Is the Heat Pump and Motor Setup Efficient?

In summary: So, in summary, the efficiency of the pump and motor setup can be calculated by finding the rate of mechanical work supplied by the motor, which is less than 5.03 kW. This efficiency is not the same as the COP of the heat pump, which is 1.75. There is some ambiguity in the problem statement regarding the heat removed from the cold reservoir, which we will assume is 8793 watts. To find the efficiency of the motor, we need to calculate the rate of mechanical work supplied to the pump by the motor, which will be less than 5.03 kW.
  • #1
lostinsauce
1
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.
 
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  • #2
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
 

What is the COP of a heat pump?

The COP (Coefficient of Performance) of a heat pump is a measure of its efficiency in providing heating or cooling. It is calculated by dividing the heat output by the energy input.

What is the COP of a heat pump compared to the COP of a refrigerator?

The COP of a heat pump is typically higher than the COP of a refrigerator. This is because a heat pump is designed to transfer heat from a colder area to a warmer area, while a refrigerator is designed to transfer heat from a warmer area to a colder area. Therefore, a heat pump requires less energy to produce the same amount of heating or cooling as a refrigerator.

How does the COP of a heat pump affect its efficiency?

The higher the COP of a heat pump, the more efficient it is. A higher COP means that the heat pump requires less energy to produce the same amount of heating or cooling, resulting in lower energy consumption and cost.

What factors can affect the COP of a heat pump?

The COP of a heat pump can be affected by various factors, including the outside temperature, the size and type of heat pump, the quality of the installation, and the maintenance of the heat pump. It is important to properly size and maintain a heat pump to ensure optimal efficiency and COP.

How can I improve the COP of a heat pump?

There are several ways to improve the COP of a heat pump, such as regularly maintaining and cleaning the heat pump, ensuring proper insulation and sealing of the building, and using a heat pump with a higher COP rating. Additionally, using a heat pump in combination with other energy-efficient systems, such as solar panels, can also improve its overall COP.

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