Consider the Carnot Power Cycle shown in the PV Diagram, below. The working fluid is air and the specific heat ratio, g, is constant. |
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Show that…
a.) V2/V1 = V3/V4
b.) (T2 / T3)g = (P2 / P3)(g-1)
c.) T2 / T3 = (V3 / V2)(g-1) |
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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. |
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(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. |
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(b) Just use the result from part (c) along with the ideal gas EOS to convert V to P. |
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Given: |
Q23 |
0 |
kJ |
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Carnot Cycle |
T1 = T2 = TH |
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Q41 |
0 |
kJ |
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T3 = T4 = TC |
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Step: |
1-2 |
Isothermal Expansion |
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2-3 |
Adiabatic Exapansion |
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3-4 |
Isothermal Compression |
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4-1 |
Adiabatic Compression |
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Find: |
Show that: |
(a) V4V2 = V1V3 |
(b) T2/T3 = (P2/P3)((g-1)/g |
(c) T2/T3 = (V3/V2)g-1 |
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Diagram: |
Given in the problem
statement. |
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Assumptions: |
1 - |
The system consists of an ideal gas. |
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2 - |
The specific heat ratio is constant (required in part (b) only). |
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3 - |
The cycle is executed in a closed system ( not required, but it makes the solution simpler). |
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4 - |
Changes in kinetic and potential energies are negligible. |
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5 - |
Boundary
work is the only type of work that crosses the system boundary. |
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6 - |
The system undergoes a Carnot cycle (reversible). |
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Equations
/ Data / Solve: |
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Part a.) |
It may be hard to
determine where to start with the proof but following the provided hints will help you. |
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Starting with the thermal effiicency: |
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Eqn 1 |
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Now we need to
determine W12 in terms of V1, V2, and TH and also W34 in terms of V3, V4, and TC. |
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PV
work done by the system during the isothermal expansion and compression processes can be evaluated as follows: |
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Eqn 2 |
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Assuming the system
consists of an ideal gas
substitute P = nRT/V: |
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Eqn 3 |
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Eqn 5 |
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Integrating: |
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Eqn 4 |
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Eqn 6 |
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Apply the 1st Law for a closed system with negligible changes in kinetic and potential energies to get Q12: |
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Eqn 7 |
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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: |
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Eqn 8 |
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We already determined W12 in Eqn 4. |
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Substituting
expressions for W12, W23 and Q12 = W12 into the thermal efficiency equation, Eqn 1, yields: |
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Eqn 9 |
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Recall that the thermal efficiency of a Carnot Cycle is: |
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Eqn 10 |
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Substituting Eqn 10 into Eqn 9 yields: |
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Eqn 11 |
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Eqn
11 simplifies to: |
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Eqn 12 |
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A little algebra
finishes the job: |
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Eqn 13 |
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Eqn 14 |
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Eqn 15 |
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Part c.) |
First we will apply
the 1st Law to adiabatic process
2-3 with no changes
in kinetic or potential energy. |
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Eqn 16 |
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Put Eqn 16 into differential form: |
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Eqn 17 |
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Substitute the definitions of boundary work and heat capacity: |
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Eqn 18 |
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Eqn 19 |
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But, for an ideal gas: |
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Eqn 20 |
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And the ideal gas EOS tells us that: |
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Eqn 21 |
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Now, plug Eqns 18 - 20 back into Eqn 17 to get : |
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Eqn 22 |
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Rearrange Eqn 22 to get : |
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Eqn 23 |
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Integrate Eqn 23 from
state 2 to state 3 : |
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Eqn 24 |
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Simplify algebraically : |
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Eqn 25 |
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Eqn 26 |
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Part b.) |
Substitute the ideal gas EOS in the form: |
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Eqn 27 |
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into the result from part (c), Eqn
26 to get: |
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Eqn 28 |
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Cancelling terms in Eqn 28 yields: |
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Eqn 29 |
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Multiply through by (T3 / T2)1-g to get : |
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Eqn 30 |
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A little more algebra
yields : |
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Eqn 31 |
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Verify: |
The assumptions made
in this solution cannot be verified with the given information. |
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Answers : |
Part a.) |
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Part b.) |
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Part c.) |
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