Example Problem with Complete Solution

8A-4 : Entropy Change For R-134a Compression in Piston-and-Cylinder Device 6 pts
Four kilograms Saturated R-134a vapor at -20oC is compressed in a piston-and-cylinder device until the pressure reaches 500 kPa. 
During the process, 26.3 kJ of heat is lost to the surroundings, resulting in an increase in the specific entropy of the surroundings of 0.095 kJ/K. 
Assuming the process is completely reversible, calculate the work for this compression process in kJ.
Read : Use the 2nd Law and the fact that the process is completely reversible to determine S2. This gives you the second intensive property you need to evaluate U2. Then, use the 1st Law to determine Wb for the compression process.
Given: m 4 kg P2 500 kPa
x1 1 kg vap/kg Q -26.3 kJ
T1 -20 oC ΔSsurr 0.095 kJ/K
P1 132.73 kPa
Find: Wb ??? kJ
Assumptions: 1 - As shown in the diagram, the system is the R-134a inside the cylinder.
2 - Boundary work is the only form of work that crosses the system boundary.
3 - Changes in kinetic and potential energies are negligible.
4 - The compression process is completely reversible, so there is no entropy generated and no entropy change of the universe.
Equations / Data / Solve:
To determine the work required we need
to apply the
1st Law for closed systems:
Eqn 1
When we assume that changes in kinetic and potential energies are negligible and we assume boundary work is the only form of work, Eqn 1 becomes:
Eqn 2
We can now solve Eqn 2 for Wb :
Eqn 3
All we need to do is determine U2 and U1 and then we can use Eqn 3 to calculate Wb and complete this problem.
Start with U1 because we were given T1 and and it is a saturated vapor, so we can immediately look-up U1 in the Saturated R-134a Table.
U1 366.99 kJ/kg
We know P2, but we need to know the value of two intensive properties before we can use the R-134a Tables to look-up U2.
Because the process is completely reversible:
Eqn 4
In this case, the universe consists of two parts: the system and the surroundings. As a result, Eqn 4 becomes:
Eqn 5
We were given ΔSsurr, so we need to consider ΔSsys further in order to use Eqn 5.
We can express the entropy change of the system in terms of the initial and final states as follows.
Eqn 6
Next, we can combine Eqn 5 and Eqn 6 and solve for S2.
Eqn 7
Eqn 8
We can evaluate S1 because we were given T1 and and it is a saturated vapor.
S1 1.7413 kJ/kg-K
Now, we can plug values into Eqn 8 to evaluate S2: S2 1.7176 kJ/kg-K
This gives us the value of a second intensive property for state 2 which allows us to calculate U2.
At P = 500 kPa : Ssat liq 1.0759 kJ/kg-K Since Ssat liq < S2 < Ssat vap, state 2 is a saturated mixture.
Ssat vap 1.7197 kJ/kg-K
Determine x2 from the specific entropy, using:
Eqn 8
x2 0.9967 kg vap/kg
Then, we can use the
quality to determine H5S, using:
Eqn 9
At P = 20 psia : Usat liq 221.10 Btu/lbm
Usat vap 386.91 Btu/lbm U2 386.36 Btu/lbm
Now, we can use U2 in Eqn 3 to evaluate Wb and finish this problem. Wb -103.80 kJ
Verify: The assumptions made in the solution of this problem cannot be verified with the given information.
Answers : Wb -104 kJ