Environmental Engineering Reference
In-Depth Information
W ater at r est
Stevin's law states: “The pressure at any point in a fluid at rest depends on the dis-
tance measured vertically to the free surface and the density of the fluid.” Stated as
a formula, this becomes
p = w × h
(14.1)
where
p = Pressure in pounds per square foot (lb/ft 2 or psf).
w = Density in pounds per cubic foot (lb/ft 3 ).
h = Vertical distance in feet (ft).
Example 14.2
Problem: What is the pressure at a point 15 ft below the surface of a reservoir?
Solution: To calculate this, we must know that the density ( w ) of water is 62.4 lb/ft 3 :
p = w × h = 62.4 lb/ft 3 × 15 ft = 936 lb/ft 2 (psf)
Water/wastewater operators generally measure pressure in pounds per square inch
rather than pounds per square foot; to convert, divide by 144 in. 2 /ft 2 (12 in. × 12 in.
= 144 in. 2 ):
936 lb/ft 2 ÷ 144 in. 2 /ft 2 = 6.5 lb/in. 2 (psi)
g auge p ressure
We defined head as the height a column of water would rise due to the pressure at its
base. We demonstrated that a perfect vacuum plus atmospheric pressure of 14.7 psi
would lift the water 34 ft. If we now open the top of the sealed tube to the atmosphere
and enclose the reservoir, then increase the pressure in the reservoir, the water will
again rise in the tube. Because atmospheric pressure is essentially universal, we
usually ignore the first 14.7 psi of actual pressure measurements and measure only
the difference between the water pressure and the atmospheric pressure; we call this
gauge pressure .
Example 14.3
Problem: Water in an open reservoir is subjected to the 14.7 psi of atmospheric pres-
sure, but subtracting this 14.7 psi leaves a gauge pressure of 0 psi. This shows that the
water would rise 0 feet above the reservoir surface. If the gauge pressure in a water
main is 100 psi, how far would the water rise in a tube connected to the main?
Solution:
100 psi × 2.31 ft/psi = 231 ft
Search WWH ::




Custom Search