7E2 :  PVT Relationships for Isentropic, IG Processes  8 pts 

Consider the Carnot Power Cycle shown in the PV Diagram, below. The working fluid is air and the specific heat ratio, g, is constant.  


Show that… a.) V_{2}/V_{1} = V_{3}/V_{4} b.) (T_{2} / T_{3})^{g} = (P_{2} / P_{3})^{(}^{g}^{1)} c.) T_{2} / T_{3} = (V_{3} / V_{2})^{(}^{g}^{1)} 

Read :  (a) Note that h = W_{cycle}/Q_{in} = (W_{12} + W_{34})/Q_{12} because Q_{23} and Q_{41} are equal and have opposite signs. Determine W_{12} in terms of V_{1}, V_{2}, and T_{H} (temperature of high temp reservoir) and also W_{34} in terms of V_{3}, V_{4}, and T_{C}. Note the relationship between Q_{12} and W_{12} determined from an energy balance during step 12. Compare h obtained this way with the Carnot cycle efficicency and you will arrive at the desired conclusion.  
(c) Easier to do part (c) before (b). Apply the 1st Law to step 23. Note that dU = m C_{V} dT, C_{V} = (R/MW)/(g1), PV = nRT and dW = P dV. You will arrive at the form (1/T) dT and (1/V) dV on both sides. Integrate to obtain the desired result.  
(b) Just use the result from part (c) along with the ideal gas EOS to convert V to P.  
Given:  Q_{23}  0  kJ  Carnot Cycle  T_{1} = T_{2} = T_{H}  
Q_{41}  0  kJ  T_{3} = T_{4} = T_{C}  
Step:  12  Isothermal Expansion  
23  Adiabatic Exapansion  
34  Isothermal Compression  
41  Adiabatic Compression  
Find:  Show that:  (a) V_{4}V_{2} = V_{1}V_{3}  (b) T_{2}/T_{3} = (P_{2}/P_{3})^{((}^{g}^{1)/}^{g}  (c) T_{2}/T_{3} = (V_{3}/V_{2})^{g}^{1}  
Diagram:  Given in the problem statement.  
Assumptions:  1   The system consists of an ideal gas.  
2   The specific heat ratio is constant (required in part (b) only).  
3   The cycle is executed in a closed system ( not required, but it makes the solution simpler).  
4   Changes in kinetic and potential energies are negligible.  
5   Boundary work is the only type of work that crosses the system boundary.  
6   The system undergoes a Carnot cycle (reversible).  
Equations / Data / Solve:  
Part a.)  It may be hard to determine where to start with the proof but following the provided hints will help you.  
Starting with the thermal effiicency:  

Eqn 1  
Now we need to determine W_{12} in terms of V_{1}, V_{2}, and T_{H} and also W_{34} in terms of V_{3}, V_{4}, and T_{C}.  
PV work done by the system during the isothermal expansion and compression processes can be evaluated as follows:  

Eqn 2  
Assuming the system consists of an ideal gas substitute P = nRT/V:  

Eqn 3 

Eqn 5  
Integrating:  

Eqn 4 

Eqn 6  
Apply the 1st Law for a closed system with negligible changes in kinetic and potential energies to get Q_{12}:  

Eqn 7  
Since the internal energy of an ideal gas depends on temperature only and the temperature is constant along Process 12, U_{2} = U_{1} and the energy balance reduces to:  

Eqn 8  
We already determined W_{12} in Eqn 4.  
Substituting expressions for W_{12}, W_{23} and Q_{12} = W_{12} into the thermal efficiency equation, Eqn 1, yields:  

Eqn 9  
Recall that the thermal efficiency of a Carnot Cycle is:  

Eqn 10  
Substituting Eqn 10 into Eqn 9 yields:  

Eqn 11  
Eqn 11 simplifies to: 

Eqn 12  
A little algebra finishes the job: 

Eqn 13  

Eqn 14  

Eqn 15  
Part c.)  First we will apply the 1st Law to adiabatic process 23 with no changes in kinetic or potential energy.  

Eqn 16  
Put Eqn 16 into differential form: 

Eqn 17  
Substitute the definitions of boundary work and heat capacity:  

Eqn 18  

Eqn 19  
But, for an ideal gas: 

Eqn 20  
And the ideal gas EOS tells us that: 

Eqn 21  
Now, plug Eqns 18  20 back into Eqn 17 to get :  

Eqn 22  
Rearrange Eqn 22 to get : 

Eqn 23  
Integrate Eqn 23 from state 2 to state 3 :  

Eqn 24  
Simplify algebraically : 

Eqn 25  

Eqn 26  
Part b.)  Substitute the ideal gas EOS in the form: 

Eqn 27  
into the result from part (c), Eqn 26 to get: 

Eqn 28  
Cancelling terms in Eqn 28 yields: 

Eqn 29  
Multiply through by (T_{3} / T_{2})^{1}^{g} to get : 

Eqn 30  
A little more algebra yields : 

Eqn 31  
Verify:  The assumptions made in this solution cannot be verified with the given information.  
Answers :  Part a.) 


Part b.) 

Part c.) 

