What is the horsepower required for a pump that moves 2500 gpm against a head of 475 feet, operating at 58% efficiency?

What is the horsepower required for a pump that moves 2500 gpm against a head of 475 feet, operating at 58% efficiency?

• A. 420 HP
• B. 600 HP
• C. 480 HP
• D. 516.8 HP

Answer: (D)

The question tests converting flow and head into horsepower and then correcting for efficiency. Start by finding the hydraulic power needed to move the flow against the head, then account for the pump’s efficiency to get the required input horsepower. Compute the hydraulic horsepower first: multiply the flow by the head and divide by 3960. 2500 gpm × 475 ft = 1,187,500; 1,187,500 / 3960 ≈ 299.9 HP. Since the pump is only 58% efficient, the input horsepower must be higher: 299.9 / 0.58 ≈ 517 HP. Equivalently, 1,187,500 / (3960 × 0.58) ≈ 516.9 HP. So the required horsepower is about 516.8–517 HP.

The question tests converting flow and head into horsepower and then correcting for efficiency. Start by finding the hydraulic power needed to move the flow against the head, then account for the pump’s efficiency to get the required input horsepower.

Compute the hydraulic horsepower first: multiply the flow by the head and divide by 3960.

2500 gpm × 475 ft = 1,187,500; 1,187,500 / 3960 ≈ 299.9 HP.

Since the pump is only 58% efficient, the input horsepower must be higher: 299.9 / 0.58 ≈ 517 HP. Equivalently, 1,187,500 / (3960 × 0.58) ≈ 516.9 HP.

So the required horsepower is about 516.8–517 HP.