0
$\begingroup$

I'm trying to calculate the efficiency of an ideal refrigeration machine ($η$) and I keep getting the right result with the wrong sign. My reasoning is as follows (I've attached a sketch for more clarity).enter image description here

Let my system be composed of a (cold) heat source and a (hot) heat sink. My machine's goal is to take heat from the cold source, and thanks to the work provided, transmit this heat to the hot sink. That is to say, following the sign convention where $\Delta W>0$ if the work is being exerted ON the system and $\Delta W<0$ if it is the system that's doing the work, I can write the first principle as follows:

$$-\Delta Q_h = \Delta W+\Delta Q_c$$

I can also define the efficiency of my refrigeration machine as:

$$η \equiv \frac{\Delta Q_c}{\Delta W}$$

Where both quantities are positive as they are "entering" the system. I can now expand the first expression, taking into account that $\Delta Q = T\cdot \Delta S$, obtaining:

$$\Delta W = -\Delta Q_h-\Delta Q_c = -T_h\cdot \Delta S_h-T_c\cdot \Delta S_c$$

Now, since I'm working with an ideal machine, I can assume the process it undergoes is reversible, which means the total increase in entropy of the system is null. Therefore: $\Delta S_T= \Delta S_h+\Delta S_s+\Delta S_c=\Delta S_h+\Delta S_c=0$, where $\Delta S_s=0$ is the machine's increase in entropy, and it's null because said machine works in cycles. Therefore, I can now rewrite my former expression as follows:

$$\Delta W = T_h\cdot \Delta S_c - T_c\cdot \Delta S_c = (T_h-T_c)\cdot \Delta S_c$$

Again, according to $\Delta Q=T\cdot \Delta S$, I can finally write:

$$\Delta W = (\frac{T_h}{T_c}-1)\cdot T_c\cdot \Delta S_c = (\frac{T_h}{T_c}-1)\cdot \Delta Q_c$$

This yields the result:

$$η = \frac{\Delta Q_c}{\Delta W} = \frac{T_c}{T_h-T_c}$$

I know this is wrong, since the efficiency of a refrigeration machine is precisely the opposite of this: $η = \frac{T_c}{T_c-T_h}$. Am I lacking a minus sign somewhere in my calculations or is my approach to the problem incorrect? Thanks for your kind answers!

$\endgroup$
0

1 Answer 1

1
$\begingroup$

Am I lacking a minus sign somewhere in my calculations or is my approach to the problem incorrect?

No. The sign is correct.

The correct term for the "efficiency" of a refrigeration cycle is the coefficient of performance (COP). For a Carnot refrigeration cycle the COP is $$COP=\frac{T_{c}}{T_{h}-T_c}$$

As in your last equation.

I have no idea why you would think it is

$$COP=\frac{T_{c}}{T_{c}-T_h}$$

The COP is defined at the desired heat transfer divided by the work required to transfer that heat. Or, in the case of any refrigeration cycle,

$$COP=\frac{Q_{c}}{W}=\frac{Q_{c}}{Q_{h}-Q_{c}}$$

For a Carnot refrigeration cycle $Q_{c}=T_{c}\Delta S$ and $Q_{h}=T_{h}\Delta S$. Thus the Carnot refrigeration COP is

$$COP=\frac{T_{c}\Delta S}{T_{h}\Delta S-T_{c}\Delta S}=\frac{T_{c}}{T_{h}-T_c}$$

Hope this helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.