Applied Thermodynamics

A steam turbine has no fixed quantity of steam in it, which is a problem, because every law in the preceding Thermodynamics course was written for a system containing a fixed quantity of matter.

The closed system runs out

The closed system is a boundary drawn around a definite parcel of matter. It may move, it may deform, it may exchange heat and work with whatever is outside it, but no molecule crosses it. That restriction was harmless while the subject was piston cylinders and sealed vessels, and everything from the First Law to the Clausius inequality was built on it.

Now look at a turbine. Steam enters through a pipe at $3$ MPa and leaves through a much larger one at $10$ kPa, and while it is inside it passes over dozens of blade rows and gives up energy to a shaft. Follow a fixed parcel of that steam and you get a system that enters, expands, accelerates, and leaves, and whose boundary is a moving smear of fluid stretching from the inlet flange to the exhaust hood. Writing $\Delta U=Q-W$ for that parcel is not wrong, it is useless: nothing about it is steady, nothing about it is measurable, and there is no instant at which the parcel coincides with the machine.

The same objection kills the closed system for a compressor, a pump, a nozzle, a boiler tube, a condenser, a throttling valve, a jet engine, and a chemical reactor. What every one of these has in common is that the interesting thing is the hardware, which sits still, while the matter streams through it. The hardware is what is bolted down, instrumented and paid for. So the sensible move is to draw the boundary around the hardware and let the matter cross it.

Drawing the boundary in space

A control volume is a region of space, chosen by whoever is doing the analysis, with a boundary called the control surface through which mass may pass. The places where it passes are ports, or inlets and outlets. Nothing about this is physics yet. It is a bookkeeping decision, and like any bookkeeping decision it can be made well or badly.

Made well, it puts the ports where the properties are known or measurable. For a turbine, the natural surface passes through the inlet flange, through the exhaust flange, along the casing, and across the shaft. Pressure and temperature are measured at both flanges, so the two states are known, and the shaft crossing the surface is where the work leaves. For a whole power station the surface can be drawn round the entire plant, cutting the fuel line, the air intake, the stack, the cooling water and the electrical connection. Both are legitimate; they answer different questions.

Made badly, the surface passes through a place where nobody knows what the fluid is doing: halfway along a boiler tube, or across a region where flow is separating and recirculating. The analysis then needs information that does not exist. Choosing the surface is the first real skill in this subject, and it is worth being deliberate about, because everything downstream inherits the choice.

One vocabulary point, since textbooks differ: a control volume with mass crossing it is often called an open system, and the two terms mean the same thing. A control volume whose ports happen to be shut is a closed system, so nothing from the previous course is being discarded here, only generalised.

How much mass crosses a port

Start with the smallest possible question. Fluid flows down a pipe of cross-sectional area $A$ at speed $V$. In a short time $\mathrm{d}t$, the fluid that gets through a given plane is exactly the fluid that was within a distance $V\mathrm{d}t$ upstream of it, a cylindrical slug of volume $AV\mathrm{d}t$. Its mass is that volume divided by the specific volume $v$ of the fluid, or equivalently multiplied by its density $\rho =1/v$. Divide by $\mathrm{d}t$ and the mass flow rate is

in kilograms per second when $A$ is in m², $V$ in m s⁻¹ and $v$ in m³ kg⁻¹. The related volume flow rate is $\dot{V}=AV=\dot{m}v$, in m³ s⁻¹. Volume flow is what a pump curve is plotted against and what a flow meter usually reads; mass flow is what conservation laws are written in. Confusing them is the commonest arithmetic error in this subject, and the two differ by a factor of over a thousand for steam.

The velocity in that formula is an average over the cross-section, and the average is not the centreline value. Real pipe flow is slower at the wall and faster in the middle, turbulent flow having a profile that is fairly flat with a thin steep layer at the wall, laminar flow a parabola whose peak is twice its mean. Writing $\dot{m}=AV/v$ means $V$ is defined as whatever makes that equation true, which is exactly the mass-averaged velocity. That is a definition, not an approximation, and it is why the formula survives the fact that no real flow is uniform.

Example. Steam at $3$ MPa and $350$ degrees Celsius, where the tables give $v=0.09056$ m³ kg⁻¹, flows down a pipe of internal diameter $200$ mm at $45$ m s⁻¹. What is the mass flow rate?

The area is $A=\pi d^2/4=\pi (0.200{)}^2/4=0.031416$ m². The volume flow is $AV=0.031416\times 45=1.4137$ m³ s⁻¹, and dividing by the specific volume gives $\dot{m}=1.4137/0.09056=15.61$ kg s⁻¹.

Now you. The same steam flows down a $150$ mm pipe at $60$ m s⁻¹. What is the mass flow rate?

The mass balance

Mass is conserved, and for a control volume that statement takes a form worth writing carefully. Over a time interval, the mass inside the region can only change by what came in through the ports minus what went out:

That is the whole law, and it has no exceptions in this course. The term on the left is what distinguishes this from the closed-system case. It is not zero in general: a compressed air receiver being charged, a boiler drum whose level is rising, a fuel tank being drained, all have $\mathrm{d}{m}_{cv}/\mathrm{d}t\ne 0$. Those cases have their own lesson later. What follows here, and what dominates the analysis of running plant, is the case where that term vanishes.

Steady flow

A process is steady when nothing at any fixed point inside the control volume changes with time. Not that nothing changes: the fluid's pressure and temperature change enormously between inlet and outlet of a turbine. What is required is that the pressure at a given point is the same now as it was a minute ago. A turbine at constant load, running for hours, satisfies this to a very good approximation, and so does a compressor, a condenser, a nozzle and a pump.

Steadiness is a strong statement and it has three consequences worth separating.

First, the mass inside the control volume is constant, so $\mathrm{d}{m}_{cv}/\mathrm{d}t=0$ and total mass in equals total mass out:

Second, for a device with one inlet and one outlet, that reduces to a single number carried through the machine, ${\dot{m}}_1={\dot{m}}_2=\dot{m}$, which combined with the definition gives the relation used constantly:

Third, the properties at each port are constant in time, so a single value of $p$, $T$, $h$ and $s$ describes each port for the whole analysis. That is what makes steady-flow problems arithmetic rather than differential equations.

Steadiness also forbids things. A steady device cannot accumulate energy, so its stored energy is constant, which is the point that makes the next lesson's energy equation as simple as it is. And a steady device cannot be starting up, shutting down, or changing load, which is exactly when real machines break.

Areas, velocities and the size of hardware

The relation $A_1{V}_1/v_1=A_2{V}_2/v_2$ looks like a triviality and is not. It is what sets the physical size of turbomachinery, and it explains a shape anyone who has seen a steam turbine has noticed.

Take the steam from the example above, $15.61$ kg s⁻¹ entering at $3$ MPa with $v=0.09056$ m³ kg⁻¹. Expand it through the turbine to $10$ kPa, at which pressure it emerges as a wet mixture of quality $0.8128$, and the saturation table gives $v_f=0.001010$ and $v_g=14.670$ m³ kg⁻¹. The mixture's specific volume is $v=v_f+x(v_g-v_f)=0.001010+0.8128\times 14.669=11.92$ m³ kg⁻¹. The steam leaves occupying over $130$ times the volume it entered with, for the same mass.

That factor has to go somewhere, and it goes into area. Holding the velocity at $45$ m s⁻¹ would need an exhaust area of $\dot{m}v/V=15.61\times 11.92/45=4.14$ m², a duct of $2.29$ m diameter for a machine whose inlet pipe is $200$ mm. Real designs let the velocity rise instead, to a few hundred metres per second, which buys back some of the area but not most of it. This is why a large steam turbine is built as a small high-pressure cylinder followed by a physically enormous low-pressure one, often split into several parallel exhaust flows, with last-stage blades over a metre long that are among the most highly stressed components in engineering.

Example. For that turbine, $15.61$ kg s⁻¹ leaving at $v=11.92$ m³ kg⁻¹, what exhaust area and diameter are needed if the velocity is allowed to reach $180$ m s⁻¹?

$A=\dot{m}v/V=15.61\times 11.92/180=1.034$ m², so $d=\sqrt{4A/\pi }=\sqrt{4\times 1.034/\pi }=1.15$ m. Four times the velocity has cut the diameter by half, and it is still nearly six times the inlet pipe.

Now you. A $600$ MW machine passes $462$ kg s⁻¹ of steam and exhausts it at the same $11.92$ m³ kg⁻¹ and $200$ m s⁻¹, split equally between six parallel low-pressure exhausts. What flow area does each need?

Two inlets, and what the balance settles on its own

Where a control volume has several ports, the mass balance alone often fixes something useful before any energy accounting begins. Consider a mixing chamber where hot and cold water streams join, or a feedwater heater in a power station where steam bled from the turbine is mixed into the condensate. Whatever the temperatures, the exit flow is the sum of the inlet flows, and the exit pipe has to be sized for it.

The same reasoning covers a splitter. A turbine passing $462$ kg s⁻¹ that bleeds $23$ per cent of it at an intermediate stage sends $0.23\times 462=106$ kg s⁻¹ to the feedwater heater and $356$ kg s⁻¹ onwards to the next stage. That extraction fraction will turn out to be the central unknown in regenerative cycles, and it is always found from a balance like this one.

Example. A mixing chamber receives $2.0$ kg s⁻¹ of water through one port and $5.0$ kg s⁻¹ through another, and discharges through a single pipe in which the specific volume is $0.001043$ m³ kg⁻¹ and the velocity is $5.0$ m s⁻¹. What exit area is needed?

Steady flow gives ${\dot{m}}_3=2.0+5.0=7.0$ kg s⁻¹. Then $A=\dot{m}v/V=7.0\times 0.001043/5.0=0.00146$ m², or $14.6$ cm², a pipe of about $43$ mm bore.

Now you. The same chamber now takes $3.0$ and $4.0$ kg s⁻¹, and the exit velocity is raised to $8.0$ m s⁻¹ with the same specific volume. What exit area is needed?

What the model quietly assumes

Three assumptions have been smuggled in, and naming them is what makes the results trustworthy.

The first is one-dimensional flow at the ports: properties are taken as uniform across each port, so one value of $p$, $T$ and $v$ describes it. Across a well-developed pipe flow this is close to true for pressure and temperature, and the velocity is handled by the mass-averaging above. Across a turbine exhaust annulus, where the flow is swirling and the outer radius sees different conditions from the hub, it is a real approximation, and it is why measured turbine performance never quite matches a one-dimensional calculation.

The second is that the ports are the only places mass crosses. Leakage past shaft seals is small in modern machines but never zero, and boiler tubes do fail.

The third is that nothing has been said about energy yet. Every result here would hold equally for water, mercury or sand, because conservation of mass does not know what is flowing. What makes the analysis thermodynamics is that the mass crossing a port carries energy with it, and that shoving it across the boundary costs work. That is the next lesson, and it produces the single equation the rest of this course runs on.

A kilogram of steam entering a turbine brings its internal energy with it, and that is not the whole story, because something had to push it through the inlet.

The question the mass balance left open

Drawing a boundary around a piece of hardware and counting the mass through its ports, as the previous lesson did, says nothing about energy. To get an energy balance we need to know what each kilogram carries as it crosses, and the naive answer is wrong in an interesting way.

The naive answer is internal energy, $u$, the thing the First Law is written in for a closed system. A kilogram of steam at $3$ MPa and $350$ degrees Celsius has $u=2844$ kJ kg⁻¹, so surely a kilogram entering a turbine delivers $2844$ kJ into the control volume. It delivers more, and the reason is mechanical rather than thermal.

Picture the inlet port with the fluid behind it. That fluid is at $3$ MPa, and it is pressing on the kilogram in front of it. As the kilogram moves through the port and into the control volume, the fluid behind does work on it, and that work enters the control volume just as surely as the internal energy does. Nobody pushed anything across the boundary of a closed system, so this term has no closed-system counterpart. It is genuinely new, and it is called flow work.

The cost of the push

Compute it. Take a plug of fluid of mass $m$, occupying volume $V=mv$, being pushed through a port of area $A$ by the pressure $p$ immediately upstream. The plug is a cylinder of length $L=V/A$. The force on its rear face is $pA$, constant while the plug goes through, and it acts through distance $L$, so the work done on it is

Per kilogram, the flow work is $pv$, in joules per kilogram when $p$ is in pascals and $v$ in m³ kg⁻¹. Notice what dropped out: the area cancelled. A narrow port needs a long plug and a wide one a short plug, and the product is the same. The push costs $pv$ per kilogram regardless of the pipework.

The numbers are not small. For steam at $3$ MPa with $v=0.09056$ m³ kg⁻¹, $pv=3000\times 0.09056=271.7$ kJ kg⁻¹, which is nearly ten per cent of the internal energy. At the turbine exhaust, $10$ kPa with $v=12.71$ m³ kg⁻¹, $pv=127.1$ kJ kg⁻¹, still substantial despite the pressure having fallen by a factor of $300$, because the volume rose by a factor of $140$.

At an outlet the same argument runs with the sign flipped: the fluid leaving has to shove the fluid downstream out of its way, so it carries $pv$ per kilogram out of the control volume. Both ports therefore contribute $pv$ alongside $u$, entering at inlets and leaving at outlets.

Example. Steam at $3$ MPa and $350$ degrees Celsius has $v=0.09056$ m³ kg⁻¹ and $h=3116.1$ kJ kg⁻¹. Find the flow work per kilogram at that port and hence the internal energy.

Flow work is $pv=3000\text{kPa}\times 0.09056{\text{m³ kg}}^{-1}=271.7$ kJ kg⁻¹, the units working because a kilopascal times a cubic metre is a kilojoule. Since $h=u+pv$ by definition, $u=3116.1-271.7=2844.4$ kJ kg⁻¹, which is what the superheated table lists.

Now you. At the turbine exhaust the steam is a wet mixture at $10$ kPa with $v=12.71$ m³ kg⁻¹ and $h=2263.5$ kJ kg⁻¹. Find the flow work and the internal energy.

Enthalpy is what flows

So the energy carried per kilogram through a port is $u+pv$, and that combination already has a name. It is the enthalpy,

introduced in the previous course as the correct bookkeeping for a constant-pressure closed process. Here it arrives for a completely different reason and lands on the same quantity. That is worth pausing on. In the closed-system setting, enthalpy looked like an algebraic convenience, a grouping that made $\Delta H=Q_p$ come out cleanly. In the open-system setting it is not a convenience at all: $u+pv$ is literally the energy per kilogram delivered by a flowing stream, internal energy plus the work of getting it through the door. Enthalpy is the natural energy of flow, and internal energy is the natural energy of a closed box. Once that is seen, the fact that every table in this subject is written in $h$ rather than $u$ stops being an editorial choice.

A stream can also carry bulk kinetic energy $V^2/2$ and potential energy $gz$ per kilogram, so the full content of a kilogram crossing a port is

with $V$ in m s⁻¹ and $z$ a height above some reference level. Watch the units: $h$ is usually tabulated in kJ kg⁻¹ while $V^2/2$ comes out in J kg⁻¹, so one of them has to be converted, and forgetting is a factor-of-a-thousand error that is at least loud enough to notice.

The steady-flow energy equation

Now assemble the balance. For a control volume, energy accumulates at a rate equal to what comes in minus what goes out, counting heat, work and the streams. Under steady flow nothing inside accumulates, so the rate of change of the control volume's energy is zero and everything must balance:

This is the steady-flow energy equation, and it is the single most used result in the subject. The sign convention is the one carried over from the previous course: $\dot{Q}$ is positive when heat flows into the control volume, $\dot{W}$ positive when work is done by it. A turbine has positive $\dot{W}$, a compressor negative, and it is usually clearer to write ${\dot{W}}_{in}$ for a compressor and keep the number positive rather than juggle minus signs.

One point about the work term deserves care. $\dot{W}$ here means shaft work, electrical work, and anything else crossing the surface, but not flow work, because flow work has already been absorbed into the enthalpies. Counting it twice is a classic mistake, and the symptom is a turbine that appears to deliver an extra $271$ kJ per kilogram.

For the common case of one inlet and one outlet, divide by $\dot{m}$ and write everything per kilogram:

That is the working form. Almost every device in this course is analysed by writing this line down and then deleting the terms that do not matter, which is the business of the next lesson.

How big are the velocity and height terms

Deleting terms requires knowing their size, and here the arithmetic is decisive. Kinetic energy per kilogram is $V^2/2$, so at $45$ m s⁻¹, a typical pipe velocity, it is ${45}^2/2=1013$ J kg⁻¹, or $1.01$ kJ kg⁻¹. Set that beside an enthalpy of $3116$ kJ kg⁻¹ and it is three parts in ten thousand.

The quadratic makes the term wake up quickly. At $100$ m s⁻¹ it is $5.0$ kJ kg⁻¹, at $200$ m s⁻¹ it is $20$ kJ kg⁻¹, and at $450$ m s⁻¹ it is $101$ kJ kg⁻¹. The useful rule is that below about $50$ m s⁻¹ the kinetic term can be dropped without thought, between $50$ and $150$ m s⁻¹ it should be checked against the enthalpy change rather than the enthalpy itself, and above that it must be carried. In a nozzle it is not a correction at all: it is the entire point of the device.

Potential energy is nearly always negligible in a machine. A height difference of $30$ m gives $gz=9.81\times 30=294$ J kg⁻¹, about $0.29$ kJ kg⁻¹. Even the $100$ m from a boiler drum to a turbine floor buys under $1$ kJ kg⁻¹. The exception is hydraulic machinery, where the height difference is the energy source and the enthalpy change is what is negligible: a hydroelectric station working under a $300$ m head has $gz=2.94$ kJ kg⁻¹ and no combustion anywhere.

Example. A turbine passes $15.61$ kg s⁻¹ of steam. It enters at $h_1=3116.1$ kJ kg⁻¹ and $45$ m s⁻¹ and leaves at $h_2=2263.5$ kJ kg⁻¹ and $200$ m s⁻¹, with negligible heat loss and no height change. What power does it deliver, and how much does the kinetic term matter?

With $q=0$ the working form gives $w=(h_1-h_2)+(V_1^2-V_2^2)/2$. The enthalpy drop is $3116.1-2263.5=852.6$ kJ kg⁻¹. The kinetic contribution is $({45}^2-{200}^2)/2=-18990$ J kg⁻¹, that is $-19.0$ kJ kg⁻¹, negative because the steam leaves faster than it arrived and takes that energy away with it. So $w=852.6-19.0=833.6$ kJ kg⁻¹ and $\dot{W}=15.61\times 833.6=13010$ kW, or $13.0$ MW. Ignoring the velocities entirely would have given $13.3$ MW, an overestimate of $2.2$ per cent.

Now you. The same turbine is fitted with a smaller exhaust so the steam leaves at $260$ m s⁻¹ instead, everything else unchanged. What power does it now deliver?

Reading the equation in both directions

The balance is used two ways, and it is worth being explicit about which.

Read forwards, it is a prediction. Given the inlet state, the outlet state and the flows, it says what power or heat duty the device must have. That is the mode of the example above, and of most textbook problems.

Read backwards, it is a measurement. Given the power on a shaft and the flow through a machine, it says what the outlet state must be, which is how plant instrumentation actually works: you cannot put a thermometer inside a turbine, so the exhaust state is inferred from a load reading and a flow reading. The same reading in reverse is how a device's condition is monitored over years. A turbine whose measured enthalpy drop at rated flow falls by two per cent has fouled or eroded blading, and nobody had to open it to find out.

Example. A compressor takes $0.50$ kg s⁻¹ of air, raising it from $295$ K to $587$ K, and its casing loses $15$ kW to the surroundings. Taking air as an ideal gas with $c_p=1.005$ kJ kg⁻¹ K⁻¹ and ignoring velocities, what shaft power does it need?

The enthalpy rise per kilogram is $c_p\Delta T=1.005\times (587-295)=293.5$ kJ kg⁻¹, so the streams carry away $0.50\times 293.5=146.7$ kW more than they brought in. The heat loss removes a further $15$ kW. The shaft must supply both: ${\dot{W}}_{in}=146.7+15=161.7$ kW. Cooling a compressor does not reduce its shaft power at a fixed outlet temperature; it increases it, because the same temperature rise now has to be paid for twice.

Now you. The same compressor is run at $0.80$ kg s⁻¹ with the same inlet and outlet temperatures, and its heat loss rises to $25$ kW. What shaft power does it need?

What this equation cannot tell you

Two limits should be stated before the equation is turned loose on hardware.

The first is that it is a balance, not a prediction of performance. Nothing in it says how much work a turbine will produce from a given inlet state, because the outlet state is an input, not an output. Given $3$ MPa and $350$ degrees Celsius at inlet and $10$ kPa at exhaust, the energy equation is equally happy with an exhaust enthalpy of $2264$ kJ kg⁻¹, which is a good turbine, or $3116$ kJ kg⁻¹, which is a turbine that produces nothing at all and simply lets the steam through. Both conserve energy exactly. Choosing between them needs the Second Law, and that is what makes the isentropic device and its efficiency, several lessons ahead, necessary rather than decorative.

The second is that steadiness has been assumed throughout, and with it the promise that nothing inside the control volume is changing. Real plant starts, stops, trips and swings load, and a boiler drum whose level is falling is storing energy in a way this equation cannot see. Handling those is the next lesson, and the answer will turn out to explain something as everyday as why a scuba cylinder is hot to the touch after filling.

Almost every component in a power station, a refrigerator or a jet engine is analysed by writing down the same equation and then deleting most of it.

One equation, six deletions

The previous lesson produced, for a control volume with one inlet and one outlet running steadily,

with $q$ and $w$ per kilogram of fluid, heat positive inwards and work positive outwards. It has five terms and no device uses all five. What distinguishes a nozzle from a turbine is not a different physical law, it is which terms survive.

The deletions are not arbitrary, and each one is a claim about the hardware that can be checked. Saying a turbine is adiabatic is a claim that its casing loses less than about one per cent of the power passing through, which is true of a lagged machine and false of an uninsulated one. Saying a throttle has no kinetic term is a claim about the pipe diameters either side. The habit worth building is to write the full equation, then justify each deletion out loud, because the deletions are where the physics is.

Nozzles and diffusers

A nozzle is a duct shaped to raise a fluid's velocity at the expense of its pressure; a diffuser is the same device used in reverse, slowing the flow and raising the pressure. Neither has any moving part, so $w=0$. The fluid is inside for a few milliseconds and the surface area is small, so $q\approx 0$. There is no height change worth counting. Every term dies except two:

Enthalpy converts to velocity, one for one. The quantity on each side, called the stagnation enthalpy, is what a nozzle conserves.

The velocities are large by construction, so this is the one device where the kinetic term is never negligible. It is also the device where the mass balance from the earlier lesson bites hardest: since $\dot{m}=AV/v$ is fixed, the area must follow $A=\dot{m}v/V$, and whether a duct that accelerates a fluid gets narrower or wider depends on whether $v$ grows faster than $V$ does. For a liquid, or for a gas below the speed of sound, $V$ wins and the duct converges. Above the speed of sound $v$ wins and the duct must diverge, which is why a rocket nozzle has a throat and then flares out. That result belongs to compressible flow rather than to thermodynamics, but the energy equation above is what it is built on.

Example. Air enters a nozzle at $200$ degrees Celsius and $30$ m s⁻¹ and leaves at $100$ degrees Celsius. Taking air as an ideal gas with $c_p=1.005$ kJ kg⁻¹ K⁻¹, what is the exit velocity?

For an ideal gas $h_2-h_1=c_p(T_2-T_1)=1.005\times (-100)=-100.5$ kJ kg⁻¹, a fall of $100500$ J kg⁻¹. So $V_2^2/2=V_1^2/2+100500=450+100500=100950$ J kg⁻¹ and $V_2=\sqrt{2\times 100950}=449$ m s⁻¹. A hundred kelvin of temperature is worth four hundred and fifty metres per second, which is the trade every jet and rocket lives on.

Now you. Air enters a nozzle at $400$ degrees Celsius and $20$ m s⁻¹ and leaves at $250$ degrees Celsius. What is the exit velocity?

Run the same equation backwards for a diffuser. An aircraft intake takes air at $250$ m s⁻¹ relative to the aeroplane and slows it to $30$ m s⁻¹ before the compressor. The temperature rise is $(V_1^2-V_2^2)/2c_p=({250}^2-{30}^2)/(2\times 1005)=30.6$ K, achieved with no machinery at all. At supersonic speeds this ram effect does a large part of the compression, which is why a ramjet needs no compressor and why it cannot work standing still.

Turbines

A turbine lets a fluid expand across rotating blade rows and takes work out through a shaft. The kinetic terms are small compared with the enthalpy drop, at least once the exhaust hood has been included in the control volume, and a lagged casing loses well under one per cent of the throughput as heat. So $q=0$, the velocity and height terms go, and

The work per kilogram is the enthalpy drop, and nothing else. A turbine passing $15.61$ kg s⁻¹ of steam from $h_1=3116.1$ to $h_2=2263.5$ kJ kg⁻¹ delivers $15.61\times 852.6=13.3$ MW, subject to the velocity correction computed in the previous lesson.

The equation is silent about how the enthalpy drop is achieved, which is exactly right: it holds for a steam turbine, a gas turbine, a hydraulic turbine and a wind turbine, and knows nothing about blades. What it does not tell you is how large the drop will be for a given inlet state and back pressure, and that gap is the reason the Second Law has to be brought in later.

Compressors, fans and pumps

Reverse the shaft and the same analysis gives a machine that raises pressure at the cost of work. The three names are the same device at different duties: a fan moves a lot of gas against a small pressure rise, a compressor raises gas pressure substantially, and a pump does the same to a liquid. Dropping the same terms,

with the sign taken care of by naming the work as an input. Compressors are sometimes deliberately cooled, in which case $q$ is not zero and must be carried, as in the worked example of the previous lesson.

The interesting comparison is between a pump and a compressor working over the same pressure range, because it explains a fact that dominates the design of power stations. Water at $10$ kPa has a specific volume of $0.001010$ m³ kg⁻¹ and saturated steam at the same pressure has $14.67$ m³ kg⁻¹, a ratio of $14500$. Raising the water from $10$ kPa to $3$ MPa costs about $3.0$ kJ kg⁻¹, which is $0.35$ per cent of the $852.6$ kJ kg⁻¹ the turbine delivers over the same pressure range. Compressing the vapour instead would cost of the same order as the turbine gives back. Pressurise the liquid, never the vapour: it is the whole reason the Rankine cycle exists in the form it does, and the reason a feed pump on a $600$ MW station absorbs about $1.4$ MW.

Throttling valves

A throttling valve is any restriction that drops pressure without any intention of extracting work: a partly open valve, an orifice plate, a porous plug, a capillary tube. There is no shaft, so $w=0$. The device is small and fast, so $q=0$. The pipes either side are usually similar, so the velocities roughly cancel, and there is no height change. Everything is gone except

Throttling is isenthalpic. That is a strange-looking result: pressure falls by a large factor and enthalpy does not move at all. What happens internally is that flow work is converted into internal energy by friction and turbulence, and the sum $u+pv$ comes out unchanged even though both parts of it change a lot.

For an ideal gas, whose enthalpy depends on temperature alone, constant enthalpy means constant temperature, and throttling does nothing thermally. For a real fluid it can do a great deal. The temperature change per unit pressure drop at constant enthalpy is the Joule-Thomson coefficient, and its sign flips at a substance's inversion temperature. Nitrogen's is $621$ K, so nitrogen throttled from room temperature cools, which is the basis of the Linde process for liquefying air. Hydrogen's is $202$ K, so hydrogen throttled from room temperature warms, and any attempt to liquefy it by throttling alone must precool it first, a fact discovered expensively.

The dramatic case is a saturated liquid, where throttling causes flash evaporation and a large temperature drop. This is what happens in the expansion valve of every refrigerator.

Example. Saturated liquid R-134a at $800$ kPa, where $h_f=95.47$ kJ kg⁻¹ and the saturation temperature is $31.31$ degrees Celsius, is throttled to $132.8$ kPa, where $h_f=25.49$ and $h_{fg}=212.91$ kJ kg⁻¹ at a saturation temperature of $-20$ degrees Celsius. What emerges?

Enthalpy is unchanged, so the exit state has $h=95.47$ kJ kg⁻¹ at $132.8$ kPa. Since that lies between $h_f=25.49$ and $h_g=238.40$, it is a wet mixture, of quality

So a third of the liquid flashes to vapour, and the mixture sits at $-20$ degrees Celsius. A valve costing a few pounds has produced a $51$ K temperature drop with no moving parts, no work and no heat. The evaporating third is what absorbs heat from the cold space, and the other two thirds are along for the ride.

Now you. The same saturated liquid at $800$ kPa is throttled to $200$ kPa instead, where $h_f=38.43$ and $h_{fg}=206.03$ kJ kg⁻¹ at $-10.1$ degrees Celsius. What quality results?

Heat exchangers

A heat exchanger brings two streams into thermal contact without mixing them: a shell full of tubes, a plate stack, a coil in a tank. Draw the control surface around the whole unit and there is no shaft, no significant velocity change and, since the outside is insulated, no heat crossing the outer boundary. What is left is that whatever one stream loses, the other gains:

The heat transferred between them does not appear, because it never crossed the surface we drew: it went from one stream to the other inside it. Draw the surface around one stream only and the same heat reappears explicitly as $\dot{Q}$. Both control volumes are correct, and choosing between them is choosing which question to answer.

Example. The condenser of the turbine above receives $15.61$ kg s⁻¹ of steam at $h=2263.5$ kJ kg⁻¹ and returns it as saturated liquid at $10$ kPa, where $h_f=191.81$ kJ kg⁻¹. Cooling water enters at $18$ degrees Celsius and may rise by $10$ K, with $c=4.18$ kJ kg⁻¹ K⁻¹. What is the heat duty and how much cooling water is needed?

Per kilogram of steam, $2263.5-191.81=2071.7$ kJ is rejected, so the duty is $15.61\times 2071.7=32340$ kW, or $32.3$ MW. The water side must absorb the same, so ${\dot{m}}_w=32340/(4.18\times 10)=774$ kg s⁻¹.

Now you. Environmental limits cut the permitted water temperature rise to $8$ K. What flow is now needed?

Scale that up and the number becomes a siting constraint. A $600$ MW steam plant passing $462$ kg s⁻¹ rejects about $957$ MW in its condenser, needing roughly $23$ m³ s⁻¹ of cooling water for a $10$ K rise. That is why large thermal stations sit on rivers, estuaries or coastlines, or else pay an efficiency penalty for cooling towers.

Mixing chambers, and choosing the surface

A mixing chamber is a heat exchanger with the wall removed: the streams join. There is no work and, if it is insulated, no heat, so the balance is enthalpy in equals enthalpy out,

alongside ${\dot{m}}_1+{\dot{m}}_2={\dot{m}}_3$ from the mass balance. Two equations, and typically two unknowns. Mixing $2.0$ kg s⁻¹ of water at $80$ degrees Celsius with $5.0$ kg s⁻¹ at $20$ degrees Celsius gives $7.0$ kg s⁻¹ at $(2\times 80+5\times 20)/7=37.1$ degrees Celsius, taking the specific heat as constant. A domestic mixer tap is exactly this calculation. So is the open feedwater heater that will turn out to raise the efficiency of a power station by several points.

That leaves the point this lesson has been circling. The equations above are not six different results; they are one result plus six defensible sets of deletions. What actually varies between problems is where the control surface goes, and drawing it in the right place is worth more than any formula. Put it around a condenser and the heat vanishes into an internal transfer. Put it around one stream and the heat reappears. Put it around a whole power station and every internal flow disappears, leaving fuel in, air in, stack out, cooling water out and electricity out, which is the right surface for asking what the plant does and the wrong one for asking which component is at fault.

Two limits are worth naming before moving on. First, every deletion above assumed steady operation, and none of these balances survives a machine that is filling, emptying or changing load. Second, and more seriously, not one of these equations decides how well a device performs. The turbine relation $w=h_1-h_2$ is satisfied by a perfect machine and by a broken one; the compressor relation is satisfied by an efficient compressor and by a heater with an impeller in it. Energy accounting alone cannot tell them apart, and the missing ingredient is entropy.

Fill a scuba cylinder and it becomes hot enough to be uncomfortable to hold, which the steady-flow energy equation has nothing to say about, because a cylinder being filled is not steady.

The assumption that just broke

Every result so far rested on the same claim: nothing inside the control volume changes with time. That let the mass balance collapse to inflow equals outflow, and it let the energy balance drop the term describing energy accumulating inside. Both collapses are what made the analysis arithmetic.

Charging and discharging break the claim on purpose. A compressed air receiver being filled has rising mass, rising pressure and rising temperature inside it. A gas cylinder being vented has all three falling. A steam accumulator absorbing a load swing exists precisely to store and release, and a boiler drum whose level moves during a load change is storing water and energy in a way no steady balance can see.

These are not exotic cases. Every plant spends its riskiest hours starting up and shutting down, safety relief devices operate only in transients, and a large fraction of pressure-vessel incidents happen during filling or emptying. The analysis needed is not much harder, but it has one extra term and one new modelling decision.

The transient balance

Keep the same control volume and stop deleting the accumulation terms. Conservation of mass, integrated over a process from state $1$ to state $2$, gives

and conservation of energy, with the energy inside the control volume written as $mu$ since it is not moving anywhere,

Both are exact. Notice which energy appears where: the contents of the control volume are counted with $u$, because nothing is pushing them anywhere, while everything crossing a port is counted with $h$, because it had to be pushed. That distinction, arbitrary-looking when enthalpy was first introduced, is doing real work here, and getting it backwards is the standard way to produce a wrong answer that looks reasonable.

The new modelling decision is what value of $h$ to use at a port, since the state at the port can change during the process. The uniform-flow model assumes it does not: the fluid crossing each port has one fixed state for the whole process, and the contents of the control volume, while changing in time, are uniform in space at any instant. For a bottle filled from a large line at fixed pressure and temperature this is very good. For a bottle discharging, the fluid leaving is at the instantaneous state inside, which is changing, and the model is worse. That case is handled below.

Filling an empty bottle

Take the simplest possible case and let it produce a result nobody expects.

A rigid, insulated, evacuated bottle is connected to a large line carrying air at fixed pressure and temperature, and the valve is opened until the pressure inside equals the line pressure. The bottle is rigid so $W=0$, insulated so $Q=0$, initially evacuated so $m_1=0$, and nothing leaves. The energy balance reduces to a single line:

and the mass balance says $m_{in}=m_2$. Cancel it:

The final internal energy inside the bottle equals the enthalpy of the fluid in the line. That is the whole result, and it is remarkable because there is no heating of any kind: nothing burned, nothing rubbed, no work crossed the boundary. What happened is that the line did flow work $pv$ per kilogram to shove the air through the valve, and that work had nowhere to go but into the internal energy of the air now sitting inside.

For an ideal gas with constant specific heats, $u_2=c_v{T}_2$ and $h_{line}=c_p{T}_{line}$, so

The gas in the bottle ends up hotter than the supply by the factor $k$, which for air is $1.400$. Air from a line at $300$ K fills a bottle at $420$ K, a rise of $120$ K, independent of the pressure, the volume and the size of the valve. Anyone who has filled a diving cylinder or a car tyre from a compressor has felt this, and much of what they felt was this rather than the compressor's own heat.

Example. A rigid, insulated, evacuated $0.500$ m³ vessel is filled from a line carrying air at $700$ kPa and $300$ K until the pressure inside reaches $700$ kPa. Find the final temperature and the mass admitted, taking $R=0.287$ kJ kg⁻¹ K⁻¹ and $k=1.400$.

The temperature follows immediately: $T_2=k{T}_{line}=1.400\times 300=420$ K, or $147$ degrees Celsius. The mass then comes from the ideal gas law at the final state, $m_2=p_{2}V/R{T}_2=(700)(0.500)/(0.287\times 420)=2.90$ kg. Had the filling somehow been isothermal at $300$ K, the bottle would have held $(700)(0.500)/(0.287\times 300)=4.07$ kg, forty per cent more. The vessel will in fact reach that mass only if it is refilled after cooling.

Now you. A rigid, insulated, evacuated $0.300$ m³ vessel is filled from a line at $1.20$ MPa and $290$ K. Find the final temperature and the mass admitted.

The corollary matters commercially. Let the filled bottle sit until it returns to $300$ K and its pressure falls to $700\times 300/420=500$ kPa, since the mass and volume are fixed. A cylinder charged to its rated pressure while hot is not full once it cools, which is why a properly filled cylinder is either filled slowly, filled in a water bath, or topped up afterwards.

Filling a bottle that is not empty

Drop the requirement that the vessel starts empty and the algebra grows one term. With $m_1$ already inside at temperature $T_1$, the energy balance is $m_2{u}_2-m_1{u}_1=(m_2-m_1)h_{line}$, and for an ideal gas that reads

The unknowns are $m_2$ and $T_2$, and the second equation needed is the ideal gas law at the final state, $m_2{T}_2=p_{2}V/R$, whose right-hand side is entirely known. Substituting turns the pair into one linear equation in $m_2$.

Example. The same rigid insulated $0.500$ m³ vessel already contains air at $100$ kPa and $300$ K. It is filled from the same line at $700$ kPa and $300$ K until the pressure inside reaches $700$ kPa. Find the final temperature and mass. Use $c_v=0.718$ and $c_p=1.005$ kJ kg⁻¹ K⁻¹.

The initial mass is $m_1=(100)(0.500)/(0.287\times 300)=0.5807$ kg. The product $m_2{T}_2$ is fixed by the final pressure: $m_2{T}_2=p_{2}V/R=(700)(0.500)/0.287=1219.5$ kg K. Put both into the balance:

The left side is $875.6-125.1=750.5$ kJ. The right side is $301.5m_2-175.1$, so $301.5m_2=925.6$ and $m_2=3.070$ kg. Then $T_2=1219.5/3.070=397$ K. The air already present, which does not have to be pushed in and so brings no flow work, drags the final temperature below the $420$ K of the empty case.

Now you. The same vessel starts at $200$ kPa and $300$ K instead. Find $m_2$ and $T_2$.

Blowing a vessel down

Discharge is the harder direction, because the fluid leaving is at the state inside the vessel, and that state is changing throughout. The uniform-flow assumption of a fixed port state is simply false.

There is a clean way through it for the commonest case. Follow the gas that remains in the vessel at the end. It never crossed the boundary, so it is a closed system. It expanded, because the gas that left made room for it. If the vessel is insulated and the expansion is slow enough to stay near equilibrium, that remaining gas has undergone a reversible adiabatic expansion, and for an ideal gas the isentropic relation applies directly:

No transient balance was needed. The trick is choosing the system: the residual gas, not the vessel's contents as a whole.

Example. The $0.500$ m³ vessel above, holding air at $700$ kPa and $420$ K, is vented to atmosphere at $100$ kPa through a valve. Taking the process as adiabatic and the remaining gas as expanding reversibly, find the final temperature and the mass left.

The pressure ratio is $100/700$ and the exponent is $(k-1)/k=0.2857$, so $T_2=420\times (1/7{)}^{0.2857}=420\times 0.5735=241$ K, which is $-32$ degrees Celsius. The mass left is $m_2=(100)(0.500)/(0.287\times 241)=0.723$ kg, against $2.90$ kg initially, so three quarters of the air has gone.

Now you. The same vessel is vented only to $300$ kPa. Find the final temperature and the mass remaining.

That $-32$ degrees Celsius is visible: vent a gas cylinder quickly and frost forms on it, and the valve can ice up hard enough to jam. It is also a hazard, since carbon steel loses toughness as it cools and a vessel designed for ambient temperature may be blown down into a range where it is brittle. Depressurisation rates in process plant are limited for exactly this reason.

Where the model earns its keep, and where it fails

The uniform-flow model is a genuine approximation and it is worth being specific about what it costs.

It assumes the contents are uniform at every instant. In a bottle being filled fast, the incoming jet is nothing like uniform, and the gas near the inlet is much hotter than the gas at the far end until mixing catches up. The final equilibrium temperature is still right, because the balance only used end states, but any statement about what the gas was doing halfway through is not.

It assumes the process is slow enough that the pressure inside is meaningful. Fast discharge through a small orifice is choked, the flow is sonic at the throat, and the gas inside is not in equilibrium with itself. The isentropic blowdown result then overestimates the cooling, because heat leaks in from the vessel walls, which have far more heat capacity than the gas and act as a reservoir. Real blowdown of a steel cylinder is nearer isothermal than isentropic for slow vents and nearer isentropic for fast ones, and both bounds are worth computing to bracket the answer.

It assumes, in the filling result, that the gas is ideal with constant specific heats. For air at these pressures that is fine. For a bottle being charged to $200$ bar it is not, and the compressibility factor has to come back.

None of these transients said anything about how much of the process was avoidable. Filling a bottle from a line at $300$ K and ending at $420$ K wastes something: that heat leaks away and the cylinder ends up holding less gas than it could. The energy balance is perfectly happy and reports no loss at all, because energy was conserved exactly. To say that the process was wasteful, and by how much, needs a quantity that increases when something irreversible happens, and applying that quantity to control volumes is the next lesson.

Given steam entering a turbine at $3$ MPa and $350$ degrees Celsius and leaving at $10$ kPa, the energy balance is equally content with a machine that produces $980$ kJ per kilogram and one that produces nothing at all.

The gap the energy balance leaves

The steady-flow energy equation says $w=h_1-h_2$ for an adiabatic turbine, and that is a true statement about any turbine. It is also a statement with two unknowns in it. Nothing in the First Law fixes $h_2$, so nothing in the First Law fixes the work. Feed the equation an exhaust enthalpy of $2136$ kJ kg⁻¹ and the turbine is excellent; feed it $3116$ kJ kg⁻¹ and the steam has passed through unchanged, producing nothing, while conserving energy perfectly.

The same hole appears everywhere. A compressor that raises air from $100$ to $800$ kPa may deliver it at $534$ K or at $700$ K, and the energy balance simply reports a larger work input in the second case without complaint. A nozzle may reach $449$ m s⁻¹ or $400$ m s⁻¹ from the same inlet state. In each case the missing statement is not about energy but about direction, and the previous course established what supplies it.

The entropy balance for a control volume

Entropy is not conserved. It is transferred with heat, carried by mass, and generated by irreversibility, and the balance says so term by term. For a control volume,

where $T_b$ is the temperature of the boundary where the heat crosses, and ${\dot{S}}_{gen}\ge 0$ always, equalling zero only for a reversible process. The second and third terms are new relative to the closed-system version, and they are simple: a kilogram crossing a port takes its entropy with it, exactly as it takes its enthalpy.

Under steady operation the left side vanishes, and for one inlet and one outlet the balance rearranges to

Now make the device adiabatic, which as the earlier lesson argued is a good description of a turbine, a compressor, a nozzle or a valve. The heat term goes and what is left is stark:

An adiabatic steady-flow device cannot lower the entropy of the fluid passing through it. Not "usually does not": cannot. Every real one raises it, and the amount by which it does is ${\dot{S}}_{gen}/\dot{m}$, a direct measure of how badly the device is behaving.

The ideal device is the isentropic one

The inequality has a boundary, and the boundary is the benchmark we were missing. The best conceivable adiabatic device is the one that generates no entropy at all, so it operates at $s_2=s_1$ and is called isentropic. Isentropic means adiabatic and reversible together, and neither half alone is enough: a throttling valve is adiabatic and hopeless.

The isentropic device is not achievable, and that does not weaken it. It is a fixed, computable target defined by the inlet state and the exit pressure alone, which is exactly what a benchmark has to be. Given $3$ MPa and $350$ degrees Celsius at inlet and $10$ kPa at exhaust, the isentropic exit state is fully determined: $s_2=s_1=6.7450$ kJ kg⁻¹ K⁻¹, which at $10$ kPa where $s_f=0.6492$ and $s_{fg}=7.4996$ gives a quality

and hence $h_{2s}=191.81+0.8128\times 2392.1=2136.2$ kJ kg⁻¹. The ideal work is $3116.1-2136.2=980.0$ kJ kg⁻¹, and no adiabatic turbine between those states can beat it.

For an ideal gas with constant specific heats the isentropic exit state comes from a formula rather than a table. Starting from $\mathrm{d}s=c_p\mathrm{d}T/T-R\mathrm{d}p/p$ and setting $\mathrm{d}s=0$,

with $k=c_p/c_v$, which for air is $1.400$ and gives an exponent of $0.2857$. This one relation carries most of the gas turbine work later in the course.

Reversible work, and why pumping liquid is cheap

Before scoring real machines, one result explains a fact that has been asserted twice already. For a reversible steady-flow process, the work per kilogram is

ignoring velocity and height changes. The derivation is two lines. The property relation $T\mathrm{d}s=\mathrm{d}h-v\mathrm{d}p$ holds for any substance between any two equilibrium states. For a reversible process $\delta q=T\mathrm{d}s$, and the energy equation in differential form is $\delta q-\delta w=\mathrm{d}h$. Substitute: $\delta w=T\mathrm{d}s-\mathrm{d}h=T\mathrm{d}s-(T\mathrm{d}s+v\mathrm{d}p)=-v\mathrm{d}p$. Integrate.

Read it as an instruction. The work of a steady-flow machine is set by the specific volume of whatever is being pushed against the pressure change. To spend little, arrange for $v$ to be small, which means compress liquids and expand vapours, never the reverse.

The numbers are brutal. Water at $10$ kPa has $v=0.001010$ m³ kg⁻¹ and is nearly incompressible, so the integral is just $v(p_2-p_1)$. Raising it to $15$ MPa costs $0.001010\times 14990=15.1$ kJ kg⁻¹. Saturated steam at the same $10$ kPa has $v=14.67$ m³ kg⁻¹, fourteen thousand times larger, and compressing it over the same range would cost of the order of the entire output of the turbine. This is why every vapour power cycle condenses its working fluid completely before raising its pressure, and it is the single most consequential line in this course.

Example. A boiler feed pump raises water from saturated liquid at $10$ kPa, where $v=0.001010$ m³ kg⁻¹, to $15$ MPa. Its isentropic efficiency is $0.85$. Find the ideal and actual work per kilogram, and the power required at $462$ kg s⁻¹.

Treating the water as incompressible, $w_s=v(p_2-p_1)=0.001010\times (15000-10)=15.14$ kJ kg⁻¹. A pump's efficiency is ideal work over actual, since work is being paid for, so $w_a=15.14/0.85=17.81$ kJ kg⁻¹. The power is $462\times 17.81=8230$ kW, or $8.2$ MW, out of a station producing several hundred.

Now you. A smaller plant pumps the same water from $10$ kPa to $5$ MPa with a pump of isentropic efficiency $0.80$. Find the ideal and actual work per kilogram.

Turbine isentropic efficiency

Now the definition. For a turbine, work is the product, so the efficiency is what you got over what you could have got, both taken between the same inlet state and the same exit pressure:

Large steam turbines reach $0.88$ to $0.92$, gas turbines $0.85$ to $0.90$, small machines much less. The definition has a subtlety worth stating: $h_{2a}$ and $h_{2s}$ are at the same pressure but not the same state. The real turbine ends at a higher entropy, and therefore at a higher enthalpy, than the ideal one.

Example. The turbine above, taking steam at $3$ MPa and $350$ degrees Celsius and exhausting to $10$ kPa, has ${\eta }_T=0.87$. Find the actual work and the actual exit state.

The ideal work was computed above as $980.0$ kJ kg⁻¹, so the actual work is $0.87\times 980.0=852.6$ kJ kg⁻¹ and the actual exit enthalpy is $h_{2a}=3116.1-852.6=2263.5$ kJ kg⁻¹. That is still inside the dome at $10$ kPa, at a quality of $(2263.5-191.81)/2392.1=0.866$, against $0.813$ for the ideal machine. The real turbine exhausts drier than the ideal one, because the losses reappear as internal energy in the fluid and re-evaporate some of the moisture. The entropy generated is $s_{2a}-s_1=7.144-6.745=0.399$ kJ kg⁻¹ K⁻¹, and a later lesson will convert that number into wasted joules.

Now you. A worn machine on the same duty manages ${\eta }_T=0.80$. Find its actual work, exit enthalpy and exit quality.

Compressors, pumps and nozzles

For a device that consumes work the ratio is inverted, so that the efficiency is again a number below one:

Typical values run from $0.75$ for a small machine to $0.90$ for a large axial compressor. Pumps are defined identically and reach $0.70$ to $0.90$.

Example. A compressor takes air at $100$ kPa and $295$ K and delivers it at $800$ kPa. Its isentropic efficiency is $0.82$. With $c_p=1.005$ kJ kg⁻¹ K⁻¹ and $k=1.400$, find the ideal work, the actual work and the delivery temperature.

The isentropic exit temperature is $T_{2s}=295\times 8^{0.2857}=295\times 1.8115=534.4$ K, so $w_s=c_p(T_{2s}-T_1)=1.005\times 239.4=240.6$ kJ kg⁻¹. The actual work is $w_a=240.6/0.82=293.4$ kJ kg⁻¹, and since the air really does absorb all of it, $T_{2a}=295+293.4/1.005=587$ K, or $314$ degrees Celsius. Fifty-two kelvin of that is pure inefficiency, and it has to be got rid of downstream.

Now you. A cheaper machine on the same duty has ${\eta }_C=0.75$. Find its actual work and delivery temperature.

A nozzle produces neither work nor heat, so its efficiency is written in kinetic energy: ${\eta }_N=V_{2a}^2/V_{2s}^2$, the ratio of actual to ideal exit kinetic energy. Well-designed nozzles are extremely good, $0.90$ to $0.97$, because the flow is accelerating and accelerating flows do not separate from walls. The air nozzle of an earlier lesson, ideally reaching $449$ m s⁻¹, manages $449\sqrt{0.95}=438$ m s⁻¹ at ${\eta }_N=0.95$.

Diffusers are the mirror image and are much worse, often $0.70$ to $0.85$, because there the flow is decelerating against a rising pressure and will separate at the smallest excuse. Anyone who has designed a duct knows that expansions are hard and contractions are easy, and this is the thermodynamic statement of it.

What the benchmark does not say

Three honest limits, because the number is used more casually than it deserves.

First, isentropic efficiencies are not comparable across devices. A turbine at $0.87$ and a compressor at $0.87$ generate different amounts of entropy on the same duty, because the two definitions divide by different things. Comparing them properly needs a common currency, which is what the exergy lesson supplies.

Second, a multistage turbine's overall isentropic efficiency exceeds the efficiency of its individual stages, which sounds impossible and is not. Each stage's losses reheat the fluid, and the reheated fluid arriving at the next stage has slightly more enthalpy available to drop, since the constant-pressure lines on an enthalpy-entropy diagram diverge as entropy rises. Some of each stage's loss is therefore recovered downstream. The effect, called the reheat factor, is worth two to four points on a large machine, and it means a stage efficiency quoted as an overall one is being flattered.

Third, the benchmark is silent on devices whose ideal work is zero. A throttling valve has $w_s=0$ and $w_a=0$, so the definition gives nothing at all, yet the valve is thoroughly irreversible: the R-134a valve of an earlier lesson generates $0.0174$ kJ kg⁻¹ K⁻¹ every time a kilogram goes through it. Isentropic efficiency measures how well a device does the job it was built for. It cannot notice a device whose job is to waste something.

With efficiencies for the individual machines in hand, the pieces can finally be chained together: the exhaust of one device becomes the inlet of the next, and the last one feeds back into the first. That closed chain is a cycle, and it is where this course starts producing power.

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: $\oint \mathrm{d}u=0$, $\oint \mathrm{d}h=0$, $\oint \mathrm{d}s=0$. Applying the energy balance around the whole loop, with the enthalpy change summing to zero,

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:

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:

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.

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, $q_{out}=\int T\mathrm{d}s=T_L\Delta s$, where $\Delta s$ 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

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 $\Delta s$ appears in both expressions. Divide:

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 $q_{in}=2921$ kJ kg⁻¹ and $\Delta s=6.7450-0.6492=6.0958$ kJ kg⁻¹ K⁻¹. So ${\bar{T}}_H=2921/6.0958=479.2$ K, which is $206$ degrees Celsius, and ${\eta }_{th}=1-318.96/479.2=0.334$. 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 $q_{in}$ to $3488$ kJ kg⁻¹ and $\Delta s$ to $6.8593$ kJ kg⁻¹ K⁻¹, change it?

The first is $1-319/479=0.334$. For the second, ${\bar{T}}_H=3488/6.8593=508.5$ K, so ${\eta }_{th}=1-318.96/508.5=0.373$. 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 $q_{in}=3376$ kJ kg⁻¹ and $\Delta s=6.0304$ kJ kg⁻¹ K⁻¹, still rejecting at $319$ K. What are the mean temperature and the efficiency?

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.

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:

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 $T_{sat}=45.81$ degrees Celsius, $v_f=0.001010$ m³ kg⁻¹, $h_f=191.81$ kJ kg⁻¹, $h_{fg}=2392.1$ kJ kg⁻¹, $s_f=0.6492$ and $s_{fg}=7.4996$ 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 $q_{in}=3116.1-194.83=2921.3$ 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

and $h_4=191.81+0.8128\times 2392.1=2136.2$ kJ kg⁻¹. The turbine therefore delivers $w_{turb}=3116.1-2136.2=980.0$ kJ kg⁻¹, and the condenser rejects $q_{out}=2136.2-191.81=1944.3$ kJ kg⁻¹.

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

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 $v_f=0.001017$, $h_f=251.42$, $h_{fg}=2357.5$, $s_f=0.8320$ and $s_{fg}=7.0752$. 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 $q_{in}=3116.1-254.45=2861.7$ 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 ${\eta }_{th}=891.4/2861.7=0.311$.

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 $v_f=0.001014$, $h_f=225.94$, $h_{fg}=2372.3$, $s_f=0.7549$ and $s_{fg}=7.2522$. Find the efficiency.

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.

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 $q_{in}=3116.1-195.36=2920.7$ 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

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, ${\eta }_T=0.90$, and a worse pump, ${\eta }_P=0.80$. 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 $q_{in}=2920.5$ kJ kg⁻¹. Net work is $882.0-3.77=878.2$ kJ kg⁻¹ and ${\eta }_{th}=878.2/2920.5=0.301$. 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 ${\eta }_T=0.78$ and ${\eta }_P=0.75$. Find the net work and efficiency.

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?

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 $46$.

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 $T_L$, the thermal efficiency is exactly $1-T_L/{\bar{T}}_H$, where ${\bar{T}}_H=q_{in}/\Delta s$ is the mean temperature at which heat is added. The baseline plant, boiling at $3$ MPa and superheating to $350$ degrees Celsius with a condenser at $10$ kPa, had ${\bar{T}}_H=479$ K and ${\eta }_{th}=0.334$.

There are only two levers. Lower $T_L$, 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 ${\bar{T}}_H$, 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 $3$ MPa the tables give $h=3457.2$ kJ kg⁻¹ and $s=7.2338$ kJ kg⁻¹ K⁻¹ at $500$ degrees Celsius, and $h=3682.8$ with $s=7.5085$ at $600$.

Run the cycle at $600$ degrees Celsius. The pump work is unchanged at $3.02$ kJ kg⁻¹, giving $h_2=194.83$ kJ kg⁻¹, so $q_{in}=3682.8-194.83=3488.0$ kJ kg⁻¹. The exhaust quality is $(7.5085-0.6492)/7.4996=0.9146$, so $h_4=191.81+0.9146\times 2392.1=2379.7$ kJ kg⁻¹ and the turbine gives $1303.1$ kJ kg⁻¹. The efficiency is $(1303.1-3.02)/3488.0=0.373$.

Both defects improved together. The efficiency rose by nearly four points, and the exhaust moisture fell from $18.7$ to $8.5$ per cent, which is the more valuable of the two, because it is what limits blade life.

Check it against the criterion: ${\bar{T}}_H=3488.0/(7.5085-0.6492)=508.5$ K, and $1-318.96/508.5=0.373$. Two hundred and fifty extra degrees of superheat moved the mean temperature by only $29$ 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 $600$ 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 $540$ and $620$ degrees Celsius, and why raising it further is a materials research programme rather than a design choice.

Example. Run the same plant at $3$ MPa and $500$ degrees Celsius, condensing at $10$ kPa, with $h_3=3457.2$ and $s_3=7.2338$. Find the efficiency, the exhaust quality and the mean temperature of heat addition.

The quality is $(7.2338-0.6492)/7.4996=0.8780$, so $h_4=191.81+0.8780\times 2392.1=2292.1$ kJ kg⁻¹ and $w_{turb}=1165.1$ kJ kg⁻¹. With $q_{in}=3457.2-194.83=3262.4$ kJ kg⁻¹, the efficiency is $(1165.1-3.02)/3262.4=0.356$. The mean temperature is $3262.4/6.5846=495.5$ K, and $1-318.96/495.5=0.356$ confirms it.

Now you. Confirm the criterion for the $600$ degree case: given $q_{in}=3488.0$ kJ kg⁻¹ and $s_3=7.5085$ kJ kg⁻¹ K⁻¹, find ${\bar{T}}_H$ and the efficiency it predicts.

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 $3$ MPa water boils at $233.85$ degrees Celsius; at $15$ MPa it boils at $342.16$.

Take the cycle to $15$ MPa and $600$ degrees Celsius, where $h_3=3583.1$ kJ kg⁻¹ and $s_3=6.6796$ kJ kg⁻¹ K⁻¹. The pump work rises to $0.001010\times 14990=15.14$ kJ kg⁻¹, so $h_2=206.95$ kJ kg⁻¹ and $q_{in}=3376.2$ kJ kg⁻¹. The turbine expands to $10$ kPa at quality $(6.6796-0.6492)/7.4996=0.8041$, giving $h_4=2115.3$ and $w_{turb}=1467.8$ kJ kg⁻¹. The efficiency is $(1467.8-15.14)/3376.2=0.430$, and ${\bar{T}}_H$ has climbed to $560$ 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 $22.06$ 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 $0.804$ means $19.6$ per cent moisture, worse than the original $350$ 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 $15$ MPa, $600$ degree cycle and reheat at $3$ MPa. The high-pressure turbine expands isentropically from $s=6.6796$, which at $3$ MPa lies between the tabulated $300$ degree entry ($h=2994.3$, $s=6.5412$) and the $350$ degree one ($h=3116.1$, $s=6.7450$). Interpolating, the fraction across is $(6.6796-6.5412)/(6.7450-6.5412)=0.679$, so the steam leaves the high-pressure turbine at $334$ degrees Celsius with $h=3077.0$ kJ kg⁻¹, having delivered $3583.1-3077.0=506.1$ kJ kg⁻¹.

Reheat at $3$ MPa to $600$ degrees Celsius, which takes $3682.8-3077.0=605.8$ kJ kg⁻¹ of extra heat and returns the steam to $h=3682.8$, $s=7.5085$. The low-pressure turbine expands to $10$ kPa at quality $0.9146$, exactly as the $3$ MPa superheated cycle did, giving $h=2379.7$ and a further $1303.1$ kJ kg⁻¹ of work.

Now total up. Turbine work is $506.1+1303.1=1809.2$ kJ kg⁻¹, pump work is $15.14$, heat input is $3376.2+605.8=3982.0$ kJ kg⁻¹, and

Both problems solved at once. The efficiency is two points above the $15$ MPa cycle without reheat, and the exhaust moisture has fallen from $19.6$ to $8.5$ per cent. The mean temperature of heat addition is now $580$ 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 $3$ MPa but only to $500$ degrees Celsius, where $h=3457.2$ and $s=7.2338$. Find the efficiency and the exhaust moisture.

The high-pressure turbine is unchanged, delivering $506.1$ kJ kg⁻¹ and leaving the steam at $3077.0$ kJ kg⁻¹. The reheat now supplies $3457.2-3077.0=380.2$ kJ kg⁻¹, and the low-pressure turbine expands to quality $0.8780$, giving $h=2292.1$ and work of $1165.1$ kJ kg⁻¹. Total turbine work is $1671.2$ kJ kg⁻¹ against a heat input of $3376.2+380.2=3756.3$ kJ kg⁻¹, so ${\eta }_{th}=(1671.2-15.14)/3756.3=0.441$. The exhaust moisture is $12.2$ 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.

Regeneration

Superheat and reheat both raise ${\bar{T}}_H$ 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 $46$ 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 $y$ be the fraction of the turbine flow that is extracted. Per kilogram entering the turbine, the heater receives $y$ kilograms of steam at $h_{ext}$ and $(1-y)$ kilograms of pumped condensate at $h_a$, and delivers $1$ kilogram of saturated liquid at $h_f$. The energy balance is $y{h}_{ext}+(1-y)h_a=h_f$, so