At steady-state, a steam turbine produces 1,050 MW by letting the pressure down from 50 bar to 1 bar.
The steam enters the turbine at 320oC with a velocity of 8 m/s
and a mass flow rate
of 150 kg/min. |
The steam leaves the turbine with a quality of 0.94 kg vapor/kg and a velocity of 65 m/s.
Determine the rate of heat loss from the turbine to 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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150 |
kg/min |
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v1 |
8 |
m/s |
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2.500 |
kg/s |
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P2 |
100 |
kPa |
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Ws |
1050 |
kW |
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x2 |
0.94 |
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P1 |
5000 |
kPa |
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v2 |
65 |
m/s |
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T1 |
320 |
oC |
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Find: |
Q |
??? |
kW |
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Diagram: |
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Assumptions: |
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 |
2986.2 |
kJ/kg |
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At P2 = 100 kPa : |
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Hsat liq |
417.50 |
kJ/kg |
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Hsat vap |
2674.95 |
kJ/kg |
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H2 |
2539.5 |
kJ/kg |
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Now, we can plug
values into Eqn 2 to evaluate Q : |
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Q |
-61.62 |
kW |
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Verify: |
None of the assumptions
made in this problem solution can be verified. |
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Answers : |
Q |
-61.6 |
kW |
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