PetroCalcHub
Interactive engineering lesson

What a Compositional Flash Calculation Solves

Follow a hydrocarbon PT flash from feed to liquid and vapor, with a six-component balance ledger, worked mole allocations and an interactive phase-split explorer.

By PetroCalcHub Team | Updated 2026-10-24

Original diagram Interactive example Worked practice 3 technical sources

Related workflow calculators

Engineering context

A compositional flash calculation asks how a specified feed distributes between the phases allowed by a thermodynamic model. In a pressure-temperature, or PT, flash, you supply overall mole fractions, absolute pressure and temperature. The solver determines a supported equilibrium state, including a liquid-vapor split when the model predicts two phases. The phase compositions are outputs, not extra feed specifications.

This lesson follows one reproducible C1–nC6 Peng–Robinson case at 280 K and 2 MPa absolute. We replayed the site's pinned CoolProp 8.0.0 implementation and retained the full-precision phase compositions for the ledger. The explorer selects a component and scales the feed amount; it does not re-solve equilibrium when you move the amount slider. Use the linked flash workflow to change temperature, pressure or composition.

Feed · overall z

100 mol

280 K · 2 MPa absolute · C1–nC6

↓ Equilibrium phase split

Liquid · composition x

34.243588 mol

Methane x = 0.097146; nC6 x = 0.143224

Vapor · composition y

65.756412 mol

Methane y = 0.709792; nC6 y = 0.001452

For every component: feed moles = liquid moles × x + vapor moles × y

A fixed PT equilibrium case splits a 100-mol feed into two phase amounts. Each phase has its own composition denominator; every feed component must be accounted for in the two allocations.
Try the idea

Change one thing and observe the result

Explore the replayed PR case at 280 K and 2 MPa absolute. Select a component, then scale the feed amount. The fixed equilibrium compositions and beta remain unchanged; changing conditions requires the linked flash solver.

Methane in feed
50.000000 mol
Methane in liquid
3.326629 mol
Methane in vapor
46.673371 mol

Total phases: L = 34.243588 mol; V = 65.756412 mol. Component balance difference: 7.1e-15 mol. Beta = 0.657564116.

Methane: three composition denominators
Feed z — per feed mole0.500000
Liquid x — per liquid mole0.097146
Vapor y — per vapor mole0.709792

All bars run from 0 to 1. Their lengths are not phase amounts; use L × x and V × y for the component allocations above.

Practical workflow

  1. 1

    Specify the supported components and verify that feed mole fractions sum to one.

  2. 2

    Set one equilibrium temperature and absolute pressure before solving the phase split.

  3. 3

    Read phase amounts separately from the composition within each phase.

  4. 4

    Reconstruct the component ledger and review equilibrium checks before using the result.

Specify a feed on one composition basis

The overall feed composition z describes component moles divided by total feed moles. Here z is 0.50 methane, 0.15 ethane, 0.15 propane, 0.10 normal butane, 0.05 normal pentane and 0.05 normal hexane. The entries sum to one. On a 100-mol feed basis, they correspond to 50, 15, 15, 10, 5 and 5 mol respectively.

The component list matters as much as the numbers. Normal butane and isobutane are different substances, and a heavy petroleum fraction is not automatically equivalent to normal hexane. This particular workflow supports the six listed hydrocarbons with zero binary interaction parameters. Its software bounds are 180–450 K and 0.1–10 MPa absolute; those bounds describe accepted inputs, not an assurance of accuracy throughout that region.

∑izi=1,nifeed=Fzi\sum_i z_i=1,\qquad n_i^{\mathrm{feed}}=Fz_i

Separate phase amount from phase composition

Liquid composition x describes moles of a component divided by all liquid-phase moles. Vapor composition y uses all vapor-phase moles as its denominator. Each phase composition vector separately sums to one. The vapor mole fraction beta instead describes total vapor moles divided by total feed moles; 1 − beta is the liquid mole fraction of the feed.

For this case beta is 0.657564116, giving 65.756412 mol of vapor and 34.243588 mol of liquid per 100 mol of feed. Methane's vapor-phase composition is about 0.70979194, but that does not mean 70.98% of the whole feed becomes methane vapor. Multiply the vapor amount by its methane composition to find that component's actual vapor allocation.

V=βF,L=(1−β)F,niL=Lxi,niV=VyiV=\beta F,\quad L=(1-\beta)F,\quad n_i^L=Lx_i,\quad n_i^V=Vy_i

Close a separate balance for every component

Every component that enters must be present in the returned phases because this PT case contains no reaction or material loss. Dividing Fzi = Lxi + Vyi by the feed amount gives zi = (1 − beta)xi + beta yi. The liquid and vapor allocation terms share the feed-mole denominator, so they can be added directly. The compositions xi and yi by themselves cannot simply be added.

The table retains the same fixed state for all six components. Composition columns use moles per mole of their own phase; allocation columns use moles per mole of feed. Display values are rounded, while the balance checks use the full-precision snapshot. The largest unrounded component residual is approximately 5.6 × 10⁻¹⁷ mol/mol feed, so small apparent rounding differences in the displayed rows are not missing material.

Replayed PR case: phase compositions and component allocations (rounded display)
ComponentFeed zLiquid xVapor yLiquid / feedVapor / feed
C10.5000000.0971460.7097920.033266290.46673371
C20.1500000.1245310.1632630.042643880.10735612
C30.1500000.2553390.0951430.087437340.06256266
nC40.1000000.2422910.0259000.082969280.01703072
nC50.0500000.1374690.0044500.047074160.00292584
nC60.0500000.1432240.0014520.049044940.00095506

Scroll the table horizontally to see every column.

See why mole, mass and volume fractions differ

A mole counts an amount of substance, while mass depends on each component's molecular mass. This vapor is enriched in lighter molecules relative to the liquid. Its fraction of total moles is therefore larger than its fraction of total mass: approximately 65.7564% of the feed moles are vapor, compared with 44.8913% of the feed mass. Both numbers can be correct for the same equilibrium split.

An in-situ phase volume fraction requires phase volumes at the stated conditions. It is not beta, and it is not a flow holdup measurement. To calculate a volume fraction you would need consistent phase amounts and molar volumes, or masses and densities. Keep the basis in every output label when carrying flash results into separator, storage or flow calculations; changing the denominator changes the quantity being reported.

Different denominators in one equilibrium result
QuantityValueDenominator
Vapor mole fraction0.657564116All feed moles
Vapor mass fraction0.448913296All feed mass
Methane in vapor, yC10.709791940All vapor moles

Understand what makes a split an equilibrium calculation

A component ledger alone permits many mathematically balanced splits. Equilibrium adds the requirement that each present component has equal fugacity in the liquid and vapor phases at the common pressure and temperature. The equation of state supplies the phase-dependent fugacity coefficients. The compositions affect those coefficients, so phase composition and phase amount must be solved together rather than assigned independently.

The workflow uses the pinned library's PT stability search and phase-split calculation, then audits its returned phase pressures, composition sums, component balances and fugacity agreement. For this snapshot the maximum log-fugacity mismatch is about 1.9 × 10⁻¹⁰ and the relative EOS pressure residual about 8.8 × 10⁻¹². These are numerical consistency diagnostics for this model; they do not measure agreement with an actual fluid sample.

fiL=fiV⟹xiφiL=yiφiVat common P,Tf_i^L=f_i^V\quad\Longrightarrow\quad x_i\varphi_i^L=y_i\varphi_i^V\quad\text{at common }P,T

Recognize the supported phase model before extending the case

The teaching example uses Peng–Robinson mixing rules with every binary interaction parameter kij set to zero. That is an explicit model choice rather than a sample-specific interaction fit. The linked C1–nC6 workflow does not include water, sour gases, heavy pseudocomponents, wax or an additional liquid phase. Those phenomena need a model and component specification that actually support them.

A returned single phase also needs careful reading: no absent-phase composition or arbitrary vapor fraction should be invented to fill a two-phase table. Near a vanishing phase or critical region, the workflow applies conservative separation and conditioning gates. Use the PT flash to study a specified state and the phase-envelope tool for its separately supported boundary task. Changing total feed moles in this ledger only rescales amounts at the same fixed equilibrium state.

Worked example

Account for methane in a 100-mol feed

At 280 K and 2 MPa absolute, use the replayed Peng–Robinson split with beta = 0.657564115965, zC1 = 0.50, xC1 = 0.097146040408 and yC1 = 0.709791940346.

  1. 1. Find methane in the feed
    100 × 0.50 = 50.000000 mol methane.

    The feed composition is applied to all 100 mol.

  2. 2. Find the two phase amounts
    V = 65.756411597 mol; L = 34.243588403 mol.

    Beta partitions feed moles; it does not partition each component by the same percentage.

  3. 3. Allocate methane within each phase
    L × xC1 ≈ 3.326629 mol; V × yC1 ≈ 46.673371 mol.

    Each phase amount multiplies its own methane composition.

  4. 4. Recombine the methane
    3.326629 + 46.673371 ≈ 50.000000 mol.

    The unrounded values close to floating-point precision; the displayed values are rounded.

Result

The methane entering in the feed is recovered in the two phase allocations, with most of this case's methane in the vapor phase.

Interpretation

The same balance must close independently for ethane through normal hexane. A high methane concentration in vapor is distinct from the total vapor mole fraction.

Test your understanding

Double the feed from 100 to 200 mol at the same composition, pressure and temperature. What happens to beta and the methane allocations?

Show the worked answer

Beta, x and y stay unchanged in this fixed equilibrium case. Vapor and liquid amounts double, as do methane allocations: approximately 6.653258 mol in liquid and 93.346742 mol in vapor, summing to 100 mol methane.

Method selection guide

Match the calculation method to the physical question and the evidence available.

ConditionUseWhy
You know feed composition, P and TUse the supported PT flashIt solves the phase state and equilibrium split for those specifications.
You need to understand an existing splitReconstruct the component ledgerPhase amounts and within-phase compositions must use the correct denominators.
You need another phase, component set or energy specificationChoose a workflow supporting those inputsA fixed C1–nC6 PT example cannot supply missing phase physics or a heat balance.

Before using the result

  • List every supported component and identify normal isomers.
  • Enter feed mole fractions that sum to one, not unconverted mass percentages.
  • Use absolute pressure and absolute temperature on the stated unit bases.
  • Keep beta separate from vapor mass fraction and phase compositions.
  • Check all component balances, phase composition sums and fugacity diagnostics.
  • Record the EOS, interaction assumptions and the difference between numerical checks and sample validation.

Start with these calculators

These links keep the guide close to the working calculator flow.

2 calculators

Continue learning

Verified sources used for this guide