Basic Units of Measurement and Conversions - Basic Mathematics for Water and Wastewater Operators - page 34

■ Example 2.71
Problem: A well pumps 410 gpm. How many million gallons per day will it pump?
Solution: Convert gpm to MGD.
410 gpm ÷ 700 = 0.59 MGD
■ Example 2.72
Problem : A channel 30 in. wide has water flowing to a depth of 1.5 ft. If the velocity
of the water is 2.5 ft/sec, what is the flow in the channel in ft 3 /sec? What is the flow
in gpm?
Solution :
30 in. ÷ 12 in./ft = 2.5 ft
Q = Width × Depth × Velocity = 2.5 ft × 1.5 ft × 2.5 ft/sec = 9.38 ft 3 /sec
9.38 ft 3 /sec × 60 sec/min × 7.48 ga l /f t 3 = 4210 gpm
■ Example 2.73
Problem : The flow through a 24-in. pipe is moving at a velocity of 5.5 ft/sec. What
is the flow rate in gallons per minute?
Solution :
24 in. ÷ 12 in./ft = 2 ft
Q = 0.785 × ( D ) 2 (ft) × Velocity (ft/sec)
Q = 0.785 × (2 ft) 2 × 5.5 ft/sec = 17.27 ft 3 /sec
17.27 f t 3 /sec × 60 sec/min × 7.48 ga l /f t 3 = 7750.78 gpm
■ Example 2.74
Problem : A channel is 48 in. wide and has water flowing to a depth of 2 ft. If the veloc-
ity of the water is 2.6 ft/sec, what is the flow in the channel in cubic feet per second?
Solution :
48 in. ÷ 12 in./ft = 4 ft
Q = 4 ft × 2 ft × 2.6 ft/sec = 20.8 ft 3 /sec
■ Example 2.75
Problem : A channel 3 ft wide has water flowing to a depth of 2.7 ft. If the velocity
through the channel is 122 ft/min, what is the flow rate in cubic feet per minute? In
million gallons per day?