Is the Heat Pump and Motor Setup Efficient?

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SUMMARY

The discussion centers on the efficiency of a heat pump and motor setup, specifically analyzing a heat pump that takes on 30,000 BTU/hr of heat while consuming 5,030 watts. The coefficient of performance (COP) for the heat pump is calculated as 2.19, but the efficiency of the motor is questioned. The calculations indicate that the mechanical work output of the motor is less than its electrical input of 5.03 kW, leading to a COP of 1.75 when considering the heat removed. Clarification on the definitions of Qh and Qc is necessary for accurate efficiency assessment.

PREREQUISITES
  • Understanding of heat pump operation and terminology
  • Knowledge of the Coefficient of Performance (COP) calculation
  • Familiarity with energy conversion between BTU/hr and watts
  • Basic principles of thermodynamics, including Carnot efficiency
NEXT STEPS
  • Research the calculation of mechanical efficiency in electric motors
  • Learn about the Carnot efficiency equation and its application to heat pumps
  • Explore the conversion factors between BTU/hr and watts for accurate energy assessments
  • Investigate the implications of COP in different operating conditions of heat pumps
USEFUL FOR

Engineers, HVAC professionals, and energy efficiency analysts seeking to optimize heat pump and motor setups for improved performance and energy savings.

lostinsauce
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Not sure if I am doing this correct and do not have answer available...
A heat pump takes on 30,000BTU/hr of heat and uses 5030 watts. However its COP as an air conditioner is 2.19. How efficient is the pump and motor setup?

I am assuming the following:
Qh=30000 BTU/hr=11.79 hp
WKin=5030 W=5.03kW/0.746kW=6.74 hp * 2545 BTU/hr = 17,153 BTU/hr

COPhp= Qh/WKin=11.79/6.74= 1.75

So is that the efficiency or should I solve for Qc as such:
Qc= Qh - WKin= 30000 BTU/hr - 17153 BTU/hr= 12,847 BTU/hr

And then use Carnot's efficiency equation? But Qc or Qh are rates and not Th or Tl...any help would be appreciated.
 
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lostinsauce said:
Not sure if I am doing this correct and do not have answer available...
A heat pump takes on 30,000BTU/hr of heat and uses 5030 watts. However its COP as an air conditioner is 2.19. How efficient is the pump and motor setup?

I am assuming the following:
Qh=30000 BTU/hr=11.79 hp
WKin=5030 W=5.03kW/0.746kW=6.74 hp * 2545 BTU/hr = 17,153 BTU/hr

COPhp= Qh/WKin=11.79/6.74= 1.75

So is that the efficiency or should I solve for Qc as such:
Qc= Qh - WKin= 30000 BTU/hr - 17153 BTU/hr= 12,847 BTU/hr

And then use Carnot's efficiency equation? But Qc or Qh are rates and not Th or Tl...any help would be appreciated.
The units are distracting. I am not sure why you are using HP but I will convert to MKS (watts). 1 BTU/hr = .2931 W. 30000 BTU/hr = 8793 watts.

First of all, there is some ambiguity in the statement of the problem. What does it mean the the heat pump "takes on 30,000 BTU/hr of heat"?. Is this the heat removed from the cold reservoir or is it the heat delivered to the warm reservoir? I will assume it is the heat removed, ie. Qc.

Second, you have to be clear on what the question is asking. It is not asking for the COP of the heat pump. It is asking for the efficiency of the motor - ie. how much mechanical work it does for each unit of energy supplied. The 5.03 kW is the (electrical) power consumption of the motor. You have to find rate of mechanical work supplied to the pump by the motor. hint: It will be less than 5.03 kW.

AM
 

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