A basic steam cycle takes most of its heat in at temperatures far below its peak, and every improvement that has ever been made to one is an attempt to fix that.
One target
The previous lessons established the criterion. For a cycle rejecting heat at a constant , the thermal efficiency is exactly , where is the mean temperature at which heat is added. The baseline plant, boiling at MPa and superheating to degrees Celsius with a condenser at kPa, had K and .
There are only two levers. Lower , which means a colder condenser, and that is set by the river or the air and is already as low as the site allows. Or raise , which is entirely a design decision.
Three modifications do it, and the whole of this lesson is the same calculation performed four times. To isolate each effect the components are taken as isentropic throughout; real machines subtract about four points from every figure below, uniformly, as the previous lesson showed.
Superheating further
The easiest move is to keep the boiler pressure and take the steam hotter. At MPa the tables give kJ kg⁻¹ and kJ kg⁻¹ K⁻¹ at degrees Celsius, and with at .
Run the cycle at degrees Celsius. The pump work is unchanged at kJ kg⁻¹, giving kJ kg⁻¹, so kJ kg⁻¹. The exhaust quality is , so kJ kg⁻¹ and the turbine gives kJ kg⁻¹. The efficiency is .
Both defects improved together. The efficiency rose by nearly four points, and the exhaust moisture fell from to per cent, which is the more valuable of the two, because it is what limits blade life.
Check it against the criterion: K, and . Two hundred and fifty extra degrees of superheat moved the mean temperature by only K. Superheating adds heat at high temperature, which helps, but it also adds a great deal of entropy, which dilutes the gain. Superheat is a moisture control first and an efficiency measure second.
The ceiling is metallurgical. Around degrees Celsius, ferritic steels give out and austenitic or nickel alloys are needed, at several times the price. That is why almost every modern steam plant in the world sits between and degrees Celsius, and why raising it further is a materials research programme rather than a design choice.
Example. Run the same plant at MPa and degrees Celsius, condensing at kPa, with and . Find the efficiency, the exhaust quality and the mean temperature of heat addition.
The quality is , so kJ kg⁻¹ and kJ kg⁻¹. With kJ kg⁻¹, the efficiency is . The mean temperature is K, and confirms it.
Now you. Confirm the criterion for the degree case: given kJ kg⁻¹ and kJ kg⁻¹ K⁻¹, find and the efficiency it predicts.
Answer
kJ kg⁻¹ K⁻¹, so K and , matching the enthalpy calculation.
Raising the boiler pressure
The stronger lever is pressure, because it raises the temperature of the boiling itself, and boiling is where sixty per cent of the heat goes. At MPa water boils at degrees Celsius; at MPa it boils at .
Take the cycle to MPa and degrees Celsius, where kJ kg⁻¹ and kJ kg⁻¹ K⁻¹. The pump work rises to kJ kg⁻¹, so kJ kg⁻¹ and kJ kg⁻¹. The turbine expands to kPa at quality , giving and kJ kg⁻¹. The efficiency is , and has climbed to K.
Nearly six points from one change, which is why boiler pressures rose steadily through the twentieth century and why modern units run supercritical, above MPa, where there is no boiling at all and the fluid passes continuously from liquid-like to vapour-like.
But look at the exhaust. Quality means per cent moisture, worse than the original degree cycle and far outside what a turbine tolerates. The reason is geometric: raising the pressure at fixed temperature lowers the entropy at turbine inlet, so the vertical expansion line lands further to the left, deeper into the dome. Pressure and dryness pull against each other, and the whole point of the next modification is to have both.
Reheat
The fix is to expand in two stages with a return to the boiler in between. Steam leaves the high-pressure turbine at an intermediate pressure, goes back to a reheater in the boiler, is brought up to a high temperature again, and then expands through the low-pressure turbine to the condenser.
Take the MPa, degree cycle and reheat at MPa. The high-pressure turbine expands isentropically from , which at MPa lies between the tabulated degree entry (, ) and the degree one (, ). Interpolating, the fraction across is , so the steam leaves the high-pressure turbine at degrees Celsius with kJ kg⁻¹, having delivered kJ kg⁻¹.
Reheat at MPa to degrees Celsius, which takes kJ kg⁻¹ of extra heat and returns the steam to , . The low-pressure turbine expands to kPa at quality , exactly as the MPa superheated cycle did, giving and a further kJ kg⁻¹ of work.
Now total up. Turbine work is kJ kg⁻¹, pump work is , heat input is kJ kg⁻¹, and
Both problems solved at once. The efficiency is two points above the MPa cycle without reheat, and the exhaust moisture has fallen from to per cent. The mean temperature of heat addition is now K.
The efficiency gain looks modest for the plumbing involved, and the moisture gain is what actually justifies it: reheat is what makes high boiler pressure usable at all. Large plants reheat once as standard and some reheat twice, with diminishing returns each time.
Example. The same plant reheats at MPa but only to degrees Celsius, where and . Find the efficiency and the exhaust moisture.
The high-pressure turbine is unchanged, delivering kJ kg⁻¹ and leaving the steam at kJ kg⁻¹. The reheat now supplies kJ kg⁻¹, and the low-pressure turbine expands to quality , giving and work of kJ kg⁻¹. Total turbine work is kJ kg⁻¹ against a heat input of kJ kg⁻¹, so . The exhaust moisture is per cent, still outside the usual limit.
Now you. For that same partial-reheat cycle, find the heat rejected in the condenser and check the efficiency from it.
Answer
kJ kg⁻¹, so , agreeing with the work route.
Regeneration
Superheat and reheat both raise by adding heat at higher temperature. Regeneration does the opposite and better: it removes the lowest-temperature heat addition from the cycle entirely.
The complaint against the basic cycle was that a quarter of its heat went into warming feedwater from degrees Celsius up to saturation, all of it at temperatures where it does the mean no good. Suppose instead that warming is done by steam bled from the turbine partway through its expansion. That steam has already produced some work, and the heat it gives up to the feedwater never passes through the boiler at all. The boiler then sees water that is already hot, and every kilojoule it supplies goes in at a high temperature.
The simplest hardware is an open feedwater heater, a mixing chamber at the extraction pressure. Condensate from the condenser is pumped up to that pressure, bled steam is mixed directly into it, and the mixture leaves as saturated liquid at the heater pressure, to be pumped again to boiler pressure. Two pumps, one vessel, no tubes.
Let be the fraction of the turbine flow that is extracted. Per kilogram entering the turbine, the heater receives kilograms of steam at and kilograms of pumped condensate at , and delivers kilogram of saturated liquid at . The energy balance is , so
Example. Fit the MPa, degree cycle with one open feedwater heater at MPa, condensing at kPa, with isentropic machines. At MPa, kJ kg⁻¹ and m³ kg⁻¹. On the expansion line at , interpolation between the degree entry (, ) and the degree one (, ) gives the extraction state. Find the extraction fraction and the thermal efficiency.
The interpolation fraction is , so kJ kg⁻¹. The condensate pump raises water from kPa to MPa, costing kJ kg⁻¹ and giving kJ kg⁻¹. So
The feed pump then raises saturated liquid from MPa to MPa, costing kJ kg⁻¹, so the boiler receives water at kJ kg⁻¹ and supplies kJ kg⁻¹. Only the unextracted reaches the condenser, at kJ kg⁻¹, so kJ kg⁻¹, and
Three and a half points above the reheat-free cycle at the same pressure, for one vessel and one extra pump.
Now you. Move the heater to MPa, where kJ kg⁻¹ and m³ kg⁻¹. The extraction state there is wet, at kJ kg⁻¹. Find and the efficiency.
Answer
The condensate pump costs kJ kg⁻¹, so kJ kg⁻¹ and . The feed pump costs kJ kg⁻¹, so the boiler receives kJ kg⁻¹ and supplies . The condenser takes kJ kg⁻¹, giving . Almost identical: the optimum extraction pressure is broad, which is a mercy for anyone designing one.
What real plants actually do
The scheme above has one heater. Real stations have six to eight, at pressures spread between the condenser and the boiler, and the gain per heater falls off sharply after the first few, with the total worth about five points over an unregenerated cycle of the same conditions.
Most of those heaters are closed, not open: the bled steam condenses on the outside of tubes carrying the feedwater, so the two streams never mix and only one feed pump is needed for the whole train. Closed heaters are slightly less effective per stage, since the feedwater cannot be heated quite to the bled steam's saturation temperature, and they are much easier to plumb. Almost every plant uses one open heater, usually called the deaerator, because direct contact with steam is what strips dissolved oxygen out of the feedwater and stops the boiler corroding. Everything else in the train is closed.
Stacking all of this gives the modern ultra-supercritical unit: steam at MPa and degrees Celsius, double reheat, eight feedwater heaters, condensing at kPa, reaching about per cent net efficiency in service after auxiliary power and boiler losses. That is roughly the practical ceiling for a steam cycle, and it has been approached asymptotically for forty years.
The reason it is a ceiling is that every lever is now against a hard stop. The condenser sits at ambient. The peak temperature sits at what steel can hold. Regeneration has taken the low-temperature heat addition out. What remains is the boiler itself, where combustion gases at perhaps K hand their heat to steam at K, and no amount of cycle rearrangement touches that, because it happens outside the working fluid.
Getting at it means using the high temperature directly, with a working fluid that goes through the combustion rather than sitting behind a tube wall. That is the gas turbine, and it is the next lesson.