Fundamental Chemistry and Hydraulics - Mathematics and Statistics for the Environment - page 368

14.2.1.2 Weight of Water
Because water must be both stored and moved in water supplies and because wastewater must be
collected, processed in unit processes, and outfalled to its receiving body we must consider some
basic relationships in the weight of water. One cubic foot of water weighs 62.4 lb and contains 7.48
gal, and 1 cubic inch of water weighs 0.0362 lb. Water 1 foot deep will exert a pressure of 0.43 psi
on the bottom area (12 in. × 0.062 lb/in. 3 ). A column of water 2 feet high exerts 0.86 psi, one 10 feet
high exerts 4.3 psi, and one 52 feet high exerts
52 ft × 0.43 psi/ft = 22.36 psi
A column of water 2.31 feet high will exert 1.0 psi. To produce a pressure of 40 psi requires a water
column that is
40 psi × 2.31 ft/psi = 92.4 ft
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-foot column of water exerts 4.3 psi. This can be called 4.3-psi pressure or 10
feet of head. Another example: If the static pressure in a pipe leading from an elevated water storage
tank is 37 psi, what is the elevation of the water above the pressure gauge? Remembering that 1 psi
= 2.31 and that the pressure at the gauge is 37 psi,
37 psi × 2.31 ft/psi = 85.5 ft (rounded)
14.2.1.3 Weight of Water Related to the Weight of Air
The theoretical atmospheric pressure at sea level (14.7 psi) will support a column of water 34 feet
high:
14.7 psi × 2.31 ft/psi = 33.957, or 34 ft
At an elevation of 1 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 = 27.72, or 28 ft).
If a tube is placed in a body of water at sea level (e.g., a glass, a bucket, water storage reservoir,
lake, pool), water will rise in the tube to the same height as the water outside the tube. The atmo-
spheric 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 0 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 ft. In practice,
it is impossible to create a perfect vacuum, so the water will rise somewhat less than 34 feet; the
distance it rises depends on the amount of vacuum created.
■ EXAMPLE 14.26
Problem: If 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?
Solution: 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 = 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 (rounded)