| 4F-1 : |
Heat and Work for a Cycle Carried Out in a Closed System |
6 pts |
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| A gas undergoes a thermodynamic cycle consisting of three quasi-equilibrium processes: |
| Process 1-2: constant volume, V = 0.028 m3, U2 - U1 = 26.4 kJ |
| Process 2-3: expansion with PV = constant, U3 = U2 |
| Process 3-1: constant pressure, P = 1.4 bar, W31 = -10.5 kJ |
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| Changes in kinetic and potential energies are negligible. |
| a.) |
Sketch the cycle on a PV Diagram. |
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| b.) |
Calculate the net work for the cycle, in kJ. |
| c.) |
Calculate the heat transfer for process 2-3, in kJ. |
| d.) |
Calculate the heat transfer for process 3-1, in kJ. |
| e.) |
Is this a power cycle or a refrigeration cycle ? |
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| Given: |
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V1 = V2 = |
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0.028 |
m3 |
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U2 - U1 = |
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26.4 |
kJ |
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Step 2-3 |
P2 V2 = P3 V3 |
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U3 = U2 |
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Step 3-1: |
P3 = P1 = |
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140 |
kPa |
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W31 = |
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-10.5 |
kJ |
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| Find: |
Sketch the cycle on a PV Diagram. |
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Wcycle = |
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kJ |
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Q23 = |
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kJ |
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Q31 = |
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kJ |
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Power or Refrigeration Cycle ? |
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| Assumptions: |
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- The gas is a closed system |
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- Boundary work is the only form of work interaction |
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- Changes in kinetic and potential energies are negligible. |
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| Part a.) |
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| Part b.) |
Since Wcycle = W12 + W23 + W31, we will work our way around the cycle and calculate each work term along the way. |
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Because the volume is constant in step 1-2: |
W12 = |
0 |
kJ |
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In step 2-3: P V = C, therefore, the definition of boundary work becomes: |
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Eqn 1 |
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But, we don't know V3 ! |
Perhaps we can use W31 to detemine V3. |
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Step 3-1 is isobaric, therefore, the definition of boundary work becomes: |
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Eqn 2 |
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Solve this equation for V3 : |
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Eqn 3 |
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V3 = |
0.103 |
m3 |
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Now, plug V3 and C = P3V3 into Eqn 1 to determine W23: |
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W23 = |
18.8 |
kJ |
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Sum the work terms for the three steps to get Wcycle: |
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Wcycle = |
8.28 |
kJ |
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| Part c.) |
Write the 1st Law for step 2-3: |
Q23 - W23 = U3 - U2 = 0 |
Eqn 4 |
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W23 = |
Q23 = |
18.8 |
kJ |
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| Part d.) |
Write the 1st Law for step 3-1: |
Q31 - W31 = U1 - U3 |
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Eqn 5 |
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But, U2 = U3 : |
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Q31 - W31 = U1 - U3 = U1 - U2 = - ( U2 - U1 ) |
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Eqn 6 |
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Solve for Q31 : |
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Q31 = W31 - ( U2 - U1 ) |
Eqn 7 |
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Plug in the given values : |
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Q31 = |
-36.9 |
kJ |
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| Part e.) |
First, we should determine Q12 from the 1st Law: |
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Q12 - W12 = U2 - U1 = 0 |
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Eqn 8 |
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Q12 = U2 - U1 |
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Eqn 9 |
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Q12 = |
26.4 |
kJ |
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Define: |
Qcycle = Q12 + Q23 + Q31 |
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Eqn 10 |
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Qcycle = |
8.28 |
kJ |
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Since Qcycle > 0 and Wcycle > 0, this is a power cycle ! |
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Notice that Qcycle = Wcycle because ΔUcycle = 0. |
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Thermal Efficiency is defined by : |
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Eqn 11 |
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hth |
18.3% |
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