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A pump is to deliver 80 GPM of water with a density of 61.83 lb/ft³ at a discharge pressure of 150 psig. The suction pressure indicates a 2 in Hg vacuum. The diameter of the suction and discharge are 5 inches and 4 inches, respectively. The pump has an efficiency of 70%, and the motor efficiency is 80%. Determine the following:

A. Water horsepower of the pump
B. Brake horsepower (BHP) of the pump motor
C. Power input to drive the motor

Answer :

A. The water horsepower of the pump is 3.03 HP.
B. The BHP of the pump motor is 3.79 HP.
C. The power input to drive the motor is also 3.79 HP.

To determine the water horsepower of the pump, we can use the following formula:

Water horsepower (WHP) = (Flow rate in GPM * Total head in ft) / (3,960 * Pump efficiency)

Given that the flow rate is 80 GPM, the total head is the difference between the discharge pressure and the suction pressure. We convert the suction pressure from in Hg to ft using the conversion factor of 1 in Hg = 1.13 ft:

Total head = (150 psig - (-2 in Hg * 1.13 ft/in Hg)) * (144 in2/ft2)

The pump efficiency is given as 70%.

To calculate the brake horsepower (BHP) of the pump motor, we use the following formula:

BHP = WHP / Motor efficiency

Given that the motor efficiency is 80%, we can calculate BHP.

The power input to drive the motor is the same as BHP, as the efficiency of the motor is already taken into account.

Now, let's calculate the values step-by-step.

Total head = (150 psig - (-2 in Hg * 1.13 ft/in Hg)) * (144 in2/ft2)
= (150 psig + 2.26 ft) * (144 in2/ft2)
= 152.26 ft

WHP = (80 GPM * 152.26 ft) / (3,960 * 0.7)
= 3.03 HP

BHP = 3.03 HP / 0.8
= 3.79 HP

Power input to drive the motor = 3.79 HP

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