Chapter 7 of 10 · 10 min read
Lost work and exergy
Turn entropy into kilowatts: the lost-work theorem, the exergy of heat, of flowing streams and of cold, an exergy balance for each piece of plant, and why every audit must state its dead state.
Entropy generated is a correct but awkward measure of loss. Few managers will act on a figure in kilowatts per kelvin. The step that makes the second law practical is to multiply by a temperature and get kilowatts of work that could have been produced, or saved, and were not. That quantity is lost work, and its counterpart, the work a flow could still produce, is exergy.
This chapter sets out the lost-work theorem, defines exergy for heat, for flowing streams and for cold, and draws up an exergy balance for each item of the running case’s plant. It shows how much the answers depend on the temperature of the surroundings, and ends with two engineers whose heat pump assessments disagreed by a factor of two for that reason alone.
The lost-work theorem
Consider any process in surroundings at temperature T₀. Compare the work it actually produces, or the work it actually needs, with what a reversible process between the same end states would produce or need, exchanging heat only with the surroundings. The difference is the lost work, and it is always
W_lost = T₀ × S_gen
where S_gen is the total entropy generated by the real process, in the system and its surroundings together. The result is known as the Gouy–Stodola theorem, after the two engineers who stated it at the end of the nineteenth century, and it is the bridge between entropy and money. Every unit of entropy generated in a plant that sits in surroundings at 293.15 K destroys 293.15 units of work potential.
The theorem needs only two things: a value for T₀, and the total entropy generated, found with the methods of Chapters 4 and 6. It applies to every irreversibility, whether heat transfer, friction, throttling, mixing or combustion, which is what makes it the common currency of a second-law audit. Different losses in different machines can be compared and added once each is in kilowatts.
The derivation is short. A reversible process generates no entropy. The real one generates S_gen, and to bring the surroundings back to the same final state as the reversible process, that much entropy would have to be removed from them as heat T₀ × S_gen, which could only be done by turning that much work into heat somewhere else. So the real process is worse than the reversible one by exactly T₀ × S_gen.
Chapter 6’s uninsulated pipe shows the theorem at work. It generated 0.00410 kW/K, so it lost 293.15 × 0.00410 = 1.20 kW of work potential. The same 1.20 kW appears if the question is put the other way round: the 10 kW of heat at 60 °C could have driven a reversible engine against the 20 °C room, producing 10 × (1 − 293.15 ÷ 333.15) = 1.20 kW.
Exergy
The exergy of something is the maximum work obtainable by bringing it into equilibrium with the surroundings, the dead state of Chapter 1. Lost work is exergy destroyed. Three forms cover the running case.
- X, x
- exergy, kW or kJ/kg
- h, s
- specific enthalpy and entropy of the stream
- Kettle steam's latent heat, 45.0 kW at 375.68 K: 45.0 × (1 − 293.15 ÷ 375.68) = 9.89 kW
- Wash water at 60 °C: 167.226 − 293.15 × 0.53472 = 10.473 kJ/kg
- Cool room, 12 kW at 275.15 K: 12 × (293.15 ÷ 275.15 − 1) = 0.785 kW
The third formula is easily overlooked. Something colder than its surroundings has exergy, because heat would flow into it from the surroundings and could drive an engine on the way. Keeping the cool room cold means continually supplying that exergy, which is why its minimum work is 0.785 kW for 12 kW of cooling.
Exergy of compressed air
For an ideal gas at the dead-state temperature, the stream exergy reduces to the reversible isothermal work of expanding it to atmospheric pressure: x = R T₀ ln(p ÷ p₀). The plant’s compressed air at 8 bar absolute carries 0.287 × 293.15 × ln(8 ÷ 1.01325) = 173.8 kJ/kg, or 10.43 kW for 0.06 kg/s, equal to the minimum compression work of Chapter 4. At 5 bar it carries 134.3 kJ/kg, or 8.06 kW. Chapter 8 follows the 2.37 kW difference between them.
Exergy balances for the plant
An exergy balance for a steady piece of plant reads: exergy in = exergy out in products + exergy carried away in wastes + exergy destroyed. Unlike energy, exergy is not conserved, and the destruction term is never negative. The balance identifies, for each item, how much of its input survives, how much leaves unused and how much is gone.
| Item | Exergy in | Product exergy | Destroyed or discarded | Second-law efficiency |
|---|---|---|---|---|
| Gas boiler, wash water | 92.9 kW | 5.24 kW | 87.7 kW | 5.6% |
| Air compressor | 22.0 kW | 10.43 kW | 11.6 kW | 47.4% |
| Cool room refrigeration | 4.8 kW | 0.785 kW | 4.0 kW | 16.4% |
| Air regulators, 8 to 5 bar | 10.43 kW | 8.06 kW | 2.37 kW | 77.3% |
| Kettle, steam to food at 90 °C | 9.89 kW | 8.68 kW | 1.21 kW | 87.8% |
The kettle’s product exergy is the heat’s exergy at the food’s temperature, 45.0 × (1 − 293.15 ÷ 363.15) = 8.68 kW, and the difference from the steam’s 9.89 kW is the 1.21 kW of Chapter 6. An exergy balance can be checked the same way as an energy balance: the terms must add up. For the compressor, 22.0 kW of electricity in equals 10.43 kW of air exergy out plus 11.6 kW destroyed in the motor, the compression and the aftercooler, whose rejected heat is at too low a temperature to carry much exergy. If a measured balance does not close, a flow or a temperature is wrong, exactly as with a first-law balance, and the error must be found before the figures are used.
The table is the first draft of the audit in Chapter 10. Already it points to the boiler as the dominant loss, an order of magnitude larger than anything else, and to the compressor as second.
Exergy that leaves unused
Not all exergy lost by a plant is destroyed inside it. Some is carried out in waste streams and destroyed later, in the drain, the stack or the atmosphere. The plant’s effluent at 45 °C carries 0.50 × 4.18 × [25 − 293.15 × ln(318.15 ÷ 293.15)] = 2.11 kW of exergy away with 52.25 kW of heat. That is exergy discarded, not destroyed, and it is the part a heat recovery scheme can still capture, which is why Chapter 6’s exchanger recovers most of it.
Avoidable and unavoidable destruction
Not all exergy destruction can be eliminated, and an audit that treats every lost kilowatt as a target sets impossible goals. A useful practice is to split each item’s destruction into two parts.
- Unavoidable destruction is what would remain with the best equipment that could reasonably be bought or built for the duty: heat exchangers with practical approach temperatures, a compressor of good modern design, a refrigeration system with sensibly sized coils. It is set by technology and cost, not by physics.
- Avoidable destruction is the rest: the part that better design, better operation or a different kind of machine could remove.
For the boiler, almost all of the 87.7 kW is avoidable in principle, because a different machine, the heat pump, does the same duty with far less. For the kettle, the 1.21 kW is mostly unavoidable: steam must be somewhat hotter than the food it heats, and a smaller temperature difference would need a larger, costlier jacket and slower heating. Splitting the figures this way keeps an audit’s attention on the items where effort can change the result.
Exergy and cost
Exergy is not price. Electricity, gas and diesel are all nearly pure exergy, but their prices per kilowatt-hour differ widely and change over time, and their emissions differ too. Two options with the same exergy destruction can have very different running costs.
The two measures answer different questions, and an audit needs both. Exergy shows where physics leaves room for improvement and how much; cost shows which improvements pay. A large exergy loss that is cheap to fix is the best opportunity in any plant. A small exergy loss that is expensive to fix is usually not worth chasing, however elegant the fix. Chapter 10 combines the two for the running case, with illustrative tariffs.
The dead state decides the numbers
Every figure in this chapter depends on T₀. The physics does not change, but the reference does, and with it the exergy of every flow near ambient temperature.
| Dead-state temperature | Exergy of 60 °C water from 20 °C | Stream exergy, 0.50 kg/s | Heat pump at 26.1 kW | Gas boiler at 92.9 kW |
|---|---|---|---|---|
| 10 °C | 15.82 kJ/kg | 7.91 kW | 30.3% | 8.5% |
| 20 °C | 10.47 kJ/kg | 5.24 kW | 20.0% | 5.6% |
| 25 °C | 7.80 kJ/kg | 3.90 kW | 14.9% | 4.2% |
The ranking of options does not change with the dead state, but the absolute second-law efficiencies do, by a factor of two across a plausible range. Three rules keep an audit honest.
- State the dead state on every exergy table, with its reason, such as an annual mean ambient or the plant’s design condition.
- Use one dead state for every item compared in the same table.
- Repeat the key comparisons at a second dead state when seasonal conditions differ widely, as they do for an air-source heat pump.
Second-law efficiency, defined carefully
Chapter 1 defined second-law efficiency as product exergy divided by exergy supplied. Two versions are in use, and they give different numbers for the same device.
- Input–output form: all exergy leaving in products, divided by all exergy supplied. Simple, but generous to devices that pass most of their input straight through, such as a regulator, whose 77% above flatters it.
- Consumed–produced form: the exergy the device adds to its product, divided by the exergy it consumes from its fuel or drive. For the regulator this form makes no sense, because it produces nothing; for a heat pump or boiler the two forms agree.
The safest practice is to report exergy destroyed in kilowatts alongside any efficiency. Kilowatts add up across a plant, can be priced and cannot be flattered by the choice of formula.
The error: two dead states, two answers
This is an illustration. Two engineers assessed the wash-water heat pump independently, one in summer and one in winter, and their reports disagreed. The summer report gave the heat pump a second-law efficiency of 14.9% and called it “barely better than the boiler”; the winter report gave 30.3% and called it “excellent”. Management asked which was right.
Both were, for their own dead states. The summer report had used 25 °C, the outdoor temperature on the day of the site visit; the winter report had used 10 °C, the mains water temperature that morning. Neither stated the dead state, so the difference looked like a disagreement about the machine. Recalculated at the plant’s agreed reference of 20 °C, both gave 20.0%, and both reports then ranked the heat pump the same way against the boiler.
The plant now keeps one line at the top of every energy assessment template: “Dead state: 20 °C, 101.325 kPa, annual reference. Seasonal check at 10 °C and 35 °C where relevant.” It costs nothing and prevents a class of arguments that are not about engineering at all.
How this applies to a small Australian business
Exergy turns a second-law calculation into kilowatts, and kilowatts can be priced. A small plant does not need an exergy balance of every pipe, but a table like the one in this chapter for its five or six largest energy users is a powerful way to see where improvement is physically possible. It also helps in conversations with suppliers and financiers: a proposal that claims to reduce exergy destruction in an item that destroys little has little to offer, however large its energy figures.
Australia’s climate varies enough between regions and seasons that the dead state matters. A plant in a hot inland town and one on a cool coast will calculate different exergies for the same hot water, and both should say which they used.
Exercise: the effluent’s exergy
The plant’s effluent leaves at 0.50 kg/s and 45 °C. Using c = 4.18 kJ/kg·K and a dead state of 20 °C, find the effluent’s exergy per kilogram and in kilowatts. What fraction of its heat, measured relative to 20 °C, is exergy? How much of that exergy does the 3 K approach exchanger of Chapter 6 leave in the effluent?
Worked answer. Per kilogram, x = c × [(T − T₀) − T₀ × ln(T ÷ T₀)] = 4.18 × [25 − 293.15 × ln(318.15 ÷ 293.15)] = 4.18 × [25 − 23.991] = 4.22 kJ/kg, or 2.11 kW for the stream. Its heat relative to 20 °C is 0.50 × 4.18 × 25 = 52.25 kW, so the exergy is 4.0% of the heat.
After the 3 K approach exchanger the effluent leaves at 23 °C. Its exergy is then 0.50 × 4.18 × [3 − 293.15 × ln(296.15 ÷ 293.15)] = 0.50 × 4.18 × 0.01525 = 0.032 kW. The exchanger captures or destroys all but about 1.5% of the effluent’s exergy. Chapter 6 found that 0.43 kW of it was destroyed in the exchanger itself, so 2.11 − 0.43 − 0.03 = 1.64 kW went into the wash water, which checks against the water’s own exergy gain from 20 °C to 42 °C: 0.50 × 4.18 × [22 − 293.15 × ln(315.15 ÷ 293.15)] = 1.64 kW.