8A4 :  Entropy Change For R134a Compression in PistonandCylinder Device  6 pts 

Four kilograms Saturated R134a vapor at 20^{o}C is compressed in a pistonandcylinder 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 S_{2}. This gives you the second intensive property you need to evaluate U_{2}. Then, use the 1st Law to determine W_{b} for the compression process.  
Diagram: 


Given:  m  4  kg  P_{2}  500  kPa  
x_{1}  1  kg vap/kg  Q  26.3  kJ  
T_{1}  20  ^{o}C  ΔS_{surr}  0.095  kJ/K  
P_{1}  132.73  kPa  
Find:  W_{b}  ???  kJ  
Assumptions:  1   As shown in the diagram, the system is the R134a 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 W_{b} : 

Eqn 3  
All we need to do is determine U_{2} and U_{1} and then we can use Eqn 3 to calculate W_{b} and complete this problem.  
Start with U_{1} because we were given T_{1} and and it is a saturated vapor, so we can immediately lookup U_{1} in the Saturated R134a Table.  
U_{1}  366.99  kJ/kg  
We know P_{2}, but we need to know the value of two intensive properties before we can use the R134a Tables to lookup U_{2}.  
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 ΔS_{surr}, so we need to consider ΔS_{sys} 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 S_{2}.  

Eqn 7 

Eqn 8  
We can evaluate S_{1} because we were given T_{1} and and it is a saturated vapor.  
S_{1}  1.7413  kJ/kgK  
Now, we can plug values into Eqn 8 to evaluate S_{2}:  S_{2}  1.7176  kJ/kgK  
This gives us the value of a second intensive property for state 2 which allows us to calculate U_{2}.  
At P = 500 kPa :  S_{sat liq}  1.0759  kJ/kgK  Since S_{sat liq} < S_{2} < S_{sat vap}, state 2 is a saturated mixture.  
S_{sat vap}  1.7197  kJ/kgK  
Determine x_{2} from the specific entropy, using: 

Eqn 8  
x_{2}  0.9967  kg vap/kg  
Then, we can use
the quality to determine H_{5S}, using: 

Eqn 9  
At P = 20 psia :  U_{sat liq}  221.10  Btu/lb_{m}  
U_{sat vap}  386.91  Btu/lb_{m}  U_{2}  386.36  Btu/lb_{m}  
Now, we can use U_{2} in Eqn 3 to evaluate W_{b} and finish this problem.  W_{b}  103.80  kJ  
Verify:  The assumptions made in the solution of this problem cannot be verified with the given information.  
Answers :  W_{b}  104  kJ 