Theory Manual

M01State Explorer

Two independent properties fix the state of a simple compressible substance. This screen accepts any admissible pair and reports the rest, with the phase region named rather than implied.

What it rests on

The state postulate: two independent intensive properties fix the state of a simple compressible substance, and every other property then follows. That is the whole content of this screen, and the qualifier "independent" is where all its difficulty lives -- pressure and temperature stop being independent the moment the substance is a liquid-vapour mixture, and no arithmetic downstream recovers from a pair that did not fix a state in the first place.

How the working relation follows

For water the properties are not fitted one by one. A single dimensionless potential is fitted per region -- the Gibbs function of reduced pressure and temperature in the liquid and vapour regions, the Helmholtz function of reduced density and temperature in the region around the critical point -- and every property reported here is a derivative of that one function. Specific volume, internal energy, enthalpy, entropy and the specific heats are therefore consistent with one another structurally, not by maintenance: they cannot disagree, because there is only one function and they are different derivatives of it.

That is also why the app carries the potentials rather than a table of numbers. A table has to be interpolated, and interpolation between tabulated enthalpies does not generally satisfy the same differential relations the enthalpies themselves do; the derivative-based identities that this screen checks would fail on interpolated data for reasons that have nothing to do with the substance.

Inside the vapour dome the state is not on either fitted surface. It is a mechanical mixture of two states that are, and the lever rule is the statement that any extensive property of the mixture is the mass-weighted average of its values at the two ends of the tie line. Writing the mixture enthalpy as the saturated-liquid value plus quality times the latent value, and reading the same relation backwards to get quality from any property and its two saturation values, is the entire two-phase treatment.

The saturated states are not separate correlations either. Saturated liquid is the liquid-region potential evaluated on the saturation line and saturated vapour is the vapour-region potential evaluated on the same line, which is how the formulation defines them -- so consistency between the phases is again structural. Above about 623 K the saturation line runs through the critical region rather than bounding it, and both of its sides are then states of the density-explicit potential; the app finds them by a bracketed root find on each side of the critical density rather than by continuing to evaluate the liquid and vapour potentials outside their own domains.

Assumptions

• The substance is simple and compressible: no surface, electrical, magnetic or elastic work modes, so two independent properties are enough.
• Equilibrium. Every state reported is one the substance would settle into and stay in; nothing here describes a substance in the middle of relaxing towards one.
• The two-phase region is a mechanical mixture at a single temperature and pressure, with no interfacial energy and no metastability. A superheated liquid or a subcooled vapour is outside the model.
• This state-explorer entry uses water. Refrigerant properties are available in the corresponding refrigeration and real-fluid tasks, not by changing the water formulation constants.
• The reported values are the international formulation's values. Its own stated uncertainty against experiment is a separate and larger quantity than the agreement between this app and the formulation.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• h = u + pv holds at every reported state, in every region.
  Verification: property test over the whole validated (p,T) rectangle
• Fixing a state, reading (p,h) from it, and re-fixing from (p,h) returns every other property to solver tolerance.
  Verification: hypothesis round trip
• T ds = du + p dv and T ds = dh - v dp hold numerically along a refined path.
  Verification: central difference on a fine grid, compared to the analytic derivatives
• At the saturation line, evaluating from the liquid side and from the vapour side gives the same T and p.
  Verification: region 1 / region 4 / region 2 agreement

Monotonic trends:

• Saturated vapour specific volume falls monotonically as saturation pressure rises.
• The latent enthalpy falls monotonically with temperature and reaches zero at the critical point.

Where it gets delicate

How it is solved

The entry provides nine pairs: p–T, p–x, T–x, p–h, p–s, T–v, p–v, h–s and u–v. The p–v inversion exposes a temperature-root choice where water's density anomaly admits alternatives; an unresolved ambiguous root is not silently chosen. The h–s and u–v searches also validate phase and property closure.

Every pair is reduced to one of the two directions the property potentials evaluate directly. Pressure and temperature is a direct evaluation once the region has been decided. Pressure with enthalpy or entropy, and the pairs that involve specific volume, are inverted by bracketed root finding on the forward relation.

The published backward equations are deliberately not used. They exist so that industrial simulators can avoid an iteration they perform billions of times; this app evaluates a state a few thousand times a second at the very most, driven by a slider. What they would cost is roughly a hundred and fifty more coefficients transcribed by hand -- and the one transcription error this project has actually found took a published nine-figure verification table to catch. Iterating on relations that are already verified to nine significant figures is slower by an amount no user can perceive and introduces nothing new to get wrong.

Two decisions inside that search are not obvious. The two-phase test comes first when inverting from pressure and enthalpy or entropy, because inside the dome the property is constant along the isobar and not invertible in temperature at all; a root finder asked to invert it there reports a bracket failure for what is in fact the commonest state in a power cycle. And the test uses a small relative window on the latent span rather than the exact saturation endpoints, because those endpoints are themselves computed: an enthalpy a few parts in ten thousand million million below the saturated-liquid value is the same state, not a different one, and without the window it fell through to a single-phase search that correctly found nothing.

In the density-explicit region the state has to be found by iteration in density. Below the critical temperature that region straddles the saturation line and an isotherm crosses the target pressure twice, once on each side. Which crossing is physical is decided by comparing the requested pressure with the saturation pressure, not by whichever root the solver reaches first -- getting that wrong returns a perfectly convergent answer for the wrong phase.

Limitations

• Outside the validated pressure and temperature rectangle the app refuses rather than extrapolating. Extrapolating a fitted potential converges, looks reasonable, and is wrong by amounts that grow quickly.
• Within about six tenths of a kelvin of the critical temperature the two-phase relations are withdrawn. The latent quantities have collapsed far enough there that quality carries no information: it becomes zero divided by zero and its condition number is unbounded.
• Quality is reported only when it is strictly between zero and one. Printing a quality of exactly 1 for superheated vapour is not merely redundant -- it teaches that superheated vapour has a quality.
• Pressure and temperature do not fix a state on the saturation line. The app says so and asks for a different pair rather than picking a quality.
• At the seam between the two liquid-side potentials, at about 623 K, the two are independently fitted and enthalpy steps by roughly one part in a hundred thousand across it. Inverting enthalpy or entropy within a few thousandths of a kelvin of the seam can land on either side. The step divided by the local slope bounds that ambiguity at about five thousandths of a kelvin, which is below the last digit the app displays.
• The published formulation is for water. Nothing on this screen generalises to another substance by changing a constant.

Reading the result

Read the phase label first. A condenser exit and a boiler exit are both "in the two-phase region" and could hardly be more different, so the label distinguishes the ends of the dome rather than lumping them; and if the label says the state is single-phase, the absence of a quality is information, not an omission.

The inputs that go dead when you choose a pair are the second thing to read. Choosing pressure and quality greys out temperature because the saturation line has already fixed it -- the app says which, in those words, because a greyed field with no explanation reads as a broken program rather than as a consequence of what you just chose.

Treat a refusal as an answer. "No state at this pressure has that enthalpy, and here is the range that is reachable" is more useful than a number, and it is what you get instead of a silently clamped one.

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M02Ideal Gas Properties

Temperature alone fixes every caloric property of an ideal gas. The screen shows the variable-specific-heat result and the constant-specific-heat result side by side, because the gap between them is the thing being taught.

What it rests on

For a gas at low enough density the molecules are far enough apart that their mutual attraction contributes nothing to the internal energy. The consequence is the one this screen is built on: temperature alone fixes every caloric property. Enthalpy, internal energy and the specific heats are functions of temperature and of nothing else, and only entropy retains a pressure dependence, which is a counting statement about how many ways the molecules can be arranged rather than anything to do with their forces.

How the working relation follows

The specific heat at constant pressure is supplied as a fitted polynomial in temperature, divided by the universal gas constant to make it dimensionless. The active data use the NASA nine-coefficient form with inverse-temperature terms and analytic enthalpy and entropy integrals. Coefficient intervals have published validity limits and joins; finite rounding of their integration constants can leave small jumps. Such a join is a property of the source fit, not a reason to extrapolate an interval or silently change its constants.

Enthalpy is the integral of that polynomial with respect to temperature, and the standard entropy is the integral of the polynomial divided by temperature. Single-state properties use the published analytic integrals. The separate two-state entropy-change calculation integrates cp/T across coefficient joins, so it must not be confused with direct subtraction of independently fitted integration constants. The integration constants are the two extra coefficients published with each fit, which is what puts the enthalpy zero at the standard formation reference rather than at zero kelvin.

Internal energy follows from enthalpy without a second fit, because the ideal-gas relation makes the difference exactly the gas constant times temperature; and the two specific heats differ by exactly the gas constant for the same reason. So of the five caloric quantities on the screen only two are fitted and three are consequences, which is why an error in the fit shows up in all of them at once and an error in the algebra shows up in one.

Entropy needs the pressure term. Between two states it is the difference of the two standard entropies minus the gas constant times the logarithm of the pressure ratio. Setting that difference to zero and rearranging gives the isentropic relation without any assumption that the specific heat ratio is constant -- and defining a relative pressure as the exponential of the standard entropy divided by the gas constant turns the whole isentropic move into a single ratio. The relative volume is the corresponding quantity for a move at fixed entropy between two volumes. Both are defined only up to a multiplicative constant, because every use of them is a ratio; the app normalises both to one at the reference temperature, which also keeps the exponential from overflowing at the top of the fitted range.

Assumptions

• The gas is ideal: no intermolecular forces and no molecular volume, so the pressure, volume and temperature relation is exact and the caloric properties depend on temperature alone.
• The species is chemically frozen. Above about 2500 K a real gas begins to dissociate and the polynomial for the intact molecule keeps returning values for a species that is no longer all there.
• Air is treated as a fixed-composition mixture of its components evaluated through the same polynomials, not as a separately fitted pseudo-species. A separate air fit would be one more thing that can drift away from its own components.
• The enthalpy and entropy zeros are the standard formation reference carried by the published coefficients, which matters the moment a calculation involves a chemical change and cancels exactly when it does not.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• c_p - c_v = R at every temperature in range.
  Verification: direct, all species
• 1 < k <= 5/3 for every species and temperature in range.
  Verification: property test
• p_r and v_r reproduce the isentropic relation: v_r(T2)/v_r(T1) = v2/v1 when s2 = s1.
  Verification: cross-check against the constant-k form in the limit of constant c_p
• The enthalpy reference is declared and is consistent throughout a calculation. Non-reacting screens may use any zero; reacting screens must use the formation reference.
  Verification: a reacting and a non-reacting module fed the same species must differ by exactly the formation enthalpy, never by anything else

Monotonic trends:

• Specific heat at constant pressure increases with temperature for every polyatomic species in range.
• The specific heat ratio decreases with temperature for every diatomic and polyatomic species below about 4800 K. Argon is monatomic and its ratio is exactly constant. Above roughly 4800 K the polynomial fits turn over -- nitrogen at 4995 K, carbon monoxide at 4970 K, carbon dioxide at 4904 K, methane at 4824 K, air as a mixture at 5181 K -- while hydrogen, water vapour and oxygen stay monotone to the top of the range. Measured, and above any temperature a gas-power screen reaches.

Where it gets delicate

How it is solved

For a single state, the variable-specific-heat branch directly evaluates the NASA functions. The constant-cp branch evaluates cp at the stated reference temperature and integrates that constant from a common formation-reference anchor. Cold-air standard uses the declared air heat-capacity ratio and is available only for air. All comparisons keep the same property reference and mass or molar basis, so changing a model is not a change of units.

The two-state entropy change is computed separately. For variable cp, the thermal contribution is the integral of cp/T, split at coefficient joins; this avoids subtracting rounded entropy integration constants across a join. Constant-cp models use the logarithmic temperature ratio. A conditioned logarithm evaluates the pressure contribution, and the screen displays the combined entropy change. The plotted heat-capacity interval follows the two entered temperatures. Out-of-range states are refused rather than extrapolated.

Limitations

• The fitted range is 200 K to 6000 K. Below or above it the app refuses rather than extrapolating.
• Nothing here knows about dissociation, ionisation or condensation. A polynomial evaluated at 5000 K for a diatomic gas returns the property of a molecule that at that temperature has substantially come apart.
• The species set is the one the coefficient data carries. An unlisted species returns nothing rather than a guess.
• Agreement with NASA coefficients checks numerical evaluation of that model. It is distinct from comparison with independent measurements or tables; there is no single blanket accuracy percentage for all species, temperatures and caloric-model choices.
• The variable model generally has a temperature-dependent heat-capacity ratio; constant-cp and cold-air-standard models deliberately remove that variation. No monotonic trend should be extrapolated beyond the coefficient range.
• The ideal-gas assumption itself is the real limit. Near the critical point of any of these species the relations on this screen are not approximately right, they are the wrong model, and the compressibility screen is where that is quantified.

Reading the result

Read the three specific-heat models against each other rather than one at a time. The screen shows the polynomial answer, the constant-specific-heat answer at a stated evaluation temperature, and the cold-air-standard constant ratio, and the gap between them is what the cold-air assumption costs -- a gap that grows with the temperature span of whatever cycle you are about to feed these numbers into.

The relative pressure and relative volume are ratios and nothing else. Their individual values carry no physical meaning at all -- a printed table picks a different normalising constant and its numbers differ from these -- but the ratio of either quantity between two temperatures is the same in both, and that ratio is the answer to an isentropic move.

Enthalpy and internal energy on this screen are absolute values on the formation reference. If you are comparing against a table whose zero is somewhere else, compare differences, not values.

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M34Real Gas: Compressibility and Cubic Equations of State

Compare four cubic equations and a branch-checked Lee-Kesler correlation with ideal gas and IF97 water, using molar-volume readings and sampled Z/volume-difference isotherms with explicit gaps at unresolved states and branch changes. The native comparison fixes water and canonical critical/acentric data; volume and EOS choice are backend capabilities, while this screen compares all eligible models from entered p and T.

What it rests on

The compressibility factor is the ratio of the actual product of pressure and molar volume to what the ideal-gas relation would give at the same temperature. It is one, exactly, for an ideal gas, so its distance from one is a direct measure of how far a state has left the model the gas screens rest on. The corresponding-states principle is the second idea: expressed in coordinates reduced by their own critical constants, substances of similar molecular shape follow nearly the same compressibility surface, which is what makes a generalised correlation possible at all.

How the working relation follows

The four cubics share p=RT/(v-b)-a(T)/[(v+epsilon b)(v+sigma b)]. Critical conditions determine their attraction and covolume constants. Multiplying the denominators produces a cubic in the dimensionless compressibility factor. This avoids dimensional coefficient scaling but does not remove all cancellation.

Three distinct roots define competing liquid-like, unstable and vapour-like branches. They do not define saturation: coexisting phases must also have equal fugacity. The outer roots can be metastable. At a critical or spinodal multiple root the stability derivative vanishes, and numerical root locations become sensitive.

Assumptions

• All curves use water at the same temperature and pressure. IF97 is a reference formulation, not experimental truth.
• The present water reference covers 273.15–1073.15 K and 0.000611213–100 MPa; Region 5 is not implemented. Saturation p,T do not specify a mixture quality. The current exact critical p,T handling also refuses a reference state.
• The Lee-Kesler equation solves its two fitted fluids separately and interpolates responses by acentric factor. The native comparison accepts only matching, resolved stable branches; equal crossing counts do not prove this.
• Molar volumes are in m³/kmol in both display-unit systems. Compressibility, reduced pressure and fractional or percent volume differences are dimensionless.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• On the dilute vapour branch, Z tends to 1 as pressure tends to zero. A liquid-like root does not obey this limit.
  Verification: limit test over all four cubics and the chart; this is the single best check that a coefficient is not mistyped
• The defining critical conditions give zero first and second volume derivatives. Floating-point cubic roots can split near the multiple root, so root-location accuracy is conditioned separately.
  Verification: Critical-condition identities and independent high-precision roots; do not assert machine-relative accuracy of every split root.
• The three roots of the cubic, when all real, satisfy the Vieta relations for its coefficients.
  Verification: algebraic identity test on the solver output
• For three distinct physical roots, the middle branch has positive dp/dv and is mechanically unstable. At a multiple root the derivative can vanish.
  Verification: property test

Monotonic trends:

• At fixed reduced temperature below about 2.5, the compressibility factor first falls below 1 and then rises above it as reduced pressure increases.
• Agreement between the cubics and the reference formulation degrades as the critical point is approached.

Where it gets delicate

How it is solved

Cubic roots use the existing closed trigonometric form for three real roots and Cardano for one, with a coefficient-scaled seam tolerance. Roots below the covolume are excluded. The comparison classifies stable outer branches using the pressure derivative and the unstable separator through the model critical volume; it leaves near-fold numerical ambiguity unresolved.

For each Lee-Kesler fitted fluid, branch verification bounds the pressure derivative on the entire connection to the dense and dilute limits. Bounds include the analytic extrema of the damped power terms. Adaptive subdivision proves a connection only when the interval upper bound is negative. Its finite iteration budget can leave a result unresolved near critical conditions; that is not evidence of a different phase. The raw correlation solver exposes the interpolated numeric response and branch metadata; this comparison applies the additional matching-branch acceptance rule.

The isotherm evaluates a logarithmic pressure grid and includes the current pressure when it is inside the plot interval. IF97 saturation pressure and nearby one-sided states are added when applicable. Segments stop at missing values, unresolved branch verification, branch changes and crossing-count changes. Volume-difference curves also stop at reference-phase changes. The chart preserves segments as separate series with one colour and legend per model. Sampling is not a continuous-domain accuracy or stability proof.

Limitations

• Low reduced pressure alone does not guarantee ideal-gas accuracy, particularly when the water reference is liquid. The screen reports actual current differences.
• A selected model root and the IF97 reference can represent different phases. Their volume difference is not a same-phase model-error measurement.
• Near a multiple root, coefficient rounding can split the numerical roots by about 1e-8. Do not read the displayed significant figures as a root-accuracy guarantee.
• At minimum reference pressure and 300 K, a tiny liquid root can lose roughly 3e-8 relative accuracy in Z and volume. Pressure reconstructed from that volume is much more ill-conditioned and can differ by tens of percent; this does not mean the volume itself is wrong by that amount.
• Cubic liquid densities are approximate and have no volume-translation correction. No model is declared globally more accurate merely because it is more elaborate.
• Lee-Kesler branch verification has a finite numerical domain and subdivision budget. Unresolved or mismatched fitted-fluid branches are omitted with a reason; the other models remain available.

Reading the result

Enter water pressure and temperature, then select the largest or smallest candidate root. Read the IF97 phase and each model’s own root count and stable-branch label. Models are compared together; none is highlighted as a preferred choice.

Choose the pressure interval and either compressibility or relative molar-volume difference for the isotherm. The horizontal axis is logarithmic reduced pressure. The current IF97 point is marked only when its reference exists and lies inside the interval. Gaps remain gaps in PNG, PDF and SVG. The table remains the source for isolated valid states that cannot form a line segment.

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M35Departure Functions

How to get an enthalpy or entropy change for a real gas: take the ideal-gas change, then correct it at each end. The correction is what the generalised charts hold, and here it is computed rather than read off a figure.

What it rests on

A departure function is the difference between a real-fluid property and its ideal-gas value at the same state coordinates. It permits ideal-gas caloric information and a nonideal equation of state to be combined without confusing a property difference with a path-dependent heat or work transfer.

How the working relation follows

Connect each real endpoint to an ideal-gas reference endpoint through an imaginary zero-density path. Since enthalpy and entropy are state properties, the complete change equals the ideal-gas change plus the difference between departure corrections at the two endpoints. The imaginary path is a device for deriving the correction; it is not a proposed physical process. Cubic-equation departures follow analytically from the equation's temperature dependence, and fugacity expresses the corresponding chemical-potential correction.

The generalized route instead evaluates the declared corresponding-states relation. It and the cubic route are alternative property models, so their difference is a model comparison. Agreement of their ideal-gas limits does not make their dense-fluid predictions interchangeable.

Assumptions

• The same species and reference convention are used at both endpoints.
• The chosen equation of state or generalized relation defines the nonideal correction; its parameter and phase ranges remain applicable.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• Both departures go to zero as pressure goes to zero, exactly.
  Verification: limit test
• The enthalpy and entropy departures obtained from a cubic satisfy the Maxwell consistency relation that connects them.
  Verification: symbolic consistency check on each cubic
• The fugacity coefficient goes to 1 as pressure goes to zero, and its logarithm is the integral that also produces the entropy departure.
  Verification: identity test
• The total change between two states is independent of the intermediate ideal-gas path used to compute it.
  Verification: path independence test; this is the property the whole construction relies on

Monotonic trends:

• The enthalpy departure grows with reduced pressure and shrinks with reduced temperature.

Where it gets delicate

How it is solved

Each endpoint is solved with the selected model before evaluating its departure. The ideal-gas change is evaluated separately from NASA caloric data, and the result displays both endpoint corrections and their net contribution. The native entry selects the gas-phase root; it has no liquid-root selector. Invalid endpoint roots are not replaced by ideal-gas values.

Limitations

• A thermodynamically consistent departure is only as accurate as the underlying equation of state. Cubic models can be poor near critical conditions and for strongly nonideal liquids.
• Property changes do not determine heat or work without specifying an actual process and energy balance.

Reading the result

Read the total together with the ideal and departure parts. A large cancellation between endpoint corrections can make the small difference sensitive even when each endpoint is finite. When comparing routes, keep the species, temperatures, pressures and property reference fixed.

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M36Thermodynamic Relations

Explore single-phase derivatives and liquid–vapour coexistence using water or a cubic equation of state. Compare available independent derivative routes, and distinguish exact relations from approximations.

What it rests on

Thermodynamic properties are linked because they are derivatives of a common potential. Maxwell relations, response coefficients and phase-boundary slopes provide alternative routes to the same quantities and expose contradictions that independently fitted property formulas can conceal.

How the working relation follows

Equality of mixed second derivatives of a thermodynamic potential produces the Maxwell relations. Combining those derivatives with the definitions of thermal expansion and isothermal compressibility gives the difference between constant-pressure and constant-volume specific heats. Applying the differential enthalpy relation at constant enthalpy gives the Joule–Thomson coefficient: its numerator compares thermal expansion with ideal-gas-like expansion, and its denominator is the actual constant-pressure heat capacity.

Along liquid-vapour coexistence, equal phase chemical potentials remain equal after a small change in temperature and pressure. Subtracting their differentials yields the Clapeyron slope from latent enthalpy and the phase volume difference. Assuming an ideal dilute vapour and negligible liquid volume gives the Clausius–Clapeyron approximation, whose restrictions are stronger than those of the original relation.

Assumptions

• Properties and derivatives come from one selected model and use consistent mass or molar units.
• Single-phase response coefficients require a resolved stable state; coexistence relations use the two saturation endpoints rather than a single-phase derivative at a mixture state.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• The latent enthalpy obtained from the slope of the saturation line equals the difference between saturated vapour and saturated liquid enthalpies.
  Verification: THE self-check of this screen. It spans the compressed-liquid region, the vapour region and the saturation line at once, so a single failure localises to one of three places.
• All four Maxwell relations hold to derivative tolerance for the reference formulation and for every cubic.
  Verification: numerical derivative check plus a symbolic proof for each cubic
• cp - cv is non-negative for every substance at every state, and reduces to R exactly in the ideal-gas limit.
  Verification: property test and limit test
• The general cp - cv relation reduces to R for an ideal gas, exactly.
  Verification: limit test
• The Joule-Thomson coefficient obtained here matches the one obtained from the throttle screen at the same state.
  Verification: cross-module consistency

Monotonic trends:

• Volume expansivity of liquid water is negative below its density maximum and positive above it, which is the one place a sign check on this screen is genuinely surprising.
• The gap between cp and cv widens as the critical point is approached.

Where it gets delicate

How it is solved

The selected water formulation or cubic model supplies the relevant derivatives and heat capacity. The task exposes both direct relations and the corresponding comparison quantities, including latent enthalpy and saturation slope. Invalid phases, degenerate denominators and unresolved critical neighbourhoods are rejected or marked unavailable.

Limitations

• Replacing real-fluid cp by ideal-gas cp changes the Joule–Thomson coefficient even when its numerator is evaluated from a real equation of state.
• A relation may be exact for a model while the model differs from experimental properties. Clausius–Clapeyron adds ideal-vapour assumptions and is not a critical-region approximation.

Reading the result

Check the model, phase and units before comparing two routes. A near-zero Joule–Thomson coefficient indicates inversion, not zero heat capacity. A diverging response close to a critical point may reflect physical conditioning; an unavailable numerical result must not be treated as a finite zero.

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M37Ideal Gas Mixtures

Composition, mixture properties, and adiabatic mixing of two streams. The entropy of mixing is the part that is forgotten, and the screen shows it as its own line so it cannot be.

What it rests on

Ideal-gas mixture properties combine species caloric properties with partial-pressure entropy. Composition fractions require a declared mass or mole basis, and mixing can generate entropy even when the entering streams have equal temperatures and no heat is exchanged.

How the working relation follows

Mole fractions weight molar heat capacities and determine apparent molecular mass; mass fractions weight mass-specific properties. Conversion between those bases therefore uses molecular masses before any averaging. Dalton's relation gives each species partial pressure, which enters its entropy logarithm. Comparing separated pure gases with the same final mixture produces the entropy-of-mixing term proportional to minus the sum of mole amounts times logarithms of mole fractions.

For adiabatic mixing of two streams, conserve each species and total enthalpy. The combined composition is known from the species balances, leaving outlet temperature to satisfy enthalpy conservation. Summing outlet entropy and subtracting inlet entropy then includes thermal equilibration, compositional mixing and any declared pressure effect. Combining identical species first avoids inventing mixing entropy merely because one substance was entered through two ports.

Assumptions

• Every component behaves as an ideal gas and retains its own temperature-dependent caloric data.
• The mixing control volume is adiabatic and uses the declared stream states and outlet pressure; no phase change or chemical reaction is inferred.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• Mole fractions sum to 1 and mass fractions sum to 1, to rounding, after any conversion.
  Verification: closure test
• Converting mole fractions to mass fractions and back returns the original composition.
  Verification: round trip
• Partial pressures sum to the mixture pressure, exactly.
  Verification: closure test
• The entropy of mixing is strictly positive for any mixture of two or more distinct species, and exactly zero for a single species.
  Verification: property test; a mixture entropy that comes out zero for a real mixture means the term was dropped
• With one component at mole fraction 1, every mixture property reduces exactly to that species' property.
  Verification: limit test
• Adiabatic mixing of two positive streams at different temperatures and equal unchanged pressure produces positive entropy. With pressure changes, thermal, composition and pressure terms sum to total entropy production; a negative total is refused.
  Verification: second-law check

Monotonic trends:

• Apparent molar mass moves monotonically between the component values as the composition varies.
• For two distinct ideal-gas species, entropy of mixing per kmol of mixture is maximised at equal mole fractions. The displayed per-kg value also divides by composition-dependent molar mass and need not attain its maximum at equal mole fractions.

Where it gets delicate

How it is solved

Fractions are normalized and converted on the selected basis. The state task evaluates mixture properties directly. The mixing task merges species inventories, brackets outlet temperature on the enthalpy balance and evaluates entropy using actual partial pressures. Thermal entropy is evaluated from each actual species at the recovered outlet temperature, composition entropy from the mole-fraction logarithms, and pressure entropy from each inlet-to-outlet pressure ratio. Invalid inputs, nonfinite balances and negative total entropy production fail explicitly.

Limitations

• An ideal-mixture result is not a real-gas mixing rule for dense gases or liquids.
• Entropy of mixing and total mixer entropy production are different quantities when temperatures or pressures differ.

Reading the result

Use the basis label before entering fractions and compare mass-specific with mass-specific outputs. At equal temperature, mixing different gases can still produce entropy; combining streams of the same gas at the same state should not. The three displayed contributions sum to the total. Pressure changes are kept separate from composition changes; a nonnegative total is necessary but does not solve momentum or establish a realizable mixer geometry.

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M03Process Path

Resolve a reversible closed-system process with an ideal-gas caloric model, water IF97, or given constant c and v; compare boundary work, heat and state changes on p-v and absolute-temperature T-s diagrams.

What it rests on

A closed-system boundary-work integral depends on the path. Internal energy, enthalpy and entropy changes depend on its endpoints. This entry resolves both endpoints and the intervening states with one selected property model, then reports specific quantities and totals for the stated mass. Heat enters the system and work leaves it.

How the working relation follows

For a reversible simple-compressible path with boundary work only, du = T ds - p dv. Thus w = integral p dv and q = delta u + w = integral T ds. A power law p v^n = constant supplies one particular mechanical path. Its n=1 member is an ideal-gas isotherm, while its n=k member is an isentrope only for an ideal gas with constant specific heats. Variable-heat-capacity gas isentropes instead hold the model entropy fixed. Water isentropes likewise solve p,s states; a water isotherm need not hold pv constant. In the wet region its pressure remains the saturation pressure as quality changes.

The user-given incompressible model holds v and c constant. Its changes are delta u = c delta T, delta h = c delta T + v delta p, and delta s = c ln(T2/T1), with pressure-volume products converted to energy units. It has zero boundary work. Temperature is constant on its isothermal and isentropic paths; pressure is constant on its isobaric and finite-exponent polytropic paths. This idealisation has no named material database and no phase-change prediction.

Assumptions

• Fixed mass, quasi-equilibrium states and an internally reversible path. Boundary work is the only work mode; kinetic and potential changes are omitted.
• Choose one of the three ideal-gas caloric models, water IF97, or user-supplied constant c and v. The cold-air model applies only to air.
• The independent initial pair and one path-specific endpoint control fix a process. Additional pressure, volume or temperature values are derived, not arbitrary simultaneous constraints.
• The incompressible model sets initial u and s to zero and retains h=u+pv. Absolute reference values need not match tables with another reference.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• Q - W = Delta-U for every path between the same two states; only Q and W depend on the path.
  Verification: two different paths between one pair of states must give the same Delta-U to rounding
• n=0 reproduces an isobar; as n grows without bound along consistent endpoints the volume displacement and boundary work approach zero.
  Verification: limit test
• For an ideal gas with constant specific heats, n = k reproduces the isentropic result and Delta-S is zero.
  Verification: limit test

Monotonic trends:

• At fixed end volumes and fixed p1, the magnitude of expansion work decreases as the polytropic exponent increases.

Where it gets delicate

How it is solved

The low-level power-law integral uses L=log(v2/v1) and x=(1-n)L, evaluating p1 v1 L expm1(x)/x with the x=0 limit equal to one. Scaled products avoid intermediate overflow. A logarithmic form handles very large positive x. Invalid or contradictory endpoints and unrepresentable nonzero work are refused. The low-level isentropic label is refused because pressure and volume alone cannot establish constant entropy.

The property-path layer samples the selected model. Isentropes solve entropy at successive pressures and use w=u1-u2, with a separately reported sampled integral. Isobars use temperature for gases and enthalpy for water, including saturation endpoints. Isochores resolve states at fixed volume. Water isotherms resolve T,v states and integrate sampled p dv; wet interval endpoints are included. Water polytropes use the explicitly selected p,v temperature root and refuse a changing root count. The same states feed both plots and the integral diagnostics.

The plots use linear pressure and absolute temperature because a geometric area on a logarithmic pressure scale, or one closed to zero Fahrenheit, is not the stated integral. The T-s diagram uses K or degR and entropy change from the initial state. A signed p-v area must still be converted from the axis product to energy per mass.

Limitations

• The displayed sampled-integral differences are diagnostics, not certified error bounds. Increase the 2–401 sample count to examine convergence. Water-isotherm work itself is a sampled integral.
• Independently fitted property regions need not satisfy cross-region identities exactly. For example, at 1 MPa the saturated IF97 heat difference and T times entropy difference differ by about 0.0078 kJ/kg; this is not purely trapezoid error. NASA interval offsets can likewise affect endpoint identities.
• Nonzero controlled steps below 64 input ulps for gas, or a conservative 2e-9 relative property-path scale for water, are refused. Water quality steps at or below 2e-9 are refused. An identical endpoint remains a zero process. Very small differences and totals cannot be reported with meaningful relative precision if they underflow.
• Water temperature-root count changes are refused. A fixed sample set is not a continuum proof of branch continuation or a guarantee that every intermediate operating point is feasible. The underlying IF97 validity and critical-region refusals still apply.
• Only ideal gases, water and the given-c/given-v model are provided here. No generic material constants, refrigerant process curves, irreversible heat-transfer mechanism or additional work mode is inferred.

Reading the result

Choose the model before setting the initial independent pair. Water can start from p,T, p,v or p,x; p,v exposes its temperature-root choice. Ideal gases use p,T or p,v, with the stated caloric approximation. An isentropic process is controlled by final pressure, an isochore by final temperature, an isobar by final temperature or water quality, and an isotherm or polytrope by final volume. The incompressible controls reflect its fixed-volume constraints.

Read the resolved state and phase labels, specific transfers, and mass totals together. The p-v area follows the boundary-work sign; the T-s area is reversible heat only under this entry’s assumptions. A refusal replaces the whole path instead of drawing a line through an unresolved sample.

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M04Closed System Energy Balance

The first law for a closed system, solved for whichever term is unknown. Work modes beyond the moving boundary are included because problems use them and a screen that only knows p dV cannot express them.

What it rests on

Energy is conserved, and for a fixed mass the only ways across the boundary are heat and work. The change in the total energy of the contents -- internal, kinetic and potential -- therefore equals the heat added minus the work done. Four quantities and one equation, which is why a screen that only ever solved for the heat would be a quarter of a screen.

How the working relation follows

Nothing needs deriving in the balance itself; what needs stating is the bookkeeping, because that is where the errors are.

Every term is a total in kilojoules, including the internal-energy change, which is the mass times the specific change and not the specific change. The distinction is silent when it is wrong: a specific change of minus fifty-five kilojoules per kilogram for four tenths of a kilogram is minus twenty-two kilojoules, and supplying the specific value instead gives an answer that is wrong by the difference with nothing on screen to say so. The app therefore attaches a note whenever a mass other than one is given, stating which reading it took.

The kinetic term is written as half the mass times the difference of the two velocities times their sum, rather than as the difference of their squares. The two are the same algebra and not the same arithmetic: squaring first and subtracting destroys the answer when the two velocities are close, which is the ordinary case in a pipe and exactly the case where the kinetic term is small enough to be interesting.

Work modes beyond the moving boundary are here because problems use them, and each is its own function with its own arguments rather than one dispatcher taking three positional numbers -- the arguments are genuinely different physical quantities, and collapsing them into slots is how a spring constant ends up in the voltage field. The list is not closed: work is a generalised force through its displacement and every named mode is an instance of that, so the generalised form is offered explicitly rather than leaving a user with an unlisted mode to conclude that the app cannot do it.

Assumptions

• The system is closed: no mass crosses the boundary. A vessel being filled or emptied is a different balance and has its own screen.
• The energy terms are totals in kilojoules, with the mass scaling only the kinetic and potential terms the app computes for itself.
• Gravity is the standard value unless another is supplied, and the potential term is linear in elevation, so this is a local field rather than an orbital one.
• Work modes add. Several may act at once and the balance carries their sum.
• The sign convention is fixed: work out of the system positive, heat into it positive. Reversing the direction of one transfer reverses exactly one sign.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• The sign convention is fixed and stated: work out of the system is positive, heat into the system is positive. Reversing the direction of a transfer reverses exactly one sign.
  Verification: sign flip test
• Around any closed cycle assembled from these processes, the sum of Delta-U is zero.
  Verification: cycle closure test
• With no velocity change and no elevation change, the kinetic and potential terms are identically zero, not merely small.
  Verification: exactness test

Monotonic trends:

• With Q fixed at zero, work out of the system and internal energy move in opposite directions.

Where it gets delicate

How it is solved

Rearrangement. The unknown is an input rather than a fixed choice, and the app solves the same equation for whichever of the heat, the work or the internal-energy change was left out. The kinetic and potential terms are always computed and always reported, even when they are exactly zero, so that a reader can see they were zero rather than assume it -- the difference between a screen that omits a term and a screen that says the term is negligible here.

There is a second mode in which every term is supplied and the app reports what the balance closes to. That residual is the answer, not an error: it is how you check a claim rather than complete one, and when it is more than about a part in a thousand million of the largest term the app says which fraction of that term it represents.

Solving for the mass is refused rather than attempted, with a reason: the mass enters through the state as well as through the balance, so it is a different problem that needs a state fixed first, and the app says which screen does that.

Limitations

• The energy balance alone does not certify the second law; use the entropy balance for that check.
• Final-water-state mode couples the first law to property inversion at a specified final pressure or specific volume. The recovered state must remain inside the supported water-property domain.
• The spring mode is linear. A hardening or softening spring is outside it.
• Electrical work is computed as a magnitude. Its sign in the balance depends on whether the system receives or delivers it.

Reading the result

Read the kinetic and potential lines even when they are zero, because that is the screen telling you it was the stationary closed system every worked problem means when it says nothing about them.

If you supplied all four numbers, the residual is the whole output. A residual that is not zero means one of the four does not belong with the other three, and its size relative to the largest term is the practical measure of how badly.

If a mass other than one kilogram is in play, read the note about the internal-energy change before reading the answer.

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M05Cycle Energy Accounting

Choose power or refrigeration/heat-pump direction; account for heat, signed work, reservoir entropy and the applicable performance ratios.

What it rests on

A complete cycle returns the working substance to its initial state. Its stored energy change therefore vanishes, leaving a balance between heat absorbed, heat rejected and signed work delivered. The direction selection identifies which heat transfer belongs to the hot reservoir and which belongs to the cold reservoir. A refrigerator and a heat pump describe the same work-consuming cycle with different useful outputs.

How the working relation follows

Write heat magnitudes as non-negative quantities and work out as positive. The first law gives work out equal to heat in minus heat out. A power cycle requires positive work out, absorbs heat from the hot reservoir and rejects heat to the cold reservoir. Its efficiency is the delivered work divided by the absorbed heat.

A reverse cycle absorbs heat from the cold reservoir, delivers more heat to the hot reservoir and consumes the difference as work input. Its refrigeration coefficient is cold-side heat divided by positive work input; its heating coefficient uses hot-side heat with the same denominator. Subtracting these coefficients gives exactly one by the first law. They must not be computed using a positive-work power-cycle denominator for the same heat pair.

The working substance also returns to its original entropy. Entropy produced equals entropy leaving with rejected heat minus entropy entering with absorbed heat, using the reservoir assigned to each direction. Requiring this difference to be non-negative gives the Carnot bounds. Zero entropy production is a reversible limit; satisfying this necessary balance does not specify a buildable machine.

Assumptions

• The two heat quantities refer to the same complete cycle and use the same total-energy basis.
• The reservoirs remain at fixed absolute temperatures. Heat into the working substance and heat out of it are non-negative magnitudes; work out is signed.
• Internal heat recovery is not an exchange across the cycle boundary and must not be added again as external supply or rejection.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• For the same refrigeration/heat-pump cycle, gamma = beta + 1. These COPs are not assigned to a power-producing cycle.
  Verification: direct algebra, all inputs
• eta never exceeds 1 - TC/TH, and beta never exceeds TC/(TH - TC). A cycle that violates either is reported as impossible, not as a number.
  Verification: property test against M12 for every generated cycle
• Q_in - Q_out - W_cycle = 0 to rounding for every accepted cycle.
  Verification: closure test

Monotonic trends:

• At fixed cold-side temperature, the coefficient of performance falls as the hot-side temperature rises.

Where it gets delicate

How it is solved

The selected direction determines the work sign and reservoir mapping before any ratio is formed. Finite non-negative heat magnitudes and finite temperatures with hot at least as warm as cold are required. A zero or poorly resolved work difference is refused: its magnitude must exceed 128 floating-point spacings of the largest entered heat divided by 10⁻⁶, so the small difference retains six resolved digits.

The entropy transfers are evaluated separately. Non-finite transfers and nonzero transfers below 10⁻³⁰⁰ kJ/K are refused to keep underflow and subnormal loss of precision out of the second-law verdict. Negative entropy production is refused beyond an absolute floor of 64 floating-point spacings of the larger transfer. A negative value within that rounding floor is shown as zero. Only performance measures belonging to the selected direction are reported.

The energy-flow outline uses the same accepted heat totals and work difference as the table. Both sides have the same total height; the separate work region is delivered on the outgoing side for power and supplied on the incoming side for a reverse cycle.

Limitations

• These are necessary energy and entropy balances, not component-state, transport or machine-design calculations.
• Near equal heat totals the work difference loses relative precision. The entry refuses an unresolved operating cycle rather than reporting a very large COP.
• Equal reservoir temperatures allow a work-consuming heater but no positive-work heat engine. Reversible reverse-cycle bounds at equal temperature are unbounded limits, not finite operating performance.

Reading the result

Select the direction first, then follow the reservoir names on the heat fields. A power result reports efficiency; a refrigerator or heat-pump result reports positive work input and both COPs. The signed work-out value is negative for the reverse cycle.

For a reverse cycle the heating COP exceeds the refrigeration COP by one. Compare the active ratio with its reversible bound and read entropy production alongside it. If the direction or second law fails, the explanatory refusal replaces both the result table and the energy-flow diagram.

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M06Nozzle and Diffuser

The one steady-flow device where the kinetic energy term is the point rather than a neglected nuisance.

What it rests on

The steady-flow energy balance says that what a stream carries in -- enthalpy, kinetic energy, potential energy -- plus whatever heat and work cross the boundary equals what it carries out. A nozzle and a diffuser are the case with no work and no heat, so the balance collapses to a trade between enthalpy and kinetic energy. This is the one device in the product where the kinetic term is the answer rather than a term dropped with a note saying it was negligible.

How the working relation follows

Set the heat, the work and the elevation change to zero in the steady-flow balance. What survives says that the sum of enthalpy and half the velocity squared is the same at the exit as at the inlet -- the stagnation enthalpy is conserved. That single conserved quantity is what makes a nozzle tractable without knowing anything about the duct: it is the right quantity to carry through the device rather than the enthalpy itself.

Solving it for the exit velocity gives the square root of the inlet velocity squared plus twice the enthalpy drop. Written that way, three things are visible at once. A nozzle needs an enthalpy drop and gets speed for it; a diffuser gives up speed and gets an enthalpy rise, which at a stated exit pressure is what pressure recovery means; and the derivative of exit speed with respect to exit enthalpy diverges as the remaining kinetic energy goes to zero. Near that stagnation limit a small enthalpy error causes a large relative velocity error.

For the nozzle, irreversibility enters as a stated efficiency defined on kinetic energy, not on velocity. The isentropic exit state is found first, at the same exit pressure with the inlet entropy; the ideal kinetic energy is the stagnation enthalpy minus that state's enthalpy; the actual kinetic energy is the efficiency times it; and the real exit state is fixed from the exit pressure and what enthalpy is left. The velocity ratio is the square root of the energy ratio and is a different number -- a nozzle at ninety-six per cent on energy is at ninety-eight on velocity -- so the two definitions are not interchangeable even though both get called nozzle efficiency.

The diffuser uses a different definition: the isentropic enthalpy rise at the specified pressure divided by the actual kinetic-energy drop. With no heat or shaft work that drop equals the actual enthalpy rise. Thus the actual rise is the isentropic rise divided by efficiency. Reusing the nozzle multiplier would give the wrong pressure recovery whenever efficiency is below one.

Mass flow closes the geometry. The one-dimensional continuity relation gives the area a stated flow needs at each end, which is what the screen is actually asked for.

Assumptions

• Steady flow: nothing accumulates inside, so one mass flow enters and the same one leaves.
• Adiabatic and doing no shaft work, which is what makes the stagnation enthalpy conserved.
• One-dimensional flow: the velocity is uniform across the section. This is the assumption that fails first in a real diffuser.
• The elevation change across the device is negligible, which for a nozzle a few centimetres long it is.
• Nozzle efficiency is an exit kinetic-energy ratio; diffuser efficiency is an ideal-to-actual enthalpy-rise ratio. Both apply to the whole device.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• The stagnation enthalpy h + V^2/2 is unchanged across an adiabatic device with no work.
  Verification: exactness test
• Mass flow computed at the inlet equals mass flow computed at the exit.
  Verification: closure test
• For a declared nozzle the exit velocity exceeds the inlet velocity; if it does not, the declaration and the data disagree and the app says so rather than reporting a diffuser.
  Verification: consistency check

Monotonic trends:

• At fixed inlet state, exit velocity increases as exit pressure falls.

Where it gets delicate

How it is solved

Fix the inlet and isentropic exit, then apply the selected device's efficiency definition. For a diffuser, first check whether inlet kinetic energy can pay for the required actual enthalpy rise. Resolve the real exit by pressure and enthalpy, then the velocity and the two areas. Efficiency one keeps the isentropic state directly. No iteration is added beyond the property inverses.

The radicand is guarded, and the guard is not tidiness. Asking for an exit enthalpy above the inlet stagnation enthalpy is asking the stream to arrive carrying more energy than it left with, and the honest answer is that no such exit exists -- which the app says in those terms, with the shortfall in kilojoules per kilogram, rather than returning nothing. Exactly at the stagnation enthalpy the velocity is exactly zero, and the guard has to get that right too: a radicand of minus a hundredth of a millionth of a millionth from rounding must give zero, not a refusal. The two are separated by scaling the tolerance against 128 ULP of the terms that were differenced. Huge velocities that cannot resolve the enthalpy change, non-finite or negative speeds, non-positive mass flows and unresolved areas are refused. The area calculation combines binary significands and exponents so an intermediate product cannot overflow when the final quotient is finite.

The declared device is an input and is then checked rather than deduced. A declared nozzle whose exit velocity came out below its inlet is a diffuser, and the app says so instead of quietly relabelling it. That is the difference between a tool and a tool that agrees with you. Pressure direction is also checked: nozzles require a fall and diffusers a rise. The graph stacks h and V²/2000 in kJ/kg (or the corresponding display unit) at both endpoints; its common top is the stagnation enthalpy. An energy closure residual and specific/rate entropy production accompany the graph.

Limitations

• The energy balance and continuity determine the required area at the stated conditions. Above Mach 0.8 a warning requests a separate geometry/choking analysis; it does not invalidate continuity or declare choking. A converging nozzle can reach Mach 1 at its exit.
• Choking, shocks and the whole compressible-duct treatment are deliberately out of scope. The warning names the limit rather than silently correcting for it.
• Inside the vapour dome the formulation defines no speed of sound, so no Mach number is reported there. The velocities and areas are unaffected -- only the sonic check is.
• The inlet area is reported as unbounded when the inlet velocity is zero, which is the honest answer: moving a finite flow arbitrarily slowly takes an arbitrarily large duct.
• An isentropic efficiency of exactly one is the reversible idealisation. The graph is an end-state energy comparison, not a nozzle contour or an area-Mach solution.

Reading the result

Read the refusals, because on this screen they carry the physics. A diffuser that cannot reach the requested pressure is told so in energy terms: slowing the stream to a standstill releases this much kinetic energy and the compression needs that much, and there is not enough speed. That sentence is the answer to the design question.

Read the Mach note as a limit on what the calculation establishes. The states, speed and required areas satisfy the balances; their realisation in a particular duct geometry has not been proved.

The stagnation enthalpy is worth watching as you move the inputs. It should not move at all when only the exit pressure changes, and if it does, the inlet velocity or the inlet state changed with it.

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M07Turbine

Work out of an expanding stream, with the exit quality checked against the blade erosion limit because a cycle can be arithmetically correct and mechanically impossible.

What it rests on

The steady-flow energy balance for an adiabatic device that produces work: the work out per unit mass is the enthalpy drop across it. Everything else on this screen is about the fact that a real expansion does not follow the reversible path, and about the one mechanical consequence -- liquid droplets in the last stages -- that makes an arithmetically valid answer mechanically impossible.

How the working relation follows

Drop the heat, the kinetic and the potential terms from the steady-flow balance and the specific work is the inlet enthalpy minus the exit enthalpy. The exit state is not free: it is fixed by the exit pressure and by how far the expansion falls short of reversibility.

The reversible reference is the isentropic exit -- the state at the exit pressure with the inlet entropy. The isentropic efficiency is the actual work over that reference work, and the ideal is in the denominator because a turbine produces work and friction can only lose some of what was available. So the actual work is the efficiency times the ideal work, the actual exit enthalpy is the inlet enthalpy minus that, and the actual exit state is fixed from the exit pressure and that enthalpy.

Two consequences fall out. The actual exit entropy is above the isentropic one, with equality only at unit efficiency, which is the second law showing up as a geometric fact on the temperature-entropy plane. And if the expansion ends inside the vapour dome in both cases, the actual exit has higher enthalpy and higher quality: the irreversibility leaves enthalpy behind, and at fixed pressure more enthalpy is more vapour. That is why a real turbine's exit is drier than the ideal one at the same pressure, which surprises people who expect losses to make everything worse.

The preceding efficiency construction is adiabatic. With measured exit or shaft-power input, a supplied heat loss is accounted explicitly: shaft work is the enthalpy drop minus outward heat per mass. Entropy production includes the positive entropy leaving with that heat at its boundary temperature.

Assumptions

• The efficiency-input mode is adiabatic. Measured-exit and shaft-power modes can include an explicitly supplied outward heat loss and its absolute boundary temperature.
• Steady flow with negligible kinetic and potential changes.
• One overall machine efficiency or measured performance describes the selected expansion; individual stages are not resolved.
• Exit pressure must be below inlet pressure. Water/steam and ideal-gas property models have separate admissible state ranges.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• For an adiabatic turbine, the isentropic efficiency lies in (0,1]. A heat-losing turbine instead checks the entropy balance including heat transfer and does not impose an adiabatic enthalpy-drop bound.
  Verification: property test
• For an adiabatic turbine the exit entropy is greater than or equal to the inlet entropy.
  Verification: second-law check
• Shaft work is positive only when enthalpy drop exceeds specified outward heat transfer per kg.
  Verification: sign test

Monotonic trends:

• At fixed inlet state, specific work increases as exit pressure falls.
• At fixed inlet state, exit quality falls as exit pressure falls.

Where it gets delicate

How it is solved

Fix the inlet state, fix the isentropic exit from the exit pressure and the inlet entropy, scale, and fix the actual exit from the exit pressure and the remaining enthalpy. The only iteration is inside the property inversions.

Every refusal is by name and with a reason. An efficiency outside its range, an exit pressure at or above the inlet, an isentropic exit that is not reachable, an actual exit that is not reachable -- each returns its own sentence rather than a shared failure, because on a screen with four inputs the useful part of a refusal is which input to change.

The same function is what the vapour-cycle screens call for their expansion leg. That is deliberate: a cycle that computed turbine work differently from the turbine screen would be two implementations of one relation, and they would drift.

Limitations

• Wet-steam expansion is evaluated thermodynamically, without a blade-erosion or droplet-dynamics model.
• A vanishing isentropic enthalpy drop makes a work-ratio efficiency indeterminate and is refused.
• Exhaust-hood losses, valve losses and stage geometry are not separately modelled.
• Mass flow converts specific shaft work to power. For nonadiabatic operation, read heat loss, entropy production and the enthalpy-drop ratio separately; the latter is not a work efficiency.

Reading the result

Read the two exit states together. The isentropic one is where the expansion would have ended; the actual one is where it did; the gap between them on the temperature-entropy plane is the whole of the irreversibility, and the gap in enthalpy is the work you did not get.

If the exit is inside the dome, the quality is the number that decides whether the machine can be built. A low quality with a high efficiency is a common and useless combination: the arithmetic is fine and the last-stage blades are not.

An efficiency the app declines to define is telling you that the expansion is too small to have one, not that something went wrong.

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M08Compressor and Pump

Liquid-water pump power, ideal-gas compression with explicit heat models, and stable real-gas pv^n compression. Every interstage thermal reset returns to inlet temperature; no final aftercooler.

What it rests on

The same steady-flow balance as the turbine with the sign of the work reversed: the work in per unit mass is the enthalpy rise. What makes this a separate screen is that the ideal now sits in the numerator of the efficiency -- a compressor consumes work, and irreversibility can only make you pay more than the reversible minimum -- and that the reversible minimum itself depends strongly on how much the fluid's volume changes while you are compressing it.

How the working relation follows

For a liquid the specific volume barely changes, so the reversible work is very nearly the specific volume times the pressure rise. For a gas it changes by orders of magnitude, so the reversible work depends on the path: an isothermal compression takes the least work, an isentropic one the most, and a polytropic path anything between. The ratio of the two extremes is why feeding a boiler costs a fraction of a per cent of what the turbine returns and why a gas turbine spends more than a third of its output driving its own compressor.

Multistage compression with intercooling is the practical consequence. Cooling between stages moves the second stage's inlet back down the volume axis, so it compresses a denser gas and pays less. With ideal intercooling back to the inlet temperature, minimising the total work over the intermediate pressures gives a result students reliably get wrong in the direction of equal pressure rises: the work is least when every stage has the same pressure ratio, so the intermediate pressures form a geometric progression. For two stages the answer is the geometric mean of the two end pressures. From one bar to a hundred bar, that is ten bar; the arithmetic mean would say 50.5.

The isentropic efficiency for a compressor or a pump is the reversible work over the work actually paid. Inverting it turns an eighty per cent machine into a hundred and twenty-five per cent one, which at least announces itself, or a hundred and twenty-five per cent machine into an eighty per cent one, which does not. The app writes the two orientations as two functions rather than one, so the flip cannot happen by passing the arguments in the wrong order.

Assumptions

• Adiabatic compression, with all the loss carried by the efficiency rather than by a heat transfer.
• For the pump shortcut, the liquid's specific volume is constant over the pressure rise. For water over a hundred atmospheres that is true to about a per cent, and the exact enthalpy difference is available beside it.
• The multistage optimum assumes ideal intercooling back to the inlet temperature and identical stages. Real intercoolers have an approach temperature and the optimum moves.
• The efficiency lies above zero and at or below one. Above one would mean the machine took less work than a reversible one.
• The exit pressure is above the inlet pressure; the app refuses the reverse rather than returning a negative work.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• For constant cp and 1 < n < k, isothermal work < polytropic work < isentropic work at the same inlet and pressure ratio.
  Verification: ordering test over the whole pressure-ratio range
• For ideal-gas n>1, identical stages with full intercooling reduce work as stage count rises; ideal-gas n=1 is independent of stage count and n<1 reverses this trend. Real-fluid behaviour must be computed from its EOS.
  Verification: limit test
• The incompressible pump shortcut and the exact enthalpy difference agree to within the stated approximation error over the validated liquid range.
  Verification: numerical comparison, with the error reported rather than assumed small

Monotonic trends:

• Compression work increases monotonically with pressure ratio.
• For ideal-gas n>1, identical stages with full intercooling reduce work as stage count rises; ideal-gas n=1 is independent of stage count and n<1 reverses this trend. Real-fluid behaviour must be computed from its EOS.

Where it gets delicate

How it is solved

Fix the inlet, fix the isentropic exit from the exit pressure and the inlet entropy, divide by the efficiency to get the actual work, and fix the actual exit from the exit pressure and the raised enthalpy. Note the direction: the turbine multiplies by the efficiency and the pump divides by it, which is the whole of the difference between them.

The polytropic work integral carries the one piece of numerical repair on this screen. Its general form has an exponent that goes to zero as the polytropic exponent approaches one, so a power raised to it approaches one and subtracting one from that destroys the answer, while the factor in front grows without bound to multiply what is left. Writing the small quantity with the standard library's exponential-minus-one function computes it directly and never forms the difference, so the general branch now joins the logarithmic one smoothly rather than merely finitely. The isothermal case is still a separate branch, because it is the exact answer to a different integral rather than a limit of this one.

That expression also carries the project's clearest illustration of why layer two exists. A missing factor of a thousand sat in it undetected, because every test it had compared polytropic work against polytropic work -- ratios between exponents, the ordering of the four models, continuity at the seam -- and a units error cancels out of every one of them. It was found by an independent implementation of the same integral disagreeing by exactly a thousand.

Limitations

• As the pressure ratio approaches one the efficiency is zero divided by zero, and the app says the ratio is too small for an efficiency rather than printing a number.
• The gas entry supports staged compression with explicit stage count and intercooling. Intermediate states, process work and the chosen ideal-gas or real-fluid model remain part of the result; the pump is a separate liquid-water calculation.
• The intercooled optimum is for identical stages with ideal intercooling. It is a design starting point rather than an answer for a real machine with real intercoolers.
• Nothing models surge, choke, or the compressor map. The efficiency is a stated number and not a function of flow.

Reading the result

Read the pump and the compressor as the same relation over two very different specific volumes. That comparison is the point of putting them on one screen, and it is what explains why the back work ratio of a vapour plant is a fraction of a per cent while a gas turbine's is a third.

For the pump, compare the shortcut against the exact enthalpy difference. The gap is the error you are accepting when you use the shortcut, and it is visible rather than assumed small.

The efficiency displayed beside a compressor result is a check, not an input being echoed: it is recomputed from the states, so if it does not match what you asked for, the exit state you supplied is not the one that efficiency implies.

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M09Heat Exchanger

Two streams exchanging energy, solvable for any one of the four terminal states or either flow rate. Direct-contact mixing is the same balance with one stream out instead of two.

What it rests on

An adiabatic exchanger transfers energy between two streams and to nowhere else, so the energy given up by the hot stream equals the energy taken up by the cold one. That is the whole first-law content. The second law then adds something the first law cannot see: the streams may not cross in temperature, and the transfer across whatever temperature difference remains destroys work potential even though nothing moves and nothing is thrown away.

How the working relation follows

Write the steady-flow balance over the whole exchanger with no work and no heat to the surroundings. Each stream contributes its mass flow times its enthalpy change, and the two contributions sum to zero. With four terminal states and two flows there are six quantities and one equation, so five have to be given and the sixth follows -- which is why the unknown is a choice on this screen rather than a fixed output.

Direct-contact mixing is the same balance with one stream fewer at the exit. That is a structural difference, not a display option: the non-mixing case has four terminal states and the mixing case has three, and they do not share an unknown. In the mixing case the exit temperature comes out of the balance rather than being asked for, and it must land between the two inlet temperatures -- which it does automatically, and which is worth asserting because it is the one case where the check is free.

The entropy production follows from the same terminal states: the entropy carried out by both streams minus the entropy carried in. A zero terminal approach alone does not imply zero total entropy production. Reversible heat transfer requires zero local temperature difference throughout the entire heat-transfer path; for nonzero duty it is an ideal limiting case.

Assumptions

• Adiabatic to the surroundings: everything the hot stream loses, the cold stream gains.
• Steady flow with no pressure drop on either side. Each stream is at one pressure from inlet to exit.
• No work, and negligible kinetic and potential changes.
• The two streams are each a single phase or a single substance the property solver can fix from pressure and enthalpy. Nothing here models fouling, an area, or a transfer coefficient.
• The non-mixing temperature path uses a counterflow heat-load coordinate. Water phase boundaries are included explicitly; the minimum between all remaining sample points is not proven.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• Energy given up by the hot stream equals energy taken up by the cold stream when the exchanger is adiabatic to its surroundings.
  Verification: closure test
• Heat must pass from locally hotter to locally colder states. The counterflow path is checked at 65 heat-load points plus water saturation breakpoints; any sampled internal temperature cross is refused, even when terminal approaches and total entropy production appear acceptable.
  Verification: property test over randomly generated terminal states
• Total entropy production across the exchanger is non-negative, and is zero only in the limit of zero temperature difference.
  Verification: second-law check

Monotonic trends:

• At fixed inlet states and fixed cold-stream flow, duty increases as hot-stream flow increases, approaching a limit.
• Entropy production increases as the terminal temperature difference widens.

Where it gets delicate

How it is solved

Fix the five supplied terminal temperatures and flow quantities, and solve the balance for the sixth -- by inverting from pressure and enthalpy where the unknown is a temperature, or by division where it is a flow.

Fixing a state from pressure and enthalpy rather than by inverting enthalpy in temperature matters here, and the mixing case is why. A direct-contact mixer fed steam and liquid water lands on the saturation line, where pressure and temperature do not fix a state at all: every quality shares the same pair. Inverting enthalpy in temperature finds the temperature and then loses the quality, and every entropy computed from it comes back as nothing. That is not an edge case for this screen -- an open feedwater heater is a direct-contact mixer whose whole job is to land on the saturation line.

A temperature cross is refused, not noted. The energy balance alone permits the cold stream to leave hotter than the hot stream entered -- energy is conserved either way -- and it is the single commonest way a heat-exchanger answer comes out plausible-looking and impossible.

The closure residual is reported rather than assumed zero. A balance that does not close is a balance with a term missing, and the number says so before any duty is read.

The accepted non-mixing balance is checked on a counterflow heat-load coordinate. At each load, both streams are reconstructed from pressure and enthalpy, preserving two-phase water states. The path includes 65 equally spaced heat-load points and the water saturated-liquid and saturated-vapour enthalpy breakpoints. A sampled internal temperature cross is refused even if both terminal differences and total entropy production appear acceptable. The graph marks the sampled minimum and reports it separately from the smaller terminal difference; finite sampling is not a continuous global minimum proof.

Direct-contact mixing requires the same fluid and pressure at both inlets. Its graph connects the two inlet temperatures to the shared outlet as a terminal-state schematic, not a spatial mixing trajectory. Known flows must be positive and finite, only the chosen unknown may be blank, and overflow or failure to close the energy balance is refused.

Limitations

• This is a thermodynamic balance, not heat-exchanger sizing: no transfer area, conductance or fouling model is inferred.
• Terminal approach and any sampled temperature path do not prove a continuous global minimum approach.
• Pressure drop is not modelled on either side.
• The non-mixing mode solves any of the four terminal temperatures or either flow. Direct mixing requires the same water/air model and common pressure, then solves the mixed outlet state.
• Entropy production is the total device balance; it is not a spatial attribution of irreversibility.

Reading the result

Read the closure line first. It should be zero to rounding; if it is not, one of the five supplied quantities does not belong with the other four.

Then read the entropy production, because it is where the irreversibility of this device hides. Nothing moves, nothing is thrown away, the energy balance closes to machine precision, and the whole terminal temperature difference is being paid for in destroyed work potential. That number is what the exergy screens will later turn into kilowatts of lost work.

A refusal on a temperature cross is a second-law refusal, not an arithmetic one. The numbers you entered conserve energy perfectly well; they just describe something that does not happen.

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M10Throttling Device

Constant enthalpy across a restriction. The same screen carries the throttling calorimeter, which turns a pressure drop into a measurement of quality, and the Joule-Thomson coefficient, whose sign decides whether throttling cools or warms.

What it rests on

A restriction does no work and, being short, exchanges no appreciable heat. The steady-flow balance then says the enthalpy is unchanged across it. That is the entire model, and everything interesting about a throttle follows from what constant enthalpy does not imply: it does not imply constant temperature, and whether the temperature falls or rises is a property of the substance rather than of the valve.

How the working relation follows

Drop the work, the heat and the mechanical terms from the steady-flow balance and the exit enthalpy equals the inlet enthalpy. The exit pressure is lower, so the exit state is fixed by the pair of pressure and enthalpy -- which for a substance near its saturation line commonly means the stream leaves partly vaporised even though nothing was heated.

The temperature change is governed by the derivative of temperature with respect to pressure at constant enthalpy. Expanding it in measurable quantities gives the specific volume over the specific heat times the quantity temperature-times-volume-expansivity minus one. The sign is the whole answer. Positive means throttling cools; negative means it warms. The condition for the sign to change is that the volume expansivity equals the reciprocal of the temperature, so the inversion temperature falls out of this expression rather than being a separate correlation.

An ideal gas has a volume expansivity of exactly the reciprocal of the temperature at every state, so the bracket vanishes identically: throttling an ideal gas does not change its temperature at all, and the whole gas is at its inversion temperature everywhere. The app returns a hard zero for that case rather than evaluating the expression numerically, because a numerically evaluated zero is whatever the round-off happened to be and invites somebody to read meaning into its sign.

The same constant-enthalpy line is what makes a throttling calorimeter work: expand a wet sample far enough that it becomes superheated, measure its downstream pressure and temperature, and the enthalpy you read there is the enthalpy it had upstream -- from which the upstream quality follows by the lever rule.

Assumptions

• Adiabatic and doing no work, which is what makes the enthalpy constant. A long insulated line with a large velocity change is not this model.
• Steady flow, with the kinetic energy change across the restriction negligible. It is small for a valve and not for an orifice sized as a flow meter.
• The exit pressure is below the inlet pressure; the app refuses the reverse.
• The exit state is an equilibrium state. The expansion itself is violently irreversible and nothing here describes what happens inside the restriction, only the state on either side of it.
• The calorimeter reading is valid only when the downstream state is superheated. Inside the dome, downstream pressure and temperature do not fix a state and the method cannot resolve anything.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• Enthalpy is unchanged across the restriction, exactly, by construction of the model.
  Verification: exactness test
• Entropy production across a throttle is strictly positive for any finite pressure drop.
  Verification: second-law check
• For an ideal gas the Joule-Thomson coefficient is identically zero, so throttling an ideal gas does not change its temperature.
  Verification: limit test; this is the check that catches a wrong sign in the departure term

Monotonic trends:

• At fixed upstream state, downstream quality increases as downstream pressure falls.

Where it gets delicate

How it is solved

Fix the inlet, then fix the exit from the exit pressure and the inlet enthalpy. That inversion goes through the two-phase test first, because inside the dome enthalpy is constant along the isobar and not invertible in temperature.

The entropy production is computed from the two states directly: a throttle transfers no heat, so the entropy carried out less the entropy carried in is entirely production. It is strictly positive for any finite pressure drop, which is the mathematical form of the statement that a throttle is the only device in this product whose exergetic efficiency is zero by construction -- it produces nothing and destroys everything it drops.

The inversion locus, where the coefficient is available from a real-gas model, is located by bracketing the sign change rather than by solving for a zero of the coefficient. Near the locus the coefficient's relative error is unbounded while its sign is not, so working with the sign is the better-conditioned problem. The search window is scanned for the sign change rather than bracketed at its two ends, because the ends are not always computable -- the caloric polynomials stop at 200 K, and an earlier version took its lower bound at 60 K, got nothing there, and reported no inversion temperature for any substance. A bracket whose endpoint cannot be evaluated is not a bracket.

Limitations

• Forward throttling supports water and the offered real fluids. Water calorimetry recovers inlet quality from a resolved downstream state. The cubic inversion map exposes its root branches; the local Joule–Thomson coefficient describes a differential pressure change.
• The restriction interior is not resolved. The inlet and outlet are equilibrium states.
• Kinetic energy change is neglected; a large outlet speed can invalidate that assumption.
• The inversion locus depends on the selected equation of state. Cubic gas/liquid roots and model boundaries must be read with the result.

Reading the result

Read the exit state, not the exit temperature. For water near saturation the interesting thing that happens across a valve is that some of the liquid flashes, and the quality is what says how much.

The entropy production is the number that makes this device's role in a cycle visible. It is why a refrigeration cycle with a throttle can never reach its reversible bound, and it is what an expander would recover.

If you are throttling a gas and expecting cooling, check which side of its inversion temperature you are on. Above it, throttling warms -- which is not a curiosity but the reason hydrogen and helium have to be pre-cooled before any liquefaction cycle will work on them at all.

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M11Transient Filling and Emptying

Charging and discharging a vessel. The adiabatic-charge result, that the final internal energy equals the supply enthalpy, is the surprise of the chapter and the screen presents it as one rather than burying it in a number.

What it rests on

The unsteady form of the energy balance for a control volume: the change in the energy stored inside equals the heat and work crossing the boundary plus the energy carried in by whatever entered minus the energy carried out by whatever left. The stream carries enthalpy, not internal energy, because pushing a parcel of fluid across the boundary takes flow work that the fluid behind it supplies -- and for a vessel being filled, that flow work is the whole surprise of the topic.

How the working relation follows

For a rigid adiabatic vessel with no work, the balance says the final mass times its internal energy minus the initial mass times its internal energy equals the mass admitted times the supply enthalpy. Start from an evacuated vessel and the initial term vanishes, the mass admitted is the final mass, and it cancels: the final specific internal energy equals the supply specific enthalpy exactly.

For the ideal-gas branch, the contents end hotter than the line even without heat input. For an ideal gas with constant specific heats the internal energy is the specific heat at constant volume times temperature and the enthalpy is the one at constant pressure times temperature, so the final temperature is the specific heat ratio times the supply temperature -- air from a 300 K line ends near 420 K in the tank. Students disbelieve it, and being able to drag the supply temperature and watch the ratio stay put is worth more than the derivation.

The reason is the flow work. Pushing each parcel into the tank takes the product of pressure and specific volume of work done by whatever is behind it in the line, and that work has nowhere to go but into the internal energy of what is already inside.

Discharging is the harder half and gets a different treatment. The escaping stream leaves at the state of the vessel, which is changing while it leaves, so the balance is a differential relation rather than an algebraic one. Treating it as a single algebraic step is not a coarser version of the same thing -- it is the assumption that the whole discharge leaves at the initial enthalpy, which is a different physical claim and gives a visibly different answer.

For the stated rigid, uniform, adiabatic discharge with no work, substitute v=V/m into d(mu)=h dm. Using h=u+pv gives du+p dv=0. The thermodynamic identity T ds=du+p dv then gives ds=0 for the remaining fluid. The final state can therefore be resolved by pressure and the initial specific entropy, including equilibrium liquid-vapor mixtures. This conclusion does not assert that the external jet, valve or surroundings are reversible.

Assumptions

• The vessel is rigid, so there is no boundary work and the volume is a constant that ties mass to specific volume at every instant.
• The supply line holds a fixed state throughout the charge. A line that draws down while filling is not this model.
• Uniform state inside the vessel at every instant. Real filling stratifies, and the model's single temperature is a mixed-cup value.
• For the discharge, the escaping stream is in equilibrium with the vessel contents at each instant.
• The heat term, when non-zero, is a total for the whole process rather than a rate.
• Discharge is adiabatic with no shaft or electrical work; the current isentropic remaining-fluid constraint depends on these assumptions. Charging accepts separately signed total Q and W.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• For adiabatic charging of an initially evacuated rigid vessel with zero work from a constant supply state, the final specific internal energy equals the supply specific enthalpy exactly.
  Verification: Independent evacuated-charge reference; this special case does not validate all nonempty, heated or discharge balances.
• Positive transferred mass equals final minus initial mass on charging and initial minus final mass on discharge.
  Verification: closure test
• For an ideal gas with constant specific heats, the evacuated adiabatic zero-work charge gives T2 = k*T_supply.
  Verification: closed-form numerical comparison

Monotonic trends:

• Final vessel temperature after adiabatic charging rises with supply temperature and is always above it.

Where it gets delicate

How it is solved

Charging uses the scaled residual u2 − h_supply − C v2 = 0, with C = [(m1/V)(u1 − h_supply) + (Q − W)/V]. This avoids losing the root when a small vessel makes every unscaled total vanish. An evacuated zero-net-transfer case directly inverts u2 = h_supply; the general water calculation also solves quality inside the saturation dome. Actual model temperature limits are used. Every root is checked again against the scaled residual, finite and resolved mass, the direction of transfer and the total energy balance. A saturation discontinuity is never accepted as a root.

Discharge resolves the remaining contents at constant specific entropy. The final state is independent of the sampling count. A separate trapezoidal integral of h dm along logarithmically spaced pressures estimates net transported energy; its signed difference from U2 − U1 is displayed. This is a convergence diagnostic, not the equation used to determine the final temperature. The native entry uses 400 segments.

Mass differences and discharge transport-energy differences must resolve against their original stored totals. A nonzero pressure change below that resolution is refused. Already equal vessel and line pressures with zero Q−W preserve the exact initial state with zero mass transfer. Rigid volume removes moving-boundary work, but does not remove electrical or shaft work: charging accepts their signed total W (positive out).

Limitations

• Uniform-state, uniform-flow. Stratification during filling and the jet at the inlet are outside the model, and both are real in a fast charge.
• Charging specifies final pressure; the solver recovers temperature or quality and obtains final mass from fixed volume. Pressure equalisation is not a mass-root search.
• Heat transfer during the process is a single supplied total, not a model. A real vessel exchanges heat with its own walls, and for a slow fill that dominates.
• The working fluids on this screen are water and air.
• The 2–5000 discharge segments control only the diagnostic h dm quadrature; no elapsed time, valve area or flow-rate history is predicted.

Reading the result

Read positive mass transferred as mass admitted on charging and mass removed on discharge. Net transported ENERGY is signed into the vessel and can be negative because enthalpy shares a model-dependent reference. The total-energy chart uses ΔU = Q − W + net transport throughout; it never mixes kJ/kg with kJ.

An evacuated adiabatic charge with zero work also shows final specific internal energy and supply specific enthalpy as two equal bars. For an ideal gas this implies a temperature rise. Water can flash and end colder while preserving the same u2 = h_supply identity.

For discharge, compare the sampled transport discrepancy with the balance-derived total. Refining the samples tests quadrature; the final-state entropy constraint remains the same.

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M12Reversible Limits

The bound every other cycle screen is measured against, and the three-way verdict on a claimed cycle: impossible, reversible, or irreversible.

What it rests on

No cycle exchanging heat with two reservoirs can be more effective than a reversible one between the same two, and every reversible cycle between the same two is equally effective. The consequence is that the bound depends on nothing but the two temperatures -- not on the working fluid, not on the hardware -- which is what makes it a bound rather than a benchmark.

How the working relation follows

Take a reversible cycle between two reservoirs. Because it is reversible, the cyclic integral of heat divided by boundary temperature is zero, which for two reservoirs is the heat supplied over the hot temperature equalling the heat rejected over the cold one. Substituting that ratio into the first-law efficiency gives one minus the temperature ratio; taking the same substitution into the two coefficients of performance gives the cold temperature over the span and the hot temperature over the span. The three are one result read three ways, and the identity that the heat-pump bound exceeds the refrigeration bound by exactly one survives from the energy balance unchanged.

Temperatures are absolute, and that is not a formality. In degrees Celsius these expressions return plausible-looking numbers that are simply wrong, and it is the commonest single error on this screen.

The Clausius residual is the more interesting quantity. For a cycle exchanging heat with two reservoirs it is the heat supplied over the hot temperature minus the heat rejected over the cold one, and it equals minus the entropy the cycle produced. Its sign is therefore a verdict rather than a measurement: positive means the claimed cycle destroys entropy and cannot exist, zero means internally reversible, negative means irreversible and says by how much.

For finite positive heat-engine inputs, this sign test and the efficiency bound are equivalent. The entropy residual additionally measures the lost work potential when multiplied by an environmental temperature. Negative heat magnitudes and reverse or zero-work cycles require a different direction convention; the actual-cycle entry refuses those inputs before giving a verdict.

Assumptions

• Two reservoirs, each large enough that exchanging heat with it does not change its temperature.
• Temperatures are absolute, and the hot reservoir is strictly warmer than the cold one.
• The heat quantities entered are magnitudes exchanged with those two reservoirs and nothing else.
• The bound is on the cycle as a whole. A cycle whose heat crosses a range of temperatures has a lower bound than its extremes suggest, and the cycle screens compute that from the temperatures heat actually crosses.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• gamma_max = beta_max + 1 exactly, for every reservoir pair.
  Verification: algebraic identity
• Every cycle produced by any other module in this canon satisfies eta <= eta_max, beta <= beta_max, gamma <= gamma_max at the same reservoir temperatures.
  Verification: THE canon-wide property test. It fails whenever a sign, a state assignment or a branch selection is wrong in ANY cycle module at ANY operating point, which is why it earns its cost.
• A cycle assembled from internally reversible processes between two reservoirs returns exactly eta_max.
  Verification: limit test

Monotonic trends:

• Maximum efficiency rises with hot reservoir temperature and falls with cold reservoir temperature.
• Maximum coefficient of performance falls as the reservoir temperature span widens.

Where it gets delicate

How it is solved

Direct evaluation, with the guards where they belong. Non-positive temperatures and a cold reservoir above the hot one return nothing rather than a number. A vanishing span makes both coefficients of performance diverge, and the kernel returns an infinity rather than whatever large finite value the rounding produced; the screen prints that as unbounded, which is a physical answer and not a failure, with the two temperatures that produced it on the same screen.

The efficiency uses the temperature difference divided by the hot temperature to avoid subtracting two nearly equal dimensionless numbers. Infinite input temperatures are not a physical infinite-reservoir limit and are refused. The graph selector shows each bound against TC/TH. The efficiency line includes its limiting endpoints; the two COP curves stop at 0.99 and explicitly disclose the divergence toward 1. A current ratio beyond that interval is identified as outside the graph, while its numeric bound is retained.

The verdict is a display decision and lives in the layer both front ends share rather than being written once in each. It is measured against the entropy the cycle takes in -- the heat supplied over the hot temperature -- which is the natural scale and makes the ruling independent of the units the heat was entered in. Where the threshold sits, what the wording is and which colour band it belongs to are all decided in one place, tested without a display, and compared across the language boundary once.

Limitations

• This is a bound and not a design. No cycle reaches it, and a cycle that comes close to it is usually one that exchanges heat over a vanishing temperature difference and therefore needs unbounded area and unbounded time.
• Two reservoirs only. Most real cycles have neither reservoir at a single temperature.
• As the two temperatures approach one another the coefficients of performance diverge and the efficiency goes to zero. These are limits, not finite-power operating conditions.
• As the cold temperature approaches absolute zero the efficiency approaches one and the refrigeration coefficient of performance approaches zero. The app states the third-law reading rather than presenting the limit as achievable.

Reading the result

The verdict is the output, and it has three values rather than two. A claimed cycle can be irreversible, which is normal; internally reversible, which is a limit; or impossible, which means the numbers you entered describe something that does not happen. This is the one place in the app where the honest answer to a user's numbers may be that no such device can exist, and it is stated plainly rather than softened into a warning.

The three bounds are worth reading together rather than one at a time. They come from the same substitution, and if a power cycle's efficiency and a heat pump's coefficient of performance at the same temperatures do not sit in the relation the screen prints, one of the temperatures is not absolute.

Entering the actual heat quantities is optional. Only a finite heat-engine claim with Qin > Qout ≥ 0 and TH > TC > 0 gets a ruling and a point on the efficiency graph. Refrigerator and heat-pump curves never borrow that point. Use Cycle Accounting for an actual reverse cycle. All three bound ratios remain visible while the graph selector changes only the plotted quantity.

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M13Entropy Balance

Entropy production is the output of this screen, not a by-product. Its sign is the answer to the question the user is really asking.

What it rests on

Entropy is not conserved. The change in the entropy of a system equals what crossed its boundary with heat plus what was produced inside it, and the production is never negative. That last clause is the second law in the form this screen uses, and the sign of the production is the answer the user is actually after -- not a by-product of computing something else.

How the working relation follows

For a closed system, integrate the heat transfer divided by the temperature of the boundary it crosses, and subtract that from the entropy change of the contents. What is left is the production. For a control volume at steady state nothing accumulates, so the entropy carried out by the streams, less the entropy carried in, less the entropy that entered with heat, is again the production.

The boundary temperature is where the mistakes are. Entropy leaves with heat at the temperature of the boundary the heat crosses -- not the temperature of the system and not the temperature of the reservoir. Those three coincide only for a reversible transfer, which is precisely the case where the production is zero and the distinction cannot be seen. Every other case, they differ, and using the wrong one moves the answer by the amount you were trying to measure.

A boundary at several temperatures is handled by summing the contributions segment by segment. A single boundary is the same calculation with one segment, so the two are one implementation rather than two branches that can drift.

The production and the entropy change are commonly comparable in size and opposite in effect, so their difference is where the significant figures go. That is why the app carries the scale of the terms that were differenced into its own verdict, instead of comparing the production against a fixed small number.

Assumptions

• Boundary heat can use a constant temperature, one to eight specified segments, or temperature linear in transferred-heat fraction. The last option evaluates the corresponding integral.
• Boundary temperatures are absolute. A non-positive one is refused by name, because that is the commonest place a Celsius value gets in.
• For the steady-flow form, the mass flows balance. An entropy balance on top of a broken mass balance would be arithmetic about nothing, and the app refuses it as such.
• The entropy change of the contents is supplied, or comes from a state the property screens fixed. Nothing on this screen re-derives it.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• Entropy production is non-negative for every process the user can construct that is not declared impossible.
  Verification: the second-law property test, run over randomly generated processes
• A process constructed as internally reversible produces exactly zero, not a small residual.
  Verification: exactness test
• For an isolated system the entropy change equals the entropy production.
  Verification: identity test
• Entropy production of two processes in series equals the sum of their individual productions.
  Verification: additivity test

Monotonic trends:

• Entropy production in heat transfer across a finite temperature difference increases as that difference widens.

Where it gets delicate

How it is solved

Sum the segment contributions, subtract, and report each part separately: the entropy change of the contents, the transfer, and the production. Reporting all three rather than only the production is what lets a reader see which one is carrying the answer.

Three verdicts are distinguished, and the thresholds are relative to the terms that were differenced. A production below the floor is reported as reversible rather than as a signed value near zero -- a genuinely reversible process rounds to a small number of either sign, and printing that number invites somebody to read meaning into it. A production clearly below zero is reported as impossible, with the three candidate causes named: a wrong boundary temperature, two states that are not both reachable, or a heat transfer entered with the wrong sign.

The steady-flow form checks the mass balance before it does anything else, and refuses with both totals shown when the flows do not match.

Limitations

• Choose a closed system or a steady control volume, with one to eight heat-transfer segments. Boundary temperature may be constant, specified per segment, or linear in transferred-heat fraction. In a steady control volume that variation is spatial; it does not turn the balance into a transient model.
• Nothing here fixes states. The entropy values are yours, and if they came from different substances or different references the balance will still close on them.
• The production is the whole system's. This screen cannot say which part of the process produced it; the exergy audit is where that question is asked of a real arrangement.
• Near zero the balance is a difference of comparable terms, so the reversible verdict has a floor scaled to those terms rather than an absolute one. A process whose production is genuinely tiny and genuinely positive is reported as reversible.

Reading the result

Read the three lines as one sentence: the contents changed by this much, this much crossed the boundary, and the difference is what the process made. If the transfer and the change are nearly equal, the production is a small difference of large numbers and its last digits are not meaningful.

The verdict tells you which of three worlds you are in. Irreversible is ordinary. Reversible is a limit, not a description of anything real, and the app says so in those words. Impossible means one of your inputs is wrong, and the note names the three usual candidates.

If you entered a boundary temperature in Celsius, the refusal you get names that possibility explicitly. It is the most frequent single error here.

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M14Isentropic Efficiency

The bridge between the ideal device and the real one, solvable in either direction: efficiency from two known states, or the exit state from a declared efficiency.

What it rests on

A reversible adiabatic device changes the fluid's state at constant entropy. That gives a reference exit state -- same entropy as the inlet, at the actual exit pressure -- against which a real device can be measured. The isentropic efficiency is the ratio of the actual to that reference, and the whole content of this screen is which of the two goes on top.

How the working relation follows

Fix the inlet. The isentropic exit is the state at the exit pressure whose entropy equals the inlet's, and it is computed by one function used by the turbine screen, the pump screen and this one, so all three measure against the same reference.

For a device that produces work the ideal is the denominator: friction can only lose some of what was available, so the ratio of actual work to reversible work is at most one. For a device that consumes work the ideal is the numerator: friction can only make you pay more than the reversible minimum, so the ratio of reversible work to actual work is at most one. Both give a number between zero and one, which is exactly why the flipped ratio is so hard to catch -- inverting an eighty per cent machine gives a hundred and twenty-five per cent, which announces itself, but inverting a hundred and twenty-five per cent machine gives eighty, which does not.

The nozzle efficiency is defined on kinetic energy rather than on velocity, and those are different numbers: the velocity ratio is the square root of the energy ratio. The diffuser's is the pressure recovery over the kinetic energy given up for it. Each is written as its own function rather than one general ratio used five ways, because the orientation is the thing that gets wrong and a shared function would hide which orientation was intended.

Read the other direction and the same relation gives the exit state from a declared efficiency, which is what every cycle screen needs: scale the isentropic enthalpy change, add or subtract it, and fix the state from the exit pressure and the result.

Assumptions

• The device is adiabatic. Every loss is carried by the efficiency; none of it leaves as heat.
• The reference is the isentropic exit at the SAME exit pressure. An efficiency quoted against a different reference pressure is a different number.
• The efficiency is a single figure for the whole device rather than per stage.
• Kinetic and potential changes are negligible for the work-producing and work-consuming devices; for the nozzle and diffuser they are the quantity being measured.
• The efficiency lies above zero and at or below one. Above one means the exit entropy fell, which no adiabatic device does, and the app reports it as impossible.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• Every isentropic efficiency lies in (0, 1]. Above 1 means the exit entropy fell and the result is reported as impossible.
  Verification: property test
• At efficiency exactly 1 the computed exit state equals the isentropic exit state to solver tolerance.
  Verification: limit test
• The actual exit entropy is greater than or equal to the isentropic exit entropy, with equality only at unit efficiency.
  Verification: second-law check
• Computing an exit state from an efficiency and then recomputing the efficiency from that state returns the original value.
  Verification: round trip

Monotonic trends:

• At fixed inlet and exit pressures, turbine work rises and compressor work falls as the efficiency rises.

Where it gets delicate

How it is solved

First fix the inlet and the isentropic exit at the selected outlet pressure. In measured-exit mode the actual state or nozzle exit speed determines the device-specific efficiency ratio. In efficiency-input mode the ratio determines actual exit enthalpy or speed and the property solver recovers the state. Water quality is an explicit alternative to ambiguous saturation p–T input.

The reference energy change must be positive and numerically resolved. Nearly equal pressures, incompatible measurements or an outlet outside the property domain produce an explanation instead of a misleading ratio.

Limitations

• Choose a water turbine, nozzle or pump, or an ideal-gas compressor. Both directions are available: infer efficiency from a measured exit, or infer the exit from a stated efficiency. Water measurements can use the active enthalpy, temperature or quality field; gas measurements do not have quality. The nozzle also supports exit speed.
• Efficiency measures aggregate loss and does not identify its location.
• An overall multistage efficiency is not a stage or polytropic efficiency.
• Nearly equal inlet and outlet pressures can leave the reference work unresolved and make the ratio undefined.

Reading the result

Read which side the ideal is on before reading the number. The screen names the device and orients the ratio for you, and that orientation is the single most common place in this subject to get a factor upside down.

Compare the two exit states rather than only the efficiency. The gap between them on the temperature-entropy plane is the irreversibility; the gap in enthalpy is the work lost or the extra work paid.

If the app declines to define an efficiency, the pressure ratio is too small for the definition to mean anything. That is an answer.

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M15Reversible Steady-Flow Work

Why pumping a liquid costs so little and compressing a gas costs so much: the same integral, over specific volumes three orders of magnitude apart.

What it rests on

For an internally reversible steady-flow device the work per unit mass is minus the integral of specific volume with respect to pressure. One integral, evaluated over specific volumes three orders of magnitude apart, is why pumping a liquid costs almost nothing and compressing a gas costs a great deal. Seeing that as one fact rather than as two unrelated formulas is what this screen is for.

How the working relation follows

Start from the steady-flow energy balance and the entropy relation for an internally reversible process. Eliminating the heat between them leaves the work as minus the integral of specific volume with respect to pressure, plus the kinetic and potential changes, which for the devices on this screen are negligible.

The relationship to boundary work is worth stating, because the two get confused constantly. Boundary work is the area under the curve of pressure against volume; reversible steady-flow work is the area beside it. They differ by the change in the product of pressure and volume, and they coincide only when the polytropic exponent is one. This screen is about the second one, and the polytropic form carries a factor of the exponent in front that the boundary-work form does not have -- writing one of those two expressions and using it for both is a common enough error that the two live in different places with different names.

Three treatments of the specific volume along the path give three answers. Take it constant and the integral is minus the volume times the pressure rise: the shortcut that makes a feedwater pump cost half a per cent of what the turbine returns, and exact to about a per cent for liquid water over a hundred atmospheres. Take a polytropic path and the integral has a closed form. Sample the path from an equation of state and integrate numerically when neither of the first two is honest.

Assumptions

• Steady, internally reversible flow with negligible kinetic and potential changes.
• A general polytropic path need not be adiabatic. Adiabatic reversible flow is the isentropic special case; an isothermal compression requires heat exchange.
• The incompressible shortcut holds specific volume constant. The polytropic expression holds its exponent constant.
• The water integration mode samples IF97 states along an isentrope at the selected pressure nodes.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• As the fluid becomes incompressible, the integrated result converges to the shortcut value.
  Verification: limit test
• For liquid water the reversible pump work is far smaller than the turbine work over the same pressure range, which is why the back work ratio of a vapour plant is small.
  Verification: ordering test, and it is also the check that catches a swapped v
• Work is negative into the device for compression and positive out for expansion, consistently with the closed-system convention.
  Verification: sign test

Monotonic trends:

• At a fixed pressure rise, required work increases in proportion to specific volume.

Where it gets delicate

How it is solved

Saturation intersections are inserted beyond the base logarithmic-pressure mesh. The two saturation-entropy branches are bracketed separately on either side of the 623.15 K IF97 fit boundary. A target entropy inside that fit's one-sided gap or overlap is refused rather than interpreted as an extra physical phase transition. A possible intersection with the existing near-critical quality exclusion band is also refused, independently of sample count. This conservative check can refuse an otherwise single-phase duty whose separation from the excluded band is not established.

The root must meet a 256-ULP entropy residual on its state scale. Supplied endpoint pressures are retained, and the displayed node count includes added intersections. Nodes split the trapezoidal integration; they do not make its remaining intervals exact or establish a rigorous quadrature-error bound.

Direct evaluation for the first two treatments. The polytropic form is the same integral the compressor screen carries, so it is one implementation and the removable singularity at unit exponent is fixed once and tested once: the isothermal branch is the exact answer to a different integral and gets its logarithm, and the general branch is written with the standard library's exponential-minus-one function so that it joins the logarithmic one smoothly rather than merely finitely. Measured in ordinary double precision against a sixty-digit reference, the naive form retains about five significant figures at an exponent one part in a million million away from one -- it has lost eleven of the sixteen it started with, and by one part in a hundred million million it retains two. The conditioned form keeps all sixteen at every one of those exponents. A continuity check a million times further out, at one part in a million, sees a relative error of a few parts in a hundred thousand million and passes comfortably, which is why nothing found this until a high-precision numerical comparison looked closer in.

The water path uses trapezoidal integration at logarithmically spaced pressure nodes. The endpoint identity h₁ − h₂ is also evaluated for this isentropic path, and the signed difference from the sampled integral is displayed. This is an observed quadrature discrepancy, not a rigorous uncertainty bound. Refine the sample count to assess convergence, especially across phase boundaries; there is no universal claim that solver error dominates quadrature error.

Limitations

• This is reversible work, not the shaft work of a real machine with losses.
• Choose an incompressible, isothermal, polytropic or numerically integrated water path. The water option uses 2–401 logarithmically spaced pressure points and trapezoidal integration, with an inlet-volume comparison. Invalid or unresolved paths are refused in full.
• A water isentrope entering the equilibrium two-phase region has a continuous mixture specific volume, although its slope and curvature can change sharply. Refine the sampling to judge convergence; do not impose a fictitious volume jump.
• The incompressible shortcut neglects variation of specific volume over the pressure interval. Its error depends on the state and pressure rise.

Reading the result

Compare the sampled water integral with the inlet-volume shortcut for the same duty. The v(p) plot outlines the area under specific volume against pressure on linear axes, with a horizontal guide for the constant inlet volume. The area magnitude is the sampled integral; compression gives negative work out, expansion positive. The separate gas formulas have their own inputs and must not be read as alternative answers to this water duty.

The sign convention is the same one the rest of the app uses: work is positive out. A compression therefore returns a negative number here, and the compressor screen reports the same quantity as a positive work in.

If you are comparing this against a boundary-work answer for the same two states, they will differ, and the difference is the change in the product of pressure and volume. They agree only when the exponent is one.

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M16Closed-System Exergy

Stored thermomechanical exergy at two states and a complete closed-system balance against one visible environment. Heat, total work and entropy production are checked independently.

What it rests on

Stored exergy measures the thermomechanical work potential of a fixed mass relative to a stated environment. The initial, final and environmental states use the same property model. Environmental pressure and temperature are visible editable inputs, initially set to the standard environment.

The result includes specific and total stored exergy, their change, signed heat and work exergy, and entropy-derived destruction. Chemical exergy is excluded.

How the working relation follows

A reversible route to the environment combines the internal-energy difference, the environmental displacement term and the entropy term. Motion adds kinetic energy relative to the environment at rest; height adds potential energy relative to one common zero-height datum. Stable thermomechanical availability is nonnegative, but the total including a negative elevation datum need not be.

For the process, first form the independent energy residual from heat, total work and the changes in internal, kinetic and potential energy. Derive volume change from the same mass and endpoint specific volumes. Then compute entropy production from total entropy change minus the boundary entropy transfer. Destruction is the environment temperature times this entropy production. It is never defined by subtracting the other exergy terms until a diagram closes.

Heat into the system and work out are positive. Heat exergy changes sign with both heat direction and boundary temperature: positive heat into a boundary colder than the environment carries negative exergy. Work exergy is total work out minus the environmental displacement work. These are signed contributions on one total-energy basis.

Assumptions

• One closed mass, one substance and one property model at both endpoints and the environmental state.
• The environment has fixed positive pressure and absolute temperature and is at rest.
• Only thermomechanical exergy is included. Chemical and composition equilibrium with the environment are outside this module.
• Heat crosses either a constant-temperature boundary or one whose temperature varies linearly with the fraction of heat transferred. This law is supplied as a model; it is not inferred from endpoint fluid temperatures.
• The given-c/v option is an incompressible idealisation with constants supplied by the user, without a material database.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• At the identical dead state, with zero speed and zero elevation, stored exergy is exactly zero.
  Verification: exactness test; this is the single best check on the whole expression
• Resolved thermomechanical availability relative to a stable equilibrium environment is nonnegative. Total stored exergy also includes the signed potential-energy datum and need not be nonnegative.
  Verification: Stable-model availability checks and an explicit negative-elevation counterexample; unresolved cancellation is refused.
• The signed destruction diagnostic is T0 times independently computed entropy production. An admissible process has nonnegative production within numerical resolution.
  Verification: identity test against M13
• For an admissible process, both the first-law and exergy residuals close within their displayed numerical scales.
  Verification: closure test

Monotonic trends:

• For the fixed-volume, constant-c model with fixed system temperature, thermal availability decreases as the environment temperature approaches that temperature.
• For fixed positive heat into the system and fixed endpoint entropy change, increasing the constant boundary temperature increases entropy production and destruction at fixed T0.

Where it gets delicate

How it is solved

Water endpoints use IF97 with an explicit pressure–temperature or pressure–quality pair. The environment uses pressure–temperature and must itself be an unambiguous supported state. The ideal-gas model offers variable polynomial heat capacity, heat capacity fixed at a stated temperature, or the cold-air standard. The given-c/v option evaluates constant-volume caloric changes directly.

Exactly identical dead states return zero at rest and at zero elevation. The constant-c thermal term and its two-state change use conditioned logarithm remainders that retain the quadratic term near equality without subtracting two large potentials. Isothermal pressure changes in this fixed-volume model retain their exact zero energy and exergy changes. Gas energy and entropy changes integrate the selected heat capacities across coefficient seams so rounded integration constants do not introduce jumps. Very small nonzero gas state steps are still refused rather than assigned false precision. IF97 potentials remain subtractive: nonzero state availability below a scale based on the terms is refused, as are unresolved two-state changes. No universal zero clamp is applied.

For constant boundary temperature, heat exergy uses the temperature difference before division and a scaled product. The linear boundary option integrates entropy transfer and splits the heat-exergy integral where it crosses the environment temperature. If opposing contributions cancel below their numerical scale, the net sign is refused.

In derive-work mode, the first law determines total work. In check-entered-work mode it provides an independent residual. Both modes still compute entropy production independently. A negative production outside its numerical scale marks an impossible process even if energy closes. Negative diagnostic values are retained. Only a process satisfying both checks and the exergy closure receives a physical transfer chart.

Limitations

• Numerical scales are not measurement uncertainty and do not prove a process reversible when entropy production is unresolved near zero.
• A finite temperature and pressure pair at water saturation is ambiguous; use the pressure–quality pair for a wet endpoint. The environmental state must be unambiguous.
• Near-dead IF97 cancellation and unresolved nonzero process changes can be refused even when their exact mathematical value exists. Change the inputs by a resolvable amount rather than reading the refusal as zero availability.
• The linear boundary law is one declared model, not a reconstruction of an arbitrary measured heat-transfer history.
• Nonzero underflow, overflow and an unresolved net sign of opposing boundary-exergy terms produce a reason instead of a physical plot.

Reading the result

Read the environment, property model and mass before comparing exergy values. Initial and final specific values are separate from their totals. The four bars are signed total ΔE, Eq, −Ew and −Ed; the three contributions sum to the stored change. They are not a nonnegative loss distribution.

Check the first-law residual, entropy production and exergy residual separately. A calculation may produce finite diagnostics while describing an impossible process. Such a case does not receive a physical transfer chart. The reported T0 times sigma remains negative when the entropy balance is negative so the conflict stays visible.

Speeds are nonnegative magnitudes; elevations share one environmental datum. Changing that datum changes initial and final total potential energy together, not the physical energy change. The stored exergy of contents colder than the environment must not be confused with the sign of heat exergy at a cold boundary.

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M17Flow Exergy and Exergetic Efficiency

The control-volume exergy balance and the per-component second-law efficiency. This is where a plant stops being a set of energy numbers and becomes a ranked list of where the work potential is actually being lost.

What it rests on

The same idea as closed-system exergy, applied to a stream. A flowing fluid carries work potential across a boundary, and the maximum work recoverable from a unit of it, as it comes to the dead state, is its flow exergy. The atmospheric-displacement term is absent because a stream does not push the atmosphere aside -- the flow work is already inside the enthalpy.

How the working relation follows

Take the enthalpy difference from the dead state, subtract the dead-state temperature times the entropy difference for the reason it was subtracted in the closed-system case, and add the kinetic and potential energies, which are wholly available. That is the specific flow exergy, and it is exactly zero at the dead state.

The control-volume balance then follows the same shape as the energy balance with one extra term. Exergy enters with the streams and with heat -- the heat carrying the reversible factor rather than itself -- leaves with the streams and as work, and the balance does not close: what is missing is destroyed, and the amount destroyed is the dead-state temperature times the entropy produced. That non-closure is the point. An energy balance always closes and therefore never tells you where anything was lost.

The exergetic efficiency is the exergy the component delivered over the exergy it was given, and what counts as the product and what counts as the input differs by component. A turbine's product is its work and its input is the drop in the stream's exergy. A compressor's product is the rise in the stream's exergy and its input is the work. A heat exchanger's product is what the cold stream gained and its input is what the hot stream gave up. Picking the wrong pairing gives a number that is not wrong arithmetically and answers a different question, which is why the app names the product and the input for whichever component is chosen rather than presenting one formula.

A throttle has an exergetic efficiency of exactly zero, by construction: it produces nothing and destroys everything it drops. That is the clearest single demonstration of what this measure asks that thermal efficiency does not.

Assumptions

• Steady flow, with no accumulation and one common environmental reference and mechanical frame.
• Ordinary components use the selected fixed-composition fluid model for thermomechanical exergy. The combustor separately includes chemical exergy of fuel, air and frozen products.
• Signed external heat crosses at the stated boundary temperature. Its exergy is added algebraically to the supplied resource; a negative heat-exergy term is deducted.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• Exergy destruction rate is non-negative in every component.
  Verification: second-law check
• The sum of component destruction rates equals the plant-level difference between exergy in and exergy out plus work, to rounding.
  Verification: closure test; this is what makes an exergy audit trustworthy
• Every exergetic efficiency lies in [0, 1].
  Verification: property test
• A throttle produces zero shaft work. Its net supplied exergy equals destroyed exergy, including signed boundary heat; total specific energy increases by q.
  Verification: exactness test
• At the same-model dead thermodynamic state, zero speed and z=z0, specific physical flow exergy is exactly zero.
  Verification: exactness test
• Adding the same elevation offset to every endpoint and z0 leaves component work and exergy unchanged.
  Verification: independent component mechanical balance

Monotonic trends:

• Specific flow exergy increases with pressure at fixed temperature, and with temperature at fixed pressure above the dead-state temperature.
• Component exergetic efficiency falls as the component's isentropic efficiency falls.

Where it gets delicate

How it is solved

Choose turbine, compressor, heat exchanger, direct mixing, throttle or combustor. The state solvers construct each actual stream and its reference state. Energy and entropy balances include signed heat, shaft work and the entered kinetic and potential energies before forming destruction and the component-specific resource/product ratio. Rate and specific results use the same declared reference-mass basis.

Direct mixing requires the same substance, caloric model and pressure in both streams. A non-mixing exchanger may use two different fluids; a sampled heat-load path checks temperature approach. A throttle is isenthalpic only when its external heat and kinetic/potential changes vanish. Otherwise those terms belong in the energy balance and change its recovered outlet enthalpy.

The combustor uses the selected CH4, H2 or CO fuel, dry-air supply and a consistent Model I or Model II chemical reference. Complete frozen products supply the reacting energy/entropy account, with explicit fuel, air and product mechanical energies. There is no shaft work or modeled pressure loss in this branch. It is not an arbitrary reaction network.

Inputs or balances without a resolved nonnegative destruction and meaningful resource denominator are refused or marked unavailable. The visible notes identify the efficiency convention; signed-heat net-input efficiency differs from counting every positive exergy resource separately.

Limitations

• A lumped external-heat balance does not establish local heat-transfer direction throughout a device.
• Mechanical energy uses the specified laboratory frame and environmental elevation datum. Do not omit it merely because an application is not a nozzle.
• The ordinary component models exclude chemical exergy. The dedicated combustor includes it but excludes dissociation, arbitrary reaction networks and finite-rate chemistry.
• Exchanger approach is checked on a sampled path, not by a continuous global proof. This entry does not size heat-transfer equipment or solve arbitrary plant topology.

Reading the result

Read the destruction, not the efficiency, if you want to know where the plant is losing. The efficiency compresses a component's behaviour into one number and the destruction says how many kilowatts of work potential that number cost.

An exergetic efficiency is almost always worse than the thermal efficiency of the same component, and that is the measure doing its job: it asks a harder question. A condenser with an excellent energy balance has a poor exergetic one, and a throttle has none at all.

If the app calls a component inactive rather than giving it an efficiency, it received no exergy. That is a statement about your inputs, not about the component.

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M21Vapour Plant Exergy Audit

Component-by-component exergy destruction for any cycle built on the previous three screens, as a ranked table and a Grassmann diagram. It answers the question the energy balance structurally cannot: not where the energy went, but where the ability to do work was lost.

What it rests on

Apply the flow-exergy balance to each component of an assembled cycle in turn and rank the results. The question this answers is the one the energy balance structurally cannot: not where the energy went -- it went out of the condenser, and that is uninteresting -- but where the ability to do work was lost.

How the working relation follows

Each component gets its own control volume. The exergy entering with its streams and with any heat, minus what leaves with its streams and as work, is what it destroyed; and by the identity from the entropy balance, that is the dead-state temperature times its entropy production. Summing over the components and adding whatever left the plant unused must return the exergy supplied minus the net work. That closure is the test for this screen: an audit that does not close is not an audit.

The answers are usually surprising, and the surprise is the pedagogy. A condenser rejects most of the energy and destroys almost none of the exergy, because it rejects it at very nearly the dead-state temperature. The steam generator destroys most of it, and it does so before the plant has produced any work at all -- the irreversibility is in the temperature difference between the flame or the source and the boiling water, not in any of the machinery.

Where the boundary is drawn decides how the condenser's loss is named, and both namings are correct. With the sink at the dead state the boundary is the environment: the whole drop from the condensing steam to the surroundings happens inside the plant, so it is destruction and the loss is zero. Give a real cooling-water temperature above the dead state and the boundary moves inward: part of that drop now happens outside, so it becomes loss -- exergy carried out rather than destroyed. The total is the same either way. The split is a bookkeeping choice and the screen has to say which one it made, which is why the sink temperature is a parameter and not a constant.

The source temperature is a parameter for a related reason. Heat at 800 K and heat at 2000 K carry very different work potential, so without a stated source temperature the exergy supplied is undefined, and defaulting it would be inventing the answer.

Assumptions

• The cycle is one the vapour-cycle screens assembled, so every station is a state the property solver fixed rather than a number typed in.
• Heat enters at a single stated source temperature. A real furnace supplies heat across a range of temperatures and its exergy input is lower than a single top temperature suggests.
• The environment is a stated dead state, and the sink temperature is stated separately so that the boundary is explicit.
• The boundary can begin at a heat source or at a stated fuel-exergy reference. The fuel-reference mode adds fuel conversion and heat-supply destruction plus calculated exhaust exergy using frozen complete products and a consistent chemical reference; it is not a general combustion-equilibrium plant model.
• Each component is adiabatic to the surroundings except where the arrangement says otherwise; casing losses are not modelled.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• Component destructions plus losses plus net work equal the exergy supplied, to rounding.
  Verification: THE test for this screen; an audit that does not close is not an audit
• Every component destruction is non-negative.
  Verification: second-law check
• Plant exergetic efficiency computed from the ranked table equals the value computed directly from inlet and outlet exergy.
  Verification: cross-check

Monotonic trends:

• The steam generator dominates the destruction table whenever the source temperature is far above the boiling temperature.
• Total destruction rises in proportion to the dead-state temperature at fixed entropy production.

Where it gets delicate

How it is solved

Assemble the cycle, walk its legs, and compute each component's destruction from its own terminal states and the transfers already attached to that leg. Rank by destruction, largest first. Everything below a stated threshold is grouped into one residual line rather than ranked, because a component whose destruction is comparable to the rounding of its own terms would otherwise change places between runs and the ranking would look unstable for reasons that have nothing to do with the plant.

The geometry of the flow diagram is computed here, not in the interface. The band widths are produced from the audit that already closed, with the closure carried alongside so that the interface can refuse to draw a diagram that does not add up. A flow diagram whose widths a front end derived for itself would eventually not conserve its own flow, and a diagram that does not conserve its flow is worse than no diagram: it looks authoritative and it is wrong.

An internal transfer -- a regenerator moving heat from one part of the cycle to another -- is marked rather than omitted. It is not a supply and not a rejection, so it must not enter the efficiency; but it still destroys exergy, so the audit has to see it.

Limitations

• The audit covers water/steam basic, single-reheat, double-reheat and regenerative cycles. Regeneration uses one to eight heaters of one selected kind throughout: open, closed with drains trapped, or closed with drains pumped. It is not an arbitrary mixed-topology plant modeller.
• The heat-source boundary excludes upstream fuel conversion. The fuel-reference boundary evaluates the chosen fuel, air supply, reactant/product temperatures and chemical reference, scales the released heat to the cycle duty, and recomputes destruction for heat transfer at the stated source temperature. High-temperature dissociation remains outside this frozen-product boundary.
• A single source temperature stands in for a heat input that in reality spans a range.
• Components below 0.5% of supplied exergy are grouped in the main ranking; the complete component table remains available. Grouping affects presentation, not the balance.
• Pressure drops and casing heat losses are not modelled, so the destruction attributed to piping is zero here and is not zero in a plant.

Reading the result

Read the ranking, then read the closure. The ranking is the answer; the closure is what says the answer is trustworthy, and if the components plus losses plus net work do not return the exergy supplied, nothing above it is worth reading.

Expect the steam generator at the top and the condenser near the bottom. If your intuition says the condenser should dominate, that intuition is about energy, and the whole purpose of this screen is the gap between the two accounts.

Read the boundary line before comparing two audits. Moving the sink temperature moves quantities between the destroyed column and the lost column without changing the total, so two audits with different sink temperatures are not comparable line by line even though their totals are.

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M18Rankine Cycle

A four-station vapour power cycle using water (IF97), R134a, ammonia, propane or R22 (Helmholtz). Resolve turbine and pump efficiencies, mass flow or power rating, property paths and sampled pressure trends. Alternatively couple a water/steam topping loop to a separate refrigerant bottoming loop through one counterflow exchanger. Split cogeneration diverts a chosen steam fraction to process heating, mixes liquid returns at a common pressure and feeds the bottoming boiler with the remaining steam.

What it rests on

A single working fluid passes through a turbine, condenser, pump and boiler. Water uses IF97; R134a, ammonia, propane and R22 use their respective published Helmholtz equations. State enthalpies/entropies and specific transfers are per kilogram. Heat, work and entropy-production rates retain their rate units. One property model supplies every actual and ideal reference state.

How the working relation follows

State 1 is saturated vapour at boiler pressure or a temperature-defined vapour or supercritical state. Find the ideal turbine exit from condenser pressure and s2s = s1, then set h2 = h1 - eta_t (h1 - h2s). The condenser returns the fluid to saturated liquid, state 3. Find the ideal pump outlet from boiler pressure and s4s = s3, then set h4 = h3 + (h4s - h3)/eta_p. The pump reference uses the same EOS; it is not silently replaced by a constant-volume liquid.

Turbine work is h1-h2, pump work is h4-h3, boiler heat is h1-h4 and condenser heat rejection is h2-h3. Net work is turbine work minus pump work. Efficiency is net work divided by boiler heat; back work ratio is pump work divided by turbine work. Their values and trends depend on the fluid and temperatures; steam-specific rules of thumb are not universal laws for organic cycles.

In mass-flow rating mode, multiply each specific transfer by the supplied kg/s to obtain kW. In net-power mode, divide the specified kW by specific net work to obtain kg/s. The same flow then determines both heat rates.

In binary mode two separate four-station loops retain independent mass flows and property references. The steam condenser supplies the entire bottoming boiler duty: r_m = q_out,top / q_in,bottom. Per kilogram of topping flow, combined net work is w_top + r_m w_bottom. Only the topping boiler supplies external heat and only the bottoming condenser rejects external heat. The internal exchanger is counted once and cancels from the overall balance. Its entropy production is mdot_top(s3-s2) + mdot_bottom(s1-s4), on each fluid's own reference basis. The streams never mix.

Split cogeneration instead sends fraction y of the turbine exhaust to a process heater. The remaining fraction (1-y) heats the refrigerant boiler and returns as saturated liquid at the specified return pressure. A second liquid stream returns from the process at that same pressure and its stated temperature. Number steam states as 1 turbine inlet, 2 exhaust/split, 3 exchanger condensate, 4 mixed feed, 5 pump outlet and 6 process return. Adiabatic mixing gives h4=(1-y)h3+y h6. The full steam flow crosses both steam machines; only the two branches are split. Thus Qprocess=mdot y(h2-h6), Qexchange=mdot(1-y)(h2-h3) and Qexternal=mdot(h1-h5). The bottoming flow follows from its boiler duty. Deduct both pump powers from both turbine powers. External boiler heat equals net electric power plus process heat plus bottoming condenser rejection. Electrical efficiency and heat-and-power utilization are different ratios; useful process heat is not additional shaft power. Mixer entropy production is mdot[s4-(1-y)s3-y s6]. Process-load exergy and entropy production require external heating temperatures not supplied by this calculation.

Assumptions

• Steady flow with negligible kinetic and potential energy changes; one common mass flow through all four devices in each ordinary loop. Split-cogeneration branch flows are explicitly weighted.
• Adiabatic turbine and pump, each described by one isentropic efficiency above zero and no greater than one.
• Ordinary loops have constant-pressure heat transfer and saturated-liquid condenser outlets. Split cogeneration permits steam-side exchanger pressure drop and fixes both returns at the common mixing pressure. No auxiliary loads or generator losses.
• External heat-source and sink temperatures are not specified. The fluid temperature-extreme bound is a consistency bound, not a full reservoir exergy audit.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• Cycle efficiency does not exceed 1 minus the minimum heat-rejection fluid temperature divided by the maximum heat-addition fluid temperature. This is an internal temperature-extreme bound, not a claim about unspecified external reservoirs.
  Verification: canon-wide bound INV-12-BOUND
• Net work equals heat added minus heat rejected, to rounding.
  Verification: closure test
• Around the closed cycle the sum of entropy changes is exactly zero, because entropy is a property.
  Verification: cycle closure; this catches a state assigned to the wrong region
• The actual cycle efficiency is strictly below the ideal cycle efficiency at the same pressures, and equals it when both isentropic efficiencies are 1.
  Verification: limit test
• Steam condenser heat rate equals bottoming boiler heat rate; combined net power equals both loop powers; only top boiler and bottom condenser enter external heat totals.
  Verification: independent balances, flow scaling and signed entropy production
• The full steam flow passes both steam machines. Only the process and exchanger branches split. External boiler heat equals combined net electric power plus process heat plus bottoming condenser rejection.
  Verification: independent balances and power scaling

Monotonic trends:

• For conventional water/steam examples on a feasible continuous branch: Thermal efficiency rises as boiler pressure rises, at fixed condenser pressure and fixed turbine inlet temperature. This is not imposed as a universal law for every nonwater fluid.
• For conventional water/steam examples on a feasible continuous branch: Thermal efficiency rises as condenser pressure falls. This is not imposed as a universal law for every nonwater fluid.
• For conventional water/steam examples on a feasible continuous branch: Thermal efficiency and turbine exit quality both rise as superheat temperature rises. This is not imposed as a universal law for every nonwater fluid.
• For conventional water/steam examples on a feasible continuous branch: At fixed turbine inlet temperature, turbine exit quality falls as boiler pressure rises. This is the trade-off the whole chapter is about. This is not imposed as a universal law for every nonwater fluid.

Where it gets delicate

How it is solved

PH and PS flashes resolve the state within the selected model and branch. Helmholtz inverse brackets end at known saturated states rather than excluding a fixed temperature interval next to coexistence. Internal single-phase flashes do not absorb a nonzero temperature step into a saturated endpoint. The numerical property residual remains 64 ULP on its stated reference scale; a bounded neighbouring-temperature search handles final floating-point noise without widening that tolerance. Unresolved states are refused explicitly.

The cycle requires positive, resolved turbine work, pump work, net work and heat transfers. It checks each adiabatic component's entropy increase and the assembled efficiency bound. A pressure span no greater than 2e-5 of boiler pressure is refused: the 1e-11 relative saturation-pressure inversion budget at two endpoints would not preserve six relative digits in that span. Enthalpy differences must exceed 128 ULP of the state enthalpy scale divided by 1e-6. These are numerical conditioning guards, not physical minimum sizes.

Property diagrams sample constant-pressure heat paths and insert saturated liquid/vapour states as exact phase-change breakpoints. Dashed machine connections show actual endpoints; separately sampled PS curves show ideal isentropic references. An efficiency determines endpoints, not a unique irreversible internal path. Chart area must not be read as actual shaft work. The local saturation envelope has two separate branches; a truncated top is not labelled as the critical point.

A pressure scan evaluates 21 evenly spaced boiler or condenser pressures at the other entered conditions. It plots efficiency, specific net work or exit quality. Rejected states and undefined single-phase qualities create gaps. The samples establish neither a continuous optimum nor unsampled feasibility.

The binary exchanger uses transferred-heat fraction as coordinate, from steam outlet/bottom inlet to steam inlet/bottom outlet. Phase-change enthalpies are explicit breakpoints. At each point a forward PT bracket must straddle the requested h beyond a 128-ULP inverse/interpolation budget. The initial temperature half-width is 1e-7 times absolute temperature; it is expanded when necessary, not assumed to be the EOS accuracy. Successful states at a model temperature limit permit a one-sided bracket. On each stable monotone isobar, hot lower temperature at the left endpoint minus cold upper temperature at the right bounds the approach over that interval. Subdivide when needed; overlapping endpoint brackets are reported as unresolved at this check's resolution, not necessarily as a physical crossing.

This is a conservative numerical check using forward-property brackets and an error budget, not verified interval arithmetic or experimental proof of EOS accuracy. The displayed lower estimate is distinct from the minimum sampled temperature difference and is not an exact pinch search.

In split cogeneration, pressure may change across the steam exchanger. Endpoint balances do not determine p(h), so the default reports endpoint approaches and leaves internal feasibility unresolved. An optional linear pressure versus transferred-heat-fraction assumption samples 33 temperatures; positive samples do not certify unsampled intervals. This is not the ordinary binary isobaric bound. Export titles and legends retain the assumption.

Zero process flow reuses the condensate as mixed feed and ignores the inactive return temperature. Nonzero mixed enthalpy contributions must exceed the property roundoff budget. Mixer and exchanger entropy production must be normal finite numbers and resolve six relative digits above their explicit roundoff budgets. Nearly isothermal mixing may therefore be refused even when physically possible; no noisy negative value is silently clamped to zero. When a machinery efficiency equals one, its solved PS state is reused rather than unnecessarily reflashed through PH, avoiding a second inverse roundoff.

Limitations

• The ordinary four-device loop has no pressure losses, reheaters, feedwater heaters, subcooled condenser exit or auxiliary loads. Split cogeneration separately permits the specified steam-exchanger pressure drop and process-return mixing.
• Saturated admission cannot be requested at or above critical pressure. A temperature-defined inlet must resolve within the selected EOS range. The boiler regime is labelled from the resolved inlet phase.
• The erosion warning is specific to water/steam. Other fluids display their calculated quality without inheriting the steam threshold.
• HEOS agreement verifies an independent implementation of the same published EOS, not a second physical model or experimental accuracy.
• Binary mode assumes the entire steam condenser duty heats one bottoming boiler, with no leakage, mixing, pressure losses or exchanger sizing model.

Reading the result

Read the selected fluid, efficiencies, rating basis and exit phase together. The station table and chart use the same kilogram basis and selected display units. A quality of zero or one is shown explicitly; quality is absent for a single-phase state. Read the signed component entropy differences alongside the numerical guard rather than treating a last-digit sign as physical proof.

Pressure-scan gaps are reasons to inspect the rejected input, not lines to join by hand. For a refused cycle, the reason identifies the state or transfer that could not be resolved; no efficiency or work diagram is certified.

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M19Rankine with Reheat

Expanding in two stages with reheat between them raises the average temperature of heat addition and keeps the exit out of the wet region. The reheat pressure is the free parameter and exit quality is what constrains it.

What it rests on

Expand in two stages and put the steam back through the boiler between them. Six stations instead of four, the same devices and the same summation. Reheat raises the average temperature at which heat is added a little, and it raises the turbine exit quality a lot -- and the second is the reason it exists.

How the working relation follows

Split the expansion at a chosen reheat pressure, return the steam to the boiler, heat it back up to a stated reheat temperature, and expand the rest of the way. The efficiency is the sum of the two expansion works minus the pump work, over the sum of the two heat additions.

The quality argument is the sharp one. A single expansion from a high boiler pressure to a deep condenser vacuum ends far inside the dome. Getting the exit dry enough by superheat alone would need a turbine inlet temperature no turbine metal survives; splitting the expansion and reheating in the middle gets there instead, because the second expansion starts from a much higher temperature at a much lower pressure.

The efficiency gain is real and modest, and it has a genuine interior maximum in the reheat pressure. Reheating just below the throttle adds almost no area to the cycle -- the second expansion is nearly the whole of it and the reheat did nothing. Reheating just above the condenser adds area at a low mean temperature. Somewhere between, the efficiency peaks, conventionally near a quarter of the boiler pressure.

The peak is shallow, and saying so is more useful than the number. A plant designer does not get to choose the reheat pressure freely, and what matters is how much is lost by taking the pressure the machinery allows instead of the optimum -- which, on a flat peak, is very little.

At either end the arrangement degenerates. As the reheat pressure approaches the boiler pressure one expansion stage vanishes and the cycle reduces exactly to the simple one, which is the limit the implementation is checked against.

Assumptions

• Everything assumed for the simple cycle: steady state, no pressure drops, saturated liquid at the condenser exit, one isentropic efficiency per machine.
• The reheat happens at constant pressure and returns the steam to a stated temperature, usually the same as the throttle temperature.
• Both expansion stages share one turbine efficiency. A real machine has different efficiencies in its high- and low-pressure sections.
• The reheat line's pressure drop is neglected, which in a real plant is a few per cent and is one of the reasons the optimum sits where it does.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• As reheat pressure approaches the boiler pressure, the cycle reduces exactly to the simple cycle of M18.
  Verification: limit test
• Efficiency never exceeds the corresponding reversible bound.
  Verification: canon-wide bound INV-12-BOUND
• Reheat raises the turbine exit quality relative to the same cycle without it.
  Verification: comparison test against M18

Monotonic trends:

• Turbine exit quality rises monotonically as reheat temperature rises.
• Efficiency has an interior maximum in reheat pressure, so the optimum is a genuine optimisation rather than a bound.

Where it gets delicate

How it is solved

The same walk as the simple cycle with two extra stations. That the summation, the bound and the closure check are untouched is the point of having a shared frame at all.

The optimum reheat pressure is found by search rather than from a derivative, for the same reason the gas-turbine optimum is: a closed form needs a constant specific heat ratio, and steam does not have one. The search is on a logarithmic grid because the answer sits around a fifth of the way up a two-decade range, and a linear grid spends almost all its points above it.

Because the objective is flat near its maximum, a naive search wanders. The app brackets the optimum and reports the flatness rather than a spurious precision -- quoting the peak to four figures would be claiming a resolution the objective does not have.

Limitations

• Choose one or two reheaters, entered pressure, maximum-efficiency pressure search, or pressure from target final turbine quality. For double reheat the first pressure stays fixed and the search varies the second. Multiple quality roots require an explicit choice; a boundary-limited efficiency is not an attained interior maximum.
• One turbine efficiency covers the expansion stages.
• The reheat line is assumed to have no pressure drop and to add no thermal inertia.
• Double reheat and its cycle path are available; compare the calculated benefit rather than assuming a universal efficiency gain.

Reading the result

Compare against the same cycle without reheat, at the same boiler and condenser pressures. The efficiency gain will be a few points; the quality gain will be the reason anyone builds it.

Move the reheat pressure and watch how little the efficiency changes over a wide range. That flatness is the practical result: it means the choice can be made on mechanical grounds without costing much.

If one of the two expansions has collapsed to nothing, the screen has reduced to the simple cycle and says so rather than solving a zero-length process.

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M20Regenerative Rankine

Bleeding steam from the turbine to heat the feedwater. The extraction fraction is not an input, it comes out of the feedwater heater energy balance, and that is the step students get wrong.

What it rests on

Bleed part of the steam out of the turbine part-way down and use it to heat the feedwater before it reaches the boiler. The bled steam does less work, but the heat it gives up is heat the boiler no longer has to supply at a low temperature -- and raising the average temperature of heat addition is the only thing that raises a cycle's efficiency without touching its extremes.

How the working relation follows

An open feedwater heater mixes the bled steam directly with the feedwater at the extraction pressure, so its exit is saturated liquid there and the arrangement needs a second pump to lift that liquid to the boiler pressure.

The extraction fraction is not an input. It comes out of the heater's own energy balance: the fraction is the enthalpy rise the feedwater needs, over the enthalpy the bled steam can give up. Every other quantity on the screen is something the user chose, which is exactly why this is the step students get wrong -- they look for a fraction to type.

Everything downstream of the extraction point carries only the remaining fraction of the flow. The cycle frame handles that by weighting each leg's transfers where the fraction is known, rather than by teaching the summation what a feedwater heater is; the regenerative cycle was the test of that design and it needed no special case in the summation.

The efficiency has an interior maximum in the extraction pressure. Extract too close to the throttle and the bled steam is valuable and there is little feedwater heating to do; extract too close to the condenser and the bled steam is nearly spent and heats the feedwater barely at all. In between, the two effects cross.

Two limits check the implementation. As the extraction fraction goes to zero the cycle reduces exactly to the simple one; and every extraction fraction must lie strictly between zero and one, with a fraction outside that range meaning the heater balance was set up backwards.

Assumptions

• One to eight heaters of one chosen type: open direct mixing, closed with trapped drains, or closed with pumped-forward drains.
• Open heater outlets and closed heater drain outlets use the prescribed saturated-liquid idealisation. Heater balances are adiabatic to the surroundings.
• Pump and turbine efficiencies apply to the corresponding stages; pressure losses and parasitic auxiliaries beyond the assembled devices are not modelled.
• Extraction pressures are strictly ordered between condenser and boiler pressures.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• Every extraction fraction lies in (0, 1) and their sum is strictly below 1.
  Verification: property test; a fraction outside this range means the heater balance was set up backwards
• As the extraction fraction goes to zero the cycle reduces exactly to M18.
  Verification: limit test
• Regeneration raises efficiency relative to the same cycle without it, at the same boiler and condenser pressures.
  Verification: comparison test against M18
• Each feedwater heater satisfies its own energy balance to rounding.
  Verification: closure test per heater

Monotonic trends:

• Efficiency has an interior maximum in extraction pressure.
• Efficiency rises with the number of heaters, with strongly diminishing returns.

Where it gets delicate

How it is solved

The single-open-heater relation above illustrates the balance. For the selected number and type, the solver constructs the corresponding feedwater, extraction and drain states, then solves the coupled heater energy and mass balances for extraction fractions. Trapped drains cascade toward lower-pressure heaters; pumped drains return according to the pumped-forward arrangement.

The cycle is assembled with each leg weighted by its actual flow. Saturated outlets are fixed by pressure and quality, avoiding ambiguous saturation p–T inversion. Extraction and remaining-main-flow fractions must be physical, and the heater and whole-cycle balances must close. Nearly coincident extraction pressures can produce a singular or ill-conditioned fraction system and are refused when the required solution is not resolved.

Limitations

• All heaters use one selected topology. Arbitrary mixtures of open, trapped-drain and pumped-drain units are outside this entry.
• Closely spaced extraction pressures can make the coupled fraction system ill-conditioned.
• More heaters do not guarantee an improvement for arbitrary user-selected pressures and losses. Compare the calculated useful work, heat input and efficiency.
• Finite temperature approaches, pressure losses and additional plant auxiliaries are not inferred by the idealised heater balances.

Reading the result

Read the extraction fraction first. It is the output that carries the physics, and it is the number a hand calculation gets wrong. If it is not strictly between zero and one, the arrangement described by your inputs does not work.

Compare the efficiency against the simple cycle at the same boiler and condenser pressures. The gain is the whole justification for the extra hardware.

Move the extraction pressure across its range and watch the efficiency rise and fall. The maximum is interior, which means the choice is a genuine optimisation and not a limit you push against.

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M22Otto Cycle

The air-standard spark-ignition cycle. Both the variable-specific-heat result and the cold-air-standard closed form are shown, because the gap between them is what the cold-air assumption costs.

What it rests on

The air-standard idealisation of a spark-ignition engine: the working fluid is air throughout, the combustion is replaced by heat addition from outside, the exhaust and intake are replaced by heat rejection, and every process is internally reversible. What survives is a four-process cycle in which the piston fixes the volume and the pressure follows -- the only screen in this app where that is the direction of the constraint.

How the working relation follows

Compress isentropically through the compression ratio, add heat at constant volume to a peak temperature, expand isentropically back through the same ratio, reject heat at constant volume back to the start. The heat added is the internal energy rise at constant volume; the heat rejected is the internal energy fall at constant volume; the efficiency is one minus their ratio.

With constant specific heats the two internal energy changes are the specific heat at constant volume times temperature differences, the two temperature ratios across the isentropic legs are equal, and everything cancels to leave one minus the compression ratio raised to one minus the specific heat ratio. Efficiency depends on the compression ratio alone -- not on the heat added, not on the peak temperature -- which is the clean result the cold-air-standard analysis exists to produce.

With variable specific heats it does not cancel and the efficiency is no longer a function of the compression ratio alone. The isentropic legs are solved through the relative volume rather than through a power relation, because a power relation is only exact when the specific heat ratio is constant. Across the temperature span of a real engine it is not, and the cold-air-standard analysis overstates the efficiency by an amount that grows with the peak temperature.

Mean effective pressure is the net work divided by the displaced volume. It is the constant pressure that, acting over one stroke, would do the same work, and it is the quantity that says how large the engine has to be for a given output.

Assumptions

• Air-standard: the working fluid is air throughout, treated as an ideal gas with the fitted temperature-dependent properties.
• Combustion is replaced by heat transfer from an external source, so nothing here knows about fuel, mixture strength, flame speed or knock.
• The exhaust and intake strokes are replaced by heat rejection at constant volume, so the cycle is closed and the same air goes round again.
• Every process is internally reversible; there is no friction, no blowby, no heat loss to the cylinder wall, and no finite combustion duration.
• The compression and expansion ratios are equal, which they are in a conventional engine and are not in an over-expanded one.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• Efficiency never exceeds 1 - T1/T3.
  Verification: canon-wide bound INV-12-BOUND
• Net work equals heat added minus heat rejected, to rounding.
  Verification: closure test
• In the limit of constant specific heats the variable-cp result converges exactly to the closed form.
  Verification: limit test; it fails if the relative-volume path is wrong

Monotonic trends:

• Thermal efficiency increases monotonically with compression ratio.
• At fixed compression ratio, cold-air-standard efficiency increases with the specific heat ratio.
• Mean effective pressure increases with heat added at fixed compression ratio.

Where it gets delicate

How it is solved

Fix the inlet state, move isentropically through the volume ratio using the relative volume, add heat at constant volume to the peak temperature, expand isentropically back through the same volume ratio, and close. Every state is a state the gas solver fixed, so the cycle plot and the station table are drawn from the same objects as the performance numbers.

The constant-volume moves are done by holding the specific volume and setting the new temperature, with the pressure following from the ideal-gas relation. That is the direction the physics runs in a piston engine, and writing it that way is what makes the relative volume the right tool rather than a convenience.

The three reciprocating cycles are one family, and the implementation is checked by the family relations rather than only by isolated reference cases: the dual cycle degenerates to this one when its cutoff ratio is one, so the same arithmetic is reached by three routes and they must agree.

Limitations

• Air-standard means no combustion. Real compression ratios in a spark-ignition engine are limited by knock, which is a chemistry constraint this model cannot see.
• Choose variable NASA specific heat or cold-air standard. Enter the selected design quantity and compare model results on the same inlet state and heat-input or peak-temperature basis.
• As the compression ratio approaches one the cycle collapses: net work and displaced volume both go to zero and the mean effective pressure is zero divided by zero. The app reports the degenerate cycle rather than a ratio of vanishing quantities.
• At very high compression ratios the peak temperature leaves the fitted polynomial range and the app says which state left it.
• No friction, no pumping loop, no heat loss. The indicated efficiency of a real engine is far below this and its brake efficiency further still.

Reading the result

Read the efficiency against the compression ratio and nothing else, if you are thinking in cold-air-standard terms -- and then notice that the screen's efficiency also moves when you change the peak temperature. That difference is the variable-specific-heat effect, and it is the honest part of the model.

Mean effective pressure is the size number. Two engines with the same efficiency and different mean effective pressures are different sizes for the same output.

The peak pressure at the top of the constant-volume heat addition is worth watching. It is what the structure has to withstand, and it rises much faster than the efficiency does.

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M23Diesel Cycle

Compression ignition, with heat added at constant pressure. Placed beside the spark-ignition result at equal compression ratio, because the comparison is the point and neither cycle means much alone.

What it rests on

The same air-standard idealisation with the heat added at constant pressure rather than at constant volume, which is what a compression-ignition engine approximates: fuel is injected into air already hot enough to ignite it, and the piston is moving while it burns. Everything else about the cycle is unchanged, and the comparison at equal compression ratio is the point.

How the working relation follows

Compress isentropically, add heat at constant pressure while the piston moves out to a cutoff ratio, expand isentropically to the original volume, reject heat at constant volume. The heat added is now an enthalpy rise, because the process is at constant pressure; the heat rejected is still an internal energy fall.

With constant specific heats the efficiency is the constant-volume result multiplied by a bracketed factor built from the cutoff ratio and the specific heat ratio. That factor is greater than one for every cutoff ratio above one, so at equal compression ratio the constant-pressure cycle is always less efficient. Real compression-ignition engines are more efficient anyway, because they run at compression ratios a spark-ignition engine cannot reach.

The bracketed factor is zero divided by zero at unit cutoff ratio and its limit there is exactly one, which is the statement that the cycle reduces to the constant-volume one when the heat addition happens instantaneously.

The cutoff ratio is an output here, not an input. It is how far the piston has travelled by the time heat addition stops, and that is fixed by the peak temperature rather than chosen independently.

Assumptions

• Air-standard, with the same list of idealisations as the constant-volume cycle: air throughout, external heat transfer instead of combustion, internally reversible processes.
• Heat addition is at exactly constant pressure. In a real engine the pressure falls during injection and the process is closer to the dual cycle.
• The expansion runs back to the original volume, so the expansion ratio exceeds the compression ratio by the cutoff ratio.
• The compression ratio is high enough for compression ignition, which this model does not check because it does not know about fuel.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• As the cutoff ratio approaches 1, the cycle reduces exactly to the spark-ignition cycle at the same compression ratio.
  Verification: limit test against M22
• At equal compression ratio, this cycle is always less efficient than the constant-volume one, because the bracketed factor exceeds 1 for every cutoff ratio above 1.
  Verification: ordering check with a symbolic derivation
• Efficiency never exceeds the corresponding reversible bound.
  Verification: canon-wide bound INV-12-BOUND

Monotonic trends:

• At fixed compression ratio, efficiency falls monotonically as cutoff ratio rises.
• At fixed cutoff ratio, efficiency rises monotonically with compression ratio.

Where it gets delicate

How it is solved

The same walk as the constant-volume cycle with the heat-addition leg at constant pressure. States are moved through the relative volume and the relative pressure as each leg requires, and the cutoff ratio falls out as the ratio of the two volumes across the heat-addition leg.

The bracketed factor's singularity at unit cutoff ratio is handled by a series expansion joined so that the two branches agree to full precision at the join and not merely on either side of it. The join itself is a high-precision comparison point, because a check placed near the join rather than on it would pass on an implementation that is wrong exactly there.

Limitations

• The native entry supports peak temperature, cutoff ratio or heat per mass as the heat-addition input, with variable NASA or cold-air-standard models. An alternative input determines the same physical cycle only when its recovered state and heat match.
• Constant-pressure heat addition is an idealisation of injection. A real compression-ignition engine's pressure trace is closer to the dual cycle, which is the next screen.
• The comparison against the constant-volume cycle is only meaningful at equal compression ratio, which is not how the two engines are actually built.
• Air-standard, so no fuel, no injection timing, no ignition delay and no soot.

Reading the result

Read this cycle beside the constant-volume one at the same compression ratio, because neither means much alone. At equal compression ratio this one is always the less efficient, and the bracketed factor is exactly how much.

The cutoff ratio is a result. If it comes out close to one, your peak temperature has made the heat addition nearly instantaneous and you have approximated the constant-volume cycle.

The efficiency falls as the cutoff ratio rises, which is to say as more heat is added. That is the opposite of the intuition that more heat means more work, and it is the characteristic behaviour of this arrangement.

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M24Dual Cycle

Heat added partly at constant volume and partly at constant pressure, which is closer to what a real engine does than either limiting cycle. It degenerates to both, and asserting those degeneracies is how the implementation is checked.

What it rests on

Heat added partly at constant volume and then at constant pressure. It is closer to what a real engine's pressure trace does than either limiting cycle, and it contains both of them: it becomes the constant-volume cycle when the cutoff ratio is one and the constant-pressure cycle when the pressure ratio is one.

How the working relation follows

Five stations instead of four. Compress isentropically; add heat at constant volume through a stated pressure ratio; add the rest at constant pressure out to the cutoff; expand isentropically to the original volume; reject heat at constant volume.

The efficiency is one minus the constant-volume heat rejection over the sum of the two heat additions -- an internal energy rise for the first and an enthalpy rise for the second. There is no single memorable closed form worth quoting, which is itself informative: the two limiting cycles have neat formulas because each has one heat-addition process, and reality has two.

At equal compression ratio the efficiency lies between the two limits, and it rises as more of the heat is added at constant volume. That ordering is the practical content: adding heat early, while the volume is still small, is worth more than adding it while the piston is already moving out.

The two degeneracies are how the implementation is checked. Set the cutoff ratio to one and the result must equal the constant-volume cycle at the same compression ratio, exactly; set the pressure ratio to one and it must equal the constant-pressure cycle. The same arithmetic reached by three routes has to agree, and that is a stronger check than any single reference point.

Assumptions

• Air-standard, with the same list of idealisations as the other two reciprocating cycles.
• The split between constant-volume and constant-pressure heat addition is fixed by the stated pressure ratio, with the remainder added at constant pressure out to the peak temperature.
• Both heat-addition processes are internally reversible and the pressure ratio is at least one.
• The expansion runs back to the original volume.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• With cutoff ratio exactly 1 the cycle reduces exactly to the constant-volume cycle.
  Verification: limit test against M22
• With pressure ratio exactly 1 the cycle reduces exactly to the constant-pressure cycle.
  Verification: limit test against M23
• At equal compression ratio the efficiency lies between the two limiting cycles.
  Verification: ordering test
• Efficiency never exceeds the corresponding reversible bound.
  Verification: canon-wide bound INV-12-BOUND

Monotonic trends:

• At fixed total heat addition, efficiency rises as more of it is added at constant volume.

Where it gets delicate

How it is solved

Choose pressure ratio plus cutoff ratio, heat fraction plus total heat addition, or pressure ratio plus peak temperature. Each pair determines the two heat-addition stages before the common cycle is assembled. The gas heat-capacity model determines how energy increments become temperature changes.

The pressure-ratio-one and cutoff-ratio-one limits reduce to the corresponding single-stage heat-addition cycles. If both stages add zero heat, thermal efficiency is indeterminate; the solver reports a degenerate cycle.

Limitations

• The heat-fraction mode divides total added heat between constant-volume and constant-pressure stages. A chosen pressure/cutoff or peak-temperature pair must satisfy the same admissible state and heat-direction constraints.
• Variable-heat-capacity and cold-air models are air-standard idealisations: no finite burn duration, fuel chemistry or wall heat loss is resolved.
• The two idealised heat-addition segments are not a measured engine pressure trace.

Reading the result

Read the two ratios as the shape of the heat release. A high pressure ratio with a cutoff near one is nearly the constant-volume cycle; the reverse is nearly the constant-pressure one; and where the two are balanced is where a real compression-ignition engine sits.

The efficiency should always land between the two limiting cycles at the same compression ratio. That bracketing is worth checking by eye, because it is the cheapest confirmation that the split you specified is the one you meant.

Peak pressure rises with the constant-volume fraction, and it is the structural constraint that stops a designer moving further in the direction efficiency prefers.

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M25Brayton Cycle

The gas turbine cycle. The pressure ratio that maximises net work is not the one that maximises efficiency, and having both curves on the same axes is the reason this screen exists rather than a formula.

What it rests on

The gas turbine cycle: compress, heat, expand, reject. It runs on the same frame as the vapour cycle and uses none of the water devices, which is the point -- the frame describes a cycle, not a steam cycle. What distinguishes it is that the compressor's work is a large fraction of the turbine's, so the net work is a difference of two big numbers and behaves quite differently from the vapour plant's.

How the working relation follows

Compress isentropically through the pressure ratio, add heat at constant pressure to the turbine inlet temperature, expand isentropically to the original pressure, reject heat at constant pressure. The efficiency is the net work over the heat added; the back work ratio is the compressor work over the turbine work, and where a vapour plant's is a fraction of a per cent, a gas turbine's is a third or more.

Two optima follow, and they are different pressure ratios. The specific net work is the turbine work minus the compressor work; raising the pressure ratio raises both, and at first the turbine gains faster. Beyond some ratio the compressor gains faster, so the net work has an interior maximum. For the ideal cycle with constant specific heats that maximum is at the pressure ratio whose isentropic temperature ratio is the square root of the overall temperature ratio.

Efficiency behaves differently. For the ideal cycle it rises monotonically with pressure ratio all the way to the limiting ratio at which the compressor exit reaches the turbine inlet temperature -- where the heat addition and the net work have both gone to zero and the efficiency has reached the reversible bound. So the ideal cycle has no interior efficiency optimum at all, and the efficiency optimum is that boundary. An interior optimum in efficiency exists only because of the losses: the shape of the efficiency curve is created by irreversibility, not by thermodynamics.

That limiting ratio is also what explains the shape of the net-work curve. Without it, the curve appears to fall off for no visible reason; with it marked, the axis has a right-hand end where the cycle stops producing anything.

Assumptions

• Air-standard and open-cycle-as-closed: the same air goes round, with heat rejection standing in for the exhaust and the fresh intake.
• Combustor and heat-rejection pressure losses are specified fractions of each exchanger’s own inlet pressure. Zero fractions recover the isobaric idealisation.
• The two turbomachines are described by isentropic efficiencies, and everything else in the cycle is reversible.
• The working fluid's properties are the fitted temperature-dependent ideal-gas ones. Combustion products differ from air, and this model does not carry that difference.
• The turbine inlet temperature is a stated metallurgical limit rather than something the model derives.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• Efficiency never exceeds 1 - T1/T3.
  Verification: canon-wide bound INV-12-BOUND
• Net work equals heat added minus heat rejected, to rounding.
  Verification: closure test
• In the limit of constant specific heats the variable-cp result converges exactly to the closed form.
  Verification: limit test
• The pressure ratio maximising specific net work is strictly below the one maximising efficiency. In the strongest form: net work has an interior maximum, while efficiency rises monotonically all the way to the limiting pressure ratio where the compressor exit reaches the turbine inlet temperature -- so the efficiency optimum IS that boundary, where heat addition and net work have both gone to zero.
  Verification: ordering check with a symbolic derivation

Monotonic trends:

• Back work ratio increases monotonically with pressure ratio.
• Ideal efficiency increases monotonically with pressure ratio.
• Specific net work has an interior maximum in pressure ratio.
• Raising the turbine inlet temperature raises net work per unit of gas strongly, and at a FIXED pressure ratio lowers thermal efficiency slightly -- measured at about one percentage point between 1200 K and 1600 K. The cold-air-standard formula makes efficiency a function of pressure ratio alone, so this is a variable-specific-heat effect and it runs opposite to what that formula suggests. What does rise with firing temperature is the ACHIEVABLE efficiency: the limiting pressure ratio moves from 172 to 571 over that range, and efficiency rises with pressure ratio all the way to it.

Where it gets delicate

How it is solved

Fix the inlet, move isentropically to the compressor exit pressure through the relative pressure rather than through a power relation, apply the compressor efficiency, heat to the turbine inlet temperature, expand the same way, apply the turbine efficiency, and close. Across a gas-turbine pressure ratio the difference between the relative-pressure route and the constant-ratio power relation is tens of kelvin.

Both optima use the same selected caloric model and bounded stationary-root search. A closed constant-specific-heat formula is not substituted into the variable-property calculation. Component efficiencies and pressure losses remain inputs to every objective evaluation.

Plant rates multiply the corresponding specific energy transfers by the stated air mass flow. This changes neither stations nor the optimum ratios. Zero flow is a valid zero-duty rating. Non-finite products or rates that cannot retain six useful digits near underflow, including after display-unit conversion, are refused rather than presented as zero or infinite power.

The numerical design search brackets the positive-heat and positive-net-work region of the selected caloric model with both pressure losses applied. It reports the lower and upper zero-work boundaries separately from the zero-heat-input boundary. Net-work and efficiency optima are refined within that region; a limiting efficiency at vanishing work is explicitly identified instead of being offered as a finite-power optimum.

Limitations

• The entry offers variable NASA or cold-air-standard properties, net-work and efficiency pressure-ratio searches, and displayed zero-work and zero-heat-input boundaries. A zero-work efficiency limit is distinguished from an interior optimum and cannot be applied as a useful operating cycle.
• Air throughout. A real gas turbine expands combustion products with a different composition and a different specific heat, and the mass flow through the turbine exceeds that through the compressor by the fuel.
• Heat-exchanger pressure drops are entered as fractions. Cooling-air bleed, shaft losses and compressor/turbine maps are not modelled.
• As the pressure ratio approaches one, compressor and turbine work both vanish and the back work ratio is zero divided by zero. The app reports the degenerate cycle rather than a ratio of vanishing quantities.
• As the back work ratio approaches one, the net work is the difference of two large numbers and loses figures. The app floors it and shows both gross terms so that the cancellation is visible rather than hidden.

Reading the result

Read the back work ratio. It is the number that separates this cycle from the vapour one, and it is why a gas turbine's performance is so sensitive to component efficiency: a few points off the compressor come straight out of a net work that is already a difference.

Raising the turbine inlet temperature raises the net work per unit of gas strongly and, at a fixed pressure ratio, lowers the thermal efficiency slightly -- measured at about one percentage point between 1200 K and 1600 K. That runs opposite to the constant-specific-heat formula, which makes efficiency a function of pressure ratio alone; it is a variable-specific-heat effect. What does rise with firing temperature is the achievable efficiency, because the limiting pressure ratio moves a long way up and efficiency rises with pressure ratio all the way to it.

If the cycle comes back with no net work, check the pressure ratio against the limiting one for your two temperatures. Past it, there is nothing left to add heat with.

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M26Brayton with Regeneration, Reheat and Intercooling

Three independent modifications on one screen, because they interact: intercooling and reheat each lower efficiency on their own and raise it once a regenerator is present. Toggling them one at a time is how that becomes visible.

What it rests on

Three modifications to the gas turbine cycle, on one screen because they interact. A regenerator moves heat from the exhaust to the compressor exit. Intercooling splits the compression and cools between stages. Reheat splits the expansion and reheats between stages. Each is simple; together they are not additive, and the direction of their effect changes depending on which of the others are present.

How the working relation follows

The regenerator's effectiveness is the enthalpy the cold stream actually gains over the most it could gain, which is set by it reaching the hot stream's inlet temperature -- an infinite-area limit. So the effectiveness is bounded by one for a reason having nothing to do with the first law, and a value above one means the two stream temperatures have been crossed somewhere.

On their own, intercooling and reheat each lower the efficiency. They raise net work by moving compression to a colder gas and expansion to a hotter one, but they also leave the compressor exit cooler and the turbine exit hotter, and without a regenerator the extra heat between the compressor exit and the turbine inlet has to come from the fuel. Add the regenerator and the same two changes now raise efficiency, because that heat comes from the exhaust instead. Toggling one at a time is how that becomes visible, and it is why the arrangement is offered as three independent choices rather than as a preset.

The staging is ideal: intercooling back to the compressor inlet temperature, reheating up to the turbine inlet temperature, with the pressure ratio split evenly between stages -- which is the optimum for equal stages and follows from the same geometric-mean argument as the intercooled compressor. In the limit of many stages with a perfect regenerator the cycle approaches the reversible bound from below, which is the check that the staging machinery is right.

The regenerator has a hard limit of its own. Above some pressure ratio the compressor exit is hotter than the turbine exit, and there is nothing left to recover: the effectiveness definition's denominator vanishes and then changes sign. That crossover is a real operating condition at high pressure ratio, not a hypothetical, and the app refuses the regenerator above it rather than reporting a transfer from the cooler stream to the hotter one.

Assumptions

• Ideal intercooling returns the gas to the compressor inlet temperature and ideal reheat returns it to the turbine inlet temperature. Real intercoolers and reheat combustors do neither exactly.
• The overall pressure ratio is split evenly between equal stages, which is the optimum for identical stages.
• The regenerator has no pressure drop on either side. A real one has a significant drop on the exhaust side and that is one of the reasons regenerators are not universal.
• All compression stages share one isentropic efficiency, and all expansion stages share another.
• Air-standard throughout, with the same idealisations as the simple gas turbine.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• With regenerator effectiveness exactly zero and one stage of each, the cycle reduces exactly to M25.
  Verification: limit test; this single assertion validates the whole configuration machinery
• Regenerator effectiveness lies in [0, 1], and effectiveness 1 makes the regenerator exit temperature equal the turbine exit temperature exactly.
  Verification: property test and exactness test
• A regenerator never transfers heat from the cooler stream to the hotter one: the arrangement is refused when the compressor exit is hotter than the turbine exit.
  Verification: second-law check; this is a real operating condition at high pressure ratio, not a hypothetical
• As the number of intercooling and reheat stages grows with a perfect regenerator, efficiency approaches the reversible bound from below.
  Verification: limit test
• Efficiency never exceeds the corresponding reversible bound.
  Verification: canon-wide bound INV-12-BOUND

Monotonic trends:

• With a regenerator, efficiency falls as pressure ratio rises, the opposite of the simple cycle.
• Intercooling raises specific net work and, without a regenerator, lowers efficiency.
• Efficiency rises with stage count and approaches a limit.

Where it gets delicate

How it is solved

Build the station list from the three choices, then hand it to the same summation as every other cycle. The regenerator's transfer is marked as internal: it moves heat within the cycle rather than across its boundary, so it is neither a supply nor a rejection. Counting it as supply is the defect that makes a regenerator appear to do nothing at all -- efficiency is net work over heat supplied, and an internal transfer added to both sides of that ratio leaves it unchanged. Marking rather than omitting it matters too, because an internal exchanger still destroys exergy and the audit has to see it.

The degenerate case is the whole test of the configuration machinery: with the regenerator effectiveness exactly zero and one stage of each, the result must equal the simple gas-turbine cycle exactly. That single assertion validates everything the three choices do.

The crossover is detected before the regenerator is applied, and the refusal names the two temperatures rather than the pressure ratio: the compressor delivers the gas at this temperature and the turbine exhausts it at that one, so a regenerator here would have to move heat from the cooler stream to the warmer one. The pressure ratio at which the two cross is not computed for this arrangement -- the simple gas-turbine cycle reports its own crossover margin as a temperature difference, and that is the quantity to watch.

Limitations

• Ideal staging. Real intercoolers have approach temperatures, real reheat combustors have pressure drops, and both move the optimum.
• The regenerator is characterised by one effectiveness with no pressure drop. On a real machine the exhaust-side pressure drop partly cancels the gain.
• Above the crossover pressure ratio the regenerator is refused rather than applied at reduced effectiveness, because there is nothing to recover.
• With a regenerator, efficiency falls as pressure ratio rises -- the opposite of the simple cycle. A design intuition carried over from the simple cycle points the wrong way here.
• Air-standard, with all the limitations that carries.

Reading the result

Toggle one modification at a time. The screen is arranged that way because the interaction is the lesson: intercooling alone lowers efficiency and raises work, and intercooling with a regenerator raises both.

With a regenerator fitted, watch what happens as you raise the pressure ratio. Efficiency falls, which is the reverse of the simple cycle, because a higher pressure ratio leaves less temperature difference for the regenerator to work with.

If the regenerator is refused, read the two temperatures in the refusal. The compressor exit is hotter than the turbine exit, so there is nothing left to recover -- which is a design answer, not a failure.

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M27Combined Gas and Steam Cycle

A gas turbine topping a vapour cycle, coupled through the heat recovery steam generator. The coupling is the whole design problem: the pinch point decides the steam flow, and the steam flow decides the output.

What it rests on

A combined cycle uses gas-turbine exhaust to supply a steam cycle through a heat-recovery generator. The two cycles have different circulating mass flows, and the exchanger must satisfy both an energy balance and a feasible temperature approach throughout heating and evaporation.

How the working relation follows

On a unit gas-flow basis, the exhaust enthalpy drop equals the steam-to-gas mass ratio times the feedwater-to-steam enthalpy rise. That balance determines steam flow for a stated stack temperature. If the minimum temperature approach is specified instead, steam flow and stack temperature must be chosen together so the two exchanger profiles remain separated.

Water can remain at nearly constant temperature while evaporating, whereas the exhaust continues cooling. Consequently, endpoint temperatures alone can miss an internal pinch or temperature crossing. The total net work is gas-cycle net work plus steam-cycle net work multiplied by the recovered mass ratio. The heat-recovery transfer is internal to the combined boundary and must not be counted again as an external heat input when computing overall efficiency.

Assumptions

• The native topping cycle uses air with temperature-dependent NASA properties and the bottoming cycle uses water/steam properties.
• The recovery generator transfers heat internally without an external heat leak. Stated turbine, compressor and pump efficiencies define the component states.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• The temperature difference between the two streams is non-negative everywhere inside the recovery generator, not merely at the ends. A profile that crosses internally while satisfying both terminal balances is the classic trap and is refused.
  Verification: profile scan across the whole exchanger, not just at the terminals
• Energy given up by the gas stream equals energy taken up by the steam stream.
  Verification: closure test
• Combined efficiency exceeds either cycle alone at the same firing temperature.
  Verification: comparison test against M18 and M25
• Overall efficiency never exceeds the reversible bound between firing temperature and condenser temperature.
  Verification: canon-wide bound INV-12-BOUND

Monotonic trends:

• Steam flow and net output both fall as the pinch-point temperature difference widens.
• Overall efficiency rises with gas turbine inlet temperature.

Where it gets delicate

How it is solved

Each component state is solved first. The coupling then uses either the entered stack temperature or a bracketed search enforcing the requested approach. Gas and water profiles are compared at sampled equal-duty positions, including water phase changes. Net power and heat rates retain the gas-flow and steam-flow bases explicitly.

Limitations

• The reported pinch is a sampled profile minimum, not a mathematical certification of every unsampled point or a prediction of exchanger area.
• Zero temperature approach is a limiting idealization requiring unbounded exchange area, not a selectable finite design. Fouling, pressure-loss distribution and detailed gas chemistry are not inferred.

Reading the result

Read the steam-to-gas ratio before interpreting steam-side power or mass flow. Compare the gas and water profiles, not only their endpoints. Overall efficiency counts the external topping-cycle heat supply once; adding recovered heat to that denominator would double-count an internal transfer.

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M28Turbojet Cycle

Diffuser, compressor, burner, turbine, nozzle. The property side only: the compressible-flow treatment of the nozzle itself is deliberately outside this app and the manual says where it lives.

What it rests on

A turbojet leaves useful energy in the flow for acceleration through a nozzle. Its turbine is sized thermodynamically to supply compressor work rather than maximize shaft output, and thrust follows the change in stream momentum under the declared nozzle boundary.

How the working relation follows

Flight kinetic energy contributes to stagnation enthalpy at the diffuser. Compression increases enthalpy, the burner adds heat to the specified turbine-inlet temperature, and the turbine removes the work required by the compressor. The remaining nozzle-inlet enthalpy can become kinetic energy on expansion. With the stated nozzle efficiency, the actual kinetic-energy rise is a fraction of the isentropic enthalpy drop.

An afterburner raises temperature after the turbine and therefore adds energy available to the nozzle without increasing the compressor's work demand. Under the fully expanded, constant working-flow approximation, specific thrust is exit velocity minus flight velocity, and total thrust multiplies that difference by mass flow. No nozzle area or pressure-thrust term can be inferred without geometry.

Assumptions

• The native working flow uses air with temperature-dependent NASA properties; burner heat addition does not introduce a separately tracked fuel mass flow.
• Turbine work balances compressor work. The reported momentum thrust assumes expansion to ambient and omits a separate pressure-thrust term.
• Component efficiencies represent thermodynamic losses; no intake, combustor or nozzle geometry is solved.

Inputs

Outputs

Choices made explicit

What always holds

Exact relations:

• Turbine work exactly balances compressor work, which is what fixes the turbine exit state in this arrangement.
  Verification: exactness test; getting this wrong is the standard error in the section
• Stagnation enthalpy is conserved across the diffuser and across the nozzle.
  Verification: exactness test
• Thrust is zero when exit velocity equals flight velocity.
  Verification: exactness test

Monotonic trends:

• Specific thrust has an interior maximum in compressor pressure ratio at fixed flight condition.
• Specific thrust rises with turbine inlet temperature.

Where it gets delicate

How it is solved

The app propagates stagnation energy through the diffuser, compressor, burner, work-balanced turbine and optional afterburner. Isentropic expansion provides the nozzle reference drop, efficiency sets exit kinetic energy, and the actual exit state is recovered from the remaining enthalpy. A pressure-ratio check uses the inlet local specific-heat ratio as a critical-ratio estimate for a converging nozzle. It is a geometry diagnostic, not a statement that the calculated fully expanded exit is sonic.

Limitations

• A choking warning means a converging-only nozzle cannot realize the assumed fully expanded endpoint. This task does not size a throat, solve area–Mach relations or determine an underexpanded jet's pressure thrust.
• Specific thrust is not propulsive efficiency, fuel consumption or a full engine performance map. A state with no positive expansion pressure difference is refused.

Reading the result

Use the nozzle warning when deciding whether the displayed idealized endpoint is physically realizable. Compare afterburner states at the same compressor and flight conditions. Increasing nozzle exit speed may raise thrust while requiring substantially more heat; this model does not supply the missing fuel-system calculation.

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M29Vapour-Compression Refrigeration

Four-station vapour compression with explicit evaporator and heat-rejection pressure losses, local outlet superheat/subcooling, and a separate transcritical discharge-pressure search.

What it rests on

Move heat from a cold space to a warm one by evaporating a fluid at a low pressure, compressing its vapour, condensing it at a high pressure, and throttling it back. The whole design rests on a fluid whose saturation temperature moves a long way with pressure: choosing the two pressures chooses the two temperatures, and everything else follows.

How the working relation follows

Four component balances determine the cycle. With zero exchanger pressure loss, evaporation and condensation follow constant-pressure paths on the pressure–enthalpy plane. With declared pressure losses those legs connect different station pressures. The valve remains isenthalpic because it performs no work and exchanges no heat. Compressor outlet enthalpy follows its isentropic comparison and stated efficiency.

The refrigerating effect per unit mass is the enthalpy rise across the evaporator, which is the evaporator exit enthalpy minus the condenser exit enthalpy -- the throttle having carried the latter across unchanged. The compressor work per unit mass is the enthalpy rise across it. The coefficient of performance is their ratio, and the condenser duty is their sum, which is the cycle-level energy balance and the check the screen reports.

The throttle is what stops this cycle reaching its reversible bound. It is the one component that produces entropy without producing anything, and replacing it by an expander -- which nobody does at this scale, because the work recovered from a liquid is small and the machine is not -- is what the bound assumes.

Above the critical pressure the cycle gains a degree of freedom. A subcritical cycle's high-side pressure is fixed by the temperature it must condense at; there is no choice to make, because saturation ties pressure to temperature. A transcritical cycle does not condense at all: it cools a supercritical fluid through a gas cooler, with no phase change and therefore no saturation temperature tying the two together. The discharge pressure becomes a free variable, and it has an optimum -- push it up and the gas cooler exit sits further left on a steeply sloping isotherm, so the refrigerating effect grows; push it up and the compressor works harder. The two cross. That optimum is the single most distinctive fact about carbon dioxide refrigeration and it does not exist in any subcritical cycle, which is why the two treatments are kept apart rather than blended.

Assumptions

• Steady state. Each exchanger pressure-loss fraction is measured against that exchanger’s own inlet pressure, not the compressor pressure ratio.
• The expansion is a throttle at constant enthalpy, not an expander.
• The compressor is adiabatic and is described by one isentropic efficiency.
• The subcritical temperature inputs define saturation pressure at each exchanger inlet. Superheat and subcooling are relative to the local outlet saturation temperature after pressure loss; gas-cooler exit temperature is entered directly.
• Superheat at the compressor inlet and subcooling at the condenser exit are stated amounts rather than consequences of a coil model.

Inputs