11.9 The Reversible Ideal

The statements of the previous section describing the second law are clearly common sense: yes, you still need to plug in your fridge, and no, you cannot skip the periodic stop at a gas station. What a surprise!

They seem to be fairly useless beyond that. For example, they say that it takes electricity to run our fridge, but they do not say it how much. It might be a megawatt, it might be a nanowatt.

Enter human ingenuity. With a some cleverness the two simple statements of the second law can be greatly leveraged, allowing an entire edifice to be constructed upon their basis.

A first insight is that if we are limited by nature’s unrelenting arrow of time, then it should pay to study devices that almost ignore that arrow. If you make a movie of a device, and it looks almost exactly right when run backwards, the device is called (almost exactly) “reversible.” An example is a mechanism that is carefully designed to move with almost no friction. If set into motion, the motion will slow down only a negligible amount during a short movie. When that movie is run backwards in time, at first glance it seems perfectly fine. If you look more carefully, you will see a slight problem: in the backward movie, the device is speeding up slightly, instead of slowing down due to friction as it should. But it is almost right: it would require only a very small amount of additional energy to speed up the actual device running backwards as it does in the reversed movie.

Dollar signs may come in front of your eyes upon reading that last sentence: it suggest that almost reversible devices may require very little energy to run. In context of the second law it suggests that it may be worthwhile to study refrigeration devices and engines that are almost reversible.

The second major insight is to look where there is light. Why not study, say, a refrigeration device that is simple enough that it can be analyzed in detail? At the very minimum it will give a standard against which other refrigeration devices can be compared. And so it will be done.

Figure 11.11: Schematic of the Carnot refrigeration cycle.
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The theoretically simple refrigeration device is called a “Carnot cycle” refrigeration device, or Carnot heat pump. A schematic is shown in figure 11.11. A substance, the refrigerant, is circulating through four devices, with the objective of transporting heat out of the fridge, dumping it into the kitchen. In the discussed device, the refrigerant will be taken to be some ideal gas with a constant specific heat like maybe helium. You would not really want to use an ideal gas as refrigerant in a real refrigerator, but the objective here is not to make a practical refrigerator that you can sell for a profit. The purpose here is to create a device that can be analyzed precisely, and an ideal gas is described by simple mathematical formulae discussed in basic physics classes.

Consider the details of the device. The refrigerant enters the fridge at a temperature colder than the inside of the fridge. It then moves through a long piping system, allowing heat to flow out of the fridge into the colder refrigerant inside the pipes. This piping system is called a heat exchanger. The first reversibility problem arises: heat flow is most definitely irreversible. Heat flow seen backwards would be flow from colder to hotter, and that is wrong. The only thing that can be done to minimize this problem as much as possible is to minimize the temperature differences. The refrigerant can be sent in just slightly colder than the inside of the fridge. Of course, if the temperature difference is small, the surface through which the heat flows into the refrigerant will have to be very large to take any decent amount of heat away. One impractical aspect of Carnot cycles is that they are huge; that piping system cannot be small. Be that as it may, the theoretical bottom line is that the heat exchange in the fridge can be approximated as (almost) isothermal.

After leaving the inside of the refrigerator, the refrigerant is compressed to increase its temperature to slightly above that of the kitchen. This requires an amount $W_C$ of work to be done, indicating the need for electricity to run the fridge. To avoid irreversible heat conduction in the compression process, the compressor is thermally carefully insulated to eliminate any heat exchange with its surroundings. Also, the compressor is very carefully designed to be almost frictionless. It has expensive bearings that run with almost no friction. Additionally, the refrigerant itself has “viscosity;” it experiences internal friction if there are significant gradients in its velocity. That would make the work required to compress it greater than the ideal $\vphantom0\raisebox{1.5pt}{$-$}$$P{\,\rm d}{V}$, and to minimize that effect, the velocity gradients can be minimized by using lots of refrigerant. This also has the effect of minimizing any internal heat conduction within the refrigerant that may arise. Viscosity is also an issue in the heat exchangers, because the pressure differences cause velocity increases. With lots of refrigerant, the pressure changes over the heat exchangers are also minimized.

Now the refrigerant is sent to a heat exchanger open to the kitchen air. Since it enters slightly hotter than the kitchen, heat will flow out of the refrigerant into the kitchen. Again, the temperature difference must be small for the process to be almost reversible. Finally, the refrigerant is allowed to expand, which reduces its temperature to below that inside the fridge. The expansion occurs within a carefully designed turbine, because the substance does an amount of work $W_T$ while expanding reversibly, and the turbine captures that work. It is used to run a high-quality generator and recover some of the electric power $W_C$ needed to run the compressor. Then the refrigerant reenters the fridge and the cycle repeats.

If this Carnot refrigerator is analyzed theoretically, {D.60}, a very simple result is found. The ratio of the heat $Q_{\rm {H}}$ dumped by the device into the kitchen to the heat $Q_{\rm {L}}$ removed from the refrigerator is exactly the same as the ratio of the temperature of the kitchen $T_{\rm {H}}$ to that of the fridge $T_{\rm {L}}$:

\begin{displaymath}
\fbox{$\displaystyle
\mbox{For an ideal cycle: }
\frac{Q_{\rm{H}}}{Q_{\rm{L}}} = \frac{T_{\rm{H}}}{T_{\rm{L}}}
$} %
\end{displaymath} (11.14)

That is a very useful result, because the net work $W$ $\vphantom0\raisebox{1.5pt}{$=$}$ $W_C-W_T$ that must go into the device is, by conservation of energy, the difference between $Q_{\rm {H}}$ and $Q_{\rm {L}}$. A “coefficient of performance” can be defined that is the ratio of the heat $Q_{\rm {L}}$ removed from the fridge to the required power input $W$:
\begin{displaymath}
\fbox{$\displaystyle
\mbox{For an ideal refrigeration cy...
...\rm{L}}}{W} = \frac{T_{\rm{L}}}{T_{\rm{H}}-T_{\rm{L}}}
$} %
\end{displaymath} (11.15)

Actually, some irreversibility is unavoidable in real life, and the true work required will be more. The formula above gives the required work if everything is truly ideal.

The same device can be used in winter to heat the inside of your house. Remember that heat was dumped into the kitchen. So, just cross out kitchen at the high temperature side in figure 11.11 and write in house. And cross out “fridge“ and write in outside. The device removes heat from the outside and dumps it into your house. It is the exact same device, but it is used for a different purpose. That is the reason that it is no longer called a refrigeration cycle but a “heat pump.” For an heat pump, the quantity of interest is the amount of heat dumped at the high temperature side, into your house. So an alternate coefficient of performance is now defined as

\begin{displaymath}
\fbox{$\displaystyle
\mbox{For an ideal heat pump: }
\...
...\rm{H}}}{W} = \frac{T_{\rm{H}}}{T_{\rm{H}}-T_{\rm{L}}}
$} %
\end{displaymath} (11.16)

The formula above is ideal. Real-life performance will be less, so the work required will be more.

It is interesting to note that if you take an amount $W$ of electricity and dump it into a simple resistance heater, it adds exactly an amount $W$ of heat to your house. If you dump that same amount of electricity into a Carnot heat pump that uses it to pump in heat from the outside, the amount of heat added to your house will be much larger than $W$. For example, if it is 300 K (27 $\POW9,{\circ}$C) inside and 275 K (2 $\POW9,{\circ}$C) outside, the amount of heat added is 300/25 = 12 W, twelve times the amount you got from the resistance heater!

Figure 11.12: Schematic of the Carnot heat engine.
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If you run the Carnot refrigeration cycle in reverse, as in figure 11.12, all arrows reverse and it turns into a heat engine. The device now takes in heat at the high temperature side and outputs a net amount of work. The high temperature side is the place where you are burning the fuel. The low temperature may be cooling water from the local river. The Kelvin-Planck statement says that the device will not run unless some of the heat from the combustion is dumped to a lower temperature. In a car engine, the exhaust and radiator are the ones that take much of the heat away. Since the device is almost reversible, the numbers for transferred heats and net work do not change much from the nonreversed version. But the purpose is now to create work, so the “thermal efficiency” of a heat engine is defined as

\begin{displaymath}
\fbox{$\displaystyle
\mbox{For an ideal heat engine: }
...
...{\rm{H}}}
= \frac{T_{\rm{H}}-T_{\rm{L}}}{T_{\rm{H}}}
$} %
\end{displaymath} (11.17)

Unfortunately, this is always less than one. And to get close to that, the engine must operate hot; the temperature at which the fuel is burned must be very hot.

(Note that slight corrections to the strictly reversed refrigeration process are needed; in particular, for the heat engine process to work, the substance must now be slightly colder than $T_{\rm {H}}$ at the high temperature side, and slightly hotter than $T_{\rm {L}}$ at the low temperature side. Heat cannot flow from colder to hotter. But since these are small changes, the mathematics is almost the same. In particular, the numerical values for $Q_{\rm {H}}$ and $Q_{\rm {L}}$ will be almost unchanged, though the heat now goes the opposite way.)

Figure 11.13: A generic heat pump next to a reversed Carnot one with the same heat delivery.
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The final issue to be resolved is whether other devices could not be better than the Carnot ones. For example, could not a generic heat pump be more efficient than the reversible Carnot version in heating a house? Well, put them into different windows, and see. (The Carnot one will need the big window.) Assume that both devices are sized to produce the same heat flow into the house. On second thought, since the Carnot machine is reversible, run it in reverse; that can be done without changing its numbers for the heat fluxes and net work noticeably, and it will show up the differences between the devices.

The idea is shown in figure 11.13. Note that the net heat flow into the house is now zero, confirming that running the Carnot in reverse really shows the differences between the devices. Net heat is exchanged with the outside air and there is net work. Enter Kelvin-Planck. According to Kelvin-Planck, heat cannot simply be taken out of the outside air and converted into useful net work. The net work being taken out of the air will have to be negative. So the work required for the generic heat pump will need to be greater than that recovered by the reversed Carnot one, the excess ending up as heat in the outside air. So, the generic heat pump requires more work than a Carnot one running normally. No device can therefore be more efficient than the Carnot one. The best case is that the generic device, too, is reversible. In that case, neither device can win, because the generic device can be made to run in reverse instead of the Carnot one. That is the case where both devices are so perfectly constructed that whatever work goes into the generic device is almost 100% recovered by the reversed Carnot machine, with negligible amounts of work being turned into heat by friction or other irreversibility and ending up in the outside air.

The conclusion is that:

All reversible devices exchanging heat at a given high temperature $T_{\rm {H}}$ and low temperature $T_{\rm {L}}$, (and nowhere else,) have the same efficiency. Irreversible devices have less.
To see that it is true for refrigeration cycles too, just note that because of conservation of energy, $Q_{\rm {L}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $Q_{\rm {H}}-W$. It follows that, considered as a refrigeration cycle, not only does the generic heat pump above require more work, it also removes less heat from the cold side. To see that it applies to heat engines too, just place a generic heat engine next to a reversed Carnot one producing the same power. The net work is then zero, and the heat flow $Q_{\rm {H}}$ of the generic device better be greater than that of the Carnot cycle, because otherwise net heat would flow from cold to hot, violating the Clausius statement. The heat flow $Q_{\rm {H}}$ is a measure of the amount of fuel burned, so the irreversible generic device uses more fuel.

Practical devices may exchange heat at more than two temperatures, and can be compared to a set of Carnot cycles doing the same. It is then seen that it is bad news; for maximum theoretical efficiency of a heat engine, you prefer to exchange heat at the highest available temperature and the lowest available temperature, and for heat pumps and refrigerators, at the lowest available high temperature and the highest available low temperature. But real-life and theory are of course not the same.

Since the efficiency of the Carnot cycle has a unique relation to the temperature ratio between the hot and cold sides, it is possible to define the temperature scale using the Carnot cycle. The only thing it takes is to select a single reference temperature to compare with, like water at its triple point. This was in fact proposed by Kelvin as a conceptual definition, to be contrasted with earlier definitions based on thermometers containing mercury or a similar fluid whose volume expansion is read-off. While a substance like mercury expands in volume very much linearly with the (Kelvin) temperature, it does not expand exactly linearly with it. So slight variations in temperature would occur based on which substance is arbitrarily selected for the reference thermometer. On the other hand, the second law requires that all substances used in the Carnot cycle will give the same Carnot temperature, with no deviation allowed. It may be noted that the definition of temperature used in this chapter is completely consistent with the Kelvin one, because all substances includes ideal gasses.