Solving Refrigerator Problem: Calculating COP and Work

  • Thread starter sportsrules
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In summary: I didn't quite understand what you were saying about the last part. Can you please clarify? First of all, the co-efficient of performance for a heat pump is defined a little differently from the co-efficient of performance for a refrigerator. For a refrigerator:cop = \frac{Q_c}{W}For a heat pump:cop = \frac{Q_h}{W}So for a heat pump:W = \frac{Q_h}{cop}
  • #1
sportsrules
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The question reads...
A heat pump is a refrigerator that uses the inside of a building as a hot-temp reservoir in the winter and the outside of the building as a hot-temp reservoir in the summer. So...
For a carnot refrigerator engine operating between 25C and 40C in the summer, what is the coefficient of performance?And, for each joule of heat trasnfer from the cooler reservoir per cycle, how many joules of work are done by the refrigerator engine? Is this work positive or negative...and, what additional information is needed to determine an appropriate heat pump for a building?

Ok, so I figured out the COP...

COP=1/ [(Th/Tc)-1]

COP=1/ [(313k/298K)-1]

COP=19.9

But thenn I get confused on the second part
I know that

COP=Qc/W
W=Qc/COP
so is the value for Qc just 1 J? THat is the part i am having trouble with, and the work is negative correct? because it is work done by the system?

And, any thouhts on the last part?
 
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  • #3
yes it was...but how about the last part...what else would I need to know? Just the size of the building?
 
  • #4
sportsrules said:
yes it was...but how about the last part...what else would I need to know? Just the size of the building?
First of all, the co-efficient of performance for a heat pump is defined a little differently from the co-efficient of performance for a refrigerator. For a refrigerator:

[tex]cop = \frac{Q_c}{W}[/tex]

For a heat pump,

[tex]cop = \frac{Q_h}{W}[/tex]

So for a heat pump:

[tex]W = \frac{Q_h}{cop}[/tex]

That is all you need to determine how much work you consume in delivering 1 joule of heat. It has nothing to do with building size.

AM
 

Related to Solving Refrigerator Problem: Calculating COP and Work

1. What is the purpose of calculating COP and work in a refrigerator?

The COP (Coefficient of Performance) and work calculations in a refrigerator can help determine the efficiency of the refrigeration cycle and the amount of work required to achieve a certain level of cooling. This information is important for understanding how well the refrigerator is functioning and identifying any potential problems.

2. How do you calculate the COP of a refrigerator?

The COP of a refrigerator can be calculated by dividing the desired cooling (or heat removal) by the work input. This can be represented by the equation COP = Q/W, where Q is the amount of heat removed and W is the work input. The higher the COP, the more efficient the refrigerator is at cooling.

3. What factors can affect the COP of a refrigerator?

The COP of a refrigerator can be affected by several factors, including the type of refrigerant used, the design and size of the refrigerator, and the temperature difference between the inside and outside of the refrigerator. Additionally, any mechanical or electrical issues within the refrigerator can also impact the COP.

4. How is work calculated in a refrigerator?

The work done by a refrigerator is calculated by multiplying the force (or pressure) applied by the distance the force is applied over. This can be represented by the equation W = F x d, where F is the force and d is the distance. In the case of a refrigerator, the work is typically calculated by measuring the compressor's power consumption.

5. How can COP and work calculations be used to troubleshoot refrigerator problems?

By calculating the COP and work of a refrigerator, any inefficiencies or issues within the refrigeration cycle can be identified. For example, a lower COP than expected may indicate a problem with the compressor or refrigerant, while a higher work input than expected may indicate a mechanical issue. These calculations can help pinpoint the source of the problem and guide repairs or maintenance.

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