Combined cycles and cogeneration

A gas turbine throws away air at nearly $800$ K, which is a better heat source than most power stations are ever offered.

Two cycles with complementary appetites

The previous two lessons produced a matched pair of complaints. The steam cycle cannot use high temperatures: metallurgy caps it near $600$ degrees Celsius, and its heat arrives from combustion gases at $1800$ K across a tube wall, which wastes most of the temperature difference. The gas turbine has the opposite problem. It uses the high temperatures beautifully and then dumps its exhaust at $790$ K, because expanding further would take the pressure below atmospheric.

Put them in series. The gas turbine runs from $1600$ K down to $790$ K. Its exhaust becomes the heat source for a steam cycle running from about $850$ K down to ambient. Between them the two cycles cover the whole temperature range from flame to river, and neither is asked to work outside the range it is good at.

The arrangement is called a combined cycle, the gas turbine is the topping cycle, the steam plant is the bottoming cycle, and the heat exchanger between them is a heat recovery steam generator, universally abbreviated to HRSG. It is an unfired boiler: the exhaust gas passes over tube banks that economise, evaporate and superheat the water, and no additional fuel is burned in most designs.

Efficiency of cycles in series

The arithmetic of stacking cycles is worth doing in symbols first. Let the topping cycle have efficiency $η_{g}$ , so per unit of fuel heat it produces $η_{g}$ of work and rejects $1 - η_{g}$ . Let a fraction $f$ of that rejected heat be captured by the HRSG, the rest going up the stack. The bottoming cycle converts a fraction $η_{s}$ of what it receives. Then

η_{c o m b} = η_{g} + (1 - η_{g}) f η_{s}

The structure explains why the result is so much better than either part. Two mediocre efficiencies in series do not multiply, they add with a discount, because the bottoming cycle is paid for by heat that was going to be wasted anyway. A gas turbine at $0.431$ and a steam cycle at $0.38$ combine, with $f = 0.80$ , to $0.431 + 0.569 \times 0.80 \times 0.38 = 0.603$ , which is far above either.

Example. A gas turbine has $η_{g} = 0.431$ and $80$ per cent of its rejected heat reaches the bottoming cycle. What combined efficiency results if the steam plant achieves $η_{s} = 0.334$ , and what if it achieves $0.40$ ?

At $η_{s} = 0.334$ : $η_{c o m b} = 0.431 + 0.569 \times 0.80 \times 0.334 = 0.431 + 0.152 = 0.583$ . At $η_{s} = 0.40$ : $η_{c o m b} = 0.431 + 0.569 \times 0.80 \times 0.40 = 0.431 + 0.182 = 0.613$ . Six and a half points of bottoming-cycle efficiency bought three points on the plant, which is why the steam side of a combined cycle is designed with as much care as a standalone station.

Now you. A cheaper machine has $η_{g} = 0.36$ and recovers $f = 0.75$ of its rejected heat into a steam cycle of $η_{s} = 0.35$ . What is the combined efficiency?

Answer

$η_{c o m b} = 0.36 + 0.64 \times 0.75 \times 0.35 = 0.36 + 0.168 = 0.528$ .

A plant, end to end

Take the gas turbine from the previous lesson at a heavier duty: air in at $300$ K, pressure ratio $18$ , turbine inlet at $1600$ K, compressor efficiency $0.88$ , turbine efficiency $0.90$ , on the cold-air-standard assumption with $c_{p} = 1.005$ kJ kg⁻¹ K⁻¹ and $k = 1.400$ .

The isentropic compression gives $T_{2 s} = 300 \times 18^{0.2857} = 685.1$ K, so the compressor absorbs $1.005 \times 385.1 / 0.88 = 439.8$ kJ kg⁻¹ and delivers air at $T_{2} = 737.6$ K. The isentropic expansion gives $T_{4 s} = 1600 / 2.2838 = 700.6$ K, so the turbine delivers $0.90 \times 1.005 \times 899.4 = 813.5$ kJ kg⁻¹ and exhausts at $T_{4} = 790.5$ K. The heat input is $1.005 \times (1600 - 737.6) = 866.7$ kJ kg⁻¹ and the net work is $813.5 - 439.8 = 373.7$ kJ kg⁻¹, so $η_{g} = 0.431$ .

The HRSG cools that exhaust from $790.5$ K to a stack temperature of $400$ K, recovering $1.005 \times 390.5 = 392.5$ kJ kg⁻¹ of air. Since the cycle rejected $866.7 - 373.7 = 493.0$ kJ kg⁻¹ in total, the recovery fraction is $f = 392.5 / 493.0 = 0.796$ . The rest, $1.005 \times (400 - 298) = 102$ kJ kg⁻¹, goes up the stack, which is $11.8$ per cent of the fuel heat.

With a bottoming steam cycle at $η_{s} = 0.38$ , the steam plant produces $0.38 \times 392.5 = 149.1$ kJ per kilogram of air, and the plant total is $373.7 + 149.1 = 522.8$ kJ kg⁻¹ against $866.7$ of fuel, so

η_{c o m b} = 0.603

Example. Scale that plant to an air flow of $600$ kg s⁻¹. Find the gas turbine output, the HRSG duty, the steam turbine output, the fuel heat and the stack loss.

The gas turbine gives $600 \times 373.7 = 224$ MW. The fuel heat is $600 \times 866.7 = 520$ MW. The HRSG duty is $600 \times 392.5 = 235$ MW, of which the steam plant converts $0.38 \times 235 = 89$ MW. Total output is $224 + 89 = 314$ MW, and the stack carries away $600 \times 102 = 61$ MW. Checking the efficiency, $314 / 520 = 0.603$ as before.

Now you. For the same plant, how much heat does the steam cycle's own condenser reject, and where does the remaining fuel energy go?

Answer

The steam cycle receives $235$ MW and converts $89$ MW, so its condenser rejects $146$ MW. The fuel's $520$ MW splits into $314$ MW of work, $146$ MW into the cooling water and $61$ MW up the stack, which sums to $521$ MW within rounding.

Why the match is so good

The reason a combined cycle works is that the temperature ranges are complementary rather than overlapping. Nothing is being done twice. The gas turbine harvests the range no steam plant can reach, and the steam plant harvests the range a gas turbine cannot exploit because its exhaust must stay above atmospheric pressure.

Real plants reach $60$ to $64$ per cent net. The Bouchain plant in France, commissioned in $2016$ , was certified at $62.2$ per cent, and the current generation of large machines is rated slightly above that. Those figures are the highest thermal efficiencies ever achieved by a heat engine of any kind, and they were reached by combination rather than by any single breakthrough.

Nothing about that required a new principle. Both cycles were mature by $1960$ , and what changed was the gas turbine's turbine inlet temperature, which pushed its exhaust hot enough to raise good steam. A combined cycle is the cheapest efficiency ever bought, because the bottoming plant is paid for with heat that was already on its way up a stack.

The pinch point

The strain is in the HRSG, and it has a specific name. Gas cools along a straight line as it gives up heat, since $c_{p}$ is nearly constant. Water does not: it warms, then boils at constant temperature absorbing enormous latent heat, then superheats. Plot both against heat transferred and the two curves approach each other most closely at the point where boiling begins, called the pinch point. The pinch is what limits how much heat can be recovered: push the steam pressure up, to make the bottoming cycle more efficient, and the boiling temperature rises, the pinch closes, and less heat can be extracted before the gas is colder than the water it is trying to heat.

This is a genuine trade-off between $f$ and $η_{s}$ in the formula above, and its resolution is the defining feature of combined-cycle design. The standard answer is to generate steam at two or three different pressures in the same HRSG, each with its own pinch, so that the composite water curve follows the gas curve more closely. A modern triple-pressure reheat HRSG stacks high-, intermediate- and low-pressure circuits and gets the stack down to about $360$ K.

There is a floor under the stack temperature that is chemical, not thermodynamic. Sulphur in the fuel makes sulphuric acid in the flue gas, which condenses on cold surfaces and destroys them, so the stack must stay above the acid dew point. For natural gas with negligible sulphur that is around $360$ K; for fuel oil it is much higher, and the recoverable heat is correspondingly less.

Cogeneration

Combined cycles chase electricity. There is another way to use rejected heat, which is to sell it.

A condensing steam plant rejects at $45.81$ degrees Celsius, a temperature at which the heat is worthless: nothing useful can be done with it and it goes into a river. But a plant does not have to condense at $45.81$ degrees Celsius. Raise the turbine exhaust pressure and the condensation temperature rises with it, until the rejected heat is hot enough to heat buildings, dry timber or run a process. Such a plant is a back-pressure or cogeneration unit, and its condenser becomes a customer's heat supply.

The trade is direct and severe.

Example. Take the $3$ MPa, $350$ degree cycle with isentropic machines, but exhaust at $200$ kPa, where $T_{s a t} = 120.21$ degrees Celsius, $h_{f} = 504.71$ , $h_{f g} = 2201.6$ kJ kg⁻¹, $s_{f} = 1.5302$ , $s_{f g} = 5.5968$ kJ kg⁻¹ K⁻¹ and $v_{f} = 0.001061$ m³ kg⁻¹. Find the electrical efficiency and the heat delivered.

The pump work is $0.001061 \times 2800 = 2.97$ kJ kg⁻¹, so $h_{2} = 507.68$ and $q_{i n} = 3116.1 - 507.68 = 2608.4$ kJ kg⁻¹. The exhaust quality is $(6.7450 - 1.5302) / 5.5968 = 0.9317$ , giving $h_{4} = 504.71 + 0.9317 \times 2201.6 = 2556.0$ kJ kg⁻¹. The turbine delivers $560.1$ kJ kg⁻¹ and the net work is $557.1$ , so the electrical efficiency is $557.1 / 2608.4 = 0.214$ . The heat delivered to the customer is $2556.0 - 504.71 = 2051.3$ kJ kg⁻¹, at $120$ degrees Celsius, which is hot enough for district heating.

Now you. The same plant with a back pressure of $500$ kPa, where $T_{s a t} = 151.83$ degrees Celsius, $h_{f} = 640.09$ , $h_{f g} = 2108.0$ , $s_{f} = 1.8604$ , $s_{f g} = 4.9603$ and $v_{f} = 0.001093$ . Find the electrical efficiency and the heat delivered.

Answer

$w_{p} = 0.001093 \times 2500 = 2.73$ kJ kg⁻¹, so $h_{2} = 642.82$ and $q_{i n} = 2473.3$ kJ kg⁻¹. The quality is $(6.7450 - 1.8604) / 4.9603 = 0.9847$ , so $h_{4} = 640.09 + 0.9847 \times 2108.0 = 2715.9$ kJ kg⁻¹, the turbine gives $400.2$ and the net work is $397.4$ kJ kg⁻¹. The electrical efficiency is $0.161$ and the heat delivered is $2075.8$ kJ kg⁻¹ at $152$ degrees Celsius. Hotter heat costs more electricity.

Electrical efficiency fell from $0.334$ to $0.214$ , and in exchange the plant now delivers $2051$ kJ kg⁻¹ of usable heat instead of dumping $1944$ kJ kg⁻¹ into a river. The utilisation factor, work plus useful heat over heat supplied, is close to $1$ for an ideal back-pressure unit and $0.80$ to $0.90$ in service once distribution losses are counted, against $0.33$ for the condensing plant.

Whether that trade is worth making is not a thermodynamic question. It depends on whether there is a customer for low-grade heat within pipe distance all year round, which is why cogeneration is common in Scandinavian cities and in chemical works and rare in temperate suburbs. It is also why the honest way to report a cogeneration plant is two numbers and not one: quoting the utilisation factor alone conceals that most of the output is heat, which is worth perhaps a fifth of electricity per joule.

That last point is the one this course has been circling. Comparing a plant that makes $0.33$ of electricity with one that makes $0.21$ of electricity and $0.79$ of hot water requires a way of pricing energy by quality and not merely by quantity. Refrigeration, in the next lesson, will make the same demand from the other direction, and the answer to both arrives at the end.