Is the Heat Pump and Motor Setup Efficient?

AI Thread Summary
The discussion centers on calculating the efficiency of a heat pump and its motor setup, given that the heat pump takes in 30,000 BTU/hr and consumes 5030 watts. The initial calculations suggest a coefficient of performance (COP) of 1.75, but there is confusion regarding whether to use this value or to derive the cooling output (Qc) for further analysis. Participants emphasize the need to clarify whether the 30,000 BTU/hr refers to heat removed or delivered, and highlight that the question pertains to the motor's efficiency rather than the heat pump's COP. It is noted that the actual mechanical work output from the motor will be less than the electrical input of 5.03 kW. The conversation concludes with a call for further clarification on the definitions and calculations involved.
lostinsauce
Messages
1
Reaction score
0
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.
 
Science news on Phys.org
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
 
I need to calculate the amount of water condensed from a DX cooling coil per hour given the size of the expansion coil (the total condensing surface area), the incoming air temperature, the amount of air flow from the fan, the BTU capacity of the compressor and the incoming air humidity. There are lots of condenser calculators around but they all need the air flow and incoming and outgoing humidity and then give a total volume of condensed water but I need more than that. The size of the...
Thread 'Why work is PdV and not (P+dP)dV in an isothermal process?'
Let's say we have a cylinder of volume V1 with a frictionless movable piston and some gas trapped inside with pressure P1 and temperature T1. On top of the piston lay some small pebbles that add weight and essentially create the pressure P1. Also the system is inside a reservoir of water that keeps its temperature constant at T1. The system is in equilibrium at V1, P1, T1. Now let's say i put another very small pebble on top of the piston (0,00001kg) and after some seconds the system...
Back
Top