| The pressure
gauge on the air in the tank shown below reads 87 kPa. Determine the manometer
reading, h2, in cm. |
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Data:
h1 = 25 cm and h3 = 65 cm
SGmercury = 13.6, SGoil = 0.75, ρwater = 1000 kg/m3 |
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| Read: |
The density of the air
is so much lover than the density of the liquids in this problem that the
weight of the air can be considered negligible in the force balances we will
write in this problem. |
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| Given: |
Pgauge |
87 |
kPa |
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hoil |
0.65 |
m |
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SGoil |
0.7 |
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hwater |
0.25 |
m |
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SGm |
13.6 |
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| Find: |
h2 |
??? |
cm |
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| Assumptions: |
1- The fluids in the
system are completely static. |
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2- The densities of the
liquids are uniform and constant. |
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3- The reference
density of water used to determine specific gravity is: |
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rwater |
1000 |
kg/m3 |
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4- The acceleration of
gravity is: |
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g |
9.8066 |
m/s2 |
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gc |
1 |
kg-m/N-s2 |
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| Equations
/ Data / Solve: |
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Begin by writing the Manometer Equation for each
interval between points A and F
on the diagram. |
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Eqn 1 |
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Eqn 3 |
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Eqn 2 |
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Eqn 4 |
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Eqn 5 |
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If we add all 5 of these
equations together we obtain : |
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Eqn 6 |
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The only unknown in this
equation is h. So, the next step is to solve the equation
for h. |
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Eqn 7 |
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Also, because Pf = Patm and the definition of gauge pressure we can use: |
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Eqn 8 |
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Pa-Pf = |
87000 |
N/m2 |
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All we need to do is
convert specific gravity into density and we are ready to plug values into Eqn 7. |
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The definition of
specific gravity is : |
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Eqn 9 |
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This helps us simplify Eqn 7 to : |
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Eqn 10 |
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Eqn 11 |
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From Eqn
9 : |
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Eqn 12 |
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rm |
13600 |
kg/m3 |
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Plugging values into Eqn 11 yields : |
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h |
0.652 |
m |
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65.2 |
cm |
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| Answers: |
h |
65.2 |
cm |
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