Basic Units of Measurement and Conversions - Basic Mathematics for Water and Wastewater Operators - page 38

Solution :

18 in. ÷ 12 in./ft = 1.5 ft

850 yd × 3 ft/yd = 2550 ft

Velocity = 2550 ft/5 min = 510 ft/min

Q = 0.785 × (1.5 ft) 2 × 510 ft/min = 900.7875 ft 3 /min

90 0.7875 ft 3 /min × 1440 min/day × 7.48 ga l /f t 3 × 1 MG/1,000,000 gal = 9.70 MGD

■ Example 2.88
Problem : A water crew is lushing hydrants on a 12-in.-diameter main. The pitot-
tube gauge indicates that 580 gpm are being lushed from the hydrant. What is the
lushing velocity (in ft/min) through the pipe?

Solution :

12 in. ÷ 12 in./ft = 1 ft

A = 0.785 × (1) 2 = 0.785 ft 2

Q = 580 gpm × 7.48 ga l /f t 3 = 4338.4 ft 3 /min

3 2

4338.4 ft /min = 0.785 ft /min × Velocity

3

4338.4 ft /min

0.785 ft /min

= Velocity

2

5226.62 ft/m in = Velocity

■ Example 2.89
Problem : Geologic studies show that water in an aquifer moves 22 ft in 55 days.
What is the average velocity of the water in feet per day?

Solution :

Velocity = 22 ft ÷ 55 days = 0.4 ft/day

■ Example 2.90
Problem : If the water in a water table aquifer moves 2.5 ft/day, how far will the water
travel in 17 days?

Solution :

2.5 ft/day × 17 days = 42.5 ft