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The ideal Brayton cycle consists of adiabatic compression, constant-pressure heat addition, adiabatic expansion, and constant-pressure heat rejection. Its ideal working substance is an ideal gas with constant heat capacities, and the insulated stages are reversible thermodynamic processes.

For pressure ratio $r_p=p_h/p_l$ and heat capacity ratio $\gamma =C_p/C_V$, a reversible ideal-gas adiabat has $T\propto p^{(\gamma -1)/\gamma }$. Thus $T_2/T_1=T_3/T_4=k=r_p^{(\gamma -1)/\gamma }$, and the constant-pressure heat transfers have $Q_{\mathrm{out}}/Q_{\mathrm{in}}=(T_4-T_1)/(T_3-T_2)=1/k$. The thermal efficiency isThis formula requires constant heat capacities and internally reversible insulated stages. An arbitrary dissipative adiabatic process does not satisfy the same pressure-temperature relation.

A Carnot cycle realizes the reversible bound in Carnot's theorem. Its four steps for an ideal gas working substance are:

The temperature versus entropy plot in the Carnot's theorem figure is therefore a rectangle traversed clockwise. Because $\delta Q_{\mathrm{rev}}=TdS$, its width is $\Delta S=Q_h/T_h$. For the illustrated engine, $\Delta S=5/3\mathrm{J}{\mathrm{K}}^{-1}$: the upper step absorbs $600\times 5/3=1000\mathrm{J}$, the lower step rejects $300\times 5/3=500\mathrm{J}$, and the enclosed area is the $500\mathrm{J}$ net work. Geometry displays the thermal efficiency without solving the gas motion. The two vertical steps are isentropic processes, not constant-temperature steps. ^{[ref] [ref]}

Reversible heat transfer is a limit of vanishing temperature differences; ordinary finite-rate transfer introduces entropy production. The Carnot cycle is a performance benchmark rather than a guarantee of useful power. Running it backward gives the ideal operation of a refrigerator or heat pump, with work supplied to move heat toward the hotter thermal reservoir. ^[ref]

No cyclic heat engine exchanging heat only with two fixed thermal reservoirs at positive absolute temperatures $T_h>T_c$ can exceed the thermal efficiency of an engine whose cycle is a reversible thermodynamic process between them. All such reversible engines have the same thermal efficiency, regardless of their working substance. Let $Q_h>0$ be heat absorbed from the hot thermal reservoir and $Q_c>0$ the magnitude rejected to the cold one. ThenThe work convention is the one in the first law of thermodynamics: $W>0$ is net output by the engine. Use absolute thermodynamic temperatures, such as values in kelvin; ratios of readings in degree Celsius are invalid because their zero is shifted.

For $T_h=600\mathrm{K}$, $T_c=300\mathrm{K}$ and $Q_h=1000\mathrm{J}$, the maximum thermal efficiency is $50\%$. A reversible engine delivers $W=500\mathrm{J}$ and rejects $Q_c=500\mathrm{J}$. The thermal reservoirs have entropy changes $-1000/600$ and $+500/300\mathrm{J}{\mathrm{K}}^{-1}$, which cancel. The engine's own entropy change is zero over its thermodynamic cycle.

A short derivation now becomes transparent: the Second law of thermodynamics gives $-Q_h/T_h+Q_c/T_c\ge 0$, and the first law of thermodynamics gives $W=Q_h-Q_c$. Together they yield the bound and, more precisely,At fixed $Q_h,T_h,T_c$, every extra unit of total entropy production reduces output by $T_c$ units of energy. ^[ref]

Fixed thermal reservoirs are essential. Finite bodies cool and warm, so their initial temperature ratio cannot be used throughout an entire extraction process. As a computed contrast, the bodies with equal constant heat capacity in the entropy example would reach $T_f=\sqrt{600\times 300}=424.26\mathrm{K}$ if a reversible engine extracted work until their temperatures matched. Zero total entropy change gives this geometric mean; the first law of thermodynamics then gives $W_{\max }=C(600+300-2T_f)=5147.19\mathrm{J}$. Direct contact reaches $450\mathrm{K}$ and extracts no work.

Carnot's 1824 book asked whether the useful output of machines driven by heat was fundamentally limited and whether changing the working substance could improve that limit. Its reversible comparisons are the historical setting of Carnot's theorem. Carnot used the then-current caloric picture, treating heat as conserved in its passage through an engine; the modern balance $Q_h-Q_c=W$ should not be read back into that text.^[ref]

For the cyclic two-side energy balance in refrigerator, define the dimensionless cooling and heating coefficients byThis identity applies even to irreversible machines, provided both coefficients refer to the same operating point, work input and boundary with no additional energy transfer or storage. It does not compare two separately operated appliances. Unlike thermal efficiency of a heat engine, a COP can exceed one: its numerator counts heat moved, including energy taken from another region. ^[ref]

To obtain temperature-only bounds, impose stronger assumptions: a cyclic machine exchanges heat only with two ideal thermal reservoirs maintained at uniform, fixed absolute thermodynamic temperatures $0<T_c<T_h$, and receives work through an ideal input carrying no entropy. Include the machine and the heat transfer interfaces in the entropy production accounting. Because its own entropy returns to its initial value each cycle, the Second law of thermodynamics givesCombining this inequality with $Q_h=Q_c+W$ givesEquality requires a reversible thermodynamic process for the entire cycle and its contacts, as in a reversed Carnot cycle. These are Carnot's theorem bounds independent of the working substance. Use kelvin values, not degree Celsius readings, in the numerators; the temperature difference can be expressed in either unit. ^{[ref] [ref]}

The diagram fixes $T_c=280\mathrm{K}$ and sets $T_h=320\mathrm{K}$, so the reversible bounds are $280/40=7$ and $320/40=8$. The stipulated machine with ${\mathrm{COP}}_{\mathrm{r}}=3$ and ${\mathrm{COP}}_{\mathrm{heat}}=4$ satisfies both. Its reservoir entropy production per cycle is $4000/320-3000/280=25/14\mathrm{J}{\mathrm{K}}^{-1}>0$. Increasing the temperature lift $\Delta T=T_h-T_c$ at fixed $T_c$ lowers both bounds. Equivalently, delivering fixed $Q_h$ needs at leastThese expressions reveal why a small temperature lift is attractive. Their divergence as $T_h-T_c\to 0^{+}$ is an ideal reversible limit, not a promise of infinite heating or cooling power. For finite heat transfer rates, the working fluid must be colder than the cold bath and hotter than the hot bath; these contact differences produce entropy and increase the internal temperature lift. Dissipative compression and the pressure-reducing restriction of vapor-compression refrigeration supply further departures from reversibility. Thus bath temperatures set an outer bound, not a prediction of real COP. ^[ref]

A heat engine uses a repeating thermodynamic cycle to turn part of an incoming heat transfer into net outgoing work. In the two-thermal reservoir model, the hot thermal reservoir supplies heat $Q_h>0$ at fixed temperature $T_h$, and the cold thermal reservoir receives heat of magnitude $Q_c>0$ at fixed temperature $T_c<T_h$. A Stirling engine makes the repeated motion visible; the abstract energy budget applies independently of its particular mechanism.^[ref]

Choose the entire cyclic device as a closed system, with no net change in bulk kinetic energy or potential energy. Count heat entering as positive and net work delivered as positive, as in the first law of thermodynamics. Thus $Q_h$ enters with a plus sign and $Q_c$, although a positive magnitude, enters with a minus sign. Assume these two heat transfers and the net work are the only transfers across the boundary. Over one complete thermodynamic cycle, all stored energy, including internal energy $U$, returns to its starting value:The device does not accumulate heat from cycle to cycle. More precisely, heat is a transfer, while internal energy is what can be stored; even a component that temporarily stores internal energy must return to its initial state for this boundary to be cyclic.^[ref]

For engine operation, $Q_h>Q_c$ and $W>0$. Its thermal efficiency isUse net work, after subtracting any work supplied during compression or required by auxiliaries included in the chosen device. For periodic operation, average power satisfies $P_{\mathrm{out}}={\dot{Q}}_h-{\dot{Q}}_c$ and $\eta =P_{\mathrm{out}}/{\dot{Q}}_h$, with the rates averaged over complete thermodynamic cycles.^[ref]

For a stipulated model at $T_h=600\mathrm{K}$ and $T_c=300\mathrm{K}$, let $Q_h=1000\mathrm{J}$ and $Q_c=600\mathrm{J}$ per thermodynamic cycle. Then $W=400\mathrm{J}$ and $\eta =2/5=40\%$. At two thermodynamic cycles per second, these become $2000\mathrm{W}$ incoming heat transfer, $1200\mathrm{W}$ outgoing heat transfer, and $800\mathrm{W}$ net power. These are constructed values, not measurements.

Different signs describe different tasks. With the same directions of heat transfer, $Q_h=Q_c$ gives $W=0$: a device transmitting heat without net work output. If $Q_c>Q_h$, net work is supplied to that device. A refrigerator or heat pump instead extracts heat from the cold side and delivers heat to the hot side, using positive input work $W_{\mathrm{in}}$ and the balance $Q_h=Q_c+W_{\mathrm{in}}$. Their useful output is heat transfer, measured by a coefficient of performance. Finally, a one-time isothermal expansion can have $Q=W$ and $\Delta U=0$ while leaving the volume changed: it is not a complete thermodynamic cycle. A warming device also fails the cyclic assumption, and a fuel-consuming device requires an accounting of matter crossing its boundary.

For two fixed thermal reservoirs at positive thermodynamic temperatures, the Second law of thermodynamics strengthens the energy balance toEquality requires the overall operation to be a reversible thermodynamic process, including its heat transfer to the thermal reservoirs. The Carnot cycle explains how the reversible benchmark is constructed, and Carnot's theorem explains its universality and the entropy production penalty. Use kelvin values, never ratios of degree Celsius readings. The example above lies below the $50\%$ bound: its total entropy production is $-1000/600+600/300=1/3\mathrm{J}{\mathrm{K}}^{-1}$ per thermodynamic cycle.

This is useful before building a machine: warmer cooling surroundings reduce the possible thermal efficiency, even if the mechanism is unchanged. A small observed thermal efficiency does not by itself test the Second law of thermodynamics; heat leaks, friction, incomplete recovery of work, and the measurement boundary can all reduce the reported ratio. A meaningful comparison requires the actual thermal reservoir temperatures, matched heat and net work measurements, and the hypotheses of Carnot's theorem. Finite bodies whose temperatures change are treated separately there.

The practical question already appears in the 1824 Réflexions sur la puissance motrice du feu et sur les machines propres à développer cette puissance by Sadi Carnot: pp. 6–8 ask whether machines driven by heat have a fundamental limit and whether changing the working substance can improve their output. Pages 9–11 describe heating and cooling in real machines, but use the historical caloric theory, rather than the modern first law of thermodynamics balance. The historical discussion under Carnot's theorem explains that distinction.

Heat-engine calorimetry measures heat transfers so that observed work can be compared with the correct energy input. Measuring motion alone cannot supply the denominator of thermal efficiency.

The institutional Laboratory 16: Stirling Engine describes a modern teaching arrangement: an ethanol lamp drives a Stirling engine coupled to an electrical output device, which supplies a resistor immersed in water in a calorimeter. The prescribed readouts include fuel mass consumed, the pressure-volume loop, electrical readings, and water temperature. The sequence separates fuel input, gas-cycle work, electrical output, and Joule heating of the water. Warming of the water is the final useful output; it is not the engine's rejected $Q_c$. The handout distinguishes warm-up from steady operation and asks students to compare the final warming with direct heating by the lamp. It gives a procedure, not a table of measured efficiencies. Its pp. 1–3 therefore establish an apparatus and measurement design without supplying a new experimental result.

For a proposed calibration, use a known input of electrical work to determine the effective heat capacity $C_{\mathrm{eff}}$ of the water, vessel and immersed parts; stir to make the measured temperature representative. A matched run with no input estimates exchange with the room. For constant $C_{\mathrm{eff}}$ and no phase transition, the resulting model iswhere $Q_{\mathrm{leak},\mathrm{out}}$ is the signed net heat lost to the room during the measurement. Here $E_{\mathrm{in}}$ is supplied electrical work when the resistor lies inside the calibration boundary; for a separate external heat input it is the received heat. These are proposed controls, not controls reported in that handout. Slow cooling, changing material properties and imperfectly matched room exchange limit this inference; the approach to thermal equilibrium and heat capacity matter as much as the thermometer reading.

To measure the actual engine-input $Q_h$, one must instrument its hot-side transfer separately, or close a calibrated energy balance around a specified heater boundary. A heater's electrical work includes changes in its stored internal energy and heat transfer that bypasses the engine. At periodic operation a cold-side flowing-water measurement can instead estimate ${\dot{Q}}_c=\dot{m}{c}_p(T_{\mathrm{out}}-T_{\mathrm{in}})$, if mass flow rate $\dot{m}$ and specific heat capacity at constant pressure $c_p$ are known, with negligible storage and other heat exchange in that measurement boundary. It measures only the rejection captured by that stream. Include all remaining transfers before using the residual to infer ${\dot{Q}}_h=P_{\mathrm{out}}+{\dot{Q}}_c$. This measurement discipline makes a comparison with Carnot's theorem meaningful.

The heat input to a cycle sums positive transfers of heat into its working substance. It is not the net heat over the entire cycle, which equals net work. A single geometric branch can contain both heat absorption and rejection, so grouping the branch into one net transfer can overestimate a heat engine's thermal efficiency.

No cyclic device can have as its sole effect the extraction of heat from one thermal reservoir and its complete conversion into net work. A putative device absorbing $Q_h>0$ and returning to its starting state with $W=Q_h$ and no other transfer would conserve energy, yet violate the Second law of thermodynamics. Rejecting heat to colder surroundings is an essential part of the ordinary two-thermal reservoir heat engine budget.^[ref]

The cyclic and sole-effect qualifications matter. A single isothermal expansion of an ideal gas can absorb heat and deliver equal work while leaving its volume changed; restoring the initial state must also be counted. Likewise, a device discharging stored internal energy or consuming fuel has another effect. Converting supplied electrical work completely to heat through Joule heating reverses the proposed conversion and is allowed. The statement also allows cyclic devices that exchange heat with one thermal reservoir while consuming net work; it forbids positive net work output with that single heat source as the only external change.

Conversely, a hypothetical complete-conversion device could take $100\mathrm{J}$ from the hot thermal reservoir and supply $100\mathrm{J}$ of work to an ideal refrigerator that extracts $100\mathrm{J}$ from the cold thermal reservoir and delivers $200\mathrm{J}$ to the hot one. At $600\mathrm{K}$ and $300\mathrm{K}$, that refrigerator is itself admissible: its reservoir entropy changes cancel. The composite would consume no external work and move a net $100\mathrm{J}$ from cold to hot, violating the Clausius statement of the second law. Together with the construction there, this proves the equivalence of the two impossibility statements.^[ref]

For the two fixed positive-temperature thermal reservoir hypotheses, Carnot's theorem gives the sharper restriction $\eta \le 1-T_c/T_h<1$. A reversible thermodynamic process reaches that bound in the ideal Carnot cycle; an ordinary device's smaller thermal efficiency reflects its entropy production and measurement boundary. An inefficient observed engine is compatible with the Second law of thermodynamics, rather than evidence that the bound has been experimentally determined.

The total work by the gas is the enclosed triangular area, $W=\frac{1}{2}{p}_0{V}_0$. The positive heat on the sloping leg isAdd $Q_{AB}=48p_0{V}_0/32$, excluding the emitting part of $BC$ and all of $CA$. HenceAs an independent conservation of energy check, the heat emitted on late $BC$ is $p_0{V}_0/32$ and on $CA$ is $80p_0{V}_0/32$; thus $Q_{\mathrm{in}}-Q_{\mathrm{out}}=16p_0{V}_0/32=W$. Using only the net heat on $BC$ would give an incorrect thermal efficiency.

Macroscopically, an adiabatic process exchanges no heat with its surroundings: $\delta Q=0$. A reversible adiabatic process is therefore isentropic, since $dS=\delta Q_{\mathrm{rev}}/T=0$. Microscopically, the system is thermally isolated, so changes in its energy arise from work done by changing external parameters rather than random energy exchange with a heat bath; in a sufficiently slow change, the populations of the corresponding microscopic states are carried along without producing entropy.

For repeated operation the working substance must execute a thermodynamic cycle. Reversible isothermal contact at $T_2$ absorbs heat $Q_2$, a reversible adiabat changes the temperature to $T_1$, reversible isothermal contact at $T_1$ rejects heat $Q_1>0$, and a second adiabat returns to the initial state. This is the Carnot cycle. Reversibility and zero net entropy change giveThe thermal efficiency is the net work output divided by the absorbed heat:ConsequentlyThe Carnot theorem shows that any irreversible cycle has smaller efficiency, so the two isotherms joined by two adiabats are optimal.

In the $(p,V)$ plane, the Otto cycle consists of:

It is a clockwise loop, so its enclosed area is the positive net work output. In the $(T,S)$ plane, the two reversible adiabats $A\to B$ and $C\to D$ are vertical constant-entropy segments. Constant-volume heating $B\to C$ moves upward and to the right, while constant-volume cooling $D\to A$ moves downward and to the left.

LetThe two adiabatic relations giveBecause heat is exchanged at constant volume,where $Q_2$ denotes the positive amount rejected. HenceThe thermal efficiency is thereforeFor a fixed gas, $\gamma$ is fixed and the idealized efficiency increases monotonically with the compression ratio $r$, so it is maximized by making $r$ as large as the physical constraints allow.

Heat enters only during the constant-pressure step, so the heat capacity at constant pressure givesThe positive magnitude of the heat rejected during constant-volume cooling isBecause the working gas returns to its initial state, the thermodynamic cycle has zero net change in internal energy and $W=Q_1-Q_2$. Its thermal efficiency is therefore

burnup refers to thermal energy per initial heavy-metal mass. Correct the electrical rating for thermal efficiency before using it, and apply the capacity factor once:Consequently the annual initial heavy-metal fuel requirement isLet $F,P,W$ be feed, enriched product and enrichment tails masses, all measured as uranium, and $x_f,x_p,x_w$ the corresponding uranium-235 mass fractions. The isotopic mass balance givesWith $P=M$, the uranium oxide concentrate requirement is thereforeThese are uranium or heavy-metal masses, not tonnes of uranium oxide. The check $327.394(0.007)=35.394(0.040)+292.000(0.003)$ closes the uranium-235 balance as well as the total mass balance.

A heat engine's thermal efficiency is net work out divided by heat absorbed,where both heat quantities are positive magnitudes. The last equality follows from the first law over a complete cycle.

Use $n$ moles of an ideal gas with constant molar $C_V,C_p$ and $\gamma =C_p/C_V>1$. For the Otto cycle, reversible adiabats give $T_2=T_1{r}^{\gamma -1}$ and $T_3/T_4=r^{\gamma -1}$. Isochoric heat input and rejection are $Q_{\mathrm{in}}=n{C}_V(T_3-T_2)$ and $Q_{\mathrm{out}}=n{C}_V(T_4-T_1)$. ThereforeThe denominator's temperature difference is $r^{\gamma -1}$ times the numerator's difference.

For the ideal Atkinson cycle, heating at fixed volume gives $T_3/T_2=r_p$. The longer expansion ends when $p_{4^{\prime }}=p_1$. Combining the adiabatic pressure–temperature relation with $p_2/p_1=r^{\gamma }$ yieldsAlso $V_{4^{\prime }}=V_1{r}_p^{1/\gamma }>V_1$ for $r_p>1$. The PDF asks explicitly for $T_3,T_{4^{\prime }}$; the transcription additionally names $T_2$, which is supplied above too.

Heat input remains $n{C}_V{T}_1{r}^{\gamma -1}(r_p-1)$, while the isobaric return rejects $n{C}_p{T}_1(r_p^{1/\gamma }-1)$. Hence the Atkinson efficiency advantage from extended expansion is quantified byCompare cycles with the same initial state, compression ratio and heat input. The extra expansion and constant-pressure return enclose positive extra work,It uses heat that the Otto cycle would reject, while $Q_{\mathrm{in}}$ is unchanged. Equivalently, concavity gives $\gamma (r_p^{1/\gamma }-1)<r_p-1$, so ${\eta }_A>{\eta }_O$ for $r_p>1$ under this matched comparison. This is an ideal-cycle result; the question's named real-engine application is not needed to establish it.

A Carnot cycle consists of two reversible isothermal branches joined by two reversible adiabatic branches. Let the hot isothermal expansion run from volume $V_a$ to $V_b$, and the cold isothermal compression from $V_c$ to $V_d$. For $n$ moles of an ideal gas, their absorbed and rejected heat magnitudes areOn a reversible ideal-gas adiabat, $T{V}^{\gamma -1}$ is constant. Applying this to the two joining branches gives $V_b/V_a=V_c/V_d$. ThereforeThe result is independent of the gas's heat-capacity ratio. This derivation presumes the usual constant-heat-capacity ideal-gas model; more generally the equal entropy changes on the two isotherms give the same heat ratio without that restriction.

To establish the upper bound, suppose an engine were more efficient between these same two reservoirs. Use its output work to drive a reversed Carnot cycle, scaled so that the work magnitudes match. For fixed work $W$, the hypothetical engine would take $W/\eta$ from the hot bath, whereas the reversed Carnot cycle would return $W/{\eta }_C>W/\eta$ to it. The combined device would transfer heat from the cold bath to the hot bath without any net external work. That contradicts the Clausius statement of the second law. Hence no such two-reservoir engine can exceed $1-T_2/T_1$.

For the triangular cycle, call the three successive vertices $A,B,C$, and measure energies in units of $p_0{V}_0$. For one mole of a monatomic ideal gas, $U=3pV/2$, and positive work done by the gas is $W=\int pdV$. The net values on the three complete branches areThus the net work is $W_{\mathrm{cycle}}=p_0{V}_0/2$, equal to the triangle's enclosed area.

There is a small but important distinction between the net heat on a branch and its total positive heat input to a cycle. On the sloping branch put $x=V/V_0$, so $p/p_0=3-x$, with $1\le x\le 2$. ThenThere is a heat-flow reversal along a thermodynamic path at $x=15/8$. The first part absorbs heat; the final part rejects it. Integrating the two signed contributions separately givesThe true total absorbed and rejected heats and thermal efficiency are thereforeFor clarity, the positive heat input on $A\to B$ is $3p_0{V}_0/2$, on $B\to C$ it is $49p_0{V}_0/32$, and on $C\to A$ it is zero. Using just the positive net heats of the complete branches would give $Q_{\mathrm{in}}=3p_0{V}_0$ and $\eta =1/6$; that shortcut misses the short heat-rejection interval on $B\to C$.

Finally, put $T_0=p_0{V}_0/R$. The minimum temperature is $T_0$ at $A$. Along $B\to C$,which has its maximum $9/4$ at $x=3/2$, not at either endpoint. Thus the corresponding two-reservoir Carnot efficiency isThe triangular cycle receives heat at a range of temperatures below $T_{\max }$ and rejects it at temperatures above $T_{\min }$. It therefore makes less work per unit input heat than a Carnot cycle using only those extrema. The geometric path alone does not prove irreversibility: it could in principle exchange heat reversibly with a continuum of matched-temperature reservoirs, while still having this lower efficiency. Restricting it to only the two extreme-temperature reservoirs would introduce additional irreversible heat transfers.

A Carnot cycle consists of four reversible stages: isothermal expansion at the hot-reservoir temperature $T_h$, adiabatic expansion to $T_c$, isothermal compression at $T_c$, and adiabatic compression back to the starting state. The adiabatic stages exchange no heat. For one mole of an ideal gas, the two isothermal stages have heats $Q_h=R{T}_h\ln (V_B/V_A)$ and $Q_c=R{T}_c\ln (V_C/V_D)$ in magnitude. The reversible adiabatic relations ensure $V_B/V_A=V_C/V_D$.

More generally, the total entropy change over any reversible cycle is zero. The two heat exchanges therefore satisfyWith work output $W=Q_h-Q_c$, its thermal efficiency isThe temperatures must be absolute. The second law makes this the maximum efficiency for an engine exchanging heat only with those two reservoirs; irreversible entropy production lowers the efficiency.

Because the working substance returns to its original thermodynamic state, its net entropy change is zero. There is no entropy change on either reversible ideal-gas adiabat; on each isothermal process the entropy transfer is the exchanged heat divided by the bath's absolute temperature. HenceThe first law of thermodynamics over a complete cycle gives $W=Q_h-Q_c$. The thermal efficiency is thereforeThe use of kelvins is essential. This result does not depend on the working substance. For the ideal gas illustration, the two reversible ideal-gas adiabats imply $V_C/V_B=V_D/V_A=(T_h/T_c{)}^{1/(\gamma -1)}$, so $\ln (V_B/V_A)=\ln (V_C/V_D)$; the explicit heat expressions in (a) give the same ratio.

An ideal Rankine cycle has four main steps: liquid water is pumped reversibly to high pressure; heat is added at high pressure to produce hot vapour or supercritical working fluid; the fluid expands approximately isentropically through a turbine to produce work; and low-pressure heat rejection in a condenser returns it to liquid. For steady flow, the main work and heat transfers are differences of specific enthalpy.

A steam implementation of the Carnot cycle would require compression of a wet liquid-vapour mixture and reversible isothermal heat transfer, both difficult to achieve with practical machinery. Complete condensation before the Rankine pump makes compression work small because liquid specific volume is small. High-pressure heating and superheating can then occur in a boiler without demanding isothermal addition of all the heat. Superheating also reduces turbine-exit moisture, although the calculated exit may still be wet. The pump, turbine, condenser and boiler can operate continuously as separate flow devices. The Rankine cycle sacrifices some ideal thermal efficiency in return for a practicable steam-plant configuration; no real engine can exceed the Carnot efficiency for the same extreme reservoir temperatures.

A vapour-power cycle using a liquid pump, high-pressure heat addition, turbine expansion and low-pressure condensation. Ideal components are reversible apart from heat transfer through finite temperature differences. With states numbered pump inlet, pump outlet, turbine inlet and turbine outlet, the net thermal efficiency is $[(h_3-h_4)-(h_2-h_1)]/(h_3-h_2)$. Liquid pumping avoids the impractical compression of a wet two-phase mixture in a steam realization of the Carnot cycle. Named after William John Macquorn Rankine.

Thermodynamics studies energy transfers and the constraints on macroscopic changes. Its vocabulary connects internal energy, entropy, temperature, pressure and volume to heat and work. The first law of thermodynamics balances energy; the Second law of thermodynamics decides which changes can occur spontaneously and limits how much work a heat engine can deliver. These principles let us bound performance before choosing a material or a detailed machine design. Applications include power generation, propulsion and refrigerators. ^[ref]

Start by choosing a system boundary. A closed system can exchange energy but no matter; an isolated system exchanges neither. A state function, such as internal energy or entropy, depends on the equilibrium state. A process specifies how that state changes. Thus a body's temperature is a property it has, whereas heat records energy crossing its boundary during a process. A body contains internal energy, rather than a stored quantity of heat.

Two insulated bodies initially at different positive temperatures approach thermal equilibrium when brought into thermal contact. Their combined energy is conserved, but their combined entropy increases. They do not spontaneously separate again into a hotter body and a colder one. This familiar asymmetry makes thermodynamics interesting: conserving energy leaves many imaginable changes available, and entropy supplies a further constraint. The same distinction underlies both entropy maximization under thermal contact and the thermal efficiency limit in Carnot's theorem.