow are computed as the average of 2 sequential values in time. The subscripts 1 and 2 denote sequential values of inflow, outflow, and storage at the beginning and end of a time interval. I1  I2 O1  O2  Δt   Δt = S2  S1 2 2

The storage - indication procedure can be used to develop an outflow hydrograph when a storage-outflow relationship is available. A storage-outflow relationship is used to develop the "storage indication curve" which is a relationship between outflow, O, and the quantity: 2  S1 Δt

 O1 or

2  S2 Δt

 O2 .

A storage - outflow curve can be developed if the plan area of the reservoir is known along as well as a functional relationship between outflow and reservoir characteristics, such as height above an outlet weir.

Assume that the storage-discharge relationship for a reservoir is given by an equation of the form: S ( O) = 1200  ( O)

1.035

The initial storage and outflow of the reservoir are both zero. The procedure used is as follows. First rewrite the basic mass balance equation in the form below.

 I1  I2  Δt   2S1  O1  Δt =  2S2  O2  Δt 1. Compute I1 + I2









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1

2





1



2

3. Read the value of O corresponding to 2S2  O  Δt from the curves below At this point the current values become the previous values and the cycle repeats





4. Read the value of 2S1  O  Δt corresponding to O from step 3, this is used in the next cycle

5. Storage corresponding to O is obtained from the S - O equation

Below we first develop the needed plots using the storage-outflow relationship i  1  400

3

Develop a vector of outflows using subscripts. Oi  i  1 

ft sec

Now compute a vector of corresponding storage values from the equation given. Note that the form of this equation depends on the nature of the outlet from the basin, pipe, weir, etc.

 Oi   ft3   sec   

Si  1200  

1.035

 ft

3

2 S  O, plot this quantity against the Δt outflow from the basin. On the same set of axis plot the storage against the outflow from the basin. Use a time interval of Δt  10  min. Compute the storage outflow function

2 Si Δt

 Oi

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Storage-Indication Curves-Small Values 6 10

500 2*S/Dt + O storage - outflow curve

5.4 10

5

400

4.8 10

5

350

4.2 10

5

300

3.6 10

5

250

3 10

450

2 Si Δt ft

 Oi 3

sec

5

Si

5

200

2.4 10

5

150

1.8 10

5

100

1.2 10

5

6 10

50 0

0

5

10

15

20

25

30

35

40

45

4

0 50

Oi ft

3

sec

The Excel spreadsheet with the required data is given below

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ft

3

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 output1   output2    output3   

2S/Dt - O

I1 0 50 100 150 200 250 300 270 240 210 180 150 120 90 60 30 0 0 0 0 0

0 0 33 122.4 249.8 402.6 572.6 754.6 890.2 941 934.8 890.4 820.4 732.8 633.6 526.4 414 298 199.8 134

2Si/Dt + O Outflow (O) 0 0 50 8.5 183 30.3 372.4 61.3 599.8 98.6 852.6 140 1122.6 184 1324.6 217.2 1400.2 229.6 1391 228.1 1324.8 217.2 1220.4 200 1090.4 178.8 942.8 154.6 783.6 128.6 616.4 101.2 444 73 298 49.1 199.8 32.9 134 22.3

New 2S/Dt - O 0 33 122.4 249.8 402.6 572.6 754.6 890.2 941 934.8 890.4 820.4 732.8 633.6 526.4 414 298 199.8 134 89.4

Storage(O) 0 12450 45810 93230 150400 213900 281400 332000 351100 348900 332000 306000 273300 236200 196300 154400 111300 74660 50080 33520 0

Now, output the inflow, outflow and storage values over time and plot the needed curves. Note that the storage is read from the right hand Y axis.

i  0  19

inflow i  output1i 

ft

3

sec

outflowi  output2i 

storagei  output3i  ft

ft

3

sec

3

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Storage Indication Procedure 1 10

4

9 10

3

8 10

3

6.3

7 10

3

5.4

6 10

3

4.5

5 10

3

9 Inflow Hydrograph outflow hydrograph storage

8.1 7.2 inflow i m

3

sec

outflow i

3

3

3.6

4 10

3

sec

2.7

3 10

3

1.8

2 10

3

0.9

1 10

3

m

0

0

0.35

0.7

1.05

1.4

1.75

2.1

2.45

2.8

3.15

storagei m

0 3.5

i  Δt hr

Note: The peak of the outflow hydrograph should be the intersection point with the recession leg of the inflow hydrograph. The curves above are a bit off off because of accumulated error in reading the plots above.

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Chin 10-35 3rd edition Muskingham Channel Routing

The flow hydrograph at the upstream end of a section of channel is given by:

 out1      out2 

Time (min) 0 30 60 90 120 150 180 210 240 270 300 330 360 390 420 450 480 510

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \routing procedures\Muskingham Routing\ 9/24/2012

Flow cms 0 12.5 22.1 15.4 13.6 12.4 11.7 10.8 9.9 8.4 8.1 7.5 4.2 0 0 0 0 0

Chin 5-75 Muskingham Routing.xmcd

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i  0  16 3

timei  out1i  min

inflow i  out2i 

m sec

Use the Muskingum Method to estimate the hydrograph 1000 meters downstream from the starting channel section. Assume that X  0.3 and K  35  min

The time step ∆t is chosen such that 2  K  X  Δt  2  K  ( 1  X) 2  K  X  21  min 2  K  ( 1  X)  49  min We will use Δt  30  min

C1 

Δt  2  K  X  0.114 2  K  ( 1  X)  Δt

C2 

Δt  2  K  X  0.646 2  K  ( 1  X)  Δt

C3 

2  K  ( 1  X)  Δt  0.241 2  K  ( 1  X)  Δt

C1  C2  C3  1 Now, spline both the inflow and outflow data, plot the data points as well as the splines 3

Outflow0  0 

m sec

j  1  16 Outflowj  C1  inflow j  C2  inflow j  1  C3  Outflowj  1 vs  lspline ( time inflow ) vs2  lspline ( time Outflow)

tt  0  min 5  min  490  min inflow2 ( tt)  interp ( vs time inflow tt) outflow2 ( tt)  interp ( vs2 time Outflow tt)

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Chin 5-75 Muskingham Routing.xmcd

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Muskingum Flow Routing 25 inflow2 ( tt) m

3

22 19

sec

inflow j m

3

16 13

sec

outflow2 ( tt) 10 m

3

7

sec

Outflowj m

3

sec

4 1 2 5

0

0.9

1.8

2.7

3.6 tt hr



4.5

timej hr



tt hr

5.4 

6.3

7.2

8.1

9

timej hr

splined inflow inflow data points splined outflow ouflow data points

Note that the splined curve drops below zero between 0 and 30 minutes.

drop_in_peak_flow 

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \routing procedures\Muskingham Routing\ 9/24/2012

22  19 22

 13.636  %

Chin 5-75 Muskingham Routing.xmcd

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Chin 10-36

Measured flows at an upstream and downstream section of a river are as follows

 out1   out2    out3   

Time (hr) Upstream Flow (m3/sec) Downstream flow (m3/sec) 0 10 10 30 10 10 60 25 12.2 90 45 23.4 120 31.3 35.1 150 27.5 32.1 180 25 28.8 210 23.8 26.2 240 21.3 24.3 270 19.4 22.1 300 17.5 20.1 330 16.3 18.3 360 13.5 16.6 390 12.1 14.4 420 10 12.6 450 10 11 480 10 10.3 510 10 10.1 540 10 10.1 570 10 10 600 10 10

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Chin 5-76 Muskingum Routing.xmcd

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i  0  20 Timei  out1i  hr 3

Inflow i  out2i 

m sec

3

Inflow 0  10 

3

outflowi  out3i 

m sec

Δt  30  hr

outflowi  10

m sec

3

m s 12.2 10

Reach Hydrographs

23.4

45

35.1

41.5

32.1 28.8

38

26.2

Inflow i

24.3 22.1

m

20.1

sec

18.3

34.5

3

31

outflow i

16.6

27.5

m

3

24

14.4 12.6

sec

20.5

...

17 13.5 10

0

60

120

180

240

300

360

420

480

540

600

Timei hr

X  .19

K  40  hr

K  Δt  K  1 3 2  K  X  Δt  2  K  ( 1  X)  1

C1 

Δt  2  K  X 2  K  ( 1  X)  Δt

 0.156

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \routing procedures\Muskingham Routing\ 5/14/2012

C2 

Δt  2  K  X 2  K  ( 1  X)  Δt

 0.477

Chin 5-76 Muskingum Routing.xmcd

C3 

[ 2  K  ( 1  X)  Δt] 2  K  ( 1  X)  Δt

 0.367

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i  1  20 3

O0  10 

m sec

Oi  C1  Inflow i  C2  Inflow i 1  C3  Oi 1

Muskingum Method 40 37 34 outflow i 31

 m3    28  sec  Oi

25

3

22

sec

19

m

16 13 10

0

60

120

180

240

300

360

420

480

540

600

Timei hr

Actual Outflow Muskingham outflow

20



 outflowi  Oi 2  0.059

i  0

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Chin 5-76 Muskingum Routing.xmcd

m s

6

2

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Chin 10-7 3rd edition The surface of a 2 hectare catchments is characterized by a runoff coefficient of 0.5, a Manning's equation for overland flow of .25, an average overland flow length of 60 m, and an average slope of 0.5%. Calculate the time of concentration using the kinematic wave equation. The drainage channel is to be sized for the peak runoff resulting from a 10 year rainfall event, and the 10 year IDF curve is given as:

150

Equation for IDF curve i =

( t  8.96)

0.78

In this equation i is the average rainfall intensity in cm/hr and t is the duration in minutes. The minimum time of concentration is 5 min. Determine the peak runoff rate.

tc = 6.99 

( n  L)

0.6

ie  S0

0.3

Discussion : Here we have to solve two equations simultaneously. The kinematic wave equation contains the rainfall intensity i BUT the rainfall intensity to be used depends on the time of concentration,....soooo we have to solve both equations simultaneously. Known values: A  2  hectare

n  .25

C  0.5

i  15 

cm hr

S0  .5% L  60  m

mm ie  10  hr

tc  25.6  min

Given

tc

   6.99         

Chin 5-52.xmcd 9/24/2012

    min  .4 ie   0.3   S0  mm   hr   n  L     m

0.6

= 1

ie = C  i

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i=

150

  tc   8.96  min 

0.78



cm hr

 tcsoln     iesoln   Find  tc ie i    isoln 

tcsoln  41.977  min

isoln 150

cm      0.78 hr   tcsoln    8.96     min  

mm isoln  69.921  hr

mm iesoln  34.96  hr

tcsoln 1

iesoln 1 C  isoln

0.6   n  L         m  min 6.99  .4  iesoln    0.3    S0   mm       hr  

1

Therefore the time of concentration is 42 minutes, the rainfall intensity is 69 mm/hr and the effective rainfall intensity is 35 mm/hr.

Chin 5-52.xmcd 9/24/2012

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Chin Problem 10-18 - Unit Hydrograph Computations Below is given a vector of UH data for a "15 minute UH". Note however, that the time interval between values for the UH data is 30 minutes. To the right I have interpolated the UH data to 15 minute time intervals so as to correspond with the duration of the event causing the original UH. The UH is to be applied to 2.1 km2 catchment

15 minute UH with 30 minute time intervals

 0     1.4   3.2     1.5   1.2     1.1   1.0  m 3 UH     .66  sec  .49     .36   .28     .25   .17   0   

 0     30   60     90   120     150   180  t     min  210   240     270   300     330   360   390   

C:\Mathcad application areas \Mathcad application areas\Fluidsopen channels-hydrology\Problems for Chins Text\Chin 3rd edition Chap 10\

15 minute UH interpolated to 15 minute time intervals

 0     .7   1.4     2.3   3.2     2.35   1.5     1.35   1.2   1.15     1.1   1.05     1.0  3 m UH15   .83     sec  .66   .58     .49   .41     .36   .32     .28   .265     .25   .21     .15   .09     0 

Chin problem 10-18 UH 15 min.xmcd

 0     .25   .50     .75   1.00     1.25   1.5     1.75   2.00   2.25     2.5   2.75     3.00  t15   3.25   hr    3.5   3.75     4.00   4.25     4.5   4.75     5.00   5.25     5.5   5.75     6.00   6.25     6.5 

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a. Verify that the unit hydrograph above is, in fact, consistent with a 1 cm rainfall over the catchment. b. estimate the runoff hydrograph for a 15 minute rainfall event of 2.8 cm c. estimate the the runoff hydrograph for a 30 min rainfall excess of 10.3 cm

Part a. Verify that the unit hydrograph above is, in fact consistent with a 1 cm rainfall over the catchment. We can do this by plotting the coordinates of the UH, integrate the area under the curve to get a volume and compare this to the volume obtained by multiplying 1 cm by an area of 2.1 km2 NOTE : We are not convolving this UH with a rainfall burst but simply multiplying each UH ordinate by 1 cm. Thus it is not necessary to interpolate the UH to 15 minute intervals. Below are plotted both sets of data.

30 minute UH data set: vs  lspline ( t UH), interp ( vs t UH time)







same data set interpolated to 15 minute intervals: vs2  lspline t15 UH15 , interp vs2 t15 UH15 time



time  0  hr 1  min  390  min interpolation value

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Chin problem 10-18 UH 15 min.xmcd

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Unit Hydrograph, 2.1 Square Km Urban Catchment 3.5 interp ( vs t UH time) m

3.15

3

2.8

sec

UH m

2.45

3

2.1

sec

UH15 m

1.75

3

1.4

sec





interp vs2 t15 UH15 time 1.05 m

3

0.7

sec

0.35 0

0

40

80

120

160 time min

200 

t



t15

min min

240 

280

320

360

400

time min

30 minute interval UH spline curve Unit Hydrograph Data Points UH data 15 minute interval 15 minute UH spline 390  min

  0hr

390  min

  0hr

6

interp ( vs t UH time) dtime  5.568  10  gal





6

interp vs2 t15 UH15 time dtime  5.531  10  gal

2

6

Check : 2.1  km  1  cm  5.548  10  gal

Result : Both data sets give essentially the same result

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Chin problem 10-18 UH 15 min.xmcd

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Part b: estimate the runoff hydrograph for a 15 minute rainfall event of 2.8 cm Here we simply multiply the coordinates of the UH by 2.8, then integrate to get the volume 30 minute UH data set: vs  lspline ( t UH), interp ( vs t UH time)







same data set interpolated to 15 minute intervals: vs2  lspline t15 UH15 , interp vs2 t15 UH15 time



time  0  hr 1  min  390  min interpolation value

Once again, since we are not convolving the UH with a hyetograph it is not necessary to interpolate the UH to 15 minute intervals. If we did we would have twice as many points

2.8 cm Excess Rainfall 9 interp ( vs t 2.8  UH time) m

8.1

3

7.2

sec

2.8  UH m

6.3

3

5.4

sec

2.8  UH15 m

4.5

3

3.6

sec





interp vs2 t15 2.8  UH15 time 2.7 m

3

1.8

sec

0.9 0

0

40

80

120

160

time min



200 t



t15

min min

240 

280

320

360

400

time min

spline for 30 minute UH data 30 minute UH data 15 minute UH data spline for 15 min UH data

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Chin problem 10-18 UH 15 min.xmcd

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The volume of excess rainfall is obtained by finding the area under the curve to get the volume: 390  min

  0hr

390  min

  0hr

7

interp ( vs t 2.8  UH time) dtime  1.551  10  gal





7

interp vs2 t15 2.8  UH15 time dtime  1.546  10  gal

2

7

Check : 2.1  km  2.8  cm  1.553  10  gal

Part c: estimate the the runoff hydrograph for a 30 min rainfall excess of 10.3 cm Definition(s) of Convolution A procedure utilizing a unit hydrograph to produce a direct runoff hydrograph from a multiperiod rainfall event. Bedient and Huber - Hydrology and Floodplain Analysis. The process by which the design storm is combined with the transfer function (unit hydrograph) to produce a direct runoff hydrograph - McCuen - Hydrologic Analysis and Design. Analytically speaking convolution is referred to as the theory of linear superpositioning. Conceptually, it is a process of multiplication, translation with time, and addition. The rainfall intensity should be logged using the same time interval as the rainfall that generated the UH. For example, if we have a 1-hr unit UH then the storm event for which the DRO hydrograph is desired should have rainfall intensities logged on a 1 hr. time interval. My explanation - We assume that each burst of rainfall occurring on a basin within an interval of time, t comes off as direct runoff in accordance with a specified unit hydrograph for that basin. That is, each rainfall burst is multiplied by every ordinate of the unit hydrograph. The unit hydrograph is then translated in time so that it's start coincides with the next burst. The process is repeated for each burst of rain. The runoff contributions of each rainfall burst are then summed with respect to the time interval in which they occurred. A good, simple example is given in McCuen, pg. 507 .

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Chin problem 10-18 UH 15 min.xmcd

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The governing equation for developing a storm hydrograph is called the convolution integral which is n

written in discrete form in Mathcad* as: Qn =



 Pi  Un i

i  0

where: n is equal to the number of rainfall pulses + number of unit hydrograph ordinates - 1 P is the amount of rain within a time interval t U is a vector of unit hydrograph ordinates taken at the same t interval. * the subscripting of the variables P and U is different in Mathcad than in most texts because the first value in a vector in Mathcad is the "zeroth" and not the first.

Because we have a 15 minute UH we need to break the rainfall into two 15 minute bursts of 5.15 cm. We assume that each burst of rainfall comes off the catchment in quantities and in time as dictated by the unit hydrograph. In addition we have to interpolate the unit hydrograph values given so as to provide readings every 15 minutes. The time interval within the UH has to match the time interval of the rainfall burst which created the UH

Finally we create two new vectors by padding the original precipitation and UH vectors out to 28 elements with zeros

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Chin problem 10-18 UH 15 min.xmcd

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 0     .7   1.4     2.3   3.2     2.35   1.5     1.35   1.2   1.15     1.1   1.05     1.0  3 m U   .83     sec  cm  .66   .58     .49   .41     .36   .32     .28   .265     .25   .21     .15   .09     0 

 .25     .5   .75     1   1.25     1.5   1.75     2   2.25     2.5   2.75   3     3.25   3.5    t   3.75   hr  4     4.25   4.5     4.75   5     5.25   5.5     5.75   6     6.25   6.5     6.75   7     7.25 

 5.15    cm  5.15 

P  

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Chin problem 10-18 UH 15 min.xmcd

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rows ( P)  2

rows ( U)  27

rows ( P)  rows ( U)  1  28

Now create two new vectors with rows ( P)  rows ( U)  1  28 elements

 5.15    cm  5.15 

P  

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 0   .7     1.4   2.3     3.2   2.35     1.5   1.35     1.2   1.15     1.1   1.05     1.0   .83  m 3 U     .66  sec  cm  .58     .49   .41     .36   .32   .28     .265   .25     .21   .17     .085   0     0 

Chin problem 10-18 UH 15 min.xmcd

 0   .25     .5   .75     1   1.25     1.5   1.75     2   2.25     2.5   2.75     3   3.25  t     hr  3.5   3.75     4   4.25     4.5   4.75   5     5.25   5.5     5.75   6     6.25   6.5     6.75 

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j  0  1

n  0  27

Below I create the new P vector with zeros after the 2 nonzero elements

Pn 

0 if n  1 Pn otherwise

n

Qn 



 Pi  Un i

i  0

rows ( Q)  28 Next we can pass smooth curve through the data points using a linear spline with interpolation

vs  lspline ( t Q)

interp ( vs t Q time)

time  0  hr .1  hr  7  hr interpolation value

31

Qn m

3

sec

interp ( vs t Q time) m

3

sec

0

27.9

1.5

24.8

3

21.7

4.5

18.6

6

15.5

7.5

12.4

9

9.3

10.5

6.2

12

3.1

13.5

0

0

0.7

1.4

2.1

2.8

3.5

4.2

4.9

5.6

6.3

Pj cm

7

tn time tj   hr hr

hr

computed direct runoff hydrograph - points splined direct runoff hydrograph Rainfall Hyetograph

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Chin problem 10-18 UH 15 min.xmcd

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Caution : When using splines of any type ALWAYS plot the spline against the data points it passes through to make sure (visually) there is good agreement. The nice thing about using a spline in Mathcad is that it can be numerically integrated by Mathcad because internally Mathcad "knows" the various equations that make it up. In this case, once we have visually checked to be sure the spline "fits" the data we can estimate the volume of direct runoff by integrating the spline.

7  hr

 7 The volume obtained by integrating the spline is:  interp ( vs t Q time) dtime  5.685  10  gal 0hr 2

7

Check : 10.3  cm  2.1  km  5.714  10  gal

7  hr

2  10.3  cm  2.1  km   interp ( vs t Q time) dtime 0hr

10.3  cm  2.1  km

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2

Chin problem 10-18 UH 15 min.xmcd

 0.503  %

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Chin Problem 10-18 - Unit Hydrograph Computations Use the 15 minute UH hydrograph for a 2.1 km2 urban catchment, given below to:

 0     1.4   3.2     1.5   1.2     1.1   1.0  m 3 UH     .66  sec  .49     .36   .28     .25   .17   0   

 0     30   60     90   120     150   180  t     min  210   240     270   300     330   360   390   

The "15 minute UH" refers to the duration of the storm that created the UH data. The time interval between sequential points within the UH is 30 minutes.

a. Verify that the unit hydrograph above is, in fact consistent with a 1 cm rainfall over the catchment. b. estimate the runoff hydrograph for a 15 minute rainfall event of 2.8 cm c. estimate the the runoff hydrograph for a 30 min rainfall excess of 10.3 cm

Part a. Verify that the unit hydrograph above is, in fact consistent with a 1 cm rainfall over the catchment. We can do this by plotting the coordinates of the UH, integrate the area under the curve to get a volume and compare this to the volume obtained by multiplying 1 cm by an area of 2.1 km2 vs  lspline ( t UH)

interp ( vs t UH time)

time  0  hr .1  hr  10  hr interpolation value

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Chin problem 10-18 UH 30 min interval.xmcd

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Unit Hydrograph, 2.1 Square Km Urban Catchment 4 3.6 3.2 interp ( vs t UH time) 2.8 m

3

2.4

sec

UH

2

3

1.6

sec

1.2

m

0.8 0.4 0

0

40

80

120

160

200 time min



240

280

320

360

400

t min

Spline curve Unit Hydrograph Data Points

390  min

  0hr

6

interp ( vs t UH time) dtime  5.568  10  gal

390  min

  0hr

4

interp ( vs t UH time) dtime  2.108  10  m

2

3

6

Check : 2.1  km  1  cm  5.548  10  gal

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Chin problem 10-18 UH 30 min interval.xmcd

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Part b: estimate the runoff hydrograph for a 15 minute rainfall event of 2.8 cmHere we simply multiply the coordinates of the UH by 2.8, then integrate to get the volume vs  lspline ( t UH)

interp ( vs t UH time)

time  0  hr .1  hr  10  hr interpolation value

2.8 cm Excess Rainfall 9 8.1 interp ( vs t 2.8  UH time) m

3

sec

2.8  UH

7.2 6.3 5.4 4.5

3

3.6

sec

2.7

m

1.8 0.9 0

0

40

80

120

160

200

time min



240

280

320

360

400

t min

The volume of excess rainfall is obtained by finding the area under the curve to get the volume: 390  min

  0hr

7

interp ( vs t 2.8  UH time) dtime  1.551  10  gal

390  min

  0hr

4

interp ( vs t 2.8  UH time) dtime  5.869  10  m

2

3

7

2.8  cm  2.1  km  1.553  10  gal

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Chin problem 10-18 UH 30 min interval.xmcd

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Definition(s) of Convolution A procedure utilizing a unit hydrograph to produce a direct runoff hydrograph from a multiperiod rainfall event. Bedient and Huber - Hydrology and Floodplain Analysis. The process by which the design storm is combined with the transfer function (unit hydrograph) to produce a direct runoff hydrograph - McCuen - Hydrologic Analysis and Design. Analytically speaking convolution is referred to as the theory of linear superpositioning. Conceptually, it is a process of multiplication, translation with time, and addition. The rainfall intensity should be logged using the same time interval as the rainfall that generated the UH. For example, if we have a 1-hr unit UH then the storm event for which the DRO hydrograph is desired should have rainfall intensities logged on a 1 hr. time interval. My explanation - We assume that each burst of rainfall occurring on a basin within an interval of time, t comes off as direct runoff in accordance with a specified unit hydrograph for that basin. That is, each rainfall burst is multiplied by every ordinate of the unit hydrograph. The unit hydrograph is then translated in time so that it's start coincides with the next burst. The process is repeated for each burst of rain. The runoff contributions of each rainfall burst are then summed with respect to the time interval in which they occurred. A good, simple example is given in McCuen, pg. 507 . The governing equation for developing a storm hydrograph is called the convolution integral which is n

written in discrete form in Mathcad* as: Qn =



 Pi  Un i

i  0

where: n is equal to the number of rainfall pulses + number of unit hydrograph ordinates - 1 P is the amount of rain within a time interval t U is a vector of unit hydrograph ordinates taken at the same t interval. * the subscripting of the variables P and U is different in Mathcad than in most texts because the first value in a vector in Mathcad is the "zeroth" and not the first.

Part c: estimate the the runoff hydrograph for a 30 min rainfall excess of 10.3 cm Because we have a 15 minute UH (15 min storm created the UH) we need to break the rainfall into two 15 minute bursts of 5.15 cm. We assume that each burst of rainfall comes off the catchment in quantities and in time as dictated by the unit hydrograph. PROCEDURE : We have a rainfall hyetograph and ordinates for a 15 minute unit hydrograph.

np  2

nord  14

j  0  1

We then create a second vector for each containing 15 elements np  nord  1  15 that is "pad" the precipitation and UH vectors with zeros out to 15 elements each.

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Chin problem 10-18 UH 30 min interval.xmcd

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 5.15     5.15   0     0   0     0   0    0   P   cm  0     0   0     0   0   0     0   0   

 0     1.4   3.2     1.5   1.2     1.1   1.0    3 .66  m U     .49  sec  cm    .36   .28     .25   .17   0     0   0   

n

Qn =



 0     .5   1   1.5     2   2.5     3  t   3.5   hr    4   4.5     5   5.5     6   6.5     7 

 Pi  Un i

i  0

n  0  14

n

Qn 



 Pi  Un i

i  0

Next we can pass smooth curve through the data points using a linear spline with interpolation

vs  lspline ( t Q)

interp ( vs t Q time)

time  0  hr .1  hr  8  hr interpolation value

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Chin problem 10-18 UH 30 min interval.xmcd

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30

0

26.9

1.5

Qn

23.8

3

3

20.7

4.5

17.6

6

14.5

7.5

11.4

9

m

sec

interp ( vs t Q time) m

3

sec

8.3

10.5

5.2

12

2.1

13.5

1

0

0.7

1.4

2.1

2.8

3.5

4.2

4.9

5.6

6.3

Pj cm

7

tn time tj   hr hr

hr

computed direct runoff hydrograph - points splined direct runoff hydrograph Rainfall Hyetograph Caution : When using splines of any type ALWAYS plot the spline against the data points it passes through to make sure (visually) there is good agreement.

The nice thing about using a spline in Mathcad is that it can be numerically integrated by Mathcad because internally Mathcad "knows" the various equations that make it up. In this case, once we have visually checked to be sure the spline "fits" the data we can estimate the volume of direct runoff by integrating the spline.

7  hr

 7 The volume obtained by integrating the spline is:  interp ( vs t Q time) dtime  5.704  10  gal 0hr 7  hr

 5 3  interp ( vs t Q time) dtime  2.159  10  m 0hr

2

7

Check : 10.3  cm  2.1  km  5.714  10  gal

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Chin problem 10-18 UH 30 min interval.xmcd

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Part a: Confirmation of UH by convolution np  1

nord  14

j  0  0

np  nord  1  14

1    0  0    0  0    0  0  P     cm 0  0    0  0    0  0  0   

 0     1.4   3.2     1.5   1.2     1.1   1.0  m 3 U     .66  sec  cm  .49     .36   .28     .25   .17   0   

 0     .5   1   1.5     2   2.5     3  t   3.5   hr    4   4.5     5   5.5     6   6.5     7 

n  0  13

n

Qn 



 Pi  Un i

i  0

Next we can pass smooth curve through the data points using a linear spline with interpolation

vs  lspline ( t Q)

interp ( vs t Q time)

time  0  hr .1  hr  8  hr interpolation value

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Chin problem 10-18 UH 30 min interval.xmcd

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3.5

0

3.15 Qn

2.8

3

2.45

m

1.5 3 4.5 6

2.1

sec

interp ( vs t Q time) m

3

sec

1.75

7.5

1.4

Pj cm

9 10.5

1.05 0.7

12

0.35

13.5

0  50

5

60

115

170

225

280

tn

time

tj

min



min



335

390

445

500

min

computed direct runoff hydrograph - points splined direct runoff hydrograph Rainfall Hyetograph

6.5  hr

 The volume obtained by integrating the spline is:  0hr

2

6

interp ( vs t Q time) dtime  5.565  10  gal

6

1  cm  2.1  km  5.548  10  gal

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Chin problem 10-18 UH 30 min interval.xmcd

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Chin Problem 10-20 Use the 15 minute UH hydrograph below (from Problem 10-18) to estimate the runoff from the rainfall hyetograph below. NOTE : The values in the UH below were collected at 30 minute intervals

15 minute UH at 30 minute collection intervals

 0     1.4   3.2     1.5   1.2     1.1   1.0  m 3 UH30_int     .66  sec  cm  .49     .36   .28     .25   .17   0   

 0-30 min  30-60 min rainfall_hyetograph =   60-90 min   90-120 min

2.4  cm 



4.5  cm  2.1  cm 



0.8  cm 

 0     .7   1.4     2.3   3.2     2.35   1.5     1.35   1.2   1.15     1.1   1.05     1.0  3 m UH15int   .83     sec  .66   .58     .49   .41     .36   .32     .28   .265     .25   .21     .15   .09     0 

O. K. we have to have the UH duration, the interval between UH ordinates and the rainfall interval to be the same.

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Chin problem 10-20 15 min interval convolution.xmcd

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Definition(s) of Convolution A procedure utilizing a unit hydrograph to produce a direct runoff hydrograph from a multiperiod rainfall event. Bedient and Huber - Hydrology and Floodplain Analysis. The process by which the design storm is combined with the transfer function (unit hydrograph) to produce a direct runoff hydrograph - McCuen - Hydrologic Analysis and Design. Analytically speaking convolution is referred to as the theory of linear superpositioning. Conceptually, it is a process of multiplication, translation with time, and addition. The rainfall intensity should be logged using the same time interval as the rainfall that generated the UH. For example, if we have a 1-hr unit UH then the storm event for which the DRO hydrograph is desired should have rainfall intensities logged on a 1 hr. time interval. My explanation - We assume that each burst of rainfall occurring on a basin within an interval of time, t comes off as direct runoff in accordance with a specified unit hydrograph for that basin. That is, each rainfall burst is multiplied by every ordinate of the unit hydrograph. The unit hydrograph is then translated in time so that it's start coincides with the next burst. The process is repeated for each burst of rain. The runoff contributions of each rainfall burst are then summed with respect to the time interval in which they occurred. A good, simple example is given in McCuen, pg. 507 . The governing equation for developing a storm hydrograph is called the convolution equation which is n

written in discrete form in Mathcad* as: Qn =



 Pi  Un i

i  0

where: n is equal to the number of rainfall pulses + number of unit hydrograph ordinates - 1 P is the amount of rain within a time interval t U is a vector of unit hydrograph ordinates taken at the same t interval. * the subscripting of the variables P and U is different in Mathcad than in most texts because the first value in a vector in Mathcad is the "zeroth" and not the first.

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Chin problem 10-20 15 min interval convolution.xmcd

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We assume that each burst of rainfall comes off the catchment in quantities and in time as dictated by the unit hydrograph. PROCEDURE : We have a rainfall hyetograph and ordinates for a unit hydrograph. Both of these data sets were collected at 30 minute intervals.

np  8

nord  27

j  0  7

We then create a second vector for each containing 11 elements np  nord  1  34 that is "pad" the precipitation and UH vectors with zeros out to 17 elements each.

 1.2     1.2   2.25   2.25     1.05   1.05     0.4   0.4     0   0     0   0     0   0     0   0    0   P   cm  0     0   0     0   0   0     0   0     0   0  C:\Mathcad application areas \Mathcad application areas\Fluidsopen channels-hydrology\Problems for Chins Text\Chin 3rd edition Chap 10\

 0     .7   1.4   2.3     3.2   2.35     1.5   1.35     1.2   1.15     1.1   1.05     1.0   .83     .66   .58    3 .49  m  UH15int    .41  sec  cm    .36   .32     .28   .265   .25     .21   .15     .09   0  Chin problem 10-20 15 min interval convolution.xmcd

 0     .25   .50     .75   1.00     1.25   1.5     1.75   2.00     2.25   2.5     2.75   3.00   3.25     3.5   3.75     4.00  t15   4.25   hr    4.5   4.75     5.00   5.25     5.5   5.75     6   6.25     6.5  3 of 7

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          



          

0  0  0 0 0 0 0

       



 6.75     7   7.25     7.5   8   8.25     8.5   8.75   

0  0 



0  0 



0  0 



0 

rows ( P)  34

 

rows t15  35





rows UH15int  34 n  0  34 Pn 

0 if n  7 Pn otherwise

UH15int  n

0 if n  26 UH15int otherwise n

n

Qn 



i  0

 

rows t15  35

 Pi  UH15int  n i 





rows UH15int  35

rows ( Q)  35

Next we can pass smooth curve through the data points using a linear spline with interpolation





vs  lspline t15 Q

C:\Mathcad application areas \Mathcad application areas\Fluidsopen channels-hydrology\Problems for Chins Text\Chin 3rd edition Chap 10\

interp ( vs t Q time)

Chin problem 10-20 15 min interval convolution.xmcd

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time  0  hr .1  hr  8  hr

25

0

22.5 Qn

20

3

17.5

m

1.5 3 4.5

15

sec



interp vs t15 Q time m

3

sec



6

12.5

7.5

10

9

7.5

10.5

5

cm

12

2.5 0

Pj

13.5 0

1.5

3

4.5

6 t15

7.5

hr

n



time hr

9 

t15

10.5

12

13.5

15

j

hr

computed direct runoff hydrograph - points splined direct runoff hydrograph Rainfall Hyetograph

Caution : When using splines of any type ALWAYS plot the spline against the data points it passes through to make sure (visually) there is good agreement. The nice thing about using a spline in Mathcad is that it can be numerically integrated by Mathcad because internally Mathcad "knows" the various equations that make it up. In this case, once we have visually checked to be sure the spline "fits" the data we can estimate the volume of direct runoff by integrating the spline.

8  hr

 7 The volume obtained by integrating the spline is:  interp vs t15 Q time dtime  5.407  10  gal 0hr





2.1  cm  2.4  cm  4.5  cm  0.8  cm  9.8  cm

2

7

2.1  km  ( 9.8  cm)  5.437  10  gal

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Chin problem 10-20 15 min interval convolution.xmcd

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390  min  6.5  hr

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Chin problem 10-20 15 min interval convolution.xmcd

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Chin Problem 10-20 Use the 15 minute UH hydrograph below (from Problem 5-59) to estimate the runoff from the rainfall hyetograph below.

15 minute UH at 30 minute collection intervals

15 minute UH at 15 minute intervals

 0     1.4   3.2     1.5   1.2     1.1   1.0  m 3 UH30_int     .66  sec  cm  .49     .36   .28     .25   .17   0   

 0-30 min  30-60 min rainfall_hyetograph =   60-90 min   90-120 min

2.4  cm 



4.5  cm  2.1  cm 



0.8  cm 

 2.4    4.5  P30int    cm  2.1     .8 

C:\Mathcad application areas \Mathcad application areas\Fluidsopen channels-hydrology\Problems for Chins Text\Chin 3rd edition Chap 10\

Chin problem 10-20 30 min interval convolution.xmcd

 0     .7   1.4     2.3   3.2     2.35   1.5     1.35   1.2   1.15     1.1   1.05     1.0  3 m UH15int   .83     sec  cm  .66   .58     .49   .41     .36   .32     .28   .265     .25   .21     .15   .09     0 

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O. K. we have to have the UH duration, the interval between UH ordinates and the rainfall interval to be the same.

Definition(s) of Convolution A procedure utilizing a unit hydrograph to produce a direct runoff hydrograph from a multiperiod rainfall event. Bedient and Huber - Hydrology and Floodplain Analysis. The process by which the design storm is combined with the transfer function (unit hydrograph) to produce a direct runoff hydrograph - McCuen - Hydrologic Analysis and Design. Analytically speaking convolution is referred to as the theory of linear superpositioning. Conceptually, it is a process of multiplication, translation with time, and addition. The rainfall intensity should be logged using the same time interval as the rainfall that generated the UH. For example, if we have a 1-hr unit UH then the storm event for which the DRO hydrograph is desired should have rainfall intensities logged on a 1 hr. time interval. My explanation - We assume that each burst of rainfall occurring on a basin within an interval of time, t comes off as direct runoff in accordance with a specified unit hydrograph for that basin. That is, each rainfall burst is multiplied by every ordinate of the unit hydrograph. The unit hydrograph is then translated in time so that it's start coincides with the next burst. The process is repeated for each burst of rain. The runoff contributions of each rainfall burst are then summed with respect to the time interval in which they occurred. A good, simple example is given in McCuen, pg. 507 . The governing equation for developing a storm hydrograph is called the convolution equation which is n

written in discrete form in Mathcad* as: Qn =



 Pi  Un i

i  0

where: n is equal to the number of rainfall pulses + number of unit hydrograph ordinates - 1 P is the amount of rain within a time interval t U is a vector of unit hydrograph ordinates taken at the same t interval. * the subscripting of the variables P and U is different in Mathcad than in most texts because the first value in a vector in Mathcad is the "zeroth" and not the first.

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Chin problem 10-20 30 min interval convolution.xmcd

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We assume that each burst of rainfall comes off the catchment in quantities and in time as dictated by the unit hydrograph.

PROCEDURE : Here we use a 15 min UH with 30 minute intervals AND a precipitation vector with 30 minute intervals

np  4

nord  14

j  0  7

We then create a second vector for each containing 17 elements np  nord  1  17 that is "pad" the precipitation and UH vectors with zeros out to 17 elements each.

 2.4     4.5   2.1     0.8   0   0     0   0    P30   0   cm  0     0   0     0   0     0   0     0 





rows P30  17

C:\Mathcad application areas \Mathcad application areas\Fluidsopen channels-hydrology\Problems for Chins Text\Chin 3rd edition Chap 10\

 0     1.4   3.2     1.5   1.2   1.1     1.0   .66    m3 UH30int   .49    .36  sec  cm    .28   .25     .17   0     0   0     0 





rows UH30int  17

Chin problem 10-20 30 min interval convolution.xmcd

 0   .50   1.00     1.5   2.00     2.5   3.00     3.5   4.00  t30     hr  4.5   5.00     5.5   6     6.5   7   7.5     8   9.5   

 

rows t30  18

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



rows UH30int  17 n  0  17 P30 

0 if n  4

n

P30 UH30int  n

n

otherwise

0 if n  12 UH30int otherwise n

n



Qn 

i  0

 P30  UH30int  i n i 





rows UH30int  18 rows ( Q)  18

Next we can pass smooth curve through the data points using a linear spline with interpolation





vs  lspline t30 Q



interp vs t30 Q time



time  0  hr .1  hr  8  hr interpolation value

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Chin problem 10-20 30 min interval convolution.xmcd

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25

0

22.5 Qn

20

3

17.5

sec

15

m



interp vs t30 Q time m

3

sec



1.5 3 4.5 6

12.5

7.5

10

9

7.5

10.5

5

j

cm

12

2.5 0

P30

13.5 0

60

120

180

240 t30

300 n

min



time min

360 

t30

420

480

540

600

j

min

computed direct runoff hydrograph - points splined direct runoff hydrograph Rainfall Hyetograph

Caution : When using splines of any type ALWAYS plot the spline against the data points it passes through to make sure (visually) there is good agreement. The nice thing about using a spline in Mathcad is that it can be numerically integrated by Mathcad because internally Mathcad "knows" the various equations that make it up. In this case, once we have visually checked to be sure the spline "fits" the data we can estimate the volume of direct runoff by integrating the spline.

8  hr

 7 The volume obtained by integrating the spline is:  interp vs t30 Q time dtime  5.416  10  gal 0hr





2.1  cm  2.4  cm  4.5  cm  0.8  cm  9.8  cm 2

7

2.1  km  ( 9.8  cm)  5.437  10  gal

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Chin problem 10-20 30 min interval convolution.xmcd

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Chin 11-1 3rd edition A four lane collector roadway is to be constructed with 3.66 m (12 ft. ) lanes, a cross slope of 1.5%, a longitudinal slope of 0.8%, and the pavement is smooth asphalt. If the roadway drainage system is to be designed for a rainfall intensity of 120 mm/hr, determine the spacing of the of the inlets.

Solution : Overall philosophy - Gutters are designed to keep drainage from accumulating in the roadway and creating unsafe driving conditions. From Table 5.37 in Chin the criteria for "collector streets" are as follows For minor events: "No curb overtopping; flow spread must leave at least one lane free of water" . Since each lane is 3.66 m the maximum top width for the flow is 3.66 m. Therefore the maximum depth at the curb is: d = T  Sx where Sx is the "cross slope" that is the slope of the pavement from the crown to the curb. T  3.66  m

Sx  1.5% n  .013 

So  0.8%

sec 1

m

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDROLOGY PROCEDURES \Gutter and Inlet Design\ 9/24/2012

d  T  Sx  5.49  cm w  3.66  m

3

5-88 Chin.xmcd

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Using the maximum curb depth and maximum top width of flow we can compute the maximum flow in the gutter 8

1

 1 d 3 S 2 Q  .375    o  n  Sx  3

Q  0.075

m s

This is the maximum flow any part of the gutter can handle without exceeding the allowable top width, T

The rainfall intensity for the event of interest is: i  120 

A=

Q C i

mm , thus the contributing area can be no larger than: hr C  1

Assuming that only the roadway contributes to the flow in the gutter and it has a runoff coefficient of 1, the area is:

A 

Q 3 2  2.247  10 m C i

The roadway is four lanes wide, 2 lanes each direction. The roadway has a gutter on each side, thus 2 lanes contribute to each gutter. The length of roadway between inlets can be computed from the roadway length and width required to give the maximum area. A = w  2 L L 

A  306.91 m w2

Thus inlets must be spaced no farther apart than 307 m to avoid excessive flooding. This solution assumes that ALL the water coming to an inlet is captured by the inlet.

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5-88 Chin.xmcd

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11-3 3rd edition - Chin 3

m . If the sec gutter flow is to be removed in a vertical sag by a 15 cm high curb inlet, determine the length of the inlet. Consider cases where (a) the width of the inlet depression is Wo  .3  m, and (b) there is no inlet depression. A roadway has a flow depth at the curb of d  9  cm and a corresponding flow in the gutter of Q  .1 

Solution : This inlet is in a sag in the road so it is important that it work properly because the water has nowhere else to drain. Water drains to the inlet from both ends and most is assumed to enter at the front. Thus the inlet is assumed to behave as a weir or an orifice.

The depth of water at the curb is less than the height of the curb inlet so the inlet will function as a weir. The equation for the flow into a depressed curb inlet is:





Q = 1.25  L  1.8  Wo  d

1.5

 Q Wo d 1.5  2.25       m3 m  m     sec   m  2.423 m 1.5 d    m

0.8   Solving for L we get:

L 

Thus, using a depressed inlet, we need an opening at least 2.43 m long.

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Chin 5-90 inlet sizing in sag.xmcd

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In the case of no depression the weir equation is Q = 1.6  L  d

1.5

0.625 

. This can be solved for L to get:

Q 3

m

sec

L 

d    m

1.5

m

L  2.315 m

We have an anomaly here, the calculations show a longer inlet length required with a depression than without. This is unlikely to be the case, therefore use an inlet length of at least 2.43 m with or without a depression.

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Chin 5-90 inlet sizing in sag.xmcd

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Chin 11-6 Sizing a Grate Inlet A roadway with a rough asphalt pavement has cross slope of 2.0%, a longitudinal slope of 2.5%, a curb height of 8 cm, and a 90-cm wide concrete gutter. If flow in the gutter is 0.07 m3/sec, determine the size and interception capacity of a reticuline grate that should be used to intercept as much of the flow as possible. 3

Q  0.07 

m sec

Sx  2  %

So  2.5%

Table 5-36, pg. 489 of Chin lists a Manning's n of n  0.016 

sec 1

m

for a rough asphalt surface

3

8

1

1  3 2 Compute the depth of flow in the gutter using: Q = .375    d  So  n  Sx





d  2  cm 8 1    1  3 2   dsoln  root Q  .375    d  So d    n  Sx  

dsoln  5.206  cm The computed depth dsoln  5.206  cm is less than the height of the curb. Thus the flow is constrained in the gutter.

The top width of gutter flow is: T 

dsoln Sx

 2.603 m

1 2 The area of flow in the gutter is given by: Aflow   dsoln  T  0.068 m 2 The resulting gutter flow velocity is: V 

Q m  1.033 Aflow s

Chin pg. 497: Grates typically consist of longitudinal and/or transverse bars oriented parallel and perpendicular to the gutter flow respectively. Grates are typically available with longitudinal dimensions dimensions in the range of 610 mm to 1220 mm (2 ft to 4 ft) and transverse dimensions from 381 mm to 914 mm (1 to 3 ft.). Frontal flow to a grate approaches the Wo dimension of the grate. Splash-over occurs when a portion of the frontal flow passes (jumps ?) the grate. The ratio of frontal flow intercepted to the total frontal flow is termed the frontal flow interception efficiency, Rf.

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Chin 5-92 Reticuline Grate design.xmcd

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



The frontal flow interception efficiency can be estimated using: Rf = 1  0.295  V  Vo where V is the flow velocity in the gutter and Vo is the splash-over velocity which is an empirical value obtained from Fig. 5-51 in Chin. Since the computed spread is 2.6 m which exceeds the largest transverse dimension available, use the largest transverse dimension, 914 mm. The length of the grate can be estimated by assuming that the computed velocity equals the splash over velocity. Entering Figure 5.51 with a splash-over velocity of 1.03 m/sec results in a reticuline grate length of approximately .4 m. Therefore, any grate longer than 0.4 m will intercept 100% of the frontal flow. To maximize the interception of the total flow because of the wide spread of the flow, use the longest grate available 1220 mm.

The ratio of the side flow intercepted by the grate to the total side flow is given by: for L  1220  mm and Wo  .914  m 1.8   V   .0828    m      sec     Rs  1    2.3 L   Sx      m 

1

 0.265

The ratio of frontal flow to total gutter flow is given by: 8

Wo   Rw  1   1   T  

3

 0.684

And finally the ratio of the intercepted flow to the total gutter flow is: Rf  1





R  Rf  Rw  Rs  1  Rw  0.768 3

The intercepted flow is therefore: R  Q  0.054

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m s

Chin 5-92 Reticuline Grate design.xmcd

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Chin 11-7 Inlet Design A roadway has a cross slope of 1.5%, a flow depth at the curb of 9 cm, and a corresponding flowrate in the gutter of 0.1 m3/sec. The gutter flow is to be removed in a vertical sag by a grate inlet that is mounted flush with the curb. Calculate the minimum dimensions of the grate. Solution : 3

m Q  0.1  sec

Sx  1.5%

d  9  cm

For depths of flow less than 12 cm the grate will act as a weir, in this problem d  9  cm Q

Weir flow equation for grates: : P =

1.5

, solve for P the perimeter of the grate not including the side

1.66  d against the curb. Note this equation, as well as most in this section are not dimensionally consistent. Q m

3

sec

P 

d 1.66     m

1.5

 m  2.231 m minimum grate perimeter

The minimum length of grate needed to intercept all frontal flow can be obtained from figure 11-7 , Chin assumes the use of a reticuline grate which is the worst case scenario, meaning that you get splash-over at low velocities. We need to compute the velocity in the gutter.

V=

Q A

The area of flow in a triangular gutter is given by : A 

V 

1 d  2  d    0.27 m thus the flow velocity in the gutter is: 2 Sx 





Q m  0.37 A s

Now, the velocity just computed is the splashover-over velocity and go to Figure 11.7 to obtain the minimum length along the gutter in order to accept all frontal flow L  .15  m Now obtain the grate width as: P = 2  w  L

w 

P 2



L  1.041 m 2

2  w  L  2.231 m

So we need a grate with of length of .15*m and a width of 1.04*m

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDRAULIC PROCEDURES \Gutter and Inlet problems\ 7/31/2012

Chin 5-93 grate inlet in a sag.xmcd

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Note that if we change any of the input values or use a different type of grate thereby obtaining a different length the problem will automatically recompute the solution.

C:\Mathcad application areas\Mathcad application areas\Fluids-open channelshydrology\HYDRAULIC PROCEDURES \Gutter and Inlet problems\ 7/31/2012

Chin 5-93 grate inlet in a sag.xmcd

2 of 2