Basic Units of Measurement and Conversions - Basic Mathematics for Water and Wastewater Operators - page 37

Solution :
24 in. ÷ 12 in./ft = 2 ft
Q = 0.785 × (2 ft) 2 × 210 ft/min = 659.4 ft 3 /min
659.4 ft 3 /min × 1440 min/day × 7.48 ga l /f t 3 = 7,102,529.2 gpd
■ Example 2.84
Problem : What would be the gpd flow rate for a 6-in. line flowing at 2.2 ft/sec?
Solution :
6 in. ÷ 12 in./ft = 0.5 ft
Q = 0.785 × (0.5 ft) 2 × 2.2 ft/sec = 0.4318 ft 3 /sec
0.4318 ft 3 /sec × 60 sec/min × 1440 min/day × 7.48 ga l /f t 3 = 279,060.24 gpd
■ Example 2.85
Problem : A 36-in. water main has just been installed. If the main is flushed at 2.8
ft/sec, how many gallons per minute of water should be flushed from the hydrant?
Solution :
36 in. ÷ 12 in./ft = 3 ft
Q = 0.785 × (3 ft) 2 × 2.8 ft/sec = 19.782 ft 3 /sec
19.782 ft 3 /sec × 60 sec/min × 7.78 ga l /f t 3 = 9234.24 gpm
■ Example 2.86
Problem : A 36-in. water main has just been installed. If the main is flowing at a
velocity of 2.2 ft/sec, how many million gallons per day will the pipe deliver?
36 in. ÷ 12 in./ft = 3 ft
Q = 0.785 × (3 ft) 2 × 2.2 ft/sec = 15.54 ft 3 /sec
15.54 ft 3 /sec × 60 sec/min × 1440 min/day
× 7.48 gal l /f t 3 × 1 MG/1,000,000 gal = 10.04 MGD
■ Example 2.87
Problem : A pipe has a diameter of 18 in. If the pipe is flowing full, and the water is
known to flow a distance of 850 yards in 5 minutes, what is the MGD flow rate for
the pipe?