What is the power input (in kW) needed to run the heat pump?

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SUMMARY

The discussion focuses on calculating the power input required to operate a heat pump maintaining a temperature of 23°C indoors while the outdoor temperature is -6.0°C. The heat pump delivers 8.8 kW of heating power with a coefficient of performance (COP) of 5.6. The calculated power input needed to run the heat pump is 1.57 kW. For an ideal heat pump scenario, the required power input is 0.86 kW after correcting temperature conversions to Kelvin.

PREREQUISITES
  • Understanding of heat pump operations and thermodynamics
  • Familiarity with the coefficient of performance (COP)
  • Knowledge of temperature conversion from Celsius to Kelvin
  • Basic proficiency in algebraic manipulation of equations
NEXT STEPS
  • Study the principles of thermodynamics related to heat pumps
  • Learn about the calculations involved in determining the coefficient of performance (COP)
  • Explore ideal versus real heat pump performance metrics
  • Investigate temperature conversion methods and their significance in thermodynamic calculations
USEFUL FOR

Students in engineering or physics, HVAC professionals, and anyone involved in the design or analysis of heating systems will benefit from this discussion.

spazmsfreak
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1. A heat pump is being used to keep the temperature inside a house at 23 C, while the tmeperature ouside is -6.0 C. The heat pump is providing heat inot the house at the rate of 8.8 kW with a coefficient of performance of 5.6.
A) What power input (in KW) is needed to run the heat pump?
B) What power input (in KW) would be needed if the heat pump were ideal and still providing 8.8 kW heating power inside the house?



Homework Equations


coefficient of performance = (QsubscriptH)/Work
For ideal heat pump:
(QsubscriptC)/(QsubscriptH)=(Temperature subcript C)/(Temperature subscript H)

W=(QsubscriptH)-(QsubscriptC)



The Attempt at a Solution


convert Temperatures to Kelvin
A) coefficient of performance = (QsubscriptH)/Work
5.6=8.8 kW/W
W=1.57 kW


Would power input be work? or would you use the power equation?

B) (QsubscriptC)/(QsubscriptH)=(Temperature subcript C)/(Temperature subscript H)
(QsubscriptC)/8.8 kW=267 K/246 K
(QsubscriptC)=7.94 kW

W=(QsubscriptH)-(QsubscriptC)
=8.8 kW -7.94 kW
=.86 kW
Are my calculations correct?
 
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A) is right. But in B, you are not converting the temperatures correctly. 23 C = 296 K and -0.6 C = 272.4 K.

AM
 
Andrew Mason said:
A) is right. But in B, you are not converting the temperatures correctly. 23 C = 296 K and -0.6 C = 272.4 K.

AM



It should be -6.0 not -0.6. But anyways thanks so much!
 

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