Environmental Engineering Reference
In-Depth Information
related to the velocity of the liquid and the cross-sectional area of the
liquid in the channel:
q
=
a
×
v
(2.12)
where:
q
= Flow, or discharge, in cubic feet per second (cfs).
a
= Cross-sectional area of the pipe or channel (ft
2
).
v
= water velocity in feet per second (fps).
Example 2.10
Problem:
A channel is 6 ft wide and the water depth is 3 ft. The velocity
in the channel is 4 fps. What is the discharge or flow rate in cubic feet
per second?
Solution:
Flow = 6 ft × 3 ft × 4 ft/s = 72 cfs
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 waste-
water treatment plants are large and are often referred to in million gal-
lons per day (MGD). The discharge or flow rate can be converted from
cubic feet per second (cfs) to other units such as gallons per minute
(gpm) or million gallons per day (MGD) by using appropriate conversion
factors.
Example 2.11
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 (Equation 2.13), we must
determine the area (
a
) of the pipe. The formula for the area of a circle is:
d
2
2
Area ()
a
=× =×
π
π
r
(2.13)
4
where:
π = the constant value 3.14159 or simply 3.14.
d
= diameter of the circle (ft).
r
= radius of the circle (ft).
Therefore, the area of the pipe is:
d
=× =
2
(1 ft)
2
2
a
π
3.14
×
=
0.785 ft
4
4
