Consider the internally reversible ammonia compressor shown below. The compression process is polytropic with
d = 1.27. 










Determine W_{s} and Q in kW. 











Read : 
The path equation given in the problem
statement tells you that the compression process follows a polytropic path with d
= 1.27. All
the properties of state 1
can be determined using the Ammonia Tables or the NIST Webbook. The polytropic path equation allows us
to determine the specific volume at state 2. This is the 2nd known intensive property at state 2
, so we can evaluate the other properties using the Ammonia Tables or the NIST Webbook. Determine the shaft
work based on the polytropic
path and then apply the 1st
Law to evaluate Q. 










Given: 
P_{1} 
140 
kPa 

Diagram: 
See the problem
statement. 

V_{1,dot} 
25 
L/s 


^{} 


x_{1} 
1 
kg vap/kg 


^{} 


P_{2} 
750 
kPa 




^{} 


d 
1.27 





^{} 









^{} 

Find: 
a.) 
(W_{s})_{int rev} 
??? 
kW 

b.) 
Q 
??? 
kW 










Assumptions: 
1  








1  
The compressor operates at steadystate. 

2  
Changes in kinetic and potential energies are negligible. 

3  
The compression is internally reversible. 

4  
The compression process follows a polytropic process path with d = 1.08. 

5  
Shaft
work and flow work are the only forms of work that cross the system boundary. 










Equations
/ Data / Solve: 

















Part a.) 
Work for an internally reversible, polytropic process is given by : 

Eqn 1 











We can determine the mass flow rate
from the volumetric
flow rate using: 



Eqn 2 










We can use the Ammonia Tables or the NIST Webbook to evaluate V_{1} because it is a saturated vapor at a known pressure of 140 kPa. 











T_{1} 
26.682 
^{o}C 



V_{1} 
0.83074 
m^{3}/kg 





H_{1} 
1409.0 
kJ/kg 









Next, plug values back
into Eqn 2 : 


m_{dot} 
0.0301 
kg/s 











Now, we need to
determine V_{2}. We can make use of the
fact that the compression process follows a polytropic process path with d
= 1.08. 


















Eqn 3 











Solve Eqn 3 for V_{2} : 




Eqn 4 











Now, we can plug
numbers into Eqn 4 and then Eqn 1 to complete this part of the
problem. 











V_{2} 
0.22156 
m^{3}/kg 



(W_{s})_{int rev} 
7.059 
kW 










Part b.) 
To determine the heat transfer rate for the compressor, we must apply the 1st Law for steadystate, SISO
processes. For this compressor, changes in kinetic and potential energies are negligible and only flow work and shaft work cross the system boundaries. The appropriate form
of the 1st Law for
this compressor is : 








Eqn 5 

We can solve Eqn 5 for the heat transfer rate: 


Eqn 6 











In part
(a) we evaluated all of the unknowns on the righthand side of
Eqn 6 except H_{2}. So, now we need to
evaluate H_{2}. 











For state
2, we know the values of two intensive
properties: P_{2} and V_{2}. Therfore, we can use
the Ammonia Tables or
the NIST Webbook to
evaluate any other properties of interest, in this case, H_{2}. 











We begin by determining
the phases present. 
At P = 750 kPa : 
V_{sat liq} 
0.0016228 
m^{3}/kg 







V_{sat vap} 
0.169798 
m^{3}/kg 

Since V_{2} > V_{sat vap}, state 2
is a superheated vapor. 












T (^{o}C) 
V (m^{3}/kg) 
H (kJ/kg) 







75 
0.21661 
1611.9 





T_{2} 
0.22156 
H_{2} 


T_{2} 
81.85 
^{o}C 


100 
0.23469 
1672.2 


H_{2} 
1628.4 
kJ/kg 











Finally, we can plug
values back into Eqn 5 to
evaluate Q and complete
this problem: 
















Q 
0.4586 
kW 

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










Answers : 
a.) 
(W_{s})_{int rev} 
7.06 
kW 







b.) 
Q 
0.459 
kW 










