COP of heat pump to COP ref


by lostinsauce
Tags: cophp/copr
lostinsauce
lostinsauce is offline
#1
Oct28-12, 10:33 PM
P: 1
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.
Phys.Org News Partner Physics news on Phys.org
Physicists design quantum switches which can be activated by single photons
'Dressed' laser aimed at clouds may be key to inducing rain, lightning
Higher-order nonlinear optical processes observed using the SACLA X-ray free-electron laser
Andrew Mason
Andrew Mason is offline
#2
Oct28-12, 11:41 PM
Sci Advisor
HW Helper
P: 6,574
Quote Quote by lostinsauce View Post
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


Register to reply

Related Discussions
Heat pump problem with heat loss? Introductory Physics Homework 2
Heat engine and Heat pump in combination Engineering, Comp Sci, & Technology Homework 3
How much energy to heat a room, and heat pump efficiency? Classical Physics 3
Heat Pump Introductory Physics Homework 6