The outer
surface of a steam
turbine is at an average temperature of 160oC and the surroundings are at 20oC. Calculate the internal, external
and total entropy generation for the turbine in kJ/kg-K. |
The operating parameters for the turbine are given in the figure
below. |
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Read : |
Apply the 1st Law to determine Q and the 2nd Law to get Sgen. Properties come from
the Steam Tables or
the NIST Webbook. The key is that the heat losses occur at the constant, average
surface temperature and this must be taken into account when
evaluating Sgen for the turbine. |
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Given: |
P1 |
2000 |
kPa |
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x2 |
1 |
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T1 |
450 |
oC |
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THT |
160 |
oC |
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P2 |
180 |
kPa |
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Tsurr |
20 |
oC |
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WS |
500 |
kJ/kg |
Find: |
(Sgen)turb |
??? |
kJ/kg-K |
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Diagram: |
The diagram in the
problem statement is adequate. |
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Assumptions: |
1 - |
The turbine operates at steady-state. |
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2 - |
Kinetic and potential energy changes
are negligible. |
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3 - |
Shaft
work and flow work are the only forms of work that cross the system boundary. |
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4 - |
Heat loss
from the turbine occurs at a constant and uniform temperature of 160oC. |
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Equations
/ Data / Solve: |
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We can determine the entropy generation from an entropy balance on the turbine. |
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The entropy balance equation for a SISO process operating at steady-state that exchanges heat only with the surroundings is: |
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Eqn 1 |
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Because we are
interested only in the entropy generation inside the turbine, the temperature at which heat transfer occurs is the surface temperature of the turbine, 160oC. If we used THT = Tsurr, we would obtain the total entropy generation for the process. This would include both the entropy generated inside
the turbine and the entropy
generated due to the irreversible nature of heat transfer through a finite temperature difference, that is between THT and Tsurr. |
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We can lookup S1 and S2 in the Steam Tables or the NIST Webbook because states 1 and 2
are completely
determined by the information given in the problem statement. |
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S1 |
7.2866 |
kJ/kg-K |
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S2 |
7.1621 |
kJ/kg-K |
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Now, use the 1st Law for a steady-state process with negligible changes in kinetic and potential energies to determine Q. |
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Eqn 2 |
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Solve Eqn 2 for Q : |
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Eqn 3 |
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We can now lookup H1 and H2 in the Steam Tables or the NIST Webbook because, again, states 1 and 2
are completely
determined by the information given in the problem statement. |
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H1 |
3358.2 |
kJ/kg |
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H2 |
2701.4 |
kJ/kg |
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Now we can plug values
into Eqn 2 to evaluate Q : |
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Q |
-156.81 |
kJ/kg |
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We can now evaluate
the entropy generation within
the turbine using Eqn 1. |
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(Sgen)turb |
0.2375 |
kJ/kg-K |
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Verify: |
The assumptions made
in the solution of this problem cannot be verified with the given
information. |
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Answers : |
Q |
-157 |
kJ/kg |
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Sgen |
0.237 |
kJ/kg-K |
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