• Author / Uploaded
• Mhorien Macatangay

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Four Reservoir Problems

PROBLEM 1. Reservoir M is at elevation 100 m above datum, supplies water to a 650 mm pipe which takes to a point at elevation 35 m, the pipe being 500 m long. Here it split into 3 pieces, 250 mm, 350 mm, and 200 mm diameter. The 250 mm pipe runs 300 m and releases to reservoir N at elevation 80 m, the 350 mm runs 450 m to reservoir O at elevation 55 m, and the 200 mm pipe runs 950 m and dispenses to reservoir P at elevation 40 m. f = 0.02.

1. Compute the rate of flow towards reservoir N. 2. Compute the rate of flow towards reservoir O. 3. Compute the rate of flow towards reservoir P.

SOLUTION: 1. Rate of flow towards reservoir N. hf =

y=

0.0826 fL Q 2 D 55

0.0826 ( 0.02 ) ( 500 ) Q12 ( 0.65 )5

Q1=0.37 √ y (eq .1) 0.0826 ( 0.02 )( 300 ) Q22 20− y= ( 0.25 )5 Q2=0.0444 √10− y (eq .2)

0.0826 ( 0.02 ) ( 450 ) Q32 45− y= ( 0.35 )5 Q3=0.0841 √ 45− y (eq .3)

60− y=

0.0826 ( 0.02 )( 950 ) Q 42 ( 0.20 )5

Q4 =0.0143 √ 60− y ( eq .4)

Q1=Q2 +Q3 +Q4

0.37 √ y=0.0444 √ 10− y +0.0841 √ 45− y +0.0143 √ 60− y y=4.709

Q2=0.0444 √10− y Q2=0.0444 √10−4.709 Q2=102 m3 /s

2. Rate of flow towards reservoir O. Q3=0.0841 √ 45− y Q3=0.0841 √ 45−4.709 Q3=0.534 m3 /s

3. Rate of flow towards reservoir P. Q4 =0.0143 √ 60− y Q4 =0.0143 √ 60−4.709 Q4 =0.106 m3 /s

PROBLEM 2 Reservoirs A and B is connected by a 200 mmØ pipe which is 950 m long. Reservoir B is 9 m below A. Another reservoirs C and D is also connected by a 320 mmØ pipe, 2200 m long with reservoir D 15 m below of C and reservoir C is also 5 m below of A. In order to enlarge the quantity of water entering reservoir D, the two pipelines AB and CD are connected by pipeline EF which is 1800 m long which splits from the 200 mmØ at point E 450 m from A and connects with the 320 mmØ pipe at F 900 meters from reservoir D. 20 liters per second of water flows through pipe EF, assume friction factor f to be 0.02 for all pipes. Consider only friction losses. 1. Determine the quantity of water flowing towards B. 2. Determine the quantity of water flowing towards D. 3. Determine also the diameter of the pipe EF.

SOLUTION: 1. Quantity of water flowing towards B. From A to B: hf 1 +hf 2 =9 m

(

fL 1 V 12 fL2 V 22 + =9 D1 2 g D2 2 g

)(

)

Q 1= A 1 V 1 Q 1= V 1=

π 2 D V 4 1 1 Q1 π 2 D 4 1

2

V1 =

Q 12 2

π D 14 4

()

1.621Q 12 V1 = D 14 2

2

V2 =

(

1.621Q 22 D 24

fL 1 1.621 Q 1 2 g D 12

)( +

fL 2 1.621 Q22 2 g D 22

)

=9

Q 1−Q 2=0.020 Q 2=Q 1−0.020 0.02(450)(1.621)Q 12 5 2 ( 9.81 )( 0.20 )

+ 0.02 ( 500 ) (1.621)¿ Q 22

¿ =9 5 ( 2 9.81 ) ( 0.20 )

2323.88 Q 12+2582.09 Q 22=9 2323.88 Q 12+2582.09 Q 22=9 0.9 Q 12+ Q 22 =0.00349 0.9 Q 12+ ( Q 1−0.020 )2 =0.00349 1.9 Q 12−0.04 Q1−0.00309 Q12−0.02 Q1−0.00162 Q1=0.052 m3 /s Q2=0.052−0.020 Q2=0.032 m 3 /s Q2=32 Liters /s ec .(flow towards B)

2. Quantity of water flowing towards D. From C to D: hf 3 +hf 4=15 m

Q4 =Q3 +0.020 0.02(1300)(1.621) Q32 0.02 ( 900 ) (1.621)Q42 + =15 5 5 2 ( 9.81 )( 0.32 ) 2 ( 9.81 )( 0.32 ) 640.24 Q32 +443.25 Q32=15

1.44 Q 32 +Q 42=0.0234 1.44 Q 32 + ( Q 3 +0.020 )2 =0.02343 2.44 Q32 +0.04 Q3 −0.02303=0 Q32 +0.016 Q3−0.00942=0 Q3=0.089m 3 / s Q4 =0.089−0.020 Q4 =0.069 m3 /s Q4 =69 Liters/ s ec .(flow towards B)

3. Diameter of pipeline EF: For line A E F D: hf 1 +hf 5 +hf 4=20

0.02( 450)(1.621)Q 12 hf 1= 5 2 ( 9.81 ) ( 0.20 ) 0.02( 450)(1.621) ( 0.052 )2 hf 1= 2 ( 9.81 )( 0.20 )5 hf 1=6.28 m

0.02 ( 900 ) (1.621)Q42 hf 4= 5 2 ( 9.81 )( 0.32 )

0.02 ( 900 )( 1.621 ) ( 0.069 )2 hf 4= 5 2 ( 9.81 ) ( 0.32 ) hf 4=2.11 m hf 1 +hf 5 +hf 4=20 6.28+hf 5+2.11=20 hf 5 =15.83 m 15.83=

0.02 ( 1800 ) (1.621 ) ( 0.020 )2 5 2 ( 9.81 ) ( D5 )

D5=0.150 m D5=150 mm

PROBLEM 3 Reservoir W, X, Y, and Z have elevations of 90 m, 110 m, 130 m and 110m, respectively. The rate of flow from P to reservoir W is 485 liters/sec. Use the properties as shown:

PIPE NO.

1

2

3

4

5

LENGTH(m)

550

550

750

250

250

DIAMETER(mm)

500

400

400

400

400

0.022

0.028

0.028

0.028

0.028

f

1. Compute the rate of flow away from reservoir X.

2. Compute the rate of flow from reservoir Y. 3. Compute the rate of flow toward reservoir Z.

SOLUTION: 1. Rate of flow away from B. Q 1=600 liters/ sec hf =

0.0826 fL Q 2 D5

0.0826 f 1 L1 Q 12 hf 1= 5 D1 0.0826 ( 0.022 ) ( 550 ) Q12 hf 1= ( 0.50 )5 hf 1=31.9827Q12 Q 1=0.6 m/ sec hf 1=31.9827 ( 0.6 )2 hf 1=11.5138