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9.3 Theory of Backflow and Backsiphonage
24˝
0.433 psig
3
62.4#/ft
12˝
12˝
12˝
Figure 9.11 Pressures exerted by (a)
1 ft (0.3048 m) of water and (b) 2 ft
Sea level Sea level
0.866 psig (0.6096 m) of water at sea level.
Conversion factors:
′′
1#= 1b = 0.454 kg; 1 = 1in. =
0.433 psig
25.4 mm; 1 psig = 6.94 kPa gauge
(a) (b) pressure.
3
The average weight of a cubic foot of water (62.4 lb/ft ) will water; atmospheric pressure, which is 14.7 psia (P absolute =
2
exert a pressure of 62.4 lb/ft (430 kPa) gauge. The base may 102 kPa), acts equally on the surface of the water within the
be subdivided into 144 square inches with each subdivision tube and on the outside of the tube.
being subjected to a pressure of 0.433 psig (P gauge = 3kPa). If, as shown in Fig. 9.12b, the tube is capped and a
3
3
Suppose another 1 ft (0.0283 m ) of water was placed vacuum pump is used to evacuate all the air from the sealed
directly on top of the first (see Fig. 9.11b). The pressure tube, a vacuum with a pressure of 0 psia (P absolute = 0kPa)
on the top surface of the first cube, which was originally is created within the tube. Because the pressure at any point
atmospheric, or 0 psig, would now be 0.433 psig (P gauge = in a static fluid is dependent on the height of that point
3 kPa) as a result of the superimposed cubic foot of water. The above a reference line, such as sea level, it follows that the
pressure of the base of the first cube would also be increased pressure within the tube at sea level must still be 14.7 psia
by the same amount of 0.866 psig (P = 6kPa), or two (P absolute = 102 kPa). This is equivalent to the pressure at the
gauge
times the original pressure. base of a column of water 33.9 ft (10.3 m) high. With the
If this process were repeated with a third cubic foot of column open at the base, water would rise to fill the column
water, the pressures at the base of each cube would be 1.299, to a depth of 33.9 ft (10.3 m). In other words, the weight of
0.866, and 0.433 psig (P = 9, 6 and 3 kPa), respectively. the atmosphere at sea level exactly balances the weight of
gauge
It is evident that pressure varies with depth below a free a column of water 33.9 ft (10.3 m) in height. The absolute
water surface; in general, each foot (0.3048 m) of elevation pressure within the column of water in Fig. 9.12b at a height
change, within a liquid, changes the pressure by an amount of 11.5 ft (3.5 m) is equal to 9.7 psia (P absolute = 67.3kPa).
equal to the weight-per-unit area of 1 ft (0.3048 m) of the liq- This is a partial vacuum with an equivalent gauge pressure
uid. The rate of increase for water is 0.433 psi/ft (9.84 kPa/m) of –5.0 psig (P gauge =−34.7kPa).
of depth. As a practical example, assume the water pressure at a
Frequently water pressure is referred to using the terms closed faucet on the top of a 100-ft (30.48-m)-high building to
pressure head or just head and is expressed in units of feet be 20 psig (P gauge = 138.8 kpa)); the pressure on the ground
of water. One foot (0.3048 m) of head would be equivalent floor would then be 63.3 psig (P gauge = 439.3kPa). If the
to the pressure produced at the base of a column of water pressure at the ground were to drop suddenly to 33.3 psig
1 ft (0.3048 m) in depth. One foot (0.3048 m) of head or 1 ft (P gauge = 231.1 kPa) due to a heavy fire demand in the area,
(0.3048 m) of water is equal to 0.433 psig (P gauge = 3kPa). the pressure at the top would be reduced to −10 psig (P gauge =
One hundred feet (30.48 m) of head is equal to 43.3 psig −69.4 kPa). If the building water system were airtight, the
(P gauge = 300 kPa). water would remain at the level of the faucet because of the
partial vacuum created by the drop in pressure. If the faucet
were opened, however, the vacuum would be broken and the
9.3.2 Siphon Theory
water level would drop to a height of 77 ft (23.47 m) above
Figure 9.12a depicts the atmospheric pressure on a water the ground. Thus, the atmosphere was supporting a column
surface at sea level. An open tube is inserted vertically into the of water 23 ft (7 m) high.