Question

You are asked to design a piping system to fill an Olympic size swimming pool. The nominal dimensions of the pool are 164 ft long, 82 ft wide and 6 ft deep, and the filling is to take place in 6 hours. The 60 F water supply is contained in a reservoir 75 ft below the nominal elevation of the pool and contains 500 feet of 6-inch diameter cast iron pipe with 4 standard 90° elbows and 1 (fully open) gate valve. The roughness of the pipe is expected to increase as e = e0 + at, where e0 is the roughness value when the pipe is new and a = 0.000168 ft/yr. What size pump (in hp) should be chosen in order for the design to last 20 years?

Answer 1

Given that the size of the pool

length = 164 ft

breadth = 82 ft

Depth = 6 ft

Volume = 80688 ft ³.

Time = 6 hours = 21600 sec

Volume flow rate = Volume / time = 80688 / 21600 = 3.74 ft³/ s

Head = 75 ft

Length of pipe = 500 ft

Number of 90 degree elbows = 4 , K value = 0.75 [ 2*.75 = 1.5]

one Gate valve fully open , k value = 0.17

h_{major_loss} = f (l / d_h) (v² / 2 g)  

Finding the value of f

We have to find the Reynolds number

Re = D *V / kinematic viscosity

Kinematic viscosity at 60F = 1.210 x10^-5 ft²/s

D = 6 inches = 0.5 ft

Velocity = Volume flow rate / Area  

Area = * (0.5/2) ² = 0.19635 ft².

Velocity = 3.74 / 0.19635 = 19.05 ft/ s

Re = 0.5* 19.05 / 1.210 x 10^-5. = 787190 flow is turbulant

k = 0.25 mm for cast iron

d = diameter = 6 inches = 152.4 mm

k/d = 0.002

f value from moodys chart 0.006

Now Hf = f ( L/D ) ( V²/2g) = 0.006 * [ (500/0.5) ( 19.05² / (2* 32.17405)] = 0.006* 5639.677 = 33.84 ft

g= 32.17405 ft/s2

H minor = k * ( V²/2g)  

Number of 90 degree elbows = 4 , K value = 0.75 [ 2*.75 = 1.5]

one Gate valve fully open , k value = 0.17

k = 1.5+0.17 = 1.67

H minor = k * ( V²/2g) = 1.67 * ( 19.05² / (2* 32.17405) = 9.45 ft

Total head loss = 33.84 + 9.45 = 43.26 ft

Required head = altitude + head loss = 75 + 43.26 = 118.26 ft

Calculation by roughness

given e = e0 + a *t

e0= 0.010 inches = 0.0008333333 ftroughness of cast iron pipe

e= 0.0008333333 + (0.000168 *20) = 0.0042

f = 0.036 for a turbulant flow the value of f varies from 0.006 to 0.06

Now Hf = f ( L/D ) ( V²/2g) = 0.036 * [ (500/0.5) ( 19.05² / (2* 32.17405)] = 0.006* 5639.677 = 203.04 ft

Total head required = 75 + 203.04 + 9.45 = 287.5 ft.