Example Problem with Complete Solution

7E-2 : 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.) V2/V1 = V3/V4
b.) (T2 / T3)g = (P2 / P3)(g-1)
c.) T2 / T3 = (V3 / V2)(g-1)

Read : (a) Note that h = Wcycle/Qin = (W12 + W34)/Q12 because Q23 and Q41 are equal and have opposite signs. Determine W12 in terms of V1, V2, and TH (temperature of high temp reservoir) and also W34 in terms of V3, V4, and TC. Note the relationship between Q12 and W12 determined from an energy balance during step 1-2. 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 2-3. Note that dU = m CV dT, CV = (R/MW)/(g-1), 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: Q23 0 kJ Carnot Cycle T1 = T2 = TH
Q41 0 kJ T3 = T4 = TC
Step: 1-2 Isothermal Expansion
3-4 Isothermal Compression
Find: Show that: (a) V4V2 = V1V3 (b) T2/T3 = (P2/P3)((g-1)/g (c) T2/T3 = (V3/V2)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 W12 in terms of V1, V2, and TH and also W34 in terms of V3, V4, and TC.
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 Q12:
Eqn 7
Since the internal energy of an ideal gas depends on temperature only and the temperature is constant along Process 1-2, U2 = U1 and the energy balance reduces to:
Eqn 8
 We already determined W12 in Eqn 4.
Substituting expressions for W12, W23 and Q12 = W12 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 2-3 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 (T3 / T2)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.)