The engine that pays back more than it takes

Carnot’s ceiling says that an engine drawing heat from a reservoir at $T_h$ and dumping it at $T_c$ can convert at best a fraction $1 - T_c/T_h$ of that heat into work. Between a boiler at 800 K and a river at 300 K that is 62%, and real engines get well under it — the ceiling is on a cycle that is reversible, and nothing that happens at a useful rate is. It is the most quoted result in thermodynamics and it is quoted almost entirely as a limitation.

Run the same cycle backwards and it becomes a licence.

Inverting the ratio #

A Carnot engine run forwards takes $Q_h$ from the hot reservoir, delivers $W$, and dumps $Q_c$, with

$$\frac{Q_h}{T_h} = \frac{Q_c}{T_c}$$

because the cycle is reversible and its total entropy change is zero. Reverse every step and the same relation holds with the arrows turned round: work $W$ goes in, $Q_c$ is taken from the cold side, and $Q_h = Q_c + W$ is delivered to the hot side.

The figure of merit depends on which of the two heats is wanted. For heating a house, what matters is $Q_h/W$:

$$\text{COP}_{\text{heat}} = \frac{Q_h}{W} = \frac{Q_h}{Q_h - Q_c} = \frac{T_h}{T_h - T_c}.$$

For cooling a fridge, what matters is $Q_c/W$:

$$\text{COP}_{\text{cool}} = \frac{T_c}{T_h - T_c},$$

and the two differ by exactly one, because $Q_h = Q_c + W$.

Note what has happened to the ceiling. The engine’s efficiency is $(T_h - T_c)/T_h$, a number between 0 and 1 that is small when the reservoirs are close together. The pump’s coefficient is its reciprocal, so it is large when the reservoirs are close together — and the closeness of the reservoirs is exactly the situation of a house in a mild winter.

The numbers for a house #

With the inside at 21 °C (294.15 K) and the outside at 7 °C (280.15 K), the difference is 14 K and

$$\text{COP}_{\text{heat}} = \frac{294.15}{14} = 21.0.$$

Twenty-one joules of heat delivered for one joule of work. That is the reversible limit and no machine approaches it; real domestic heat pumps at that temperature deliver about 4 to 5, which is a fifth to a quarter of Carnot. At $-7$ °C the ideal falls to 10.5 and a real machine to about 2.6.

Two things are worth extracting from those numbers.

Even a bad heat pump beats a perfect heater. A resistive heater turns one joule of electricity into exactly one joule of heat, and that is not a shortcoming that could be engineered away — it is the definition of resistive heating. The comparison is not 100% against some lower efficiency; it is a coefficient of 1 against a coefficient of 3 or 4.

The advantage shrinks as it gets colder, which is when it is wanted. The coefficient falls as $T_h/(T_h - T_c)$, so the coldest days — the ones with the largest heating demand — are the ones on which the machine performs worst. Sizing a heat pump is therefore a different problem from sizing a boiler: the output falls just as the requirement rises, which is why a heat pump is sized against a design temperature and a boiler against a peak load.

Where the “extra” energy comes from #

The uncomfortable feeling about a coefficient greater than one is worth confronting directly, because it is a good instinct pointed at the wrong quantity.

Nothing is created. The heat delivered indoors is heat that was outdoors, plus the work that moved it. Energy is conserved to the joule, and the machine’s books balance:

$$Q_h = Q_c + W.$$

What is being bought with the work is not heat but transport. Heat does not flow from cold to hot by itself — that is what the second law says, as arithmetic on a count of arrangements — so making it do so costs something, and the minimum cost is set by entropy bookkeeping. Removing $Q_c$ from a reservoir at $T_c$ lowers its entropy by $Q_c/T_c$; delivering $Q_h$ to one at $T_h$ raises entropy by $Q_h/T_h$. The second requirement is that the total not fall:

$$\frac{Q_h}{T_h} \ge \frac{Q_c}{T_c},$$

and substituting $Q_h = Q_c + W$ and rearranging gives exactly the coefficient above, with equality for a reversible machine. The ceiling is an entropy ceiling.

Where the “extra” energy comes from is the question the whole subject turns on, and the answer is that there is no extra energy. A heat pump does not create the heat it delivers; it moves heat that was already outside into a place that is warmer, and the work is what pays for moving it uphill. The delivered heat is the work plus the heat moved, which is why the ratio of delivered heat to work exceeds one without anything being manufactured.

Why a joule of warm heat is worth a twentieth of a joule of work #

There is a way of stating the resistive-heater comparison that removes the surprise entirely, and it is worth having because it generalises to every energy conversion there is.

Ask what a joule of heat delivered at 21 °C is worth, in a world whose surroundings are at 7 °C. Worth means: how much work could be got back out of it by the best possible engine. The answer is Carnot’s, with the room as the hot reservoir and the outdoors as the cold one:

$$1 - \frac{280.15}{294.15} = 4.8\ \text{per cent}.$$

A joule of heat at room temperature is worth about five hundredths of a joule of work. It is a very low-grade form of energy, because it is barely above the temperature of everything around it and there is almost nowhere for it to flow to.

Now the comparison is not surprising at all. A resistive heater takes a joule of work — the highest grade of energy there is, convertible into anything at a hundred per cent — and turns it into a joule of the lowest-grade heat in the building. It has destroyed about ninety-five per cent of what it was given, in the sense that ninety-five per cent of the ability to do work has gone. The device is a hundred per cent efficient by the energy accounts and about five per cent efficient by the only account that measures usefulness.

And the two numbers are exactly reciprocal. The ideal coefficient of performance was 21.0, and 1/21.0 is 4.8 per cent. That is not a coincidence: the maximum coefficient is the reciprocal of the Carnot factor between the same two temperatures, by construction, because the ideal machine is the engine run backwards. A heat pump is a device that converts a joule of work into twenty-one joules of a commodity worth a twenty-first as much — which is a fair trade rather than a free lunch, and the arithmetic closes exactly.

Stated that way the whole subject becomes one question: what grade of energy is being consumed, and what grade is wanted? Burning gas at 1900 °C to warm a room to 21 °C is the same waste in a different device, and it is the reason the accounting quantity that matters for buildings is not energy at all but the fraction of it that could still do work.

How the machine is actually built #

A real heat pump does not run a Carnot cycle, because the isothermal steps of one are impossibly slow. It uses a phase change instead, and the reason is worth stating: a boiling or condensing fluid absorbs or releases large amounts of heat at constant temperature, which is the closest practical thing to an isothermal reservoir contact.

The useful step is a phase change, and the reason is the flat stretches. Heating a kilogram of water through melting and boiling shows two regions where the temperature does not move at all while the energy pours in — 334 kJ to melt it, 2,260 kJ to boil it. A refrigerant is chosen so that those flat stretches fall at the two temperatures the machine has to work between, because heat crossing at a constant temperature is heat crossing with no wasted difference.

The cycle is: evaporate the refrigerant outdoors at low pressure, absorbing latent heat from the cold air; compress the vapour, which is where the work goes in and where the temperature rises; condense it indoors at high pressure, releasing the latent heat plus the work; expand it through a valve back to low pressure. The compressor is the only part that consumes energy and the expansion valve is where the reversibility is lost — a step chosen deliberately, because a turbine small enough for a domestic machine would cost more than the work it recovered. Where the heat goes during the two flat stretches is the whole reason a phase change is used rather than a gas cycle.

The design lives on a boiling curve. A refrigerant’s boiling point depends on its pressure, so choosing two pressures chooses two temperatures — and a machine’s whole operating range is a pair of points on one line. That is why carbon dioxide systems run at a hundred atmospheres, its line sitting far above water’s, and why a water-based machine cannot be built at all for domestic temperatures: the pressures it would need are below anything a compressor can usefully handle.

The same argument, counted in entropy #

There is a way of seeing the coefficient that needs no cycle at all and makes the size of it obvious.

Heat $Q_c$ leaving a reservoir at $T_c$ carries entropy $Q_c/T_c$ away with it. That entropy has to go somewhere, and the only place available is the warm reservoir, which will accept entropy at the rate $Q_h/T_h$ per joule delivered. Since $T_h > T_c$, each joule delivered to the warm side accepts less entropy than a joule taken from the cold side carried — so extra energy has to be added to make up the difference, and that extra energy is the work.

The amount required is fixed by the ratio of the temperatures and by nothing else. Where the two are close the shortfall is small and the work required is small; where they are far apart the shortfall is large. The coefficient is therefore not a property of any machine but an accounting identity about entropy, and every machine is somewhere below it.

Counted in entropy, the same argument becomes an accounting identity rather than a bound. Heat leaving the cold outside carries entropy away from it; heat arriving indoors carries entropy in; and the work carries none, because work is energy with no entropy attached to it. The second law requires the entropy delivered to be at least the entropy removed, and rearranging that requirement gives the coefficient of performance directly — which is why the ceiling depends only on the two temperatures and on nothing whatever about the machine.

Read backwards, the same identity is Carnot’s ceiling: an engine cannot convert all of $Q_h$ to work because the entropy that came in with the heat has to be dumped somewhere, and dumping it requires sending some heat to the cold side. The heat that must be discarded and the work that must be supplied are the same entropy debt seen from either end.

Why the ground beats the air #

The falling curve has a direct engineering consequence, and it is the reason two kinds of heat pump exist rather than one.

The problem with drawing heat from outdoor air is that the cold reservoir’s temperature is exactly the thing that varies, and it varies in the wrong direction: coldest when the demand is highest. A machine whose coefficient is 4.5 in October and 2.4 in February is a machine whose worst performance coincides with its heaviest duty, and the electrical supply has to be sized for the second.

A few metres below the surface the ground does not do that. Its temperature is close to the annual average air temperature and it varies through the year by a couple of degrees rather than by thirty, because the soil’s thermal diffusivity is low enough that the seasonal wave is attenuated and delayed with depth. Draw heat from there instead and the cold reservoir sits near 10 °C in midwinter, when the air is at $-5$: a temperature lift of eleven kelvin rather than twenty-six, and an ideal coefficient roughly twice as high on exactly the days it matters.

The cost is the hole. A ground loop is a substantial excavation or a borehole, and it is why the choice between the two kinds of machine is almost always decided by the site rather than by the physics. What the physics says is only that the whole advantage is in the flatness of the curve rather than in its height, and a comparison made on a mild day will not show it.

There is a limit to how much the ground will give, and it is a rate rather than a total. Drawing heat from a borehole cools the rock around it, and the heat has to be resupplied by conduction from further out, which is slow. A loop run too hard depresses its own source temperature over a season, and the machine’s coefficient falls for a reason that has nothing to do with the weather.

The machine with no moving parts #

One variant is worth knowing about, because it makes the same identity work with a different input.

Everything above assumes the machine is driven by work — a compressor, an electric motor. It need not be. A machine can be driven by heat at a high temperature instead, in which case the arrangement is an engine and a heat pump in one box: heat arrives at some hot temperature, drives an engine, and the engine’s work runs the pump. The three reservoirs — hot source, cold source, warm sink — make the analysis a two-step version of the one above, and the ceiling is the product of the engine’s Carnot factor and the pump’s coefficient.

Built, that becomes an absorption cycle. Instead of compressing a vapour mechanically, the refrigerant is dissolved into a liquid absorbent, the solution is pumped to high pressure — which costs very little, because pumping a liquid is nearly free compared with compressing a gas — and then the refrigerant is boiled back out of it by the heat source. The rest of the cycle is unchanged.

The result is a refrigerator with no compressor and, in the domestic version, no moving parts at all, driven by a gas flame or by waste heat. Its coefficient is much lower than a compression machine’s, which is the honest price of the extra conversion. Where it wins is where work is expensive and heat is free: on waste heat from an industrial process, on solar heat, or in any application where silence and the absence of a motor are worth more than efficiency.

Where the model stops #

Carnot is unreachable and not by a small margin. Real machines reach 25–50% of the ideal coefficient. The losses are the compressor’s inefficiency, the temperature drops needed to push heat through the two heat exchangers at a useful rate, and the throttling valve, which is a frankly irreversible step chosen because a turbine at that scale would cost more than it recovered.

The reservoirs are not reservoirs. A ground loop cools the ground it draws from and an air unit ices up its own evaporator, so $T_c$ falls as the machine runs and the coefficient falls with it. Defrost cycles run the machine backwards periodically, which is pure loss.

The reservoirs are at one temperature each. They are not: outdoor air arrives at the evaporator, is cooled by several kelvin, and leaves. The effective cold reservoir is therefore colder than the outdoor air by an amount that depends on how fast the machine is running, and pushing more heat through the same exchanger costs a larger temperature drop. That is the mechanism behind most of the gap between the reversible limit and reality, and it means the coefficient of a real machine falls as its output rises — a trade-off the Carnot expression has no term for.

The comparison with a heater assumes the electricity is free of context. Comparing a coefficient of 3.5 against 1 is a comparison of delivered heat per unit of electricity. If the electricity came from a thermal power station at 40% efficiency, the heat pump delivers 1.4 units of heat per unit of fuel and a gas boiler delivers about 0.9 — still a win, and a much smaller one than the coefficient alone suggests.

What the pictures cannot show #

The coefficient curves are steady-state and say nothing about capacity. A machine with a coefficient of 4 may still be unable to keep a house warm, because the coefficient is a ratio and not a rate; the quantity that sizes the equipment is the kilowatts it can deliver, and that also falls in cold weather.

The curves also rise without limit as the temperature difference goes to zero, which is correct and useless. At a difference of one kelvin the ideal coefficient is 294 and no real machine gets near it, because the heat exchangers themselves need several kelvin of difference to move heat at all. The reversible limit becomes unreachable exactly where it becomes attractive.

And nothing here shows time. Every quantity is per cycle or per joule, and the second law’s real content — that the total entropy of the universe increases whenever anything happens at a finite rate — appears only as the difference between the drawn ideal and the machines that exist.

Cooling as the same machine #

Nothing in the arithmetic distinguishes a heat pump from a refrigerator or from an air conditioner. All three take heat from a colder place, add work, and deliver the total to a warmer one; what differs is which of the two heats is called the product and which is called the waste.

That has an odd consequence worth stating plainly. A domestic fridge is a heater. It removes heat from its interior, adds the compressor’s work, and rejects the sum through the coils at the back — so a kitchen with a fridge running in it gets warmer, by exactly the food’s heat plus the electricity consumed. Opening the door does not cool the room; it makes the fridge work harder, and warms the room faster.

The same identity is why an air conditioner cannot be used as a fan. Every joule it consumes ends up as heat somewhere, and the somewhere is outside; a unit with no outside — a portable one with its hose disconnected — is a heater with a compressor attached, and delivers rather more heat than a bar fire of the same rating.

Where the ladder goes next #

The rung below establishes that there is a ceiling and where it comes from. This one runs the argument backwards and finds that the same expression, inverted, licenses something that looks impossible and is not.

The rungs above are about the gap between the ideal and the real. Endoreversible analysis asks what the best coefficient is for a machine that has to move heat at a finite rate, and gets a different and lower answer that depends on the heat exchangers rather than only on the temperatures. Beyond that lies the general question of what work can be extracted from a system that is out of equilibrium with its surroundings — a quantity called availability, of which both the engine’s ceiling and the pump’s floor are special cases.

The habit is one worth practising on any inequality. When a result is presented as a limit, ask what it says when the process runs the other way. The ceiling on an engine and the floor under a heat pump are the same sentence, and one of them reads as a disappointment and the other as a bargain.