What is the pump efficiency of a 5 Hp pump used to pump 3 MGD of wastewater 8 feet?

To determine the pump efficiency of a 5 horsepower (Hp) pump being used to pump 3 million gallons per day (MGD) of wastewater to a height of 8 feet, it is important to understand how efficiency is calculated in the context of pumping systems.

First, horsepowers must be converted into a usable output form considering both the flow rate and the head (height). The hydraulic power required to lift water can be calculated using the formula:

Hydraulic Power (HP) = (Flow Rate (GPM) x Head (ft) x 8.33) / 33,000

1. Convert 3 MGD to gallons per minute (GPM): 3 MGD = 3,000,000 gallons/day ÷ 1,440 minutes/day = approximately 2,083 GPM.

2. Next, calculate the hydraulic power required: Hydraulic Power = (2,083 GPM x 8 ft x 8.33) / 33,000.

This calculation results in a specific value for the hydraulic power, which can then be compared with the actual power being supplied by the pump (5 Hp).

3. To find efficiency, you use the formula: Efficiency (%) = (