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ES202 – Fluid
& Thermal Systems – Winter 2007-2008 Homework Answer and Hints Updated 6:20 PM -- 15 January 2008 |
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Set |
Problem |
Answer |
Hint |
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4 |
2-40 |
110
psia |
Apply
conservation of linear momentum to the pistons |
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2-62 |
3.39
kPa |
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2-69 |
562.5
kgm3 |
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5 |
2-96 |
12.0
kPa; 2.0 m |
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|
10-16 |
30,900
lbf |
Use
hydrostatic force concepts to find force of water on gate and then use this
information along with cons. of linear momentum to analyze the gate. |
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6 |
10-34 |
1050
N |
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10-45 |
9.64
x 108 N; 17.1 m; 8.35
x 108 N |
This is the
total force (sum of the force due to the atmospheric pressure AND the water).
The answer would be different if you just used the force due to the water
alone. If you were attempting to
study the dam to see if the water will move
the dam, the forces due to atmospheric pressure would just cancel out.
HOWEVER, if there was a very small leak under
the dam, the hydrostatic force pushing up vertically on the dam would
increase the possibility of tipping the dam. |
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7 |
12-27 |
2.43
m/s |
Apply
Bernoulli between the free surface of the gasoline and the leak opening in
the tank wall. |
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|
12-35 |
0.52
in |
Apply
Bernoulli (correctly) along any single streamline between section 1 and section
2. Section 1 and section 2 are located directly over the opening of the
manometer. Recognizing that the pressure variation across parallel
streamlines is hydrostatic. |
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8 |
12-41 |
6.24
cm |
Apply Bernoulli
(correctly) along any single streamline between section 1 and section 2.
Section 1 and section 2 are located directly over the opening of the
manometer. Recognizing that the pressure variation across parallel
streamlines is hydrostatic. |
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|
12-42 |
4.48
ft3/s |
Apply
Bernoulli (correctly) along any single streamline between section 1 and
section 2. Section 1 and section 2 are located directly over the opening of
the manometer. Recognizing that the pressure variation across parallel
streamlines is hydrostatic. |
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12-46 |
33.8
m/s |
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9 |
12-61 |
68.7
kPa |
Apply
mechanical energy balance between two locations where information is known so
that you have one equation and one unknown. |
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|
12-64 |
0.204
m, 40 W |
The useful
pumping power is just the pumping power that would be required to overcome
the losses in the fitting. Physically this is the size of a pump required to
circulate a fluid at the specified rate in a closed loop that consists of
only the fitting and a pump. |
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12-56 |
2.4
W; 6.9 cm; 40 Pa |
Fan-motor
efficiency is the ratio of the (Power into the air from the fan)/(Electrical power into the fan motor). Apply mechanical
energy balance. |
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11 |
12-65 |
55
kW |
Mechanical
energy balance. |
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11 |
12-69 |
50.9
m/s; 201 kW |
Mechanical
energy balance. |
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12 |
12-63 |
Pump
1150 kW; Turbine 530 kW |
Mechanical
energy balance |
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12 |
14-38 |
69.9
kPa; 4.40 kW |
Friction
factor relation. Is the flow laminar OR turbulent? It makes a difference |
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13 |
14-42 |
14.2
lbf/ft2; 0.37 W |
Friction
factor relation. Is the flow laminar OR turbulent? It makes a difference. |
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13 |
14-45 |
124
Pa |
Friction
factor relation. Is the flow laminar OR turbulent? It makes a difference. |
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14 |
14-87 |
217
W |
Write MEB
between the pump inlet and the pump inlet. You are left with an expression
for w_pump in terms of the gh_loss. |
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14 |
14-75 |
0.00595
m3/s |
If you
write the MEB between 1 and 2, you end up with a single equation that ONLY
depends on the velocity in one of the pipes ---- that means if I gave you the
velocity you could solve for everything else in the equation. HOWEVER, only
one unique value of the velocity will satisfy the equation! So this requires
an iterative solution like EXAMPLE 14-5 in the textbook. Think of moving
everything to one side of the equal sign … 0 = F(V)
where there is only one V value that will make this true --- V is the root of
the equation. So you can plot (solve) for F(V)
versus V and see when F(V)=0. |
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Updated
5 December 2007 by D. E. Richards