MECH3260 Chap.2 Exergy and Irreversibility
Exergy and Irreversibility
Week 2 introduces exergy, also called availability: the property that measures how much useful work a system could still deliver on its way to equilibrium with its environment. That equilibrium point is the dead state, and it is where exergy is defined to be zero.
The chapter separates three quantities that are easy to run together, namely total work, the useful part of it, and the reversible maximum, then defines irreversibility as the gap between the last two. From there it builds non-flow and flow exergy, the exergy transfer terms that carry work, mass and heat across a boundary, and the decrease-of-exergy principle that ties destroyed exergy directly to generated entropy.
The second-law efficiency defined here is the third efficiency in this unit and is reused in the mixtures, psychrometry, combustion and power-cycle chapters.
What this chapter covers
- 01
The dead state (T0, P0) -- equilibrium with the environment, where exergy is zero
- 02
Non-flow (closed-system) exergy phi = (u-u0) + P0(v-v0) - T0(s-s0), derived via a Carnot-engine bookkeeping argument
- 03
Flow exergy psi = (h-h0) - T0(s-s0) + V^2/2 + gz for a steady-flow device
- 04
Exergy destruction mechanisms: finite-temperature-difference heat transfer, mixing, free expansion, friction and turbulence
- 05
The decrease-of-exergy principle X_destroyed = T0 * S_gen, linking destroyed exergy directly to entropy generation
- 06
Second-law (exergetic) efficiency eta_II = W_u/W_rev = 1 - X_destroyed/X_supplied
Exergy of an expanding air charge
- +1Convert to kelvin: T1=500K, T2=360K, T0=290K.
- +1Internal-energy change: delta u = cv(T2-T1) = 0.718(360-500) = -100.5 kJ/kg.
- +2Entropy change delta s = cv ln(T2/T1) + R ln(v2/v1) = 0.0154 kJ/kg.K.
- +2Non-flow exergy change delta phi = delta u + P0 delta v - T0 delta s = -64.8 kJ/kg; for 0.08 kg, delta Phi = -5.18 kJ.
- +2Entropy generated S_gen = m delta s + Qout/T0 = 0.01503 kJ/K, so X_destroyed = T0 S_gen = 4.36 kJ.
Key terms
- Exergy (availability)
- The maximum useful work obtainable as a system moves to equilibrium with its environment. Unlike energy, exergy can be destroyed, and quantifying that destruction is the point of the whole chapter.
- Dead state
- The condition of complete equilibrium with the surroundings: the same temperature and pressure, and no motion or elevation relative to them. A system there has energy but no work potential, so its exergy is zero.
- Useful work
- Total work minus the work spent pushing the atmosphere out of the way, which is the ambient pressure times the volume change. It is zero for a rigid boundary and cancels over a complete cycle.
- Reversible work
- The largest useful work out, or the smallest work in, that is possible between a given pair of end states. Because it depends only on the end states it equals the change in exergy, which makes it a fair benchmark for any proposed design.
- Irreversibility
- The difference between reversible work and the useful work actually delivered. It is the same quantity as destroyed exergy and it is always positive, since no real process can beat the reversible result.
- Second-law efficiency
- Useful work divided by reversible work when work leaves the system, and the reciprocal when work enters. It scores a device against the best possible outcome for its own end states rather than against a perfect machine.
- Exergy transfer with heat
- The Carnot fraction of a heat flow, equal to one minus the ratio of the environment temperature to the boundary temperature, times the heat. It shrinks as heat crosses cooler and cooler surfaces, and that shrinkage is the destruction.
Exergy and Irreversibility FAQ
Why is exergy not conserved the way energy is?
Energy never disappears; the 1st law guarantees it. Exergy is a statement about potential rather than quantity, and potential is fragile. Friction, mixing, and heat crossing a finite temperature difference all leave the joules intact while moving them into a form no device can extract work from.
The decrease-of-exergy principle puts a number on that loss by making it the environment temperature multiplied by the entropy generated.
My device has a high thermal efficiency but a poor second-law efficiency. Is that a mistake?
No, and the pair of numbers is telling you something. Thermal efficiency asks what fraction of the heat input became work. Second-law efficiency asks how close the output came to the reversible maximum for the same end states. A well-insulated electric resistance heater turns every joule of electricity into heat, so its 1st-law figure is perfect, while it destroys nearly all the work potential of that electricity.
Quote whichever one the question asked for.
Where should I draw the boundary for an exergy balance?
Extend it outward until it sits at the environment temperature. The heat transfer term carries a factor of one minus the environment temperature over the boundary temperature, so a boundary at the environment temperature makes that factor zero and deletes the term entirely. What remains is the change in system exergy and the destruction, so the destruction becomes the only unknown.
This is the standard trick when a question asks for total destruction rather than a component-by-component breakdown.
Can I get destroyed exergy without doing a full exergy balance?
Usually yes, and it is often faster. Every destruction mechanism is also an entropy-generation mechanism, and the two are related exactly by the environment temperature. So set up the entropy balance over the system and the surroundings that receive its heat, find the total entropy generated, and multiply. The reverse also works when an exergy balance is the easier of the two to write.
Exam move
Start by getting the three work quantities apart on paper, because most errors on this topic are a total work value used where a useful work value belonged. Write the definition of useful work once and note the two cases that delete the surroundings term.
Next learn the closed-system and flow-system exergy formulas as one family rather than two results: they differ only by the substitution of enthalpy for internal energy and by the motion and elevation terms a stream carries. Then make the entropy route your default for destruction, since an entropy balance is usually easier to set up than a direct exergy balance and the two give the same number.
Finally, practise writing a one-sentence comment that names the dominant destruction mechanism and ties it to a temperature or pressure difference in the question, because comment marks are the ones most often left on the table.
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