| 4F-3 : | Coefficient of Performance of a Refrigeration Cycle | 3 pts |
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| A refrigeration cycle has a COP of β = 2.5. For the cycle, QH = 2000 kJ. Determine QC and Wcycle, each in kJ. |  |
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| Read : | This one is a straightforward application of the definition of the the 1st Law and the efficiency of a power cycle. | ||||||||
| Given: | β = | 2.5 |
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| QH = | 2000 |
kJ | |||||||
| Find: | QC = | ??? |
kJ | ||||||
| Wcycle = | ??? |
kJ | |||||||
| Assumption: | - The cycle only exchanges heat with the two thermal reservoirs. | ||||||||
| Equations: | 1st Law applied to the refrigerator: |  |
Eqn 1 | ||||||
| Definition of COP for a refrigerator (b) : |  |
Eqn 2 | |||||||
| Degree of freedom analysis: 2 eqns in 2 unk | |||||||||
| Solve equations: | Solve Eqn 2 for QC and use the result to eliminate QC from Eqn 1 : | ||||||||
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Eqn 3 |  |
Eqn 4 | ||||||
| Next, solve Eqn 4 for Wcycle in terms of the known quantities QH and β. | |||||||||
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Eqn 5 | ||||||||
| Plug numbers into Eqn 5 : | Wcycle = | 571 |
kJ | ||||||
| Now, use this value for Wcycle and the given value of β in Eqn 3 to evaluate QC : | |||||||||
| QC = | 1429 |
kJ | |||||||
| Verify assumption: | The only assumption cannot be verified. | ||||||||
| Answer questions: | QC = | 1429 |
kJ | Wcycle = | 571 |
kJ | |||
Example Problem with Complete Solution
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