| An industrial refrigerator rejects heat at a rate of 24,750 kJ/min to the surroundings. If the refrigeration cycle has a COP of b = 3.3, determine QC and Wcycle, each in kJ/min. |
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| Read : |
This one is a
straightforward application of the definition of the the 1st
Law and the COP of a refrigeration cycle. |
Diagram : |
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| Given: |
COP = b |
3.3 |
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Find: |
QC |
??? |
kJ/min |
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QH |
24,750 |
kJ/min |
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Wcycle |
??? |
kJ/min |
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| Assumption: |
- The cycle only exchanges heat with the two thermal reservoirs. |
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| Equations
/ Data / Solve: |
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1st
Law applied to the refrigerator: |
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Eqn 1 |
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Definition of COP for a refrigerator : |
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Eqn 2 |
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Degree
of freedom analysis: 2
eqns in 2 unknowns: QC and Wcycle. |
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Solve Eqn 2 for QC and use the result to eliminate QC from Eqn 1 : |
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Eqn 3 |
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Eqn 4 |
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Next, solve Eqn 4 for Wcycle in terms of the known quantities QH and b. |
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Eqn 5 |
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Plug numbers into Eqn 5 : |
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Wcycle |
5755.8 |
kJ/min |
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Now, use this value
for Wcycle and the given value of b in Eqn 3
to evaluate QC : |
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QC |
18994.2 |
kJ/min |
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| Verify: |
The only assumption
cannot be verified. |
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| Answers : |
QC
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19000 |
kJ/min |
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Wcycle = |
5760 |
kJ/min |
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