Water Hydraulics - Mathematics and Statistics for the Environment - page 594

where
p = Pressure at a cross-section.
A = Pipe cross-sectional area (cm 2 , in. 2 ).
V = Mean velocity.

■ EXAMPLE 22.8
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

22.2.5.6 Head and Pressure
If the head is known, the equivalent pressure can be calculated by

Head (ft)
2.31 ft/psi

Pressure (psi) = (22.11)

■ EXAMPLE 22.9
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

22.3 FLOW AND DISCHARGE RATE: 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 in a waterworks and distribution system
and in a wastewater treatment plant and collection system is nearly always in motion. 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 (22.12)

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, ft/sec).

■ EXAMPLE 22.10
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/sec = 72 cfs