Illustrated Physics
physics, drawn
Thermodynamics

The temperature an engine really takes its heat at

Carnot's ceiling is set by two temperatures, and no engine that burns fuel takes its heat in at one temperature or gives it out at another. It takes heat over a range, from the moment combustion starts to the moment it ends. For any reversible cycle there is an exact replacement for Carnot's two numbers: the average temperature at which heat arrives and the average at which it leaves, each weighted by the entropy the heat carries. The gap between a real cycle and Carnot is a gap between those averages and the extremes.
16 min read 5 figures The same equation againApproximations that lie

Assumes: The ceiling on every engine, set before it was designed · Entropy is a count, and the arrow of time is arithmetic

The ceiling on every engine is one formula. An engine taking in heat at a temperature ThT_h and rejecting it at TcT_c can convert at most 1−Tc/Th1 - T_c/T_h of the heat into work, whatever it is made of and however cleverly it is built. The argument is complete, and it applies to an engine that does something no actual engine does: takes all of its heat in at one temperature and gives all of it out at another.

A petrol engine takes its heat in over a range. The charge is compressed to a few hundred degrees, the spark fires, and the temperature climbs through two thousand kelvin while heat is added; the exhaust leaves hot and cools in the air. A gas turbine heats its air steadily from the compressor’s outlet to the turbine’s inlet. A steam plant warms water, boils it, and superheats the steam, each at a different temperature. Applying Carnot’s formula to any of them means choosing which temperature to call ThT_h, and the obvious choice, the hottest point in the cycle, gives a ceiling that a real engine misses by thirty points and more.

There is an exact replacement, and it is not a correction to Carnot but the same result stated for heat that arrives over a range.

The rectangle every cycle is equal to. An ideal Otto cycle for air on a temperature–entropy diagram: compression ratio 9, intake at 300 K, 1400 kJ/kg added at constant volume. Compression takes the charge to 722 K, combustion to 2672 K, expansion back to 1110 K, and the exhaust cools at constant volume. The shaded loop is the work; the region under the lower curve is the heat rejected. Heat enters over a range of temperatures, and its entropy-weighted mean — heat divided by the entropy it brings — is 1491 K; the heat leaves at a mean of 619 K. The dashed rectangle between those two temperatures has exactly the loop's area, and 1 − 619/1491 = 58.5%, which is the Otto efficiency, checked to rounding. A Carnot engine between the coldest and hottest points of the same cycle would reach 88.8%.
Fig. 1 An ideal Otto cycle for air on temperature–entropy axes: compression ratio 9, intake 300 K, 1,400 kJ/kg added. Heat enters between 722 and 2,672 K at an entropy-weighted mean of 1,491 K and leaves at a mean of 619 K. The dashed rectangle between those temperatures has the loop’s area, and 1 − 619/1,491 = 58.5%, the Otto efficiency. Carnot between 300 and 2,672 K would allow 88.8%.

A diagram where heat is an area

The figure plots temperature against entropy, and on those axes heat has a shape. For a reversible process the heat added is ∫T dS\int T\,dS, the area under the curve that the process traces. A cycle that returns to where it started has taken in the area under its upper edge and given out the area under its lower edge, and the work it delivers is the difference: the area enclosed by the loop.

The Otto cycle is four steps. The charge is compressed without heat exchange, which leaves its entropy unchanged, so the step is a vertical line from 300 K up to 722 K. Heat is added at constant volume, and the temperature and entropy rise together along an exponential curve to 2,672 K. The gas expands without heat exchange, straight down to 1,110 K. And the exhaust gives its heat away at constant volume, down the lower exponential back to the start.

Entropy is a count explains what the horizontal axis measures. What matters here is only that heat entering at a high temperature carries little entropy per joule and heat leaving at a low temperature carries a lot, and that a reversible cycle must give out exactly the entropy it takes in.

The mean that makes the formula exact

That last requirement is the whole argument. Call the entropy taken in ΔS\Delta S. The heat taken in can always be written as Qin=Tˉin ΔSQ_\text{in} = \bar T_\text{in}\,\Delta S, which defines Tˉin\bar T_\text{in}: the average temperature of the heat input, weighted by the entropy each joule brings. The heat given out is Qout=Tˉout ΔSQ_\text{out} = \bar T_\text{out}\,\Delta S with the same ΔS\Delta S, because a reversible cycle creates no entropy. So the efficiency is

η=1−QoutQin=1−TˉoutTˉin,\eta = 1 - \frac{Q_\text{out}}{Q_\text{in}} = 1 - \frac{\bar T_\text{out}}{\bar T_\text{in}},

exactly, for any reversible cycle at all.

On the diagram, Tˉin\bar T_\text{in} and Tˉout\bar T_\text{out} are the heights of two horizontal lines that enclose a rectangle of the same width as the loop and the same area. The rectangle is a Carnot cycle — heat in at one temperature, out at another — doing the same work with the same entropy, and the figure checks that its area matches the loop’s. For the Otto cycle drawn, heat enters at a mean of 1,491 K and leaves at a mean of 619 K, and 1−619/1,4911 - 619/1{,}491 is 58.5 per cent, the textbook Otto efficiency to every digit.

For heat added to a gas at constant volume or constant pressure the mean has a closed form: the difference of the two end temperatures divided by the logarithm of their ratio, the logarithmic mean. Between 722 and 2,672 K it is 1,491 K, well below the arithmetic mean of 1,697 K, because the heat added at the low end of the range carries more entropy per joule and so counts for more.

The gap to Carnot, located

The gap between Otto and Carnot is a gap in mean temperature. Ideal efficiency against compression ratio from 2 to 16, with 1400 kJ/kg of heat added per cycle. The solid curve is the Otto efficiency 1 − r¹⁻ᵞ; the dots, one per whole ratio, are 1 − T̄out/T̄in computed from the mean temperatures of each cycle, and sit on it exactly. The upper curve is a Carnot engine between the intake temperature and the peak temperature the same cycle reaches. At a ratio of 9 the Otto cycle reaches 58.5% and the Carnot ceiling 88.8%: the 30.3% between them is the difference between receiving heat at 2672 K and receiving it at a mean of 1491 K.
Fig. 2 Ideal efficiency against compression ratio from 2 to 16, with the same heat added. The solid curve is the Otto efficiency; the dots, one per whole ratio, are 1 − T̄out/T̄in from each cycle’s mean temperatures and sit on it. The dashed curve is Carnot between the intake and the cycle’s own peak. At a ratio of 9 the two are 58.5% and 88.8%.

Plotted against compression ratio, the Otto curve rises steeply and then flattens, and the Carnot curve for the same cycle’s extremes sits far above it everywhere. The mean temperatures say exactly what the thirty-point gap at a ratio of 9 is made of. The peak of 2,672 K is reached only at the very end of combustion, and most of the heat went in at lower temperatures; the exhaust leaves at 1,110 K and most of its heat is rejected well above the 300 K the Carnot figure assumes. Measured by the temperatures at which heat actually crosses the boundary, the engine is a Carnot engine between 1,491 and 619 K, and it does exactly as well as one.

The same reading applies to the Diesel cycle, which adds its heat at constant pressure while the piston is already moving down. At the same compression ratio it is less efficient than the Otto cycle, because the heat added during expansion leaves the cycle with a hotter exhaust and a higher Tˉout\bar T_\text{out}. Diesel engines are nonetheless the more efficient engines in practice, because compressing air alone, with no fuel in it to knock, lets them run at compression ratios of fifteen to twenty where a petrol engine is held near ten — and the mean temperature of heat input rises with the compression.

That changes what the ceiling is for. Carnot’s formula between the extremes answers the question of what an engine could do if all its heat went in at the top and came out at the bottom. The mean-temperature formula answers the question of what this cycle does, and it points at what would improve it: anything that raises Tˉin\bar T_\text{in} or lowers Tˉout\bar T_\text{out}, with no reference to the peak at all.

A stack of identical Carnot engines

An Otto cycle is a stack of identical Carnot engines. The same Otto cycle cut into 12 thin vertical strips of equal entropy width. Each strip is a Carnot cycle of its own, taking heat in at the upper curve and rejecting it at the lower. For this cycle the two curves are the same exponential scaled by the temperature ratio of the compression, so every strip has the same efficiency, 58.48%, equal to the whole cycle's to twelve decimal places. The strips carry different amounts of heat — the first 60 kJ/kg, the last 198 — but convert it at one rate, which is why an Otto engine's ideal efficiency depends on nothing but its compression ratio. The 12 strips hold the loop's work to 0.05%.
Fig. 3 The same cycle cut into twelve vertical strips of equal entropy width, each a thin Carnot cycle between its own top and bottom temperatures. Every strip has the same efficiency, 58.48%, equal to the whole cycle’s; they carry different amounts of heat, from 60 kJ/kg in the first to 198 kJ/kg in the last, but convert it at one rate.

There is a second way to see the same result, and for the Otto cycle it has a surprise in it. Any reversible cycle can be sliced into thin vertical strips on the temperature–entropy diagram, and each strip is a small Carnot cycle: a sliver of heat in at the temperature of the top edge, the same sliver of entropy out at the temperature of the bottom edge. The whole cycle’s work is the sum of the strips’, and its efficiency is their heat-weighted average.

For the Otto cycle every strip has the same efficiency. The top edge is the exponential the constant-volume heating follows, the bottom edge is the same exponential lower by the factor the compression raised the temperature, and so at every entropy the ratio of bottom to top is the same number: r1−γr^{1-\gamma}, with rr the compression ratio and γ\gamma the ratio of the gas’s heat capacities. Twelve strips, twelve efficiencies of 58.48 per cent, agreeing with the whole cycle to twelve decimal places.

That is why an ideal Otto engine’s efficiency depends on its compression ratio and on nothing else. Adding more fuel lengthens the loop along the entropy axis and adds more strips, each as efficient as the last. Burning hotter does not help unless it comes with more compression, and the engine’s efficiency is set the moment the piston reaches the top of its stroke. In a real engine the limit on compression is knock, the charge igniting by itself before the spark, which is why the fuel’s resistance to it is rated by a number printed on the pump.

Moving where the heat is paid for

A gas turbine’s cycle is the Brayton cycle: compress without heat exchange, heat at constant pressure, expand, cool at constant pressure. On the temperature–entropy diagram it looks much like the Otto cycle with slightly flatter curves, and its ideal efficiency has the same form, set by the pressure ratio.

A regenerator raises one mean and lowers the other. An ideal gas-turbine cycle for air, pressure ratio 12, intake 300 K, turbine inlet 1500 K, on temperature–entropy axes. Left, the simple cycle: heat enters from 610 to 1500 K at a mean of 989 K and leaves from 737 to 300 K at a mean of 486 K, for 50.8%. Right, an ideal regenerator passes the exhaust's heat above 610 K to the compressed air, drawn dotted: now heat is bought only from 737 K up, at a mean of 1074 K, and thrown away only below 610 K, at a mean of 437 K, for 59.3%. The loop is the same; only where its heat is paid for has moved. Both efficiencies from the mean temperatures match the closed forms, and a Carnot engine between 300 and 1500 K would reach 80.0%.
Fig. 4 An ideal gas turbine at pressure ratio 12, intake 300 K and turbine inlet 1,500 K. Left: heat enters at a mean of 989 K and leaves at 486 K, for 50.8%. Right: a regenerator passes the exhaust’s heat above 610 K to the compressed air, so heat is bought only from 737 K up, at a mean of 1,074 K, and discarded only below 610 K, at 437 K, for 59.3%. The loop is the same.

It has a feature the Otto cycle does not. The exhaust leaves the turbine at 737 K and the air leaves the compressor at 610 K, so the exhaust is hotter than the air about to be heated, and a heat exchanger can pass heat from one to the other. That is a regenerator, and the figure shows exactly what it does. The loop is unchanged. What moves is which part of its heat is bought from the fuel and which is thrown away: the fuel now supplies only the heat above 737 K, where its mean temperature is 1,074 K instead of 989 K, and the surroundings receive only the heat below 610 K, at a mean of 437 K instead of 486 K. Both means move the right way at once, and the efficiency goes from 50.8 per cent to 59.3 per cent without a single change to the turbine.

The mean-temperature reading also says when a regenerator stops helping. As the pressure ratio rises the compressor’s outlet gets hotter and the turbine’s exhaust cooler, and once they cross there is nothing for the exchanger to pass: at a pressure ratio of about 16.6 for these temperatures they are equal, and above it a regenerator would have to heat the exhaust with the compressed air, which lowers the mean at which heat enters. Regenerated gas turbines are built with low pressure ratios for exactly this reason.

A second engine on the exhaust

A second engine on the exhaust lowers the mean it rejects at. The simple gas-turbine cycle again, pressure ratio 12 and 1500 K at the turbine, converting 50.8% of its heat. Its exhaust leaves at 737 K, and a second ideal engine, drawn as the lower shaded region, takes heat from that exhaust as it cools to 380 K and rejects it at the 300 K surroundings. That heat arrives at a mean of 539 K, so the second engine converts 44.4% of it. Together they turn 68.7% of the fuel's heat into work. The same number comes from the mean temperatures of the plant as a whole: heat in at 989 K, out at 310 K, with the entropy taken in and rejected equal to one part in a billion. Real combined-cycle plants, with losses everywhere this model has none, reach about sixty per cent.
Fig. 5 The simple gas turbine again, converting 50.8%, with a second ideal engine taking heat from its exhaust as it cools from 737 K to a 380 K stack and rejecting it at 300 K. That heat arrives at a mean of 539 K and the second engine converts 44.4% of it. Together they convert 68.7% of the fuel’s heat: in at a mean of 989 K, out at 310 K.

The other way to lower Tˉout\bar T_\text{out} is to stop rejecting heat at 737 K at all. A combined-cycle power station puts a steam plant on the gas turbine’s exhaust: the exhaust boils water, the steam drives a second turbine, and only then does the heat go to the cooling towers. In the figure the second engine is ideal. It takes the exhaust’s heat as the gas cools from 737 K to a stack temperature of 380 K, at a mean of 539 K, and converts 44.4 per cent of it; the pair converts 68.7 per cent of the fuel’s heat.

The plant as a whole obeys the same formula. Its heat enters at a mean of 989 K, from the gas turbine’s combustor, and leaves at a mean of 310 K, partly through the stack and mostly through the second engine’s condenser; the figure checks that the entropy taken in and given out agree to a part in a billion, and 1−310/9891 - 310/989 is the 68.7 per cent. A combined-cycle plant is not a more efficient engine in any mysterious sense. It is one whose rejected heat leaves at a temperature much closer to the surroundings’.

Real combined-cycle plants, with friction in the turbines, temperature differences across every heat exchanger and a steam cycle that cannot follow the exhaust’s cooling curve, reach about sixty per cent: the best ratio of electricity to fuel of any thermal power station.

The same arithmetic, run backwards

A heat pump is an engine run in reverse, and the mean temperatures work for it too. The engine that pays back more than it takes found that an ideal heat pump delivers Th/(Th−Tc)T_h/(T_h - T_c) joules of heat for each joule of work, and that the ratio is large when the two temperatures are close. For a heat pump that warms water over a range, the ThT_h in that formula is the entropy-weighted mean of the delivery temperatures, not the final one.

Take heat from outside air at 5 °C and heat water from 30 to 50 °C. Delivering all of the heat at 50 °C would allow an ideal coefficient of 7.18. But the water is at 30 °C when heating begins, and its mean delivery temperature is 313.0 K, about 40 °C; the ideal coefficient for the actual job is 8.97, a quarter higher. The same logic that penalises an engine for taking heat in at low temperatures rewards a heat pump for delivering it at low ones, and it is why heat pumps are best suited to underfloor heating and large radiators, which run cool, and why they struggle to produce water hot enough for an old system of small ones.

The steam plant, read the same way

The same reading explains the shape of every steam cycle ever built. Water is heated as a liquid, boiled, and the steam is expanded through a turbine and condensed. Boiling happens at a single temperature, set by the pressure — a boiling point is a pressure — and most of the heat goes in there, as the latent heat that changes no temperature. So the heat input is close to the isothermal ideal Carnot wanted, but at a low temperature: water boiling at 60 atmospheres is at 276 °C.

Everything done to steam plants since then raises Tˉin\bar T_\text{in}. Higher boiler pressures raise the boiling temperature. Superheating the steam after it boils adds heat at temperatures above the boiling point. Reheating it partway through the turbine adds more heat at high temperature. And above 221 bar water no longer boils at all, so a supercritical plant heats it smoothly past 600 °C. Each step pulls the mean temperature of heat input upwards, and the efficiency of the best coal and nuclear stations has followed it from under twenty per cent to about forty-five.

Where the model stops

Every cycle is reversible. The formula η=1−Tˉout/Tˉin\eta = 1 - \bar T_\text{out}/\bar T_\text{in} uses the fact that the entropy out equals the entropy in. A real cycle generates entropy in friction, in turbulence and in heat flowing across finite temperature differences, so it gives out more entropy than it takes in, and its efficiency is lower by the rejection temperature times the entropy generated, per joule of heat input. The mean temperatures still locate the ideal; the entropy generated measures the distance from it.

The working fluid is ideal. The figures use air with constant heat capacities; real combustion products have heat capacities that rise with temperature and change composition, which moves the numbers by a few per cent and the argument not at all.

Heat is added, not generated. An internal-combustion engine does not have heat delivered to it; its working fluid reacts. Treating combustion as heat added at constant volume is the air-standard idealisation, and it captures what the compression ratio does while leaving out what the chemistry does.

And an engine is not run for efficiency alone. A car engine is sized for the power it must deliver when grip hands over to power and spends most of its life far below it, where friction and pumping losses take a larger share, and real ones deliver a quarter to a third of their fuel’s heat as work. Even the smallest engine anyone has proposed obeys the same accounting: the engine a fluctuation cannot run is held below Carnot by exactly the entropy its leaks generate. The engine that has to finish found that the most power comes at a lower efficiency, and every design here trades efficiency against size, cost and the temperatures its materials survive. Turbine inlet temperatures are limited by blades, not by thermodynamics.

What the pictures cannot show

A temperature–entropy diagram has no time in it. The Otto cycle’s four steps take a few hundredths of a second together, and the diagram cannot show that combustion is not at constant volume in any real engine, because the piston is moving while the flame spreads. It shows the ideal the engine is measured against.

Nor can it show the heat exchangers. The regenerator and the steam boiler are drawn as transfers along a curve, and in a real plant they are large pieces of equipment in which the two streams must differ in temperature for heat to flow at all; that difference is entropy generated, and its cost is why they are as large as they are.

Still open: how hot a turbine blade can be made to run

For the gas turbine, and for every combined-cycle plant built on one, the mean temperature of heat input is limited by the turbine inlet, and the turbine inlet is limited by the first row of blades, which sit in a gas stream hotter than the melting point of the nickel alloys they are made of. They survive because they are single crystals, coated with ceramic, and cooled by air bled from the compressor through passages inside them — air that is then not doing work, which costs efficiency back.

Whether the next increase comes from ceramic matrix composites that tolerate higher temperatures without cooling, from better coatings, or from cycles that burn hydrogen or use supercritical carbon dioxide as the working fluid, and how the gains trade against durability over tens of thousands of hours, is an open engineering question rather than a thermodynamic one. The thermodynamics is settled: each kelvin added to the mean at which heat arrives is worth a calculable fraction of a per cent.

The habit worth keeping is the one the rectangle teaches. When a limit is stated for an ideal, find the ideal’s equivalent for the real case before measuring the shortfall. Carnot between the extremes says an Otto engine wastes thirty points; Carnot between its mean temperatures says it wastes nothing, and that the thirty points are the price of receiving heat where it does.

Part 5 of 8

This essay is one argument about Heat engines. The others:

One temperature, four pawls, and nothing gained. The net rate of a ratchet whose gas and whose pawl are at the same temperature, against the load, for notches 2, 5, 10, 20 times the thermal energy deep. Every curve passes through zero at zero load and is negative everywhere else. The device is not merely unable to lift a weight; under any load at all it turns the wrong way and lets the weight down, converting its potential energy into heat in the gas. Making the notch deeper slows everything down — an exponential in the depth — and does not change the sign anywhere. That is the second law arriving as a mechanism rather than as a prohibition. Nothing was assumed about entropy; the pawl was simply allowed to be as warm as everything else, and its own fluctuations undo exactly the rectification it was there to provide. Any rectifier small enough for thermal noise to matter has this problem, and the rectifier being clever does not help, because the same noise reaches the cleverness. The engine a fluctuation cannot run Part 4 — A ratchet lets a shaft turn one way and not the other. Put a paddle in a gas on the same shaft and molecular collisions appear to become a lifted weight — an engine running on one reservoir. It does not work, and following exactly why turns the second law from a prohibition into a mechanism: the pawl is as warm as the gas, and it lifts whenever it is asked to. Two bodies an engine draws together. Two equal bodies of 4.186 kJ/K — a kilogram of water each — one at 90.0 °C and one at 10.0 °C, against the heat drawn from the hot one. The solid curves are the best possible engine running between them, a reversible one, which leaves the product of the two temperatures unchanged and brings both to the geometric mean, 320.7 K (47.5 °C). It draws 177.8 kJ from the hot body and delivers 20.8 kJ of work, C(√T₁ − √T₂)², checked against the heat balance. The dashed lines are the same bodies simply touching: they meet at the arithmetic mean, 323.1 K, having exchanged 167.4 kJ and delivered nothing. The 2.5 K between the two endpoints is the work, left behind as heat. The work left in two buckets of water Part 6 — Carnot's ceiling assumes reservoirs so large that taking heat from one and giving it to the other changes neither temperature. Two buckets of water are not reservoirs. Run the best possible engine between a hot one and a cold one and both temperatures move, the efficiency available shrinks as they do, and the engine stops when they meet — at the geometric mean of the starting temperatures, not the ordinary one. The work it delivered is exactly the difference between those two meeting points, and it is far less than the starting temperatures promise.

Where this goes next

A Carnot cycle on pressure–volume axes. Two isothermal steps joined by two adiabatic ones, forming a closed loop. The gas expands 9.6-fold in reaching the cold reservoir at 0.50 of the hot one, and the area enclosed is the net work done over one cycle. The ceiling on every engine, set before it was designed There is a maximum efficiency no heat engine can exceed, and it depends on nothing but two temperatures. Not the fuel, not the working substance, not the cleverness of the engineer. The ceiling, the estimate, and three power stations. Two efficiencies against the ratio of the cold reservoir's temperature to the hot one. The upper curve is Carnot's 1 − Tc/Th, which is a ceiling on the work per unit of heat and is reached only by an engine that runs infinitely slowly, because a reversible heat flow needs a vanishing temperature difference to drive it and therefore infinite time. The lower curve is 1 − √(Tc/Th), the efficiency of an engine with finite thermal contact run for the most power rather than the most work. Three measured plants are marked: West Thurrock, coal runs at 36 per cent against a ceiling of 64 and a finite-time estimate of 40; CANDU, nuclear runs at 30 per cent against a ceiling of 48 and a finite-time estimate of 28; Larderello, geothermal runs at 16 per cent against a ceiling of 33 and a finite-time estimate of 18. Every one of them is closer to the lower curve — within 4.4 points at worst, against 16.5 at best from the ceiling. The second law is not what limits a working power station. What limits it is that somebody wants the electricity this year. The engine that has to finish Carnot's ceiling is exact and it is reached only by an engine that takes for ever, because a reversible heat flow needs a vanishing temperature difference to drive it. Ask instead for the most power rather than the most work per joule of heat, and the answer is a different function of the same two temperatures — and three measured power stations sit on it rather than on the ceiling. Heating 1 kg of water from -20°C to 130°C. Temperature against heat added for 1 kilogram of water taken from -20 to 130 degrees Celsius. The two flat stretches are the melting and the boiling, where 334 and 2260 kilojoules go in and the temperature does not move. Melting costs as much as warming the water by 80 degrees; boiling costs as much as warming it by 541, which is 73 per cent of the whole journey. The heat that changes no temperature, and where it actually goes A kettle reaches a hundred degrees in a minute and takes five more to boil dry. The heat going in during those five minutes changes nothing a thermometer can see, and it is most of the energy in the whole process.

What links here

Essays that reach for this one mid-argument — the half of a link its own author cannot write down.

The objects named here

The third axis, after the field and the reading path: the things themselves, and every essay that touches each one.

Brayton cycleThe Carnot cycleCombined-cycleEfficiencyEntropyHeat engineMean temperatureOtto cycleRegeneratorReversibility