Attractlet attractlet model · § 7.4

The Carnot Cycle

The most famous loop in physics is usually drawn rather than run. Run it, and two things turn up that are not on the diagram: it produces no power at all, and it is not an attractor.

Give an engine a hot source and a cold sink, and Carnot’s cycle tells you the most work you can ever get out of the heat: a fraction 1 − T_c/T_h of it. That ceiling is real. What the diagram does not tell you is how long reaching it takes. Every stroke has to be slow enough that the gas never lags behind the reservoir it is touching, which means infinitely slow. An engine that takes forever to finish one cycle delivers work at a rate of zero.

The Loop, Actually Running

an ideal gas, a piston on a schedule, two reservoirs

reservoirs,

Carnot efficiency,

efficiency at maximum power, predicted,

efficiency now,

power now,

work per cycle,

how much of a nudge survives one cycle,

starting

Nothing in this panel is numerically integrated. Inside each leg of the cycle the conductance and the rate of change of log-volume are both constant, so the gas obeys T′ = A − BT with A and B fixed, and the end temperature, the integral of T over the leg, the work and the heat all have closed forms. The engine is computed to machine precision. The same fact makes the whole cycle an affine map on temperature, which is why the panel can solve for the repeating cycle directly instead of waiting for it, and why the number in the last readout is exact rather than fitted.

Two Things the Diagram Leaves Out

both fixed by the same knob

The textbook picture is the dashed rectangle in the right-hand plot: two isotherms at T_h and T_c, two vertical adiabats, and an efficiency of 1 − T_c/T_h. To draw it you assume the gas is always at exactly the temperature of whatever it is touching, which requires heat to cross a contact at zero temperature difference, which requires infinite time.

First thing left out: the power is zero. Drag the cycle-time slider to the right and the efficiency climbs towards the Carnot value while the power drains away. The lower plot draws that trade-off. Carnot’s own point is the grey dot at the far right, sitting on the floor.

Second thing left out: it is not an attractor. With no heat crossing a temperature difference there is no dissipation anywhere in the cycle, so nothing in the physics distinguishes the intended loop from its neighbours. Press Insulate it and watch the loop collapse to a line: with both conductances at zero the gas simply follows the piston adiabatically, the enclosed area goes to zero, and a nudged gas lands on a parallel line and stays there. One setting removes the work and the basin together.

The Basin, Which Has a Closed Form

this card’s claim, earned

Each leg of the cycle is linear in the gas temperature, so going once round is an affine map, T → aT + b. The single number a decides the stability:

a = exp( −(k_h t_hot + k_c t_cold) / nC_v )

Every displacement from the repeating cycle shrinks by that factor each time round. At the default setting a = 8.09 × 10⁻³, so press Nudge the gas and a swing of 132 K is down to 1.07 K after a single cycle, 8.6 millikelvin after two, and 70 microkelvin after three.

Nothing appears in that formula except the two conductances and the two contact times. Contact with the reservoirs is the whole source of the basin. Set the conductances to zero and a is exactly 1: a displacement never shrinks, every temperature is its own closed loop, and there is no basin. Lengthen the contact or raise the conductance and a falls towards zero: the cycle grips harder, the engine slows to quasi-static, and the power drains away.

One dissipation does three jobs here. Heat crossing a temperature difference is what makes the cycle attracting, what holds the engine below the Carnot efficiency, and what lets it produce any power at all.

Where the Power Is Greatest

a result the panel measures rather than looks up

So the question worth asking is what efficiency an engine has when it is run flat out for power. Curzon and Ahlborn answered that in 1975, for this exact model:

η at maximum power = 1 − √(T_c/T_h)

The panel never uses that formula. It sweeps the cycle time, finds the best the engine can do at each one, and marks the peak with a red dot. The red dashed line is where the formula says the dot should land.

hot reservoir	Carnot	1 − √(T_c/T_h)	measured at peak power	error
350 K	14.29%	7.418%	7.420%	+0.023%
500 K	40.00%	22.540%	22.538%	−0.010%
600 K	50.00%	29.289%	29.287%	−0.009%
900 K	66.67%	42.265%	42.260%	−0.012%
1500 K	80.00%	55.279%	55.281%	+0.005%

All against a cold reservoir at 300 K. At power-station temperatures the efficiency available at full output is a little over half the Carnot number. The shortfall has nothing to do with friction or leaks. Heat only moves down a temperature difference, so an engine that moves heat fast has to waste some of it.

Why This Is an Attractlet

the plainest kind of supplied structure

Everything here is delivered. Something outside the picture holds the two reservoirs at their temperatures, the piston follows a schedule somebody wrote, and the gas contributes nothing but its equation of state. Take away the temperature difference and the cycle, the work and the basin all stop at once.

This is the clearest case in the set of a structure that is entirely a framework. A scientist builds it in order to think about engines, and it lasts exactly as long as somebody runs it. Its basin is real and measurable and has a closed form, and every term in that closed form was put there by the person who set the problem up.

What This Model Is Not

the limits

The piston is on a schedule, not a shaft. Volume is prescribed as a function of time, so the only thing free to move is the gas temperature. A real engine has a crank, a flywheel and a load, and its cycle lives in a bigger space than one number. The basin measured here belongs to the gas alone.

Newton’s law of cooling. Heat crosses each contact in proportion to the temperature difference. That assumption is Curzon and Ahlborn’s, and it is where the square root comes from. Other heat-transfer laws give other answers.

No friction, no leakage, no heat loss anywhere else. The engine is internally reversible, which is what “endoreversible” means: every irreversibility in the model lives in the two heat exchangers. Real engines lose heat everywhere.

The square-root result has been argued about ever since. The Carnot efficiency is a bound on every engine between two reservoirs. This one is only what a particular family of engines does when it is optimised, and how far it generalises is still discussed.

No real engines are quoted here. Curzon and Ahlborn’s paper famously compared their formula against three working power stations, and those numbers are widely reproduced. This page leaves them out because the original could not be read directly, and figures copied second-hand are not a comparison. Every number here came out of this panel.

Curzon, F. L. & Ahlborn, B. (1975). Efficiency of a Carnot engine at maximum power output. American Journal of Physics, 43(1), 22–24. Source of the endoreversible model and the efficiency at maximum power.
The many avatars of Curzon-Ahlborn efficiency, arXiv:1903.04381 (fetched and read 2026-09-13). Source of the statement of the model used here: Newtonian heat transfer Q_h = α(T_h − T₁)t₁, the internal reversibility condition Q_h/T₁ = Q_c/T₂, and the result.
True nature of the Curzon-Ahlborn efficiency, arXiv:1612.07922 (fetched and read 2026-09-13). Source of the caveats in the limits above.

← Back to Attractlets