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The thermal efficiency of a reversible heat engine operating between two given thermal reservoirs is 0.4. The device is used either as a refrigerator or as a heat pump between the same reservoirs. Then the coefficient of performance as a refrigerator (COP)R and the co-efficient of performance as a heat pump (COP)HP are
- a)(COP)R = (COP)HP = 0.6
- b)(COP)R = 2.5; (COP)HP = 1.5
- c)(COP)R = 1.5; (COP)HP = 2.5
- d)(COP)R = (COP)HP = 2.5
Correct answer is option 'C'. Can you explain this answer?
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The thermal efficiency of a reversible heat engine operating between t...
The coefficient of performance (COP) is a measure of the efficiency of a refrigeration or heat pump system. It is defined as the ratio of the desired output to the required input. In this case, we need to determine the COP as a refrigerator (COP_R) and the COP as a heat pump (COP_HP) for a reversible heat engine with a given thermal efficiency of 0.4.
Reversible Heat Engine Efficiency:
The thermal efficiency of a reversible heat engine is given by the formula:
η = 1 - (Tc/Th)
where η is the thermal efficiency, Tc is the temperature of the cold reservoir, and Th is the temperature of the hot reservoir.
Given that the thermal efficiency is 0.4, we can rearrange the formula to solve for Tc/Th:
0.4 = 1 - (Tc/Th)
Tc/Th = 1 - 0.4
Tc/Th = 0.6
Coefficient of Performance as a Refrigerator:
The coefficient of performance as a refrigerator (COP_R) is given by the formula:
COP_R = Qc/W
where COP_R is the coefficient of performance as a refrigerator, Qc is the heat removed from the cold reservoir, and W is the work input.
In a refrigerator, the desired output is the heat removed from the cold reservoir. Therefore, Qc is positive, and W is the required input.
Since the heat removed from the cold reservoir is equal to the heat rejected to the hot reservoir, we can use the formula for the thermal efficiency to relate Qc and Qh:
Qc/Qh = Tc/Th
Qc = (Tc/Th) * Qh
Substituting this into the equation for COP_R, we get:
COP_R = (Tc/Th) * Qh/W
Since the thermal efficiency is equal to 0.4, we can substitute Tc/Th = 0.6:
COP_R = 0.6 * Qh/W
Coefficient of Performance as a Heat Pump:
The coefficient of performance as a heat pump (COP_HP) is given by the formula:
COP_HP = Qh/W
In a heat pump, the desired output is the heat supplied to the hot reservoir. Therefore, Qh is positive, and W is the required input.
Using the equation for the thermal efficiency, we can relate Qc and Qh:
Qc/Qh = Tc/Th
Since Qc is equal to the heat removed from the cold reservoir, we can substitute Qc = COP_R * W:
COP_R * W/Qh = Tc/Th
Solving for Qh, we get:
Qh = (Th/Tc) * COP_R * W
Substituting this into the equation for COP_HP, we get:
COP_HP = (Th/Tc) * COP_R
Given that COP_R = 1.5, we can substitute this value into the equation for COP_HP:
COP_HP = (Th/Tc) * 1.5
Since Th/Tc = 1/(Tc/Th) = 1/0.6 = 1.6667, we can calculate COP_HP:
COP_HP = 1.6667 * 1.5 = 2.5
Therefore, the correct answer
Reversible Heat Engine Efficiency:
The thermal efficiency of a reversible heat engine is given by the formula:
η = 1 - (Tc/Th)
where η is the thermal efficiency, Tc is the temperature of the cold reservoir, and Th is the temperature of the hot reservoir.
Given that the thermal efficiency is 0.4, we can rearrange the formula to solve for Tc/Th:
0.4 = 1 - (Tc/Th)
Tc/Th = 1 - 0.4
Tc/Th = 0.6
Coefficient of Performance as a Refrigerator:
The coefficient of performance as a refrigerator (COP_R) is given by the formula:
COP_R = Qc/W
where COP_R is the coefficient of performance as a refrigerator, Qc is the heat removed from the cold reservoir, and W is the work input.
In a refrigerator, the desired output is the heat removed from the cold reservoir. Therefore, Qc is positive, and W is the required input.
Since the heat removed from the cold reservoir is equal to the heat rejected to the hot reservoir, we can use the formula for the thermal efficiency to relate Qc and Qh:
Qc/Qh = Tc/Th
Qc = (Tc/Th) * Qh
Substituting this into the equation for COP_R, we get:
COP_R = (Tc/Th) * Qh/W
Since the thermal efficiency is equal to 0.4, we can substitute Tc/Th = 0.6:
COP_R = 0.6 * Qh/W
Coefficient of Performance as a Heat Pump:
The coefficient of performance as a heat pump (COP_HP) is given by the formula:
COP_HP = Qh/W
In a heat pump, the desired output is the heat supplied to the hot reservoir. Therefore, Qh is positive, and W is the required input.
Using the equation for the thermal efficiency, we can relate Qc and Qh:
Qc/Qh = Tc/Th
Since Qc is equal to the heat removed from the cold reservoir, we can substitute Qc = COP_R * W:
COP_R * W/Qh = Tc/Th
Solving for Qh, we get:
Qh = (Th/Tc) * COP_R * W
Substituting this into the equation for COP_HP, we get:
COP_HP = (Th/Tc) * COP_R
Given that COP_R = 1.5, we can substitute this value into the equation for COP_HP:
COP_HP = (Th/Tc) * 1.5
Since Th/Tc = 1/(Tc/Th) = 1/0.6 = 1.6667, we can calculate COP_HP:
COP_HP = 1.6667 * 1.5 = 2.5
Therefore, the correct answer
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The thermal efficiency of a reversible heat engine operating between two given thermal reservoirs is 0.4. The device is used either as a refrigerator or as a heat pump between the same reservoirs. Then the coefficient of performance as a refrigerator (COP)R and the co-efficient of performance as a heat pump (COP)HP area)(COP)R = (COP)HP = 0.6b)(COP)R = 2.5; (COP)HP = 1.5c)(COP)R = 1.5; (COP)HP = 2.5d)(COP)R = (COP)HP = 2.5Correct answer is option 'C'. Can you explain this answer?
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The thermal efficiency of a reversible heat engine operating between two given thermal reservoirs is 0.4. The device is used either as a refrigerator or as a heat pump between the same reservoirs. Then the coefficient of performance as a refrigerator (COP)R and the co-efficient of performance as a heat pump (COP)HP area)(COP)R = (COP)HP = 0.6b)(COP)R = 2.5; (COP)HP = 1.5c)(COP)R = 1.5; (COP)HP = 2.5d)(COP)R = (COP)HP = 2.5Correct answer is option 'C'. Can you explain this answer? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about The thermal efficiency of a reversible heat engine operating between two given thermal reservoirs is 0.4. The device is used either as a refrigerator or as a heat pump between the same reservoirs. Then the coefficient of performance as a refrigerator (COP)R and the co-efficient of performance as a heat pump (COP)HP area)(COP)R = (COP)HP = 0.6b)(COP)R = 2.5; (COP)HP = 1.5c)(COP)R = 1.5; (COP)HP = 2.5d)(COP)R = (COP)HP = 2.5Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The thermal efficiency of a reversible heat engine operating between two given thermal reservoirs is 0.4. The device is used either as a refrigerator or as a heat pump between the same reservoirs. Then the coefficient of performance as a refrigerator (COP)R and the co-efficient of performance as a heat pump (COP)HP area)(COP)R = (COP)HP = 0.6b)(COP)R = 2.5; (COP)HP = 1.5c)(COP)R = 1.5; (COP)HP = 2.5d)(COP)R = (COP)HP = 2.5Correct answer is option 'C'. Can you explain this answer?.
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