Question

Water is drawn from a reservoir and pumped an equivalent length of 2 miles through a horizontal, circular duct of 10-in. i.d. (wall roughness k = 0.01 ft). At the end of this duct, the flow is divided into a 4- and a 3-in. schedule 40 steel pipe. The 4-in. line has an equivalent length of 200 ft. and discharges to the atmosphere at a point 50 ft. above the surface of the water in the reservoir. This flow must be maintained at 1000 gal/min. The 3-in. line discharges to the atmosphere at a point 700 ft. from the junction at the reservoir level. The pump you are considering has an efficiency of 70. Kinetic energy effects can be ignored. Determine how much horsepower is needed for the system.

Answer 1

Since, Flow is not given in 3 inch line, Considering flow in 3 inch pipe = flow in 4 inch line * (cross sectional area of 3 inch pipe)/(Cross sectional area of 4 inch pipe)

Therefore, Flow in 3 inch pipe = 1000*(9/16) = 562.5 gal/min

Therefore, flow in 10 inch pipe = flow inch 3 inch pipe + flow in 4 inch pipe = 1562.59 gal/min

Considering wall roughness to be same for all the pipe, c = 0.01 ft

Reynold number, Re = P*V*D/u/A = P*V*D/u/(3.14/4*D^2)  

for 10 inch pipe, Re₁₀ = 1000*1569.59*3.78*10^-3/60*10*2.54/100/(8.9*10^-4)/(3.14/4*(10*2.54/100)^2) = 557228

for 4 inch pipe, Re4 = 1000*1000*3.78*10^-3/60*4*2.54/100/(8.9*10^-4)/(3.14/4*(4*2.54/100)^2) = 887538

for 3 inch pipe, Re3 = 1000*569.53*3.78*10^-3/60*3*2.54/100/(8.9*10^-4)/(3.14/4*(4*3.54/100)^2) = 195175

Since, Re in all 3 pipes > 3000, therefore, flow in all the pipes are turbulent

head loss in pipe using Darcy friction factor formula given below

For 10 inch pipe, relative roughness, c/d = .01*12/10 = .012

Using Moody's chart, Darcy friction factor, F₁₀ = 0.038

L = 2 miles = 3218 meter

v = Volumetric flowrate/area = 1569.69*3.28*10^-3/60/(3.14/4*(10*2.54/100)^2) =1.69 m/s

g = 10 m2/s

Therefore friction Head loss in 10 inch Pipe, HF₁₀ = .038*3218*1.69^2/2/(10*2.54/100)/10 =68.75m

Power required = Density*g*Head*flowrate

Therefore Power required in 10 inch pipe, P₁₀ = 68.75*1569.69*3.28*10^-3/60*1000*10 = 58900 W = 58.9 KW

For 4 inch pipe, relative roughness, c/d = .01*12/4 = .03

Using Moody's chart, Darcy friction factor, F4 = 0.058

L = 200 ft = 60.96 meter

v = Volumetric flowrate/area = 1000*3.28*10^-3/60/(3.14/4*(4*2.54/100)^2) = 6.74 m/s

Therefore friction Head loss in 4 inch Pipe, HF₄ = 0.058*60.96*6.74^2/2/(4*2.54/100)/10 =79 m

4 inch pipe outlet is at 50 Ft height, therefore Potential head, PE₄ = 50 ft = 15.24 meter

Therefore Power required in 4 inch pipe, P4 = (79+15.24)*1000*3.28*10^-3/60*1000*10 = 51517 W = 51.52 KW

For 3 inch pipe, relative roughness, c/d = .01*12/3 = .04

Using Moody's chart, Darcy friction factor, F3 = 0.07

L = 700 ft = 213.96 meter

v = Volumetric flowrate/area = 569*3.28*10^-3/60/(3.14/4*(3*2.54/100)^2) = 6.82 m/s

Therefore friction Head loss in 3 inch Pipe, HF₃ = 0.07*213.96*6.82^2/2/(3*2.54/100)/10 =457.1 m

Therefore Power required in 3 inch pipe, P₃ = (457.1)*1000*3.28*10^-3/60*1000*10 = 249881 W = 249.8KW

Therefore, Total power required by fluid = P₁₀+P₄+P₃ = 58.9+51.52+249.8 = 360.22 KW

Efficiency of Pump = 70%

therefore, power required by pump = Power required by fluid/effciency = 360.22/.7 =514.6 KW = 514.6*1.3 horsepower = 669 HP

Power required by pump = 669 HP