Topic 12 · 13 min · 5 figures

Vapour power cycles

Most of the world's electricity is still made by boiling water, and the cycle that does it looks nothing like the reversible rectangle that sets its ceiling. The Carnot cycle, drawn honestly for a condensing vapour, asks for a pump that meters a two-phase mixture to an exact quality: a machine nobody has built. The Rankine cycle is what you get when you refuse that machine and build something practical instead.

The previous topic built three engines from air that never condenses, and every temperature in them was free to be whatever the compression ratio made it. A vapour cycle gives that freedom up on purpose, because boiling and condensing a fluid, rather than merely heating and cooling a gas, is what lets a huge amount of heat move at close to a single, fixed temperature, which is exactly the trade a Carnot engine wants. What it costs to get that trade is the subject of this topic.

Why a vapour cycle

Draw a Carnot cycle honestly for a substance that condenses, rather than for an ideal gas, and something changes for the better before it changes for the worse. A reversible isothermal process is a horizontal line on a T–s diagram, and inside the saturation dome a horizontal line is also a constant-pressure line, because temperature and pressure are locked together while two phases coexist. Boiling water at a fixed pressure boils at a fixed temperature, so the two hardest-to-arrange legs of a Carnot cycle (heat added and heat rejected, both isothermally) become the two easiest processes in mechanical engineering: boil the fluid, and condense it. No gas cycle offers that for free.

The trouble sits in the other two legs. A Carnot cycle needs isentropic compression to close the loop, from the state after heat rejection back to the state where heat addition begins. Draw those four corners entirely inside the dome (which is where a Carnot cycle drawn against a real saturation curve wants to sit, since that is where the isotherms are flat) and every one of the four states is a wet mixture, quality strictly between 0 and 1. The compression leg is therefore asking for a pump that takes in a spray of liquid droplets suspended in vapour and delivers, at a higher pressure, a mixture of some other exact, specified quality.

No pump does that. Handling two-phase flow at all is difficult. Liquid droplets erode blades and unbalance rotating machinery, and a centrifugal or reciprocating pump built for liquid cavitates badly the moment vapour appears in the suction line. Metering the outcome to a precise quality, rather than simply moving whatever arrives, is not a control problem anyone has solved economically. The theoretical cycle is not wrong. It is simply not hardware.

The fix keeps everything that made the vapour cycle attractive and discards only the one leg that could not be built. Instead of stopping heat rejection at some partial quality, run the condenser all the way to saturated liquid: quality zero, a single phase, nothing left to meter. Compressing a liquid is a solved problem; it is what every pump in the world already does, and because a liquid’s specific volume barely changes with pressure, the isentropic relation from isentropic processes and efficiency collapses to almost nothing:

\[w_p \approx v_1(P_2 - P_1)\]

Trading a precisely-metered two-phase compression for a full condensation followed by an ordinary liquid pump is the single substitution that turns an impossible reversible ideal into the Rankine cycle, and it is worth noticing what was and was not given up in making it. The isothermal heat addition is gone. A real boiler has to heat the compressed liquid up to its boiling point before any boiling starts, which is not at the fixed temperature of the eventual boiling, so the Rankine cycle is no longer a Carnot cycle and no longer touches the Carnot ceiling exactly. What survives is the part that made a vapour worth using in the first place: a condenser that rejects a large quantity of heat at close to one fixed low temperature, cheaply, by simply condensing.

A Carnot cycle that stays inside the dome — step 1 of 4

Two isotherms, two isentropes — a Carnot rectangle drawn entirely under the dome.

\(1 \to 2 \to 3 \to 4 \to 1\)
\(4 \to 1: \quad x_4 \to x_1,\ \text{both wet}\)
\(x = 0 \text{ at the condenser exit}\)
\(w_{\text{pump}} \approx v_1(P_2 - P_1)\)
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The Rankine cycle

Four devices, four processes, and (because the whole cycle is a steady flow through fixed hardware rather than a piston repeating a stroke) every one of them is analysed with the steady-flow energy equation reduced exactly as it was for the individual devices in that topic. Saturated liquid leaves the condenser at state 1 and is pumped, nearly isentropically, to boiler pressure at state 2:

\[w_p = h_2 - h_1 \approx v_1(P_2 - P_1)\]

The boiler adds heat at constant pressure, first raising the compressed liquid to its new saturation temperature, then boiling it, commonly continuing on to superheat it well clear of the dome. Steam leaving wet is a problem for the turbine, addressed below, so real boilers are run to avoid it. All of that happens at one pressure, so the heat added is a single enthalpy difference:

\[q_{\text{in}} = h_3 - h_2\]

The turbine expands the steam, ideally isentropically, delivering the cycle’s work exactly as any turbine does:

\[w_t = h_3 - h_4\]

and the condenser rejects the remaining heat at constant pressure, returning the fluid to saturated liquid and closing the loop:

\[q_{\text{out}} = h_4 - h_1\]

Thermal efficiency is built from these four numbers exactly as it was for every heat engine since the second law:

\[\eta_{\text{th}} = \frac{w_{\text{net}}}{q_{\text{in}}} = \frac{w_t - w_p}{q_{\text{in}}}\]

A worked example fixes the scale of these numbers. Take a boiler at 3 MPa producing saturated vapour, a condenser at 75 kPa, and an isentropic turbine. From the steam tables, \(h_1 = 384.4\) kJ/kg and \(v_1 = 0.001037\) m³/kg at the condenser exit, so \(w_p = 3.0\) kJ/kg and \(h_2 = 387.5\) kJ/kg. At 3 MPa, \(h_3 = 2803.4\) kJ/kg and \(s_3 = 6.187\) kJ/kg·K; following that entropy down to 75 kPa lands on a quality of \(x_4 \approx 0.80\) and \(h_4 \approx 2199.5\) kJ/kg. That gives \(q_{\text{in}} = 2416\) kJ/kg, \(w_t = 604\) kJ/kg, and

\[\eta_{\text{th}} = \frac{604 - 3}{2416} = 0.248\]

Two things about that arithmetic are worth carrying forward. The pump work is small enough to be almost a rounding error next to the turbine work, a striking contrast with the Brayton cycle’s compressor, which routinely consumes 40 to 80 per cent of its turbine’s output, precisely because compressing an almost-incompressible liquid is cheap in a way compressing a gas never is. And the turbine exit quality, 0.80, is low enough to matter: a stream that is one-fifth liquid droplets by mass erodes real turbine blades, which is the practical constraint the next section is built around.

Four processes, one loop — step 1 of 4

A pump raises the condensed liquid to boiler pressure, for almost no work.

\(1 \to 2: \quad w_p = h_2 - h_1\)
\(2 \to 3: \quad q_{\text{in}} = h_3 - h_2\)
\(3 \to 4: \quad w_t = h_3 - h_4\)
\(\eta_{\text{th}} = \frac{w_t - w_p}{q_{\text{in}}}\)
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Boiler and condenser pressure

Two pressures are free to choose in the cycle above, and both move efficiency the same direction for the same underlying reason: heat added at a higher average temperature, or rejected at a lower one, narrows the gap the Carnot argument always cared about, \(1 - T_L/T_H\), even though a real Rankine cycle is not itself a Carnot cycle. Raise the boiler pressure and the saturation temperature at which boiling happens rises with it (at 3 MPa water boils at 233.9 °C; at 15 MPa, 342.2 °C), so more of the heat is added at a temperature closer to \(T_H\), and the cycle’s efficiency climbs.

The same pressure change has a second, less welcome consequence. Raising the boiler pressure while holding the condenser pressure fixed steepens the drop the turbine has to make, and the isentrope from a higher, narrower point on the dome lands further inside it at the same condenser pressure. Taking the turbine inlet as saturated vapour again, at 15 MPa the exit quality falls to roughly 0.74, down from 0.80 at 3 MPa: a real gain in efficiency, bought at the cost of a wetter exhaust.

Lowering the condenser pressure buys the same kind of efficiency gain through the other end of the ratio, and it is the cheaper lever in practice: condenser pressure is set by the temperature of whatever cooling water or air is available, and pushing it down towards a vacuum, well below atmospheric, costs comparatively little hardware, which is why every large steam condenser runs under vacuum rather than at atmospheric pressure or above. It exacts the same quality penalty as raising the boiler pressure does, for the same geometric reason: the isentrope from a fixed turbine inlet state reaches further into the dome as the exit pressure falls.

Both trades run up against the same wall. Liquid droplets in the low-pressure turbine stages erode the blades (moisture impacting the trailing rows at speed is a genuine mechanical failure mode, not a theoretical nicety), and the conventional design floor is an exit quality of about 0.90. Push either pressure to chase efficiency and that floor is what stops you, which is exactly why neither of the next two sections tries to push the pressures further. Both instead find efficiency somewhere else, so that the pressures can be chosen for good exit quality rather than in spite of it.

Taller cycle, wetter exhaust — step 1 of 4

At 3 MPa, the turbine exit is a mixture around 88 per cent vapour.

\(P_{\text{boiler}} = 3\ \text{MPa}, \quad x_4 \approx 0.88\)
\(P_{\text{boiler}} = 15\ \text{MPa}, \quad x_4 \approx 0.74\)
\(T_L \downarrow \;\Rightarrow\; \eta_{\text{th}} \uparrow\)
\(x_4 \gtrsim 0.90 \text{, by convention}\)
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Reheat

Reheat answers the moisture problem by refusing to let the turbine finish the expansion it started. Steam leaves the boiler and expands through a first turbine stage down to some intermediate pressure, not all the way to condenser pressure, and is then routed back through the boiler a second time, where it is reheated at that intermediate pressure back up towards its original temperature. Only then does it enter a second turbine stage and finish expanding to condenser pressure.

Nothing about the energy balance is new; there are simply two heat additions and two turbine work terms where there was one of each before:

\[q_{\text{in}} = (h_3 - h_2) + (h_5 - h_4), \qquad w_t = (h_3 - h_4) + (h_5 - h_6)\]

with efficiency built from the same ratio as always, now with both heat-addition terms in the denominator. What reheat buys is not primarily a large efficiency number (the gain from reheat alone is typically a few percentage points) but the moisture fix, and it is worth seeing why the fix is so effective. The second expansion starts from a state that is hotter and drier than the point the single, unbroken expansion would have passed through at the same pressure, because reheating added enthalpy back in before the second stage began. Expanding from there down to the same final condenser pressure can land on a dry saturated state or even a state still in the superheated region (no moisture at all in the last stage of the turbine, precisely where blade erosion does the most damage) instead of the same wet mixture a single expansion would have produced.

Reheat and the boiler-pressure trade of the previous section are not competitors; they are usually run together deliberately. A high boiler pressure buys efficiency at the cost of moisture, and reheat buys back the moisture at a modest additional efficiency gain of its own, which is why real power stations operate at both high boiler pressures and with one, sometimes two, stages of reheat rather than choosing between the two ideas.

Expand, reheat, expand again — step 1 of 4

A single expansion to condenser pressure leaves the steam wet.

\(3 \to 4: \quad x_4 \approx 0.74\)
\(3 \to 4'\)
\(4' \to 3': \quad q_{\text{reheat}} = h_{3'} - h_{4'}\)
\(3' \to 4: \quad x_4 \ge 1\)
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Regeneration

The remaining inefficiency worth chasing sits at the cold end of the boiler. Feedwater arrives at the boiler inlet close to condenser temperature (perhaps 40 or 90 °C, depending on the condenser pressure chosen), and the first stretch of heating, up to the saturation temperature, adds a large amount of heat at a comparatively low average temperature, dragging down exactly the average heat-addition temperature the whole of this topic has been trying to raise.

Regeneration, or feedwater heating, raises that average by warming the feedwater before it reaches the boiler, using steam that has already done some of its work in the turbine rather than burning more fuel to do it. A small fraction of the flow, \(y\), is bled from an intermediate stage of the turbine and diverted to a feedwater heater, while the remaining \(1 - y\) continues on through the rest of the turbine, the condenser, and the condensate pump as before. In the simplest arrangement, an open feedwater heater, the two streams are allowed to mix directly, and an energy balance on the heater (steady flow, no work, no heat loss) fixes the extraction fraction needed to bring the mixture out as saturated liquid at the bleed pressure:

\[y = \frac{h_{\text{sat}} - h_w}{h_e - h_w}\]

where \(h_e\) is the enthalpy of the extracted steam, \(h_w\) the enthalpy of the returning condensate before mixing, and \(h_{\text{sat}}\) the saturated-liquid enthalpy at the extraction pressure. A second pump then raises this warmer feedwater the rest of the way to boiler pressure, and the boiler starts its work from a substantially hotter inlet than it would have without the extraction.

The trade is a familiar one by now: the fraction \(y\) of the steam no longer passes through the rest of the turbine, so the work per unit of boiler steam falls a little, but the heat that no longer has to be added at the cold end of the boiler falls by more, and \(q_{\text{in}}\) shrinks faster than \(w_{\text{net}}\) does. A single open feedwater heater typically lifts thermal efficiency by two to four percentage points; real power stations use several, each bleeding a smaller fraction at a different pressure, approximating, without ever quite reaching, a boiler inlet that is already hot all the way along, which is the direction every idea in this topic, from reheat to regeneration, has been pushing the cycle since the first section admitted that a real vapour cycle would never be Carnot’s rectangle.

Feedwater heating — step 1 of 4

Bleed a fraction of the steam before it finishes expanding.

\(y = \dot m_{\text{extracted}} / \dot m_{\text{total}}\)
\(1 - y \text{, from the condenser}\)
\(y = \frac{h_{\text{sat}} - h_w}{h_e - h_w}\)
\(T_{\text{feed}} \uparrow \;\Rightarrow\; \eta_{\text{th}} \uparrow\)
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