Saturated
ammonia vapor at 10^{o}C is compressed in an insulated pistonandcylinder device until the pressure reaches 750 kPa. Assuming the process is internally
reversible, calculate the work for this process in kJ/kg. 











Read : 
The key to solving
this problem is to recognize that any process that is both adiabatic
and reversible is ISENTROPIC. This means that S_{2} = S_{1} and this allows you to
fix state 2 and evaluate U_{2}. Use U_{2} in the 1st Law to evaluate W. 










Given: 
T_{1} 
10 
^{o}C 


Find: 
W 
??? 
kJ/kg 

P_{2} 
750 
kPa 
















Diagram: 











Assumptions: 
1  
Process is internally reversible. 


2  
Changes in kinetic and potential energies are negligible. 


3  
Boundary
work is the only form of work that crosses the system boundary. 










Equations
/ Data / Solve: 


















Begin by applying the 1st Law to the process, assuming changes in kinetic and potential energies are negligible: 








Eqn 1 











The process is adiabatic so Eqn 1 can be simplified to : 




















Eqn 2 











Use the NIST Webbook to obtain properties
for state 1, saturated vapor at 10^{o}C : 











P_{1} 
290.71 
kPa 



U_{1} 
1309.9 
kJ/kg 







S_{1} 
5.4701 
kJ/kgK 











Because the process is
both reversible and adiabatic, it is isentropic. 











Therefore : 




S_{2} 
5.4701 
kJ/kgK 











At this point we know
values of two intensive variables for state 2, so we can use the NIST Webbook to determine the value
of any other property. In this case, we need U_{2}. First we need to
determine the phases
that exist at state 2. 











At P_{2} : 
T_{sat} 
54.05 
^{o}C 


S_{sat vap} 
4.7209 
kJ/kgK 







S_{sat liq} 
1.5744 
kJ/kgK 











Because S_{2} > S_{sat vap} at P_{2}, we can conclude that state 2 is a superheated vapor. We could have reached
the same conclusion after careful consideration of a TS Diagram. 











We can get the
following data from the superheated Ammonia Tables or from the NIST Webbook : 











At 750
kPa : 
T (^{o}C) 
S (kJ/kgK) 
U (kJ/kg) 







50 
5.4388 
1401.6 







T_{2} 
5.4701 
U_{2} 







75 
5.6233 
1449.4 















Interpolation yields : 



T_{2} 
54.25 
^{o}C 







U_{2} 
1409.73 
kJ/kg 











Now, we can plug
values back into Eqn 2 : 

W 
99.80 
kJ/kg 










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










Answers : 
W 
99.8 
kJ/kg 















