Pumping Calculations - Basic Mathematics for Water and Wastewater Operators - page 216

■ Example 14.60
Problem: During a 30-min pumping test, 3760 gal are pumped into a tank, which
has a diameter of 12 ft. The water level before the pumping test was 2 ft. What is the
gpm pumping rate?

Solution:

3760 gal ÷ 30 min = 125.3 gpm

■ Example 14.61
Problem: A 60-ft-diameter tanks has water to a depth of 8 ft. The inlet valve is
closed and a 3-hr pumping test is begun. If the water level in the tank at the end of
the test is 2.5 ft, what is the gpm pumping rate?

Solution:

2 3

Volume = 0.785 × (60ft) ×− = ( 825 . ft) 15,543 ft

3 3

15,543 ft × 748 . gal/ft = 116 261 64 , . gal

116 , 261 64 . al

= 645.9 gpm

180 min

■ Example 14.62
Problem: A tank has a length of 12 ft, a depth of 12 ft, and a width of 12 ft, and it
has water to a depth of 8 ft. If the tank can be emptied in 1 hour, 45 minutes, what is
the gpm pumping rate?

Solution:

2 3

Volume = (12ft) × (8 ft) × 7.48 gal/ft = 8616.9 g al

8616.9 gal
105 min

= 82.07 gpm

■ Example 14.63
Problem: During a pumping test, water was pumped into an empty tank 10 ft by 10
ft by 8 ft deep. The tank completely filled with water in 12 minutes. Calculate the
gpm pumping rate.

Solution:

2 3

Volume = (10ft) (8

5984 gal
12 min

× ft) × 7.48 gal/ft = 5984 gal

= 498.7 gpm