Basic Units of Measurement and Conversions - Basic Mathematics for Water and Wastewater Operators - page 35

Solution :

Q = 3 ft × 2.7 ft × 122 ft/sec = 988.2 ft 3 /sec

Using the flow chart in Figure 2.1 to convert:

Q = 10.64 MGD

■ Example 2.76
Problem: A channel is 3 ft wide and has water flowing at a velocity of 1.3 ft/sec. If
the flow through the channel is 8.0 ft 3 /sec, what is the depth of the water in the chan-
nel in feet?

Solution:

3

8.0 ft /sec =× × 3ftDepth 1.3 ft/sec

3

8.0 ft /se c

= Depth

3ft × 1.3 ft/sec

2.1 ft = Depth

■ Example 2.77
Problem : The flow through a 2-ft-diameter pipeline is moving at a velocity of 3.3 ft/
sec. What is the flow rate in cubic feet per second?

Solution:

Q = 0.785 × (2 ft) 2 × 3.3 ft/sec = 0.785 × 4 ft 2 × 3.3 ft/sec = 10.36 ft 3 /sec

■ Example 2.78
Problem : The flow through a pipe is 0.6 ft 3 /sec. If the velocity of the flow is 3.5 ft/sec,
and the pipe is flowing full, what is the diameter of the pipe in inches?

Solution :

3 2

0.6 ft /sec = 0 785 . × D × 3.5 ft/sec

3

0.6 ft /sec

2

= D

0 .785 3.5 ft/sec

0.2181 ft

×

2

= D

0.2181 ft
0

= D

.47 ft = D

0.47 ft × 12 in./ft = 5.64 in.