Training The Carnot Cycle — Thermodynamic Limits & Irreversibility Irreversibility, Lost Work & the Gouy-Stodola Theorem
3 / 6

Irreversibility, Lost Work & the Gouy-Stodola Theorem

30 min The Carnot Cycle — Thermodynamic Limits & Irreversibility

Every deviation from reversibility destroys work potential. The Gouy-Stodola theorem quantifies this destruction precisely: the rate of lost work equals the product of the dead-state temperature and the rate of entropy generation. This lesson develops the full exergy (availability) framework from the Gouy-Stodola theorem, applies it to heat transfer across a finite temperature difference, friction, and mixing, and shows how second-law efficiency pinpoints where a real system loses the most work.

Irreversibility and the Gouy-Stodola Theorem

Gouy-Stodola Theorem

The rate of lost (destroyed) work due to irreversibilities is:

$$\dot{W}_{\text{lost}} = T_0 \dot{S}_{\text{gen}}$$

where $T_0$ is the dead-state (environment) temperature and $\dot{S}_{\text{gen}} \geq 0$ is the total entropy generation rate of the system plus its surroundings. This is the single most powerful equation in engineering thermodynamics for identifying where energy is wasted.

Sources of Entropy Generation
  • Heat transfer across finite $\Delta T$: $\dot{S}_{\text{gen}} = \dot{Q}(1/T_L - 1/T_H) > 0$
  • Viscous friction: mechanical energy dissipated to thermal energy irreversibly.
  • Free (throttled) expansion: $\Delta P > 0$ work potential destroyed.
  • Mixing of different substances or temperatures: entropy of mixing is always positive.
  • Chemical reactions (non-equilibrium): affinity drives irreversible entropy production.
Exergy (Availability) of Heat

The maximum work extractable from a heat transfer $Q$ at temperature $T$ to a dead-state at $T_0$ is:

$$\text{Ex}_Q = Q\left(1 - \frac{T_0}{T}\right) = Q \cdot \eta_{\text{Carnot}}(T, T_0)$$

This is the exergy of heat — the Carnot factor $\psi = 1 - T_0/T$ converts thermal energy to work potential.

Example 1 — Lost Work in a Heat Exchanger

In a heat exchanger, 5 kW is transferred from a stream at $T_H = 600$ K to a stream at $T_L = 350$ K. The dead-state temperature is $T_0 = 300$ K. Find the rate of entropy generation and the lost work rate.

  1. $\dot{S}_{\text{gen}} = \dot{Q}\left(\frac{1}{T_L} - \frac{1}{T_H}\right) = 5000\left(\frac{1}{350} - \frac{1}{600}\right) = 5000(0.002857 - 0.001667) = 5000 \times 0.001190 = 5.95$ W/K
  2. $\dot{W}_{\text{lost}} = T_0 \dot{S}_{\text{gen}} = 300 \times 5.95 = 1{,}785$ W = 1.785 kW
  3. Exergy of heat at 600 K: $5(1 - 300/600) = 2.5$ kW
  4. Exergy of heat at 350 K: $5(1 - 300/350) = 0.714$ kW
  5. Exergy destroyed = $2.5 - 0.714 = 1.786$ kW ✓
Example 2 — Second-Law Efficiency

A real heat engine produces 3 MW of work from a source at 900 K to a sink at 300 K. The fuel supplies 8 MW. What is the first-law efficiency, the second-law efficiency, and the exergy destruction rate?

  1. First-law efficiency: $\eta_I = W/Q_H = 3/8 = 37.5\%$
  2. Carnot efficiency: $\eta_C = 1 - 300/900 = 66.7\%$
  3. Second-law efficiency: $\eta_{II} = \eta_I / \eta_C = 37.5/66.7 = 56.2\%$ — the engine captures only 56% of the available work potential.
  4. Entropy generation rate: From Gouy-Stodola:
  5. $\dot{W}_{\text{Carnot}} = 8 \times 0.667 = 5.33$ MW
  6. so $\dot{W}_{\text{lost}} = 5.33 - 3 = 2.33$ MW.
  7. $\dot{S}_{\text{gen}} = \dot{W}_{\text{lost}}/T_0 = 2.33 \times 10^6/300 = 7{,}767$ W/K

Practice Problems

1. Heat flows at 2 kW from a body at 800 K to one at 400 K. $T_0 = 298$ K. Find $\dot{S}_{\text{gen}}$ and $\dot{W}_{\text{lost}}$.
2. An adiabatic turbine receives steam with exergy 1.2 MJ/kg and produces 0.9 MJ/kg of work. Find $\eta_{II}$ and specific exergy destruction.
3. Show that for reversible heat transfer ($T_L \to T_H$), $\dot{S}_{\text{gen}} \to 0$ and $\dot{W}_{\text{lost}} \to 0$.
4. A heat engine has $\eta_{II} = 0.70$. Its source is 700 K, sink is 300 K. What is its actual efficiency?
5. Mixing 1 kg of water at 80°C with 1 kg at 20°C (cp = 4186 J/(kg·K), $T_0 = 293$ K). Find the entropy generated and lost work.
6. Why does free expansion of an ideal gas generate entropy even though no heat is transferred and no work is done?
Show Answer Key

1. $\dot{S}_{\text{gen}} = 2000(1/400 - 1/800) = 2000(0.0025 - 0.00125) = 2.5$ W/K; $\dot{W}_{\text{lost}} = 298 \times 2.5 = 745$ W

2. $\eta_{II} = 0.9/1.2 = 75\%$; Exergy destroyed $= 1.2 - 0.9 = 0.3$ MJ/kg

3. As $T_L \to T_H$, $(1/T_L - 1/T_H) \to 0$, so $\dot{S}_{\text{gen}} \to 0$ and $\dot{W}_{\text{lost}} = T_0 \dot{S}_{\text{gen}} \to 0$. Reversible heat transfer approaches zero temperature difference (infinite heat exchanger area).

4. $\eta_C = 1 - 300/700 = 57.1\%$; $\eta_I = \eta_{II} \cdot \eta_C = 0.70 \times 57.1 = 40\%$

5. Equilibrium temp: $T_f = (353 + 293)/2 = 323$ K. $\Delta S = m c_p[\ln(T_f/T_H) + \ln(T_f/T_L)] = 1(4186)[\ln(323/353) + \ln(323/293)] = 4186(-0.0879 + 0.0972) = 4186(0.00926) = 38.8$ J/K. $W_{\text{lost}} = 293 \times 38.8 = 11{,}368$ J ≈ 11.4 kJ.

6. Free expansion of an ideal gas: $\Delta U = 0$ (no heat, no work), so $T$ is unchanged, but $V$ increases. The gas occupies more microstates — $\Delta S = nR\ln(V_2/V_1) > 0$. The process is irreversible because the gas cannot spontaneously recompress.

Proof of Carnot's Theorem & The Clausius Inequality Endoreversible Theory — Maximum Power at Finite Speed