Training The Carnot Cycle — Thermodynamic Limits & Irreversibility Endoreversible Theory — Maximum Power at Finite Speed
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Endoreversible Theory — Maximum Power at Finite Speed

30 min The Carnot Cycle — Thermodynamic Limits & Irreversibility

A Carnot engine produces maximum efficiency only at zero power — because reversible heat transfer requires an infinitesimally small temperature difference and hence infinitely slow operation. Real engines must run at finite speed and accept irreversible heat exchange with their reservoirs. Curzon and Ahlborn (1975) solved the endoreversible engine problem: find the efficiency at maximum power when the engine interior is reversible but the heat exchangers are not. Their result, $\eta_{CA} = 1 - \sqrt{T_L/T_H}$, fits observed efficiencies of large power plants remarkably well.

Endoreversible Engine Theory (Curzon-Ahlborn)

The Endoreversible Engine Model

The engine interior is a reversible (Carnot) sub-cycle operating between an internal hot temperature $\tau_H < T_H$ and an internal cold temperature $\tau_L > T_L$. Heat flows through finite-conductance exchangers:

$$\dot{Q}_H = K_H(T_H - \tau_H), \qquad \dot{Q}_L = K_L(\tau_L - T_L)$$

The internal sub-cycle is reversible, so $\dot{Q}_H/\tau_H = \dot{Q}_L/\tau_L$ (Carnot condition on internal temps). The net power is $\dot{W} = \dot{Q}_H - \dot{Q}_L$.

Curzon-Ahlborn Efficiency at Maximum Power

Optimizing $\dot{W}$ over the free parameters $\tau_H, \tau_L$ (subject to the Carnot constraint on the interior) yields:

$$\boxed{\eta_{CA} = 1 - \sqrt{\frac{T_L}{T_H}}}$$

This lies exactly halfway between 0 and the Carnot limit on the efficiency axis: $\eta_{CA} = \eta_C/2 + O(\eta_C^2)$. The corresponding maximum power is:

$$\dot{W}_{\max} = K_{\text{eff}}\left(\sqrt{T_H} - \sqrt{T_L}\right)^2, \qquad \frac{1}{K_{\text{eff}}} = \frac{1}{K_H} + \frac{1}{K_L}$$

Example 1 — Coal Power Plant Benchmark

A coal plant operates between $T_H = 838$ K (steam) and $T_L = 298$ K (condenser). Compute the Carnot efficiency, the Curzon-Ahlborn efficiency at maximum power, and compare to a typical observed efficiency of 36%.

  1. $\eta_C = 1 - 298/838 = 64.4\%$
  2. $\eta_{CA} = 1 - \sqrt{298/838} = 1 - \sqrt{0.3556} = 1 - 0.5963 = 40.4\%$
  3. Observed: 36% (UK power station average cited by Curzon & Ahlborn).
  4. The CA prediction of 40% is far closer to reality than the 64% Carnot limit, because the CA model correctly accounts for finite-rate heat exchange.
  5. The gap (40% vs. 36%) is explained by additional irreversibilities: friction in turbine/pump, pipe heat losses, and combustion irreversibility not modelled by the endoreversible framework.
Example 2 — Optimal Internal Temperatures

For symmetric heat conductances ($K_H = K_L = K$), find the optimal internal temperatures $\tau_H^*$ and $\tau_L^*$ that maximise power.

  1. The optimality conditions give:
  2. $$\tau_H^* = \frac{T_H(\sqrt{T_H} + \sqrt{T_L})}{2\sqrt{T_H}} \cdot \frac{2\sqrt{T_H}}{\sqrt{T_H} + \sqrt{T_L}} \cdot \frac{\sqrt{T_H}}{\sqrt{T_H} + \sqrt{T_L}} \cdot \ldots$$
  3. For $K_H = K_L$: $\tau_H^* = T_H(\sqrt{T_H} + \sqrt{T_L})/(2\sqrt{T_H})\cdot 2\sqrt{T_H}/(\sqrt{T_H}+\sqrt{T_L})$.
  4. Simplifying: $\tau_H^* = (T_H + \sqrt{T_H T_L})/2\cdot\ldots$
  5. More simply: $\tau_H^* = T_H^{1/2}(T_H^{1/2} + T_L^{1/2})/2$ and $\tau_L^* = T_L^{1/2}(T_H^{1/2} + T_L^{1/2})/2$.
  6. Their geometric mean: $\sqrt{\tau_H^* \tau_L^*} = (T_H^{1/2} + T_L^{1/2})^2/4$, which exceeds $\sqrt{T_H T_L}$ — internal temperatures straddle the geometric mean of the reservoirs.

Practice Problems

1. A nuclear plant: $T_H = 600$ K, $T_L = 300$ K. Compute $\eta_C$, $\eta_{CA}$, and $\eta_{CA}/\eta_C$.
2. Show that $\eta_{CA} \approx \eta_C/2$ for small temperature differences ($T_H - T_L \ll T_L$).
3. Why does a real engine's observed efficiency sit closer to $\eta_{CA}$ than to $\eta_C$?
4. What efficiency does the endoreversible engine approach as $K_H, K_L \to \infty$ (infinite heat exchanger area)?
5. A refrigerator (endoreversible) operating at maximum COP between 250 K and 300 K: find the Curzon-Ahlborn COP analogue.
6. Explain why the CA formula $1 - \sqrt{T_L/T_H}$ is independent of the heat conductances $K_H, K_L$.
Show Answer Key

1. $\eta_C = 50\%$; $\eta_{CA} = 1 - \sqrt{0.5} = 29.3\%$; ratio = 58.6%

2. Let $\epsilon = (T_H - T_L)/T_L \ll 1$. Then $\sqrt{T_L/T_H} = (1 + \epsilon)^{-1/2} \approx 1 - \epsilon/2$. So $\eta_{CA} \approx \epsilon/2 \approx \eta_C/2$ to leading order.

3. Carnot assumes reversible heat transfer (zero temperature difference → zero power). Real engines must transfer heat at finite rate → finite $\Delta T$ → internal irreversibility captured by the CA model.

4. As $K \to \infty$, $\tau_H \to T_H$ and $\tau_L \to T_L$, so the internal temperatures approach the reservoir temperatures and $\eta \to \eta_C$. Maximum power $\to \infty$ (but in practice limited by other factors).

5. For the endoreversible refrigerator at maximum COP: $\text{COP}_{CA} = \sqrt{T_L/T_H} \cdot T_L/(T_H - T_L)^{-1}\cdot\ldots$ → $\text{COP}_{CA} = \sqrt{T_L}/(\sqrt{T_H} - \sqrt{T_L})$. For 250 K / 300 K: $\sqrt{250}/(\sqrt{300} - \sqrt{250}) = 15.81/(17.32 - 15.81) = 15.81/1.51 = 10.47$.

6. The CA efficiency depends only on the temperature ratio because maximizing power over $\tau_H, \tau_L$ eliminates $K_H, K_L$ from the optimality condition — the ratio $K_H/K_L$ shifts where the maximum occurs but not the efficiency at that maximum when $K_H = K_L$ (symmetric case). For asymmetric conductances the efficiency at maximum power does depend weakly on $K_H/K_L$.

Irreversibility, Lost Work & the Gouy-Stodola Theorem Interactive Carnot Explorer