Availability, exergy and irreversibility

Dead state and exergy of closed systems, flowing streams and heat, reversible work, irreversibility via the Gouy–Stodola theorem, exergy balances and second-law efficiency.

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Why it matters

Energy is always conserved, so an energy balance cannot tell you where a turbine, combustor or heat exchanger is wasting its potential. Exergy (availability) can: it measures how much useful work a system or stream could deliver relative to the environment, and it is destroyed wherever irreversibility occurs. Exergy analysis shows that in a gas-turbine engine the combustor, not the exhaust, is the largest destroyer of work potential.

Key ideas

Environment and dead state. The environment is a large surrounding medium at T₀ and p₀ (typically 298 K or 300 K and 1 atm) whose intensive properties do not change. A system at T₀, p₀, at rest and at zero elevation relative to the environment is in the dead state; its exergy is zero.

Exergy (availability). The maximum useful work obtainable as a system is brought reversibly from its state to the dead state, exchanging heat only with the environment. "Useful" excludes the work done pushing back the atmosphere, p₀(V₀ − V) in the closed-system expression. Exergy is never negative, depends on both the system state and the environment, and is not conserved.

Exergy of heat. Heat Q available at a constant temperature T is worth at most Q·(1 − T₀/T) of work — the Carnot fraction. The same 100 kJ is worth far more at 1500 K than at 350 K. Heat below T₀ also has exergy (a cold space can drive an engine with the environment as source).

Reversible work, irreversibility and the Gouy–Stodola theorem.

  • W_rev is the maximum useful work (or minimum work input) between two given states.
  • Irreversibility (exergy destroyed) I = W_rev − W_useful for work-producing devices, and I = W_useful,in − W_rev,in for work-consuming ones.
  • Gouy–Stodola: I = T₀·S_gen,total ≥ 0, where S_gen includes the system and its immediate surroundings. Zero only for reversible processes.

Exergy balance. Exergy in − exergy out − exergy destroyed = change in exergy stored. Exergy leaves with useful work, with heat (at its boundary temperature), and with exiting streams; it is destroyed by friction, heat transfer across finite ΔT, throttling, mixing and combustion.

Second-law (exergetic) efficiency. η_II = useful work / W_rev for engines and turbines, W_rev,in / W_actual,in for compressors and refrigerators, or η_th / η_rev for heat engines. A 40% efficient engine working from a 1000 K source with T₀ = 300 K reaches 0.40/0.70 = 57% of its potential.

Formulas

φ = (u − u₀) + p₀·(v − v₀) − T₀·(s − s₀) (closed system, per kg, KE and PE neglected) ψ = (h − h₀) − T₀·(s − s₀) + V²/2 + g·z (flowing stream, per kg) X_Q = Q·(1 − T₀/T) (exergy of heat at constant T) I = T₀·S_gen W_rev = W_actual + I (work-producing device) η_II = W_actual / W_rev

  • T₀, p₀: environment (dead-state) temperature, K, and pressure, kPa
  • u, h, s, v: specific properties at the given state; subscript 0 at the dead state
  • φ, ψ: specific closed-system and flow exergy, kJ/kg
  • S_gen: total entropy generated, kJ/K (or kW/K for rates); I: exergy destroyed, kJ (kW)

Worked examples

Example 1 (standard). (a) An engine receives 1000 kJ from a source at 1000 K and produces 400 kJ; T₀ = 300 K. Find the exergy of the heat supplied, the second-law efficiency and the exergy destroyed plus lost. (b) Find the exergy of 1 kg of liquid water at 100 °C with the environment at 25 °C (c = 4.18 kJ/(kg·K), treat as incompressible).

  1. (a) X_Q = Q·(1 − T₀/T) = 1000 × 0.7 = 700 kJ
  2. η_II = W/X_Q = 400/700 = 0.571 (η_th = 0.40)
  3. Exergy not converted = 700 − 400 = 300 kJ (destroyed inside the engine or lost with its rejected heat).
  4. (b) For an incompressible liquid with v ≈ v₀: φ = c·(T − T₀) − T₀·c·ln(T/T₀)
  5. φ = 4.18 × 75 − 298.15 × 4.18 × ln(373.15/298.15) = 313.5 − 279.6 = 33.9 kJ/kg

Answer: (a) 700 kJ, η_II = 57.1%, 300 kJ; (b) about 34 kJ — only about 11% of the 313.5 kJ of "sensible heat" is work potential.

Example 2 (GATE level). Air expands adiabatically in a turbine from 800 kPa, 900 K to 100 kPa, 550 K. T₀ = 300 K; cp = 1.005, R = 0.287 kJ/(kg·K); neglect KE and PE. Find, per kg, the actual work, the entropy generated, the exergy destroyed, the reversible work and the second-law efficiency. Also find the isentropic efficiency.

  1. w = cp·(T₁ − T₂) = 1.005 × 350 = 351.75 kJ/kg
  2. s_gen = s₂ − s₁ = cp·ln(T₂/T₁) − R·ln(p₂/p₁) (adiabatic, so all Δs is generated)
  3. = 1.005 × ln(550/900) − 0.287 × ln(0.125) = −0.4949 + 0.5968 = 0.1019 kJ/(kg·K)
  4. I = T₀·s_gen = 300 × 0.1019 = 30.56 kJ/kg
  5. w_rev = w + I = 382.3 kJ/kg; η_II = 351.75/382.3 = 0.920
  6. Isentropic exit: T₂s = 900 × (1/8)^(0.4/1.4) = 496.8 K, so η_T = 350/(900 − 496.8) = 0.868

Answer: w = 351.8 kJ/kg, s_gen = 0.102 kJ/(kg·K), I = 30.6 kJ/kg, w_rev = 382.3 kJ/kg, η_II = 92.0%, η_T = 86.8%

The two efficiencies differ: η_T compares with an isentropic turbine reaching the same pressure, while η_II also credits the extra exergy still in the hotter exhaust.

Common mistakes

  • Treating exergy as conserved; only energy is.
  • Forgetting p₀(v − v₀) in closed-system exergy, or including it for a flowing stream (it is already inside h).
  • Using T₀ in °C, or using the system temperature instead of T₀ in I = T₀·S_gen.
  • Computing S_gen from the system alone when heat crosses to the surroundings.
  • Confusing isentropic efficiency with second-law efficiency.
  • Getting a negative exergy — it signals inconsistent data or an arithmetic error.

For GATE AE

Expect numericals on the exergy of heat from a reservoir, the irreversibility of a turbine, compressor or heat exchanger via I = T₀·S_gen, available and unavailable energy of heat, and second-law efficiency. Conceptual questions test the dead state, why exergy is destroyed, and the difference between energy and exergy efficiency. Practise computing S_gen carefully first; the rest follows from it.

Quick check

  1. What is the exergy of a system in the dead state?
  2. 500 kJ of heat is available at 800 K with T₀ = 300 K. Find its exergy.
  3. A process generates 10 J/K of entropy with T₀ = 300 K. Find the exergy destroyed.
  4. An engine receives 500 kJ at 600 K and rejects 300 kJ at 300 K (= T₀). Find the irreversibility.
  5. Can exergy be negative?

Answers: 1. Zero. 2. 312.5 kJ. 3. 3000 J. 4. 50 kJ. 5. No.

Interview questions

All Engineering Thermodynamics interview questions

Try answering each one aloud before you open it.

  1. 1.What is exergy and how is it different from energy?Concept

    Exergy is the maximum useful work obtainable as a system or stream is brought reversibly into equilibrium with its environment (the dead state at T₀, p₀). Energy is conserved in every process, but exergy is destroyed by irreversibilities such as friction, throttling, mixing and heat transfer across finite temperature differences, at the rate T₀·S_gen. Energy measures quantity; exergy measures quality — 100 kJ of heat at 1500 K has far more exergy than 100 kJ at 350 K.

  2. 2.Define availability in the context of thermodynamics.Concept

    Availability, also known as exergy, is the measure of a system's potential to perform work when it is brought into equilibrium with its surroundings. It represents the useful work potential of a system and is a function of the system's state and the environment.

  3. 3.Explain the concept of irreversibility in thermodynamic processes.Concept

    Irreversibility refers to the loss of exergy due to factors such as friction, unrestrained expansion, heat transfer across a finite temperature difference, and mixing of different substances. It is the reason why real processes are less efficient than ideal processes. Irreversibility results in the generation of entropy and a decrease in the system's ability to perform useful work.

  4. 4.Why is exergy analysis important in engineering applications?Application

    Exergy analysis is important because it helps identify where and how energy is being wasted in a system. By understanding the exergy destruction and losses, engineers can design more efficient systems, reduce energy consumption, and improve sustainability. It provides insights into the quality of energy transformations and helps optimize processes.

  5. 5.What happens to the exergy of a system when it reaches equilibrium with its surroundings?Application

    When a system reaches equilibrium with its surroundings, its exergy becomes zero. This is because there is no longer any potential to perform work, as the system and surroundings are at the same state. At equilibrium, all available energy has been degraded to a form that cannot do work.

  6. 6.How does the second law of thermodynamics relate to exergy?Concept

    The second law of thermodynamics states that entropy of an isolated system always increases over time. This law implies that exergy is destroyed in real processes due to irreversibilities. The second law provides the basis for understanding why exergy is not conserved and highlights the importance of minimizing irreversibilities to improve efficiency.

  7. 7.In what ways can irreversibility be minimized in a thermodynamic process?Application

    Irreversibility can be minimized by reducing friction, ensuring heat transfer occurs over small temperature differences, avoiding sudden expansions or compressions, and minimizing mixing of different substances. Using more efficient components and optimizing process conditions can also help reduce irreversibility.

  8. 8.Calculate the exergy of 1 kg of water at 100°C and 1 atm, assuming the environment is at 25°C and 1 atm.Numerical

    Use φ = (u − u₀) + p₀(v − v₀) − T₀(s − s₀); for liquid water v ≈ v₀, so φ ≈ (h − h₀) − T₀(s − s₀). From steam tables (saturated liquid values), h = 419.17 and h₀ = 104.83 kJ/kg, s = 1.3072 and s₀ = 0.3672 kJ/(kg·K), so φ = 314.34 − 298.15 × 0.9400 ≈ 34.1 kJ/kg. Treating water as incompressible with c = 4.18 kJ/(kg·K) gives about 33.9 kJ/kg — only about 11% of the 314 kJ/kg of sensible heat is work potential.

  9. 9.A heat engine receives 500 kJ of heat from a reservoir at 600 K and rejects 300 kJ to a sink at 300 K, which is also the environment temperature. Calculate the irreversibility.Numerical

    The engine works in a cycle, so its own entropy change is zero and S_gen comes from the reservoirs: S_gen = −500/600 + 300/300 = −0.8333 + 1.0 = 0.1667 kJ/K. Irreversibility I = T₀·S_gen = 300 × 0.1667 = 50 kJ. Check: a reversible engine would give 500(1 − 300/600) = 250 kJ, the actual engine gives 500 − 300 = 200 kJ, and the 50 kJ difference is the work potential destroyed.

  10. 10.Explain how exergy efficiency differs from energy efficiency.Concept

    Exergy efficiency is the ratio of useful exergy output to the exergy input, reflecting how well a system converts available energy into useful work. Energy efficiency, on the other hand, is the ratio of useful energy output to energy input, not accounting for the quality of energy. Exergy efficiency provides a more accurate measure of a system's performance by considering irreversibilities and the potential to do work.

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