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## Q. 2.18

For a gage reading at A of -2.50 psi, determine (a) the elevations of the liquids in the open piezometer columns EF, and G and (b) the deflection of the mercury in the U-tube gage in Fig, 2-12.

## Verified Solution

(a)    Since the unit weight of the air (about 0.08 lb/ft$^{3}$) is very small compared with that of the liquids, the pressure at elevation 49.00 may be considered to be -2.50 psi without introducing significant error in the calculations.

For column E:

The elevation at L being assumed as shown, we have, in psf gage,

$p_{K}=p_{L}$

Then

$p_{H}+\gamma h=0$

or

-2.50 x 144 + (0.700 x 62.4)h = 0   and    h=8.24 ft.

Hence the elevation at L is 49.00 – 8.24 = 40.76 ft.

For column F:

pressure at El. 38.00 = pressure at El. 49.00 + pressure of 11 ft of liquid of sp gr 0.700

$=-2.50+\frac{(0.700\times 62.4)(49.00-38.00)}{144}=0.837\ psi$

which must equal the pressure at M. Thus the pressure head at M is $\frac{0.837\times 144}{62.4}=1.93$ ft of water, and column F will rise 1.93 ft above M or to elevation 39.93 at N.

For column G:

pressure at El. 26.00 = pressure at El. 38.00 + pressure of 12 ft of water

or

$p_{O}=0.837+\frac{62.4\times 12}{144}=6.04$ psi

which must be the pressure at R. Then the pressure head at R is $\frac{6.04\times 144}{1.600\times 62.4}=8.71$ ft of the liquid, and column G will rise 8.71 ft above R or to elevation 34.71 at Q.

(b)    For the U-tube gage, using feet of water units,

13.57$h_{1}$ = pressure head at El. 38.00 + pressure head of 24.00 ft of water
13.57$h_{1}$ = 1.93 + 24.00
from which $h_{1}$ = 1.91 ft.