Example Problem with Complete Solution

7B-1 : Reversible Adiabatic Compression of R-134a 5 pts
Saturated ammonia vapor at -10oC is compressed in an insulated piston-and-cylinder 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 S2 = S1 and this allows you to fix state 2 and evaluate U2.  Use U2 in the 1st Law to evaluate W.
Given: T1 -10 oC Find: W ??? kJ/kg
P2 750 kPa
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 -10oC :
P1 290.71 kPa U1 1309.9 kJ/kg
S1 5.4701 kJ/kg-K
Because the process is both reversible and adiabatic, it is isentropic.
Therefore : S2 5.4701 kJ/kg-K
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 U2.  First we need to determine the phases that exist at state 2.
At P2 : Tsat 54.05 oC Ssat vap 4.7209 kJ/kg-K
Ssat liq 1.5744 kJ/kg-K
Because S2 > Ssat vap at P2, 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 (oC) S (kJ/kg-K) U (kJ/kg)
50 5.4388 1401.6
T2 5.4701 U2
75 5.6233 1449.4
Interpolation yields : T2 54.25 oC
U2 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