Geoscience Reference
In-Depth Information
EXAMPLE 14. 33
Problem: A tank 50 ft in diameter is filled with water to depth of 4 ft. To conduct a pumping test,
the outlet valve to the tank is closed, and the pump is allowed to discharge into the tank. After 80
minutes, the water level is 5.5 ft. What is the pumping rate in gallons per minute?
Solution: We must first determine the volume pumped in cubic feet:
Volume pumped = Area of circle × depth = 0.785 × 50 ft × 50 ft × 1.5 ft = 2944 ft 3 (rounded)
Now convert the cubic-foot volume to gallons:
2944 ft 3 × 7.48 gal l /f ft3 3 = 22,021 gal (rounded)
The pumping test was conducted over a period of 80 minutes. Using Equation 14.20, calculate the
pumping rate in gallons per minute:
22,021 gal
80 min
Pumpingrate
=
=
275.3 gpm (ro
unded)
14.2.3 C alCulating h ead l oss
Pump head measurements are used to determine the amount of energy a pump can or must impart
to the water; they are measured in feet. One of the principle calculations used in pumping problems
is determining head loss . The following formula is used to calculate head loss:
H f = K ( V 2 /2 g )
(14.22)
where
H f = Friction head.
K = Friction coefficient.
V = Velocity in pipe.
g = Gravity (32.17 ft/sec/sec).
14.2.4 C alCulating h ead
For centrifugal pumps and positive displacement pumps, several other important formulae are used
in determining head . In centrifugal pump calculations, the conversion of the discharge pressure to
discharge head is the norm. Positive displacement pump calculations often leave given pressures in
psi. In the following formulas, the specific weight of a liquid is expressed in pounds per cubic foot.
For water at 68°F, the specific weight is 62.4 lb/ft 3 . A water column 2.31 ft high exerts a pressure of
1 psi on 64°F water.
Use the following formulas to convert discharge pressure in psi to head in feet:
• Centrifugal pumps
Pressure (psig)
×
ity
2.31
Head (ft)
=
(14.23)
Specificgrav
• Positive displacement pumps
Pressure (psig) 144
Specificweigh
×
Head (ft)
=
(14.24)
t
 
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