Environmental Engineering Reference
In-Depth Information
Example 4.8
Problem: The pressure gauge on the discharge line from the influent pump reads
72.3 psi. What is the equivalent head in feet?
Solution:
Head = 72.3 × 2.31 ft/psi = 167 ft
Head and Pressure
If the head is known, the equivalent pressure can be calculated as follows:
Head (ft)
2.31 ft/psi
Pressure (psi)
=
(4.10)
Example 4.9
Problem: A tank is 22 ft deep. What is the pressure in psi at the bottom of the tank
when it is filled with water?
Solution:
22 ft
2.31 ft/psi
Pressure
=
=
9.52 psi (rounde
d)
Flow and Discharge Rates: Water in Motion
The study of fluid flow is much more complicated than that of fluids at rest, but it
is important to have an understanding of these principles because the water used in
hydropower applications is nearly always in motion (e.g., the water is used to propel
turbine blades). Discharge (or flow) is the quantity of water passing a given point in
a pipe or channel during a given period. Stated another way for open channels, the
flow rate through an open channel is directly related to the velocity of the liquid and
the cross-sectional area of the liquid in the channel:
Q = A × V
(4.11)
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 or ft/s).
Example 4.10
Problem: A channel is 6 ft wide, and the water depth is 3 ft. The velocity in the chan-
nel 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
 
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