Thermodynamics- Refrigerator Problem

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SUMMARY

The refrigerator problem involves a 0.25 hp motor operating to maintain an interior temperature of 2.00 degrees Celsius while the room temperature is 35 degrees Celsius. The coefficient of performance (COP) is calculated to be 5.68, which is 50% of the maximum theoretical value. The maximum heat leak that can be tolerated is determined using the equations for COP and heat transfer, leading to a calculated Qhot of 1246.16 watts. If the heat leak exceeds this value, the refrigerator will fail to maintain the desired temperature, resulting in potential overheating and inefficiency.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically the coefficient of performance (COP).
  • Familiarity with heat transfer equations, including Qcold and Qhot.
  • Basic knowledge of unit conversions, particularly between horsepower and watts.
  • Ability to perform calculations involving temperature in Kelvin.
NEXT STEPS
  • Study the principles of thermodynamics, focusing on the coefficient of performance (COP) in refrigeration cycles.
  • Learn about heat transfer calculations and their applications in real-world refrigeration systems.
  • Explore the effects of temperature variations on refrigeration efficiency and performance.
  • Investigate common issues in refrigeration systems, including heat leaks and their impact on system performance.
USEFUL FOR

This discussion is beneficial for students studying thermodynamics, engineers designing refrigeration systems, and anyone interested in understanding the efficiency of cooling mechanisms.

jessedevin
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Homework Statement


A refrigerator is operated by a 0.25 hp (1 hp=746 watts) motor. If the interior is to be maintained at 2.00 degrees Celsius and the room temperature of the room is 35 degrees C, what is the maximum heat leak in watts that can be tolerated? Assume that the coefficient of preformance is 50% of the maximum theoretical vale. What happens if the leak is greater than your calculated maximum value.

Homework Equations


\eta= Tcold/(Thot-Tcold)
\eta=Qcold/w
Qhot=Qcold+W.

The Attempt at a Solution


What I am first doing is finding the coefficient of performance \eta
\eta=375K/(408K-375K)=11.36
Then it says that the coefficient of preformance is 50% of the the max theoretical value, so \eta= 11.36/2=5.68
Then I find the rate of Qcold = w*\eta
Qcold=(746w/4)(5.68)=1059.66w
Lastly I used Qhot=Qcold+W,
Qhot=1059.66w+ 746w/4=1246.16w

But the answer in my book says its 779 watts, so did I miss a step or do something wrong. And what happens if the leak is greater than your calculated maximum value?
 
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First, check your conversion to Kelvin temperature. It is not correct.

Second, your method is correct except that you do not calculate Qh. You use Qc, since this is the heat that has to be removed.

What do you think happens if the refrigerator is not able to remove the heat as fast as it is entering the refrigerator?

AM
 

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