Engines and Refrigerators

A heat engine absorbs heat and converts part of it into work. An isothermal expansion turns all the absorbed heat into work, but it can’t continue: to run periodically, an engine must operate in cycles, returning to its original state, and that forces it to dump some waste heat into its environment. (An internal-combustion engine counts too, treating the burned fuel as a heat source even though it burns inside.) It absorbs \(Q_h\) from a hot reservoir at \(T_h\) (something so large its temperature doesn’t change noticeably) and expels \(Q_c\) to a cold one at \(T_c\), so by the first law \(Q_h = Q_c + W\) and its efficiency is

Entropy settles the maximum. The engine returns to its starting state, so it must expel at least as much entropy as it absorbs: \(Q_c/T_c \ge Q_h/T_h\), which gives

Between 500 K and 300 K, no engine can exceed 40%. The first law says you can’t win (\(W\) can’t exceed \(Q_h\)); the second says you can’t even break even (\(e = 1\) only if \(T_c = 0\) or \(T_h = \infty\)). A less efficient engine produces extra entropy during its operation, which it must dispose of by dumping extra heat into the cold reservoir.

An engine that does reach the limit needs a working substance, such as a gas, that absorbs heat while only slightly cooler than \(T_h\) and expels it while only slightly warmer than \(T_c\), so that no new entropy is made. That is the Carnot cycle: (1) isothermal expansion at \(T_h\), absorbing \(Q_h\); (2) adiabatic expansion, cooling to \(T_c\); (3) isothermal compression at \(T_c\), expelling \(Q_c\); (4) adiabatic compression, back to \(T_h\). For an ideal gas (Problem 4.5), \(Q_h = Nk_{\mathrm{B}}T_h\ln(V_2/V_1)\) and \(Q_c = Nk_{\mathrm{B}}T_c\ln(V_3/V_4)\), while \(T_h^{f/2}V_2 = T_c^{f/2}V_3\) and \(T_c^{f/2}V_4 = T_h^{f/2}V_1\) give \(V_2/V_1 = V_3/V_4\), so

The catch: every step must be quasistatic, so it takes forever to get a significant amount of work out. Very efficient, horribly impractical. The cycle explorer below runs it, along with two real engines.

Problem 4.6, the Curzon–Ahlborn “endoreversible” engine. To get real power, let heat flow at a finite rate: the gas works at \(T_{hw} < T_h\) and \(T_{cw} > T_c\), with \(Q_h/\Delta t = K(T_h - T_{hw})\) and \(Q_c/\Delta t = K(T_{cw} - T_c)\), and a reversible Carnot cycle in between. Entropy balance fixes \(T_{cw}\) for each \(T_{hw}\). Power is zero both when \(T_{hw} = T_h\) (no heat flows) and when the working temperatures meet (no work), so it peaks in between, at \(T_{hw} = \tfrac12(T_h + \sqrt{T_hT_c})\), where

Going further. The same efficiency at maximum power had already been found in 1957 by Chambadal and Novikov, so this engine is also called the Chambadal–Novikov–Curzon–Ahlborn engine. A supplementary page works through it in more detail: the derivation step by step, the actual working cycle on a P–V diagram, where the entropy is produced, a comparison with real power plants, what changes when the two conductances differ, and remarks on a quantum-dot heat engine. Open it in English or 한국어.

A refrigerator is a heat engine run in reverse: work \(W\) (from electricity, say) moves heat \(Q_c\) out of the cold reservoir and dumps \(Q_h = Q_c + W\) into the hot one. Its figure of merit is the coefficient of performance, which can exceed 1:

since the entropy dumped into the hot reservoir must be at least as much as the entropy taken from the cold: \(Q_h/T_h \ge Q_c/T_c\). A kitchen refrigerator between 298 K and 255 K has \(\mathrm{COP} \le 5.9\): each joule of electricity can move up to 5.9 J out of the freezer and dump 6.9 J into the kitchen. The COP is largest when the temperatures are close, and poor when \(T_c\) is very low, such as 4 K for liquid helium. Use the machine toggle above to explore it. The best refrigerator is a Carnot cycle run backward, with the gas slightly hotter than \(T_h\) while expelling heat and slightly colder than \(T_c\) while absorbing it; it, too, is far too slow to be practical, and Section 4.4 describes a real one.

This is why an air conditioner goes in a window, not in the middle of a room, and why you can’t cool your kitchen by leaving the refrigerator door open: it dumps \(Q_h = Q_c + W\), more heat than it removes, into the same room (Problems 4.7 and 4.8).

Early versions of the second law, built on experience, stated that every heat engine must produce some waste heat and every refrigerator needs some work input. Carnot and others realized that these follow from the limit \(e \le 1 - T_c/T_h\): an engine that beat it could drive a Carnot refrigerator and, together, move heat from cold to hot with no work at all (Problem 4.16). Carnot also recognized a quantity associated with heat that flows in and out of an ideal engine in equal amounts; Rudolf Clausius made it precise in 1865 as \(Q/T\) and named it entropy, and Ludwig Boltzmann worked out what it actually is by 1877. Claim an efficiency and see what follows:

The familiar gasoline engine follows, in idealized form, the Otto cycle: adiabatic compression, ignition (heat in at constant volume), the adiabatic power stroke, and exhaust (heat out at constant volume). With compression ratio \(V_1/V_2\) and \(\gamma = (f+2)/f\) (\(7/5\) for air, mostly N₂ and O₂), Problem 4.18 gives

less than the Carnot efficiency between its highest and lowest temperatures, and even less in practice because of friction. A higher compression ratio helps, but risks preignition; the Diesel engine avoids this by compressing only air and then spraying in the fuel. The Stirling engine (Problem 4.21) is a true external-heat engine of two isotherms and two constant-volume steps, during which the gas passes through a regenerator that stores heat from one step and returns it in the other. With a perfect regenerator its efficiency equals Carnot’s; without one,