Water is to be moved from one large reservoir to another at

Water is to be moved from one large reservoir to another at a higher elevation as shown in the below figure. The loss in available energy from sections (1) to (2) is 61 V^2/2g feet, where V is the average velocity of the water in the 8-in. inside-diameter piping involved. The volumetric flow rate is 2.5 P/s. Determine the amount of shaft power required in horsepower. Draw the hydraulic grade line and energy grade line between points (1) and (2) for this system. Determine the power required to drive the pump in horsepower if the pumps efficiency is 85%.

Solution

Cross-section area of pipe, A = 3.14/4*(8/12)^2 = 0.3489 ft^2

Velocity V = Q / A

= 2.5 / 0.3489

V = 7.166 ft/s

Head loss = 61*V² / (2g)

= 61*7.166² / (2*32.2)

= 48.64 ft

Shaft power = rho*Q*H

= 62.4*2.5*(50 + 48.64)......where water density has been taken as 62.4 lb/ft^3

= 15387.84 ft-lb/s

= 15387.84*60 ft-lb/min

= 923270.4 ft-lb/min

= 28 hp

Power required to drive the pump = 28 / 0.85 = 32.94 hp