The animation below illustrates a heat pump operating between a cold reservoir a

Question

The animation below illustrates a heat pump operating between a cold reservoir and a hot reservoir.

http://www.webassign.net/serway/af/AF_1212.swf

Instructions: Use the slider to adjust the coeffiecient of performance (COP) and click start to begin the animation. You can click the H/C button to switch between heating and cooling mode.
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A heat pump can heat your house in the winter and cool your house in the summer. It does this by running "in reverse" compared to a heat engine. Work is supplied to the heat pump so that energy flows from the cold reservoir to the hot reservoir.

Heating Mode

In this case, the hot reservoir is inside your house, and the cold reservoir is outside. The heat pump removes heat from the cold reservoir outside, and moves it inside.

Good heat pumps have a high coefficient of performance (COP). In heating mode, the COP is given by

COP(heating) = Qh/W

In words, in heating mode, the COP is equal to the heat Qh delivered to the inside of your house divided by the work required to run the heat pump.

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Cooling Mode

COP(cooling) = Qc/W

In this case, the hot reservoir is outside and the cold reservoir is inside. In cooling mode, the COP measures the heat energy Qc extracted from indoors divided by the amount of work necessary to achieve this:



Exercise 18.AF.7

A certain heat pump has a COP of 6.17 while operating in both heating and cooling mode.

Suppose the heat pump operates in heating mode on a cold day.

For every 1000 W of power it takes to run the heat pump, how much power is delivered to the inside of the house?
_____ W

How much power is removed from the outside?
_____ W

The same heat pump now operates in cooling mode on a hot day.
For every 1000 W of power it takes to run the heat pump, how much power is removed from the inside of the house?
_____ W

How much power is expelled to the outside?
_____ W
Increasing the COP makes the heat pump work

Explanation / Answer

1.)    (6.17)(1000)= 6170 J per sec. 

2.)     6170-1000= 5170 J per sec.

3.)     (6.17)(1000)= 6170 J per sec. 

4.)     (6170)(1000)=7170 J per sec. 

 

It is quite easy actually. I'm sorry that I couldn't get to you sooner but I hope you did well! 

 

If you need clarification, yes, #'s 1 and 3 are both the same since "A certain heat pump has a COP of 6.17 while operating in BOTH heating and cooling mode."

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