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Thermodynamics · Chapter 8 of 10 · Advanced

Exergy

The first law counts energy; exergy counts the useful work hidden inside it. Measured against the environment, exergy reveals how much potential each stream really has, and how much irreversibility quietly destroys.

01

Readiness check

This chapter combines the first and second laws. Tick only what you can do closed-notes.

  • State the Carnot efficiency in terms of two temperatures.
  • Compute entropy generation for a heat-transfer process.
  • Use absolute temperature throughout.
  • Form a ratio as an efficiency.
  • Track heat, work, and a reference environment.
0 or 1 weak itemsContinue with this chapter.
2 weak itemsRevisit entropy generation in Chapter 7.
3 or more weak itemsReview the Carnot limit in Chapter 6.
02

The core idea

Exergy is the maximum useful work a stream or a quantity of heat could deliver as it comes into balance with the environment. Every irreversibility destroys exergy in proportion to the entropy it generates.

Xheat = Q(1 − T₀/T)Xdestroyed = T₀ SgenηII = exergy recovered / exergy supplied

Energy is conserved, but its quality is not. A joule of work is pure exergy; a joule of heat at the environment temperature has none. The reference is the dead state, where a system is at the environment's temperature and pressure and can do no more work. Exergy measures the distance from that state. Because exergy can be destroyed but never created, the destroyed amount, T₀ times the entropy generated, is the cleanest single measure of waste.

The skill works when: you fix the dead state and track exergy supplied, recovered, and destroyed.
The skill breaks down when: energy and exergy are treated as the same, or the environment temperature is left out.
The concept. Exergy supplied splits into the part recovered as useful work and the part destroyed by irreversibility. The destroyed share equals the environment temperature times the entropy generated.
03

The skills, taught in order

Five skills define the dead state and exergy, give the exergy of heat, connect lost work to entropy generation, and turn it all into a second-law efficiency.

8.1 The dead state and exergy

A system can do work only while it differs from its surroundings. When it reaches the environment's temperature T₀ and pressure P₀ it is at the dead state and its exergy is zero. Exergy (also called availability) is the maximum useful work obtainable as the system is brought reversibly to the dead state.

8.2 The exergy of heat

Heat Q supplied at temperature T carries only the work a Carnot engine could extract from it: Xheat = Q(1 − T₀/T). The hotter the source relative to the environment, the more exergy each joule carries; heat at T₀ carries none. This is the quality of heat made quantitative.

Source temperature TExergy fraction 1 − T₀/T (T₀ = 300 K)
1000 K0.70
800 K0.625
500 K0.40
350 K0.143

The same joule of heat is worth far more from a hot source than a warm one.

8.3 Reversible work and irreversibility

The reversible work is the most a process could deliver between the same end states. The shortfall is the irreversibility, I = Xdestroyed = T₀ Sgen, the Gouy-Stodola result. Because Sgen ≥ 0, exergy is always destroyed, never created, by friction, mixing, and heat transfer across a temperature gap.

8.4 Flow exergy

A flowing stream carries flow exergy ψ = (h − h₀) − T₀(s − s₀) + V²/2 + gz, measured from the dead-state values h₀ and s₀. This is the open-system counterpart of the stream's enthalpy, and it is what a perfect turbine could extract from the flow.

8.5 Second-law efficiency

The second-law efficiency compares what was recovered with what was available: ηII = exergy recovered / exergy supplied. Unlike first-law efficiency, it reaches one only for a reversible device, so it shows how much room for improvement truly remains.

Engineering connection: exergy analysis pinpoints where in a power plant or refrigeration cycle the largest losses occur, guiding which component to improve first.

04

Worked example 1: exergy of heat and what is destroyed

A furnace supplies 1000 kJ of heat at 800 K to an engine in an environment at 300 K. The engine produces 400 kJ of work. Find the exergy of the supplied heat, the second-law efficiency, and the exergy destroyed, then confirm it equals T₀ Sgen.

Figure 1. Of the 625 kJ of work potential in the supplied heat, 400 kJ is captured and 225 kJ is destroyed by irreversibility, the same 225 kJ as T₀ times the entropy generated.
  1. ProblemFind the heat exergy, ηII, and the exergy destroyed for the engine in Figure 1.
  2. Given / findQH = 1000 kJ, TH = 800 K, T₀ = 300 K, W = 400 kJ. Find Xheat, ηII, Xdestroyed.
  3. AssumptionsHeat supplied at a steady 800 K; environment at 300 K is the dead state; rejected heat goes to the environment.
  4. ModelThe exergy of the heat is its Carnot work; second-law efficiency is work over that exergy; the rest is destroyed, which should match T₀ Sgen.
  5. EquationsXheat = QH(1 − T₀/TH)ηII = W/XheatSgen = QL/T₀ − QH/TH
  6. SolveXheat = 1000(1 − 300/800) = 625 kJ. ηII = 400/625 = 0.64. Xdestroyed = 625 − 400 = 225 kJ. Check: QL = 600 kJ, Sgen = 600/300 − 1000/800 = 2 − 1.25 = 0.75 kJ/K, so T₀ Sgen = 300 × 0.75 = 225 kJ.
  7. CheckThe destroyed exergy from the balance equals T₀ Sgen exactly, the Gouy-Stodola relation. The second-law efficiency 0.64 matches the ratio of real to Carnot efficiency from Chapter 6.
  8. ConclusionExergy turns the vague idea of waste into a number, and ties it directly to the entropy generated. That is what makes it the engineer's improvement compass.
Result. Xheat = 625 kJ, ηII = 0.64, Xdestroyed = 225 kJ = T₀ Sgen.
05

Worked example 2: first-law versus second-law efficiency

A power plant receives 1000 kW of heat from a source at 1000 K and produces 350 kW of electricity, in an environment at 300 K. Find the first-law efficiency, the exergy supplied, the second-law efficiency, and the exergy destroyed.

Figure 2. First-law efficiency compares the work to the heat; second-law efficiency compares it to the exergy of that heat, which is smaller, so ηII is higher and more honest.
  1. ProblemFind ηI, the exergy supplied, ηII, and the exergy destroyed for the plant in Figure 2.
  2. Given / findQ̇ = 1000 kW, T = 1000 K, Ẇ = 350 kW, T₀ = 300 K. Find ηI, Ẋin, ηII, Ẋdestroyed.
  3. AssumptionsSteady operation; heat supplied at a fixed 1000 K; environment at 300 K.
  4. ModelFirst-law efficiency is work over heat; the exergy supplied is the heat exergy; second-law efficiency is work over exergy supplied.
  5. EquationsηI = Ẇ/Q̇Ẋin = Q̇(1 − T₀/T)ηII = Ẇ/Ẋin
  6. SolveηI = 350/1000 = 0.35. Ẋin = 1000(1 − 300/1000) = 700 kW. ηII = 350/700 = 0.50. Ẋdestroyed = 700 − 350 = 350 kW.
  7. CheckThe second-law efficiency exceeds the first-law value because the benchmark, the exergy of the heat, is smaller than the heat itself. Half of the available work potential is still being destroyed.
  8. ConclusionA plant that looks only 35 percent efficient by the first law is actually capturing half of what is thermodynamically possible. The second-law view sets a fairer and more useful target.
Result. ηI = 0.35, Ẋin = 700 kW, ηII = 0.50, Ẋdestroyed = 350 kW.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Energy equals exergyTreating all heat as fully convertible"At what temperature is this heat available?"Only the fraction 1 − T₀/T of heat is exergy.
Forgetting the environmentExergy computed with no T₀"What is the dead-state temperature?"Exergy is always measured relative to T₀ and P₀.
Expecting exergy to be createdNegative exergy destroyed"Is Sgen non-negative?"Exergy is destroyed, never created; Xdestroyed = T₀ Sgen ≥ 0.
Confusing the two efficienciesηII set equal to ηI"Am I dividing by heat or by exergy?"First-law uses energy in; second-law uses exergy in.
07

Practice ladder

Level 1 · Direct skill

Find the exergy of 500 kJ of heat supplied at 600 K in an environment at 300 K.

Show answer

X = Q(1 − T₀/T) = 500(1 − 300/600) = 500 × 0.5 = 250 kJ. Half of this warm heat is work potential.

Level 2 · Mixed concept

For the engine of Worked Example 1, what would the exergy destroyed be if the engine were reversible?

Show answer

A reversible engine generates no entropy, so Xdestroyed = T₀ Sgen = 0. It would produce the full 625 kJ of work, with ηII = 1.

Level 3 · Independent problem

Hot water gives up 200 kJ of heat across a wall to a room. The water side is at 350 K and the room at 300 K. Find the exergy destroyed in the heat transfer.

Show answer

Sgen = Q/Tcold − Q/Thot = 200/300 − 200/350 = 0.6667 − 0.5714 = 0.0952 kJ/K. Xdestroyed = T₀ Sgen = 300 × 0.0952 = 28.6 kJ. Heat transfer across a gap always destroys exergy.

Level 4 · Transfer to real engineering

Explain why using a high-temperature flame just to heat a room to 20 °C is a large exergy waste, even though it can be 100 percent efficient by the first law.

What good work looks like

The flame's heat carries high exergy (large 1 − T₀/T), but warming a room needs only low-exergy heat near T₀; the mismatch destroys most of the work potential, which a heat pump would instead exploit.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that my exergy of heat uses the source temperature and T₀ correctly."
"Give me four heat sources; I will rank them by exergy per joule."
"How much work can this give?" Computing Q(1 − T₀/T) yourself is the skill.
"What is the second-law efficiency?" Dividing work by exergy supplied is the point.

Portfolio task

Pick a real energy device. Identify the dead state, compute the exergy supplied, find the useful output, and report both the second-law efficiency and the exergy destroyed.

Must include: a dead state, an exergy supplied, a recovered output, and an exergy-destroyed value tied to entropy generation.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. What is the dead state?

The state in equilibrium with the environment (T₀, P₀), where a system can do no more useful work and its exergy is zero.

2. Write the exergy of heat.

Xheat = Q(1 − T₀/T), the Carnot work obtainable from heat at temperature T.

3. Relate exergy destroyed to entropy.

Xdestroyed = T₀ Sgen, which is never negative.

4. Define second-law efficiency.

Exergy recovered divided by exergy supplied; it equals one only for a reversible device.

5. Why can ηII exceed ηI?

Because the benchmark, the exergy of the heat, is smaller than the heat itself.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive the exergy of heat and the destroyed exergy from a blank page.
+3 daysCompute exergy destruction for two new processes.
+7 daysCarry exergy into the gas power cycles, Chapter 9.
+30 daysUse exergy to find the largest loss in a full cycle.
10

Textbook mapping

ItemMapping
Primary sourceBorgnakke and Sonntag, Fundamentals of Thermodynamics, Chapter 10 (Irreversibility and Availability)
Cross-referenceÇengel and Boles, Ch. 8 · Moran, Ch. 7
Core topics8.1 Dead state and exergy · 8.2 Exergy of heat · 8.3 Reversible work and irreversibility · 8.4 Flow exergy · 8.5 Second-law efficiency
Engineering connectionExergy accounting finds the largest losses in power and refrigeration systems.
Read nextChapter 9: Gas Power Cycles.