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 f^3/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

Given data:

Volumetric flow rate Q=2.5 ft3/s = 1122.08 gpm

Diameter of pipe d = 8 in=0.667 ft

Efficiency of pump = 85%

Head loss from section(1) to section (2) hf = 61V^2/2g

Gravity = 9.81m/s^2= 32.185 ft/s^2

Cross sectional area of pipe A = Pid2/4 = 22*0.667^2/7*4

A=0.35 ft^2

Continuity equation

Q=AV

2.5=0.35 V

V=7.14 ft/s

Head loss = 61V^2/2g = 61*7.14^2/2*32.185 =48.3 ft

Net head = 50+48.3 = 98.3 ft

Power require to drive pump = QH/3960*

P = 1122.08*98.3/3960*0.85 =32.8HP