5C-2 : |
Heat Losses form a Steam Turbine |
5 pts |
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Steam enters a turbine operating at steady-state with a mass flow rate of 4600 kg/h. The turbine develops a power output of 1000 kW. In the feed, the pressure is 60 bar, the temperature is 400oC and the velocity is 10 m/s. For the effluent, the pressure is 0.1 bar, the quality is 90% and the velocity is 50 m/s. Calculate the rate of heat transfer between the turbine and the surroundings in kW. |
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Read : |
Apply the steady-state form of the 1st Law for open systems and solve for Q. Assume changes in potential energy are negligible. We know the values of two intensive variables for state 1, so we can look up H1. We know the pressure and quality for state 2, so we can also determine H2. Then, just plug back into the 1st Law to get Q ! |
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Given : |
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4600 |
kg/h |
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v1 |
10 |
m/s |
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1.278 |
kg/s |
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P2 |
10 |
kPa |
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Ws |
1000 |
kW |
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x2 |
0.90 |
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P1 |
6000 |
kPa |
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v2 |
50 |
m/s |
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T1 |
400 |
oC |
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Find : |
Q |
??? |
kW |
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Diagrams : |
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Assumptions: |
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1 - |
The turbine operates at steady-state. |
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2 - |
The change in the potential energy of the fluid from the inlet to the outlet is negligible. |
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Equations / Data / Solve : |
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Let's begin by writing the steady-state form of the 1st Law for open systems. |
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Eqn 1 |
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Solve Eqn 1 for Q : |
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Eqn 2 |
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We must use the Steam Tables to determine H2 and H1 : |
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Eqn 3 |
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H1 |
3178.2 |
kJ/kg |
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At P2 = 10 kPa : |
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Hsat liq |
191.81 |
kJ/kg |
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Hsat vap |
2583.9 |
kJ/kg |
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H2 |
2344.7 |
kJ/kg |
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Now, we can plug values into Eqn 2 to evaluate Q : |
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Q |
-63.51 |
kW |
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Verify : |
None of the assumptions made in this problem solution can be verified. |
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Answers : |
Q |
-63.5 |
kW |
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