Example

Problem: What is the pressure at a point 20 ft below the surface of a reservoir?

Solution: To calculate this, we must know that the density of water (w) is 62.4 lb/ft3. Thus, p = w x h = 62.4 lb/ft3 x 20 ft = 1248 psf

Waterworks operators generally measure pressure in pounds per square inch rather than pounds per square foot; to convert, divide by 144 in.2/ft2 (12 in. x 12 in. = 144 in.2):

1248 lb/ft2 2

144 in2/ft2

9.5 GAUGE PRESSURE

Recall that head is the height a column of water will rise due to the pressure at its base. Also, a perfect vacuum plus atmospheric pressure of 14.7 psi will lift the water 34 feet. If we now open the top of the sealed tube to the atmosphere and enclose the reservoir and

Key Point: Gauge pressure r

+ atmoSpheric pressure = then increase the pressure in the reservoir, the water absolute pressure. will again rise in the tube.

In actual pressure measurements, we usually ignore the first 14.7 psi (since atmospheric pressure is essentially universal) and measure only the difference between the water pressure and the atmospheric pressure; we call this gauge pressure. For example, water in a lake is subjected to the 14.7 psi of atmospheric pressure, but subtracting this 14.7 psi leaves a gauge pressure of 0 psi. This shows that the water would rise 0 feet above the lake surface. If the gauge pressure in a water main is 120 psi, the water would rise in a tube connected to the main:

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