Optimizing Refrigerator Efficiency: Solving for Necessary Power Draw

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SUMMARY

This discussion focuses on calculating the necessary power draw for a refrigerator experiencing a heat leak of 300W. Using the coefficient of performance (COP) formula, the maximum COP is determined to be 5.2 based on the operating temperatures of 298K and 250K. Consequently, the power required from the wall is calculated as 57.69W, confirming the efficiency of the refrigerator's operation. The discussion emphasizes the importance of algebraic manipulation for accuracy in calculations.

PREREQUISITES
  • Understanding of thermodynamics, specifically the concepts of heat transfer and efficiency.
  • Familiarity with the coefficient of performance (COP) in refrigeration systems.
  • Basic algebraic manipulation skills for solving equations.
  • Knowledge of temperature scales, particularly Kelvin.
NEXT STEPS
  • Research the principles of thermodynamics related to refrigeration cycles.
  • Explore advanced refrigeration efficiency metrics beyond COP.
  • Learn about real-world factors affecting refrigerator performance, such as insulation and ambient temperature.
  • Investigate the impact of different refrigerants on system efficiency.
USEFUL FOR

Engineers, HVAC professionals, and students studying thermodynamics or refrigeration systems will benefit from this discussion, particularly those interested in optimizing energy efficiency in cooling appliances.

patrickmoloney
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Homework Statement


Suppose that heat leaks into your kitchen refrigerator at an average rate of 300W. Assuming ideal operation, how much power must it draw from the wall.

Homework Equations


\eta = \dfrac{T_C}{T_H - T_C}

W= \dfrac{Q}{COP}

The Attempt at a Solution



A typical refrigerator works between 298 \, K 250 \, K. The maximum possible coefficient of performance is

COP = \dfrac{250}{298-250}=5.2

W = \dfrac{Q}{COP}= \dfrac{300 \, W}{5.2} = 57.69 \, W
 
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patrickmoloney said:

Homework Statement


Suppose that heat leaks into your kitchen refrigerator at an average rate of 300W. Assuming ideal operation, how much power must it draw from the wall.

Homework Equations


\eta = \dfrac{T_C}{T_H - T_C}

W= \dfrac{Q}{COP}

The Attempt at a Solution



A typical refrigerator works between 298 \, K 250 \, K. The maximum possible coefficient of performance is

COP = \dfrac{250}{298-250}=5.2

W = \dfrac{Q}{COP}= \dfrac{300 \, W}{5.2} = 57.69 \, W
Looks right.
The are many benefits in working algebraically, only plugging in numbers right at the end. One is improved accuracy. In the present case you would have got 57.6W exactly.
 
Yeah I think I'll do it algebraically so until the final line. Thanks very much.
 

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