7D3 :  Work, Efficiency and the TS Diagram for an Ideal Gas Power Cycle  8 pts 

Air
contained in a pistonandcylinder device undergoes a power cycle made up of three internally reversible processes. Step 12: Adiabatic compression from 20 psia and 570^{o}R to 125 psia 

Step 23: Isothermal expansion to 20 psia Step 31: Isobaric compression a.) Sketch the process path for this power cycle on both PV and TS diagrams b.) Calculate T_{3} in ^{o}F 

c.) Calculate the boundary work in Btu/lb_{m} d.) Calculate the thermal efficiency of the power cycle 

Read :  For part (a) sketch the cycle first to get a better understanding of the processes.  
For part (b) recall that the Process 23 is isothermal and therefore T_{3} = T_{2}. Determine S^{o}(T_{2}) and look it up in the Ideal Gas Entropy Table for air to determine T_{2}.  
For part (c) determine the net work by determining the work for each process and then adding them together.  
For part (d) determine the thermal efficiency as the ratio of the net work to the heat going into the system.  
Given:  T_{1}  570  ^{o}R  P_{2}  125  psia  
P_{1}  20  psia  P_{3}  20  psia  
^{}  
Find:  Part (a)  Sketch PV and TS diagrams  Part (c)  W_{cycle}  ?  Btu/lb_{m}  
Part (b)  T_{3}  ?  ^{o}R  Part (d)  h  ?  
^{}  
Diagram: 





Assumptions:  1   The system is the air inside the cylinder.  
2   The air is modeled as an ideal gas.  
3   Each process is internally reversible.  
4   Boundary work is the only form of work that crosses the system boundary.  
5   There is no change in kinetic or potential energy for either of the two processes.  
Equations / Data / Solve:  
Part b.)  Since Process 23 is isothermal: 

Eqn 1  
So we should work on determining T_{2}.  
Let's apply the 2nd Gibbs
Equation for ideal gases to Process 12 : 

Eqn 2  
We can determine T_{2} and thus T_{3} from: 

Eqn 3  
Lookup S^{o}(T_{1}) in the Ideal Gas Entropy Tables:  S^{o}(T_{1})  0.014381  Btu/lb_{m}^{o}R  
Now, plug values into Eqn 3 to determine S^{o}(T_{2}):  
R  1.987  Btu/lbmol^{o}R  S^{o}(T_{2})  0.14007  Btu/lb_{m}^{o}R  
MW  28.97  lb_{m}/lbmol  
Now that we know the value of S^{o} at T_{2}, we can interpolate on the air Ideal Gas Property Table to determine T_{2}.  
T (^{o}R)  S^{o} (Btu/lb_{m}^{o}R)  
950  0.13939  
T_{2}  0.14007  Interpolation yields :  T_{2} = T_{3} =  952.60  ^{o}R  
960  0.14202  492.93  ^{o}F  
Part c.)  The net work is the sum of the work done during each process: 

Eqn 4  
We need to determine the work involved in each process. Begin with Process 12.  
Since Process 12 is adiabatic, and changes in internal and kinetic energies are negligible, the appropriate form of the 1st Law is :  

Eqn 5 

Eqn 6  
Because we know both T_{1} and T_{2}, we can look up the U's in the Ideal Gas Property Table:  
At T_{1} = 570 ^{o}R, no interpolation is required:  U_{1}  5.6705  Btu/lb_{m}  
T (^{o}R)  U^{o }(Btu/lb_{m})  
950  72.806  
952.60  U_{2}  Interpolation yields :  U_{2}  73.281  Btu/lb_{m}  
960  74.631  
Now, we can plug values back into Eqn 6 to determine W_{12} :  W_{12}  67.610  Btu / lb_{m}  
Boundary work done by the system during Process 23 can be calculated from the definition of boundary work :  

Eqn 7  
For an ideal gas substitute P = nRT/V into Eqn 7 : 

Eqn 8  
Integrate Eqn 2 (the process is isothermal, T_{2} = T_{3} ) : 

Eqn 9  
Since P_{2}V_{2} = nRT_{2} and P_{3}V_{3} = nRT_{3} and T_{2} = T_{3}, we conclude that P_{2} V_{2} = P_{3} V_{3} , or : 

Eqn 10  
Combining Eqn 10 and Eqn 9 yields : 

Eqn 11  
We can plug numbers into Eqn 11 to evaluate W_{23} :  W_{23}  119.74  Btu / lb_{m}  
Boundary work done by the system during Process 31 can be calculated from the definition of boundary work :  

Eqn 12  
Since Process 31 is isobaric (P=constant), Eqn 12 simplifies to : 

Eqn 13  
Since : 

Eqn 14 

Eqn 15  
And since P_{1} = P_{3}, Eqns 13, 14 & 15 can be combined to obtain : 

Eqn 16  
Now, we can plug values into Eqn 16 to evaluate W_{31} :  W_{31}  26.24  Btu / lb_{m}  
Now, we can calculate W_{cycle} from the sum of the work terms for each step, using Eqn 4 :  
W_{cycle}  25.88  Btu / lb_{m}  
Part d.)  The thermal efficiency of the cycle is defined by : 

Eqn 17  
We know W_{cycle}, so we need to determine Q_{in}. We also know that Q_{12} = 0 (adiabatic process) and from the TS Diagram it can be concluded that Q_{23} > 0 and Q_{31} < 0. Therefore, Q_{in} = Q_{23}.  

Eqn 18  
Now we need to determine the heat transferred into the cycle during Process 23. Start from the definition of entropy :  

Eqn 19  
Because Process 23 is internally reversible, we can integrate Eqn 19 to get: 

Eqn 20  
Now, because Process 23 is isothermal, the T pops out of the integral and Eqn 20 is easy to integrate: 

Eqn 21  
Now, we can again apply the 2nd Gibbs Equation for ideal gases to Process 23 to evaluate DS:  

Eqn 22  
Since the process is isothermal: S^{o}(T_{2}) = S^{o}(T_{3}) and Eqn 22 simplifies to: 

Eqn 23  
When we substitute Eqn 23 into Eqn 21 we get : 

Eqn 24  
DS_{23}  0.12569  Btu/lb_{m}^{o}R  Q_{23}  119.74  Btu / lb_{m}  
Finally, we plug values back into Eqn 18 to evaluate the thermal efficiency of the cycle :  
h  21.62%  
Verify:  The ideal gas assumption needs to be verified.  
We need to determine
the specific volume at each state and check if: 


Air can be considered a diatomic gas.  
Solving the Ideal Gas EOS for molar volume yields : 


Use :  R  10.7316  psiaft^{3} / lbmol^{o}R  
V_{1}  305.85  ft^{3}/lbmol  
V_{2}  81.78  ft^{3}/lbmol  V_{3}  511.15  ft^{3}/lbmol  
The specific volume is greater than 80 ft^{3}/lbmol for all states so the ideal gas assumption is valid.  
Answers :  a.)  See diagrams above.  c.)  W_{cycle}  25.9  Btu/lb_{m}  
b.)  T_{3}  953  ^{o}R  d.)  h  21.6% 