AZING! How does the Carnot cycle affect refrigerator efficiency?

[nightingale]

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.

 

[Andrew Mason]

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 Q_L

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

AM

 

[nightingale]

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

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

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.

 

[Andrew Mason]

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

 

[Andrew Mason]

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, Q_L, per kW of input work and the heat rejected to the condenser, Q_H (per kW of input work). So W is 1 kW.

AM

 

[nightingale]

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.

 

[Andrew Mason]

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