










Assume air behaves as an ideal gas and heat losses to the surroundings are negligible.
a.) Calculate the power requirement for the compressor and the required cooling water mass flow rate for the heat exchanger.

b.) Calculate the entropy production rate for the compressor and for the heat exchanger separately. 











Read : 
Use the ideal gas EOS and the volumetric flow rate to determine
the mass flow rate. 

Use an energy balance to determine the work for the compressor in kW. 

To determine the water flow
rate, draw the control
volume enclosing the heat
exchanger. This control
volume has four mass flows entering
or leaving but no Q or W.
An energy balance on
this control volume
yields the water flow rate. 

Very important point:
the air and the water DO NOT MIX in the heat exchanger! 

For the enthalpy of the cooling water, use H(T)
for saturated liquid water from the Steam Tables. 

Part
(b) Don't forget about the cooling water when you calculate the entropy
generated. Use S(T) for saturated liquid water
from the Steam Tables. 










Given: 
T_{A} 
15 
^{o}C 



P_{2} 
255 
kPa 

T_{B} 
35 
^{o}C 



T_{2} 
143 
^{o}C 

P_{1} 
104 
kPa 



P_{3} 
255 
kPa 

T_{1} 
30 
^{o}C 



T_{3} 
65 
^{o}C 

m_{air} 
35 
kg/min 
















Find: 
Part (a) 
W_{S} 
??? 
kW 

Part (b) 
S_{gen, comp} 
??? 
kW/K 


m_{cw} 
??? 
kg/s 


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










Diagram: 

Diagram already
provided in the problem statement. 










Assumptions: 
1  
Both the compressor and heat exchanger operate at steadystate. 


2  
Heat
exchange between the equipment in this process and the surroundings is negliqible. 


3  
There is no shaft
work in the heat
exchanger. 


4  
Kinetic and potential energy changes
are negligible. 


5  
The air behaves as an ideal gas. 


6  
The properties of the cooling water are the same as the properties of saturated liquid water
at the same temperature. 










Equations
/ Data / Solve: 













Part a.) 
R 
8.314 
kJ/kmol K 







MW 
28.97 
kg/kmol 



m_{air} 
0.5833 
kg/s 











We can determine the
properties of air at all three states by interpolating on the Ideal Gas Property Tables for air
because all three temperatures are given in the problem statement. 











T_{1} 
303.15 
K 
T (K) 
H^{o}
(kJ/kg) 
S^{o} (kJ/kgK) 



T_{2} 
416.15 
K 
300 
87.410 
0.0061681 




T_{3} 
338.15 
K 
310 
97.396 
0.038914 
H_{1} 
90.556 
kJ/kg 













410 
198.63 
0.32178 






420 
208.88 
0.34649 
H_{2} 
204.93 
kJ/kg 












330 
117.45 
0.10159 





340 
127.51 
0.13163 
H_{3} 
125.65 
kJ/kg 










Now, we can plug
values back into Eqn 1 to evaluate W_{s}: 

W_{S} 
66.72 
kW 











The mass flow rate of the cooling water can be determined by an energy
balance on the heat
exchanger.
For a steadystate process with negligible heat transfer, kinetic
and potential energy changes and no shaft
work: 




Eqn 3 

Eqn
3 can be simplified because no shaft work crosses the system boundary and when we use the entire HEX as
the system, there is no heat
transfer across the system
boundary either. 












Eqn 4 











Now, we can solve Eqn
4 for m_{cw} : 





Eqn 5 











We do not know the pressure of the cooling water,
so we cannot look up
its properties. Therefore, we assume that the enthalpy of the cooling water
is the same as the enthalpy of saturated liquid water at the same temperature. This assumption is accurate as long as water is nearly an incompressible liquid. The properties of saturated liquid water were determined from NIST WebBook: 











Saturated
liquid water at T_{A}: 



H_{A} 
62.981 
kJ/kg 

Saturated
liquid water at T_{B}: 



H_{B} 
146.63 
kJ/kg 











Now, we can plug
values into Eqn 5 and evaluate
m_{cw}: 

m_{cw} 
0.5529 
kg/s 










Part b.) 
Entropy
generation is defined by: 



Eqn 6 











No heat transfer crosses the system boundary for either the compressor or the HEX, so Eqn 6
simplifes to: 


















Eqn 7 











We can determine the entropy change for air
(ideal gas) in the compressor from the 2nd Gibbs Equation: 















Eqn 8 











Therefore, the rate at which entropy is generated in the compressor is: 
















Eqn 9 











We can determine the properties of air at all three states by interpolating on the Ideal Gas Property Tables for air
because all three temperatures are given
in the problem statement. 









S^{o}(T_{1}) 
0.016483 
kJ/kgK 






S^{o}(T_{2}) 
0.33698 
kJ/kgK 


S^{o}(T_{3}) 
0.12607 
kJ/kgK 











Now, we can plug
values back into Eqn 9 : 


S_{gen, comp} 
0.03681 
kW/K 











The entropy generated in the HEX must take into account the entropy change
of BOTH the air and the cooling water. 



Eqn 10 











Again, we assume that the entropy of the cooling water is the same as the entropy of saturated liquid water
at the same temperature. The properties of saturated liquid water were determined from NIST WebBook: 











Saturated
liquid water at T_{A}: 



S(T_{A}) 
0.22446 
kJ/kgK 

Saturated
liquid water at T_{B}: 



S(T_{B}) 
0.50513 
kJ/kgK 





Now, we can plug value into Eqn 10 : 

S_{gen, HEX} 
0.03215 
kW/K 





Verify: 
Check the Ideal Gas Assumption: 


V_{1}
= 
24.23 
L/mole 





V_{2}
= 
13.57 
L/mole 





V_{3}
= 
11.03 
L/mole 











Since air can be considered to be a diatomic gas and all three molar volumes are greater than 5 L/mole,
it is accurate to
treat the air as an ideal gas. 










Answers : 
Part a.) 
W_{S} 
66.7 
kW 

Part b.) 
S_{gen, comp} 
0.0368 
kW/K 


m_{cw} 
0.553 
kg/s 

S_{gen, HEX} 
0.0321 
kW/K 










