Calculating COP of Refrigerator: Solve with W=P(delta)t

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SUMMARY

The discussion focuses on calculating the Coefficient of Performance (COP) of a refrigerator using the formula COP = Qc/W. The scenario involves cooling 8.43 kg of water from 21.1°C to 4°C using a 0.1655 hp motor over a duration of 9 minutes. The work done by the motor is calculated as W = P(delta)t, resulting in 66670.02 J. The participants emphasize the need to determine the heat removed from the cold reservoir to finalize the COP calculation.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically the Coefficient of Performance (COP).
  • Familiarity with the concept of work and energy in physics.
  • Knowledge of horsepower conversion to watts (1 hp = 746 W).
  • Basic skills in unit conversion and calculations involving temperature changes.
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  • Calculate the heat removed (Qc) from the water during the cooling process.
  • Review the principles of thermodynamics related to refrigeration cycles.
  • Explore the implications of COP in energy efficiency for refrigeration systems.
  • Learn about different types of refrigerants and their impact on COP.
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Students in thermodynamics, engineers working with refrigeration systems, and anyone interested in energy efficiency calculations in cooling technologies.

jdog6
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A vessel containing 8.43 kg of water at 21.1C is put into a refrigerator. The 0.1655 hp motor runs for 9 min to cool the liquid to the refrigerator's low temperature 4C.

What is the COP of the refrigerator?

W=P(delta)t
= 123.463Wx540s
= 66670.02 J

COP = Qc/W

Please Help.

Thank you.
 
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jdog6 said:
A vessel containing 8.43 kg of water at 21.1C is put into a refrigerator. The 0.1655 hp motor runs for 9 min to cool the liquid to the refrigerator's low temperature 4C.
What is the COP of the refrigerator?
W=P(delta)t
= 123.463Wx540s
= 66670.02 J
COP = Qc/W
Please Help.
Thank you.
The Coefficient of Performance for a refrigerator is defined as the ratio of the heat removed from the cold reservoir to the work added to the system. What is the work or energy input? (I think you have correctly worked this out based on one hp = 746 J/sec). What is the heat removed (in J.) ?

AM
 
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