The outer
surface of a air
compressor is at an average temperature of 485 K
and the surroundings
are at 300 K. The
operating parameters for the compressor are given in the figure below. 





a.) Calculate the internal, external and total
entropy generation rates
for the compressor in kW/K.
b.) Calculate the internal, external and total
lost work for the compressor in kW.

c.) Show that the total lost
work determined in part
(b) is the difference
between the power
requirement of a completely
reversible compressor that accomplishes the same change in state of the air stream and the actual work for the 
real compressor. 











Read : 
Equations for
evaluating total and external entropy
generation can be obtained from Lesson 8D. You could just take the difference between
these to determine the internal entropy generation. But it is useful to understand how the position of the system boundary can be manipulated to directly yield an
equation for the internal entropy generation. 











The keys are to use
the 1st Law to determine Q and to use the 2nd Gibbs Equation and the Ideal Gas Entropy Function to
evaluate DS. 











Once the entropy generation rates are known,
it is easy to determine the lost work rates. 











In order to verify the
definition of lost work, we must determine the
amount of work a reversible compressor would require to accomplish the same compression
process. Q_{rev} must be different from Q_{act}. The key to this
part of the problem is that total entropy generation rate for the reversible compressor must be zero. 










Given: 
P_{1} 
96 
kPa 



T_{1} 
310 
K 

P_{2} 
1200 
kPa 



T_{2} 
640 
K 

m 
1.30 
kg/s 



T_{surf} 
485 
K 

W_{s,act} 
555 
kW 



T_{surr} 
300 
K 










Find: 
a.) 
S_{gen,int} 
??? 
kW/K 

b.) 
W_{lost,int} 
??? 
kW 


S_{gen,ext} 
??? 
kW/K 


W_{lost,ext} 
??? 
kW 


S_{gen,tot} 
??? 
kW/K 


W_{lost,tot} 
??? 
kW 











c.) 
Show that : 




Eqn 1 










Assumptions: 
1  
The surface temperature of the compressor remains constant. Therefore, it can be treated as a thermal
reservoir. 


2  
The compressor operates at steadystate with negligible changes in kinetic and potential energies. 


3  
Air behaves as an ideal gas. 










Diagram: 
The diagram in the
problem statement is adequate. 










Equations
/ Data / Solve: 

















Part a.) 
We can use the
following equations to evaluate the internal and total entropy generation
rates. 

















Eqn 2 











If we place the system boundary far from the surface of the compressor, then ALL of the irreversibilities are inside the system because the temperature at the system boundary is the same as the temperature of the surroundings. So, heat exchange between this big system and the surroundings is reversible. 



This explains why
using T_{surr} in Eqn 2
tells us that we are computing the TOTAL entropy generation rate. Because with this big system, there are no irreversibilities OUTSIDE
the system boundary. Notice that heat
transfer from this system does
occur, but it occurs with zero temperature driving force, so it is reversible ! 

















Eqn 3 











Using our usual system
boundary, right at the surface of the compressor, heat exchange with the surroundings is irreversible. So, by using Eqn 3, with T_{surf} instead
of T_{surr}, we have excluded the external irreversibility due to heat transfer through a finite temperature difference. So, this entropy generation equation gives us
the INTERNAL entropy generation only ! 























The external entropy
generation rate is equal to the rate at which the entropy of the universe (compressor and surroundings) changes
due only to the heat transfer from the compressor to the surroundings. 















Eqn 4 











The minus sign appears in Eqn 4 because the sign of Q
from the perspective of the system is negative, but from the perspective of the surroundings, Q > 0
because heat is entering the surroundings ! 











Eqn
4 can be rearranged to give us : 


Eqn 5 











Let's begin by
determing DS for the working fluid using the 2nd
Gibbs Equation and the Ideal
Gas Property Table for air. 

















Eqn 6 











S^{o}_{1} 
0.038914 
kJ/kgK 



S^{o}_{2} 
0.78826 
kJ/kgK 











R 
8.314 
J/moleK 







MW 
28.97 
g/mole 



DS 
0.02450 
kJ/kgK 











Next, we need to
evaluate Q, the rate of heat loss
from the compressor.
We can do this using the 1st Law for open systems operating at steadystate with negligible changes
in kinetic and potential energies. 

















Eqn 7 











We can solve Eqn 7 for Q : 



Eqn 8 











Now, we can use the Ideal Gas Property Table for air to evaluate H_{1} and H_{2}. 











H^{o}_{1} 
97.396 
kJ/kg 



H^{o}_{2} 
439.98 
kJ/kg 











Plug values back into Eqn 7 to evaluate Q : 


Q 
109.64 
kW 











Now, we are able to
plug values back into Eqns 2, 3 & 5 to evaluate the entropy generation rates. 











S_{gen,int} 
0.25791 
kW/K 







S_{gen,ext} 
0.13941 
kW/K 



S_{gen,tot} 
0.39731 
kW/K 











Double check your
calculations using : 






















Eqn 9 










Part b.) 
Once we have completed
part (a), part (b) is a straightforward
application of the relationship between lost work and entropy generation. 


















Eqn 10 











Eqn
10 applies for internal, external
and total entropy generation and lost work. 











W_{lost,int} 
77.37 
kW 







W_{lost,ext} 
41.82 
kW 



W_{lost,tot} 
119.19 
kW 










Part c.) 
Here we must verify that our answer from part (b) is consistent with the defintion of lost work. 

















Eqn 11 











In order to do this,
we must evaluate W_{S,rev}. 











The key to doing this
is to understand that the reversible process still
operates between state 1
and state 2. 











The values of Q and W_{S} are different from those for the real compressor. 











The fact that ties
this part of the problem together and allows us to determine Q and W_{S} for the reversible compressor is that : 


















Eqn 12 











We can use this fact
and solve Eqn 2 for Q_{rev} : 


Eqn 13 

















Q_{rev} 
9.55 
kW 











Next, we can apply the
1st Law to the reversible compressor to evaluate W_{S,rev}. 











We can solve Eqn
7 for W_{S,rev}. The result is : 


Eqn 14 











Plugging values into Eqn 14 yields : 


W_{S,rev} 
435.81 
kW 











Finally, put values
into the righthand side of Eqn 11 : 

119.19 
kW 











This matches our result for lost work from part (b). So, we have confirmed the relationship between reversible work, actual work and lost work. 










Verify: 
None of the assumptions
made in this problem solution can be verified. 













Answers : 
a.) 
S_{gen,int} 
0.258 
kW/K 

b.) 
W_{lost,int} 
77.4 
kW 


S_{gen,ext} 
0.139 
kW/K 


W_{lost,ext} 
41.8 
kW 












S_{gen,tot} 
0.397 
kW/K 


W_{lost,tot} 
119.2 
kW 















c.) 
W_{S,rev}  W_{S,act} = 
119.2 
kW 









