Solutions Manual to Accompany Introduction to Hydraulics and Hydrology with Applications for Stormwater Management Four
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Solutions Manual to Accompany
Introduction to Hydraulics and Hydrology with Applications for Stormwater Management Fourth Edition
John E. Gribbin, P.E. Essex County College
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Solutions Manual to Accompany Introduction to Hydraulics and Hydrology with Applications for Stormwater Management, 4E John E. Gribbin Vice President, Editorial: Dave Garza Director of Learning Solutions: Sandy Clark Senior Acquisitions Editor: James DeVoe Managing Editor: Larry Main Senior Product Manager: John Fisher Editorial Assistant: Aviva Ariel Vice President, Marketing: Jennifer Ann Baker Marketing Director: Deborah Yarnell
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Contents Solutions: Chapter 1 - Hydraulics and Hydrology in Engineering...............................................1 Chapter 2 - Fluid Mechanics.......................................................................................4 Chapter 3 - Fundamental Hydrostatics ......................................................................6 Chapter 4 - Fundamental Hydrodynamics ...............................................................23 Chapter 5 - Hydraulic Devices .................................................................................39 Chapter 6 - Open Channel Hydraulics . ...................................................................51 Chapter 7 - Uniform Flow in Channels ....................................................................58 Chapter 8 - Varied Flow in Channels .......................................................................67 Chapter 9 - Culvert Hydraulics . ..............................................................................75 Chapter 10 - Fundamental Hydrology .......................................................................81 Chapter 11 - Runoff Calculations...............................................................................97 Chapter 12 - Storm Sewer Design.............................................................................126 Chapter 13 - Culvert Design ....................................................................................140 Chapter 14 - Stormwater Detention..........................................................................150 Chapter 15 - Detention Design.................................................................................156
iii
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C h a p t e r
1 Hydraulics and Hydrology in Engineering
1. A 5 A5
pd2 4 p(3.04)2 4
A 5 7.26 ft2 (3 sig. figures) (Answer) 2. 1.0 mi 3
5280 ft 5 5280 ft 1.0 mile
5 (L)(W)(T) V V 5 (5280)(22.0)(.650) V 5 75,504 ft3 V 5 2796 yd3 V 5 2800 yd3 (2 sig. figures) (Answer) 3. 6 1/2s 5 13/2s (2 sig. figures) 13/2 in 3
1 ft 5 13/24 ft 5 0.5417 ft 12 in
Depth 5 0.54 ft (2 sig. figures) (Answer) 4. V 5 V 5
pd 2L 4
p(1.30)2(60) 4
V 5 79.64 ft3 V 5 80 ft3 (2 sig. figures) (Answer) 1
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2 CHAPTER 1 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
5. 12.25 151 25.0 188.25 Answer is 188 ft. 6. D 5 36 in 3
.02540 m 5 0.9144 m 1 in
A 5 pD2/4 5 p(.9144)2/4 A 5 0.66 m2 (2 sig. figures) (Answer) 7. a) A 5 (45.00)(125.00) 5 5625 ft2 (4 sig. figures) 1 acre 5 0.1291 acres (4 sig. figures) b) A 5 5625 ft2 3 43560 ft2 .0929 2 c) A 5 5625 ft2 3 m 5 522.6 m2 (4 sig. figures) 1 ft2 1 acre .40469 hectare d) A 5 5625 ft2 3 5 .05226 hectare (4 sig. figures) 2 3 1 acre 43560 ft 8. Design a connector roadway. Design considerations:
1. Intersection configurations
2. Width of road based upon expected traffic
3. Type of curb
4. Amount of crown
5. Grade of profile
6. Grading of adjacent ground
7. Sight distances
8. Property lines
9. Pavement thickness
10. Drainage
11. Stop signs and painted stop lines
12. Signalization (consideration)
13. Consideration of sidewalks
14. Consideration of guide rails
15. Consideration of snow removal
9. Design a pedestrian walkway. Design considerations:
1. Size and location of walkway to consider amount of pedestrian traffic and convenience of traffic routing
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 3
2. Consideration of conflicts with existing utilities, trees, etc.
3. Grading to provide drainage
4. Handicap access
5. Concrete mixture and thickness
6. Crushed stone base course
7. Expansion and control joints
8. Driveway treatment: reinforcement, added thickness
10. Design a culvert pipe. Design considerations:
1. Material of pipe
2. Diameter based upon expected flood flow
3. Cover over pipe
4. Gradient for gravity flow
5. Erosion protection at inlet and outlet
6. Headwalls
7. Trench: stone bedding, shoring if needed
8. Alignment of culvert to match stream
9. Consideration of possible permits for wetlands, etc.
10. Traffic control
11. Flooding during construction
12. Fish habitat
13. Upstream water level after completion
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C h a p t e r
2 Fluid Mechanics
1. W 5 gV (Equation 2-1) W 5 (62.4)(1.0) W 5 62.4 lb. (Answer) 2. W 5 gV (Equation 2-1) But W 5 mg Thus, gV 5 mg Density 5
m V
Substituting, Density 5 Therefore, Density 5
3. Dimensions of can 5 V 5
pd2 h 4
V 5
p(.333)2 (.50) 4
g g
62.4 5 1.94 slugs/ft3 (Answer) 32.2 4.0 6.0 ft dia. 3 ft high 12 12
V 5 0.0428 ft2 g 5
W 2.0 5 5 46.77 lb/ft2 V .0428
g 5 47 lb/ft3 (Answer)
4
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 5
4. Dimensions of container 5 0.100 m 3 0.200 m 3 0.150 m V 5 (.100)(.200)(.150) V 5 0.00300 m3 g5
W 450 5 5 150,000 N/m3 V .00300
g 5 150 kN/m3 (Answer) 5. S.G. 5 S.G. 5
6. S.G. 5 S.G. 5
S.W. 62.4 lb/ft3 57.4 5 0.92 (Answer) 62.4 S.W. 9.81 kN/m3 7.85 3 103 5 0.800 (Answer) 9.81 3 103
7. The water rises higher in the 2.0 mm diameter tube because it is the narrower tube. 8. t 5
t5
F (Equation 2-4) A 1.5 3 10 24 .75
t 5 2.0 3 10 24 lb d m 5 t (Equation 2-8) v m5
(2.0 3 10 24)(.0417) 10
ad 5
.50 ftb 12
m 5 8.3 3 10 27 lb-s/ft2 (Answer) 9. S.W. 5 45 lb/ft3 Density 5
g 45 5 1.40 slugs/ft3 5 g 32.2
v5
m r
v5
2.2 3 10 25 1.40
v 5 1.57 3 10 25 ft2/s (Answer) 10. S.W. 5 7.85 3 103 N/m3 Density 5 v5
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g 7.85 3 103 5 0.800 3 103 kg/m3 5 g 9.81
m 2.4 3 10 23 5 5 3.00 3 10 26 m2/s (Answer) r .800 3 103
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C h a p t e r
3 Fundamental Hydrostatics
1. p 5 gz (Equation 3-2) p 5 162.42 13502
p 5 21,840 lb/ft2 p 5 21,800 lb/ft2 (Answer) 2. a) p 5 gz (Equation 3-2)
b) p 5 gz (Equation 3-2)
p 5 162.42 14.02
p 5 162.42 112.02
2
p 5 749 lb/ft2 (Answer)
p 5 250 lb/ft (Answer)
3. z 5 82.5 2 38.0 z 5 44.5 ft p 5 gz (Equation 3-2) p 5 162.42 14.02
p 5 2776.8 lb/ft2 p 5 2780 lb/ft2 (Answer) 4. p 5 gz (Equation 3-2) p 5 162.42 18.502 p 5 530.4 lb/ft2
p 5 530 lb/ft2 (Answer)
6
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 7
5. Resultant force: FR 5
gz2 w (Equation 3-4) 2
FR 5
162.42 18.02 2 110.02 2
FR 5 19,968 lb
FR 5 20,000 lb (Answer) Location of resultant force: 1 yR 5 z (Equation 3-5) 3 1 yR 5 18.02 3 yR 5 2.67 ft yR 5 2.7 ft (Answer) A
FR yR
B
© Cengage Learning 2014
8.0′
6. Resultant force: FR 5 gzlw (Equation 3-3) FR 5 162.42 18.02 112.02 110.02 FR 5 59,904 lb
FR 5 60,000 lb (Answer) Location of resultant force: FR located at centroid of pressure xR 5
12.0 2
xR 5 6.0 ft (Answer)
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8 CHAPTER 3 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
xR
B
7. u 5 tan21
C
© Cengage Learning 2014
FR
5.0 8.0
u 5 32.0° FH 5 gz1lwcos u (Equation 3-8) FH 5 162.42 13.02 110.02 19.4342 cos 32.0° FH 5 14,976 lb
FV 5 FV 5
gw 1z1 1 z2 2 l sin u (Equation 3-9) 2
(62.4)(10.0) 3 3.0 1 8.0 4 (9.434) sin 32.0° 2
FV 5 17,160 lb
W5
gwl2 sin u cos u (Equation 3-10) 2
W5
162.42 110.02 19.4342 2 sin 32.0° cos 32.0° 2
W 5 12,480 lb yR 5 yR 5
Iy 2z1 1 z2 c d (Equation 3-7) 3 z1 1 z2 5.0 (2)(3.0) 1 8.0 c d 3 3.0 1 8.0
yR 5 2.12 ft
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 9
xR FH
3.0′ D
8.0′ W
W = 10.0′
FV
© Cengage Learning 2014
yR C xW 8.0′
FR
C
lR
3′
9.4
© Cengage Learning 2014
D
xR 5 one-half horizontal projection 1 xR 5 l cos u 2 1 xR 5 (9.434) cos 32.0° 2 xR 5 4.00 ft xW 5 one-third horizontal projection 1 xW 5 (9.434) cos 32.0° 3 xW 5 2.67 ft
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10 CHAPTER 3 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
FR 5 "Fv2 1 (FH 1 W)2 (Equation 3-11)
FR 5 "17,1602 1 (14,976 1 12,480)2 FR 5 32,377 lb
FR 5 32,000 lb (Answer) lR 5 lR 5
1 3 F (yR) 1 W(xW) 1 FH(xR) 4 (Equation 3-12) FR v
1 3 (17,160)(2.12) 1 (12,480)(2.67) 1 (14,976)(4.00) 4 32,377
lR 5 4.00 ft (Answer)
8. Since the length of the dam is not indicated, the force will be computed per linear foot of length (w 5 1 ft). u 5 tan21
1 3
u 5 18.435° FH 5 0 FV 5
gw (z1 1 z2) l sin u (Equation 3-9a) 2
FV 5
(62.4)(1) (0 1 15)(47.434) sin 18.435° 2
FV 5 7,020 lb/L.F. W5
gwl2 sin u cos u (Equation 3-10) 2
W5
(62.4)(1)(47.434) sin 18.435° cos 18.435° 2
W 5 21,060 lb/L.F. xR 5 0 yR 5
ly 2z1 1 z2 c d (Equation 3-7) 3 z1 1 z2
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 11
θ
W
15.0′ Fv
© Cengage Learning 2014
yR
xW
45.0′
θ
FR
© Cengage Learning 2014
lR
4′
.43
47
yR 5
15 (2)(0) 1 15 c d 3 0 1 15
yR 5 5.0 ft
xW 5 one-third horizontal projection xW 5 15.0 ft
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12 CHAPTER 3 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
FR 5 "FV2 1 (FH 1 W)2 (Equation 3-11) FR 5 "7,0202 1 21,0602
FR 5 22,199 lb/L.F.
FR 5 22,200 lb/L.F. (Answer) lR 5
lR 5
1 3 F (y ) 1 W(xw) 1 FH(xR) 4 (Equation 3-12) FR V R 1 3 (7,020)(5.0) 1 (21,060)(15.0) 1 0 4 22,199
lR 5 15.81 ft
lR 5 15.8 ft (Answer) 9. For this problem, w 5 2.5 ft.
7.5′
Fv yR
FR 5 FR 5
g ly w 2
© Cengage Learning 2014
11.5′
(z1 1 z2) (Equation 3-6)
(62.4)(4.0)(2.5) (7.5 1 11.5) 2
FR 5 5,928 lb (Answer)
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 13
yR 5 yR 5
ly 2z1 1 z2 d (Equation 3-7) c 3 z1 1 z2 4.0 2(7.5) 1 11.5 c d 3 7.5 1 11.5
yR 5 1.86 ft (Answer) 10. For this problem, w 5 16 ft, f 5 90°. xH FH
z1
z2
B
Fv W © Cengage Learning 2014
yv A xW r = 2.0′ B
ϕR
A
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FR
© Cengage Learning 2014
ϕ=90˚
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14 CHAPTER 3 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
FH 5 gz1wrsin f (Equation 3-13) FH 5 (62.4)(1.5)(16)(2.0) sin 90° FH 5 2995 lb
FV 5
gw (z1 1 z2)(r 2 r cos f) (Equation 3-14) 2
FV 5
(62.4)(16) (1.5 1 3.5)(2.0 2 2.0 cos 90°) 2
FV 5 4992 lb
W 5 gw c
f r2 pr2 2 sin f cos fd (Equation 3-15) 360 2
(2.0)2 90 2 W 5 (62.4)(16) c p(2.0) 2 sin 90 cos 90 d 360 2 W 5 3137 lb
xH 5 Centroid of rectangle xH 5 1.0 ft
yV 5 Centroid of trapezoid
yV 5
yV 5
ly 2z1 1 z2 c d 3 z1 1 z2
2.0 2(1.5) 1 3.5 c d 3 1.5 1 3.5
yV 5 0.867 ft
xW 5 Centroid of wedge of water xW 5
4r 3p
xW 5
(4)(2.0) 3p
xW 5 0.849 ft
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 15
FR 5 "Fv2 1 (FH 1 W)2 (Equation 3-11)
FR 5 "49922 1 (2995 1 3137)2 FR 5 7,907 lb
FR 5 7,900 lb (Answer) fR 5 sin21 e
fR 5 sin21 e
1 3 F (y ) 1 W(xW) 1 FH(xH) 4 f rFR v v
1 3 (4992)(.867) 1 (3137)(.849) 1 (2995)(1.0) 4 f (2.0)(7907)
fR 5 sin21(.6315) fR 5 39.16°
fR 5 39° (Answer) 11. For this problem, w 5 14.0 ft. The centroid of the wedge of water must be determined by approximate graphical methods. FH 5 0 FV 5
gz2 w (Equation 3-4) 2
FV 5
(62.4)(9.4)2 (14.0) 2
FV 5 38,596 lb W 5 gV where V is the volume of the wedge of water. W 5 gAw where A is the cross sectional area of the wedge. W 5 (62.4)(22.7)(14.0) W 5 19,831 lb
FR 5 "FV 2 1 (FH 1 W)2 (Equation 3-11) FR 5 "38,5962 1 (0 1 19,831)2
FR 5 43,393 lb (Answer)
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16 CHAPTER 3 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
9.4′ W
Fv
Fv ϕR
W
FR
ϕ
© Cengage Learning 2014
FR
fR 5 tan21
W FV
fR 5 tan21
19,831 43,393
fR 5 24.6° (Answer) Note: In this problem, fR represents only the orientation of FR and not its actual position in space. To find the point where FR acts on the ogee surface, locate graphically the point where an angle of 24.6° is perpendicular to a tangent to the surface. 12. For this problem, w 5 8.0 ft. The area and centroid of the wedge of water can be estimated graphically to avoid a more complex mathematical analysis. FH 5 0 gz2 FV 5 w (Equation 3-4) 2
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 17
FV 5
(62.4)(4.24) (8.0) 2
FV 5 4,487 lb W 5 gV where V is the volume of the wedge of water. W 5 gAw where A is the cross sectional area of the wedge. A 5 2.76 ft2 W 5 (62.4)(2.76)(8.0) W 5 1,378 lb
r
z Fv
W
yv
Fv ϕR
W
6.0′ FR
ϕR
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© Cengage Learning 2014
4.24′
FR
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18 CHAPTER 3 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
FR 5 "FV 2 1 (FH 1 W)2 (Equation 3-11)
FR 5 "44872 1 (0 1 1378)2 FR 5 4,694 lb
FR 5 4,700 lb (Answer)
fR 5 tan21
W FV
fR 5 tan21
1378 4487
fR 5 17.1° (Answer) 13. Buoyant force equals weight of water displaced. FB 5 gV (Equation 3-17) a) 2½s 3 5s 5 .208r 3 .417r p(.208)2 pd2 L 5 (.417) 5 0.0142 ft3 4 4
Vol. 5
FB 5 (62.4)(.0142) 5 0.886 lb (Answer)
b) Since the weight of the can is less than the buoyant force, the can will float. 14. Buoyant force equals weight of water displaced. FB 5 gV (Equation 3-17) Vol. 5 AL where A 5 area of cross section, L 5 length 1 A 5 (.866)(1) 5 0.433 in2 5 0.00301 ft2 2 L 5 12.75 in 5 1.0625 ft Vol. 5 (.00301)(1.0625) 5 0.00320 ft3 FB 5 (62.4)(.00320) 5 0.20 lb Since the weight of the scale (0.25 lb) is greater than the buoyant force (0.20 lb), the scale will sink. 15. Buoyant force equals weight of water displaced. FB 5 gV (Equation 3-17) Vol. 5 (8.0)(8.0)(6.0) 5 384 ft3 FB 5 (62.4)(384) 5 23,962 lb 5 12 tons Since the weight of the chamber (10 tons) is less than the buoyant force (12 tons), the chamber will float.
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 19
16. Buoyant force equals weight of water displaced. FB 5 gV (Equation 3-17) Vol. 5 1½s 3 3½s 3 2.0r 5 (.125)(.292)(2.0) 5 0.0729 ft3 FB 5 (62.4)(.0729) 5 4.55 lb Since the weight of the lumber is less than the buoyant force, the lumber will float. The volume below water Vr is W 4.0 5 5 0.0641 ft3 Vr 5 g 62.4 Depth below water is computed from the lumber dimensions. Vr 5 (.292)(2.0)(depth) Depth 5
0.0641 5 0.1098 ft (.292)(2.0)
Height above water is 0.125 2 0.1098 5 0.0152 ft (Answer) 17. Buoyant force equals weight of water displaced. FB 5 gV (Equation 3-17) Vol. 5 (4.0)(4.0)(4.0) 5 64 ft3 FB 5 (62.4)(64) 5 3994 lb 5 2.00 tons Since the weight of the chamber is less than the buoyant force, the chamber will float. The volume below water Vr is W 3000 5 5 48.1 ft3 Vr 5 g 62.4 Depth below water is computed from the chamber dimensions. Vr 5 (4.0)(4.0)(depth) Depth 5
48.1 5 3.00 ft (4.0)(4.0)
Height above water is 4.0 2 3.0 5 1.0 ft (Answer) 18. Buoyant force equals weight of water displaced. FB 5 gV (Equation 3-17) Vol. 5 p(.03125)2(.146) 5 0.000448 ft3 FB 5 (62.4)(.000448) 5 0.0280 lb 5 0.45 ounces Since the weight of the cork is less than the buoyant force, the cork will float. The volume below water Vr is Vr 5
W .0125 5 5 0.00020 ft3 g 62.4
Depth below water is computed using the formula for the segment of a circle. See sketch.
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20 CHAPTER 3 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Closed valve
Elev. 525.0
© Cengage Learning 2014
Elev. 550.0
The shaded area in the sketch is a segment of the circle comprising a cross section of the cork. Call the area ar. r2 up ar 5 c 2 sin u d 2 180
where u is measured in degrees. Bisecting the angle u gives a right triangle in which r2d u cos 5 r 2 The area ar is computed from the volume Vr. Vr 5 arl 0.00020 5 ar (.146) ar 5 0.00137 ft2 Since r 5 0.03125 ft, the segment equation becomes 0.00137 5
(.03125)2 up c 2 sin u d 2 180
which reduces to
2.81 5 .0175 u 2 sin u The solution to this equation is u 5 170.0°. Then depth d is computed as follows u r2d cos 5 r 2 cos
170 .03125 2 d 5 2 .03125
d 5 0.0285 ft. Height above water 5 2r 2 d 5 2(.03125) 2 .0285 Height 5 0.034 ft 5 0.41 inches (Answer)
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 21
19. Buoyant force equals weight of water displaced. FB 5 gV (Equation 3-17) Vol. 5 p(1.0)2(3.0) 5 9.42 ft3 FB 5 (62.4)(9.42) 5 588 lb Since the weight of the tank is less than the buoyant force, the tank will float. The volume below water Vr is Vr 5
W 100 5 5 1.60 ft3 g 62.4
Depth below water is computed using the formula for the segment of a circle. See sketch.
r
Volume below water
r–d d, depth below water
© Cengage Learning 2014
θ
The shaded area in the sketch is a segment of the circle comprising a cross section of the tank. Call the area ar. ar 5
r2 up 2 sin u 2 180
where u is measured in degrees. Bisecting the angle u gives a right triangle in which u r2d cos 5 r 2 The area ar is computed from the volume Vr. Vr 5 arl 1.60 5 ar (3.60) ar 5 0.533 ft2 Since r 5 1.0 ft, the segment equation becomes 0.533 5
(1.0)2 up c 2 sin u d 2 180
which reduces to
1.067 5 .0175 u 2 sin u The solution to this equation is u 5 113.4°.
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22 CHAPTER 3 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Then depth d is computed as follows r2d u cos 5 r 2 cos
113.4 1.0 2 d 5 2 1.0
d 5 0.451 ft. Height above water 5 2r 2 d 5 2(1.0) 2 .451 Height 5 1.55 ft (Answer) 20. Buoyant force equals weight of water displaced. FB 5 gV (Equation 3-17) 4 Vol. 5 p (.5)3 5 0.5236 ft3 3 FB 5 (62.4)(.5236) 5 32.67 lb 5 0.45 ounces Since the weight of the ball is less than the buoyant force, the ball will float. The volume below water Vr is Vr 5
W 25 5 5 0.401 ft3 g 62.4
Depth below water d is computed using the formula for the partial volume of a sphere. See sketch.
d
r
© Cengage Learning 2014
c
The volume below water Vr can be expressed as Vr 5
p d 3 3d(2r 2 d) 1 d2 4 6
Therefore, 0.401 5
p d 3 3d(2r 2 d) 1 d2 4 6
which reduces to
0.765 5 d 3 3d(1.0 2 d) 1 d2 4
The solution to this equation is d 5 0.69 ft. Height above water 5 2r 2 d 5 2(.5) 2 .69 Height 5 0.31 ft. (Answer)
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C h a p t e r
4 Fundamental Hydrodynamics
1. NR 5 NR 5
Dv where y 5 1 3 1025 ft2/s. y
12.02 16.212 1 3 1025
NR 5 12.42 3 105 1Answer2
NR 5 1,242,000
Since NR . 10,000, flow is turbulent. (Answer) 2. NR 5
Dv where y 5 1 3 1025 ft2/s. y
D 5 2.5 in 5 0.208 ft a5
p1.2082 2 pD2 5 0.0341 ft2 4 4
v5
Q .150 5 4.40 ft/s a .0341
NR 5
1.2082 14.402 0.915 3 105 1 3 1025
NR 5 91,500
1Answer2
Since NR . 10,000, flow is turbulent. (Answer) 3. NR 5
Dv y
D 5 4 inches 5 0.333 ft a 5 pr2 5 p1.1672 2 5 0.0872 ft2 v5
Q 4.0 3 1023 5 5 0.0459 ft/s A .0872
23
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24 CHAPTER 4 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
NR 5
1.3332 1.04592 5 0.0153 3 105 1 3 1025 1Answer2
NR 5 1,530
Since NR , 2,000, flow is laminar. (Answer) 4. NR 5
Dv y
where y 5 9.29 3 1027 m2/s.
D 5 300 mm 5 0.300 m p1.3002 2 pD2 a5 5 5 0.0707 m2 4 4 v5
Q .250 5 5 3.54 m/s a .0707
NR 5
1.3002 13.542 7 27 5 0.114 3 10 9.29 3 10 1Answer2
NR 5 1,140,000
Since NR . 10,000, flow is turbulent. (Answer) 5. Velocity head 5 Velocity head 5 6. Pressure head 5 Pressure head 5 7. Velocity head 5 Velocity head 5
y2 2g 16.212 2 5 0.60 ft 2132.22 p g
1Answer2
22.1 5 0.35 ft. 1Answer2 62.4 v2 2g 14.252 2 2132.22
Velocity head 5 0.28 ft 1Answer2 8. Friction head loss 5 f
L v2 D 2g
Friction head loss 5 1.02152
11152 17.492 2 5 3.23 ft. 1Answer2 .67 2132.22
9. e 5 0.001 1Table 4-12 e .001 5 5 0.00033 D 3.0 NR 5
13.02 112.52 Dv 5 3.75 3 106 5 y 1 3 1025
f 5 0.015 1from Moody Diagram2
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 25
Friction head loss 5 f
L y2 D 2g
12252 112.52 2 5 2.73 ft. 1Answer2 Friction head loss 5 1.0152 3.0 2132.22
10. e 5 0.000005 1Table 4-12 e .000005 5 5 0.00003 D .167 NR 5
1.1672 11.352 Dv 5 2.25 3 104 5 y 1 3 1025
f 5 0.026 1from Moody Diagram2 Friction head loss 5 f
L y2 D 2g
Friction head loss 5 1.0262 11. h1 1
1252 11.352 2 5 0.110 ft. 1Answer2 1.1672 2132.22
p2 v1 2 v2 2 1 5 h2 1 1Equation 4-72 g 2g 2g
v1 5 0, p2 5 0 410 1 0 5 372.5 1 0 1
v2 2 2132.22
v22 5 2132.22 1410 2 372.52 5 2415 v2 5 49.14 ft/s 1Answer2 Q 5 v2a 1Equation 4-32 a5
p1.6672 2 pd2 5 5 0.3491 ft2 4 4
Q 5 149.142 1.34912
Q 5 17.15 cfs 1Answer2 410.0
8″ Dia
372.5 50′
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© Cengage Learning 2014
.
11/12/12 8:43 AM
26 CHAPTER 4 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
12. h1 1
p2 v2 2 v1 2 1 5 h2 1 1 he 1 hf g 2g 2g
v1 5 0 due to negligible water movement in the reservoir p2 5 0 no pressure at free discharge he 5 0 neglect entrance losses v2 2 v2 2 1 f 1L/D2 2g 2g v2 2 v2 2 1 f 150/.672 410 5 372.5 1 2132.22 2132.22
h1 5 h2 1
37.5 5
v2 2 11 1 75f 2 64.4
2415 5 v22 11 1 75f 2
Solve this equation by use of the Moody diagram. e 5 .00085 1Table 4-12 e .00085 5 5 .00127 D .67 Trial 1: f 5 0.02 1assumed value of f 2
v2 2 5
2415 1 1 751.022
v22 5 966 v2 5 31.1 ft/s NR 5
131.12 1.672 5 2.08 3 106 1from Equation 4-12 1025
f 5 0.021 1from Moody Chart2 Trial 2: f 5 0.021 v2 2 5
2415 1 1 751.0212
v22 5 937.9 v2 5 30.6 ft/s NR 5
130.62 1.672 5 2.04 3 106 1025
f 5 0.021 (from Moody Chart) v2 5 30.6 ft/s (Answer) Q 5 v2a (Equation 4-3) a5
p1.672 2 pd2 5 5 0.349 ft2 4 4
Q 5 130.62 1.3492
Q 5 10.7 cfs 1Answer2
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 27
13. Locate Station 1 in the upper reservoir and Station 2 at the point where the pipe connects to the lower reservoir. p2 v2 2 v1 2 1 h1 1 5 h2 1 1 he 1 hf g 2g 2g v1 5 0 due to negligible water movement in the reservoir he 5 0 neglect entrance losses p2 by h2 where h2 is measured to the free surface of the lower reservoir instead g of the center of the pipe.
Replace h2 1
h1 5 h2 1
v2 2 L v2 2 1f 2g D 2g
v22 425 v22 1132.0 5 1079.5 1 1f 2132.22 1.0 2132.22
52.5 5
v22 11 1 425f 2 64.4
3381 5 v22 11 1 425f 2
Solve this equation by use of the Moody Diagram. e 5 0.001 1Table 4-12 e .001 5 5 0.001 D 1.0 Trial 1: f 5 0.02 (assumed value of f ) v2 2 5
3381 1 1 4251.022
v22 5 355.9 v2 5 18.87 ft/s NR 5
11.02 118.872 5 1.908 3 106 1 3 1025
f 5 0.0195 (from Moody Diagram) Trial 2: f 5 0.0195 v2 2 5
3381 1 1 4251.01952
v22 5 364.0 v2 5 19.08 ft/s NR 5
11.02 119.082 5 1.908 3 106 1 3 1025
f 5 0.0195 (from Moody Diagram) (OK)
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28 CHAPTER 4 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Q 5 v2a2 a2 5
p11.02 pD2 5 5 0.785 ft2 4 4
Q 5 119.082 1.7852 Q 5 14.98 cfs
Q 5 15 cfs (Answer) 14. First, find the velocity at the discharge end of the pipe. h1 1
p2 v2 2 v12 1 5 h2 1 1 he 1 hf g 2g 2g
v1 5 0 due to negligible water movement in the reservoir p2 5 0 no pressure at free discharge he 5 0 neglect entrance losses h1 5 h2 1
v2 2 L v2 2 1f 2g D 2g
h2 5 221.12 1 2.50 5 223.62 ft 268.0 5 223.62 1
v22 1425 v22 1f 2132.22 5.0 2132.22
2858 5 v22 11 1 285 f 2
Solve this equation by use of the Moody Diagram. e 5 0.010 (Table 4-1) .010 e 5 5 0.002 D 5.0 Trial 1: f 5 0.02 (assumed value of f ) v2 2 5
2858 1 1 2851.022
v22 5 426.6 v2 5 20.65 ft/s NR 5
15.02 120.652 5 1.033 3 107 1 3 1025
f 5 0.022 (from Moody Diagram) Trial 2: f 5 0.024 v2 2 5
2858 1 1 2851.0242
v22 5 364.5 v2 5 19.1 ft/s NR 5
15.02 119.12 5 9.5 3 106 1 3 1025
f 5 0.024 (from Moody Diagram) (OK)
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 29
Therefore, velocity at discharge is 19.1 ft/s. But velocity is the same throughout the pipe, including Point A. h1 5 h2 1
p2 v2 2 L v2 2 1 1f g 2g D 2g
Find h2: Slope of pipe 5
242.5 2 221.12 5 0.0150 ft/ft 1425
h2 5 234.25 1 2.50 5 236.75 ft. 268.0 5 236.75 1 31.25 5
119.12 2 p2 550.0 119.12 2 1 1 1.0242 62.4 2132.22 5.0 2132.22
p2 1 5.66 1 14.95 62.4
p2 5 664 lb/ft2 (Answer) 15. Water in the piezometer at Point B rises to the level of the HGL. The HGL is a distance p/g above the center of the pipe. Find p/g at Point B. Velocity, from Problem 14, is 19.1 ft/s. Find h2: Slope of pipe, from Problem 14, is 0.0150 ft/ft. Invert elevation at B 5 242.5 2 8211.01502 5 230.19 ft. h2 5 230.19 1 2.50 5 232.69 ft. p2 v2 2 L v2 2 1 h1 5 h2 1 1f g 2g D 2g
ELEV. 268.0 PIEZOMETER
INV. 221.12
HG
L
B
60″ 821′ 1425′
92690_ch4_ptg01_p023-038.indd 29
© Cengage Learning 2014
INV. 242.5
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30 CHAPTER 4 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
268.0 5 232.69 1 35.31 5
119.12 2 p2 821 119.12 2 1 1 1.0242 g 2132.22 5.0 2132.22
p2 1 5.66 1 22.32 g
p1 5 7.33 ft g
Elevation in piezometer 5 232.69 1 7.33 5 240.02 ft. (Answer) 16. Velocity from Problem 14 is 19.1 ft/s. Since velocity is the same throughout the pipe, velocity at Point A is 19.1 ft/s. (Answer) 17. h1 1
p2 v1 2 v2 2 1 5 h2 1 1 he 1 hf g 2g 2g
p2 5 0 no pressure at free discharge he 5 0 neglect entrance losses h1 5 h2 1
v2 2 L v2 2 1f 2g D 2g
Find h2: h2 5 489.20 1
2 12
h2 5 489.37 ft. D5
4 5 0.33 ft 12
515.50 5 489.37 1 2
26.13 5
v2 2 175 v22 1f 2132.22 .33 2132.22
v2 11 1 530f 2 64.4
1683 5 v22 11 1 530f 2
Solve this equation by use of the Moody Diagram. e 5 0.00085 (Table 4-1) e .00085 5 5 0.0026 D .33 Trial 1: f 5 0.02 (assumed value of f ) v2 2 5
1683 1 1 5301.022
v22 5 145 v2 5 12.05 ft/s NR 5
1.332 112.052 5 4.0 3 105 1 3 1025
f 5 0.026 (from Moody Diagram) Trial 2: f 5 0.026 v2 2 5
1683 1 1 5301.0262
v22 5 113.9
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 31
v2 5 10.67 ft/s NR 5
1.332 110.672 5 3.55 3 105 1 3 1025
f 5 0.026 (from Moody Diagram) (OK) Therefore, v2 5 10.67 ft/s. (Answer) Q 5 v2a2 p1.332 2 pD2 a2 5 5 5 0.087 ft2 4 4 Q 5 110.672 1.0872
Q 5 0.929 cfs (Answer) 18. h1 5 h2 1
v2 2 v2 2 1 f 1L/D2 2g 2g
v22 v2 2 1 f 12250/102 2132.22 2132.22
825.0 5 700.0 1
8050 5 v22 11 1 225f 2
Solve this equation by use of the Moody diagram. e 5 0.001 (Table 4-1) .001 e 5 5 0.0001 D 10 Trial 1: f 5 0.02 v2 2 5
8050 1 1 2251.022
v22 5 1464 v2 5 38 ft/s NR 5
1382 1102 5 3.8 3 107 (From Equation 4-1) 1 3 1025
f 5 0.012 (From Moody chart) Trial 2: f 5 0.012 v2 2 5
8050 1 1 2251.0122
v22 5 2176 v2 5 47 ft/s NR 5
1472 1102 5 4.7 3 107 (From Equation 4-1) 1 3 1025
f 5 0.012 (From Moody chart) (OK) v2 5 47 ft/s (Answer) Q 5 v2 a (Equation 4-3)
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32 CHAPTER 4 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
a 5 pr2 5 p152 2 5 78.5 ft2 Q 5 1472 178.52
Q 5 3,690 cfs (Answer) 19. h1 5 h2 1
v2 2 v2 2 1 f 1L/D2 2g 2g
v22 v2 2 615.0 5 602.25 1 1 f 15000/.52 2132.22 2132.22
821.1 5 v22 11 1 10000f 2
Solve this equation by use of the Moody diagram. e 5 0.0005 (Table 4-1) .0005 e 5 5 0.001 D .5 Trial 1: f 5 0.02 v2 2 5
821.1 2 1 100001.022
v22 5 4.085 v2 5 2.02 ft/s NR 5
12.022 1.52 5 1.01 3 105 (From Equation 4-1) 1 3 1025
f 5 0.0225 (From Moody chart) Trial 2: f 5 0.0225 v2 2 5
821.1 2 1 100001.02252
v22 5 3.63 v2 5 1.91 ft/s NR 5
11.912 1.52 5 .95 3 105 (From Equation 4-1) 1 3 1025
f 5 0.0225 (From Moody chart) (OK) v2 5 1.91 ft/s (Answer) Q 5 v2 a (Equation 4-3) a 5 pr2 5 p1.252 2 5 0.196 ft2 Q 5 11.912 1.1962
Q 5 0.38 cfs (Answer)
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 33
20. See sketch below.
© Cengage Learning 2014
EGL HGL
21. a) V elocity head is constant throughout the pipe. Velocity head 5 2.4 ft measured by scale. Velocity is computed as v2 2g
Velocity head 5
2.4 5
v 5 12.4 ft/s (Answer)
v2 2132.22
b) Pressure head at the midway point in the pipe is measured by scale to be 4.0 ft. Pressure is computed as p g
Pressure head 5
4.0 5
p 5 249.6 lb/ft2
p 5 250 lb/ft2 (Answer)
p 62.4
c) Entrance loss is the drop in the EGL at the beginning of the pipe and is measured by scale to be 0.8 ft. (Answer) d) Friction loss head at 10.0 feet is the drop of the EGL at a point 10.0 feet into the pipe and is measured as 5.7 ft. (Answer) 22. a) V elocity head is constant throughout the pipe. Velocity head 5 1.5 ft measured by scale. Velocity is computed as
92690_ch4_ptg01_p023-038.indd 33
v2 2g
Velocity head 5
1.5 5
v 5 9.83 ft/s (Answer)
v2 2132.22
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34 CHAPTER 4 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
b) Pressure head at the entrance to Reservoir 2 is measured by scale to be 5.5 ft. Pressure is computed as p g
Pressure head 5
5.5 5
p 5 343.2 lb/ft2
p 5 343 lb/ft2 (Answer)
p 62.4
c) Velocity head is constant throughout the pipe. Velocity head 5 1.5 ft measured by scale. Velocity is computed as v2 2g
Velocity head 5
1.5 5
v 5 9.83 ft/s (Answer)
v2 2132.22
d) Pressure head at Point A is measured by scale to be 8.5 ft. Pressure is computed as Pressure head 5 8.5 5
p g
p 62.4
p 5 530.4 lb/ft2 p 5 530 lb/ft2 (Answer) 23. Divide cross section into Sectors A, B, C, D, and E. Using a scale, measure the areas. Area (ft2)
Sector
A
B
16.0
C
21.25
D
12.25
E
2.75
4.25
Calculate average velocities. 0 1 .86 vA 5 5 0.43 ft/s 2 vB 5
.86 1 3.15 5 2.005 ft/s 2
vC 5
3.15 1 2.59 5 2.87 ft/s 2
vD 5
2.59 1 .46 5 1.525 ft/s 2
vE 5
.46 1 0 5 0.23 ft/s 2
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 35
Calculate discharges using Equation 4-3. QA 5 1.432 14.252 5 1.83 cfs
QB 5 12.0052 116.02 5 32.08 cfs
QC 5 12.872 121.252 5 60.99 cfs
QD 5 11.5252 112.252 5 18.68 cfs QE 5 1.232 12.752 5 0.63 cfs
Sum the discharges. Q 5 114.21 cfs
Q 5 114 cfs (Answer) 24. Divide cross section into Sectors A, B, C, D, E and F. Using a scale, measure the areas. Area (ft2)
Sector
A
30.4
B
90.0
C
174.7
D
212.5
E
150.0
F
25.5
Calculate average velocities. vA 5
0 1 .75 5 0.375 ft/s 2
vB 5
.75 1 1.91 5 1.33 ft/s 2
vC 5
1.91 1 5.21 5 3.56 ft/s 2
vD 5
5.21 1 4.78 5 5.00 ft/s 2
vE 5
4.78 1 1.34 5 3.06 ft/s 2
vF 5
1.34 1 0 5 0.67 ft/s 2
Calculate discharges using Equation 4-3. QA 5 1.3752 130.42 5 11.4 cfs
QB 5 11.332 190.02 5 119.7 cfs
QC 5 13.562 1174.72 5 621.9 cfs
QD 5 15.002 1212.52 5 1062.5 cfs QE 5 13.062 1150.02 5 459.0 cfs QF 5 10.672 125.52 5 17.1 cfs
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36 CHAPTER 4 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Sum the discharges. Q 5 2,291.6 cfs Q 5 2,290 cfs (Answer) 25. Using scale, h 5 0.80 ft. h5
v2 2g
.80 5
v2 2132.22
v2 5 1.802 122 132.22 v 5 7.18 ft/s
v 5 7.2 ft/s. (Answer) 26. Using scale, hr 5 0.80 ft. hr 5 .43
v2 2g
.80 5 .43 v2 5
v2 2132.22
.80164.42 .43
v2 5 119.8 v 5 10.95 ft/s v 5 11 ft/s. 1Answer2 27. W 5 9.0 inches, H 5 712 inches 5 0.625 ft. Q 5 3.0H1.53 (Table 4-3) Q 5 3.01.6252 1.53 Q 5 1.46 cfs Q 5 1.5 cfs (Answer) 28. W 5 2.50 ft, H 5 1.8 ft. Q 5 4 WH1.522W
0.026
(Table 4-3)
Q 5 412.502 11.82 1.522W Q 5 1102 11.82 3.805
0.026
0.026
Q 5 1102 11.82 1.035 Q 5 18.38 cfs
Q 5 18 cfs (Answer)
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 37
29. D1 5 8.00 inches 5 0.667 ft D2 5 4.00 inches 5 0.333 ft p1 5 1.64 ft g p2 5 0.88 ft g a2 5
p1.3332 2 5 0.0871 ft2 4 2ga
Q 5 ca2
p2 p1 2 b g g
D2 4 12a b ï D1
Q 5 c1.08712
Q 5 c1.62922 v5
c1.62922 .0871
ã
(Equation 4-12)
2132.22 11.64 2 .882 .333 4 12a b .667
v 5 c17.2242 Solve by trial and error. Trial 1: v 5 5 ft/s (Assumed value of v) c 5 0.957 (from Table 4-2) v 5 1.9572 17.2242 v 5 6.91 ft/s
Trial 2: v 5 6.91 ft/s (Assumed value of v) c 5 0.9576 (from Table 4-2) v 5 1.95762 17.2242 v 5 6.92 ft/s
Trial 3: v 5 6.92 ft/s (Assumed value of v) c 5 0.9576 (from Table 4-2) v 5 1.95762 17.2242
v 5 6.92 ft/s (OK) Therefore, v 5 6.92 ft/s, c 5 0.9576 Q 5 16.922 1.08712
Q 5 0.603 cfs (Answer)
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38 CHAPTER 4 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
30. D1 5 24.00 inches 5 2.00 ft D2 5 10.00 inches 5 0.833 ft p1 5 2.36 ft g p2 5 0.63 ft g a2 5
p1.8332 2 5 0.5454 ft2 4 2ga
Q 5 ca2
p2 p1 2 b g g
D2 4 12 a b ï D1
Q 5 c1.54542
Q 5 c15.8452 v5
c15.8452 .5454
ã
(Equation 4-12)
2132.22 12.36 2 .632 12a
.833 4 b 2.00
v 5 c110.7182 Solve by trial and error. Trial 1: v 5 5 ft/s (Assumed value of v) c 5 0.9645 (Interpolated from Table 4-2) v 5 1.96452 110.7182 v 5 10.34 ft/s
Trial 2: v 5 10.34 ft/s (Assumed value of v) c 5 0.9762 (Interpolated from Table 4-2) v 5 1.97622 110.7182 v 5 10.46 ft/s
Trial 3: v 5 10.46 ft/s (Assumed value of v) c 5 0.9763 (Interpolated from Table 4-2) v 5 1.97632 110.7182
v = 10.46 ft/s (OK) Therefore, v 5 10.46 ft/s, c 5 0.9763 Q 5 110.462 1.54542
Q 5 5.70 cfs (Answer)
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C h a p t e r
5 Hydralic Devices
1. a 5 pd2/4 5 p11.252 2/4 5 1.227 ft2 Q 5 ca"2gh (Equation 5-3) Q 5 (.62)(1.227)"2(32.2)(7.5) Q 5 17 cfs (Answer) 2. Elevation of center of orifice: 905.25 1 0.5 5 905.75
h 5 928.75 2 905.75 5 23.0 ft
Area of orifice: a 5 (1.0)(1.0) 5 1.0 ft2 Q 5 ca"2gh (Equation 5-3) Q 5 (.62)(1)"2(32.2)(23) Q 5 24 cfs (Answer)
928.75
905.25
© Cengage Learning 2014
12″ x 12″ Orifice
39
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40 CHAPTER 5 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
3. h 5 296.85 2 293.59 5 3.26 ft
(The elevation of the center of the orifice is not used.)
Area of the orifice: a 5
p(.8333)2 pd2 5 5 0.5454 ft2 4 4
Q 5 ca"2gh (Equation 5-3) Q 5 (.62)(.5454)"2(32.2)(3.26) Q 5 4.90 cfs (Answer)
296.85
293.59
289.12
© Cengage Learning 2014
10″ Dia. Orifice
4. h 5 472.00 2 467.50 5 4.5 ft.
p(.25)2 pd2 Area of the orifice: a 5 5 5 0.0491 ft2 4 4 Q 5 ca"2gh (Equation 5-3) Q 5 (.62)(.0491)"2(32.2)(4.5) Q 5 0.52 cfs (Answer)
5. h 5 472.00 2 470.00 5 2.0 ft.
Area of the orifice: a 5 0.0491 ft2 (from prob. 4) Q 5 (.62)(.0491)"2(32.2)(2.0) Q 5 0.35 cfs (Answer)
6. a 5
p(.150)2 5 0.01767 m2 4
h 5 79.25 2 66.10 5 13.15 m Q 5 ca"2gh Q 5 (.62)(.01767)"2(9.81)(13.15) Q 5 0.176 m3/s (Answer)
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 41
7. a 5 0.01767 m2 (from Problem 6)
h 5 79.25 2 71.98 5 7.27 m
Q 5 ca"2gh
Q 5 (.62)(.01767)"2(9.81)(7.27)
Q 5 0.131 m3/s (Answer) p(.333)2 5 0.0871 ft2 4 h 5 10.0 2 1.00 5 9.0 ft
8. a) a 5
Q 5 ca"2gh Q 5 (.62)(.0871)"2(32.2)(9.0) Q 5 1.30 cfs (Answer) b) Divide the tank into five horizontal slices with 1-foot thickness as shown below. Compute discharge at each slice and then compute the time to drain each slice.
1.0′
9.0′
1.0′
10.0′
92690_ch5_ptg01_p039-050.indd 41
10-foot level:
Q 5 1.30 cfs (from Part a)
9-foot level:
Q 5 (.62)(.0871)"2(32.2)(8.0) 5 1.23 cfs
8-foot level:
Q 5 (.62)(.0871)"2(32.2)(7.0) 5 1.15 cfs
7-foot level:
Q 5 (.62)(.0871)"2(32.2)(6.0) 5 1.06 cfs
6-foot level:
Q 5 (.62)(.0871)"2(32.2)(5.0) 5 0.969 cfs
5-foot level:
Q 5 (.62)(.0871)"2(32.2)(4.0) 5 0.867 cfs
© Cengage Learning 2014
4″ ORIFICE
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42 CHAPTER 5 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Time to drain first (top) slice: Average Q 5
1.30 1 1.23 5 1.265 cfs 2
Volume of each slice 5 (10.0)(10.0)(1.0) 5 100 ft3 Time 5
Vol. 100 5 5 79.1 s Q 1.265
Time to drain second slice: Average Q 5 Time 5
1.23 1 1.15 5 1.19 cfs 2
100 5 84.0 s 1.19
Time to drain third slice: Average Q 5 Time 5
1.15 1 1.06 5 1.105 cfs 2
100 5 90.5 s 1.105
Time to drain fourth slice: Average Q 5 Time 5
1.06 1 .969 5 1.0145 cfs 2
100 5 98.6 s 1.0145
Time to drain fifth slice: Average Q 5 Time 5
.969 1 .867 5 0.918 cfs 2
100 5 108.9 s .918
Total time 5 79.1 1 84.0 1 90.5 1 98.6 1 108.9 5 461.1 s Total time 5 461.1 s 5 7.69 min (Answer) 9. h 5 7.5 in 3
1 ft 5 0.625 ft 12 in
c 5 2.5
u Q 5 c tan H 5/2 (Equation 5-7) 2
Q 5 2.5(tan 45)(.625)5/2
Q 5 0.77 cfs (Answer)
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 43
10. h 5 6.25 in 3
1 ft 5 0.521 ft 12 in
c 5 2.5
Q 5 c tan u H5/2 (Equation 5-7)
Q 5 2.5(tan 60)(.521)5/2
Q 5 0.85 cfs (Answer)
11. H 5 0.55 ft
n 5 1 (one contraction)
L 5 Lr 2 0.1 nH (Equation 5-5)
L 5 4.00 2 0.1(1)(.55)
L 5 3.945 ft
H c 5 3.27 1 0.40 (Equation 5-6) P .55 c 5 3.27 1 0.40 1.25 c 5 3.446
Q 5 cLH3/2
Q 5 (3.446)(3.945)(.55)3/2
Q 5 5.55 cfs (Answer)
12. H 5 121.32 2 120.00 5 1.32 ft
n 5 0 (No contractions)
L 5 Lr 5 8.50r
H c 5 3.27 1 0.40 (Equation 5-6) P 1.32 c 5 3.27 1 .40 1.50 c 5 3.62
Q 5 cLH3/2 (Equation 5-4)
Q 5 (3.62)(8.50)(1.32)3/2
Q 5 46.7 cfs (Answer)
Elev. 121.32
Channel Inv. 118.50 8.50′
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© Cengage Learning 2014
Crest Elev. 120.00
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44 CHAPTER 5 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
13. H 5 0.82 ft, Lr 5 2.50 ft, P 5 1.75 ft, n 5 2.
L 5 Lr 2 0.1 nH (Equation 5-5)
L 5 2.50 2 (.1)(2)(.82)
L 5 2.336 ft
H c 5 3.27 1 0.40 (Equation 5-6) P (.82) c 5 3.27 1 .40 1.75 c 5 3.457
Q 5 cLH3/2 (Equation 5-4)
Q 5 (3.457)(2.336)(.82)1.5
Q 5 5.996 cfs
Q 5 6.00 cfs (Answer)
14. H 5 121.32 2 120.00 5 1.32 ft
c 5 2.90 (From Appendix A-5)
Q 5 cLH3/2 (Equation 5-4)
Q 5 (2.90)(8.50)(1.32)3/2
Q 5 37.4 cfs (Answer)
15. First, calculate Q1 for the primary (lower) crest.
H 5 525.14 2 522.75 5 2.39 ft
c 5 3.02 (From Appendix A-5)
Q1 5 cLH3/2 (Equation 5-4)
Q1 5 (3.02)(1.50)(2.39)3/2
Q1 5 16.7 cfs
Next, calculate Q2 for the secondary crest.
L 5 10.00 2 1.50 5 8.50 ft
H 5 525.14 2 522.75 2 2.00
H 5 0.39 ft
c 5 2.61 (From Appendix A-5)
Q2 5 cLH3/2 (Equation 5-4)
Q2 5 (2.61)(8.50)(.39)3/2
Q2 5 5.4 cfs
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 45
Finally, find total Q.
Q 5 Q1 1 Q2
Q 5 16.7 1 5.4
Q 5 22.1 cfs (Answer) Elev. 525.14
2.00′ 1.50′
Elev. 522.75
© Cengage Learning 2014
10.00′
16. Breadth 5 1.0 ft, L 5 12.0 ft, H 5 2.0 ft.
c 5 3.30 (Figure A-5)
Q 5 cLH3/2 (Equation 5-4)
Q 5 (3.30)(12.0)(2.0)1.5
Q 5 112 cfs (Answer)
17. Breadth 5 7.0 ft (by scale)
a) L1 5 15 ft (by scale) H1 5 322.10 2 320.0 5 2.1 ft c1 5 2.65 (Interpolated from Figure A-5) Q1 5 (2.65)(15)(2.1)1.5 Q1 5 121 cfs (Answer)
b) L1 5 15 ft H1 5 324.17 2 320.0 5 4.17 ft c1 5 2.68 (Interpolated from Figure A-5) Q1 5 (2.68)(15)(4.17)1.5 Q1 5 342 cfs L2 5 130 ft (by scale) H2 5 324.17 2 324.0 5 0.17 ft c2 5 2.40 (Interpolated from Figure A-5) Q2 5 (2.40)(130)(.17)1.5 Q2 5 21.9 cfs Q 5 342 1 21.9 5 364 cfs (Answer)
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46 CHAPTER 5 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
c) L1 5 15 ft H1 5 325.24 2 320.0 5 5.24 ft c1 5 2.76 (Interpolated from Figure A-5) Q1 5 (2.76)(15)(5.24)1.5 Q1 5 497 cfs L2 5 130 ft H2 5 325.24 2 324.0 5 1.24 ft c2 5 2.67 (Interpolated from Figure A-5) Q2 5 (2.67)(130)(1.24)1.5 Q2 5 479 cfs L2 5 56 ft (by scale) H2 5 325.24 2 325.0 5 0.24 ft c2 5 2.42 (Interpolated from Figure A-5) Q2 5 (2.42)(56)(.24)1.5 Q2 5 15.9 cfs Q 5 497 1 479 1 15.9 Q 5 992 cfs Q 5 990 cfs (Answer)
18. a 5 1/ 2(4.00)(6.25) 5 12.50 ft2 h 5 12.72 2 c 5 0.77
1 6.25 a b 5 11.1575 ft 2 2
Q 5 ca"2gh (Equation 5-3) Q 5 (.77)(12.50)"2(32.2)(11.1575) Q 5 258 cfs (Answer) p2 v22 v12 1 5 h2 1 1 he 1 h f 19. a) h1 1 g 2g 2g v1 5 0 due to negligible water movement in the pool p2 5 0 no pressure at free discharge he 5 0 neglect entrance losses h1 2 h2 5
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v22 L v22 1f 2g D 2g
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 47
10.0 5
v22 v22 50 1f 2(32.2) .083 2(32.2)
644 5 v22(1 1 600f )
Solve this equation by use of the Moody Diagram. e 5 0.000005 (Table 4-1) e .000005 5 5 0.00006 D .083 Trial 1: f 5 0.02 (assumed value of f ) v22 5
644 1 1 600(.02)
v22 5 49.54 v2 5 7.04 ft/s NR 5
(.083)(7.04) 5 5.86 3 104 1 3 1025
f 5 0.02 (from Moody Diagram) (OK) v2 5 7.04 ft/s a5
p(.083)2 pD2 5 5 0.00544 ft2 4 4
Q 5 v2a Q 5 (7.04)(.00544) Q 5 0.038 cfs (Answer)
b) Divide the pool into four horizontal 1-foot thick slices as shown below. Compute discharge at each slice and then compute the time to drain each slice.
6.0′ 2.0′
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© Cengage Learning 2014
1.0′
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48 CHAPTER 5 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
6-foot level: Q 5 0.038 cfs (from Part a) 5-foot level: v22 v22 50 9.0 5 1f 2(32.2) .083 2(32.2)
579.6 5 v22(1 1 600f )
f 5 0.02 v2 5 6.68 ft/s Q 5 (6.68)(.00544) Q 5 0.036 cfs 4-foot level: 8.0 5
v22 v22 50 1f 2(32.2) .083 2(32.2)
515.2 5 v22(1 1 600f )
f 5 0.021 v2 5 6.15 ft/s Q 5 (6.15)(.00544) Q 5 0.033 cfs 3-foot level: v22 v22 50 1f 7.0 5 2(32.2) .083 2(32.2)
450.8 5 v22(1 1 600f )
f 5 0.0215 v2 5 5.69 ft/s Q 5 (5.69)(.00544) Q 5 0.031 cfs 2-foot level: v22 v22 50 1f 6.0 5 2(32.2) .083 2(32.2)
386.4 5 v22(1 1 600f )
f 5 0.022 v2 5 5.22 ft/s Q 5 (5.22)(.00544) Q 5 0.028 cfs
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 49
Time to drain first (top) slice: Average Q 5
.038 1 .036 5 0.037 cfs 2
Volume of each slice 5 (8.00)(12.00)(1.00) 5 96.0 ft3 Time 5
Vol. 96.0 5 5 2595 s Q .037
Time to drain second slice: Average Q 5 Time 5
.036 1 .033 5 0.0345 cfs 2
96.0 5 2783 s .0345
Time to drain third slice: Average Q 5 Time 5
.033 1 .031 5 0.032 cfs 2
96.0 5 3000 s .032
Time to drain fourth slice: Average Q 5 Time 5
.031 1 .028 5 0.0295 cfs 2
96.0 5 3254 s .0295
Total time 5 2595 1 2783 1 3000 1 3254 5 11,632 s Total time 5 193.9 min Total time 5 3.23 hours (Answer) 20. h1 5 h2 1
f(L) v22 v22 1 2g D 2g
1125.0 5 (1032.0 1 7.5) 1
f(1230) v22 v22 1 2(32.2) 15 2(32.2)
85.5(64.4) 5 v22(1 1 82f )
5506 5 v22(1 1 82f )
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50 CHAPTER 5 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Solve this equation by use of the Moody diagram. e 5 0.001 (Table 4-1) e .001 5 5 0.000067 D 15 Trial 1: f 5 0.02 v22 5
5506 1 1 82(.02)
v22 5 2086 v2 5 45.7 ft/s NR 5
(15)(45.7) 5 6.9 3 107 (From Equation 4-1) 1 3 1025
f 5 0.011 (From Moody chart) Trial 2: f 5 0.011 v22 5
5506 1 1 82(.011)
v22 5 2895 v2 5 53.8 ft/s NR 5
(15)(53.8) 5 8.1 3 107 (From Equation 4-1) 1 3 1025
f 5 0.011 (From Moody chart) (OK) v2 5 53.8 ft/s (Answer) Q 5 v2a (Equation 4-3) a 5 pr2 5 p(7.5)2 5 176.7 ft2 Q 5 (53.8)(176.7) Q 5 9507 cfs Q 5 9500 cfs (Answer)
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C h a p t e r
6 Open Channel Hydraulics
1. slope 5 slope 5
drop length 3.75 200
slope 5 0.01875 ft/ft slope 5 1.88% (Answer) 2. slope 5
drop length
slope 5
423.92 2 422.05 150 2 100
slope 5
1.87 50
slope 5 0.0374 ft/ft slope 5 3.74% (Answer) 3. (a) a 5 (6.0)(2.45) 5 14.7 ft2 (Answer) (b) p 5 (2)(2.45) 1 6.0 5 10.9 ft (Answer) (c) R 5 4. a 5 a5
a 14.7 5 5 1.35 ft (Answer) p 10.9
Q (Equation 4-3) v 210 5.45
a 5 38.5 ft2 (Answer) 51
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52 CHAPTER 6 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
5. (a) a 5
8.5 2 4.5 (1.50) 5 9.75 ft2 (Answer) 2
(b) p 5 (2)(2.5) 1 4.5 5 9.5 ft (Answer)
(c) R 5
6. (a) a 5
a 9.75 5 5 1.03 ft (Answer) p 9.5 8.0 1 5.0 (3.0) 5 19.5 ft2 (Answer) 2
(b) p 5 "(1.5)2 1 (3.0)2 1 5.0 1 "(1.5)2 1 (3.0)2 5 11.7 ft (Answer) (c) R 5
a 19.5 5 5 1.67 ft (Answer) p 11.7
7. (a) Dn 5 2 1 0.5 5 2.5 ft (Answer)
(b) a 5 (2)(.5) 1 c
(d) R 5
10 1 22 d (2) 5 33 ft2 (Answer) 2
(c) p 5 "22 1 62 1 4 1 .5 1 2 1 .5 1 2 1 .5 1 4 1 "22 1 62 5 23.6 ft (Answer)
8. (a) a 5
a 33 5 5 1.40 ft (Answer) p 23.6
24.0 1 4.00 (5.0) 5 70 ft2 (Answer) 2
(b) p 5 (2)"(5.0)2 1 (10.0)2 1 4.00 5 26 ft (Answer)
(c) R 5
a 70 5 5 2.69 ft (Answer) p 26
9. (a) Dn 5 3.8 ft (by scale) (Answer)
(b) a 5 (3.8)(8.0) 5 30.4 ft2 (Answer)
(c) p 5 3.8 1 8.0 1 3.8 5 15.6 ft (by scale) (Answer)
(d) R 5
a 30.4 5 5 1.95 ft (Answer) p 15.6
10. (a) Dn 5 4.5 ft (by scale) (Answer)
(b) a 5 c
(d) R 5
23.7 1 6.0 d (4.5) 5 66.8 ft2 (Answer) 2
(c) p 5 9.8 1 6.0 1 9.8 5 25.6 ft (by scale) (Answer) a 66.8 5 5 2.61 ft (Answer) p 25.6
11. (a) Dn 5 3.7 ft (by scale) (Answer)
(b) a 5 1/2(18.3)(3.7) 5 33.9 ft2 (Answer)
(c) p 5 5.2 1 14.9 5 20.1 ft (by scale) (Answer)
(d) R 5
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a 33.9 5 5 1.69 ft (Answer) p 20.1
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 53
12. (a) Dn 5 5.15 ft (by scale) (Answer)
(b) a 5 1/2(8.75)(.9) 1 (8.75)(.85) 1 c
(d) R 5
16.0 1 20.0 d (3.4) 5 73 ft2 (Answer) 2
(c) p 5 3.4 1 4.0 1 .85 1 4.5 1 4.5 1 .85 1 3.3 1 5.0 5 26.4 ft (Answer) a 73 5 5 2.75 ft (Answer) p 26.4
13. (a) Dn 5 5.6 ft (by scale) (Answer)
(b) a 5 c
18.2 1 22.0 d (5.6) 1 (6.8)(3.2) 5 134 ft2 (Answer) 2
(c) The inclined sides cannot be measured directly using an exaggerated scale. Use the Pythagorean Theorem. Left side: length 5 "(5.6)2 1 (25)2 5 25.6 ft
Middle side: length 5 "(2.4)2 1 (6.0)2 5 6.5 ft
Right side: length 5 "(3.2)2 1 (7.5)2 5 8.2 ft
p 5 25.6 1 18.2 1 6.5 1 6.8 1 8.2 5 65.3 ft (Answer)
(d) R 5
a 134 5 5 2.05 ft (Answer) p 65.3
14. (a) Dn 5 4.8 ft (by scale) (Answer)
(b) To measure area use CAD software or divide the area into vertical strips and measure each incremental area. Following is measurement of eight strips. (since the channel is symmetrical, four strips will be measured and the area doubled.
strip 1: a 5 1(2.0)(2.0) 5 2.0 ft2
strip 2: a 5 c
2.0 1 3.5 d 12.02 5 5.5 ft2 2
strip 4: a 5 c
4.3 1 4.8 d (2.0) 5 9.1 ft2 2
strip 3: a 5 c
3.5 1 4.3 d (2.0) 5 7.8 ft2 2
a 5 2(2.0 1 5.5 1 7.8 1 9.1) 5 48.8 ft2 (Answer)
(c) To estimate wetted perimeter, measure the length of each strip. (Make four measurements and double the length.)
strip 1: p 5 2.8 ft
strip 2: p 5 2.4 ft
strip 3: p 5 2.2 ft
strip 4: p 5 2.0 ft
p 5 2(2.8 1 2.4 1 2.2 1 2.0) 5 18.8 ft (Answer)
(d) R 5
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a 48.8 5 5 2.60 ft (Answer) p 18.8
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54 CHAPTER 6 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
15. T 5 8.0 ft, Q 5 125 cfs. Dc 5 c
Dc 5 c
Q2 1/3 d (Equation 6-3) T2g 1/3 1252 d 18.02 2(32.2)
Dc 5 2.0 ft (Answer)
16. Q 5 100 cfs. Find Dc using Equation 6-2 solved by trial and error. Q2 (100)2 5 310.6 5 g 32.2 Trial
Dc (ft)
a (ft2)
a3
T (ft)
a3/T
1
1.0
5.5
166
6.0
28
2
2.0
12.0
1728
7.0
247
3
3.0
19.5
7415
8.0
927
4
2.14
12.99
2192
7.14
307
(OK)
Therefore, Dc 5 2.14 ft (Answer) 17. Q 5 200 cfs. Find Dc using Equation 6-2 solved by trial and error. Q2 (200)2 5 1242 5 g 32.2 Trial
Dc (ft)
a (ft2)
T (ft)
a
a3/T
1
1.0
6.0
216
8.0
27
2
2.0
16.0
4096
12.0
341
3
3.0
30.0
27000
16.0
1688
4
2.78
26.58
18772
15.12
1242
3
(OK)
Therefore, Dc 5 2.78 ft (Answer) Note: Check this answer using Chart 17 in Appendix A-3. 18. Q 5 350 cfs. Find Dc using Equation 6-2 solved by trial and error. Q2 (350)2 5 3804 5 g 32.2 Trial
Dc (ft)
a (ft2)
a3
T (ft)
a3/T
1
2.0
10.2
1061
10.2
104
2
3.0
22.8
11852
15.2
780
3
4.0
40.0
64000
20.0
3200
4
4.2
43.3
80958
20.6
3930
(OK)
Therefore, Dc 5 4.2 ft (Answer)
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 55
19. Q 5 650 cfs. Find Dc using Equation 6-2 solved by trial and error. Q2 (650)2 5 13,121 5 g 32.2 Trial
Dc (ft)
1
2.0
2
a (ft2)
a3
T (ft)
a3/T
28
21952
18.0
1220
3.0
48
110592
22.0
5027
3
4.0
72
373248
26.0
14356
4
3.90
69.4
334544
25.6
13068 (OK)
Therefore, Dc 5 3.90 ft (Answer) Note: Check this answer using Chart 23 in Appendix A-3. 20. Q 5 150 cfs. Find Dc using Equation 6-2 solved by trial and error. Q2 (150)2 5 699 5 g 32.2 Trial
Dc (ft)
a (ft2)
a3
T (ft)
a3/T
1
1.0
7.0
343
9.0
38
2
2.0
18.0
5832
13.0
449
3
3.0
33.0
35937
17.0
2114
4
2.25
21.4
9766
14.0
698
(OK)
Therefore, Dc 5 2.25 ft (Answer) Note: Check this answer using Chart 18 in Appendix A-3. 21. Q 5 115 cfs. Find velocity using Equation 4-10. Find the Froude number using Equation 6-4. a 5 18 ft2 v5 F5
Q 115 5 5 6.4 ft/s (Answer) a 18 v "gD
5
6.4 "(32.2)(2.0)
5 0.80 (Answer)
Flow is subcritical (F , 1) (Answer) 22. Q 5 180 cfs. Find the Froude number using Equation 6-4. a 5 (15.0)(1.2) 5 18 ft2 v5 F5
Q 180 5 5 10 ft/s a 18 v "gD
5
10 "(32.2)(1.2)
5 1.6 (Answer)
Flow is supercritical (F . 1) (Answer)
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56 CHAPTER 6 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
23. Q 5 300 cfs. Find the Froude number using Equation 6-5. a 5 19.1 ft2 v5
Q 300 5 5 15.7 ft/s a 19.1
Dh 5 F5
a 19.1 5 5 1.47 ft T 13.0 v
"gDh
5
15.7 "(32.2)(1.47)
5 2.28 (Answer)
Flow is supercritical (F . 1) (Answer) 24. Q 5 450 cfs. Find the Froude number using Equation 6-5. a 5 37.5 ft2 v5
Q 450 5 5 10.7 ft/s a 37.5
Dn 5 F5
a 37.5 5 5 1.88 ft T 20 v
"gDn
5
10.7 "(32.2)(1.88)
5 1.38 (Answer)
Flow is supercritical (F . 1) (Answer)
Dn (ft)
y (ft/s)
E (ft)
0.5 1.0 1.5 2.0 2.1 2.2 2.3 2.4 2.5 3.0 4.0 5.0
40.0 20.0 13.3 10.0 9.5 9.1 8.7 8.3 8.0 6.7 5.0 4.0
25.3 7.2 4.2 3.6 3.5 3.49 3.47 3.48 3.5 3.7 4.4 5.2
© Cengage Learning 2014
25. See Table 1 and Figure 1.
TABLE 1 Values of Dn vs. E for problem 25.
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D (ft)
HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 57
5 4 3
Dc = 2.3′ 2
1
2
3
4
5
6 E (ft)
7
8
9
10
© Cengage Learning 2014
1
FIGURE 1 Graph of Dn vs. E for problem 25.
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C h a p t e r
7 Uniform Flow in Channels
1. Dn 5 1.25 ft, so 5 2.50%. Find discharge using Equation 7-2. a 5 15.02 11.252 5 6.25 ft2
p 5 15.02 1 211.252 5 7.5 ft
R5
a 6.25 5 5 0.833 ft p 7.5
1a2 Q 5 16.252
1.49 1.8332 2/3 1.0252 1/2 .014
Q 5 93.1 cfs (Answer)
1b2 v 5
Q 93 5 5 14.9 ft/s (Answer) a 6.25
2. Dn 5 2.9 ft, so 5 0.47%. Find discharge using Equation 7-2. a 5 13.752 12.92 5 10.88 ft2
p 5 13.752 1 212.92 5 9.55 ft
R5
a 10.88 5 5 1.14 ft p 9.55
1a2 Q 5 110.882
1.49 11.142 2/3 1.00472 1/2 .016
Q 5 76 cfs (Answer)
(b) v 5
Q 76 5 5 7.0 ft/s (Answer) a 10.88
58
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 59
3. Dn 5 2.35 ft, so 5 1.1%. Find discharge using Equation 7-2. a 5 21.03 ft2 (Trapezoid) p 5 14.252 1 215.252 5 14.76 ft
R5
a 21.03 5 5 1.42 ft p 14.76
1a2 Q 5 121.032
1.49 11.422 2/3 1.0112 1/2 .027
Q 5 154 cfs (Answer)
1b2 v 5
Q 154 5 5 7.31 ft/s (Answer) a 21.03
4. Dn 5 3.0 ft, so 5 4.2%. Find discharge using Equation 7-2. a 5 46.5 ft2 (Trapezoid) p 5 18.02 1 218.082 5 24.16 ft
R5
a 46.5 5 5 1.93 ft p 24.16
1a2 Q 5 146.52
1.49 11.932 2/3 1.0422 1/2 .024
Q 5 916 cfs (Answer)
1b2 v 5
Q 916 5 5 19.7 ft/s (Answer) a 46.5
© Cengage Learning 2014
5. Channel cross section is shown below:
D
20.0′
Find Q using Equation 7-2 solved by trial and error. 1.49 1/2 s 5 9.926 n o
1
D 1ft2
1.0
a 1ft2 2
20
p 1ft2
2
2.0
40
24
1.67
1.41
558
3
1.5
30
23
1.30
1.19
356
4
1.49
29.8
23
1.30
1.19
352 (OK)
Trial
22
R 1ft2
.909
R2/3 .938
Q 1cfs2 186
D 5 1.49 ft (Answer)
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60 CHAPTER 7 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
6. Channel cross section is shown below: 1
1
2 © Cengage Learning 2014
2
D 3.00′
Find D using Equation 7-2 solved by trial and error. 1.49 1/2 1.49 s 5 1.012 1/2 5 6.208 n o .024
1
D 1ft2
2.0
a 1ft2 2
14
p 1ft2
11.9
R 1ft2
1.18
1.11
Q 1cfs2
2
3.0
27
16.4
1.65
1.39
234
3
2.2
16.28
12.8
1.27
1.17
119
4
2.26
17.00
13.11
1.30
1.19
125 (OK)
Trial
2/3
R
97
D 5 2.26 ft (Answer) 7. Dn 5 2 ft, so 5 1.75%. Find discharge using Equation 7-2. n 5 0.013 (Appendix A-1) a 5 32 1 1.0 5 33 ft2 (Trapezoid and rectangle) p 5 216.322 1 2142 1 21.52 1 2 5 23.65 ft R5
a 33 5 5 1.40 ft p 23.65
1a2 Q 5 1332
1.49 11.402 2/3 1.01752 1/2 .013
Q 5 626 cfs (Answer)
1b2 v 5
Q 626 5 5 19 ft/s (Answer) a 33
8. Q 5 56 ft, Dn 5 1.50 ft. Find average velocity using Equation 4-3. a 5 16.02 11.502 5 9.0 ft2 v5
Q 56 5 5 6.2 ft/s (Answer) a 9.0
9. Q 5 210 ft, v 5 5.45 ft/s. Find cross sectional area using Equation 4-3. a5
Q 210 5 5 38.5 ft2 v 5.45
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 61
10. so 5 0.62%. Find discharge using Equation 7-2. n 5 0.013 (Appendix A-1) 16.0 1 20.0 a 5 12 18.752 1.92 1 18.752 1.852 1 c d 13.42 5 73 ft2 2
p 5 3.4 1 4.0 1 .85 1 4.5 1 4.5 1 .85 1 3.3 1 5.0 5 26.4 ft
R5
a 73 5 5 2.75 ft p 26.4
1a2 Q 5 1732
1.49 12.752 2/3 1.00622 1/2 .013
Q 5 1,294 cfs
Q 5 1,300 cfs (Answer) 1b2 v 5
Q 1294 5 5 18 ft/s (Answer) a 73
11. Q 5 225 cfs, so 5 2.0%. Find normal depth using Equation 7-2 solved by trial and error. 1.49 1/2 1.49 s 5 1.0202 1/2 5 17.56 n o .012
1
Dn 1ft2
1.0
a 1ft2 2
7.0
p 1ft2
R 1ft2
0.739
0.817
Q 1cfs2
2
2.0
18.0
13.9
1.29
1.19
375
3
1.5
12.0
11.7
1.02
1.02
214
4
1.54
12.44
11.86
1.05
1.03
226 (OK)
Trial
9.47
R2/3
100
Therefore, Dn 5 1.54 ft (Answer) Note: Check this answer using Chart 18 in Appendix A-3. 12. Dn 5 2.25 ft, v 5 7.32 ft/s. Find discharge using Equation 4-3. a 5 19.13 ft2 Q 5 va 5 17.322 119.132 5 140 cfs (Answer)
13. a 5 134 ft2 p 5 65.3 ft
R 5 2.05 ft n 5 0.0325 (Appendix A-1) so 5 2.25% (a) To find Q use Manning’s equation. Q 5 11342
1.49 12.052 2/3 1.02252 1/2 .0325
Q 5 1,487 cfs (Answer)
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62 CHAPTER 7 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
(b) v 5 v5
Q a 1487 5 11.1 ft/s (Answer) 134
14. a 5 48.8 ft2 p 5 18.8 ft R 5 2.60 ft n 5 0.028 (Appendix A-1) so 5 3.00% (a) To find Q use Manning’s equation. Q 5 148.82
1.49 12.602 2/3 1.03002 1/2 .028
Q 5 851 cfs (Answer) (b) v 5 v5
Q a 851 5 17.4 ft/s (Answer) 48.8
15. Q 5 300 cfs, so 5 1.25%. Find normal depth using Equation 7-2 solved by trial and error. n 5 0.013 (Appendix A-1) 1.49 1/2 1.49 s 5 1.01252 1/2 5 12.81 n o .013
1
Dn 1ft2
1.0
a 1ft2 2
6.75
14.2
2
2.0
22.75
20.5
Trial
p 1ft2
R 1ft2
.475 1.11
R2/3 .609 1.07
Q 1cfs2 53
312 (OK)
Therefore, Dn 5 2.0 ft (Answer) 16. Q 5 600 cfs, so 5 2.0%. Find normal depth using Equation 7-2 solved by trial and error. n 5 0.013 (Appendix A-1) 1.49 1/2 1.49 s 5 1.0202 1/2 5 16.21 n o .013
1
Dn 1ft2
2.0
a 1ft2 2
15.5
18.7
2
3.0
32.8
21.1
1.55
1.34
714
3
2.8
29.3
20.6
1.42
1.26
601 (OK)
Trial
p 1ft2
R 1ft2
.829
R2/3 .882
Q 1cfs2 222
Therefore, Dn 5 2.8 ft (Answer)
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 63
17. a 5 1/2
pd2 p32 5 1/2 4 4
a 5 3.534 ft2 v5
Q (Equation 4-3) a
v5
15 3.534
v 5 4.2 ft/s (Answer) 18. Use Manning’s Equation for full flow. a5
p11.252 2 pd 2 5 5 1.227 ft2 4 4
p 5 pd 5 p11.252 5 3.927 ft R5
a 1.227 5 5 0.312 ft (Equation 6-1) p 3.927
Q 5 11.2272
1.49 1.3122 2/3 1.01252 1/2 (Equation 7-2) .015
Q 5 6.27 cfs (Answer)
19. Use Manning’s Equation for full flow. a5
p132 2 pd 2 5 5 7.069 ft2 4 4
p 5 pd 5 p132 5 9.425 ft R5
a 7.069 5 5 0.750 ft p 9.425
Q 5 17.0692
1.49 1.7502 2/3 1.0082 1/2 (Equation 7-2) .012
Q 5 64.8 cfs (Answer)
20. Use Manning’s Equation for full flow. p1.672 2 pd 2 5 5 0.349 ft2 a5 4 4 p 5 pd 5 p1.672 5 2.09 ft R5
a .349 5 5 0.167 ft p 2.09
Q 5 1.3492
1.49 1.1672 2/3 1.0252 1/2 .009
Q 5 2.77 cfs (Answer)
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64 CHAPTER 7 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
21. Q 5 75 cfs, so 5 1.20%. Find normal depth using Equation 7-2 solved by trial and error. Find cross sectional areas and wetted perimeters using Figure 7-3. Area of 60-inch pipe 5 p12.52 2 5 19.63 ft2 Perimeter of 60-inch pipe 5 p152 5 15.71 ft 1.49 1/2 1.49 s 5 1.01202 1/2 5 13.60 n o .012
1
Dn 1ft2
1.0
a 1ft2 2
2.75
4.56
2
2.0
7.26
6.91
3
1.7
5.89
6.28
Trial
p 1ft2
R 1ft2
R2/3
.603 1.05
.714 1.03
.938
.958
Q 1cfs2 27 102 77 (OK)
Therefore, Dn 5 1.7 ft (Answer) Note: Check this answer using Chart 47 in Appendix A-4. 22. Q 5 50 cfs, so 5 1.90%. Find normal depth using Equation 7-2 solved by trial and error. Find cross sectional areas and wetted perimeters using Figure 7-3. Area of 48-inch pipe 5 p12.02 2 5 12.57 ft2 Perimeter of 48-inch pipe 5 p152 5 12.57 ft 1.49 1/2 1.49 s 5 1.01902 1/2 5 17.12 n o .012
1
Dn 1ft2
1.0
a 1ft2 2
2.39
4.15
2
2.0
6.29
6.29
3
1.3
3.52
4.90
Trial
p 1ft2
R 1ft2
R2/3
.576 1.00
.692 1.00
.718
.802
Q 1cfs2 28 108 48 (OK)
Therefore, Dn 5 1.3 ft (Answer) Note: Check this answer using Chart 45 in Appendix A-4. 23. Q 5 2.0 cfs. so 5 4.50%. Find normal depth using Equation 7-2 solved by trial and error. Find cross sectional areas and wetted perimeters using Figure 7-3. Area of 15-inch pipe 5
p11.252 2 5 1.227 ft2 4
Perimeter of 15-inch pipe 5 p11.252 5 3.927 ft
0.25
a 1ft2 2
.175
p 1ft2
1.16
R 1ft2
R
1
Dn 1ft2
.151
.415
Q 1cfs2
2
0.50
.458
1.71
.268
.283
4.0
3
0.35
.281
1.39
.202
.344
2.0 (OK)
Trial
2/3
1.0
Therefore, Dn 5 0.35 ft (Answer) Note: Check this answer using Chart 36 in Appendix A-4.
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 65
24. Find elevation using Manning’s Equation separately for channel and overbanks. Solve by trial and error. Left Overbank:
n 5 0.060 (From Appendix A-1)
1.49 1/2 s 5 0.8236 n o
Channel:
n 5 0.0425 (From Appendix A-1)
1.49 1/2 s 5 1.163 n o
Right Overbank: same as left overbank To solve Manning’s Equation, find a and p by scaling distances on the cross section (drawn to scale). Computations are shown in Table 2. Elevation 5 235.6 (Answer) Channel R.O.B. L.O.B. p R Q a p R Q Trial Elev. D a p R Q a 1ft2 1ft2 (ft2) (ft) (ft) (cfs) (ft2) (ft) (ft) (cfs) (ft2) (ft) (ft) (cfs) 1
230.0 3.7 –
–
–
85 –
–
–
–
2
232.0 5.7 3
6
.50 2 73.5 20 3.68 204 10 10.2 .98
3
234.0 7.7 24.8 16.8 1.47 26 104.5 20 5.23 366 41 20.8 1.97 53 445
4
236.0 9.7 65
26.5 2.45 97 135.5 20 6.78 565 96 32.5 2.95 163 825
5
235.6 9.3 55
25.5 2.16 76 129.3 20 6.47 522 84 30.5 2.75 136 734 (OK)
85
8 214
© Cengage Learning 2014
– 42.5 19 2.24
TOTAL Q (cfs)
TABLE 2 Summary of computations of normal depth for the stream in Problem 24. 25. Find elevation using Manning’s Equation separately for channel and overbanks. Solve by trial and error. Left Overbank:
n 5 0.070
1.49 1/2 s 5 1.72 n o
Channel:
n 5 0.060
1.49 1/2 s 5 2.00 n o
Right Overbank:
n 5 0.135
1.49 1/2 s 5 0.890 n o
To solve Manning’s Equation, find a and p by scaling distances on the cross section (drawn to exaggerated scale). Computations are shown in Table 3. Graph is shown in Figure 2. Elevation 5 1208.80 1Dn 5 7.0 ft2 (Answer)
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66 CHAPTER 7 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
1 4.0 2 5.0 3 6.0 4 7.0 5 8.0
– – – – 12.5 25 .500 14 49 49 1.00 84 110 73 1.51 145 192 96 2.00 524
72 98 126 154 182
28 30 30 30 30
2.57 3.27 4.20 5.13 6.07
270 432 656 917 1212
– – 65 170 285
– – – – – – 102 .637 43 110 1.55 202 118 2.42 457
TOTAL Q (cfs) 270 446 783 1264 2193
© Cengage Learning 2014
L.O.B. Channel R.O.B. Trial D a p R Q a p R Q a p R Q (ft) (ft) (ft) (ft) (cfs) (ft) (ft) (ft) (cfs) (ft) (ft) (ft) (cfs)
TABLE 3 Summary of computations of normal depth for the stream in Problem 25.
10.0
Depth (ft)
8.0 6.0 4.0
500
1000 1500 Discharge (cfs)
2000
2500
© Cengage Learning 2014
2.0
FIGURE 2 Stream rating curve for Problem 25.
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C h a p t e r
8 Varied Flow in Channels
1. Q 5 225 cfs, so 5 0.50%. Find normal depth using Equation 7-2 solved by trial and error. Find Fraude number using Equation 6-4. 1.49 1/2 1.49 s 5 1.0052 1/2 5 7.02 n o .015
1
Dn 1ft2
2.0
a 1ft2 2
12.0
p 1ft2
10.0
R 1ft2
1.20
1.13
Q 1cfs2
2
3.0
18.0
12.0
1.50
1.31
166
3
4.0
24.0
14.0
1.71
1.43
242
4
3.8
22.8
13.6
1.68
1.41
226 (OK)
Trial
R2/3
95
Therefore, Dn 5 3.8 ft Note: Check this result using Chart 5 in Appendix A-3. v5 F5
Q 226 5 5 9.9 ft/s a 22.8 v "gD
5
9.9 "132.22 13.82
5 0.89
Flow is subcritical 1F , 12 (Answer)
2. Q 5 225 cfs, so 5 1.0%. Find normal depth using Equation 7-2 solved by trial and error. Find Fraude number using Equation 6-4. 1.49 1/2 1.49 s 5 1.0052 1/2 5 9.93 n o .015
67
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68 CHAPTER 8 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
1
Dn 1ft2
2.0
a 1ft2 2
12.0
p 1ft2
10.0
R 1ft2
1.20
1.13
Q 1cfs2
2
3.0
18.0
12.0
1.50
1.31
234
3
2.9
17.4
11.8
1.47
1.30
224 (OK)
Trial
R2/3
135
Therefore, Dn 5 2.9 ft (Note: Check this result using Chart 5 in Appendix A-3.) v5 F5
Q 224 5 5 12.87 ft/s a 17.4 v "gD
5
12.87 "132.22 12.92
5 1.33
Flow is supercritical 1F . 12 (Answer)
3. Q 5 50 cfs, so 5 1.0%. Find normal depth using Equation 7-2 solved by trial and error. Find Fraude number using Equation 6-4. 1.49 1/2 1.49 s 5 1.0052 1/2 5 5.96 n o .025
1
Dn 1ft2
1.0
a 1ft2 2
6.0
p 1ft2
2
2.0
16.0
12.94
3
1.3
Trial
8.47
8.58
9.81
R 1ft2
.708 1.24 .874
R2/3 .795 1.15 .914
Q 1cfs2 28
110 47 (OK)
Therefore, Dn 5 1.3 ft v5
Q 47 5 5 5.5 ft/s a 8.58
Dh 5 F5
a 8.58 5 5 0.933 ft T 9.2 v
"gDh
5
5.5 "132.22 1.9332
5 1.00
Flow is critical 1F 5 12 (Answer)
4. Q 5 800 cfs. First, determine if flow is subcritical or supercritical. Dc 5 4.27 ft (Equation 6-3) Dn 5 6.0 ft (Equation 7-2, trial and error) Flow is subcritical 1Dn . Dc 2. Therefore, control section is at downstream end of channel reach. Set cross section locations at 200-ft intervals. Assume channel bottom elevation of 100.00 ft at Station 0 1 0. See Table 4. See Figure 3. (Answer)
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 69
5. Q 5 200 cfs. First, determine if flow is subcritical or supercritical. Dn 5 2.8 ft
(Chart 21, Appendix A-3)
Dc 5 2.25 ft
(Chart 21, Appendix A-3)
Flow is subcritical 1Dn . Dc 2. Therefore, control section is at the downstream end of the channel reach. Set cross section locations at 100-ft intervals. See Table 5.
See Figure 4. (Answer) Note: If the cross sections had been set at 200-ft intervals, the velocity would have varied by more than 20%. 6. First, determine whether the flow in the emergency spillway channel is subcritical or super- critical. For Q 5 50 cfs, Dc 5 0.67 ft and Dn 5 1.7 ft. For Q 5 300 cfs, Dc 5 2.25 ft and Dn 5 2.4 ft. Therefore, subcritical. Use Equation 7-2 and Equation 8-5 to find values of Dr. Dn (ft)
Q (cfs)
Dr (ft)
0.5
47
1.03
1.0
142
2.23
1.5
270
3.46
7. First, determine whether the flow in the emergency spillway channel is subcritical or super- critical. For Q 5 50 cfs, Dc 5 0.67 ft and Dn 5 0.56 ft. For Q 5 200 cfs, Dc 5 1.7 ft and Dn 5 1.3 ft. Therefore, supercritical. Use Equation 6-2 and Equation 8-6 to find values of Dr. Dc (ft)
Q (cfs)
Dr (ft)
0.5
32
0.75
1.0
91
1.50
1.5
167
2.25
2.0
257
3.00
8. First, determine whether the flow in the emergency spillway channel is subcritical or supercritical. For Q 5 50 cfs, Dc 5 0.74 ft and Dn 5 1.4 ft. For Q 5 200 cfs, Dc 5 2.1 ft and Dn 5 2.9 ft. Therefore, subcritical. Use Equation 7-2 and Equation 8-5 to find values of Dr.
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92690_ch8_ptg01_p067-074.indd 70
8.0 8.0 7.8 7.77 7.54 7.53 7.30 7.10 6.90 6.73 6.58 6.46 6.36 6.28
108.0 108.4 108.2 108.17 108.34 108.33 108.50 108.70 108.90 109.13 109.38 109.66 109.96 110.28
010
210
210
210
410
410
610
810
1010
1210
1410
1610
1810
2010
© Cengage Learning 2014
(3) D (ft)
(2) Elev. (ft)
(1) Station
100.5
101.8
103.4
105.3
107.7
110.4
113.6
116.8
120.5
120.6
124.3
124.8
128
128
(4) a (ft2)
7.96
7.86
7.74
7.60
7.43
7.25
7.04
6.85
6.64
6.63
6.44
6.41
6.25
6.25
(5) v (ft/s)
0.98
0.96
0.93
0.90
0.86
0.82
0.77
0.73
0.68
0.68
0.64
0.64
0.61
0.61
(6) v2/2g (ft)
111.26
110.92
110.59
110.28
109.99
109.72
109.47
109.23
109.01
109.02
108.81
108.84
109.01
108.61
(7) H (ft)
3.52
3.54
3.58
3.61
3.66
3.70
3.76
3.82
3.88
3.88
3.94
3.95
4.0
4.0
(8) R (ft)
5.35
5.40
5.46
5.54
5.63
5.73
5.85
5.96
6.09
6.09
6.22
6.24
6.35
6.35
(9) R4/3
.0017
.0017
.0016
.0015
.0014
.0013
.0012
.0012
.001
.001
.001
.001
.0009
.0009
(10) s
.00170
.00165
.00155
.00145
.00135
.00125
.0012
.0011
.001
.001
.001
.001
.0009
—
(11) s
200
200
200
200
200
200
200
200
200
200
200
200
200
—
(12) Dist (ft)
0.34
0.33
0.31
0.29
0.27
0.25
0.24
0.22
0.20
0.20
0.20
0.20
0.18
—
(13) hf (ft)
0
0
0
0
0
0
0
0
0
0
0
0
0
—
(14) he (ft)
111.26
110.92
110.5
110.2
109.9
109.7
109.4
109.23
109.0
109.0
108.8
103.8
103.7
108.6
(15) H (ft)
70 CHAPTER 8 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
TABLE 4 Computation of backwater profile for Problem 4.
11/12/12 9:33 AM
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© Cengage Learning 2014
Datum 98 ft 20 + 0
18 + 0
16 + 0
12 + 0
10 + 0
Channel Profile Scale: Horiz. 1″ = 300′ Vert. 1″ = 3′
14 + 0
8+0
om
el Bott
Chann
6+0
h Line
l Dept
Critica
h Line
l Dept
Norma
Water Surface
4+0
2+0
Elev. 100.0
0+0
8.0′
HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 71
FIGURE 3
11/12/12 9:33 AM
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5.00 4.60 4.55 4.10 3.85 3.65 3.68 3.23 3.33 3.10 3.06 2.84 2.89
205.00 205.10 205.05 205.10 205.35 205.15 205.18 205.28 205.33 205.60 205.56 205.84 205.89
010
110
110
210
310
310
310
410
410
510
510
610
610
© Cengage Learning 2014
(3) D (ft)
(2) Elev. (ft)
(1) Station
39.8
38.85
43.2
44.0
48.8
47.8
56.52
55.85
60.45
66.42
77.81
79.12
90.0
(4) a (ft2)
5.02
5.15
4.63
4.54
4.10
4.19
3.54
3.58
3.31
3.01
2.57
2.53
2.22
(5) v (ft/s)
.39
.41
.33
.32
.26
.27
.19
.20
.17
.14
.103
.099
.077
(6) v2/2g (ft)
206.28
206.25
205.89
205.92
205.59
205.55
205.37
205.35
205.52
205.24
205.15
205.20
205.08
(7) H (ft)
1.90
1.88
1.99
2.01
2.13
2.11
2.31
2.30
2.40
2.52
2.74
2.77
2.96
(8) R (ft)
2.36
2.31
2.51
2.54
2.74
2.70
3.05
3.03
3.21
3.43
3.84
3.89
4.26
(9) R4/3
.0043
.0047
.0035
.0033
.0025
.0026
.0017
.0017
.0014
.0011
.00070
.00067
—
(10) s
.0039
.0041
.003
.029
.0021
.0022
.0014
.0014
.00125
.0009
.00070
.00067
—
(11) s
100
100
100
100
100
100
100
100
100
100
100
100
—
(12) Dist (ft)
0.39
0.41
0.30
0.29
0.21
0.22
0.14
0.14
0.13
0.09
0.07
0.07
—
(13) hf (ft)
0
0
0
0
0
0
0
0
0
0
0
0
—
(14) he (ft)
206.28
206.30
205.89
205.88
205.59
205.60
205.38
205.38
205.37
205.24
205.15
205.15
205.08
(15) H (ft)
72 CHAPTER 8 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
TABLE 5 Computation of backwater profile for Problem 5.
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© Cengage Learning 2014
Datum 198 ft 6+0
5+0
4+0
2+0
Channel Profile Scale: Horiz. 1″ = 100′ Vert. 1″ = 2′
3+0
ottom
nel B
Chan
al De pth L ine Critic al De pth L ine
Norm
1+0
Elev. 200.0
0+0
5.0′
HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 73
FIGURE 4
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74 CHAPTER 8 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Dn (ft)
Q (cfs)
Dr (ft)
0.5
18
0.67
1.0
27
1.08
2.0
95
2.18
9. First, determine whether the flow in the emergency spillway channel is subcritical or super- critical. For Q 5 50 cfs, Dc 5 0.74 ft and Dn 5 0.6 ft. For Q 5 200 cfs, Dc 5 2.1 ft and Dn 5 1.4 ft. Therefore, supercritical. Use Equation 6-2 and Equation 8-6 to find values of Dr. Dc (ft)
Q (cfs)
Dr (ft)
0.5
20
0.71
1.0
61
1.40
2.0
190
2.71
10. Q 5 525.0 cfs, Dn 5 0.6 ft. Find D1 and D2 using Equation 8-7 and Equation 6-4. D2 must match the sequent depth of 3.20 ft. Trial
D1 (ft)
a1 (ft2)
v1 (fts)
F1
D2 (ft)
1
0.5
10.0
52.5
13.1
4.4
2
1.0
20.0
26.3
4.6
2.8
3
0.75
15.0
35.0
7.1
3.4
4
0.83
16.6
31.6
6.1
3.2 (OK)
Therefore, the initial depth is 0.83 ft and the sequent depth is 3.2 ft. The height of the jump is 2.37 ft. The location of the jump is 3.0 ft left of Point A. (Answer)
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C h a p t e r
9 Culvert Hydraulics
1. Q 5 20 cfs HW 5 1.48 D
(Appendix B-1, Chart 2, Scale 1)
HW 5 11.482 12.02 HW 5 2.96 ft
HW 5 3.0 ft (Answer) 2. Q 5 45 cfs HW 5 1.76 D
(Appendix B-1, Chart 2, Scale 3)
HW 5 11.762 12.52
HW 5 4.4 ft (Answer) 3. Q 5 285 cfs, twin 48-inch pipes Consider
Q 5 143 cfs for a single 48-inch pipe. 2
HW 5 2.0 D
(Appendix B-1, Chart 2, Scale 1)
HW 5 12.02 14.02
HW 5 8.00 ft (Answer)
75
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76 CHAPTER 9 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
4. Q 5 125 cfs Q 125 5 5 20.83 cfs/ft B 6 HW 5 1.32 D
(Appendix B-1, Chart 1, Scale 1)
HW 5 11.322 13.02
HW 5 3.96 ft (Answer) 5. Q 5 225 cfs ke 5 0.5
(Appendix B-3)
H 5 1.40 ft
(Appendix B-2, Chart 8)
Dc 5 2.9 ft
(Appendix A-3, Chart 7)
TWr 5
Dc 1 D 2.9 1 4.0 5 5 3.45 ft 2 2
Since TWr . TW, use TWr to find headwater depth. HW 5 TWr 1 H 2 Ls HW 5 3.45 1 1.40 2 1802 1.0062 HW 5 4.37 ft (Answer)
6. Q 5 1620 cfs, triple 5r 3 12r box culvert Consider
Q 5 540 cfs for a single 5r 3 12r box culvert 3
ke 5 0.4
(Appendix B-3)
H 5 2.1 ft
(Appendix B-2, Chart 8)
Dc 5 4.0 ft
(Appendix A-3, Chart 10)
TWr 5
Dc 1 D 4.0 1 5.0 5 5 4.5 ft 2 2
Since TWr . TW, use TWr to find headwater depth. HW 5 TWr 1 H 2 Ls HW 5 4.5 1 2.1 2 11052 1.00752 HW 5 5.81 ft (Answer) 7. Q 5 200 cfs Assume inlet control HW 5 1.52 D
(Appendix B-1, Chart 1)
HW 5 11.522 15.02 HW 5 7.6 ft
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 77
Assume outlet control ke 5 0.5
(Appendix B-3)
H 5 3.7 ft
(Appendix B-2, Chart 9)
Dc 5 4.0 ft
(Appendix A-4, Chart 47)
TWr 5 4.0 1 5.0 5 4.5 ft 2 Since TWr . TW, use TWr to find headwater depth. HW 5 TWr 1 H 2 Lso HW 5 4.5 1 3.7 2 12502 1.00552 HW 5 6.8 ft
Since 7.6 ft . 6.8 ft, culvert operates under inlet control, HW 5 7.6 ft (Answer) 8. Q 5 25 cfs, twin 18-inch pipe Consider
Q 5 12.5 cfs for a single 18-inch pipe. 2
ke 5 0.5
(Appendix B-3)
H 5 1.7 ft
(Appendix B-2, Chart 9)
Dc 5 1.3 ft
(Appendix A-4, Chart 37)
TWr 5
Dc 1 D 1.3 1 1.5 5 5 1.4 ft 2 2
Since TWr , TW, use TW to find headwater depth. HW 5 TW 1 H 2 Ls HW 5 3.33 1 1.7 2 1502 1.00882 HW 5 4.59 ft (Answer) 9. Assume inlet control Q 65 5 5 13 cfs/ft B 5
(Appendix B-1, Chart 1)
HW 5 0.90 (Scale (1)) D HW 5 1.902 132 5 2.7 ft
Now, assume outlet control
92690_ch9_ptg01_p075-080.indd 77
n 5 0.012
(concrete culvert)
ke 5 0.4
(Appendix B-3)
H 5 0.52 ft
(Appendix B-2, Chart 8)
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78 CHAPTER 9 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Next find TWr and compare to TW. Dc 5 1.7 ft Therefore, TWr 5
(Appendix A-3, Chart 4) Dc 1 D 1.7 1 3 5 5 2.35 ft 2 2
Since TWr . TW, use TWr to find headwater depth. HW 5 TWr 1 H 2 Lso (Equation 9-2b) 5 2.35 1 0.52 2 0.11 5 2.76 ft Since 2.76 . 2.7, the culvert operates under outlet control and the headwater depth is 2.76 ft (Answer) 10. Since tailwater is at the downstream crown, the culvert must operate under outlet control. n 5 0.012
(concrete culvert)
ke 5 0.4
(Appendix B-3)
H 5 1.71 ft
(Appendix B-2, Chart 9)
HW 5 TW 1 H 2 Lso 5 4.00 1 1.71 2 0.25 5 5.46 ft (Answer) 11. Q 5 20 cfs HW 5 1.4 D
(Appendix B-1, Chart 7, Entrance Type A)
HW 5 11.42 12.02
HW 5 2.8 ft (Answer) Note: Headwater is 0.2 ft lower than the culvert in problem 1. 12. Assume inlet control. HW 5 1.03 ft D
(Appendix B-1, Chart 2, Scale (3))
HW 5 11.032 15.02 5 5.15 ft Now, assume outlet control. n 5 0.012 ke 5 0.5 H 5 1.42 ft
(concrete culvert) (Appendix B-3) (Appendix B-2, Chart 9)
Next find TWr and compare to TW. Dc 5 3.3 ft Therefore, TWr 5
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(Appendix A-4, Chart 47) Dc 1 D 3.3 1 5 5 5 4.15 ft 2 2
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 79
Since TWr . TW, use TWr to find headwater elevation. HW 5 TWr 1 H 2 Lso 5 4.15 1 1.42 2 11002 1.0062 5 4.97 ft
(Answer)
Since 5.15 ft . 4.97 ft, the culvert operates under inlet control, HW 5 5.15 ft (Answer) 13. Q 5 285 cfs, beveled entrance. Consider
Q 5 143 cfs for a single 48-inch pipe. 2
HW 5 1.8 D
(Appendix B-1, Chart 7, Entrance Type A)
HW 5 11.82 14.02
HW 5 7.2 ft (Answer) Note: Headwater is 0.8 ft lower than the culvert in problem 3. 14. Q 5 140 cfs, 60-inch CMP Assume inlet control HW 5 1.17 D
(Appendix B-1, Chart 5, Scale 3)
HW 5 11.172 15.02 HW 5 5.85 ft
Assume outlet control ke 5 0.9 n 5 0.024 H 5 2.55 ft
(Appendix B-2, Chart 11)
TWr 5 4.15 ft
(same as problem 12)
Since TWr . TW, use TWr to find headwater depth. HW 5 TWr 1 H 2 Lso HW 5 4.15 1 2.55 2 11002 1.0062 HW 5 6.1 ft
Since 6.1 ft . 5.85 ft, culvert operates under outlet control, HW 5 6.1 ft (Answer) Note: Compared to problem 12, the use of a CMP resulted in higher HW and outlet control instead of inlet control.
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80 CHAPTER 9 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
15. In this problem, we must first find TW by finding normal depth in the downstream channel by the use of Manning’s Equation. 1.49 1/2 s 5 1.904 n o
1
D 1ft2
2.0
a 1ft2 2
28
2
3.0
3
3.5
4
3.45
Trial
20.6
R 1ft2
1.36
1.23
Q 1cfs2
51
27
1.89
1.53
148
64.75
30
2.16
1.67
206
29.8
2.12
1.65
199 (OK)
63.3
p 1ft2
R2/3
65
TW 5 3.45 ft Now, assume inlet control. Q 200 5 5 25 cfs/ft B 8
(Appendix B-1, Chart 1)
HW 5 1.05 (Scale (1)) D HW 5 11.052 14.02 5 4.2 ft
Now, assume outlet control. n 5 0.012
(concrete culvert)
ke 5 0.4
(Appendix B-3)
H 5 0.94 ft
(Appendix B-2, Chart 8)
Next, find TWr and compare to TW. (Appendix A-3, Chart 7)
Dc 5 2.75 ft Therefore, TWr 5
Dc 1 D 2.75 1 4 5 5 3.38 ft 2 2
Since TW . TWr, use TW to find headwater elevation. HW 5 TW 1 H 2 Lso 5 3.45 1 0.94 2 1602 1.0022 5 4.27 ft
Since 4.27 ft . 4.2 ft, the culvert operates under outlet control, HW 5 4.27 ft (Answer)
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C h a p t e r
10 Fundamental Hydrology
1. a) A 5 45.0 acres b) A 5 182,000 m2 2. a) A 5 1.03 acres b) A 5 4170 m2 3. a) 11.4 acres b) 46,100 m2 4. a) 1.40 acres b) 5650 m2 5. For delineation, see Figure 5. A 5 55.0 acres. 6. For delineation, see Figure 6. A 5 128 acres. 7. For delineation, see Figure 7. A 5 1,670 acres. 8. For delineation, see Figure 8.
Watershed A: A 5 675 acres.
Watershed B: A 5 233 acres.
81
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82 CHAPTER 10 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
ROA D
eK
cre
11 00
1100
BM K AN
E
974
RD
RO AD
Point of Analysis
1000
1100
Gravel Pit
140 0 120 0
15 0
13 0
0
0
WOODS
14 00
Basin Divide 89
S T A T E 0
0 15 14 00
P R E S E R V E 1300 1400
TOPOGRAPHIC MAP SCALE: 1" = 500' CONTOUR INTERVAL 20'
© Cengage Learning 2014
SI
FIGURE 5 Catchment area delineation for Problem 5.
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 83
TOPOGRAPHIC MAP SCALE: 1" = 500' CONTOUR INTERVAL 5'
Basin Divide
© Cengage Learning 2014
Point of Analysis
FIGURE 6 Catchment area delineation for Problem 6.
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84 CHAPTER 10 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Point of Analysis
TOPOGRAPHIC MAP SCALE: 1" = 2000' CONTOUR INTERVAL 20'
© Cengage Learning 2014
Basin Divide
FIGURE 7 Watershed area delineation for Problem 7.
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 85
Point of Analysis A
Basin Divide
Point of Analysis B
TOPOGRAPHIC MAP SCALE: 1" = 2000' CONTOUR INTERVAL 20'
© Cengage Learning 2014
Basin Divide
FIGURE 8 Watershed area delineations for Problem 8.
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86 CHAPTER 10 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
9. For hydraulic path, see Figure 9. Overland flow: L 5 100 ft, s 5 1.5%. t1 5 14.5 min
(Appendix C-2)
Shallow concentrated flow: L 5 515 ft, s 5 4.2%. v 5 3.3 ft/s t2 5
(Figure 10-9, unpaved)
L 515 5 5 156 s 5 2.6 min v 3.3
Stream flow: L 5 930 ft, s 5 2.8%. 12′
n = 0.032 6′
a5
© Cengage Learning 2014
3′
112 1 62 132 5 27 ft2 2
p 5 14.4 ft R5
a 27 5 5 1.875 ft p 14.4
v5
1.49 2/3 1/2 R s n
v5
1.49 11.8752 2/3 1.0282 1/2 n
v 5 11.8 ft/s t3 5
L 930 5 5 79 s 5 1.3 min v 11.8
tc 5 t1 1 t2 1 t3 tc 5 14.5 1 2.6 1 1.3 tc 5 18.4 min. (Answer)
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Basin Divide
Point of Analysis
Scale: 1" = 200'
© Cengage Learning 2014
Contour Interval 2'
FIGURE 9 Hydraulic path for watershed in Problem 9.
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88 CHAPTER 10 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
10. For hydraulic path, see Figure 10. Overland flow: L 5 100 ft, s 5 6%. t1 5 10.5 min
(Appendix C-2)
Shallow concentrated flow: L 5 200 ft, s 5 7%. v 5 4.3 ft/s t2 5
(Figure 10-9, unpaved)
L 200 5 5 47 s 5 0.8 min v 4.3
Gutter flow: L 5 210 ft, s 5 5.5%. v 5 4.8 ft/s t3 5
(Figure 10-9, paved)
L 210 5 5 44 s 5 0.7 min v 4.8
tc 5 t1 1 t2 1 t3 tc 5 10.5 1 0.8 1 0.7 tc 5 12 min. (Answer) 11. For hydraulic path, see Figure 11. Overland flow: L 5 100 ft, s 5 10%. t1 5 9.5 min
(Appendix C-2)
Shallow concentrated flow: L 5 750 ft, s 5 15%. v 5 6.25 ft/s t2 5
(Figure 10-9, unpaved)
L 750 5 5 120 s 5 2.0 min v 6.25
Stream flow: L 5 1,750 ft, s 5 19%. a5
112 1 62 132 5 27 ft2 2
p 5 14.4 ft R5
a 27 5 5 1.875 ft p 14.4
v5
1.49 2/3 1/2 R s n
v5
1.49 (1.875)2/3 (.19)1/2 .032
v 5 31 ft/s t3 5
L 1750 5 5 56 s 5 0.94 min v 31
tc 5 t1 1 t2 1 t3 tc 5 9.5 1 2.0 1 0.94 tc 5 12.4 min. (Answer)
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TOPOGRAPHIC MAP Scale: 1" = 100' (1:1200) Contour interval 2'
Inlet (point of analysis)
© Cengage Learning 2014
Basin divide
FIGURE 10 Hydraulic path for watershed in Problem 10.
92690_ch10_ptg01_p081-096.indd 89
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90 CHAPTER 10 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
ROA D
eK
cre
11 00
1100
BM K AN
E
974
RD
RO AD
Point of Analysis
1000
1100
Gravel Pit
140 0 120 0
15 0
13 0
0
0
WOODS
14 00
Remotest Point 89
S T A T E 0
0 15 14 00
P R E S E R V E 1300 1400
TOPOGRAPHIC MAP SCALE: 1" = 500' CONTOUR INTERVAL 20'
© Cengage Learning 2014
SI
FIGURE 11 Hydraulic path for watershed in Problem 11.
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 91
12. For hydraulic path, see Figure 12. Overland flow: L 5 100 ft, s 5 1.8 %. t1 5 13.0 min
(Appendix C-2)
Shallow concentrated flow: L 5 1400 ft, s 5 15.6%. v 5 6.25 ft/s t2 5
(Figure 10-9, unpaved)
L 1400 5 5 224 s 5 3.7 min v 6.25
Gutter flow: L 5 850 ft, s 5 2.0%. v 5 2.9 ft/s t3 5
(Figure 10-9, paved)
L 850 5 5 293 s 5 4.9 min v 2.9
Shallow concentrated flow: L 5 1200 ft, s 5 2.3%. v 5 2.4 ft/s t4 5
(Figure 10-9, unpaved)
L 1200 5 5 500 s 5 8.3 min v 2.4
Stream flow: L 5 580 ft, s 5 1.7%. a5
112 1 62 132 5 27 ft2 2
p 5 14.4 ft R5
a 27 5 5 1.875 ft p 14.4
v5
1.49 2/3 1/2 R s n
v5
1.49 11.8752 2/3 1.0172 1/2 .032
v 5 9.2 ft/s t5 5
L 580 5 5 63 s 5 1.1 min v 9.2
tc 5 t1 1 t2 1 t3 1 t4 1 t5 tc 5 13.0 1 3.7 1 4.9 1 8.3 1 1.1 tc 5 31.0 min. (Answer)
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92 CHAPTER 10 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
TOPOGRAPHIC MAP SCALE: 1" = 500' CONTOUR INTERVAL 5'
Point of Analysis
Basin Divide
Remotest Point
ulic
dra
Hy © Cengage Learning 2014
Path
FIGURE 12 Hydraulic path for watershed in Problem 12.
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 93
13. For hydraulic path, see Figure 13.
Remotest Point
Point of Analysis
TOPOGRAPHIC MAP SCALE: 1" = 2000' CONTOUR INTERVAL 20'
© Cengage Learning 2014
Basin Divide
FIGURE 13 Hydraulic path for watershed in Problem 13.
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94 CHAPTER 10 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
14. For resultant hydrograph, see Figure 14. Peak 5 56 cfs. 100
Runoff (cfs)
75
Resultant Hydrograph
50
0 0
2
4
6 8 Time (hours)
10
12
© Cengage Learning 2014
25
FIGURE 14 Resultant hydrograph for Problem 14. 15. For resultant hydrograph, see Figure 15. Peak 5 70 cfs. 100
Runoff (cfs)
75
Resultant Hydrograph
50
0 0
4
8
12 16 Time (hours)
20
24
© Cengage Learning 2014
25
FIGURE 15 Resultant hydrograph for Problem 15.
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 95
16. Determine the volume of the runoff by measuring the area under the hydrograph. 18.02 136002 11502 5 2.16 3 106 ft3 (Answer) Volume 5 2 To compute the depth of rainfall excess, use the following equation: Volume 5 Depth 3 Area 3 43560 Depth 5 Depth 5
Volume 1Area2 1435602 2.16 3 106 11652 1435602
Depth 5 0.30 ft 5 3.6 inches (Answer) 17. Determine the volume of runoff by measuring area under the hydrograph. Volume 5 2.70 3 106 ft3 (Answer) To compute depth of rainfall excess, use the following equation: Volume 5 Depth 3 Area 3 43560 Depth 5 Depth 5
Volume 1Area2 1435602 2.70 3 106 12502 1435602
Depth 5 0.248 ft 5 3.0 inches (Answer)
Time (h) 0 1 2 3 4 5 6 7 8 9 10 11 12
Hydrograph 1 (cfs)
Hydrograph 2 (cfs)
Hydrograph 3 (cfs)
Hydrograph 4 (cfs)
Sum of Four Hydrographs (cfs)
0 0 0 0 14 50 39 19 12 8 6 5 4
0 0 0 0 0 56 200 156 76 48 32 24 20
0 0 0 0 0 0 42 150 117 57 36 24 18
0 0 0 0 0 0 0 14 50 39 19 12 8
0 0 0 0 14 106 281 339 255 152 93 65 50
© Cengage Learning 2014
18. Using one hour as the unit time, there are four components of rainfall excess having depths of 0.5, 2.0, 1.5 and 0.5 inches, respectively. Using a lag of 2.5 hours, the unit hydrographs are listed in Table 6. The derived hydrograph is shown in the last column of Table 6. The peak runoff is 339 cfs.
TABLE 6 Derived Hydrograph for Problem 18.
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96 CHAPTER 10 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
19. Using two hours as the unit time, there are three components of rainfall excess having depths of 1.5, 1.0, and 0.5 inches, respectively. Using a lag of 2.5 hours, the unit hydrographs are listed in Table 7. The derived hydrograph is shown in the last column of Table 7. The peak runoff is 217 cfs. Hydrograph 1 (cfs)
Hydrograph 2 (cfs)
Hydrograph 3 (cfs)
Sum of Three Hydrographs (cfs)
0 0 0 0 42 150 117 57 36 24 18 15 12
0 0 0 0 0 28 100 78 38 24 16 12 10
0 0 0 0 0 0 0 14 50 39 19 12 8
0 0 0 0 42 178 217 139 124 87 53 39 30
0 1 2 3 4 5 6 7 8 9 10 11 12
© Cengage Learning 2014
Time (h)
TABLE 7 Derived Hydrograph for Problem 19. 20. For resultant hydrograph, see Figure 16. Peak runoff for Watershed A is 63 cfs and for Watershed B is 42 cfs. Peak runoff for the combined (resultant) watersheds is 90 cfs. (Answer) Note: Peak runoff values were scaled from the hydrographs. Also, the peak resultant runoff is less than the arithmetic sum of the peak runoffs for Watersheds A and B.
100
Watershed A 50
Watershed B
5
10 Time (h)
© Cengage Learning 2014
Runoff (cfs)
Resultant Watershed
FIGURE 16 Runoff hydrographs for Problem 20.
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C h a p t e r
11 Runoff Calculations
1. Assume 25-year storm or lower. First choose c-values for individual cover conditions using Appendix C-1 or Table 11-1. Impervious: c 5 0.90 Grass: c 5 .30 Woods: c 5 0.20 Convert all area units to acres. Impervious
0.12 acres
Grass
0.45 acres
Wooded
1.21 acres
c5
1.122 1.902 1 1.452 1.302 1 11.212 1.202 5 0.27 (Answer) 1.78
2. Assume 25-year storm or lower. First choose c-values for individual cover conditions using Appendix C-1 or Table 11-1. Downtown business area: c 5 0.95 Residential (attached multi-units): c 5 0.78 Cemetery: c 5 0.20 Compute weighted average: c5
18.52 1.952 1 115.02 1.782 1 13.22 1.202 5 0.76 (Answer) 26.7
97
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98 CHAPTER 11 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
3. Assume 25-year storm or lower. First choose c-values for individual cover conditions using Appendix C-1 or Table 11-1. Houses (use impervious): c 5 0.95 Driveways (use impervious): c 5 0.95 Lawns-sandy soil: c 5 0.086 Road (use impervious): c 5 0.95 Compute areas in acres: 8(25)(40) 5 0.184 acres 43560
Houses: Area 5
Driveways: Area 5 Lawns: Area 5 Road: Area 5
8(20)(35) 5 0.129 acres 43560
8(75)(35) 5 0.482 acres 43560
(15)(600) 5 0.207 acres 43560
Compute weighted average: c5
1.1842 1.952 1 1.1292 1.952 1 1.4822 1.0862 1 1.2072 1.952 5 0.46 (Answer) 1.163
4. Overland: average grass, L 5 80 ft, s 5 2.5% t1 5 11 min.
(Appendix C-2)
Shallow concentrated flow: L 5 450 ft, s 5 4.8% v 5 3.5 ft/s t2 5
(Figure 10-9, “unpaved”)
L 450 5 5 129 sec. 5 2.1 min. v 3.5
Gutter flow: L 5 200 ft, s 5 1.9% v 5 2.8 ft/s t3 5
(Figure 10-9, “paved”)
L 200 5 5 71.4 sec. 5 1.2 min. v 2.8
tc 5 t1 1 t2 1 t3 tc 5 11 1 2.1 1 1.2 tc 5 14 min. (Answer)
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 99
5. Overland: average grass, L 5 100 ft, s 5 3.5% t1 5 11.5 min.
(Appendix C-2)
Shallow concentrated flow: L 5 200 ft, s 5 3.5% v 5 3.0 ft/s
(Figure 10-9, “unpaved”)
200 5 66.7 sec. 5 1.1 min. 3.0
t2 5
Swale flow: L 5 500 ft, s 5 2.0% Find v using Manning’s Equation assuming full bank flow. a 5 4.0 ft2 p 5 16.1 ft R5
4.0 5 0.248 ft 16.1
v5
1.49 (.248)2/3(.020)1/2 .025
v 5 3.33 ft/s t3 5
500 5 150 sec. 5 2.5 min. 3.33
tc 5 t1 1 t2 1 t3 tc 5 11.5 1 1.1 1 2.5 tc 5 15.1 min. tc 5 15 min. (Answer) 6. Overland flow: average grass, L 5 100 ft, s 5 1.6% n 5 0.24
(Table 11-3)
P2 5 2.0 in. Tt 5 Tt 5
0.0071nL2
(Appendix D-3) 0.8
P20.5s0.4
(Equation 11-9)
.007 5 1.242 11002 6 0.8 12.02 .5 1.0162 .4
Tt 5 0.33 hour.
Shallow concentrated flow: L 5 680 ft, s 5 3.2% v 5 2.9 ft/s t2 5
92690_ch11_ptg01_p097-125.indd 99
(Figure 10-9, unpaved)
L 680 5 5 234 sec. 5 3.9 min. 5 0.065 h. v 2.9
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100 CHAPTER 11 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Stream flow: L 5 2950 ft, so 5 0.61%
4′
n = 0.035
12′
© Cengage Learning 2014
20′
Find v using Manning’s Equation assuming full bank flow. 120 1 122 4 5 64 ft2 2
a5
P 5 23 ft R5
a 64 5 5 2.78 ft P 23
v5
1.49 2/3 1/2 R so n
v5
1.49 12.782 2/3 1.00612 1/2 .035
v 5 6.58 ft/s t3 5
L 2950 5 5 448 sec. 5 7.5 min. 5 0.125 h. v 6.58
tc 5 t1 1 t2 1 t3 tc 5 .33 1 .065 1 .125 tc 5 0.52 hour (Answer) 7. Overland: woods (light underbrush), L 5 100 ft, s 5 3.0% n 5 0.40
(Table 11-3)
P2 5 3.5 in. Tt 5
(Appendix D-3)
0.007 3 1.402 11002 4
0.8
(3.5)0.5 (.030)0.4
Tt 5 0.29 hour
Shallow concentrated flow: L 5 1,200 ft, s 5 4.0% v 5 3.0 ft/s t2 5
(Figure 10-9, “unpaved”)
200 5 66.7 sec. 5 0.019 hour 3.0
Stream flow: L 5 1,500 ft, s 5 0.75% Find v using Manning’s Equation assuming full bank flow. a 5 64 ft2 p 5 23 ft
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 101
R 5 2.78 ft v5
1.49 (2.78)2/3(.0075)1/2 .035
v 5 7.29 ft/s t3 5
1500 5 206 sec. 5 0.057 hour 7.29
tc 5 t1 1 t2 1 t3 tc 5 0.29 1 0.019 1 0.057 tc 5 0.366 hour tc 5 0.37 hour (Answer) 8. i 5 5.0 in/h.
(Appendix C-3, IDF for NJ)
9. i 5 1.6 in/h.
(Appendix C-3, IDF for Orange Co.)
10. i 5 1.3 in/h.
(Appendix C-3, IDF for Atlanta)
11. First, find tc. Overland flow: average grass, L 5 100 ft, s 5 5.0% t1 5 10.5 min.
(Appendix C-2)
Shallow concentrated flow: L 5 50 ft, s 5 7.0% v 5 4.25 ft/s t2 5
(Figure 10-9, unpaved)
L 50 5 5 11.8 sec. 5 0.2 min. v 4.25
tc 5 t1 1 t2 tc 5 10.5 1 0.2 tc 5 10.7 min. Next, find composite c. Impervious: c 5 0.90 Grass: c 5 0.30 Woods: c 5 0.20 c5
1.062 1.92 1 16.52 1.32 1 17.52 1.22 14.06
c 5 0.25
Next, find rainfall intensity. i 5 5.25 in/h.
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(Appendix C-3, IDF for Phoenix, P1 5 1.75 in)
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102 CHAPTER 11 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Then, (Equation 11-2)
Qp 5 Aci Qp 5 114.062 1.252 15.252
Qp 5 18 cfs (Answer) Qp 5 0.53 m3/s (Answer) 12. tc 5 10.7 min.
(see Problem 11)
Find composite c. Impervious: c 5 0.95 Grass: c 5 0.35 Woods: c 5 0.25 c5
1.062 1.952 1 16.52 1.352 1 17.52 1.252 14.06
c 5 0.30
i 5 6.8 in/h.
(Appendix C-3, IDF for Phoenix, P1 5 2.25 in.)
Qp 5 Aci
(Equation 11-2)
Qp 5 (14.06)(.30)(6.8) Qp 5 29 cfs (Answer) Qp 5 0.83 m3/s (Answer) 13. Compute composite CN. Bartley Loam: Hyd. Soil Group C Parker Loam: Hyd. Soil Group B
(Appendix D-2)
Percentage of each hyd. soil group: 45 3 100 5 64.3% 70 25 3 100 5 35.7% B: 70
C:
CN (Appendix D-1) Cover
B (35.7%)
C (64.3%)
Impervious
98
98
98
Residential (1/3-acre)
72
81
77.8
Wooded (fair)
60
73
68.4
Grass (fair)
69
79
75.4
CN
Product
Cover Impervious
Area (acres) 2.0
Interpolated CN
98
196
Residential
20
77.8
1556
Wooded
28
68.4
1915
Grass
20
75.4
1508
70
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5175
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 103
5175 5 73.9 70 CN 5 74 (Answer) Weighted CN 5
14.
Hyd. Soil Group Maplecrest
B
Hamel
C
Vinsad Alluvial land
C C
(Appendix D-2)
Percentage of each hyd. soil group: 120 3 100 5 57.4% 209 89 3 100 5 42.6% C: 209
B:
Cover
B (57.4%)
C (42.6%)
Impervious
98
98
98
Wooded
60
73
65.5
Disturbed soil
86
91
88.1
Brush (poor)
67
77
71.3
CN
Product
Cover Impervious
Area (acres)
Interpolated CN
1.0
98
98
Wooded
100
65.5
6550
Disturbed soil
23
88.1
2026
Brush
85
71.3
6061
209 Weighted CN 5
14735
14735 5 70.5 209
CN 5 71 (Answer) 15. P 5 5.7 in. Q 5 2.7 in. 16. P 5 4.3 in. Q 5 1.3 in. 17. P 5 12.0 in
(Appendix D-3) (Figure 11-9) (Answer) (Appendix D-3) (Figure 11-9) (Answer) (Appendix D-3)
Q 5 6.7 in
(Figure 11-9)
Rainfall distribution Type III
(Appendix D-4)
Ia 5 1.268
(Table 11-2)
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104 CHAPTER 11 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Then,
Ia 1.268 5 5 0.11 P 12
qu 5 160 csm/in
(Appendix D-5, Chart 4)
qp 5 quAmQ
(Equation 11-10)
qp 5 11602 11.2502 16.72
qp 5 1340 cfs (Answer) 18. P 5 0.8 in.
(Appendix D-3)
Q 5 4.25 in.
(Figure 11-9)
Rainfall distribution Type II
(Appendix D-4)
Ia 5 0.941 in.
(Table 11-2)
Then,
Ia .941 5 5 0.12 P 8.0
qu 5 195 csm/in.
(Appendix D-5, Chart 3)
qp 5 qu Am Q
(Equation 11-10)
qp 5 11952 11.4132 14.252
qp 5 1170 cfs (Answer) 19. Drainage area is measured on Figure 17. A 5 1.14 acres Cover conditions are found on Figure 17 as follows: Impervious
0.33 acres
Lawn
0.81 acres
Choose c-value for each cover condition. Impervious: c 5 0.90 Lawn: c 5 0.30 c5
1.332 1.92 1 1.812 1.32 5 0.47 1.14
Time of concentration: Hydraulic path is delineated in Figure 17. Overland flow: L 5 100 ft, s 5 1.5% t1 5 13.5 min.
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(Appendix C-2)
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 105
Basin divide
© Cengage Learning 2014
Inlet (point of analysis)
FIGURE 17 Hydraulic path for watershed in Problem 19.
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106 CHAPTER 11 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Shallow concentrated flow: L 5 15 ft, s 5 10% v 5 5.1 ft/s L 15 5 2.9 sec 5 0.05 min. t2 5 5 v 5.1
(Figure 10-9, unpaved)
Gutter flow: L 5 450 ft, s 5 0.56% v 5 1.5 ft/s L 450 5 300 sec 5 5.0 min. t3 5 5 v 1.5
(Figure 10-9, paved)
Therefore, tc 5 t1 1 t2 1 t3 tc 5 13.5 1 0.05 1 5.0 tc 5 18.6 min. i 5 2.7 in/h
(Chart 5, Appendix C-3)
Qp 5 Aci
(Equation 11-2)
Qp 5 11.142 1.472 12.72
Qp 5 1.45 cfs (Answer) 20. Step 1: Watershed area is measured in Figure 18. Am 5 0.711 s.m. Step 2: Find composite CN. Hydrologic soil group: C Cover conditions:
Impervious (roads, houses): 2.0 acres
Woods: 453 acres
Cover Impervious Woods Weighted CN 5
Area (acres)
CN
Product
2.0
98
196
453
73
33069
33265
33265 5 73.1 455
CN 5 73 Step 3: Find Q. P 5 4.5 in
(Appendix D-3)
Q 5 1.9 in
(Figure 11-9)
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 107
Watershed divide
Point of analysis
Project location
SCALE: 1" = 2000' CONTOUR INTERVAL 20'
© Cengage Learning 2014
PENNSYLVANIA
FIGURE 18 Hydraulic path for watershed in Problem 20.
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108 CHAPTER 11 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Step 4: Find time of concentration, tc. Hydraulic path is delineated in Figure 18. Overland flow: L 5 100 ft, s 5 10% n 5 0.40
(Table 11-3)
P2 5 2.7 in
(Appendix D-3)
Tt 5 Tt 5
0.0071nL2 0.8 P20.5s0.4
(Equation 11-9)
0.007 5 1.42 11002 6 0.8
2.70.5.100.4
Tt 5 0.20 h Shallow concentrated flow: L 5 1900 ft, s 5 9% v 5 4.8 ft/s t2 5
(Figure 10-9, unpaved)
L 1900 5 5 396 sec 5 6.6 min 5 0.11 h v 4.8
Stream flow: L 5 6800 ft, so 5 1.76%
4′
n = 0.035
12′
3 20 1 12 4 4 5 64 ft2 2
a5
p 5 23 ft
R5
a 64 5 5 2.78 ft p 23
v5
1.49 2/3 1/2 R s n
v5
v 5 11.2 ft/s
t3 5
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© Cengage Learning 2014
20′
1.49 12.782 2/3 1.01762 1/2 .035 L 6800 5 5 607 sec 5 0.17 h v 11.2
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 109
Therefore, tc 5 t1 1 t2 1 t3 tc 5 0.20 1 0.11 1 0.17 tc 5 0.48 h Step 5: Find qu Rainfall distribution Type II
(Appendix D-4)
Ia 5 0.740 in
(Table 11-2)
Then,
Ia .740 5 5 0.16 P 4.5
qu 5 520 csm/in
(Appendix D-5, Chart 3)
Step 6: Compute qp. qp 5 qu Am Q
(Equation 11-10)
qp 5 15202 1.7112 11.92
qp 5 700 cfs (Answer) 21. Watershed area is measured on Figure 19. Am 5 0.0184 s.m. Hydrologic soil group: B Cover condition: wooded CN 5 60
(Appendix D-1)
P 5 4.0 in
(Appendix D-3)
Q 5 0.75 in
(Figure 11-9)
Time of concentration: Hydraulic path is delineated in Figure 19. Overland flow: L 5 100 ft, s 5 4% n 5 0.40
(Table 11-3)
P2 5 1.8 in
(Given)
Tt 5
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0.007 1nL2 0.8 P21/2s0.4
(Equation 11-9)
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110 CHAPTER 11 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Basin divide
© Cengage Learning 2014
Point of analysis
FIGURE 19 Hydraulic path for watershed in Problem 21.
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 111
Tt 5
10.0072 5 1.42 11002 6 0.8 11.82 0.5 1.042 0.4
Tt 5 0.36 hour
Shallow concentrated flow: L 5 470 ft, s 5 14% v 5 6.0 ft/s t2 5
(Figure 10-9, unpaved)
L 470 5 5 78 sec 5 .022 h v 6
Stream flow: L 5 720 ft, s0 5 9%
4′
n = 0.035
12′
a5
© Cengage Learning 2014
20′
3 20 1 12 4 4 5 64 ft2 2
p 5 23 ft R5
a 64 5 5 2.78 ft P 23
v5
1.49 2/3 1/2 R s n
v5
1.49 12.782 2/3 1.092 1/2 .035
v 5 25 ft/s t3 5
L 720 5 5 29 sec 5 .008 h v 25
Therefore, tc 5 t1 1 t2 1 t3 tc 5 .36 1 .022 1 .008 tc 5 0.39 h Rainfall distribution Type I
(Appendix D-4)
Ia 5 1.333 in
(Table 11-2)
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112 CHAPTER 11 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Ia 1.333 5 5 0.33 P 4.0 qu 5 165 csm/in
(Appendix D-5, Chart 1)
qp 5 qu Am Q
(Equation 11-10)
Then,
qp 5 11652 1.01842 1.752
qp 5 2.3 cfs (Answer) 22. Drainage area was determined in Problem 3, Chapter 10. A 5 11.4 acres c 5 0.20 Time of concentration: Hydraulic path is delineated in Figure 9. Overland: L 5 100 ft, s 5 1.5%
t1 5 14.5 min.
(Appendix C-2)
Shallow concentrated flow: L 5 515 ft, s 5 4.2%
v 5 3.3 ft/s
t2 5
(Figure 10-9)
L 515 5 5 156 sec 5 2.6 min. v 3.3
Stream flow: L 5 930 ft, so 5 2.8%
3′
n = 0.032
6′
3 12 1 6 4 3 5 27 ft2 2
a5
P 5 14.4 ft
R5
a 27 5 5 1.875 ft P 14.4 1.49 2/3 1/2 v5 R s n 1.49 (1.875)2/3 1.0282 1/2 .032
v5
v 5 11.8 ft/s
© Cengage Learning 2014
12′
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 113
L 930 5 5 79 sec 5 1.3 min. v 11.8
t3 5
tc 5 t1 1 t2 1 t3
tc 5 14.5 1 2.6 1 1.3
tc 5 18.4 min.
i 5 4.6 in/h
(Appendix C-3)
Qp 5 Aci
(Equation 11-2)
Qp 5 111.42 1.202 14.62
Qp 5 10.5 cfs. (Answer)
23. Step 1: Watershed area is measured in Figure 20. Am 5 1.29 s.m. Step 2: Find composite CN. First, determine the percentage of each hydrologic soil group.
B—65% C—25% D—10%
Next, determine the area of each cover condition from the topograhic map.
Impervious (roads, houses): 6.6 acres
Woods: 819 acres
ext determine CN for each cover condition by interpolating among B, C, and D values N using Appendix D-1. CN Cover
Interpolated CN
B (65%)
C (25%)
D (10%)
Impervious
98
98
98
98
Woods
60
73
79
65.15
Finally, create a table as follows: Cover Impervious Woods
Area (acres) 6.6 819 825.6
Weighted CN 5
CN
Product
98
647
65.15
53358 54005
54005 5 65.4 825.6
CN 5 65
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114 CHAPTER 11 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Watershed divide
Point of analysis
Poject location
SCALE: 1" = 2000' CONTOUR INTERVAL 20'
© Cengage Learning 2014
NEW YORK
FIGURE 20 Hydraulic path for watershed in Problem 23.
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 115
Step 3: Find Q P 5 5.0 in
(Appendix D-3)
Q 5 1.65 in
(Figure 11-9)
Step 4: Find time of concentration, tc. Hydraulic path is delineated in Figure 20. Overland flow: L 5 100 ft, s 5 4.0% n 5 0.40
(Table 11-3)
P2 5 2.7 in.
(Appendix D-3)
Tt 5 Tt 5
0.007 1nL2 0.8 P20.5s0.4
(Equation 11-9)
.007 5 1.42 11002 6 0.8
12.72 0.5 1.042 0.4
Tt 5 0.30 h
Shallow concentrated flow: L 5 2300 ft, s 5 2.7% v 5 2.4 ft/s t2 5
(Figure 10-9, unpaved)
L 2300 5 5 958 sec 5 0.27 h v 2.4
Stream flow: L 5 9000 ft, so 5 3.4%
4′
n = 0.035
12′
3 20 1 12 4 4 5 64 ft2 2
a5
p 5 23 ft
R5
a 64 5 5 2.78 ft P 23
v5
1.49 2/3 1/2 R s n
v5
v 5 15.5 ft/s
t3 5
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© Cengage Learning 2014
20′
1.49 12.782 2/3 1.0342 1/2 .035 L 9000 5 5 581 sec 5 0.61 h v 15.5
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116 CHAPTER 11 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Therefore, tc 5 t1 1 t2 1 t3 tc 5 .30 1 .27 1 .16 tc 5 0.73 h Step 5: Find qu Rainfall distribution Type II
(Appendix D-4)
(Table 11-2)
Ia 5 1.077 in
Then,
Ia 1.077 5 5 0.22 P 5.0
qu 5 380 csm/in
(Appendix D-5, Chart 3)
Step 6: Compute qp. qp 5 qu Am Q
(Equation 11-10)
qp 5 13802 11.292 11.652
qp 5 809 cfs (Answer) 24. Drainage area was determined in Problem 4, Chapter 10. A 5 1.40 acres Cover conditions are found on Figure 10 as follows:
Impervious
0.31 acres
Lawns
0.88 acres
Woods
0.21 acres
Choose c-values for each cover condition. Impervious: c 5 0.90 Lawn: c 5 0.35 Woods: c 5 0.25 c5
1.312 1.902 1 1.882 1.352 1 1.212 1.252 5 0.46 1.40
Time of concentration: Hydraulic path is delineated in Figure 10. Overland flow: L 5 100 ft, s 5 6% t1 5 10.5 min.
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(Appendix C-2)
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 117
Shallow concentrated flow: L 5 200 ft, s 5 7% v 5 4.3 ft/s t2 5
(Figure 10-9, unpaved)
L 200 5 5 47 sec 5 0.8 min. v 4.3
Gutter flow: L 5 210 ft, s 5 5.5% v 5 4.8 ft/s t3 5
(Figure 10-9, paved)
L 210 5 5 44 sec 5 0.7 min v 4.8
Therefore, tc 5 t1 1 t2 1 t3 tc 5 10.5 1 0.8 1 0.7 tc 5 12 min i 5 2.5 in/h
(Appendix C-3)
Qp 5 Aci
(Equation 11-2)
Qp 5 11.42 1.462 12.52
Qp 5 1.6 cfs (Answer) 25.
P 5 8.0 in
(Appendix D-3)
Q 5 5.2 in
(Figure 11-9)
Ia 5 0.632
(Table 11-2)
Then,
Ia .632 5 5 0.079 P 8.0
Rainfall distribution Type II
(Appendix D-4)
Find unit hydrograph from Appendix D-6 for: Rainfall Type II tc 5 1.5 h Ia /P 5 0.10 Tt 5 0 The unit hydrograph is found on Chart 5, line 1 of Appendix D-6 and is reproduced in Table 3 as columns 1 and 2. q 5 qt Am Q
(Equation 11-11)
q 5 qt 1.8922 15.22
q 5 4.64 qt
Each q value becomes a value in the third column in Table 8 and is the desired hydrograph and answer to the problem.
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(1) Hydrograph Time (h)
(2) Unit Discharge (csm/in)
(3) Hydrograph Ordinate (cfs)
11.0 11.3 11.6 11.9 12.0 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8 13.0 13.2 13.4 13.6 13.8 14.0 14.3 14.6 15.0 15.5 16.0 16.5 17.0 17.5 18.0 19.0 20.0
9 11 15 21 25 31 41 58 82 112 147 184 216 255 275 236 198 159 129 98 76 57 43 35 30 25 23 21 18 16
42 51 70 97 116 144 190 269 380 520 682 853 1002 1183 1276 1095 918 738 598 455 353 264 199 162 139 116 107 97 83 74
© Cengage Learning 2014
118 CHAPTER 11 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
TABLE 8 Runoff hydrograph for Problem 25.
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 119
26. P 5 6.0 in
(Appendix D-3)
Rainfall distribution Type II
(Appendix D-4)
Determine appropriate unit hydrographs for sub-areas 1, 2, and 3 and use them to derive hydrographs at the point of analysis. Parameters are as follows: SubArea
tc (h)
1
1.0
.439 5 .07 6.0
0.26
2
2.0
0.26
3
0.6
1.175 5 .20 6.0 .817 5 .14 6.0
Ia/P
Tc (h)
0.00
Unit hydrographs are found in Appendix D-6 as follows: Sub- Area
Chart
Line
1 2
3 6
4 4
3
2
1
Values of unit hydrograph ordinates are shown in Table 9 as columns 2, 4, and 6. Values of derived hydrograph ordinates are found by multiplying by the quantity q 5 Am Q and are reproduced in columns 3, 5, and 7 of Table 9. Values of Am Q for each sub-area follow. Values of Q are obtained from Figure 11-9. SubArea 1 2 3
Am 1s.m.2
1.012 0.761 0.550
Q 1in2 4.0 2.2 2.9
AmQ 1s.m. 2 in2 4.048 1.674 1.595
The answer to the problem is the hydrograph shown in column 8 of Table 9 which is com puted by adding values in columns 3, 5, and 7 for each time value in column 1.
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120 CHAPTER 11 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
(1)
(2) (3) Sub-Area 1
(4) (5) Sub-Area 2
(6)
(7) Sub-Area 3
9 12 16 22 24 28 35 48 70 105 152 205 256 323 310 254 193 146 113 81 61 46 36 31 27 24 22 20 18 16
6 7 9 12 14 15 18 21 27 35 45 59 76 117 159 191 211 208 196 163 128 95 68 51 40 33 28 25 20 18
17 23 32 57 94 170 308 467 529 507 402 297 226 140 96 74 61 53 47 41 36 32 29 26 23 21 20 19 16 14
(8)
11.0 11.3 11.6 11.9 12.0 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8 13.0 13.2 13.4 13.6 13.8 14.0 14.3 14.6 15.0 15.5 16.0 16.5 17.0 17.5 18.0 19.0 20.0
36 49 65 89 97 113 142 194 283 425 615 830 1036 1308 1255 1028 781 591 457 328 247 186 146 125 109 97 89 81 73 65
10 12 15 20 23 25 30 35 45 59 75 99 127 196 266 320 353 348 328 273 214 159 114 85 67 55 47 42 33 30
27 37 51 91 150 271 491 745 844 809 641 474 360 223 153 118 97 85 75 65 57 51 46 41 37 33 32 30 26 22
73 98 131 200 270 409 663 974 1172 1293 1331 1403 1523 1727 1674 1466 1231 1024 860 666 518 396 306 251 213 185 168 153 132 117
© Cengage Learning 2014
HydroUnit Hydrograph Unit Hydrograph Unit Hydrograph Hydrograph Time Discharges Ordinates Discharges Ordinates Discharges Ordinates graph (csm/in) (cfs) (csm/in) (cfs) (csm/in) (cfs) (h) (cfs)
TABLE 9 Runoff hydrographs for sub-areas for Problem 26.
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 121
27. P 5 5.4 in
(Appendix D-3)
Rainfall distribution Type II
(Appendix D-4)
Determine appropriate unit hydrographs for sub-areas 1, 2 and 3 and use them to derive hydrographs at the points of analysis. Parameters are as follows: SubArea
tc (h)
1
0.50
.857 5 .16 5.4
0.75
2
1.0
.381 5 .07 5.4
0.75
3
1.50
1.333 5 .25 5.4
0.00
Ia/P
Tc (h)
Unit hydrographs are found in Appendix D-6 as follows: Sub- Area
Chart
Line
1 2
2 3
7 7
3
5
13
Values of the unit hydrograph ordinates are shown in Table 10 as columns 2, 4 and 6. Values of the derived hydrograph ordinates are found by multiplying the quantity q 5 Am Q and are reproduced in columns 3, 5 and 7 of Table 10. Values of AmQ for each sub-area follow. Values of Q are obtained from Figure 11-9. SubArea 1 2 3
Am 1s.m.2
0.786 1.255 0.923
Q 1in2
2.35 3.60 1.55
AmQ 1s.m. 2 in2 1.847 4.518 1.431
The answer to the problem is the hydrograph shown in column 8 of Table 10 which is computed by adding values in columns 3, 5 and 7 for each value in column 1.
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122 CHAPTER 11 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
(1)
(2) (3) Sub-Area 1
(4) (5) Sub-Area 2
(6)
(7) Sub-Area 3
(8)
11.0
9
17
7
32
0
0
49
11.3
11
20
8
36
0
0
56
11.6 11.9 12.0 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8 13.0 13.2 13.4 13.6 13.8 14.0 14.3 14.6 15.0 15.5 16.0 16.5 17.0 17.5 18.0 19.0 20.0
14 19 21 24 27 31 37 49 74 118 182 319 374 328 244 169 117 76 56 43 35 31 28 25 22 21 18 16
26 35 39 44 50 57 68 91 137 218 336 589 691 606 451 312 216 140 103 79 65 57 52 46 41 39 33 30
11 14 16 17 19 21 25 30 38 53 76 146 228 284 293 256 208 143 99 66 46 36 31 27 24 22 19 17
50 63 72 77 86 95 113 136 172 239 343 660 1030 1283 1324 1157 940 646 447 298 208 163 140 122 108 99 86 77
0 0 0 1 6 15 31 53 80 112 144 193 225 208 186 157 134 108 89 70 56 48 42 37 34 31 28 25
0 0 0 1 9 21 44 76 114 160 206 276 322 298 266 225 192 155 127 100 80 69 60 53 49 44 40 36
76 98 111 122 145 173 225 303 423 617 885 1525 2043 2187 2041 1694 1348 941 677 477 353 289 252 221 198 182 159 143
© Cengage Learning 2014
Hydrograph Unit Hydrograph Unit Hydrograph Unit Hydrograph HydroTime Discharges Ordinates Discharges Ordinates Discharges Ordinates graph (h) (cfs) (csm/in) (cfs) (csm/in) (cfs) (csm/in) (cfs)
TABLE 10 Runoff hydrographs for sub-areas for Problem 27.
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 123
28. Drainage area was found in Problem 5, Chapter 10. A 5 55.0 acres. Cover conditions are shown in Figure 10-28 to be woods. Choose c-value for woods: c 5 0.20 Time of concentration was found in Problem 11, Chapter 10. tc 5 12.4 min i 5 6.6 in/h
(Figure C-3, Chart 3)
Qp 5 Aci
(Equation 11-2)
Qp 5 155.02 1.202 16.62 Qp 5 72.6 cfs
Qp 5 73 cfs (Answer) 29. Drainage area was found in Problem 6, Chapter 10. A 5 128 acres. Cover conditions are found on Figure 10-29 as follows: Impervious 1.2 acres Lawn
30 acres
Woods
96.8 acres
Choose c-value for each cover condition. Impervious c 5 0.90 Lawn
c 5 0.30
Woods
c 5 0.20
c5
11.22 1.92 1 1302 1.32 1 196.82 1.22 5 0.23 128
Time of concentration was found in Problem 12, Chapter 10. tc 5 31.0 min i 5 3.7 in/h
(Figure C-3, Chart 4)
Qp 5 Aci
(Equation 11-2)
Qp 5 11282 1.232 13.72 Qp 5 108.9 cfs
Qp 5 110 cfs (Answer) 30. Watershed area was found in Problem 7, Chapter 10. A 5 1,670 acres Am 5
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1670 5 2.61 s.m. 640
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124 CHAPTER 11 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Cover conditions are found on Figure 10-30 as follows: Impervious 10 acres Woods
1,660 acres
Choose CN-value for each cover condition. Soil type: Jonesville Hydrologic soil group: B Impervious CN 5 98 Woods CN 5
CN 5 60
1102 1982 1 116602 1602 5 60.2 1670
(Use 60)
P 5 5.3 in
(Appendix D-3, Map 1)
Q 5 1.5 in
(Figure 11-9)
Time of concentration: Hydraulic path was delineated in Problem 13, Chapter 10. Overland flow: L 5 100 ft
s 5 2.0%
(estimated)
n 5 0.24
(estimated)
P2 5 2.6 in Tt 5 Tt 5
0.007 1nL2 0.8 1P2 2 0.5s0.4
(.007) 5 1.242 11002 6 .8
12.62 .5 1.022 .4
(Appendix D-3, Map 1) (Equation 11-9)
5 0.26 h
t1 5 0.26 h
Shallow concentrated flow: L 5 2100 ft, s 5 7.6%. v 5 4.4 ft/s t2 5
(Figure 10-9, unpaved)
L 2100 5 5 477 s 5 0.13 h v 4.4
Stream flow: L 5 12,000 ft, s 5 4.3%. a 5
3 12 1 6 4 132 5 27 ft2 2
p 5 14.4 ft R 5
a 27 5 5 1.875 ft p 14.4
v 5
1.49 2/3 1/2 R s n
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 125
v 5
1.49 11.8752 2/3 1.0432 1/2 .032
v 5 14.7 ft/s t3 5
L 12000 5 5 816 s 5 0.23 h. v 14.7
tc 5 t1 1 t2 1 t3 tc 5 0.26 1 0.13 1 0.23 tc 5 0.62 h. Rainfall distribution Type II Ia 5 1.333 in
(Appendix D-4) (Table 11-2)
Ia 1.333 5 5 0.25 P 5.3 qu 5 410 csm/in
(Appendix D-5, Chart 3)
qp 5 qu Am Q qp 5 14102 12.612 11.52 qp 5 1605 cfs
qp 5 1600 cfs. (Answer)
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C h a p t e r
12 Storm Sewer Design
1. First, determine outfall velocity. v 5 9.0 ft/s
(Appendix A-4, Chart 39)
9.0 ft/s exceeds all permissible velocities shown in Appendix A-2. Therefore, a protective lining is required. Next, determine stone size. TW 5 Normal depth 5 1.35r
(Appendix A-4, Chart 39)
.02 1Q/D0 2 4/3 TW
d50 5
(Equation 12-1)
.02 120/2.02 4/3 1.35
d50 5
d50 5 0.32 ft 5 3.8 in use 4-inch stone (Answer) Next, determine apron length, La. Since TW 5 1.35 ft, which is greater than 1/2 D0, use Equation 12-2. La 5
3Q D03/2
La 5
132 1202
(Equation 12-2)
12.02 3/2
La 5 21.2 ft
use La 5 22 ft (Answer) Next, since there is no channel downstream of the outfall, determine apron width, W. Since TW . 1/2 D0, use Equation 12-4. 126
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 127
W 5 3 D0 1 0.4 La W 5 3 12.02 1 .4 1222 W 5 14.8 ft use W 5 15 ft (Answer) 2. First, determine outfall velocity. v 5 9.2 ft/s
(Appendix A-4, Chart 47)
9.2 ft/s exceeds all permissible velocities shown in Appendix A-2. Therefore, a protective lining is required. Next, determine stone size. TW 5 Normal depth 5 2.1 ft d50 5 d50 5
.02 1Q/D0 2 4/3 TW
(Appendix A-4, Chart 47) (Equation 12-1)
.02 185/52 4/3 2.1
d50 5 0.42 ft 5 5.0 in use 6s stone (Answer) Next, determine apron length, La. Since TW 5 2.1 ft which is less than 1/2 D0, use Equation 12-3. La 5 La 5
1.80 1 7D0 D03/2
(Equation 12-3)
1.80 1 172 152 53/2
La 5 35.2 ft
use La 5 35 ft (Answer) Finally, since a channel exists downstream of the outflow, the riprap will line the channel. It will cover the bottom and extend up the sides to a vertical height of 3.1 ft, which is TW plus 1-foot freeboard.
3. First, determine outfall velocity. v 5 9.5 ft/s
(Appendix A-4, Chart 9)
9.5 ft/s exceeds all permissible velocities shown in Appendix A-2. Therefore, a protective lining is required.
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128 CHAPTER 12 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Next, determine stone size. a5
Q v
a5
215 9.5
(Equation 4-3)
a 5 22.63 ft2 TW 5
22.63 8
TW 5 2.83 ft
Therefore, .02 1Q/D0 2 4/3 TW
d50 5
(Equation 12-1)
.02 1215/82 4/3 2.83
d50 5
d50 5 0.56 ft 5 6.8 in use 8s stone (Answer)
Next, determine apron length, La. Since TW 5 2.83 ft, which is greater than 1/2 D0, use Equation 12-2. La 5 La 5
3Q D03/2
(Equation 12-2)
312152 83/2
La 5 28.5 ft use La 5 30 ft (Answer) Finally, since a channel exists downstream of the outfall, the riprap will line the channel. It will cover the bottom and extend up the sides to a vertical height of 3.83 ft, which is TW plus 1-foot freeboard. 4. v 5 13 ft/s
(Appendix A-4, Chart 45)
Exceeds permissible velocity
(Appendix A-2)
D 5 1.6 ft
(Appendix A-4, Chart 45)
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 129
Therefore, TW 5 1.6 ft d50 5 d50 5
.02 1Q/D0 2 4/3 TW
(Equation 12-1)
.02 175/42 4/3 5 0.62 ft 5 7.5 in 1.6
use 8s stone (Answer) TW , 1/2 D0 La 5
1.80 1 7D0 D03/2
(Equation 12-3)
1.80 1 7142 5 28.2 ft 43/2 use La 5 28 ft (Answer)
La 5
W 5 3D0 1 La W 5 3142 1 28 5 40 ft (Answer) 5. See Figure 21. 6. See Figure 22. 7. Drainage area delineations and hydraulic paths are shown on Figure 23.
Composite c-value results are shown in Table 11. Time of concentration: Drainage Basin
1
tc 5 6 min (by inspection)
2
Overland: L 5 100 ft s 5 13%
t1 5 9 min
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Shallow Conc.: L 5 160 ft s 5 8% t2 5 0.6 min. Gutter: L 5 350 ft s 5 2.3% t3 5 1.9 min tc 5 9 1 0.6 1 1.9 5 11.5 min
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FIGURE 21 Storm sewer computation for Problem 5.
© Cengage Learning 2014
85 50 125 30 130 65 140 108 25 50 45
2.5 2.1 1.1 2.4 1.0 2.8 3.5 1.0 2.2 3.0 1.0
12 12 12 12 15 12 12 12 12 15 18
6.3 6.0 4.1 6.3 7.0 6.7 7.5 3.9 6.0 12.5 12.0
6.1 6.4 4.2 7.2 5.8 6.6 7.2 4.5 8.0 10.1 —
.23 .13 .50 .07 .37 .16 .32 .40 .05 .08 —
Pipe Segment
1.4 1.9 1.0 2.5 4.2 1.5 1.6 1.4 4.3 8.3 8.3
A—Increm. Area (acres) 3.6 3.6 3.2 3.2 3.2 5.3 4.1 5.3 4.3 3.1 3.1
C—Runoff Coefficient 16 16.2 18.5 19.0 19.1 6.0 12 6.0 10 19.5 19.6
(4)
A.c—Increm. .40 .52 .32 .79 1.31 .28 .38 .26 .99 2.68 2.68
(5)
A.c—Cummul.
.40 .12 .32 .47 — .28 .10 .26 .73 — —
(6)
tc—Time of Conc. (min.)
.62 .34 .28 .26 — .89 .65 .90 .66 — —
(7)
i—Rainfall Intensity (in/h)
.65 .34 1.16 1.79 — .31 .15 .29 1.10 — —
(8) Qp—Peak Runoff (cfs)
2 5 4 5 10 7 10 9 10 11 12
(9)
Pipe Length (ft)
1 2 3 4 5 6 7 8 9 10 11
(10)
Slope (%)
To
(11)
Size (in)
From
(14)
(13)
(12)
Capacity (full) (cfs)
(3)
0.3 0.1 —
Velocity (fps) (Design Flow)
(2)
7.0 6.2 —
Travel Time in Pipe (min.)
(1)
© Cengage Learning 2014
5.5 6.9 12.5
12 15 21
Pipe Segment
1.8 0.9 0.5
(2)
A—Increm. Area (acres) 140 30 75
(3)
C—Runoff Coefficient 3.2 6.1 8.4
(4)
A.c—Increm. 5.6 5.5 5.5
(5)
A.c—Cummul. 18.0 18.3 18.4
(6)
tc—Time of Conc. (min.)
0.57 1.10 1.52
(7)
i—Rainfall Intensity (in/h)
0.57 0.53 0.42
(8) Qp—Peak Runoff (cfs)
0.46 0.62 0.80
(9)
Pipe Length (ft)
1.24 .86 .52
(10)
Slope (%)
2 3 4
(11)
Size (in)
1 2 3
(12)
Capacity (full) (cfs)
To
(13)
Velocity (fps) (Design Flow)
From
(14)
Travel Time in Pipe (min.)
(1)
130 CHAPTER 12 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
FIGURE 22 Storm sewer computation for Problem 6.
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 131
Plan of Proposed Storm Sewer
SCALE: 1" = 100'
CONTOUR INTERVAL 2'
7 6
5 4 3
© Cengage Learning 2014
1
2
FIGURE 23 Drainage areas and hydraulic paths used for Problem 7.
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Drainage Basin No.
Total Area (Acres)
Imperv. Area (Acres)
Lawn Area (Acres)
Wooded Area (Acres)
c
1 2 3 4 5 6
0.17 1.50 2.05 0.14 — 0.54
.10 .25 .30 .14 — .19
.04 .50 .45 — — .33
.03 .75 1.30 — — .02
0.64 0.35 0.32 0.90 — 0.51
© Cengage Learning 2014
132 CHAPTER 12 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
TABLE 11 Summary of composite c computations for Problem 7. Drainage Basin
3
Overland: L 5 100 ft s 5 11%
t1 5 9.3 min
tc 5 9.3 1 2.0 1 0.6 5 11.9 min
4
tc 5 6 min (by inspection)
5
N.A.
6
tc , 11.9 min (by inspection)
Shallow flow: L 5 450 ft s 5 5.8% v 5 3.8 ft/s t2 5 2.0 min Gutter: L 5 90 ft s 5 1.7% v 5 2.65 ft/s t3 5 0.6 min
Design computations shown in Figure 24. 8. System layout is shown at a larger scale in Figure 25. Drainage areas are delineated in Figure 26. Profiles are shown in Figure 27. Notice that Manholes 5 and 6 were added to connect roof drains from the building. Also, notice Manholes 16 and 17 were added to reduce the pipe slope and therefore the velocity of flow. Composite c-values are computed in Table 12.
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.20 .21 .13 .16 .30 — 3.7 7.8 5.4 3.4 14 8.0 4.9 6.3 12.5 3.9 28 13.0 12 12 18 12 18 18 © Cengage Learning 2014
Pipe Segment
1.5 2.5 1.0 1.0 6.0 1.2
(2)
A—Increm. Area (acres)
45 100 42 32 255 45
(3)
C—Runoff Coefficient
0.86 3.4 8.3 1.0 9.2 10.8
(4)
A.c—Increm.
7.8 6.5 6.4 7.8 6.4 6.3
(5)
A.c—Cummul.
6.0 11.5 11.9 6.0 12.0 12.3
(6)
tc—Time of Conc. (min.)
.11 .53 1.30 .13 1.43 1.71
(7)
i—Rainfall Intensity (in/h)
.11 .53 .66 .13 — .28
(8)
Qp—Peak Runoff (cfs)
.64 .35 .32 .90 — .51
(9)
Pipe Length (ft)
.17 1.50 2.05 .14 — .54
(10)
Slope (%)
3 3 5 5 6 7
(11)
Size (in)
1 2 3 4 5 6
(12)
Capacity (full) (cfs)
To
(13)
Velocity (fps) (Design Flow)
From
(14)
Travel Time in Pipe (min.)
(1)
HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 133
FIGURE 24 Storm sewer computations for Problem 7.
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134 CHAPTER 12 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Proposed Storm Sewer Layout SCALE: 1" = 100'
PARKING ROAD
1 2 3
7 9
PARKING
10
4 6
11 12
RO
AD
5
8
13
14
15 16
17 © Cengage Learning 2014
18
FIGURE 25 Storm sewer layout for Problem 8.
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 135
© Cengage Learning 2014
Site Plan SCALE: 1" = 200' CONTOUR INTERVAL 2'
FIGURE 26 Drainage areas used for Problem 8.
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136 CHAPTER 12 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Profiles SCALE: Horiz. 1" = 100' Vert. 1" = 10'
INLET 4
390
INLET 1 INLET 2 INLET 3
400
MH 12
MH 6
380
370
HW 18
350
© Cengage Learning 2014
MH 17
INLET 15 MH 16
360
INLET 14
INLET 12
INLET 13
370
INLET 11
380
INLET 7 INLET 8 INLET 9
390
FIGURE 27 Storm sewer profiles used in Problem 8.
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Drainage Basin No.
Total Area (Acres)
Imperv. Area (Acres)
Lawn Area (Acres)
Wooded Area (Acres)
c
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
0.51 0.19 0.06 0.43 0.47 0.36 1.48 0.12 0.05 1.21 0.52 — 0.56 0.46 0.08 — —
.24 .07 .06 .36 .47 .36 .35 .05 .05 .10 .40 — .03 .30 .08 — —
.08 .08 — .07 — — .14 .07 — .17 .12 — .53 .16 — — —
.19 .04 — — — — .99 — — .94 — — — — — — —
0.59 0.54 0.95 0.85 0.95 0.95 0.42 0.58 0.95 0.32 0.80 — 0.35 0.73 0.95 — —
© Cengage Learning 2014
HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 137
TABLE 12 Summary of composite c computations for Problem 8.
Key times of concentration are computed below: Inlet 1: Overland flow: L 5 100 ft s 5 18%
t1 5 8.7 min
(Appendix C-2)
Shallow conc. flow: L 5 45 ft s 5 18%
t2 5 0.1 min
Gutter flow: L 5 270 ft s 5 4% v 5 4.1 ft/s
(Figure 10-9)
t3 5 1.1 min
tc 5 8.7 1 0.1 1 1.1 5 9.9 min Inlet 7: Overland flow: L 5 100 ft s 5 8.0%
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t1 5 9.8 min
(Appendix C-2)
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138 CHAPTER 12 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Shallow flow: L 5 350 ft s 5 11% v 5 5.3 ft/s
t2 5 1.1 min
Sheet flow: L 5 200 ft s 5 4.2% v 5 4.2 ft/s
(Figure 10-9)
t3 5 0.8 min
Gutter flow: L 5 50 ft s 5 6% v 5 5.0 ft/s
(Figure 10-9)
(Figure 10-9)
t4 5 0.2 min
tc 5 9.8 1 1.1 1 0.8 1 0.2 5 11.9 min Inlet 10: Overland flow: L 5 100 ft s 5 11%
t1 5 9.2 min
Shallow flow: L 5 300 ft s 5 16% v 5 6.5 ft/s
(Figure 10-9)
t2 5 0.8 min
Gutter flow: L 5 200 ft s 5 4% v 5 4.1 ft/s
(Appendix C-2)
(Figure 10-9)
t3 5 0.8 min
tc 5 9.2 1 0.8 1 0.8 5 10.8 min Hydraulic design is shown in Figure 28. To finish this design, an outfall apron at headwall 18 would be designed.
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© Cengage Learning 2014
2.1 2.8 3.3 5.9 3.8 11.3 4.3 4.8 5.1 2.6 10.4 21.2 22.6 22.2 24.5 24.5 24.5
15 20 70 155 255 40 20 20 75 105 100 115 25 70 15 85 25
Pipe Length (ft)
7.1 7.1 7.1 7.1 8.5 7.0 6.9 6.9 6.9 6.8 6.7 6.7 6.7 6.6 6.6 6.6 6.6
2.0 2.0 3.0 8.0 1.0 1.0 2.0 2.0 4.0 1.0 4.0 3.0 4.0 4.0 4.0 6.0 1.0
12 12 12 12 12 18 12 12 12 12 15 21 21 21 21 21 24
5.7 5.7 7.0 12.5 3.9 12.0 5.7 5.7 8.0 3.9 14.5 30 35 35 35 42 25
6.7 7.0 8.3 14 5.8 7.4 7.8 7.9 10 5.2 13 13 15 15 16 18 9
.04 .05 .14 .18 .73 .09 .04 .04 .13 .34 .13 .15 .03 .08 .02 .08 —
(10)
Slope (%)
9.9 9.9 10.0 10.1 6.0 10.3 11.9 11.9 12.0 12.1 12.5 12.6 12.7 12.8 12.8 12.9 12.9
(11)
Size (in)
.30 .40 .46 .83 .45 1.62 .62 .69 .74 .39 1.55 3.17 3.37 3.71 3.79 3.79 3.79
(14)
(13)
(12)
Capacity (full) (cfs)
.30 .10 .06 .37 .45 .34 .62 .07 .05 .39 .42 — .20 .34 .08 — —
A—Increm. Area (acres)
Pipe Segment
(9)
Velocity (fps) (Design Flow)
.59 .54 .95 .85 .95 .95 .42 .58 .95 .32 .80 — .35 .73 .95 — —
(3)
C—Runoff Coefficient
(2)
Travel Time in Pipe (min.)
.51 .19 .06 .43 .47 .36 1.48 .12 .05 1.21 .52 — .56 .46 .08 — —
(4)
A.c—Increm.
2 3 4 6 6 12 8 9 11 11 12 13 14 15 16 17 18
(5)
A.c—Cumul.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
(6)
tc—Time of Conc. (min.)
To
(7)
i—Rainfall Intensity (in/h)
From
(8) Qp—Peak Runoff (cfs)
(1)
HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 139
FIGURE 28 Storm sewer computations for Problem 8.
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C h a p t e r
13 Culvert Design
1. See Figure 26. Cost for solution A: Culvert 55 LF @ $2000/LF 5 $110,000 Channel 50 LF @ 250/LF 5
12,500
TOTAL 5 $122,500 Cost for solution B: Culvert 50 LF @ $2000/LF 5 $100,000 Channel 90 LF @ 250/LF 5
22,500
TOTAL 5 $122,500 S ince costs are equal, solution A may be preferred because it alters fewer linear feet of stream channel. 2. a) Circular concrete pipe
Assume n 5 0.012
Trial 1: 60s RCP culvert
Assume inlet control.
HW/D 5 4.4
HW 5 14.42 152 5 22 ft
Trial 2: Twin 60s RCP culvert
Assume inlet control.
HW/D 5 1.64
HW 5 11.642 152 5 8.2 ft
(Appendix B-1, Chart 2)
Headwater elev. 5 80 1 22 5 102 (exceeds A.H.E.)
Headwater elev. 5 80 1 8.2 5 88.2 (exceeds A.H.E.)
140
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30′
HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 141
Proposed road
A
Plan SCALE: 1" = 20'
Exis
ting
strea m
© Cengage Learning 2014
B
FIGURE 29 Alternate culvert layouts for use in Problem 1.
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142 CHAPTER 13 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Trial 3: Triple 60s RCP culvert
Assume inlet control.
HW/D 5 1.1
HW 5 11.12 152 5 5.5
Assume outlet control.
Dc 5 3.0 ft
(Appendix A-4, Chart 47)
TW 5 4.1 ft
(use TW)
TWr 5 13 1 52/2 5 4.0 ft ke 5 0.5
(Appendix B-3)
H 5 1.35 ft
(Appendix B-2, Chart 9)
Headwater elev. 5 79.60 1 4.1 1 1.35 5 85.05 (does not exceed A.H.E.)
Headwater elev. 5 80 1 5.5 5 85.5 (does not exceed A.H.E)
Therefore, culvert operates under inlet control and upstream water level is 85.50, which does not exceed A.H.E. Culvert: Triple 60s RCP. b) Concrete Box Culvert
Assume n 5 0.012
Trial 1: 5r 3 10r box
Assume inlet control.
Q 420 5 5 42 cfs/ft B 10
HW/D 5 1.25
HW 5 11.252 152 5 6.25 ft
Trial 2: 5r 3 12r box
Assume inlet control.
Q 420 5 5 35 cfs/ft B 12
HW/D 5 1.07
HW 5 11.072 152 5 5.35 ft
Assume outlet control.
Dc 5 3.3 ft
TWr 5 13.3 1 52/2 5 4.15 ft
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(Appendix B-1, Chart 1)
Headwater elev. 5 80.00 1 6.25 5 85.25 (exceeds A.H.E.)
Headwater elev. 5 80.00 1 5.35 5 85.35 (does not exceed A.H.E.) (Appendix A-3, Chart 10)
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 143
1use TWr2
TW 5 4.1 ft
ke 5 0.4
(Appendix B-3)
H 5 1.18 ft
(Appendix B-2, Chart 8)
Headwater elev. 5 79.60 1 4.15 1 1.18 5 84.93 (does not exceed A.H.E.)
Therefore, culvert operates under inlet control and upstream water level is 85.35, which does not exceed A.H.E. Culvert: 5r 3 12r box Choice of culvert: Both culverts have the correct height. The triple pipe culvert requires about 18 ft width while the box culvert requires only about 14 ft width (outside dimensions). However, the triple pipe culvert probably is less expensive and easier to install. 3. A.H.E 5 7.5 ft so 5 0.19%
(measured by scale)
(measured by scale)
approximate max. height of culvert opening is 5.5 ft. This leaves 2.0 ft clearance to the finished grade to account for culvert and road thickness. a) Circular concrete pipe
Assume n 5 0.012.
Trial 1: Twin 60s RCP culvert
Assume inlet control 750 5 375 cfs 2
Use Q 5
HW/D 5 3.6
HW 5 13.62 15.02 5 18.0 ft (exceeds A.H.E.)
Trial 2: Triple 60s RCP culvert
Assume inlet control 750 5 250 cfs 3
Use Q 5
HW/D 5 2.0
HW 5 12.02 15.02 5 10.0 ft (exceeds A.H.E.)
Trial 3: Quadruple 60s RCP culvert
Assume inlet control 750 5 188 cfs 4
Use Q 5
HW/D 5 1.43
HW 5 11.432 15.02 5 7.15 ft (less than A.H.E.) (OK)
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144 CHAPTER 13 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Assume outlet control
Dc 5 4.0 ft
TWr 5
TW 5 4.0 ft
(use TWr)
Ke 5 0.5
(Appendix B-3)
L 5 120 ft
(measured by scale)
H 5 2.8 ft
(Appendix B-2, Chart 9)
HW 5 TWr 1 H 2 Lso
HW 5 4.5 1 2.8 2 11202 1.00192
4.0 1 5.0 5 4.5 ft 2
HW 5 7.07 ft
Therefore, culvert operates under inlet control. HW 5 7.15 ft which does not exceed A.H.E. Culvert: Quadruple 60s RCP b) Concrete box culvert
Assume n 5 0.012
Trial 1: 5r 3 10r box
Assume inlet control
Q 750 5 5 75 cfs/ft B 10
HW/D 5 2.4
HW 5 12.42 15.02 5 12.0 ft (exceeds A.H.E.)
Trial 2: 5r 3 12r box
Assume inlet control
Q 750 5 5 62.5 cfs/ft B 12
HW/D 5 1.9
HW 5 11.92 15.02 5 9.5 ft (exceeds A.H.E.)
Trial 3: 5r 3 15r box
Assume inlet control
Q 750 5 5 50 cfs/ft B 15
HW/D 5 1.47
HW 5 11.472 15.02 5 7.35 ft (less than A.H.E.) (OK)
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 145
Assume outlet control
Dc 5 4.3 ft
TWr 5
TW 5 4.0 ft
(use TWr)
Ke 5 0.4
(Appendix B-3)
L 5 120 ft
(measured by scale)
H 5 2.45 ft
(Appendix B-2, Chart 9)
HW 5 TWr 1 H 2 Lso
HW 5 4.65 1 2.45 2 11202 1.00192
4.3 1 5.0 5 4.65 ft 2
HW 5 6.88 ft
(less than A.H.E.) (OK)
Therefore, culvert operates under inlet control. HW 5 7.35 ft which does not exceed A.H.E. Culvert: 5r 3 15r box Choice of culvert: Both culverts have the correct height. The quadruple pipe culvert requires about 24 feet width while the box culvert requires only about 17 feet width (outside dimensions). However, the pipe culvert is probably less expensive and easier to install. 4. Evaluate adequacy:
Assume inlet control.
Q 860 5 5 43 cfs/ft B 20
HW/D 5 1.79
HW 5 11.792 142 5 7.16 ft
Headwater elev. 5 179.0 1 7.16 5 186.16 (exceeds A.H.E.)
Therefore, existing culvert is inadequate.
Design replacement culvert.
Trial 1: 5r 3 20r box culvert 1n 5 0.0122
Assume inlet control.
Q 860 5 5 43 cfs/ft B 20
HW/D 5 1.25
HW 5 11.252 152 5 6.25 ft
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(Appendix B-1, Chart 1)
Headwater elev. 5 179.0 1 6.25 5 185.25 (exceeds A.H.E.)
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146 CHAPTER 13 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Trial 2: Twin 5r 3 12r box culvert
Assume inlet control.
Q 430 5 5 35.8 cfs/ft B 12
HW/D 5 1.09
HW 5 11.092 152 5 5.45 ft
Assume outlet control.
Dc 5 3.4 ft
(Appendix 4-3, Chart 10)
TW 5 3.9 ft
(use TWr)
TWr 5 13.4 1 52/2 5 4.2 ft ke 5 0.4
(Appendix B-3)
H 5 1.22 ft
(Appendix B-2, Chart 8)
Headwater elev. = 177.4 1 4.2 1 1.22 5 182.82 (does not exceed A.H.E.)
Headwater elev. 5 179.0 1 5.45 5 184.45 (does not exceed A.H.E.)
Therefore, culvert operates under inlet control and upstream water level is 184.45, which does not exceed A.H.E. Culvert: Twin 5r 3 12r box in same location as existing. Outlet apron:
a 5 1122 13.92 5 46.8 ft2
d50 5
d50 5
d50 5 0.61 ft 5 7.26 in
use d50 5 8 in
La 5
La 5
La 5 31 ft
use 31 ft
v5
Q 430 5 5 9.2 ft/s (exceeds permissible velocities in Appendix A-2) a 46.8 .02 1Q/D0 2 4/3 TW
(Equation 12-1)
.02 1430/122 4/3 3.9
3Q D03/2
(Equation 12-2)
314302 123/2
Riprap to line channel to vertical height of 4.9 ft, which is TW plus 1-foot freeboard.
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 147
5. Evaluate adequacy of existing triple 30s RCP.
Assume outlet control since tailwater is above downstream culvert crown.
ke 5 0.5
H 5 3.2 ft
TW 5 3.25 ft
Headwater elev. 5 316.75 1 3.25 1 3.2 5 323.2 (exceeds A.H.E.)
(Appendix B-3) (Appendix B-2, Chart 9)
Design replacement culvert.
Trial 1: 2.5r 3 10r box culvert
Assume outlet control.
ke 5 0.4
(Appendix B-3)
H 5 0.90 ft
(Appendix B-2, Chart 8)
TW 5 3.25 ft
Headwater elev. 5 316.75 1 3.25 1 0.9 5 320.9 (does not exceed A.H.E.)
Therefore, culvert operates under outlet control and upstream water level is 320.9, which does not exceed A.H.E. Culvert: 2.5r 3 10r box culvert. Outlet apron:
v5
Q a
v5
152 5 6.08 ft/s (exceeds permissible velocities in Appendix A-2) 25
d50 5
d50 5
.02 1Q/D0 2 4/3 TW
(Equation 12-1)
.02 1152/102 4/3 2.5
Note: When TW is greater than the culvert height, use culvert height as TW in Equation 10-1.
d50 5 0.30 ft 5 3.6 in
use d50 5 4 in
La 5
La 5
La 5 14 ft
use La 5 15 ft
3Q D03/2 311522
(Equation 12-2)
103/2
Riprap to line channel to vertical height of 4.25 ft, which is TW plus 1-foot freeboard.
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148 CHAPTER 13 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
6. Trial 1: 3r 3 10r box culvert
Assume outlet control since downstream water level is at downstream crown.
ke 5 0.4
H 5 3.20 ft
TW 5 3.0 ft
Headwater elev. 5 209.6 1 3.00 1 3.20 5 215.8
A.H.E. 5 213.40 1 0.20 5 213.60
Therefore, u/s elev. exceeds A.H.E.
(Appendix B-3) (Appendix B-2, Chart 8)
Trial 2: 3r 3 12r box culvert
ke 5 0.4
H 5 2.10 ft
TW 5 3.0 ft
Headwater elev. 5 209.6 1 3.00 1 2.10 5 214.7 (exceeds A.H.E.)
Trial 3: 3r 3 14r box culvert
ke 5 0.4
H 5 1.60 ft
TW 5 3.0 ft
Headwater elev. 5 209.6 1 3.00 1 1.60 5 214.2 (exceeds A.H.E.)
Trial 4: 3r 3 18r box culvert
ke 5 0.4
H 5 0.91 ft
TW 5 3.0 ft
u/s elev. 5 209.6 1 3.00 1 0.91 5 213.51 (does not exceed A.H.E.)
Therefore, culvert is 3r 3 18r box. Outlet protection: Q 340 5 5 6.30 ft/s (exceeds permissible velocities in Appendix A-2) a 54
v5
d50 5
.02 1Q/D2 4/3 TW
d50 5
.02 1340/182 4/3 3.0
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(Equation 12-1)
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 149
d50 5 0.34 ft 5 4.0 in
use d50 5 4 in
La 5
La 5
La 5 13.4 ft
use La 5 15 ft
3Q D03/2 313402
(Equation 12-2)
183/2
Riprap to line channel to vertical height of 4.0 ft, which is TW plus 1-foot freeboard.
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C h a p t e r
14 Stormwater Detention
1. Use the orifice equation, Equation 5-3, where c 5 0.62. Remember that the head is the water level above the center of the orifice. Assume the orifice is discharging freely. See table below. 60 Orifice Elevation (ft)
H (ft)
Q (cfs)
200
0
0
201
0.75
0.85
202
1.75
1.29
203
2.75
1.62
204
3.75
1.89
205
4.75
2.13
2. Use Equation 5-4 for a broad crested weir with a breadth of 1.0 foot. For c-values, use Appendix A-5. See Table below. 59 Weir Elevation (ft)
H (ft)
c
100 101 102 103 104
0 1.0 2.0 3.0 4.0
— 2.98 3.30 3.32 3.32
Q (cfs) 0 14.9 46.7 86.3 133
150
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 151
3. Use Equation 5-3 for the orifice and Equation 5-4 for the broad crested weir with a breadth of 0.5 foot. At each water level considered, compute the headwater, HW, for the outflow pipe to check for submerged conditions for the orifice. 50 Orifice Elevation (ft)
Total
2.29 Weir
H (ft)
Q (cfs)
H (ft)
c
Q (cfs)
Q (cfs)
HW (ft)
100
0
0
0
—
0
0
0
101
0.79
0.60
0
—
0
0.60
*
102
1.79
0.91
0
—
0
0.91
*
103
2.79
1.13
0
—
0
1.13
*
104
3.79
1.32
0
—
0
1.32
*
105
3.46**
1.26
1.0
3.32
7.3
8.56
1.54
106
2.70**
1.11
2.0
3.32
20.7
21.8
3.30
* Discharge too small to find HW on the culvert chart. ** H computed by subtracting culvert headwater elevation from reservoir water surface elevation.
4. First, determine whether the emergency spillway flow will be subcritical or supercritical. For Q 5 78 cfs, for example, F 5 1.18. Therefore, flow is supercritical. Next, choose a list of Dc-values and find their corresponding Q-values using Equation 6-2. Then, using Equation 8-6, find corresponding Dr-values. Dc (ft) 0.5 1.0 1.5
Q (cfs) 21.7 66.5 132
Dr (ft) 0.72 1.41 2.07
Elevation (ft) 220.72 221.41 222.07
The Dr-values represent reservoir water levels above the emergency spillway crest and are converted to elevations by adding 220.00 ft to each value.
5. First, compute the inflow hydrograph using the given parameters. See Figure 30.
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152 CHAPTER 14 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
5
0.5 Time (h)
1.0
© Cengage Learning 2014
Q (cfs)
10
FIGURE 30 Inflow hydrograph for Problem 5. Next, determine a time interval for the routing and prepare a table of inflow values. Use Dt 5 3 min. Time (h)
Inflow (cfs)
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65
0 2.0 4.0 5.9 7.9 9.9 8.7 7.5 6.3 5.2 4.0 2.8 1.6 0.4
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 153
Next, using the given detention basin parameters, create a table of values of outflow, O, versus 2S/Dt 2 O and versus 2S/Dt 1 O. O (cfs)
2S Dt
2 O
(cfs)
0 2.5 3.5 4.2 4.9 5.5
0 8.6 18.7 40.2 61.8 83.4
2S Dt
1 O
(cfs) 0 13.6 25.7 48.6 71.6 94.4
A graph of this table is shown in Figure 31.
6 2S Δt
−O 2S Δt
+O
O (cfs)
4
50 2S Δt
− O (cfs)
100 2S Δt
+ O (cfs)
© Cengage Learning 2014
2
FIGURE 31
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154 CHAPTER 14 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Finally, compute the routing using the form shown in Figure 14-9. Time (h) 0 .05 .10 .15 .20 .25 .30 .35 .40 .45 .50 .55 .60 .65
I1 (cfs) 0 2.0 4.0 5.9 7.9 9.9 8.7 7.5 6.3 5.2 4.0 2.8 1.6 0.4
I1 1 I2 (cfs) 2.0 6.0 9.9 13.8 17.8 18.6 16.2 13.8 11.5 9.2 6.8 4.4 2.0 —
2S Dt
2 O
(cfs) 0 1.5 5.0 10.0 17.0 27.5 38.0 45.8 50.2 52.8 53.0 50.4 46.4 40.4
2S Dt
1 O
(cfs) 0 2.0 7.5 14.9 23.8 34.8 46.1 54.2 59.6 61.7 62.0 59.8 54.8 48.4
O2 (cfs) 0 0.3 1.3 2.7 3.4 3.8 4.1 4.4 4.5 4.6 4.6 4.5 4.4 4.2
Peak inflow 5 9.9 cfs. Peak outflow 5 4.6 cfs. (Answer) The outflow hydrograph, together with the inflow hydrograph, is shown in Figure 32. Note that the hydrograph can also be computed using application software. 6. First, compute the inflow hydrograph using the given parameters. The inflow hydrograph was computed using Intelisolve software and is shown below. Time (h)
Inflow (cfs)
Time (h)
Inflow (cfs)
11.8 11.9 12.0 12.1 12.2 12.3 12.4
0.36 1.59 3.75 6.03 8.10 8.89 8.37
12.5 12.6 12.7 12.8 12.9 13.0 13.1
7.65 6.72 5.64 4.45 3.30 2.60 2.35
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 155
10
Outflow Hydrograph
5
© Cengage Learning 2014
Q (cfs)
Inflow Hydrograph
1.0
0.5 Time (h)
FIGURE 32 Inflow and outflow hydrographs for Problem 5. Next, using the given detention basin parameters, compute the routing. The inflow and outflow hydrographs are shown in Figure 33. Peak inflow 5 8.89 cfs. Peak outflow 5 5.48 cfs. (Answer)
10 8
Q (cfs)
Inflow 6 Outflow
4
0 0.0
2.5
5.0
7.4
9.9
12.4 14.9 Time (hr)
17.4
19.8
22.3
24.8
© Cengage Learning 2014
2
FIGURE 33 Inflow and outflow hydrographs for Problem 6.
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C h a p t e r
15 Detention Design
Elev. (ft)
Area (ft2)
Incr. Volume (ft3)
Cumm. Volume (ft3)
100 101 102 102.5 104 106
0 5984 12062 12470 15770 20905
0 2992 9023 6133 21180 36675
0 2992 12015 18148 39328 76003
© Cengage Learning 2014
1. See Table 13.
TABLE 13 Summary of storage volume computations for Problem 1.
180 Weir
40 Orifice Elev. (ft) 100 101 102 102.5 104 106
H (ft) 0 .83 1.83 2.33 3.83 5.83
Q (cfs) 0 .40 .59 .66 .85 1.05
219 Weir
H (ft)
Q (cfs)
H (ft)
Q (cfs)
Total Outflow (cfs)
— — — — 1.5 3.5
— — — — 9.0 32.6
— — — — — 2.0
— — — — — 197
0 .40 .59 .66 9.0 231.0
© Cengage Learning 2014
2. See Table 14.
TABLE 14 Summary of outflow computations for Problem 2. 156
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 157
3. See Table 15 and Figure 34. Dt 5 0.25 h. To assist in computing Table 15, it may be helpful to sketch a graph of Elevation vs. Storage and Elevation vs. Outflow using the information in Tables 13 and 14. 2S 1 O Dt (cfs)
2S 2O Dt (cfs)
.4 .6 .7 3.5 10
7.0 27.3 40.7 72.4 97
6.2 26.1 39.3 65.4 77
TABLE 15 Values of O vs.
© Cengage Learning 2014
O (cfs)
2S 2S 1 O and O vs. 2 O used in Problem 3. Dt Dt
10
–O
2S Dt
50 2S Dt
FIGURE 34 Graph of O vs.
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+O
5
+ O, cfs;
100 2S Dt
– O, cfs
© Cengage Learning 2014
O, cfs
2S Dt
2S 2S 1 O and O vs. 2 O used in Problem 3. Dt Dt
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158 CHAPTER 15 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Time (h)
I (cfs)
I1 1 I2 (cfs)
2S 2O Dt (cfs)
0 .25 .50 .75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00
1.2 1.5 2.1 3.2 11.6 22.1 9.8 4.5 3.2 2.7 2.4 2.0 1.8
2.7 3.6 5.3 14.8 33.7 31.9 14.3 7.7 5.9 5.1 4.4 3.8 —
0 2.7 6.3 11.6 26.4 54 71 71 68 65 63 61 —
2S 1O Dt (cfs)
O2 (cfs)
0 2.7 6.3 11.6 26.4 60.1 85.9 85.3 78.7 73.9 70.1 67.4 64.8
— .2 .4 .5 .6 2.0 6.5 6.5 4.8 3.8 3.2 2.8 2.6
© Cengage Learning 2014
4. See Table 16. Peak outflow is 6.5 cfs. Maximum water level elevation is determined from Table 14 to be 103.8.
TABLE 16 Summary of routing computations for Problem 4. 5. First, compute the peak runoff for existing conditions. Using Intelisolve software, the following values were computed. Storm
qp (cfs)
100 10
43.0 21.8
These values are now used as the maximum allowable runoff for proposed conditions. The detention basin size and outflow is chosen by trial and error. Maximum depth is chosen to be 6 feet so that the basin is not too deep. A two-stage outflow structure is chosen with an orifice as the primary stage and a weir as the second stage. The outflow structure is shown in Figure 35. The orifice diameter was given as 4 inches. (A 4-inch orifice is used for water quality control. If water quality was not part of the design, the orifice could have been much larger and the detention basin smaller.) The weir size is chosen as a 4.5-ft broad crested weir so that outflow is approximately equal to the 100-year storm runoff (43.0 cfs) at
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 159
4.5′ Weir Elev. 656
Elev. 654
© Cengage Learning 2014
4″ Orifice
Elev. 650
FIGURE 35 Outflow structure for Problem 5. elevation 656 ft which is the maximum design elevation. Outflow values are shown in the following table. 40 Orifice
4.59 Weir
Elevation (ft)
H (ft)
Q (cfs)
H (ft)
650 652 654 656
0 1.83 3.83 5.83
0 0.59 0.85 1.05
0 0 0 2
c
Q (cfs)
Total Q (cfs)
— — — 3.3
0 0 0 42
0 0.59 0.85 43
The detention basin is an open cut, grass-lined basin with bottom slope of 2 percent and side slopes of 3 horizontal to 1 vertical. After a few trials, the size is chosen as the smallest size that results in the required outflows and a maximum water level less than elevation 656 ft. The following table shows the basin size.
Elevation (ft) 650 652 654 656
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Area (s.f.) 0 15,600 21,700 28,700
Incremental Volume (c.f.) 0 15,600 37,300 50,400
Cumulative Volume (c.f.) 0 15,600 52,900 103,300
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160 CHAPTER 15 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
Finally, rout the inflow hydrographs through the detention basin. Results of the routing are shown below.
Peak Runoff Existing Conditions (cfs) 43.0 21.8
Storm 100 10
Peak Inflow (cfs) 71.9 42.2
Detention Basin Peak Maximum Outflow Water Level (cfs) (ft) 42.2 655.92 17.2 654.76
For both the 100-year and 10-year storms, peak outflow from the detention basin is less than peak runoff under existing conditions. Also, maximum water level is less than the maximum design water level. Inflow and outflow hydrographs are shown in Figure 36. 80
Q (cfs)
60
Inflow
40 Outflow 20
0 0.0
2.4
4.9
7.3
9.8
12.2 14.6 Time (hr)
17.1
19.5
22.0
24.4
19.6
22.1
24.5
(a) 100-year storm 50
Inflow
30 20
Outflow
10 0 0.0
2.5
4.9
7.4
9.8
12.3 14.7 Time (hr)
(b) 10-year storm
17.2
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Q (cfs)
40
FIGURE 36 Inflow and outflow hydrographs for Problem 5.
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6. First, compute the peak runoff for existing conditions. Using Intelisolve software, the following values were computed. Storm
Qp (cfs)
100 10
15.8 11.5
These values are now used as the maximum allowable runoff for proposed conditions. The detention basin size and outflow is chosen by trial and error. Maximum depth is chosen to be 5 feet so that the basin is not too deep. A two-stage outflow structure is chosen with an orifice as the primary stage and a weir as the second stage. The outflow structure is shown in Figure 37. The orifice diameter was given as 4 inches. (A 4-inch orifice is used for water quality control. If water quality was not part of the design, the orifice could have been much larger and the detention basin smaller.) The weir size is chosen as a 4.5-ft broad crested weir so that outflow is approximately equal to the 100-year storm runoff (15.8 cfs) at elevation 405 ft which is the maximum design elevation. Outflow values are shown in the following table. 40 Orifice
4.59 Weir
Elevation (ft)
H (ft)
Q (cfs)
H (ft)
c
Q (cfs)
Total Q (cfs)
400 402 404 405
0 1.83 3.83 5.83
0 0.59 0.85 1.05
0 0 0 1
— — — 3.32
0 0 0 15
0 0.59 0.85 16
4.5′ Weir Elev. 405 Elev. 404
Elev. 400
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4″ Orifice
FIGURE 37 Outlet structure for Problem 6.
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The detention basin is an open cut, grass-lined basin with bottom slope of 2 percent and side slopes of 3 horizontal to 1 vertical. After a few trials, the size is chosen as the smallest size that results in the required outflows and a maximum water level less than elevation 405 ft. The following table shows the basin size.
Elevation (ft) 400 402 404 405
Area (s.f.) 0 4,500 6,250 7,220
Incremental Volume (c.f.) 0 4,500 10,750 13,470
Cumulative Volume (c.f.) 0 4,500 15,250 28,720
Finally, rout the inflow hydrographs through the detention basin. Results of the routing are shown below.
Storm 100 10
Peak Runoff Existing Conditions (cfs) 15.8 11.5
Detention Basin Peak Maximum Outflow Water Level (cfs) (ft) 15.7 404.98 9.6 404.58
Peak Inflow (cfs) 36.4 26.3
For both the 100-year and 10-year storms, peak outflow from the detention basin is less than peak runoff under existing conditions. Also, maximum water level is less than the maximum design water level. Inflow and outflow hydrographs are shown in Figure 38. 7. Delineation of watershed tributary to detention basin is shown in Figure 39. Use same watershed for pre-development and post-development conditions. Runoff hydrographs and routing are computed using computer software produced by Intelisolve. Compute parameters to use with software. Pre-development conditions:
1. Watershed Area:
Am 5 17.0 acres (includes area of detention basin)
2. Curve Number: Cover Condition Impervious Lawn Wooded
Area (acres) 0.3 1.7 15
CN 98 61 60
Product 29 104 900 1033
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40
Q (cfs)
30
Inflow
20 Outflow 10
0 0.0
0.2
0.5
0.7
0.9
1.1 1.4 Time (hr)
1.6
1.8
2.0
2.3
4.1
4.7
5.3
5.8
(a) 100-year storm 30 25
Q (cfs)
20
Inflow
15 10
Outflow
0 0.0
0.6
1.2
1.8
2.3
2.9 3.5 Time (hr)
(b) 10-year storm
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5
FIGURE 38 Inflow and outflow hydrographs for Problem 6. Weighted CN 5
1033 5 60.8 17
CN 5 61
Note: Impervious area includes existing off-site houses.
3. 24-hour precipitation: P 5 6.0 in
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(Appendix D-3)
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SCALE: 1" = 200' CONTOUR INTERVAL 2'
© Cengage Learning 2014
TOPOGRAPHIC MAP
FIGURE 39 Watershed delineation used in Problem 7.
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4. Time of concentration: Overland flow: L 5 100 ft s 5 12%
n 5 0.40
Tt 5 Tt 5
P2 5 2.75 in .0071nL2 P2.5 s.4
(Table 11-3) (Appendix D-3)
.8
.007 5 1.42 11002 6 .8 12.752 .5 1.122 .4
(Equation 11-5) 5 0.19 h
t1 5 0.19 h
Shallow concentrated flow: L 5 1000 ft s 5 7% v 5 4.3 ft/s t2 5
(Figure 10-9, unpaved)
d 1000 5 5 3.9 min 5 0.06 h v 4.3
tc 5 t1 1 t2 5 .19 1 .06 tc 5 0.25 h
5. Rainfall distribution: Type II
(Appendix D-4)
Peak runoff under existing conditions is 35.1 cfs. This value is now used as the maximum allowable runoff for proposed conditions. The detention basin size and outflow is chosen by trial and error. Maximum depth is chosen to be about 4 feet so that the basin is not too deep. A two-stage outflow structure is chosen with an orifice as the primary stage and a weir as the second stage. The outflow structure is shown in Figure 40. 3′ Weir Elev. 186
12″ Orifice Elev. 182.5
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Elev. 184
FIGURE 40 Outflow structure for Problem 7.
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The orifice diameter was chosen to be 12 inches. The weir size was chosen as a 3-ft broad crested weir so that outflow is approximately equal to the 100-year storm runoff (35.1 cfs) at elevation 186 ft which is the maximum design elevation. Outflow values are shown in the following table. 120 Orifice
39 Weir
Elevation (ft)
H (ft)
Q (cfs)
H (ft)
c
182.5 183 184 185 186
0 0.25 1.0 2.0 3.0
0 1.9 3.8 5.3 6.6
0 0 0 1 2
— — — 3.32 3.32
Q (cfs) 0 0 0 10.0 28.2
Total Q (cfs) 0 1.9 3.8 15.3 34.8
Developed conditions:
1. Watershed area:
Am 5 17.0 acres
2. Curve Number: Cover Condition Impervious Lawn Wooded
Area (acres) 3.7 7.1 6.2
CN 98 61 60
Product 393 433 372 1168
1168 5 68.7 17 CN 5 69
Weighted CN 5
3. 24-hour precipitation:
P 5 6.0 in
(Appendix D-3)
4. Time of concentration: Overland flow: L 5 100 ft s 5 12%
n 5 0.40 P2 5 2.75 in
(Table 11-3) (Appendix D-3)
t1 5 0.16 in (See pre-developed)
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Shallow concentrated flow: L 5 500 ft s 5 12% v 5 5.6 ft/s t2 5
(Table 10-9, unpaved)
d 500 5 5 89 sec 5 0.02 h v 5.6
Gutter flow: L 5 150 ft s 5 1.5%
v 5 2.5 ft/s t3 5
(Table 10-9, paved)
d 150 5 5 60 sec 5 0.02 h v 2.5
Pipe flow (estimated): L 5 650 ft s 5 4.0%
v 5 9 ft/s t4 5
(Appendix A-4, Chart 36)
d 650 5 5 72 sec 5 0.02 h v 9
tc 5 .16 1 .02 1 .02 1 .02 tc 5 0.22 h
5. Rainfall distribution: Type II
(Appendix D-4)
For hydrograph, see Figure 41. The detention basin is an open cut, grass-lined basin with bottom slope of 2 percent and side slopes of 3 horizontal to 1 vertical. After a few trials, the size is chosen as the smallest size that results in the required outflows and a maximum water level less than elevation 186 ft. The following table shows the basin size.
Elevation (ft) 182.5 183 184 185 186
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Area (s.f.) 0 3,600 13,000 14,200 15,500
Incremental Volume (c.f.) 0 900 8,300 13,600 14,850
Cumulative Volume (c.f.) 0 900 9,200 22,800 37,650
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© Cengage Learning 2014
168 CHAPTER 15 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL
FIGURE 41 Grading plan showing detention basin design for Problem 7.
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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 169
A grading plan showing the detention basin is shown in Figure 41. Finally, rout the inflow hydrograph through the detention basin. Results of the routing are shown below.
Peak Runoff Existing Conditions (cfs) 35.1
Storm 100
Peak Inflow (cfs) 50.1
Detention Basin Peak Maximum Outflow Water Level (cfs) (ft) 32.8 185.90
Peak outflow from the detention basin is less than peak runoff under existing conditions. Also, maximum water level is less than the maximum design water level. Inflow and outflow hydrographs are shown in Figure 42.
60 50
Inflow
Q (cfs)
40 30 20
Outflow
0 0.0
2.4
4.9
7.3
9.7
12.1 14.6 Time (hr)
17.0
19.4
21.9
24.3
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10
FIGURE 42 Inflow and outflow hydrographs for Problem 7.
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