Example Problem with Complete Solution

8B-2 : Entropy Generation in a Steam Turbine 5 pts
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.
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.
Given: P1 2000 kPa x2 1
T1 450 oC THT 160 oC
P2 180 kPa Tsurr 20 oC
WS 500 kJ/kg
Find: (Sgen)turb ??? kJ/kg-K
Diagram: The diagram in the problem statement is adequate.
Assumptions: 1 - The turbine operates at steady-state.
2 - Kinetic and potential energy changes are negligible.
3 - Shaft work and flow work are the only forms of work that cross the system boundary.
4 - Heat loss from the turbine occurs at a constant and uniform temperature of 160oC.
Equations / Data / Solve:
We can determine the entropy generation from an entropy balance on the turbine.
The entropy balance equation for a SISO process operating at steady-state that exchanges heat only with the surroundings is:
Eqn 1
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.
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.
S1 7.2866 kJ/kg-K S2 7.1621 kJ/kg-K
Now, use the 1st Law for a steady-state process with negligible changes in kinetic and potential energies to determine Q.
Eqn 2
Solve Eqn 2 for Q :
Eqn 3
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.
H1 3358.2 kJ/kg H2 2701.4 kJ/kg
Now we can plug values into Eqn 2 to evaluate Q : Q -156.81 kJ/kg
We can now evaluate the entropy generation within the turbine using Eqn 1.
(Sgen)turb 0.2375 kJ/kg-K
Verify: The assumptions made in the solution of this problem cannot be verified with the given information.
Answers : Q -157 kJ/kg Sgen 0.237 kJ/kg-K