1E-1 : |
Pressure Measurement Using a Multi-Fluid Manometer |
6 pts |
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The water in a tank is pressurized by air, and the pressure is measured by a multi-fluid manometer, as shown in the figure. Determine the gage pressure of air in the tank if h1 = 0.2 m, h2 = 0.3 m and h3 = 0.46 m. |
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Take the densities of water, oil and mercury to be 1000 kg/m3, 850 kg/m3 and 13,600 kg/m3, respectively. |
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Read : |
Use the barometer equation to work your way through the different fluids from point 1 to point 2. |
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Remember that gage pressure is the difference between the absolute pressure and atmospheric pressure. |
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Given : |
h1 |
0.20 |
m |
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ρw |
1000 |
kg/m3 |
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h2 |
0.30 |
m |
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ρoil |
850 |
kg/m3 |
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h3 |
0.46 |
m |
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ρHg |
13600 |
kg/m3 |
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P2 |
101.325 |
kPa |
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Find : |
P1,gage |
??? |
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Solution : |
Gage pressure is defined by : |
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Eqn 1 |
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If we assume that P2 is atmospheric pressure, then Eqn 1 becomes : |
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Eqn 2 |
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The key equation is the Barometer Equation : |
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Eqn 3 |
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Now, apply Eqn 1 repeatedly to work our way from point 1 to point 2. |
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Some key observations are: |
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Eqn 4 |
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Eqn 5 |
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These are true because the points are connected by open tubing, the fluid is not flowing in this system and no change in the composition of the fluid occurs between A & B or C & D or D & E. |
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PA > P2, therefore : |
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Eqn 6 |
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PE > P1, therefore : |
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Eqn 7 |
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PB > PC, therefore : |
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Eqn 8 |
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Combine Eqns 2, 5 & 6 to get : |
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Eqn 9 |
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Use Eqns 3 & 5 to eliminate PC from Eqn 7 : |
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Eqn 10 |
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Now, solve for P1 - P2 : |
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Eqn 11 |
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Combining Eqns 10 & 2 yields : |
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Eqn 12 |
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Plugging values into Eqn 11 yields : |
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g |
9.8066 |
m/s2 |
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P1,gage |
56888 |
Pa gage |
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gC |
1 |
kg-m/N-s2 |
P1,gage |
56.89 |
kPa gage |
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Answers : |
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P1,gage |
56.9 |
kPa gage |
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If you are curious : |
P1 |
158.21 |
kPa |
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P2 |
101.325 |
kPa |
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PA = PB |
162.68 |
kPa |
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PC = PD = PE |
160.17 |
kPa |
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