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Mechanical Properties of Fluids

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Question : 22 of 31
Marks: +1, -0
A manometer reads the pressure of a gas in an enclosure as shown in figure. (a). When a pump removes some of the gas, the manometer reads as in figure (b). The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury.
(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury.
(b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) are poured into the right limb of the manometer? (ignore the small change in the volume of the gas).
Solution:  
Here, atmospheric pressure, P=76P = 76 cm of mercury.
(a) In figure (a) pressure head, h=+20 cmh = +20 \, \text{cm}
∴ Absolute pressure =P+h=76+20=96= P + h = 76 + 20 = 96 cm of mercury
Gauge pressure =h=20= h = 20cm of mercury
In figure (b) pressure head, h=−18h = -18 cm.
Absolute pressure =P+h=76+(−18)=58= P + h = 76 + (-18) = 58 cm of mercury
Gauge pressure =h=−18= h = -18 cm of mercury
(b) Here 13.6 cm of water added in right limb is equivalent to 13.613.6=1\frac{13.6}{13.6}=1 = cm of mercury column.
i.e. h′=1h' = 1 cm of mercury column.
Now pressure at AA,
Pa=P+h′=76+1=77 cmPa = P + h' = 76 + 1 = 77 \, \text{cm}
Let the difference in mercury levels in the two limbs be h1h_1, then pressureat b,b,
Pb=58+h1P_{b}=58+h_{1}
As, Pa=Pb′P_{a}=P_{b'}
∴77=58+h1\therefore 77=58+h_{1} or h1=77−58=19h_{1}=77-58=19 cm of mercury column.
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