11.01.06 · stat-mech-physics / thermodynamics

Thermodynamic cycles: Carnot efficiency, Otto and Diesel cycles, and entropy production

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Anchor (Master): Landau & Lifshitz, Statistical Physics, Part 1, 3rd ed. (1980), §19; Fermi, Thermodynamics (1937), Ch. 4

Intuition Beginner

A heat engine converts thermal energy into mechanical work. It absorbs heat from a hot reservoir, converts some of it to work, and dumps the rest into a cold reservoir. No engine can convert all the absorbed heat into work — some heat must always be wasted. The efficiency $η$ is the fraction of absorbed heat that becomes work.

The Carnot cycle is the most efficient possible engine operating between two temperatures. It uses only reversible steps (slow, quasi-static processes), and its efficiency depends only on the two temperatures:

η_{Carnot} = 1 - \frac{T _{c}}{T _{h}} .

A Carnot engine working between $T_{h} = 500$ K and $T_{c} = 300$ K has efficiency $1 - 300/500 = 40%$ . No real engine can beat this. Real engines (Otto, Diesel) use irreversible steps (rapid combustion, sudden expansion) and are less efficient, but they produce more power per cycle.

Visual Beginner

The Carnot cycle on a $P$ - $V$ diagram consists of four steps: (1) isothermal expansion at $T_{h}$ (absorbing heat), (2) adiabatic expansion (cooling from $T_{h}$ to $T_{c}$ ), (3) isothermal compression at $T_{c}$ (dumping heat), (4) adiabatic compression (heating from $T_{c}$ to $T_{h}$ ). The cycle forms a closed loop; the enclosed area is the net work done per cycle.

Worked example Beginner

The Otto cycle (model of a petrol engine) has four steps: (1) adiabatic compression ( $V_{1} \to V_{2}$ ), (2) constant-volume heat addition ( $V_{2}$ , combustion), (3) adiabatic expansion ( $V_{2} \to V_{1}$ , power stroke), (4) constant-volume heat rejection ( $V_{1}$ , exhaust). The compression ratio is $r = V_{1} / V_{2}$ . For an ideal gas with heat capacity ratio $γ$ :

η_{Otto} = 1 - \frac{1}{r ^{γ - 1}} .

For $r = 10$ and $γ = 1.4$ (air): $η = 1 - 1/1 0^{0.4} = 1 - 0.398 = 60.2%$ . Real petrol engines achieve about 25–35% due to friction, heat loss, and non-instantaneous combustion.

Check your understanding Beginner

Formal definition Intermediate+

Carnot cycle

The Carnot cycle consists of two reversible isothermal processes and two reversible adiabatic processes:

Isothermal expansion at $T_{h}$ : absorb heat $Q_{h}$ from hot reservoir.
Adiabatic expansion: temperature drops from $T_{h}$ to $T_{c}$ .
Isothermal compression at $T_{c}$ : reject heat $Q_{c}$ to cold reservoir.
Adiabatic compression: temperature rises from $T_{c}$ to $T_{h}$ .

For an ideal gas, the work done in each step can be computed explicitly. The net work is $W = Q_{h} - Q_{c}$ and the efficiency is:

η_{Carnot} = \frac{W}{Q _{h}} = 1 - \frac{Q _{c}}{Q _{h}} = 1 - \frac{T _{c}}{T _{h}} .

The equality $Q_{c} / Q_{h} = T_{c} / T_{h}$ follows from the fact that entropy change is zero over a complete reversible cycle: $Q_{h} / T_{h} - Q_{c} / T_{c} = 0$ .

Clausius inequality and entropy production

For any cyclic process (reversible or irreversible):

\oint \frac{δ Q}{T} \leq 0.

For an irreversible cycle, the entropy change of the universe (system plus reservoirs) is positive:

Δ S_{universe} = - \frac{Q _{h}}{T _{h}} + \frac{Q _{c}}{T _{c}} \geq 0.

The entropy production is $σ = Δ S_{universe} \geq 0$ , with $σ = 0$ only for reversible processes.

Key result: Carnot theorem

Theorem (Carnot). No heat engine operating between two temperatures $T_{h}$ and $T_{c}$ can have efficiency greater than $1 - T_{c} / T_{h}$ . Equality holds only for a reversible engine.

The proof uses the second law: if an engine had $η > η_{Carnot}$ , you could couple it with a Carnot refrigerator to produce a perpetual-motion machine of the second kind (extracting net work from a single temperature).

Bridge. The Carnot theorem builds toward the theory of irreversible thermodynamics 11.02.04, where the entropy production rate is related to the fluxes (heat flow, particle flow) and forces (temperature gradients, chemical potential gradients) by linear response coefficients. This is exactly the content of the Onsager reciprocal relations. The foundational reason the Carnot efficiency is a universal bound is that it follows from the second law alone, independent of the working substance; putting these together with the fluctuation-dissipation theorem 11.02.04, irreversibility and entropy production are the macroscopic shadows of microscopic fluctuations.

Exercises Intermediate+

Exercise 5 (hard, symbolic).

Prove the Carnot theorem using the second law: show that no engine operating between $T_{h}$ and $T_{c}$ can exceed $η_{Carnot} = 1 - T_{c} / T_{h}$ .

η_{X} > η_{Carnot}

. Then

W_{X} / Q_{h} > 1 - T_{c} / T_{h}

, so

Q_{c}^{X} = Q_{h} - W_{X} < Q_{h} T_{c} / T_{h}

Now couple engine X with a Carnot refrigerator C operating between the same two reservoirs. Engine X produces work $W_{X}$ which drives the refrigerator. The Carnot refrigerator absorbs $Q_{c}^{C}$ from the cold reservoir and dumps $Q_{h}^{C} = Q_{c}^{C} + W_{X}$ to the hot reservoir, with $Q_{h}^{C} / T_{h} = Q_{c}^{C} / T_{c}$ .

Net heat extracted from the cold reservoir: $Q_{c}^{C} - Q_{c}^{X} > 0$ (since $Q_{c}^{X} < Q_{h} T_{c} / T_{h} = Q_{c}^{Carnot}$ and the Carnot refrigerator removes more). Net heat deposited in the hot reservoir: $Q_{h}^{C} - Q_{h}$ .

The combined system extracts heat from the cold reservoir and deposits it in the hot reservoir using no external work — a refrigerator requiring no energy input. This violates the Clausius statement of the second law (heat cannot spontaneously flow from cold to hot). Contradiction. Therefore $η_{X} \leq η_{Carnot}$ .

Exercise 6 (hard, symbolic).

An ideal gas undergoes a throttling (Joule-Thomson) expansion from $(P_{1}, T_{1})$ to $(P_{2}, T_{2})$ through a porous plug. Show that enthalpy is conserved ( $H_{1} = H_{2}$ ) and compute the entropy production $Δ S = C_{P} ln (T_{2} / T_{1}) - R ln (P_{2} / P_{1})$ .

Q = 0

) and no work (

W = 0

). The first law for steady flow gives

H_{1} = H_{2}

(enthalpy conservation). For an ideal gas,

H

depends only on

T

, so

T_{1} = T_{2}

(no temperature change for an ideal gas).

The entropy change: $Δ S = n C_{P} ln (T_{2} / T_{1}) - n R ln (P_{2} / P_{1})$ . For an ideal gas ( $T_{2} = T_{1}$ ): $Δ S = - n R ln (P_{2} / P_{1}) > 0$ since $P_{2} < P_{1}$ . The process is irreversible — entropy is produced even though enthalpy is conserved.

For a real gas (van der Waals), $H$ depends on both $T$ and $P$ , so throttling does change the temperature. The Joule-Thomson coefficient $μ_{J T} = (\partial T / \partial P)_{H} = (1/ C_{P}) [T (\partial V / \partial T)_{P} - V]$ determines whether the gas cools ( $μ_{J T} > 0$ ) or heats ( $μ_{J T} < 0$ ).

Exercise 8 (hard, symbolic).

For a general cyclic process with working substance at temperatures $T (t)$ that varies over the cycle, show that the entropy production per cycle is $σ = \oint (1/ T_{res} (t) - 1/ T (t)) δ Q \geq 0$ where $T_{res} (t)$ is the reservoir temperature.

Δ S_{system} = 0

. The entropy change of the reservoirs is

Δ S_{res} = - \oint δ Q / T_{res}

. The total entropy production is:

$σ = Δ S_{system} + Δ S_{res} = - \oint \frac{δ Q}{T _{res}}$ .

Now, the entropy change of the working substance per cycle is also $Δ S_{system} = \oint δ Q / T + σ_{int} = 0$ , where $σ_{int} \geq 0$ is the internal entropy production from irreversibilities within the working substance.

Combining: $σ = \oint \frac{δ Q}{T} - \oint \frac{δ Q}{T _{res}} = \oint (\frac{1}{T} - \frac{1}{T _{res}}) δ Q \geq 0$ .

This is the general Clausius inequality. For a reversible process, $T = T_{res}$ at every point, so $σ = 0$ . For an irreversible process, $σ > 0$ .

Lean formalization Intermediate+

The Carnot theorem is a consequence of the second law and the definition of thermodynamic temperature. Formalising it requires a thermodynamic-cycle framework with path integrals for heat and work, a definition of efficiency, and a proof that no cycle exceeds the Carnot bound. The mathematical content is elementary (algebraic manipulation of heat and work integrals), but the physical framework (distinguishing reversible from irreversible processes) requires domain-specific infrastructure.

Advanced results Master

Endoreversible thermodynamics

The Curzon-Ahlborn model (1975) treats the engine as internally reversible but with finite-rate heat transfer to the reservoirs. The working fluid temperatures are $T_{h w}$ (hot side) and $T_{c w}$ (cold side), with heat flow rates $\dot{Q}_{h} = α (T_{h} - T_{h w})$ and $\dot{Q}_{c} = β (T_{c w} - T_{c})$ . Maximising power output gives $η_{C A} = 1 - T_{c} / T_{h}$ , which agrees well with real power plants.

Entropy production minimisation

Bejan (1996) proposed entropy generation minimisation as a design principle for thermal systems. The entropy production rate $\overset{σ}{˙}$ quantifies the thermodynamic losses in a system. Minimising $\overset{σ}{˙}$ at fixed power output is equivalent to maximising the second-law efficiency. This approach is used in the design of heat exchangers, cooling systems, and chemical reactors.

Stirling and Ericsson cycles

The Stirling cycle (two isotherms connected by two isochors, with regenerative heat storage) and the Ericsson cycle (two isotherms connected by two isobars, with regeneration) both achieve Carnot efficiency when the regeneration is perfect. In practice, imperfect regeneration reduces their efficiency below Carnot, but both offer advantages in specific applications (Stirling engines for solar thermal power, Ericsson cycles for gas turbines with intercooling and reheat).

Synthesis. Thermodynamic cycles build toward the microscopic theory of irreversibility [11.02.03, 11.02.04], where entropy production is related to the collision operator in the Boltzmann equation and the fluctuation-dissipation theorem. The central insight is that the Carnot efficiency is a universal bound that follows from the second law alone, independent of the working substance or the cycle details. This is dual to the microscopic irreversibility encoded in the H-theorem: entropy production is positive for any irreversible process, whether it is a macroscopic engine cycle or a microscopic collision cascade. The foundational reason the Carnot bound holds is that heat flows from hot to cold, and extracting work from a single temperature would violate this directionality. Putting these together, the Carnot cycle is the idealisation that separates the thermodynamic content (the universal efficiency bound) from the engineering details (the specific cycle shape), and this separation generalises to the Onsager reciprocal relations 11.02.04 where the linear-response coefficients of all irreversible processes are constrained by symmetry.

Full proof set Master

Proposition. The efficiency of any reversible engine operating between two reservoirs at $T_{h}$ and $T_{c}$ is $η = 1 - T_{c} / T_{h}$ , independent of the working substance.

Proof. For a reversible engine, the total entropy change over one cycle is zero. The engine absorbs $Q_{h}$ from $T_{h}$ and rejects $Q_{c}$ to $T_{c}$ . The entropy change of the hot reservoir is $- Q_{h} / T_{h}$ and of the cold reservoir is $Q_{c} / T_{c}$ . The entropy change of the working substance is zero (cyclic process). So:

\frac{Q _{c}}{T _{c}} - \frac{Q _{h}}{T _{h}} = 0 ⟹ \frac{Q _{c}}{Q _{h}} = \frac{T _{c}}{T _{h}} .

Efficiency: $η = 1 - Q_{c} / Q_{h} = 1 - T_{c} / T_{h}$ . The result depends only on the reservoir temperatures, not on the working substance. Any two reversible engines operating between the same temperatures have the same efficiency. $□$

Connections Master

11.01.01 The first and second laws of thermodynamics provide the foundation for all cycle analysis.
11.01.02 Thermodynamic potentials (Helmholtz, Gibbs, enthalpy) appear naturally in the analysis of different cycle steps.
11.02.04 Entropy production in irreversible cycles connects to the fluctuation-dissipation theorem and the Onsager relations.
11.02.03 The Boltzmann H-theorem provides the microscopic mechanism for entropy production: collisions drive the distribution function toward equilibrium.
11.04.04 The ideal gas partition function provides the equation of state used in computing the work and heat for ideal-gas cycles.

Historical and philosophical context Master

Sadi Carnot published his Reflexions sur la puissance motrice du feu in 1824, before the concept of energy (first law) was formalised. His insight was that the efficiency of a heat engine depends only on the temperature difference, not on the working substance — a result he derived using the caloric theory of heat. Clausius (1850) and Thomson (Kelvin, 1851) rederived Carnot's result using the first and second laws, establishing the absolute temperature scale and the concept of entropy.

The Otto cycle (Nikolaus Otto, 1876) and Diesel cycle (Rudolf Diesel, 1892) model the thermodynamic processes in real internal combustion engines. Neither achieves Carnot efficiency because they involve irreversible processes (rapid combustion, finite-rate heat transfer), but they produce far more power per cycle than a Carnot engine operating at the same temperatures.

Bibliography Master

@book{carnot1824,
  author = {Carnot, Sadi},
  title = {R\'eflexions sur la puissance motrice du feu},
  publisher = {Bachelier},
  address = {Paris},
  year = {1824}
}

@book{callen1985,
  author = {Callen, Herbert B.},
  title = {Thermodynamics and an Introduction to Thermostatistics},
  edition = {2nd},
  publisher = {Wiley},
  year = {1985}
}

@book{reif1965,
  author = {Reif, Frederick},
  title = {Fundamentals of Statistical and Thermal Physics},
  publisher = {McGraw-Hill},
  year = {1965}
}

@book{fermi1937,
  author = {Fermi, Enrico},
  title = {Thermodynamics},
  publisher = {Dover},
  year = {1937}
}

Prerequisites

11.01.01
11.01.02

Tier anchors

beginner: Schroeder, An Introduction to Thermal Physics (2000), Ch. 4 (Engines and refrigerators)
intermediate: Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd ed. (1985), Ch. 4–5; Reif, Fundamentals of Statistical and Thermal Physics (1965), Ch. 5
master: Landau & Lifshitz, Statistical Physics, Part 1, 3rd ed. (1980), §19; Fermi, Thermodynamics (1937), Ch. 4

References

Carnot 1824 — Reflexions sur la puissance motrice du feu · Bachelier (Paris). Carnot showed that the maximum efficiency of a heat engine depends only on the temperatures of the hot and cold reservoirs: eta = 1 - T_c/T_h.
Callen — Thermodynamics and an Introduction to Thermostatistics, 2nd ed. (Wiley, 1985) · Ch. 4–5 Thermodynamic engines, Carnot cycle, efficiency bounds, and entropy production in irreversible processes.
Fermi — Thermodynamics (Dover, 1937) · Ch. 4 Thermodynamic potentials and the Carnot cycle. The Clausius inequality and entropy production in irreversible cycles.
Reif — Fundamentals of Statistical and Thermal Physics (McGraw-Hill, 1965) · Ch. 5 Simple applications of thermodynamics. Carnot cycle, Otto cycle, Diesel cycle, and the second-law efficiency.

Estimated time

beginner: 15 min
intermediate: 35 min
master: 60 min