Cycles and the mean temperature

A turbine on its own is not a power station, because after one pass the working fluid has been used up and thrown away.

Closing the loop

Each device analysed so far takes a fluid from one state to another and stops. Expand steam through a turbine and you have low-pressure wet steam and a shaft turning; expand it again and nothing happens, because it is already at the exhaust pressure. To keep producing work indefinitely the fluid has to be brought back to its starting state and sent round again, and the sequence of processes that does that is a cycle.

That is not merely a convenience about not wasting water. A cycle is what makes continuous power possible from a finite quantity of working fluid, and closing the loop imposes conditions of its own that shape every real plant. The fluid leaves the turbine at low pressure, and something has to raise it again. Raising a vapour's pressure is ruinously expensive, as the reversible work integral showed, so the vapour is condensed first and a pump does the job instead. That single decision determines the layout of every steam plant on earth.

For a cycle, the working fluid returns to its initial state, so every property returns with it: $\oint d u = 0$ , $\oint d h = 0$ , $\oint d s = 0$ . Applying the energy balance around the whole loop, with the enthalpy change summing to zero,

w_{n e t} = q_{i n} - q_{o u t}

The net work equals the net heat, per kilogram of fluid circulated. Nothing is stored anywhere.

Two numbers that describe any cycle

The first is the thermal efficiency, work delivered over heat paid for:

η_{t h} = \frac{w_{n e t}}{q_{i n}} = 1 - \frac{q_{o u t}}{q_{i n}}

The second is the back work ratio, the fraction of the gross work output that the cycle has to feed back into its own compression:

bwr = \frac{w_{c o m p r e s s i o n}}{w_{e x p a n s i o n}}

Efficiency gets all the attention and back work ratio decides whether a machine is buildable. A cycle with a back work ratio of $0.9$ delivers a tenth of its gross work to the outside world, so a two per cent slip in either machine's efficiency wipes out a fifth of the output. It is also why gas turbines could not be built until compressor design improved: the concept is Victorian, the hardware is not.

Example. A steam plant supplies $2921$ kJ of heat and rejects $1944$ kJ per kilogram of steam. Its turbine delivers $980.0$ kJ kg⁻¹ and its feed pump absorbs $3.02$ kJ kg⁻¹. Find the net work, the thermal efficiency and the back work ratio.

The net work is $2921 - 1944 = 977$ kJ kg⁻¹, which also equals $980.0 - 3.02 = 977.0$ kJ kg⁻¹, and the two routes agreeing is the check that the cycle balances. The efficiency is $977 / 2921 = 0.334$ . The back work ratio is $3.02 / 980.0 = 0.0031$ , three parts in a thousand, which is the enormous structural advantage of pumping a liquid.

Now you. A gas turbine's compressor absorbs $311.8$ kJ kg⁻¹ and its turbine delivers $715.3$ kJ kg⁻¹, with $793.8$ kJ kg⁻¹ of heat supplied. Find the net work, the efficiency and the back work ratio.

Answer

$w_{n e t} = 715.3 - 311.8 = 403.5$ kJ kg⁻¹, $η_{t h} = 403.5 / 793.8 = 0.508$ , and the back work ratio is $311.8 / 715.3 = 0.436$ . The efficiency is far better than the steam plant's and the back work ratio is a hundred and forty times worse.

The bound, and the size of the gap

The previous course established the Carnot limit: no cycle exchanging heat with reservoirs at $T_{H}$ and $T_{L}$ can beat $1 - T_{L} / T_{H}$ . A steam plant with metal at $600$ degrees Celsius rejecting to a condenser at $45.81$ degrees Celsius has a bound of $1 - 318.96 / 873.15 = 0.635$ . Real plants of that description reach about $0.45$ .

The gap is not one thing. Some of it is component inefficiency, the turbines and pumps of the previous lesson falling short of isentropic. Some is heat transfer across finite temperature differences, in the boiler, in the condenser and in every feedwater heater. But a large part of it is neither, and it survives even in a cycle whose every component is perfect. Understanding that part requires a sharper statement of the limit than "Carnot".

The mean temperature of heat addition

Here is the sharpening, and it is exact rather than approximate.

Consider any cycle whose heat rejection happens at a single constant temperature $T_{L}$ . That is not a contrived case: a steam plant condenses at constant pressure inside the saturation dome, and constant pressure inside the dome means constant temperature. Over the rejection process, $q_{o u t} = \int T d s = T_{L} Δ s$ , where $Δ s$ is the entropy change of the fluid across the condenser.

Heat addition is not at a single temperature: the water enters the boiler cold, warms, boils, and superheats, so $T$ climbs throughout. But the integral is still an integral, so define a mean temperature of heat addition by

{\overline{T}}_{H} = \frac{q_{i n}}{Δ s}

which is nothing more than the average of $T$ weighted by entropy change. Since the cycle is closed, the entropy change across the boiler equals the entropy change across the condenser in magnitude, and the same $Δ s$ appears in both expressions. Divide:

η_{t h} = 1 - \frac{q_{o u t}}{q_{i n}} = 1 - \frac{T_{L} Δ s}{{\overline{T}}_{H} Δ s} = 1 - \frac{T_{L}}{{\overline{T}}_{H}}

A cycle with reversible components has exactly Carnot's efficiency, evaluated not at its peak temperature but at the mean temperature at which it actually takes heat in. That is the sharpening. The Carnot bound compares against the hottest metal in the plant, which is irrelevant if only a tenth of the heat goes in there.

Test it on real numbers. A steam cycle taking water from saturated liquid at $10$ kPa to steam at $3$ MPa and $350$ degrees Celsius has $q_{i n} = 2921$ kJ kg⁻¹ and $Δ s = 6.7450 - 0.6492 = 6.0958$ kJ kg⁻¹ K⁻¹. So ${\overline{T}}_{H} = 2921 / 6.0958 = 479.2$ K, which is $206$ degrees Celsius, and $η_{t h} = 1 - 318.96 / 479.2 = 0.334$ . That is precisely the efficiency computed above from enthalpies, by a route that never mentioned entropy.

The lesson in the number is that $479$ K is a long way below the $623$ K of the superheated steam. The cycle spends a great deal of its heat input warming subcooled water from $46$ to $234$ degrees Celsius, and that heat goes in at low temperature and drags the mean down. Every improvement in the next lessons is an attack on this one number.

Example. A cycle takes heat at a mean temperature of $479$ K and rejects at $319$ K. What is the best efficiency it can have, and how does raising the superheat to $600$ degrees Celsius, which lifts $q_{i n}$ to $3488$ kJ kg⁻¹ and $Δ s$ to $6.8593$ kJ kg⁻¹ K⁻¹, change it?

The first is $1 - 319 / 479 = 0.334$ . For the second, ${\overline{T}}_{H} = 3488 / 6.8593 = 508.5$ K, so $η_{t h} = 1 - 318.96 / 508.5 = 0.373$ . Two hundred and fifty degrees of extra superheat moved the mean temperature by only $29$ K, because superheating adds heat at high temperature but also adds a lot of entropy.

Now you. Raising the boiler pressure to $15$ MPa at $600$ degrees Celsius gives $q_{i n} = 3376$ kJ kg⁻¹ and $Δ s = 6.0304$ kJ kg⁻¹ K⁻¹, still rejecting at $319$ K. What are the mean temperature and the efficiency?

Answer

${\overline{T}}_{H} = 3376 / 6.0304 = 560$ K, so $η_{t h} = 1 - 318.96 / 560 = 0.430$ . Pressure is a more effective lever than superheat, because raising the pressure raises the temperature at which the boiling itself happens.

Why nobody builds a Carnot cycle

If the Carnot cycle is the best possible, an obvious question is why every power station is not one. The answer is entirely practical, and it is worth going through because each objection explains a feature of the cycle that gets built instead.

Take the Carnot cycle executed with steam entirely inside the saturation dome, where isothermal heat transfer is automatic: boil at $T_{H}$ , expand isentropically, condense partially at $T_{L}$ , compress isentropically back to saturated liquid. Four objections, in order of severity.

The heat addition is isothermal, so it must happen inside the dome, which caps the top temperature at water's critical temperature of $373.95$ degrees Celsius. Even at that cap the bound is only $0.51$ , and worse, the latent heat vanishes as the critical point is approached, so a cycle running near it circulates enormous quantities of water for very little heat. The metallurgical limit of a modern boiler is well above $600$ degrees Celsius, and a cycle that cannot use it is throwing away its best asset.

The isentropic expansion runs from saturated vapour down to $T_{L}$ , ending deep in the wet region. Boiling at $3$ MPa, where $s_{g} = 6.1856$ kJ kg⁻¹ K⁻¹, the exhaust quality at $10$ kPa is $(6.1856 - 0.6492) / 7.4996 = 0.738$ , and pushing the boiling pressure to $15$ MPa to chase efficiency makes it $0.622$ . A third of the mass arriving at the last blade row as liquid is not a thermodynamic inconvenience, it is destruction: droplets travelling at hundreds of metres per second erode blade leading edges, and practice limits the exhaust moisture to about $10$ per cent.

The compression stage is worse. It takes a two-phase mixture and compresses it isentropically back to saturated liquid, and no machine exists that handles a mixture of liquid and vapour at high pressure ratio. Pumps cavitate on vapour and compressors are destroyed by liquid.

Finally, the partial condensation must be stopped at exactly the quality that makes the subsequent isentropic compression land on the saturated liquid line. Controlling a condenser to a precise intermediate quality is not something a plant operator can do.

The gas version fares no better. A Carnot cycle in a gas needs isothermal compression and isothermal expansion, meaning heat transfer at a vanishing temperature difference and therefore infinite heat exchanger area, and the isentropic legs between $300$ K and $1400$ K alone demand a pressure ratio of $(1400 / 300)^{3.5} = 220$ . The enclosed area on the pressure-volume diagram is small compared with the swept volume, so the machine is enormous for the power it makes.

The repair

Every objection points the same way, and the fixes are all corrections to the Carnot cycle rather than replacements for it.

Condense the vapour completely to saturated liquid rather than partially, which removes the impossible two-phase compressor and replaces it with a pump. That costs a little, because heat must then be added to warm subcooled liquid at low temperature, dragging the mean temperature down. It is worth it a hundred times over: the back work ratio falls from a substantial number to $0.003$ .

Add heat at constant pressure rather than constant temperature, so that the cycle may leave the dome and superheat as far as the metallurgy allows. That gives up the exact isothermal ideal in exchange for a much higher peak temperature, and moves the exhaust away from the wet region.

What is left after those two changes is the Rankine cycle: pump, boiler, turbine, condenser. It is not the most efficient cycle imaginable and it is the one that can be built, and it generates the large majority of the world's electricity, whether the heat comes from coal, gas, uranium or concentrated sunlight. Analysing one completely, with real steam properties and real component efficiencies, is the next lesson.