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.59}, 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 cycl...
..._{\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: }
\bet...
..._{\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: }
\e...
...{Q_{\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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...kebox(0,0)[tl]{$W_{{\rm Carnot}}$}}
\end{picture} }
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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.