# Does a heat engine inside of a refrigerator increase efficiency?

Consider a normal refrigeration cycle, except a heat engine runs off of the temperature difference between the evaporator and the condenser, as shown below.

This heat engine powers the compressor, so that we can have liquid for throttling (and therefore cooling via evaporation) again.

State 1 is a liquid, which is throttled and therefore cooled by phase change to state 2, which then proceeds through an evaporator that absorbs heat from the heat engine. State 3 is then full vapor, which goes through a compression to increase the pressure to hotter vapor at state 4. This hot vapor then runs through a condenser that supplies heat to the heat engine, and the refrigerant condenses back to the liquid state 1.

Would this design allow for a more efficient refrigerator?

• "Disregarding any irreversibilities, would this design allow for a perpetual refrigeration machine?" - disregarding the existence of gravity, can I walk in mid-air between the tops of two neighbouring skyscrapers? – Emilio Pisanty Nov 19 '18 at 17:04
• @EmilioPisanty No, but what's the point of your rhetoric? The answer to my original question is also "no". My proposed refrigeration cycle will not work even if all the components are completely reversible, due to Carnot's theorem. – Drew Nov 19 '18 at 18:29
• Regardless of how well-posed the rest of the question is, that statement has no place in a serious physics discussion the way you phrased it. – Emilio Pisanty Nov 19 '18 at 23:31
• @EmilioPisanty I agree... I should have rephrased it as "Will this increase the efficiency of refrigeration cycle?" or "Why is this not a perpetual refrigerator?", as that was more of what I was asking. – Drew Nov 20 '18 at 0:30
• If you agree that the post needs to be fixed, then edit it and fix it. – Emilio Pisanty Nov 20 '18 at 8:00

The heat engine that runs the compressor can never be perfectly efficient. In fact, its efficiency is limited by the Carnot efficiency, $$\eta=1-\frac{T_C}{T_H}$$, so it usually can't even be close to perfectly efficient. As it runs, some waste heat is always produced, and the work performed on the fluid by the compressor is therefore less than would be necessary to extract the same amount of heat from the condenser than was put into the evaporator (in other words, $$Q_H). As such, in each subsequent cycle, less and less heat will be extracted from the condenser, which means less and less heat will be input to the evaporator. This means that the temperature difference between the evaporator and the condenser will eventually decrease to zero, at which point the heat engine and compressor completely stop working.

• In fact, the Carnot efficiency of the device in the drawing is SO low (e.g. 3%-5%) that it would very quickly come to a halt. This device would also violate the 2nd law of thermodynamics if it worked. – David White Nov 18 '18 at 18:58
• @DavidWhite How are you getting the Carnot efficiency of this device? Wouldn't you have to know which fluid is being used as the working fluid to get $T_H$ and $T_C$? – probably_someone Nov 18 '18 at 19:02
• No. You just need to know the source and sink temperatures. If I recall correctly, the efficiency is calculated as 1 - Tc/Th. – David White Nov 18 '18 at 21:29
• @DavidWhite That's exactly my point. Where are you getting the source and sink temperatures from? – probably_someone Nov 18 '18 at 22:53
• In reality, the source temperature would have to be the temperature of the condenser, but the OP doesn't realize that you can't use the sink temperature of the evaporator. If the evaporator just removed heat from some small environment, you DON'T want to put that heat back into that environment. I see your point, and the OP should see it too. This means that the only sink temperature available is the ambient temperature, such as the outside air temperature. – David White Nov 19 '18 at 15:40

Congratulations! You have discovered the thought experiment that powers a lot of classical thermodynamics arguments.

So if the heat engine or the refrigerator are too efficient, the system will spontaneously describe a flow of heat from a cold reservoir to a hot one; this is taken as experimentally impossible: nobody has ever seen such spontaneous flows.

What is the maximum efficiency? This is given by a perfectly reversible engine/fridge, one that can be run in both directions with the same heat transfers and work. The reason it is maximal is just that it could fit into either slot here: if you had a more efficient engine you could use it as a fridge to get a spontaneous flow from cold to hot; if you had a more efficient fridge you could use it as an engine to do the same.

Here comes the part that our textbooks somewhat gloss over: these ideas were originally available to Sadi Carnot but with some other insights, they only specify that the transfer of heat for such a reversible engine must depend somehow on temperature: they don't say exactly how. And they can't because up to this point temperature is ambiguous: it is measured by the expansion and contraction of some complicated fluid, say, but does not have a clean microscopic description. It turned out that we needed Thomson to come by and invent an absolute temperature scale before we could make further progress.

Taking the later work of science for granted, we can for example build a reversible fridge out of a piston of ideal gas: such a gas is also, at constant pressure, a thermometer of absolute temperature. We can determine that in terms of the “Carnot cycle” that emerges here of isothermal expansion/compression interleaved with adiabatic expansion/compression, the heat exchanged must obey $$Q_\text{hot}/Q_\text{cold}=T_\text{hot}/T_\text{cold}$$ in terms of this absolute temperature. This gives a concrete basis for speaking about this efficiency directly, but it was also the doorway to Clausius thinking that there had to be something important about this ratio $$Q_\text{hot}/T_\text{hot}=Q_\text{cold}/T_\text{cold}$$ and to define entropy changes as $$\delta S=\delta Q/T.$$

You do not want to do that. Your heat engine will be pumping heat back to the refrigerator, thus reducing its efficiency. You want intead to release the heat to the outside. The hot source will be the condenser. The hotter the condenser the less efficient the refrigerator, so you do not want a condenser that is too hot.

If you calculate the total work, $$W_T$$ spent to substract a given $$Q_c$$ from the fridge, you see that increasing the temperature to make the carnot cycle more efficient makes you spend more work, because the decreases in the refrigerator's performance is larger than the gains produced by the heat engine ($$T_a$$ and $$Q_a$$ are the outside temperature and released heat, respectively):

$$W_T=W-W_{carnot}=(Q_h-Q_c)-(Q_h-Q_A)=(\frac{T_A}{T_c}-1)T_hQ_c$$

Of course, I assumed both, the refrigerator and the heat engine, are made of reversible carnot engines.

• This assumes a particular refrigerant and type of lubricating oil. There may be industry standards on this sort of thing, but even so, I would be curious as to which particular substances your numbers apply to. – probably_someone Nov 20 '18 at 0:30