Exergy

A power station rejects seventy per cent of its heat through the condenser, which is why almost everyone who looks at one for the first time concludes that the condenser is where the problem is.

The question the scores cannot answer

This course has produced a set of numbers that cannot be compared with each other. A steam plant at $0.29$ thermal efficiency. A gas turbine at $0.43$ . A cogeneration unit at $0.21$ electrical and a utilisation factor near $1$ . A heat pump at $4.29$ . A turbine at $0.87$ isentropic efficiency and a compressor at $0.80$ .

Worse, none of them locates a loss. The First Law says energy is conserved, so it can only ever report where energy went, and energy goes into the cooling water in enormous quantities whether the plant is good or bad. The Second Law says entropy is generated, which does locate irreversibility, but in units of joules per kelvin that no accountant recognises.

What is needed is a single currency: the amount of useful work a given quantity of energy could still deliver. Energy at $1800$ K is worth a great deal, the same energy at $46$ degrees Celsius is worth almost nothing, and a measure that prices the difference makes every loss in a plant comparable, and comparable with the output as well.

The dead state

Work cannot be extracted from a system in equilibrium with its surroundings, so the surroundings define the zero of the scale. The dead state is the state a system reaches when it has come to thermal and mechanical equilibrium with its environment: temperature $T_{0}$ , pressure $p_{0}$ , zero velocity, zero elevation relative to the reference. Ordinary values are $T_{0} = 298.15$ K and $p_{0} = 100$ kPa, and the choice matters, so it must be stated with any exergy figure.

Exergy, also called availability, is the maximum useful work obtainable as a system is brought reversibly to the dead state, interacting only with the environment. It is not a property of the system alone: it depends on the environment too, which is what makes it a measure of quality rather than of quantity.

For heat, the maximum work extractable from a quantity $q$ supplied at temperature $T$ is what a Carnot engine between $T$ and $T_{0}$ would give:

{ex}_{h e a t} = (1 - \frac{T_{0}}{T}) q

For a flowing stream, the derivation runs through a reversible device taking the stream from $(h, s)$ to $(h_{0}, s_{0})$ while exchanging heat only with the environment, and gives the flow exergy

ψ = (h - h_{0}) - T_{0} (s - s_{0})

ignoring velocity and elevation. Work is pure exergy: a joule of shaft work is a joule of work potential, no discount.

Example. A boiler supplies $2921$ kJ per kilogram of steam. How much of it is work potential if it comes from combustion gases at $1600$ K, and how much if the same quantity arrives from a geothermal source at $500$ K? Take $T_{0} = 298.15$ K.

At $1600$ K the factor is $1 - 298.15 / 1600 = 0.8137$ , so the exergy is $0.8137 \times 2921 = 2377$ kJ kg⁻¹. At $500$ K it is $1 - 298.15 / 500 = 0.4037$ , giving $1179$ kJ kg⁻¹. The same energy, less than half the work potential, which is exactly why geothermal and solar-thermal plants have low thermal efficiencies without being badly designed.

Now you. The condenser rejects $2072$ kJ kg⁻¹ at $319$ K. What is its work potential?

Answer

$1 - 298.15 / 319 = 0.0654$ , so the exergy is $0.0654 \times 2072 = 135$ kJ kg⁻¹, six and a half per cent of the energy. Nearly all of that huge heat rejection is worthless.

Exergy destroyed

Exergy, unlike energy, is not conserved. Every irreversible process destroys some, and the amount is given by a result already met in the preceding Thermodynamics course:

{ex}_{d e s t} = T_{0} S_{g e n}

the Gouy-Stodola theorem. It converts entropy generation, in joules per kelvin, into lost work potential, in joules, at the exchange rate set by the environment's temperature. Every irreversibility identified anywhere in this course now has a price: a turbine falling short of isentropic, a throttling valve, a heat exchanger with a temperature difference across it, friction in a pipe.

The R-134a expansion valve of the previous lesson generated $0.0174$ kJ kg⁻¹ K⁻¹, so at $T_{0} = 298.15$ K it destroys $5.19$ kJ kg⁻¹ of work potential, against the $4.41$ kJ kg⁻¹ a perfect turbine would have recovered there. The two figures are close because they are describing the same loss from two directions.

Example. The turbine at $3$ MPa and $350$ degrees Celsius, exhausting to $10$ kPa with $η_{T} = 0.87$ , enters at $s_{1} = 6.7450$ and leaves at $s_{2} = 7.1444$ kJ kg⁻¹ K⁻¹ having delivered $852.6$ kJ kg⁻¹. Find the exergy destroyed and the second-law efficiency.

The destruction is $T_{0} (s_{2} - s_{1}) = 298.15 \times 0.3994 = 119.1$ kJ kg⁻¹. A reversible device between the same two end states would have produced $852.6 + 119.1 = 971.6$ kJ kg⁻¹, which is also the drop in flow exergy $ψ_{1} - ψ_{2}$ across the machine. The second-law efficiency is what was produced over what was available:

η_{I I} = \frac{852.6}{971.6} = 0.877

Now you. A worn turbine on the same duty has $η_{T} = 0.80$ , delivering $784.0$ kJ kg⁻¹ and exhausting at $s_{2} = 7.3595$ kJ kg⁻¹ K⁻¹. Find its exergy destroyed and second-law efficiency.

Answer

${ex}_{d e s t} = 298.15 \times (7.3595 - 6.7450) = 183.2$ kJ kg⁻¹. The exergy available was $784.0 + 183.2 = 967.2$ kJ kg⁻¹, so $η_{I I} = 784.0 / 967.2 = 0.811$ .

Note that the isentropic efficiency and the second-law efficiency are close but not equal: $0.87$ against $0.877$ , and $0.80$ against $0.811$ . They differ because the isentropic benchmark compares against a different exit state while the second-law efficiency compares against the actual exit state, and the difference is exactly the reheat effect described earlier.

Auditing a whole plant

Now do the thing that justifies the whole apparatus. Take the real Rankine plant analysed earlier: boiler at $3$ MPa and $350$ degrees Celsius, condenser at $10$ kPa, turbine of isentropic efficiency $0.87$ , pump taken as ideal, heat supplied from combustion gases treated as a reservoir at $1600$ K, surroundings at $298.15$ K.

The cycle numbers are already known: $q_{i n} = 2921.3$ , $q_{o u t} = 2071.7$ , $w_{n e t} = 849.5$ kJ kg⁻¹, thermal efficiency $0.291$ . The exergy supplied with the heat is $0.8137 \times 2921.3 = 2376.9$ kJ kg⁻¹. Now charge every component.

The boiler takes heat at $1600$ K and delivers it to water whose entropy rises from $0.6492$ to $6.7450$ kJ kg⁻¹ K⁻¹. The entropy generated by that transfer is $6.7450 - 0.6492 - 2921.3 / 1600 = 4.270$ kJ kg⁻¹ K⁻¹, so it destroys $298.15 \times 4.270 = 1273.1$ kJ kg⁻¹.

The turbine destroys $119.1$ kJ kg⁻¹, computed above.

The condenser takes steam at $s = 7.1444$ and returns saturated liquid at $s = 0.6492$ , dumping $2071.7$ kJ kg⁻¹ into surroundings at $298.15$ K. Its entropy generation is $0.6492 - 7.1444 + 2071.7 / 298.15 = 0.4534$ kJ kg⁻¹ K⁻¹, so it destroys $135.2$ kJ kg⁻¹. The same number comes from the flow exergies: the steam arrives carrying $ψ = 138.1$ kJ kg⁻¹ and leaves carrying $2.9$ , and the difference goes nowhere useful.

The pump, taken as ideal, destroys nothing.

Destination	kJ kg⁻¹	Share of exergy supplied
Net work out	$849.5$	$35.7$ %
Destroyed in the boiler	$1273.1$	$53.6$ %
Destroyed in the condenser	$135.2$	$5.7$ %
Destroyed in the turbine	$119.1$	$5.0$ %
Total	$2376.9$	$100$ %

The audit closes exactly, which it must: exergy in equals work out plus exergy destroyed. The plant's second-law efficiency is $849.5 / 2376.9 = 0.357$ , meaning it captures $36$ per cent of the work potential it was handed.

What the audit changes

Set the exergy column against the energy column and the two tell opposite stories.

The condenser carries away $70.9$ per cent of the energy and destroys $5.7$ per cent of the exergy. It is the largest energy flow in the plant by a wide margin and very nearly the smallest loss. Every intuition that says to insulate it, recover from it, or make it smaller is chasing a rounding error, because the heat it rejects is at $46$ degrees Celsius and is worth almost nothing.

The boiler loses nothing at all on an energy basis, if it is well lagged, and destroys $53.6$ per cent of the exergy. Combustion gases at $1600$ K are handing their heat to water that is between $46$ and $350$ degrees Celsius, and the temperature difference across that transfer is where a majority of the plant's work potential disappears, silently, in a component the First Law reports as perfect.

That single result restates everything in the second half of this course. Superheat, reheat and regeneration all raise the mean temperature of heat addition, which narrows the gap across the boiler and reduces exactly this destruction. The combined cycle is the same idea taken to its conclusion: it puts a gas turbine in the temperature range where the steam cycle was losing everything, so the heat is used at $1600$ K instead of being degraded to $600$ K before anything is done with it. Both of those were justified earlier by the mean-temperature argument. Exergy explains why that argument was the right one.

Example. For the plant above, express the condenser's loss both ways and say which figure should drive a decision to spend money.

On energy: $2071.7 / 2921.3 = 70.9$ per cent of the heat supplied leaves through the condenser. On exergy: $135.2 / 2376.9 = 5.7$ per cent of the work potential is destroyed there. The exergy figure is the one that matters, because it is the one denominated in work that could have been produced and was not.

Now you. The same plant is fitted with a better turbine, $η_{T} = 0.90$ , which reduces the turbine's exergy destruction to about $92$ kJ kg⁻¹. Roughly how much extra net work does that buy per kilogram, and what does it do to the boiler's share?

Answer

The turbine's destruction falls by about $27$ kJ kg⁻¹, and since exergy in is unchanged that appears almost entirely as extra work: net work rises from $850$ to $879$ kJ kg⁻¹. The boiler destroys the same $1273$ kJ kg⁻¹ and its share is essentially unmoved, which is the point: improving the turbine does not touch the plant's dominant loss.

Second-law efficiency as a common currency

The other use of exergy is to make incomparable machines comparable. Define, generally,

η_{I I} = \frac{exergy recovered}{exergy supplied}

and every device in this course gets a score on the same scale.

A refrigerator with ${COP}_{R} = 3.29$ , moving heat from $- 10$ to $25$ degrees Celsius where the reversible coefficient is $7.52$ , has $η_{I I} = 3.29 / 7.52 = 0.44$ . The same machine judged as a heat pump, with ${COP}_{H P} = 4.29$ against a reversible $8.52$ , scores $0.50$ , because the work it consumes is itself delivered to the warm space and counts as recovered. The Rankine plant above scores $0.357$ . A domestic gas boiler burning methane to make hot water at $60$ degrees Celsius scores about $0.09$ , despite a first-law efficiency above $0.90$ , because it is destroying a fuel worth $1900$ K to produce heat worth $333$ K. That last figure is the one that changed the argument about how to heat buildings, and no first-law analysis can produce it.

Where exergy stops

Three honest limits.

Exergy is relative to a chosen dead state, and quoting a figure without stating $T_{0}$ is meaningless. A plant in Siberia and the same plant in Kuwait have different exergy inputs from the same fuel. Worse, some processes are analysed against a dead state that varies during the day, which makes seasonal comparisons delicate.

The treatment above counts only thermal and mechanical exergy. A fuel that has not yet been burned also carries chemical exergy, the work obtainable by reacting it reversibly with the environment, and a full plant audit must include it. For methane the chemical exergy is about $1.04$ times the lower heating value, so treating the fuel's heating value as its exergy is a good approximation and not an exact one. For hydrogen the ratio is $0.83$ , and for carbon monoxide $1.07$ , so the approximation cannot be assumed.

Finally, exergy is a thermodynamic price and not an economic one. It says the domestic gas boiler is a poor machine, and it does not say that the heat pump replacing it costs six times as much to install and needs a different distribution system. Exergy tells you where the thermodynamic loss is, which is the necessary first step and never the whole argument. Combining it with capital cost is a discipline of its own, called thermoeconomics, and it starts exactly where this course ends.

Between them, the twelve lessons here have taken the laws established in the preceding course and pointed them at hardware: a boundary drawn in space, an energy equation for streams, six devices, an entropy benchmark, and then the cycles that generate and refrigerate the world, ending with the accounting that says which part of any of them to fix first.