Water Hydraulics - Mathematics and Statistics for the Environment - page 595

Discharge or flow can be recorded as gal/day (gpd), gal/min (gpm), or cubic feet per second (cfs).
Flows treated by many waterworks or wastewater treatment plants are large and are often referred
to in million gallons per day (MGD). The discharge or flow rate can be converted from cfs to other
units such as gpm or MGD by using appropriate conversion factors.

■ EXAMPLE 22.11
Problem: A 12-in.-diameter pipe has water flowing through it at 10 fps. What is the discharge in (a)
cfs, (b) gpm, and (c) MGD?

Solution: Before we can use the basic formula, we must determine the area ( A ) of the pipe. The
formula for the area of a circle is

2

D
=× A π (22.13)

4

where
π = Constant value 3.14159.
D = Diameter of the circle in feet.

Thus, the area of the pipe is

2 2

D
=× =× =

() 1ft

0.785 ft 2

A π 314 .

4 4

Now, we can determine the discharge in cubic feet per second for part (a):

Q = V × A = 10 ft/sec × 0.785 ft 2 = 7.85 ft 3 /sec (cfs)

For part (b), we need to know that 1 cfs is equal to 449 gpm, so 7.85 cfs × 449 gpm/cfs = 3525 gpm.
Finally, for part (c), 1 MGD is equal to 1.55 cfs, so

7.85 cfs
1.55 cfs/MGD

= 5.06 MGD

22.3.1 a rea and v eloCity
The law of continuity states that the discharge at each point in a pipe or channel is the same as the
discharge at any other point (if water does not leave or enter the pipe or channel). That is, under the
assumption of steady-state flow, the flow that enters the pipe or channel is the same flow that exits
the pipe or channel. In equation form, this becomes

Q 1 = Q 2 or A 1 V 1 = A 2 V 2 (22.14)

■ EXAMPLE 22.12
Problem: A pipe 12 in. in diameter is connected to a 6-in.-diameter pipe. The velocity of the water
in the 12-in. pipe is 3 fps. What is the velocity in the 6-in. pipe?

Solution: Using the equation A 1 V 1 = A 2 V 2 , we need to determine the area of each pipe: