Pumping Calculations - Basic Mathematics for Water and Wastewater Operators - page 199

Pumping rate (gpm) = Gallons ÷ Minutes (14.4)

Pumping rate (gph) = Gallons ÷ Hours (14.5)

■ Example 14.6
Problem: The meter on the discharge side of the pump reads in hundreds of gallons.
If the meter shows a reading of 110 at 2:00 p.m. and 320 at 2:30 p.m., what is the
pumping rate expressed in gallons per minute (gpm)?

Solution : The problem asks for pumping rate in gallons per minute, so we use
Equation 14.4:

Pumping rate (gpm) = Gallons ÷ Minutes

To solve this problem, we must first find the total gallons pumped (determined from
the meter readings):

32,000 gal - 11,000 gal = 21,000 gal

The volume was pumped between 2:00 p.m. and 2:30 p.m., for a total of 30 minutes.
From this information, calculate the gallons-per-minute pumping rate:

Pumping rate = 21,000 gal ÷ 30 min = 700 gpm

■ Example 14.7
Problem: During a 15-minute pumping test, 16,400 gal were pumped into an empty
rectangular tank. What is the pumping rate in gallons per minute?

Solution : The problem asks for the pumping rate in gallons per minute, so again we
use Equation 14.4:

Pumping rate = Gallons ÷ Minutes = 16,400 gal ÷ 15 min = 1093 gpm

■ Example 14.8
Problem: A tank 50 ft in diameter is filled with water to a depth of 4 ft. To conduct
a pumping test, the outlet valve to the tank is closed and the pump is allowed to dis-
charge into the tank. After 80 minutes, the water level is 5.5 ft. What is the pumping
rate in gallons per minute?

Solution: We must first determine the volume pumped in cubic feet:

Volume pumped = Area of circle × Depth = 0.785 × (50 ft) 2 × 1.5 ft = 2944 ft 3

Now convert the cubic-foot volume to gallons:

2944 ft 3 × 7.48 gal l /f ft3 3 = 22,021 gal