Calculating the Cost of Heating with a Less Efficient Heat Pump

In summary, the question asks about the cost of heating a home using a heat pump with a lower coefficient of performance than ideal electric heaters. The given information includes the temperature of the heat exchanger and baseboard radiators, as well as the cost of heating with ideal electric heaters for one winter. The formula for COP is mentioned and it is suggested to convert temperatures to Kelvin and consider the given inefficiency. Other important factors to consider are Qh and Win, and their respective units, and how they can be used to solve the problem.
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Homework Statement


Assume that you heat your home with a heat pump whose heat exchanger is at Tc=2∘C, and which maintains the baseboard radiators at Th=47∘C. If it would cost $1000 to heat the house for one winter with ideal electric heaters (which have a coefficient of performance of 1), how much would it cost if the actual coefficient of performance of the heat pump were 75% of that allowed by thermodynamics?

Homework Equations


Th/(Th-Tc)=K
U=W+Q
K=Qh/Win

The Attempt at a Solution


i converted to kelvin and then plugged the temperatures into the equation and set it equal to Qh/Win but i am unsure how to proceed. any help is greatly appreciated.
 
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  • #2
It's best to type out your working so we can see exactly what you've done.

I assume 'K' is COP (Coefficient of performance), did you account for the given inefficiency here?

What does Qh represent? what are it's units? Is it the same for both the heater and the heat pump? How can you use that information?

What does W_in represent? what are it's units? Is it the same for both the heater and the heat pump? How can you use that information?
 

Related to Calculating the Cost of Heating with a Less Efficient Heat Pump

1. How do heat pumps and refrigerators work?

Heat pumps and refrigerators work by using a compressor and refrigerant to transfer heat from one location to another. The compressor compresses the refrigerant, causing it to turn into a liquid and release heat. The liquid refrigerant then travels through coils to the desired location, where it absorbs heat and turns back into a gas. The gas is then compressed again and the cycle continues, allowing the heat to be moved from one place to another.

2. What is the difference between a heat pump and a refrigerator?

The main difference between a heat pump and a refrigerator is the direction of heat transfer. A heat pump is designed to move heat from a cooler location to a warmer one, while a refrigerator moves heat from a warmer location to a cooler one. This is why a heat pump can provide heating and cooling for a space, while a refrigerator only provides cooling.

3. Are heat pumps and refrigerators energy efficient?

Yes, heat pumps and refrigerators are generally considered to be energy efficient. They use electricity to power the compressor, but the majority of the heat transfer process is done through the refrigerant, which does not require much energy. However, the efficiency of a heat pump or refrigerator can vary depending on factors such as the size, age, and maintenance of the unit.

4. Can a heat pump or refrigerator be used in any climate?

Heat pumps and refrigerators can be used in most climates, but they may not be as effective in extreme temperatures. In very cold climates, a heat pump may struggle to extract heat from the outside air, and a refrigerator may struggle to keep items cold in very hot temperatures. However, there are models specifically designed for different climates that can still be effective.

5. How often should a heat pump or refrigerator be serviced?

It is recommended to have a heat pump or refrigerator serviced once a year to ensure it is working properly and efficiently. This includes checking the refrigerant levels, cleaning coils, and inspecting other components for any potential issues. Regular maintenance can also help extend the lifespan of the unit and prevent costly repairs in the future.

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