The Rankine cycle

Most of the world's electricity is produced by boiling water, expanding the steam through a turbine, condensing it and pumping it back, and the cycle that describes this has four processes and no spare parts.

Four devices, four processes

The layout comes straight out of the repairs made to the Carnot cycle in the previous lesson. Name the states going round the loop and each device is one of the steady-flow machines already analysed.

State $1$ is saturated liquid leaving the condenser, at the lowest pressure in the plant. The feed pump raises it to boiler pressure, ideally isentropically, reaching state $2$ : compressed liquid, still cold. The boiler adds heat at constant pressure, warming the water to its saturation temperature, boiling it, and superheating the vapour to state $3$ . The turbine expands it, ideally isentropically, to the condenser pressure at state $4$ . The condenser rejects heat at constant pressure and temperature, returning the fluid to saturated liquid at state $1$ .

The ideal Rankine cycle is this loop with an isentropic pump, an isentropic turbine, and no pressure drop anywhere. It is not the Carnot cycle and does not pretend to be. Heat is added at constant pressure rather than constant temperature, so the process is externally irreversible against a hot source, and the cycle's efficiency is Carnot's evaluated at the mean temperature of heat addition rather than the peak.

Each device is analysed with the steady-flow energy equation and no velocity or height terms:

w_{p u m p} = h_{2} - h_{1} = v_{1} (p_{2} - p_{1}), q_{i n} = h_{3} - h_{2}, w_{t u r b} = h_{3} - h_{4}, q_{o u t} = h_{4} - h_{1}

The pump form uses the incompressibility of liquid water, which is accurate to better than a per cent and avoids needing a compressed-liquid table.

A complete ideal cycle

Take a plant with boiler conditions of $3$ MPa and $350$ degrees Celsius and a condenser at $10$ kPa. The saturation table at $10$ kPa gives $T_{s a t} = 45.81$ degrees Celsius, $v_{f} = 0.001010$ m³ kg⁻¹, $h_{f} = 191.81$ kJ kg⁻¹, $h_{f g} = 2392.1$ kJ kg⁻¹, $s_{f} = 0.6492$ and $s_{f g} = 7.4996$ kJ kg⁻¹ K⁻¹. The superheated table at $3$ MPa and $350$ degrees Celsius gives $h_{3} = 3116.1$ kJ kg⁻¹ and $s_{3} = 6.7450$ kJ kg⁻¹ K⁻¹.

Work round the loop. At state $1$ , $h_{1} = 191.81$ kJ kg⁻¹. The pump work is $v_{1} (p_{2} - p_{1}) = 0.001010 \times (3000 - 10) = 3.02$ kJ kg⁻¹, so $h_{2} = 194.83$ kJ kg⁻¹. The boiler supplies $q_{i n} = 3116.1 - 194.83 = 2921.3$ kJ kg⁻¹.

The turbine is isentropic, so $s_{4} = s_{3} = 6.7450$ kJ kg⁻¹ K⁻¹. At $10$ kPa that entropy is below $s_{g} = 8.1488$ , so the exhaust is wet, with quality

x_{4} = \frac{6.7450 - 0.6492}{7.4996} = 0.8128

and $h_{4} = 191.81 + 0.8128 \times 2392.1 = 2136.2$ kJ kg⁻¹. The turbine therefore delivers $w_{t u r b} = 3116.1 - 2136.2 = 980.0$ kJ kg⁻¹, and the condenser rejects $q_{o u t} = 2136.2 - 191.81 = 1944.3$ kJ kg⁻¹.

Net work is $980.0 - 3.02 = 977.0$ kJ kg⁻¹, which must and does equal $q_{i n} - q_{o u t} = 2921.3 - 1944.3$ . The thermal efficiency is

η_{t h} = \frac{977.0}{2921.3} = 0.334

and the back work ratio is $3.02 / 980.0 = 0.0031$ .

Example. The condenser of that plant is fouled and its pressure rises from $10$ kPa to $20$ kPa, where $v_{f} = 0.001017$ , $h_{f} = 251.42$ , $h_{f g} = 2357.5$ , $s_{f} = 0.8320$ and $s_{f g} = 7.0752$ . Boiler conditions are unchanged. What happens to the efficiency?

The pump work becomes $0.001017 \times 2980 = 3.03$ kJ kg⁻¹ and $h_{2} = 254.45$ kJ kg⁻¹, so $q_{i n} = 3116.1 - 254.45 = 2861.7$ kJ kg⁻¹. The exhaust quality is $(6.7450 - 0.8320) / 7.0752 = 0.8357$ , giving $h_{4} = 251.42 + 0.8357 \times 2357.5 = 2221.7$ kJ kg⁻¹ and turbine work $894.4$ kJ kg⁻¹. Net work is $891.4$ kJ kg⁻¹ and $η_{t h} = 891.4 / 2861.7 = 0.311$ .

Ten kilopascals of condenser pressure, a change of one tenth of an atmosphere at the coldest point in the plant, has cost $2.3$ efficiency points, about seven per cent of the output. This is why condenser vacuum is watched obsessively, and why air in-leakage is a serious fault rather than a nuisance.

Now you. Suppose instead the plant is built on a hot river and the condenser must run at $15$ kPa, where $v_{f} = 0.001014$ , $h_{f} = 225.94$ , $h_{f g} = 2372.3$ , $s_{f} = 0.7549$ and $s_{f g} = 7.2522$ . Find the efficiency.

Answer

$w_{p} = 0.001014 \times 2985 = 3.03$ kJ kg⁻¹ so $h_{2} = 228.97$ and $q_{i n} = 2887.1$ kJ kg⁻¹. The quality is $(6.7450 - 0.7549) / 7.2522 = 0.8260$ , giving $h_{4} = 225.94 + 0.8260 \times 2372.3 = 2185.4$ kJ kg⁻¹ and $w_{t u r b} = 930.7$ kJ kg⁻¹. Net work is $927.7$ kJ kg⁻¹ and $η_{t h} = 927.7 / 2887.1 = 0.321$ , between the other two as expected.

Where the heat actually goes in

The boiler duty is one number in the efficiency calculation and three physically distinct jobs, and separating them explains the whole shape of the cycle.

Stage	Temperature range	Heat, kJ kg⁻¹	Share
Warming liquid to saturation	$46$ to $234$ °C	$812.9$	$27.8$ %
Boiling at $3$ MPa	$234$ °C	$1794.9$	$61.5$ %
Superheating	$234$ to $350$ °C	$312.9$	$10.7$ %

Nearly twenty-eight per cent of the heat goes into warming subcooled water between $46$ and $234$ degrees Celsius, all of it at temperatures far below the peak. That is what holds the mean temperature of heat addition down to $479$ K when the steam leaves at $623$ K, and it is the target of the regeneration scheme in the next lesson. The boiling, at a single temperature of $234$ degrees Celsius, is where most of the heat goes, which is why raising the boiler pressure and therefore the boiling temperature is the strongest lever available. The superheat, despite adding the highest-temperature heat in the cycle, is only a tenth of the total.

Real components

Replace the two ideal machines with real ones. A turbine of isentropic efficiency $0.87$ and a pump of $0.85$ change the numbers as follows, with the isentropic values as the reference.

The pump now absorbs $3.02 / 0.85 = 3.55$ kJ kg⁻¹, so $h_{2} = 195.36$ kJ kg⁻¹ and $q_{i n} = 3116.1 - 195.36 = 2920.7$ kJ kg⁻¹, barely changed. The turbine delivers $0.87 \times 980.0 = 852.6$ kJ kg⁻¹, so $h_{4} = 3116.1 - 852.6 = 2263.5$ kJ kg⁻¹ at a quality of $0.866$ , and the condenser must now reject $2263.5 - 191.81 = 2071.7$ kJ kg⁻¹. Net work is $852.6 - 3.55 = 849.0$ kJ kg⁻¹ and

η_{t h} = \frac{849.0}{2920.7} = 0.291

Thirteen per cent of the ideal cycle's output has gone, essentially all of it in the turbine. Notice where it reappears: the condenser duty has risen from $1944$ to $2072$ kJ kg⁻¹. Energy is conserved, so work not taken by the shaft leaves through the cooling water instead.

Example. A plant with the same boiler and condenser conditions has a better turbine, $η_{T} = 0.90$ , and a worse pump, $η_{P} = 0.80$ . Find the net work and thermal efficiency.

The turbine gives $0.90 \times 980.0 = 882.0$ kJ kg⁻¹ and the pump absorbs $3.02 / 0.80 = 3.77$ kJ kg⁻¹, so $h_{2} = 195.58$ and $q_{i n} = 2920.5$ kJ kg⁻¹. Net work is $882.0 - 3.77 = 878.2$ kJ kg⁻¹ and $η_{t h} = 878.2 / 2920.5 = 0.301$ . Three points of turbine efficiency bought one point of cycle efficiency, and the pump's five-point loss cost nothing measurable, which is the back work ratio speaking.

Now you. An old machine on the same duty has $η_{T} = 0.78$ and $η_{P} = 0.75$ . Find the net work and efficiency.

Answer

Turbine work is $0.78 \times 980.0 = 764.4$ kJ kg⁻¹ and pump work is $3.02 / 0.75 = 4.03$ kJ kg⁻¹, giving $h_{2} = 195.84$ and $q_{i n} = 2920.3$ kJ kg⁻¹. Net work is $760.4$ kJ kg⁻¹ and $η_{t h} = 760.4 / 2920.3 = 0.260$ .

What the plant looks like at full size

Efficiency is dimensionless and hides the scale, so convert. A plant on the real cycle above, $849.0$ kJ kg⁻¹ net, producing $100$ MW needs a steam flow of $100000 / 849.0 = 118$ kg s⁻¹. Its boiler duty is $118 \times 2920.7 = 344$ MW and its condenser duty is $118 \times 2071.7 = 244$ MW.

That condenser number deserves a moment. It is nearly two and a half times the electrical output, and it is rejected at $45.81$ degrees Celsius, a temperature at which nobody wants heat. At a $10$ K cooling water rise it requires $244000 / (4.18 \times 10) = 5840$ kg s⁻¹, about $5.9$ m³ s⁻¹. Half the mass of a large river passes through a modest power station.

The fuel follows too. At $29.1$ per cent efficiency, $100$ MW of electricity needs $344$ MW of heat into the boiler, and boiler and combustion losses would add more in a real plant. Coal of $25$ MJ kg⁻¹ heating value would be consumed at $13.8$ kg s⁻¹, which is $1190$ tonnes per day, a train a day for a plant that is small by modern standards.

Example. The same cycle is scaled to $250$ MW. What steam flow, condenser duty and cooling water flow does it need at a $10$ K rise?

The flow is $250000 / 849.0 = 294$ kg s⁻¹. The condenser duty is $294 \times 2071.7 = 610$ MW, and the cooling water flow is $610000 / 41.8 = 14600$ kg s⁻¹, or about $14.6$ m³ s⁻¹.

Now you. A $60$ MW combined heat and power unit runs the same cycle. What steam flow and condenser duty does it have?

Answer

$\dot{m} = 60000 / 849.0 = 70.7$ kg s⁻¹, and the condenser duty is $70.7 \times 2071.7 = 146$ MW.

How a real plant departs further

The analysis above still idealises in ways worth naming, because each one shows up as a discrepancy against plant data.

There is pressure drop everywhere. Friction in boiler tubes, superheater headers and connecting pipework means the pump must deliver above the nominal boiler pressure, often by ten per cent, and the steam arrives at the turbine below it. The extra pump work is negligible; the lost turbine inlet pressure is not.

Condensate is usually subcooled a few degrees below saturation, deliberately, to stop the feed pump cavitating on flashing vapour at its suction. That subcooling is heat rejected and then paid for again in the boiler, and it costs a fraction of a point.

Steam leaks past turbine glands, air leaks into the condenser, and both degrade performance quietly. Air is the more damaging: it does not condense, so it accumulates on tube surfaces and destroys heat transfer, which is why condensers run continuous air ejectors.

Finally, the boiler is not the heat source. Combustion gases enter at perhaps $1800$ K and leave the stack at $400$ K, and the transfer from gas to steam happens across enormous temperature differences. That is by far the largest irreversibility in the plant, and nothing in the cycle analysis so far can see it, because it happens outside the control volume drawn around the working fluid. Making it visible is the job of exergy, at the end of this course.

Two defects worth fixing

The completed analysis leaves the cycle with two clear faults, and both have the same cure.

The efficiency is low. Twenty-nine per cent against a Carnot bound of $48.8$ per cent between the same extreme temperatures, and the reason is now precisely known: the mean temperature of heat addition is only $479$ K, dragged down by the $28$ per cent of the heat spent warming subcooled water.

The exhaust is too wet. At quality $0.866$ the real cycle is just inside the practical limit of about ten per cent moisture, and every attempt to improve efficiency by raising boiler pressure makes it worse, because higher pressure means lower entropy at turbine inlet and therefore a wetter exhaust.

Three modifications attack both faults at once, and each is a direct assault on the mean temperature of heat addition. They are the subject of the next lesson, and together they take this same plant from $29$ per cent to $46$ .