Environmental Engineering Reference
In-Depth Information
Discharge or flow can be recorded as gal/day (gpd), gal/min (gpm), or cubic
feet per second (cfs). Flows treated by many hydropower systems are large and are
often expressed 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.
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Example 4.11
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
Area (
A
) = π × (
D
2
/4) = π ×
r
2
(4.12)
where
π = 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
A
= π × (
D
2
/4) = 3.14 × (1 ft
2
/4) = 0.785 ft
2
(a) Now we can determine the discharge in cfs:
Q
=
V
×
A
= 10 fps × 0.785 ft
2
= 7.85 ft
3
/s (cfs)
(b) We need to know that 1 cfs is 449 gpm, so 7.85 cfs × 449 gpm/cfs = 3525 gpm
(rounded).
(c) Finally, 1 million gallons per day is 1.55 cfs, so
7.85 cfs ÷ 1.55 cfs/MGD = 5.06 MGD
Area and Velocity
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
(4.13)
■
Example 4.12
Problem:
A pipe 12 inches 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?
