The outer
surface of a steam
turbine is at an average temperature of 160^{o}C and the surroundings are at 20^{o}C. Calculate the internal, external
and total entropy generation for the turbine in kJ/kgK. 
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 S_{gen}. 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 S_{gen} for the turbine. 










Given: 
P_{1} 
2000 
kPa 



x_{2} 
1 


T_{1} 
450 
^{o}C 



T_{HT} 
160 
^{o}C 

P_{2} 
180 
kPa 



T_{surr} 
20 
^{o}C 







W_{S} 
500 
kJ/kg 
Find: 
(S_{gen})_{turb} 
??? 
kJ/kgK 
















Diagram: 
The diagram in the
problem statement is adequate. 










Assumptions: 
1  
The turbine operates at steadystate. 


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 160^{o}C. 










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 steadystate 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, 160^{o}C. If we used T_{HT} = T_{surr}, 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 T_{HT} and T_{surr}. 











We can lookup S_{1} and S_{2} in the Steam Tables or the NIST Webbook because states 1 and 2
are completely
determined by the information given in the problem statement. 











S_{1} 
7.2866 
kJ/kgK 



S_{2} 
7.1621 
kJ/kgK 











Now, use the 1st Law for a steadystate 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 H_{1} and H_{2} 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. 











H_{1} 
3358.2 
kJ/kg 



H_{2} 
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. 

















(S_{gen})_{turb} 
0.2375 
kJ/kgK 










Verify: 
The assumptions made
in the solution of this problem cannot be verified with the given
information. 










Answers : 
Q 
157 
kJ/kg 



S_{gen} 
0.237 
kJ/kgK 









