Availability, irreversibility and exergy

Why it matters

Energy is never lost, yet an engine still wastes most of its fuel. Exergy (availability) explains this by measuring the useful work potential of energy, and it is destroyed by every irreversibility. Exergy analysis tells an engineer where the real losses in an engine, turbocharger or air-conditioning system are, and how much could in principle be recovered, for example from exhaust heat.

Key ideas

Dead state and environment. The surroundings are modelled as an environment at T₀ and P₀ (often 298 K and 101.325 kPa). A system in thermal and mechanical equilibrium with it, at rest and at zero elevation, is at the dead state and has zero exergy.

Exergy (availability). The maximum useful work obtainable as a system is brought reversibly from its given state to the dead state, exchanging heat only with the environment. It depends on both the state of the system and the environment. Unlike energy, exergy is not conserved: it is destroyed by irreversibilities.

Useful work. When a closed system expands, part of its boundary work only pushes back the atmosphere, P₀(V₂ − V₁). Useful work W_u = W − P₀(V₂ − V₁).

Reversible work and irreversibility. The reversible work W_rev is the maximum useful work (or minimum work input) for a process between two given states. Irreversibility I = W_rev − W_u (work-producing devices) or I = W_u,in − W_rev,in (work-consuming devices). It is always ≥ 0.

Gouy–Stodola theorem. I = T₀ S_gen, where S_gen is the total entropy generated (system plus surroundings). This links the second law's entropy accounting to lost work.

Exergy of heat. Heat Q available at temperature T carries exergy Q(1 − T₀/T), the Carnot work it could yield. Heat at a temperature close to ambient has little value; heat at high temperature has a lot. Transferring heat across a temperature difference keeps the energy but destroys exergy.

Second-law efficiency. η_II compares actual performance with the reversible ideal between the same states: for an engine η_II = η_th/η_rev = W_u/W_rev; for a compressor η_II = W_rev,in/W_actual,in; for a refrigerator η_II = COP/COP_rev. It answers "how good is this device compared with the best possible", which first-law efficiency cannot.

Sources of exergy destruction in a vehicle. Combustion (the largest), heat transfer from hot gas to cylinder walls and coolant, throttling of intake air in SI engines, friction, and exhaust leaving at high temperature (an exergy loss rather than destruction, partly recoverable by turbocharging or waste-heat recovery).

Formulas

φ = (u − u₀) + P₀(v − v₀) − T₀(s − s₀) + V²/2 + g z — specific exergy of a closed system (non-flow availability), kJ/kg.

• u, v, s: state values; subscript 0: dead state; P₀ in kPa, T₀ in K.

ψ = (h − h₀) − T₀(s − s₀) + V²/2 + g z — specific flow exergy of a stream, kJ/kg.

X_Q = Q (1 − T₀ / T) — exergy of heat Q transferred at a constant temperature T (K).

W_u = W − P₀ (V₂ − V₁) — useful work of a closed system.

I = T₀ S_gen — irreversibility (exergy destroyed), kJ; S_gen in kJ/K.

w_rev = ψ₁ − ψ₂ — reversible work per unit mass of a steady-flow adiabatic device with one inlet and one outlet (negative for a compressor, meaning minimum input).

ψ₂ − ψ₁ = (h₂ − h₁) − T₀ (s₂ − s₁) — change in flow exergy; kinetic and potential terms neglected.

η_II = W_u / W_rev (work-producing) or η_II = W_rev,in / W_in (work-consuming).

x = c [(T − T₀) − T₀ ln(T/T₀)] — exergy per unit mass of an incompressible liquid or solid (pressure term neglected).

Worked examples

Example 1 (standard: exergy of heat and its destruction). 1000 kJ of heat leaves a furnace at 1000 K and enters a body at 500 K. The environment is at 300 K. Find the exergy of the heat before and after the transfer, and the exergy destroyed.

1. Exergy at 1000 K: X_Q = Q(1 − T₀/T) = 1000 × (1 − 300/1000) = 700 kJ.
2. Exergy at 500 K: 1000 × (1 − 300/500) = 400 kJ.
3. Exergy destroyed = 700 − 400 = 300 kJ.
4. Check with Gouy–Stodola: S_gen = Q/T_L − Q/T_H = 1000/500 − 1000/1000 = 1.0 kJ/K, so I = T₀ S_gen = 300 × 1.0 = 300 kJ. Consistent. Energy (1000 kJ) is conserved; work potential is not.

Example 2 (GATE level: second-law efficiency of a compressor). Air is compressed adiabatically in steady flow from 100 kPa, 300 K to 500 kPa, 500 K. Take c_p = 1.005 kJ/kg·K, R = 0.287 kJ/kg·K, T₀ = 300 K, and neglect kinetic and potential energy changes. Find the actual work, the irreversibility, the minimum (reversible) work and the second-law efficiency.

1. Actual work input: w_in = c_p (T₂ − T₁) = 1.005 × 200 = 201.0 kJ/kg.
2. Entropy change: s₂ − s₁ = c_p ln(T₂/T₁) − R ln(P₂/P₁) = 1.005 ln(500/300) − 0.287 ln 5 = 0.5134 − 0.4619 = 0.0515 kJ/kg·K.
3. Adiabatic, so S_gen = s₂ − s₁; irreversibility i = T₀ s_gen = 300 × 0.05147 = 15.4 kJ/kg.
4. Minimum work input = increase in flow exergy: ψ₂ − ψ₁ = (h₂ − h₁) − T₀(s₂ − s₁) = 201.0 − 15.4 = 185.6 kJ/kg.
5. η_II = w_rev,in / w_in = 185.6/201.0 = 0.923 (92.3%).

Note: actual = reversible + destroyed (201.0 = 185.6 + 15.4).

Common mistakes

• Using T in °C in (1 − T₀/T) or in T₀ S_gen.
• Forgetting the P₀(v − v₀) term in closed-system exergy, or wrongly including it in flow exergy (flow work is already inside h).
• Taking I = T₀ ΔS_system for a non-adiabatic process; S_gen must include the entropy change of the surroundings or reservoirs.
• Confusing exergy loss (exergy carried out with exhaust, still recoverable) with exergy destruction (gone for good).
• Expecting exergy to be conserved. Only energy is; exergy balance has a destruction term.
• Assuming a reversible process keeps the system's exergy constant. The system's exergy can change by transferring work or heat; what is zero is the exergy destroyed.

For GATE ME

Questions typically ask for the available and unavailable parts of heat supplied at a given temperature, the maximum work obtainable from a hot body or a gas in a tank, irreversibility via T₀ S_gen for heat transfer, throttling, mixing or adiabatic compression and expansion, and second-law efficiency. Practise keeping the entropy generation of system plus surroundings, and know the closed and flow exergy expressions by heart.

Quick check

1. What is the exergy of 500 kJ of heat at 600 K if T₀ = 300 K?
2. A process generates 10 J/K of entropy with T₀ = 300 K. How much exergy is destroyed?
3. What is the exergy of a system at the dead state?
4. Air is throttled adiabatically from 500 kPa to 100 kPa at a constant 300 K (ideal gas). What is the irreversibility per kg if T₀ = 300 K and R = 0.287 kJ/kg·K?

Answers: 1. 500 × 0.5 = 250 kJ. 2. 3000 J. 3. Zero. 4. s_gen = R ln 5 = 0.462 kJ/kg·K, so i = 300 × 0.462 ≈ 138.6 kJ/kg.