Availability and irreversibility

Why it matters

Two plants can have the same first-law efficiency yet waste very different amounts of useful work. Availability (exergy) analysis tells you how much of the energy in a stream or a heat source could actually become work, and where in the plant that potential is being destroyed — the boiler, the throttle valve or the heat exchanger. It is the second law turned into a design tool for deciding where an improvement is worth the money.

Key ideas

Dead state and environment. Work potential is always measured against the surroundings, taken as a large reservoir at T₀ and p₀ (typically 25 °C or 300 K and 1 atm, as the problem states). A system in thermal and mechanical equilibrium with this environment is at the dead state (subscript 0): it can produce no more work.

Availability (exergy). The maximum useful work obtainable as a system is brought reversibly from its state to the dead state, exchanging heat only with the environment. It is a property of the system and the environment together. Unlike energy, exergy is not conserved: every irreversibility destroys some.

Available and unavailable energy of heat. Heat Q drawn from a source at constant T can at best drive a Carnot engine rejecting heat to T₀. The available energy is Q(1 − T₀/T); the unavailable part T₀·Q/T must be rejected to the environment. When heat flows across a finite temperature difference from T₁ to a lower T₂, the energy is conserved but its available part shrinks — this is the "degradation" of energy.

Useful work. A closed system that expands must push back the atmosphere, doing work p₀(V₂ − V₁) that nobody can use. Useful work W_u = W − p₀(V₂ − V₁). For a steady-flow device this term does not arise.

Reversible work and irreversibility.

• W_rev is the work a device would produce (or the minimum it would need) if it worked reversibly between the same end states.
• Irreversibility (exergy destroyed) I = W_rev − W_u for work-producing devices, and I = W_u,in − W_rev,in for work-consuming ones. It is always ≥ 0 and zero only for a reversible process.
• Gouy–Stodola theorem: I = T₀·S_gen, where S_gen is the entropy generated (system plus surroundings). This is the most useful way to compute it.

Second-law (exergetic) efficiency. First-law efficiency compares energy out with energy in; second-law efficiency compares actual performance with the reversible limit, so it shows how much room for improvement really exists. A reversible device has η_II = 1. Note that η_II differs from isentropic efficiency: both use the same actual work, but isentropic efficiency compares with an isentropic process to the same exit pressure, while η_II compares with a reversible process to the same exit state.

Where exergy is lost in practice. Combustion and heat transfer across large ΔT (boilers) destroy the most; throttling, mixing, friction and heat rejection at above-ambient temperature also contribute. This is why regeneration and reheat, which reduce temperature differences, raise plant efficiency.

Formulas

AE = Q (1 − T₀ / T) — available energy of heat Q from a source at constant T. UAE = T₀ Q / T — unavailable energy.

• Q: heat (kJ), T: source temperature (K), T₀: environment temperature (K).

φ = (u − u₀) + p₀ (v − v₀) − T₀ (s − s₀) + V²/2 + g z — specific non-flow (closed-system) exergy, kJ/kg. Φ = (U − U₀) + p₀ (V − V₀) − T₀ (S − S₀) — closed system at rest, kJ.

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

W_rev = ṁ (ψ₁ − ψ₂) — steady-flow device with one inlet and one outlet, heat exchanged only with the environment (kW).

W_u = W − p₀ (V₂ − V₁) — useful work of a closed system.

I = W_rev − W_u = T₀ S_gen — irreversibility (kJ or kW). S_gen = ΔS_system + ΔS_surroundings (kJ/K or kW/K).

η_II = W_u / W_rev (work-producing), η_II = W_rev / W_u (work-consuming), η_II = η_th / η_Carnot (heat engine), η_II = COP / COP_Carnot (refrigerator).

Worked examples

Example 1 (standard, degradation of heat). 1000 kJ of heat is available from a reservoir at 800 K; the environment is at 300 K. Find the available energy. The heat is first passed to a body at 500 K. How much available energy is lost?

1. At 800 K: AE = Q (1 − T₀/T) = 1000 × (1 − 300/800) = 1000 × 0.625 = 625 kJ.
2. At 500 K: AE = 1000 × (1 − 300/500) = 1000 × 0.4 = 400 kJ.
3. Loss = 625 − 400 = 225 kJ.
4. Check by Gouy–Stodola: S_gen = −1000/800 + 1000/500 = −1.25 + 2.00 = 0.75 kJ/K; I = T₀ S_gen = 300 × 0.75 = 225 kJ. ✓

AE = 625 kJ; available energy lost = 225 kJ

Example 2 (GATE level, compressor). Air (c_p = 1.005 kJ/kg·K, R = 0.287 kJ/kg·K) is compressed adiabatically and steadily from 100 kPa, 300 K to 500 kPa, 520 K. Environment: 300 K. Neglect KE and PE. Find the actual work, the irreversibility, the minimum (reversible) work, the second-law efficiency and, for comparison, the isentropic efficiency.

1. Actual work input: w = c_p (T₂ − T₁) = 1.005 × 220 = 221.1 kJ/kg.
2. Entropy change: Δs = c_p ln(T₂/T₁) − R ln(p₂/p₁) = 1.005 ln(520/300) − 0.287 ln 5 = 0.55279 − 0.46190 = 0.09089 kJ/kg·K.
3. Adiabatic, so s_gen = Δs: i = T₀ s_gen = 300 × 0.09089 = 27.27 kJ/kg.
4. Minimum work = flow-exergy rise: w_rev = (h₂ − h₁) − T₀ (s₂ − s₁) = 221.1 − 27.27 = 193.83 kJ/kg.
5. η_II = w_rev / w = 193.83 / 221.1 = 0.877.
6. Isentropic exit temperature: T₂s = 300 × 5^(0.4/1.4) = 300 × 1.5838 = 475.1 K; η_c = (475.1 − 300)/(520 − 300) = 175.1/220 = 0.796.

w = 221.1 kJ/kg, i ≈ 27.3 kJ/kg, w_rev ≈ 193.8 kJ/kg, η_II ≈ 87.7 % (isentropic efficiency ≈ 79.6 %)

The two efficiencies differ because the hotter actual exit air still carries exergy that the isentropic comparison ignores.

Example 3 (closed system). A system has U = 500 kJ, V = 2 m³, S = 1.5 kJ/K; at the dead state U₀ = 300 kJ, V₀ = 2.5 m³, S₀ = 1.2 kJ/K, with p₀ = 100 kPa and T₀ = 300 K.

1. Φ = (U − U₀) + p₀ (V − V₀) − T₀ (S − S₀) = 200 + 100 × (−0.5) − 300 × 0.3 = 200 − 50 − 90.

Φ = 60 kJ

Common mistakes

• Writing exergy as U + pV − T₀S with no dead-state terms. Exergy is always measured relative to the dead state.
• Using ΔS of the system alone for I = T₀ S_gen. You must include the entropy change of the surroundings (zero only for an adiabatic process).
• Using the source temperature or the mean temperature instead of the environment temperature T₀ in the Gouy–Stodola theorem.
• Forgetting the p₀ΔV term for a closed system that expands against the atmosphere.
• Treating second-law efficiency and isentropic efficiency as the same quantity.
• Thinking energy is "lost" in a heat exchanger. Energy is conserved; exergy is destroyed.

For GATE ME

Expect available and unavailable energy of heat from a constant-temperature source (or from a body cooling at varying temperature, using T₀ΔS), loss of available energy when heat flows across a temperature difference, irreversibility of adiabatic compressors and turbines via T₀·S_gen, flow exergy of a stream, and second-law efficiency. Practise doing the entropy-generation bookkeeping carefully; nearly every question reduces to it.

Quick check

1. What is the available energy of 500 kJ of heat at 600 K when T₀ = 300 K?
2. State the Gouy–Stodola theorem.
3. What is the irreversibility of a reversible process?
4. A heat engine has η = 30 % between 900 K and 300 K. Find η_II.
5. Why is exergy not conserved even though energy is?

Answers: 1. 250 kJ. 2. Irreversibility I = T₀ × entropy generated. 3. Zero. 4. η_Carnot = 0.667, so η_II = 0.30/0.667 = 0.45. 5. Every real process generates entropy, and each unit of entropy generated destroys T₀ units of work potential.