The Carnot cycle and heat engine efficiency

Engineering · Senior school and first-year engineering · about 8 min

Every power station and car engine turns heat into work, and every one of them throws some heat away. The Carnot cycle is the ideal, perfect engine. It sets the best possible efficiency for any engine working between two temperatures, which makes it a powerful reality check.

The key idea

An engine taking heat $Q_H$ from a hot reservoir at $T_H$ and rejecting heat to a cold one at $T_C$ (temperatures in kelvin) can never beat

\eta_{Carnot} = 1 - \frac{T_C}{T_H}

and for any engine, $\eta = \dfrac{W}{Q_H}$ with $W = Q_H - Q_C$ .

A vertical block diagram. A wide flat rectangle at the top for the hot reservoir and a wide flat rectangle at the bottom for the cold reservoir. A circle in the middle for the engine. A thick dark red arrow runs down from the hot reservoir into the engine. A thin dark blue arrow runs down from the engine to the cold reservoir. A dark red arrow leaves the engine to the right. — A heat engine takes in heat, turns some into work and rejects the rest.

Why some heat must be thrown away

The second law of thermodynamics says you can't turn heat into work in a cycle without rejecting some heat to something colder. That rejected heat $Q_C$ is why car radiators and power-station cooling towers exist, and why $\eta$ is always below $1$ .

The four steps of the Carnot cycle

Isothermal expansion at $T_H$ : heat $Q_H$ flows in.
Adiabatic expansion: the gas cools from $T_H$ to $T_C$ with no heat flow.
Isothermal compression at $T_C$ : heat $Q_C$ flows out.
Adiabatic compression: back up to $T_H$ .

Every step is reversible, which is what makes it the upper limit. Real engines have friction and fast, irreversible heat transfer, so they do worse.

A pressure against volume graph with one closed loop of four connected curves, each with a small arrow, running clockwise. State 1 is at the upper left. From 1 to 2: a gently falling curve along the top. From 2 to 3: a steeper falling curve down the right. From 3 to 4: a gentle curve along the bottom, running left. From 4 back to 1: a steep rising curve up the left. An arrow points into the top curve from above, and an arrow points away from the bottom curve downwards. — The Carnot cycle runs clockwise on a pressure-volume diagram.

Worked example: checking an engine claim

An engine runs between $600$ K and $300$ K and takes in $1000$ J of heat per cycle. What is the most work it could produce? A company claims an engine between the same temperatures that is $60\%$ efficient. Is that possible?

Show the solution

Carnot efficiency:

\eta_{Carnot} = 1 - \frac{300}{600} = 0.5 = 50\%

Maximum work per cycle:

W_{max} = \eta\,Q_H = 0.5 \times 1000 = 500 \text{ J}

The other $500$ J at least must be rejected to the cold reservoir.

The claim: $60\% > 50\%$ , so it's impossible. No engine between $600$ K and $300$ K can beat the Carnot limit.

Common mistakes

Using degrees Celsius in the efficiency formula

The formula needs absolute temperatures. Add $273.15$ to convert °C to K.

Expecting a real engine to reach the Carnot efficiency

Carnot is the ceiling for a perfect, reversible engine. Real engines are always lower.

Thinking better insulation could make it 100% efficient

The heat rejected at $T_C$ is required by the second law, not a leak you can plug.

Check your understanding

1. What is the Carnot efficiency between $500$ K and $400$ K?

2. To raise the maximum possible efficiency, you should...

3. An engine takes in $800$ J and does $200$ J of work. How much heat does it reject?

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