Statistical Physics 1, supplementary material: beyond the Carnot engine
The Chambadal–Novikov–Curzon–Ahlborn engine
A Carnot engine that has to deliver power in finite time. Drag the working temperature below and watch the trade-off between efficiency and power play out, then follow the derivation step by step.
Setting up the model
A Carnot engine run quasi-statically reaches \(\eta_C = 1 - T_c/T_h\), but only because each isothermal stroke takes forever, so its power is zero. To get heat in and out at a finite rate, the working substance must be colder than the hot reservoir (\(T_{hw} < T_h\)) and hotter than the cold reservoir (\(T_{cw} > T_c\)). Following Schroeder, the heat flows obey a linear conduction law with the same conductance \(K\) on both sides and the same duration \(\Delta t\) for both isotherms:
\[ \frac{Q_h}{\Delta t} = K\,(T_h - T_{hw}), \qquad \frac{Q_c}{\Delta t} = K\,(T_{cw} - T_c). \]Everything inside the engine is reversible; the only irreversibility lives in the two conducting walls. That is why the model is called endoreversible.
A note on the name. The result \(\eta = 1 - \sqrt{T_c/T_h}\) is widely known as the Curzon–Ahlborn efficiency after their 1975 paper, which is also the basis of Problem 4.6 in Schroeder's textbook. The same expression had appeared in 1957 in the work of Pierre Chambadal and Igor Novikov on the efficiency of nuclear power stations, so this page uses the name Chambadal–Novikov–Curzon–Ahlborn for the engine. The symbol \(\eta_{CA}\) is kept below to match the cited papers.
The working cycle on a P–V diagram
Inside the walls, the working substance runs an ordinary Carnot cycle, only between \(T_{hw}\) and \(T_{cw}\) instead of \(T_h\) and \(T_c\). The diagram shows this cycle for a monatomic ideal gas (\(\gamma = 5/3\)), in units where \(Nk = 1\) and the volume at the start of the hot stroke is \(V_1 = 1\). Pressure is written \(p\) here to keep it apart from the power \(P\).
The cycle has four strokes: isothermal expansion at \(T_{hw}\) (1→2), adiabatic expansion (2→3), isothermal compression at \(T_{cw}\) (3→4), and adiabatic compression (4→1). Each isothermal stroke lasts the same fixed \(\Delta t\), so the heat absorbed is \(Q_h = K\Delta t\,(T_h - T_{hw}) = NkT_{hw}\ln(V_2/V_1)\), which sets the expansion ratio. The adiabats then fix \(V_3/V_2 = V_4/V_1 = (T_{hw}/T_{cw})^{1/(\gamma-1)}\).
Because \(\Delta t\) is held fixed, the enclosed area is proportional to the power, \(P = W/2\Delta t\). The loop is therefore largest exactly at the maximum-power point, and it collapses in both limits. In the short-circuit limit the two isotherms merge and the loop flattens onto a single isotherm: lots of heat passes through, but no area. In the quasi-static limit \(Q_h \to 0\) at fixed \(\Delta t\), so the isotherms shrink to points and the loop collapses onto an adiabat. The textbook Carnot cycle keeps a finite area in that limit only because it takes \(\Delta t \to \infty\).
Deriving the efficiency at maximum power
Entropy balance
Over one cycle the working substance returns to its initial state, so its entropy change is zero. Since it is reversible internally, the entropy it absorbs at \(T_{hw}\) must equal the entropy it rejects at \(T_{cw}\):
\[ \frac{Q_h}{T_{hw}} = \frac{Q_c}{T_{cw}}. \]Substituting the conduction laws (the factors \(K\Delta t\) cancel):
\[ \frac{T_h - T_{hw}}{T_{hw}} = \frac{T_{cw} - T_c}{T_{cw}} \quad\Longleftrightarrow\quad \boxed{\;\frac{T_h}{T_{hw}} + \frac{T_c}{T_{cw}} = 2\;} \]Solving for the cold-side working temperature gives the form used below,
\[ T_{cw} = \frac{T_c\,T_{hw}}{2T_{hw} - T_h}. \]A working engine needs \(T_{cw} < T_{hw}\), which translates into \(\tfrac12(T_h + T_c) < T_{hw} \le T_h\). At the lower end \(T_{cw} = T_{hw}\) (short circuit); at the upper end \(T_{hw} = T_h\) and \(T_{cw} = T_c\) (quasi-static Carnot).
Power output
With negligible time for the adiabats, one cycle lasts \(2\Delta t\). By the first law \(W = Q_h - Q_c\), so
\[ P = \frac{Q_h - Q_c}{2\Delta t} = \frac{K}{2}\Big[(T_h - T_{hw}) - (T_{cw} - T_c)\Big]. \]Eliminating \(T_{cw}\) with the second-law relation from step 1:
Maximizing the power
Hold \(T_h\) and \(T_c\) fixed and differentiate. Using \(\dfrac{d}{dT_{hw}}\dfrac{T_{hw}}{2T_{hw}-T_h} = \dfrac{-T_h}{(2T_{hw}-T_h)^2}\),
\[ \frac{dP}{dT_{hw}} = \frac{K}{2}\left[-1 + \frac{T_h T_c}{(2T_{hw} - T_h)^2}\right] = 0 \quad\Longrightarrow\quad 4T_{hw}^2 - 4T_h T_{hw} + T_h^2 - T_h T_c = 0. \]The quadratic gives \(T_{hw} = \tfrac12\big(T_h \pm \sqrt{T_h T_c}\big)\). The minus root lies below \(\tfrac12(T_h+T_c)\) and would make \(T_{cw}\) negative, so it is unphysical. Since \(d^2P/dT_{hw}^2 = -2K T_hT_c/(2T_{hw}-T_h)^3 < 0\), the plus root is a maximum:
Each working temperature sits halfway between its reservoir and the geometric mean \(\sqrt{T_hT_c}\). Plugging back in, the maximum power is
\[ P_{\max} = \frac{K}{4}\left(\sqrt{T_h} - \sqrt{T_c}\right)^2, \]which matches Eq. (19) of Curzon and Ahlborn with \(\alpha = \beta = K\).
Efficiency at maximum power
The efficiency follows from the first law and the entropy balance of step 1:
\[ \eta = 1 - \frac{Q_c}{Q_h} = 1 - \frac{T_{cw}^*}{T_{hw}^*} = 1 - \frac{\sqrt{T_c}\,(\sqrt{T_c} + \sqrt{T_h})}{\sqrt{T_h}\,(\sqrt{T_h} + \sqrt{T_c})} \quad\Longrightarrow\quad \boxed{\;\eta_{CA} = 1 - \sqrt{T_c/T_h}\;} \]For a coal-fired steam turbine, \(T_h = 600\,^\circ\mathrm{C} = 873.15\ \mathrm K\) and \(T_c = 25\,^\circ\mathrm{C} = 298.15\ \mathrm K\):
\[ \eta_{CA} = 1 - \sqrt{\tfrac{298.15}{873.15}} \approx 0.416, \qquad \eta_C = 1 - \tfrac{298.15}{873.15} \approx 0.659. \]At maximum power the working temperatures are \(T_{hw}^* \approx 691.7\ \mathrm K\) (418.5 °C) and \(T_{cw}^* \approx 404.2\ \mathrm K\) (131.0 °C). The Curzon–Ahlborn value of about 42% is far closer to the roughly 40% achieved by real coal plants than the Carnot bound of 66%. Real plants are built and run to deliver power, not to approach the quasi-static limit.
Where the entropy comes from
The working substance's entropy returns to its initial value every cycle, but the reservoirs do not. The hot reservoir loses \(Q_h/T_h\), which is less than the \(Q_h/T_{hw}\) the engine absorbs, and the cold reservoir gains \(Q_c/T_c\), which is more than the \(Q_c/T_{cw}\) the engine rejects. The total entropy produced per cycle is therefore
\[ \Delta S_{\rm tot} = \frac{Q_c}{T_c} - \frac{Q_h}{T_h} > -\frac{Q_h}{T_{hw}} + \frac{Q_c}{T_{cw}} = 0. \]Combining this with the first law gives a compact budget (the Gouy–Stodola relation), which is what the colored bar in the demo shows:
\[ W = \eta_C\,Q_h - T_c\,\Delta S_{\rm tot}. \]Every bit of entropy generated in the walls costs \(T_c\) worth of work. The two limits of the slider make the trade-off concrete:
Quasi-static limit (\(T_{hw}\to T_h,\ T_{cw}\to T_c\)). The temperature gaps vanish, no entropy is produced, and \(\eta \to \eta_C\). But the conduction law forces \(\Delta t \to \infty\) to move a finite amount of heat, so the power goes to zero.
Short-circuit limit (\(T_{hw}\to T_{cw}\)). Heat flows as fast as possible, but the internal Carnot engine now runs between two equal temperatures, extracts no work, and simply passes heat from hot to cold, like a wire shorting a battery. Power and efficiency both vanish.
The maximum-power point is the compromise between these two zeros, and its efficiency happens to depend on nothing but the reservoir temperatures.
Comparison with real power plants
Curzon and Ahlborn's Table I compared their formula with operating plants. The values below are recomputed from the listed temperatures, with Schroeder's example of a typical coal-fired turbine added at the top.
| Power source | \(T_{c}\) (°C) | \(T_{h}\) (°C) | \(\eta_{C}\) | \(\eta_{CA}\) | Observed \(\eta\) |
|---|
Observed efficiencies are taken from Curzon and Ahlborn (1975), Table I; the 40% for a typical coal plant is the figure quoted by Schroeder. Small differences from the printed table come from rounding the Celsius-to-kelvin conversion.
Do the simplifying assumptions matter?
Schroeder's footnote says that neither equal conductances nor equal stroke times are needed for the final efficiency. Curzon and Ahlborn allow different conductances \(\alpha\) (hot side) and \(\beta\) (cold side) and let the stroke times \(t_1, t_2\) float. Maximizing over both working temperatures gives
\[ T_{hw}^* = C\sqrt{T_h},\quad T_{cw}^* = C\sqrt{T_c},\quad C = \frac{\sqrt{\alpha T_h} + \sqrt{\beta T_c}}{\sqrt{\alpha} + \sqrt{\beta}},\quad \frac{t_1}{t_2} = \sqrt{\frac{\beta}{\alpha}}, \]so \(T_{cw}^*/T_{hw}^* = \sqrt{T_c/T_h}\) again. With \(\alpha = \beta\) the optimal stroke times come out equal, which is why Schroeder's equal-\(\Delta t\) assumption costs nothing in that case. The widget below also checks what happens when you impose equal stroke times even though the conductances differ: the optimum shifts and the power drops, yet the efficiency at maximum power stays at \(\eta_{CA}\).
Remarks: a quantum-dot heat engine
The Curzon–Ahlborn result started a long line of work on efficiency at maximum power. Near equilibrium (small \(\eta_C\)), Van den Broeck showed using linear irreversible thermodynamics that any "tightly coupled" engine, one whose heat flux is strictly proportional to its work-producing flux, should have \(\eta_{\rm op} \approx \tfrac12\eta_C\). The Curzon–Ahlborn engine obeys this, since \(\eta_{CA} = \tfrac12\eta_C + \tfrac18\eta_C^2 + \tfrac1{16}\eta_C^3 + \cdots\).
A 2018 study by S. H. Lee (your instructor), J. Um, and H. Park asked how robust this "\(\tfrac12\)-universality" really is, using one of the simplest engines imaginable: a single quantum-dot level at energy \(E_{\rm QD}\) (measured from \(\mu_1\)) connecting a hot lead (\(T_1\)) to a cold lead (\(T_2\)) held at a higher chemical potential, \(\Delta\mu = \mu_2 - \mu_1\). Each electron that hops through carries heat \(E_{\rm QD}\) out of the hot lead and delivers work \(\Delta\mu\), so
\[ J = \tfrac12\left[\frac{1}{e^{E_{\rm QD}/T_1}+1} - \frac{1}{e^{(E_{\rm QD}-\Delta\mu)/T_2}+1}\right],\qquad \dot W = J\,\Delta\mu,\qquad \eta = \frac{\Delta\mu}{E_{\rm QD}} \](with \(k_B = 1\)). The engine is tightly coupled by construction. The role played by \(T_{hw}\) in the endoreversible engine is now played by two knobs: the gate voltage sets \(E_{\rm QD}\), and the source–drain voltage sets \(\Delta\mu\). Which knobs you are allowed to turn turns out to matter a great deal.
Both knobs free (global optimization). \(\eta_{\rm op} = \tfrac12\eta_C + \tfrac18\eta_C^2 + 0.0775\,\eta_C^3 + \cdots\), agreeing with Curzon–Ahlborn through second order and departing from the CA coefficient \(1/16\) only at third order.
Gate fixed, \(\Delta\mu\) tuned. The \(\tfrac12\) survives, but the second-order coefficient now depends on \(E_{\rm QD}\): \(\eta_{\rm op} = \tfrac12\eta_C + \tfrac{E_{\rm QD}}{16T_2}\tanh\!\big(\tfrac{E_{\rm QD}}{2T_2}\big)\eta_C^2 + \cdots\).
Bias fixed, gate tuned. Here the universality breaks: \(\eta_{\rm op} \approx \eta_C - (T_2/\Delta\mu)\,\eta_C^2\), with linear coefficient 1 rather than \(\tfrac12\). The expansion point \(E_{\rm QD}\to\infty\) is a zero-flux point that is not an equilibrium, so linear response does not apply there. The mechanical flux vanishes with an essential singularity instead of linearly. If the flux behaved as \(J_2 \propto (X_2 + \xi' X_t)X_2^{\,n}\), one would find \(\eta_{\rm op} = \frac{n+1}{n+2}\eta_C\); the ordinary linear case \(n=0\) gives \(\tfrac12\), and the essential singularity acts like \(n\to\infty\), giving 1. A quantum-dot experiment by Josefsson et al. (2018) tuning the gate voltage reported efficiencies at maximum power well above \(\eta_{CA}\), consistent with this picture.
The curves below are computed live by numerically maximizing \(\dot W\) (with \(T_2 = 1\)).
The moral connects back to the main lecture: "efficiency at maximum power" is not a single number attached to an engine. It depends on the transport law (linear conduction gives Curzon–Ahlborn), on which parameters are optimized, and on whether the regime being expanded around is a genuine equilibrium. In the Curzon–Ahlborn engine, even forcing equal stroke times leaves \(\eta_{CA}\) intact; in the quantum dot, choosing the gate over the bias changes the answer at leading order.