Heat Engines and the Carnot Cycle

A steam locomotive on the Darjeeling Himalayan Railway chuffs up the ridge toward Ghum, 2 258 m above sea level. Inside its boiler, coal burns and turns water into pressurised steam at perhaps 450 K. That steam rushes into a cylinder, pushes a piston, spins a wheel, lifts the 50-tonne locomotive against gravity, and is finally vented into the thin mountain air at perhaps 370 K. Each cycle, some of the coal's chemical energy becomes mechanical work (the climb), some becomes useless heat (the exhaust), and the water — after condensing somewhere down the line — is recycled and heated again. The whole thing runs as a cycle: water starts as liquid, becomes steam, does work, returns to liquid.

That locomotive is a heat engine. So is the 2.5 MW gas turbine at a Bengaluru industrial park that runs on natural gas. So is the 4-stroke petrol engine in a Maruti Swift, the diesel engine in a Tata Ace, the 660 MW supercritical unit at NTPC Sipat, and the solar thermal tower at Dhursar in Rajasthan that boils water using concentrated sunlight. All of them share one diagram — heat in at the top, heat out at the bottom, work out to the side — and all of them are bound by one theorem. No matter how clever the engineering, the fraction of heat that a cyclic engine can convert to work is bounded above by a number that depends only on the two temperatures between which it operates. That number is the Carnot efficiency.

This article is the full construction. Start from the first and second laws (already derived in first law of thermodynamics and second law of thermodynamics). Build a generic heat engine and write down its efficiency. Design the Carnot cycle — two isothermals at T_H and T_C, joined by two adiabatics — and compute, line by line, the heat absorbed, the heat rejected, and the work done in each leg. Divide and watch the working substance, the details of the gas, the cylinder dimensions all cancel. What remains is 1 - T_C/T_H. Then prove (in the going-deeper section) that Carnot's engine is the best possible — any engine that beat it could be used, in series with a reversed Carnot engine, to make heat flow spontaneously from cold to hot, violating Clausius. Finally, explain why a real Indian Railways diesel locomotive or a Kudankulam nuclear reactor falls well short of the Carnot ceiling, and what engineers do to edge closer.

The generic heat engine — efficiency from the first law

Forget, for the moment, what is inside the engine. Draw a box. Label the box "engine". Draw a hot reservoir above (temperature T_H) and a cold reservoir below (T_C). In one full cycle, heat Q_H flows from hot reservoir into the engine, heat Q_C flows from engine into cold reservoir, and work W is delivered by the engine to an external mechanical load (a wheel, a generator, a piston). All three quantities are positive magnitudes; the directions are set by the arrows.

Because the engine is cyclic, the working substance returns to its starting state after each cycle, so \Delta U_{\text{engine}} = 0. The first law (\Delta U = Q - W) applied to the engine across one full cycle gives

where Q_H - Q_C is the net heat absorbed (positive inflow minus positive outflow) and W is the work done by the engine on its surroundings. Rearrange:

Why: over one cycle, every joule of heat that went in must either come out as heat or leave as work, because the engine itself cannot store energy (it returned to its initial state). This is first-law book-keeping, nothing more.

The efficiency \eta of the engine is defined as "what you get divided by what you pay for":

Why: you pay for the heat Q_H (that is the coal you burn, the uranium you fission, the sunlight you concentrate); you get the work W (the train pulled, the electricity generated). The ratio is what fraction of your input became useful output.

Already this is telling you something concrete. If Q_C = 0 — no waste heat — then \eta = 1, a perfect engine. If Q_C = Q_H — all heat in equals heat out — then \eta = 0, no useful work. Real engines sit between these limits. Since the Kelvin-Planck statement of the second law outright bans Q_C = 0 in a cycle, \eta = 1 is unreachable. Some heat must always be dumped to a cold reservoir.

All of the above holds for any cyclic engine — steam, petrol, diesel, gas turbine, nuclear, solar-thermal, Stirling. The first law does not care what is inside. The differences between engines are entirely in the size of Q_C/Q_H — that is, in how much waste heat they generate per joule of work.

Sadi Carnot's question — what is the best possible engine?

In 1824, a 28-year-old French military engineer named Sadi Carnot asked the sharpest question one can ask about heat engines: given two reservoirs at temperatures T_H and T_C, what is the maximum fraction of the heat absorbed at T_H that can be converted to work? It is a question about limits. It does not care whether your engine is steam, petrol, or a cleverness yet uninvented. It asks: what does the physics allow?

Carnot reasoned that the answer would come from an idealised engine, one that suffered no losses from friction, no losses from finite-rate heat transfer across non-zero temperature gaps, no losses from turbulence or leakage. Such an engine would be reversible: you could run it backwards, cycle after cycle, with no accumulated effect on the universe. The forward engine takes Q_H from hot, rejects Q_C to cold, and does work W. The reverse engine takes Q_C from cold, uses external work W, and rejects Q_H to hot — a refrigerator.

Carnot designed a specific reversible cycle, now called the Carnot cycle, from exactly four reversible processes: two isothermal and two adiabatic. Every step is slow enough to be quasi-static, every heat exchange happens across zero temperature difference, and every adiabatic compression or expansion is frictionless. A reader who has followed thermodynamic processes — isothermal and adiabatic already knows what each step looks like; here the job is to string them together into a closed loop.

The four strokes of the Carnot cycle

Work with a fixed amount of ideal gas (say n moles) enclosed in a cylinder with a frictionless piston. The cylinder can be placed in thermal contact with a hot reservoir at T_H, a cold reservoir at T_C, or insulated from everything. Label four states — A, B, C, D — on a pressure-volume diagram. The cycle goes A \to B \to C \to D \to A.

Step A \to B: isothermal expansion at T_H

Place the cylinder in contact with the hot reservoir at T_H. Let the gas expand infinitely slowly from volume V_A to V_B, with pressure dropping along the hyperbola PV = nRT_H.

Along this leg, temperature is constant, so for an ideal gas \Delta U_{AB} = 0 (internal energy depends only on T, and T did not change). The first law gives Q_{AB} = W_{AB}. The work done by the gas is the area under the isotherm (see PV diagrams and work):

Why: an ideal isothermal expansion has P = nRT_H/V; the integral of dV/V is \ln V, so the area under the isotherm from V_A to V_B is nRT_H\,\ln(V_B/V_A). This is derived in full in the isothermal process article.

Since V_B > V_A, this is positive: work out. And because \Delta U_{AB} = 0, the heat absorbed from the hot reservoir exactly matches:

Why: every joule the gas delivers as work during the isothermal expansion must be replaced by heat from the reservoir, because the internal energy cannot change at constant T. The reservoir supplies heat; the piston delivers an equal amount of work.

Step B \to C: adiabatic expansion, T_H \to T_C

Now isolate the cylinder (remove it from the hot reservoir, surround it with insulation). Let the gas continue to expand from V_B to V_C. With no heat flow, Q_{BC} = 0. The first law becomes \Delta U_{BC} = -W_{BC}: the gas does work at the expense of its own internal energy, so its temperature drops. Along an adiabat, PV^{\gamma} = \text{constant} and equivalently TV^{\gamma - 1} = \text{constant} (see adiabatic process). The gas cools from T_H to T_C.

The link between V_B, V_C and the temperatures is

Step C \to D: isothermal compression at T_C

Place the cylinder in contact with the cold reservoir at T_C. Slowly compress the gas from V_C to V_D. Temperature stays at T_C, so \Delta U_{CD} = 0 again. Work is done on the gas (the piston pushes in), so W_{CD} < 0, and heat flows out of the gas into the cold reservoir, so Q_{CD} < 0. The magnitudes:

Since V_D < V_C, \ln(V_D/V_C) < 0, so Q_{CD} < 0 and W_{CD} < 0. The heat rejected to the cold reservoir (a positive magnitude) is

Why: -\ln(V_D/V_C) = \ln(V_C/V_D), and since V_C > V_D, this is positive — the gas dumps heat into the cold reservoir while being compressed at constant temperature.

Step D \to A: adiabatic compression, T_C \to T_H

Finally, isolate the cylinder again and compress from V_D back to V_A. With Q_{DA} = 0, the work of compression goes into raising the gas's internal energy, warming it from T_C back to T_H. Along this adiabat,

The gas is now back at state A. One full cycle complete.

The efficiency collapses to a ratio of temperatures

Add up the heats and works:

• Q_H = nRT_H \ln(V_B/V_A) (into the gas during A \to B).
• Q_C = nRT_C \ln(V_C/V_D) (out of the gas during C \to D).
• The two adiabatic legs have Q = 0, so they contribute no heat to the ledger.

The efficiency is

Now the miracle. Use (\star) and (\star\star) — the two adiabatic relations linking the four volumes and the two temperatures.

From (\star): T_H V_B^{\gamma - 1} = T_C V_C^{\gamma - 1}, so \dfrac{V_C}{V_B} = \left(\dfrac{T_H}{T_C}\right)^{1/(\gamma - 1)}.

From (\star\star): T_H V_A^{\gamma - 1} = T_C V_D^{\gamma - 1}, so \dfrac{V_D}{V_A} = \left(\dfrac{T_H}{T_C}\right)^{1/(\gamma - 1)}.

The two right-hand sides are equal, so \dfrac{V_C}{V_B} = \dfrac{V_D}{V_A}, which rearranges to

Why: cross-multiplying V_C/V_B = V_D/V_A gives V_C V_A = V_D V_B, and then V_C/V_D = V_B/V_A. The two logarithms in \eta are therefore equal.

Plug this identity back into \eta. The logarithms cancel:

Why: the \ln factor was the only place where the geometry of the cylinder — the specific volumes V_A, V_B, V_C, V_D — entered. Once it cancels, nothing remains of the working substance, the cylinder size, or the amount of gas. Only the two reservoir temperatures matter.

Two features of this result deserve emphasis.

1. Temperatures are in kelvin. The formula uses absolute temperatures, not Celsius. At T_C = 27 °C = 300 K and T_H = 227 °C = 500 K, \eta_{\text{Carnot}} = 1 - 300/500 = 40 \%, not 1 - 27/227.
2. Only reservoir temperatures matter. Nothing about the gas, the cylinder, the piston material, the mass of working substance, the chemistry of the fuel. Two Carnot engines running between the same T_H and T_C — one using helium, one using hot steam — have identical efficiency. This is an outrageously strong statement, and it is the content of Carnot's theorem.

Worked examples

Exploring the ceiling — a slider for T_C/T_H

Before you read the going-deeper section, drag the temperature ratio below and watch the Carnot efficiency respond. Fix T_H = 800 K (roughly a boiler temperature) and vary T_C from 50 K (liquid nitrogen-ish) to 780 K (almost as hot as the boiler). The efficiency is \eta = 1 - T_C/T_H.

Why no real engine reaches Carnot efficiency

The Carnot cycle is an idealisation stacked on idealisations. Strip away the assumptions one at a time and watch the efficiency fall.

Quasi-static processes require infinite time. The isothermal expansion at T_H is done so slowly that the gas is always exactly at T_H; any faster, and a finite temperature difference develops between the gas and the reservoir, so heat flows across a gap — and that is irreversible. A real engine runs at hundreds of RPM, not infinitely slowly. The time saved is traded against efficiency lost.

Heat transfer across finite \Delta T is irreversible. In a real boiler, the combustion gas is at, say, 1800 K, but the steam inside the boiler tubes is only at 870 K. Heat flows across this 900 K gap, and every joule that flows does so irreversibly — entropy is generated (see entropy introduction). The engine then operates between 870 K and, say, 310 K, not between 1800 K and 310 K. The Carnot ceiling it can actually pursue is 1 - 310/870 = 64.4\%, not 1 - 310/1800 = 82.8\%.

Friction in real pistons and turbines. A real cylinder has friction between piston and wall; a real turbine has friction in its bearings and turbulent drag on the blades. Every joule lost to friction reappears as heat — heat that is now at a lower temperature than T_H, and so a smaller fraction of which can be converted back to work.

Adiabatic processes leak heat. A "perfectly insulated" adiabatic leg does not exist. A real cylinder conducts heat through its walls during the compression and expansion strokes, which makes the process neither purely adiabatic nor perfectly isothermal.

Non-ideal working substances. Real steam does not obey PV = nRT exactly, especially near the saturation curve. Real air has a slightly varying \gamma. These cause the adiabats and isotherms to deviate from their ideal shapes.

Leakage and incomplete combustion. A real piston rings let some combustion gas slip past them on every stroke. A real cylinder does not burn every droplet of fuel. Each leak and each unburned droplet is energy that never entered the engine's thermodynamic cycle at all.

Put all of these together and the gap between \eta_{\text{Carnot}} and \eta_{\text{actual}} opens up. A modern combined-cycle gas turbine at the GMR Power Plant near Chennai can reach \eta \approx 60\% — about 80 % of its Carnot ceiling, impressive engineering. A typical 1950s-era Indian Railways steam locomotive reached perhaps \eta \approx 8\% — because its peak temperature was modest, its boiler leaked, its pistons had friction, and its cylinder walls radiated heat. The second law set the ceiling; engineering set the floor.

Common confusions

• "Efficiency can be more than 1 if the engine is very clever." No. For a cyclic heat engine, \eta > 1 would require W > Q_H, which would mean the engine is producing more work than the heat it took in — violating the first law (energy conservation). Even the Carnot engine, the theoretical maximum, has \eta < 1 whenever T_C > 0 K — and T_C = 0 K is unreachable (the third law of thermodynamics).

• "The Carnot cycle is the ideal because the gas is ideal." The specific derivation used PV = nRT at two isothermal steps, but the result \eta = 1 - T_C/T_H holds for any reversible cycle between two reservoirs — not just an ideal gas. Carnot's theorem (in the going-deeper section) proves this. The working substance cancels out because the second law forces it to.

• "T must be in Celsius because that's what the kitchen thermometer reads." No. Absolute temperatures (kelvin) are required because the derivation involves PV = nRT where T is the absolute temperature. Using Celsius would give nonsense: at T_C = 0 °C, the formula would read \eta = 1 - 0/T_H = 1, claiming 100 % efficiency from an engine whose cold reservoir is merely at freezing, which is absurd. Always kelvin.

• "Carnot efficiency ignores the work done to cool the cold reservoir." For a passive cold reservoir (the Hasdeo river, the Arabian Sea, the night sky), this work is zero — the environment is naturally cold. But if you are artificially cooling a reservoir (running a refrigerator to create the cold side), then the work needed to maintain T_C eats into your net gain. Carnot efficiency is the engine's local efficiency and does not account for cascaded systems.

• "A heat engine must involve steam or gas." Not necessarily. The Carnot cycle is a recipe — isothermal, adiabatic, isothermal, adiabatic — and any system with a "working substance" that can exchange heat and do work on a piston (or equivalent) can in principle run it. Thermoelectric generators (based on the Seebeck effect) are heat engines that use the electron gas in a conductor as the working substance; their efficiency is bounded by the same Carnot formula.

• "If I can just find two reservoirs with a huge T_H/T_C ratio, I'll get close to 100 %." You can — the sun (5800 K) and deep space (3 K) would give \eta_{\text{Carnot}} = 1 - 3/5800 = 99.95\%. The catch: getting heat from the sun to your engine at 5800 K requires focusing sunlight to that temperature, which requires nearly ideal mirrors over enormous areas. Every practical engineering step eats into the ceiling.

Where this leads next

• Entropy — Introduction — the state function hinted at by \oint dQ_{\text{rev}}/T = 0, and the second law in its cleanest form.
• Refrigerators and Heat Pumps — what happens when you run a heat engine backwards, with a work input instead of a work output.
• Second Law of Thermodynamics — the Kelvin-Planck and Clausius statements, already used here.
• Thermodynamic Processes — Isothermal and Adiabatic — detailed derivations of the four legs of the Carnot cycle.
• First Law of Thermodynamics — the energy-conservation law used throughout this article to link Q, W, and \Delta U.