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## Q. 2.17

For a gage pressure at A of -10.89 kPa, find the specific gravity of the gage liquid B in Fig. 2-11. ## Verified Solution

pressure at C = pressure at D

$p_{A}+ \gamma h=p_{D}$

-10.89 + (1.60 x 9.79)(3.200 – 2.743) = -3.73 kPa

Now $p_{G} =p_{D} = -3.73\ kPa$, since the weight of 0.686 m of air can be neglected without introducing significant error. Also $p_{E} = p_{F} = 0$.

Thus,

pressure at G = pressure at E – pressure of (3.429 – 3.048) m of gage liquid

or $p_{G} = p_{E} – (sp\ gr\times 9.79)(3.429 – 3.048)$

-3.73 = 0 – (sp gr x 9.79)(0.381) and sp gr = 1.00