R134a vapor enters an adiabatic compressor at 20^{o}C and leaves at 700 kPa. If the R134a is saturated when it enters the compressor, determine the minimum shaft work required by the compressor in kJ/kg. 











Read : 
We must apply the 1st Law to the compressor. We can get H_{1} from the R134a tables or the NIST Webbook, but we do not know H_{2}. The key to solving
this problem is that a process that requires the minimum shaft work is an isentropic process. Knowing that S_{2} = S_{1} gives us the value of
a 2nd intensive variable for state 2. This allows us to use the R134a tables or NIST Webbook to detemine H_{2}. We can then plug H_{2} into the 1st Law to determine the work requirement per kg of R134a. 










Given: 
T_{1} 
20 
^{o}C 

Find: 

W_{S} 
??? 
kJ/kg 

x_{1} 
1 
kg vap/kg 







P_{2} 
700 
kPa 
















Diagram: 











Assumptions: 
1  
The compressor is isentropic. 


2  
The compressor operates at steadystate. 


3  
Changes in kinetic and potential energies are negligible. 


4  
Shaft
work and flow work are the only types of work that cross the system boundary. 










Equations
/ Data / Solve: 


















Apply the 1st Law to the compressor to determine the shaft work requirement. 











For a steadystate, singleinlet, single outlet system with no heat
transfer and negligible kinetic
and potential energy changes, the 1st Law is: 

















Eqn 1 











We can get H_{1} from the R134a tables or the NIST Webbook because we know the temperature and we know it is a saturated
vapor: 







H_{1} 
386.6 
kJ/kg 











The compressor is isentropic, so S_{2} = S_{1} and we can get S_{1} from the R134a tables or the NIST Webbook. 











S_{1} 
1.7413 
kJ/kgK 



S_{2} 
1.7413 
kJ/kgK 











Now, we know the
values of two intensive properties at state 2, so we can use the R134a tables or the NIST Webbook to evaluate any other properties by interpolation. Here, we are
interested in H_{2}. 











At P
= 700 kPa: 


















T (^{o}C) 
H (kJ/kg) 
S^{o}
(kJ/kgK) 







30 
416.60 
1.7269 







T_{2} 
H_{2} 
1.7413 



T_{2} 
34.39 
^{o}C 

40 
426.72 
1.7598 



H_{2} 
421.0 
kJ/kg 











Now, we can plug
values back into Eqn 1 : 


W_{S} 
34.49 
kJ/kg 










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










Answers : 
W_{S} 
34.5 
kJ/kg 















