Water Hydraulics - Mathematics and Statistics for the Environment - page 588

Another relationship is also important:
1 gal H 2 O = 8.34 lb
At standard temperature and pressure, 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
Weight of 1 gal of water = 62.4 lb ÷ 7.48 gal = 8.34 lb/gal
Thus,
1 gal H 2 O = 8.34 lb
■ EXAMPLE 22.1
Problem: Find the number of gallons in a reservoir that has a volume of 855.5 ft 3 .
Solution:
855.5 ft 3 × 7.48 gal l /f ft3 3 = 6399 gal (rounded)
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 referred to as 4.3 psi pressure
or 10 ft of head. Another example: If the static pressure in a pipe leading from an elevated water
storage tank is 45 pounds per square inch (psi), what is the elevation of the water above the pressure
gauge?
Remembering that 1 psi = 2.31 feet and that the pressure at the gauge is 45 psi,
45 psi × 2.31 ft/psi = 104 ft
When demonstrating the relationship of the weight of water to the weight of air we can say, theo-
retically, that the atmospheric pressure at sea level (14.7 psi) will support a column of water 34 feet
high:
14.7 psi × 2.31 ft/psi = 34 ft
At an elevation of one mile above sea level, where the atmospheric pressure is 12 psi, the column of
water would be only 28 feet high: 12 psi × 2.31 ft/psi = 28 ft (rounded).
If a glass or clear plastic tube is placed in a body of water at sea level, the water will rise in the
tube to the same height as the water outside the tube. The atmospheric pressure of 14.7 psi will push
down equally on the water surface inside and outside the tube. However, if the top of the tube is
tightly capped and all of the air is removed from the sealed tube above the water surface, forming a
perfect vacuum , the pressure on the water surface inside the tube will be zero psi. The atmospheric
pressure of 14.7 psi on the outside of the tube will push the water up into the tube until the weight of
the water exerts the same 14.7 psi pressure at a point in the tube even with the water surface outside
the tube. The water will rise 14.7 psi × 2.31 ft/psi = 34 feet.
In practice, it is impossible to create a perfect vacuum, thus the water will rise somewhat less
than 34 feet; the distance it rises depends on the amount of vacuum created. If, for example,
enough air was removed from the tube to produce an air pressure of 9.7 psi above the water in the
tube, how far will the water rise in the tube? To maintain the 14.7 psi at the outside water surface
level, the water in the tube must produce a pressure of 14.7 psi - 9.7 psi = 5.0 psi. The height of
the column of water that will produce 5.0 psi is
5.0 psi × 2.31 ft/psi = 11.5 ft