Second Law of Thermodynamics - heat pumps and air conditioners

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SUMMARY

The discussion focuses on calculating the effective coefficient of performance (COP) for a heat pump operating as an air conditioner, which draws 20,000 Btu/hr between temperatures of 21°C and 41°C. The COP is given as 35% of the Carnot COP, which is calculated using the temperatures provided. The user initially calculated the ideal COP as 1.05 but encountered difficulties in determining the effective COP and the power required for the compressor motor. Accurate calculations are essential for understanding the performance of heat pumps in thermodynamic applications.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically the Second Law of Thermodynamics.
  • Familiarity with the concept of Coefficient of Performance (COP) in heat pumps and air conditioners.
  • Knowledge of unit conversions, particularly between Btu and joules.
  • Basic proficiency in using thermodynamic equations related to heat transfer and work done.
NEXT STEPS
  • Calculate the ideal COP using the formula COPideal = TL / (TH - TL) with the given temperatures.
  • Determine the effective COP by applying the 35% factor to the ideal COP.
  • Compute the power required for the compressor motor using the formula W = QL / COP.
  • Explore the implications of COP values on the efficiency of heat pumps in practical applications.
USEFUL FOR

Students studying thermodynamics, HVAC engineers, and professionals involved in the design and analysis of heat pumps and air conditioning systems.

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Homework Statement


A heat pump operates as an air conditioner, drawing 20000 Btu per hr from a building. It is operating between 21 degrees Celsius and 41 degrees Celsius.

a) If the COP (coefficient of performance) is 35% that of a Carnot air conditioner, what is the value of the effective COP (coefficient of performance)?

b) What is the value of the power (Watts) required for the compressor motor?

Homework Equations


QL = heat removed from the low temperature area (inside the building)
QH = heat removed from the high temperature area (outside the building)
W = work done to remove the heat
TL = temperature of system at lower temperature (inside the building)
TH = temperature of system at higher temperature (outside the building)

COP = QL/W

COP = QL/(QH - QL)

COPideal = TL / (TH - TL)

The Attempt at a Solution


I first converted Btu (British thermal unit) to joules: 1Btu = 1055.056 J. Then I tried finding the ideal COP by using the above equation with the given temperatures and got 1.05, but I'm pretty sure the ideal COP is not calculated this way. I tried multiplying 1.05 by 0.35 but got the wrong answer for part a. Thanks for any help.
 
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