Recycling heat

Leans on Entropy is missing information.

Can heat that is on its way out still do some work?

Picture a water wheel. It turns only while water falls past it. Set the same wheel in the middle of a pond, where the water has nowhere lower to go, and nothing happens, however much water there is.

A heat engine similarly requires a temperature difference. It absorbs heat from a hotter source, converts part into work, and rejects the rest to a colder sink. No cyclic engine can extract net work from heat at a single temperature. Sadi Carnot used the waterfall analogy in 1824.

The Vigil has a big drop built in. Its computers run warm, at 300 K, room temperature. Its radiators run at 30 K, ten times colder, for reasons that fill the next page, Cooling is the economy. Every watt the computers use turns into heat at the warm end, and all of that heat has to travel to the cold end to leave. Engines recover work during that transfer.

Here is what they do with one petawatt of supply. The computers use it, and all of it becomes heat. On its way to the radiators the heat passes through the engines, which catch a share of it, 0.36 of it in the Vigil's design, and turn that share back into electricity. The electricity goes straight back into the computers. They use it, and it becomes heat again, which falls through the engines again, and a share of that is caught again. Each round is smaller than the one before it, and the rounds add up to a limit.

For one petawatt in, the computers end up doing the work of 1.5625 petawatts. More than half again as much thinking, from the same supply.

Guess first The Vigil feeds in one petawatt and its computers use more than that. Where does the extra come from? And how much heat leaves at the radiators?

The apparent extra is energy used more than once: a joule runs the computers, turns to heat, is partly caught by an engine, and runs the computers again. What leaves at the cold end is exactly 1 petawatt, as much as went in. The energy balance includes both the external supply and the recirculated flow.

In a steady state, the Vigil takes in a petawatt and radiates a petawatt. Internal recovery increases the energy flowing through its computers without changing that external balance.

Lower radiator temperatures increase the fraction the engines recover from each pass. They also lower the minimum erasure cost (see The cost of forgetting). Rejecting the remaining heat at low temperature requires more radiator area, reaching the surface area of many Earths in the Vigil's design.

The cascade as flows. The computer's heat feeds an engine that hands part of it back as work, so the supply buys more computing than it carries, and exactly as much heat leaves as came in. Arrow width is power.

How much does the cascade add up to?

Start with the best an engine can do. An ideal engine working across a drop from a hot end at temperature T_hot to a cold end at T_cold, both in kelvin, can turn this share of the heat passing through it into work:

ideal share = 1 − T_cold ÷ T_hot

This is Carnot's limit, and nothing can beat it. The cold end matters a great deal:

Cold end	Ideal share turned into work
300 K, same as the computers	none
150 K	0.50
30 K, the Vigil's radiators	0.90
2.7 K, the afterglow of the Big Bang in 2540	0.99

Real engines fall short. The Vigil's ledger assumes they reach 40 percent of the ideal design, so each pass recovers

r = 0.40 × 0.90 = 0.36

Next, follow the petawatt round the loop. The supply is the first pass. Each later pass is 0.36 of the one before:

Pass	Power added (PW)	Running total (PW)
Supply	1.00	1.000
1	0.360	1.360
2	0.130	1.490
3	0.0467	1.536
4	0.0168	1.553
5	0.00605	1.559
6	0.00218	1.561

The running total creeps toward a limit. After 4 passes it is within one percent. A sum like this, each term a fixed fraction of the last, has a closed form. (The passes are not really taken one after another. In a steady state all of them are happening at once.)

computing = supply ÷ (1 − r)

With r = 0.36 and a supply of one petawatt, that is 1 ÷ 0.64 = 1.5625 petawatts of computing. Of it, 1 petawatt is the supply and 0.5625 petawatt is recovered. The heat that finally leaves is the computing power minus what the engines caught, which is 1 petawatt, the supply, exactly.

The ledger uses this relation to calculate how engine performance affects computing power:

Engines reach	Recovered per pass	Computing from 1 PW (PW)
20 percent of ideal	0.18	1.22
40 percent of ideal (the Vigil)	0.36	1.5625
60 percent of ideal	0.54	2.17
Perfect engines	0.90	10

The last row is the ceiling: perfect engines stretch the supply by exactly the ratio of the two temperatures, 300 ÷ 30. No real engine gets there, and the folds below say why. A colder end raises the ceiling, to 110 with the cold end at the temperature of the sky of 2540, but only on paper: a radiator as cold as the sky sheds no net heat into it. The heat that does leave the Vigil leaves at 30 K, and one petawatt of it needs 43 Earths of radiating surface. The radiator page calculates the resulting area and mass.

With JavaScript on, this is a calculator: set the temperature of the computers and the radiators and how good the engines are, and it shows the passes adding up, the computing carried by a petawatt of supply, and the heat that leaves. The tables above give the same numbers for the Vigil.

Open everything

Why is Carnot's share the limit, and why does the series add up?An entropy balance, a geometric series and a maximum-power engine

Where does 1 − T_c/T_h come from?

An engine takes heat $Q_h$ from a hot reservoir at $T_h$ , gives heat $Q_c$ to a cold one at $T_c$ , and puts out work $W$ . Over one full cycle the engine is back where it started, so energy conservation says

$W = Q_h - Q_c$

The second law adds the entropy balance. As Entropy is missing information shows, heat $Q$ at temperature $T$ carries entropy $Q/T$ . The hot reservoir loses $Q_h/T_h$ of it, the cold one gains $Q_c/T_c$ , and the engine, back in its starting state, holds no more than it did. Entropy cannot fall, so

$\frac{Q_c}{T_c} - \frac{Q_h}{T_h} \ge 0 \quad\Rightarrow\quad Q_c \ge Q_h\,\frac{T_c}{T_h}$

Put that into the work:

$W = Q_h - Q_c \le Q_h\left(1 - \frac{T_c}{T_h}\right) = \eta_C\, Q_h$

Equality needs the entropy balance to hold exactly, which means no entropy is made anywhere in the engine: a reversible cycle.

The entropy carried per joule explains the result. A joule of heat at 300 K carries 0.0033 J/K of entropy, and the engine must hand that entropy on. A joule of heat at 30 K carries 0.033 J/K, 10 times as much per joule, so 0.1 of a joule at the cold end is enough to carry it away. That leaves $1 - 0.1 = 0.9$ of the joule free to leave as work. A colder sink receives more entropy per joule, leaving a larger fraction of the incoming energy available as work.

Why does the cascade add up?

Let $P$ be the power the computers use and $P_0$ the supply. In a steady state all of $P$ becomes heat at the hot end. The engines turn a share $r = f\,\eta_C$ of that heat back into work, where $f$ is the share of the ideal they reach. The work $rP$ returns to the computers, so

$P = P_0 + rP \quad\Rightarrow\quad P = \frac{P_0}{1 - r}$

The same answer comes out of the picture of passes. The supply is $P_0$ , the first recovery is $rP_0$ , the next $r^2P_0$ , and so on:

$P = P_0\left(1 + r + r^2 + \dots\right) = P_0 \lim_{n\to\infty}\frac{1 - r^{n+1}}{1 - r} = \frac{P_0}{1 - r}$

The limit exists because $r < 1$ , so $r^{n+1} \to 0$ . The heat that leaves is what is not recovered:

$P_{\text{out}} = (1 - r)\,P = P_0$

The first law is satisfied exactly, for any $r$ . With $f$ = 0.4 and $\eta_C$ = 0.9, $r = 9/25$ , so $1 - r = 16/25$ and the ratio is $25/16 = 1.5625$ .

With perfect engines $r = \eta_C$ , so $1 - r = T_c/T_h$ and the stretch is exactly $T_h/T_c$ , which is 10 for the Vigil's two temperatures. The sum blows up only as $r \to 1$ , which needs a perfect engine and a cold end at absolute zero. Neither exists: absolute zero cannot be reached, and a perfect engine is a reversible one, which delivers no power in any finite time. The second claim is the next section.

How fast can an engine be and still be good?

A reversible cycle approaches its ideal efficiency as the operating rate tends to zero. To deliver power, an engine must let heat flow across a gap, and a flow needs a temperature difference. In 1975 Frank Curzon and Boye Ahlborn put the simplest version of this into a model. Their engine's working fluid is at $x$ while it takes heat from a source at $T_h$ , and at $y$ while it gives heat to a sink at $T_c$ , with $x < T_h$ and $y > T_c$ . Heat crosses each gap at a rate proportional to the temperature difference, with the same conductance $K$ on both sides. Inside the engine everything is reversible, so $Q_h/x = Q_c/y$ , and the strokes between the hot and cold stages take no time.

The time to absorb $Q_h$ is $Q_h/\big(K(T_h-x)\big)$ and the time to reject $Q_c$ is $Q_c/\big(K(y-T_c)\big)$ . Write $\rho = y/x$ , so that $Q_c = \rho Q_h$ and $y = \rho x$ . The power is work over cycle time:

$P = \frac{Q_h - Q_c}{t} = \frac{K(1-\rho)}{\dfrac{1}{T_h - x} + \dfrac{\rho}{\rho x - T_c}}$

For a fixed $\rho$ , make the denominator as small as possible. Setting its derivative with respect to $x$ to zero gives $\rho x - T_c = \rho(T_h - x)$ , so $x = (T_h + T_c/\rho)/2$ . Put that back and the denominator becomes $4\rho/(\rho T_h - T_c)$ , so

$P = \frac{K}{4}\,(1-\rho)\left(T_h - \frac{T_c}{\rho}\right)$

Next choose $\rho$ . Multiplied out, the product is $T_h + T_c - \rho T_h - T_c/\rho$ , and its derivative $T_c/\rho^2 - T_h$ vanishes at $\rho = \sqrt{T_c/T_h}$ . The efficiency of the engine is $1 - \rho$ , and the maximum power is

$\eta_{CA} = 1 - \sqrt{\frac{T_c}{T_h}}, \qquad P_{\max} = \frac{K}{4}\left(\sqrt{T_h} - \sqrt{T_c}\right)^2$

The efficiency does not depend on $K$ , and it comes out the same if the two gaps have different conductances. For 300 K and 30 K it is 0.68, which is 0.76 of Carnot's 0.90. The working fluid sits at 197 K and 62 K, inside the span from 30 K to 300 K. Curzon and Ahlborn also showed that the engines of their day were well described by it.

An engine like this, reversible inside and lossy only at its borders, is called endoreversible, and the square root is exact for it when heat flows in proportion to the temperature difference; other laws of heat flow give other values. Nor is it a ceiling: the same engine can run closer to Carnot, at less power.

Two things follow for the Vigil.

The ledger's 40 percent of Carnot, a recovery of 0.36, sits well below the maximum-power efficiency. That is a conservative design choice design, with room for the friction and heat leaks this model leaves out. As a what-if only: if every pass recovered the Curzon-Ahlborn share, one petawatt would carry 3.2 petawatts of computing, 2.0 times the ledger's figure. The Vigil's books do not count on it.
This is a different trade from the one on Cooling is the economy, where the engine is perfect and the limit comes from how brightly a cold surface glows. Here the engine is the limit, because heat has to cross finite gaps into and out of it. A real plant faces both.

Can an engine reach Carnot at a useful power, and what does the ledger really count?Bounds on power and efficiency, and exergy

The Curzon-Ahlborn efficiency is not universal

The square-root formula depends on the model of heat transfer. In 2010 Massimiliano Esposito, Ryoichi Kawai, Katja Lindenberg and Christian Van den Broeck studied engines that reach Carnot's efficiency when run reversibly and lose a little to dissipation when run fast. They found that, for low dissipation, the efficiency at maximum power is pinned between $\eta_C/2$ and $\eta_C/(2-\eta_C)$ , the two extremes of how the dissipation divides between the hot and cold stages. When the dissipation is the same in both, the Curzon-Ahlborn value comes back. Between 300 K and 30 K the limits are 0.45 and 0.82, and the Curzon-Ahlborn value of 0.68 lies between them. The Vigil's 0.36 sits below even the lower limit, providing margin for losses omitted from the model.

Carnot at finite power is ruled out

Is there some cleverer engine that reaches the full ideal share and still delivers real power? In 2016 Naoto Shiraishi, Keiji Saito and Hal Tasaki proved there is not. For any heat engine described as a Markov process, however far from equilibrium, they bounded the dissipation from below by the square of the heat current, so any flowing current dissipates. That gives a universal trade-off between power and efficiency, and a theorem: an engine with nonvanishing power never attains the Carnot efficiency. In 2018 Patrick Pietzonka and Udo Seifert added a third side for steady-state engines: of large power, efficiency near Carnot's and small fluctuations in the output, only two are compatible.

For the Vigil the 0.9 is therefore a ceiling that real engines approach only as their power falls toward zero. The trade-off is about power per engine, though. A plant can win efficiency back by spreading the heat over more machinery, each unit running slower and closer to the ideal, at the price of building more of it. This is a design inference from those results design.

What does the ledger actually count?

The petawatt figures count energy, and energy alone does not say how much work a flow of heat can still do. The quantity that does is exergy: the most work obtainable from a form of energy, given an environment at temperature $T_0$ . For heat $Q$ at temperature $T$ :

$\text{exergy} = Q\left(1 - \frac{T_0}{T}\right)$

With the radiators at 30 K as the environment, a joule of heat at 300 K carries 0.9 joule of exergy: Carnot's share, seen from the other side. A joule at the radiators' own temperature carries none, and no engine can use it. The ledger takes 40 percent of that exergy, 0.36, and passes it back. The rest, 0.54 joule of exergy for each joule of heat that goes through, is destroyed.

Destroyed exergy has an exact meaning. Adrian Bejan's review of exergy analysis states the link: exergy destruction is proportional to entropy generation, the factor being the temperature of the environment. Check it against the ledger. One joule leaves the computers and 0.64 joule arrives at 30 K, so the entropy made is $0.64/30 - 1/300 = 0.018$ J/K. Multiplied by $T_0$ = 30 K, that is 0.54 joule, the same number. The two calculations agree. Energy conservation balances the total flow, while exergy accounting quantifies the loss of capacity to do work.