A pump delivers 175 gallons per minute against a pressure of 55 psi; the suction pressure is 15 psi. If the pump is 80% efficient, what is the horsepower needed?

A pump delivers 175 gallons per minute against a pressure of 55 psi; the suction pressure is 15 psi. If the pump is 80% efficient, what is the horsepower needed?

• A. 3.3 HP
• B. 5.1 HP
• C. 7.2 HP
• D. 9.4 HP

Answer: (B)

The main idea is to determine the energy the pump must add to move water from suction to discharge, using the head the pump must develop and then account for efficiency. First find the head the pump must overcome: the discharge pressure minus the suction pressure = 55 psi − 15 psi = 40 psi. Convert this pressure difference to feet of head: 40 psi × 2.31 ft/psi = 92.4 ft. Calculate the hydraulic power (water horsepower) with head in feet and flow in GPM: WHP = (GPM × head_ft) / 3960 = (175 × 92.4) / 3960 ≈ 4.08 HP. Account for efficiency to get the input (shaft) horsepower: HP_input = WHP / efficiency = 4.08 / 0.80 ≈ 5.10 HP. So about 5.1 HP is required.

The main idea is to determine the energy the pump must add to move water from suction to discharge, using the head the pump must develop and then account for efficiency.

First find the head the pump must overcome: the discharge pressure minus the suction pressure = 55 psi − 15 psi = 40 psi. Convert this pressure difference to feet of head: 40 psi × 2.31 ft/psi = 92.4 ft.

Calculate the hydraulic power (water horsepower) with head in feet and flow in GPM: WHP = (GPM × head_ft) / 3960 = (175 × 92.4) / 3960 ≈ 4.08 HP.

Account for efficiency to get the input (shaft) horsepower: HP_input = WHP / efficiency = 4.08 / 0.80 ≈ 5.10 HP.

So about 5.1 HP is required.