AZING! How does the Carnot cycle affect refrigerator efficiency?

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Homework Help Overview

The discussion revolves around the efficiency of a refrigerator operating on the reversed Carnot cycle, specifically focusing on the calculation of the coefficient of performance (COP), the refrigerant effect per kW of input work, and the heat rejected to the condenser. The temperatures provided are -6°C for the evaporator coil and 22°C for the condenser coil.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of COP using the temperatures provided, with some expressing uncertainty about how to find the input work and heat rejected. There are attempts to clarify relationships between heat removed, work done, and the heat rejected.

Discussion Status

Several participants have provided insights into the relationships between the variables involved, including the application of the first law of thermodynamics. There is ongoing clarification regarding the definitions of QL and QH, as well as the signs associated with heat flow and work done.

Contextual Notes

Participants note that only the high and low temperatures are given, which limits the information available for calculations. There is also a discussion about the cyclical nature of the system and its implications for internal energy changes.

nightingale
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Homework Statement


The temperature in a refrigerator evaporator coil is -6oC and that in the condenser coil is 22oC. Assuming that the machine operates on the reversed Carnot cycle, calculate the COPref , the refrigerant effect per kW of input work, and the heat rejected to the condenser.
Answers: 9.54, 9.54 kW, 10.54 kW

Homework Equations





The Attempt at a Solution


COPref = 1/((TH/TL)-1)
CoPref = 1/ (28/627)
COPref = 9.53

I have no idea how to find the the input of work and the heat rejected. Could you please help? Thank you.
 
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nightingale said:

Homework Statement


The temperature in a refrigerator evaporator coil is -6oC and that in the condenser coil is 22oC. Assuming that the machine operates on the reversed Carnot cycle, calculate the COPref , the refrigerant effect per kW of input work, and the heat rejected to the condenser.
Answers: 9.54, 9.54 kW, 10.54 kW

Homework Equations


The Attempt at a Solution


COPref = 1/((TH/TL)-1)
CoPref = 1/ (28/627)

I have no idea how to find the the input of work and the heat rejected. Could you please help? Thank you.
I think you meant COPref = 1/(28/267) = 9.536 = 9.54

For a refrigerator, COP = heat removed from cold/work done

You are given W and you have calculated COP. You just need to find QL

Once you have QL you should be able to find QH. What is the relationship between QH, QL and W?

AM
 
Andrew Mason said:
For a refrigerator, COP = heat removed from cold/work done

You are given W and you have calculated COP. You just need to find QL

AM

Thank you Andrew. But I was only given the high and low temperatures, not the W. Could you please explain further? I think that OL = QH + W. Is this correct? Thank you in advance.
 
nightingale said:
Thank you Andrew. But I was only given the high and low temperatures, not the W. Could you please explain further? I think that OL = QH + W. Is this correct? Thank you in advance.
Not quite. Apply the first law to the system: Q = ΔU + W where Q is the total (net) heat flow for the system, W is the work done BY the system and ΔU is the change in internal energy of the system. Hint: Since the system is cyclical, what can you say about the change in U over any number of complete cycles?

AM
 
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nightingale said:
Thank you Andrew. But I was only given the high and low temperatures, not the W. Could you please explain further?
You are asked to find the refrigerant effect, QL, per kW of input work and the heat rejected to the condenser, QH (per kW of input work). So W is 1 kW.

AM
 
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Andrew Mason said:
Not quite. Apply the first law to the system: Q = ΔU + W where Q is the total (net) heat flow for the system, W is the work done BY the system and ΔU is the change in internal energy of the system. Hint: Since the system is cyclical, what can you say about the change in U over any number of complete cycles?

AM

The ΔU equals to zero in complete cycle? Does this mean the QL = 0 + W? QL = W, I do get the answer 9.54 kw.
QL = W + QH
9.54 = 1 + QH
QH = 8.54kW, which is the wrong answer. Is the heat rejected QL or QH?
Thank you very much Andrew.
 
nightingale said:
The ΔU equals to zero in complete cycle? Does this mean the QL = 0 + W? QL = W, I do get the answer 9.54 kw.
QL = W + QH
9.54 = 1 + QH
QH = 8.54kW, which is the wrong answer. Is the heat rejected QL or QH?
Thank you very much Andrew.
The heat rejected is QH. But you have to be careful with the signs.

Since heat is absorbed by the hot reservoir, the heat flow from system to the hot reservoir is negative: QH<0. Since heat is removed from the cold reservoir, the heat flow from the cold reservoir to the system is positive, QL > 0. The ΔU of the system = 0.

So from the perspective of the system, where W = work done BY the system: QH + QL = ΔU + W = W. Using absolute values (QH = -|QH|): |QL|-|QH| = W

So |QH| = |QL| - W. Since positive work is done ON the system, W<0 so W = -|W|. In other words:

|QH| = |QL| + |W|. Normally this is written QH = QL + W where the absolute values are implicitly understood.

If 9.54 kW is removed from the cold reservoir for each 1.00 kW of work done ON the system, then what is QH (i.e. |QH|)?

AM
 

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