Science of Water : Concepts and Applications

Water Hydraulics 47
Another relationship is also important:
1 gal H 2 O = 8.34 lb
At STP, 1 ft 3 of water contains 7.48 gal. With these two relationships, we can determine the
weight of 1 gal of water. This is accomplished by
Thus,
62.4 lb
Wt. of gallon of water 8.34
lb/gal
7.48 gal
1 gal H 2 O = 8.34 lb
√ Important Point: Further, this information allows cubic feet to be converted into gallons by
simply multiplying the number of cubic feet by 7.48 gal/ft 3 .
Example 3.1
Problem:
Find the number of gallons in a reservoir that has a volume of 855.5 ft3 .
Solution:
855.5 ft 3 × 7.48 gal/ft 3 = 6399 gal (rounded)
√ Important Point: The term head is used to designate water pressure in terms of the height of a
column of water in feet. For example, a 10-ft column of water exerts 4.3 psi. This can be called
4.3-psi pressure or 10 ft of head.
STEVIN’S LAW
Stevin’s law deals with water at rest. Specifi cally, it states: “The pressure at any point in a fl uid at
rest depends on the distance measured vertically to the free surface and the density of the fl uid.”
Stated as a formula, this becomes
where
p = Pressure in pounds per square foot, psf
w = Density in pounds per cubic foot, lb/ft 3
h = Vertical distance in feet
Example 3.2
p wh
Problem:
What is the pressure at a point 18 ft below the surface of a reservoir?
Solution:
√ Note: To calculate this, we must know that the density of the water, w, is 62.4 lb/ft 3 .
(3.1)