Heat pump problem with heat loss?

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SUMMARY

The discussion revolves around calculating the power used by a heat pump designed to maintain a house temperature of 70°F against an outside temperature of 5°F, with a total heat loss of 10,000 BTU/hr. The efficiency of the heat pump is stated as 45% of the theoretical maximum. The relevant equations include the coefficient of performance (COP) defined as e = Q_H/W and the ideal efficiency equation e = 1/(1-T_C/T_H). The challenge lies in integrating the heat loss into the calculations to determine the required work or power of the heat pump.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically heat pumps and refrigeration cycles.
  • Familiarity with the coefficient of performance (COP) and its calculations.
  • Knowledge of heat transfer concepts, particularly heat loss in buildings.
  • Basic proficiency in using BTU as a unit of energy measurement.
NEXT STEPS
  • Study the calculations for the coefficient of performance (COP) in heat pumps.
  • Learn about the theoretical maximum efficiency of heat pumps and how to apply it in practical scenarios.
  • Research methods to calculate heat loss in residential buildings, including insulation factors.
  • Explore the relationship between temperature differentials and energy consumption in heating systems.
USEFUL FOR

This discussion is beneficial for students studying thermodynamics, HVAC engineers, and anyone involved in designing or optimizing heating systems using heat pumps.

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Homework Statement


Consider a heat pump used to heat a house to 70°F when the outside temp is 5°F. The total heat loss from the house is 10000 BTU/hr. Efficiency of the heat pump is 45% of the theoretical maximum. What will be the power used by the pump?


Homework Equations


e (or cop) = Q_H/W
ideally, e = 1/(1-T_C/T_H)


The Attempt at a Solution



I am given to understand that a heat pump is a kind of refrigerator, which is a heat engine run backwards...what I don't understand is what to do with the heat loss I am given. I have absolutely no idea where the heat lost from the house would fit into the equations. Which leaves me stumped on where to even begin to start finding the work. Which is what I'm assuming is meant by "power"; however, I am not sure about that, either.
 
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The heat from the house is lost into the atmosphere. Consequently the heat pump has to pump as much heat back into the house to keep the temperature inside constant.
 
To maintain equilibrium (eg constant temperature) the heat coming in must equal the heat going out.
 

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