8A5 :  Entropy Production for the Adiabatic Compression of Air  6 pts 

Air is compressed in an adiabatic pistonandcylinder device, as shown below, from 200 kPa and 360 K to 800 kPa.  


a.) Calculate the final temperature, T_{2}, and the boundary work if
the process is internally reversible. 

b.) Calculate T_{2} and the entropy generation if a real pistonandcylinder device requires 15% more work than the internally reversible device.  
Read :  Assume ideal gas behavoir for air. Apply an energy balance and an entropy balance.  
Notice that in part (a) the problem asks for the "work required", therefore our answer will be positive.  
To get S and U data, use the 2nd Gibbs Equation in terms of the Ideal Gas Entropy Function.  
Given:  T_{1}  360  K  m  2.9  kg  
P_{1}  200  kPa  Q  0  KJ  
P_{2}  800  kPa  Part (b)  W_{part (b) }=  15%  > W_{part (a)}  
Find:  Part (a)  T_{2S}  ???  K  Part (b)  T_{2}  ???  K  
W_{b}  ???  kJ  S_{gen}  ???  kJ/K  
W_{lost}  ???  kJ  
^{}  
Diagram: 


Assumptions:  1   As shown in the diagram, the system is the air inside the cylinder.  
2   Air is modeled as an ideal gas.  
3   No heat transfer occurs.  
4   Boundary work is the only form of work that crosses the system boundary.  
5   Changes in kinetic and potential energies are negligible.  
6   For Part (a), there is no entropy generated.  
Equations / Data / Solve:  
Part a.)  To determine the work required we need to apply the 1st Law for closed systems: 

Eqn 1  
Because the process is adiabatic and 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  or: 

Eqn 3  
So, in order to answer part (a), we need to determine U for both the initial and final states.  
Use the Ideal Gas Property Table for air to evaluate U_{1} and U_{2}, but first we must know T_{1} and T_{2}.  
Because this process is both adiabatic and internally reversible, the process is isentropic.  
In this problem, we have air and we assume it behaves as an ideal gas.  
We can solve this problem using the ideal gas entropy function.  
The 2nd Gibbs Equation in terms of the S^{o} is: 

Eqn 4  
Since part (a) is an isentropic process, Eqn 4 becomes: 

Eqn 5  
The final temperature, T_{2}, can be determined from determining S^{o}(T_{2}) and then interpolating on the Ideal Gas Properties Table for air.  
Solving Eqn 5 for S^{o}(T_{2}) yields: 

Eqn 6  
Properties for state 1 are determined from Ideal Gas Properties Table for air.  
S^{o}(T_{1})  0.189360  kJ/kgK  
U(T_{1})  44.3940  kJ/kg  
Now, we can plug values into Eqn 3:  R  8.314  kJ/kmol K  
MW  28.97  kg/kmol  
S^{o}(T_{2})  0.58721  kJ/kgK  
Now, we can go back to the Ideal Gas Properties Table for air and determine T_{2} and U_{2} by interpolation.  
T (K)  U^{o} kJ/kg  S^{o} kJ/kgK  
520  163.42  0.56803  
T_{2}  U_{2}  0.58721  T_{2}  529.59  K  
530  171.04  0.58802  U_{2}  170.73  kJ/kg  
Put values into Eqn 3 to finish this part of the problem:  W_{b}  366.38  kJ  
Part b.)  The actual work is 15% greater than the work determined in Part (a): 

Eqn 7  
W_{b}  421.33  kJ  
In this part of the problem, we know the actual work, but we don't know T_{2} or U_{2}.  
We can solve Eqn 3 for U_{2} interms of the known variables m, W_{b} and U_{1} : 

Eqn 8  
Plugging values into Eqn 7 yields :  U_{2}  189.68  kJ/kg  
Next, we can determine T_{2} and S^{o}(T_{2}) by interpolating on the Ideal Gas Properties Table for air.  
U^{o} kJ/kg  T (K)  S^{o} kJ/kgK  
186.36  550  0.62702  
189.68  T_{2}  S^{o}(T_{2})  T_{2}  554.3  K  
194.05  560  0.64605  S^{o}(T_{2})  0.63524  kJ/kgK  
The 2nd Law in terms of the entropy generated is: 

Eqn 9  
Since there is no heat transfer in this problem: 

Eqn 10  
The entropy change can be determined from Eqn 4: 

Eqn 11  
Now, we can plug values into Eqns 9 & 10 to determine the entropy generated:  
S_{gen}  0.04803  kJ/K  
Verify:  The ideal gas assumption needs to be verified. 

Eqn 12  
We need to determine the specific volume at each state and check if : 

Eqn 13  
V_{1}  14.97  L/mol  
V_{2A}  5.50  L/mol  V_{2B}  5.76  L/mol  
The specific volume at each state is greater than 5 L/mol for all states and the working fluid can be treated as a diatomic gas, so the ideal gas assumption is valid.  
Answers :  Part a.)  T_{2S}  530  K  Part b.)  T_{2}  554  K  
W_{b}  366  kJ  S_{gen}  0.0480  kJ/K  