Engineering lesson
Phase fraction versus phase composition in a flash result
Phase fraction tells how much of each phase exists. Phase composition tells what each phase contains. A vapor mole fraction β is not the methane mole fraction in the vapor.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Separate amount from composition
For total feed F, vapor moles are βF and liquid moles are (1 − β)F. Vapor component i is βFyi and liquid component i is (1 − β)Fxi.
Each phase's composition sums to one independently. Neither xi nor yi is multiplied by the overall feed amount until a component inventory is requested.
Use a component allocation ledger
The recorded 280 K, 2 MPa PR case has β about 0.65756, while methane's vapor composition is about 0.70979. These are different fractions with different denominators.
This lesson expands the component allocation exercise rather than repeating the introductory flash solver. The explorer lets readers inspect feed, liquid and vapor contributions component by component.
Mole and mass fractions need another conversion
Phase molecular weights differ, so vapor mass fraction generally differs from vapor mole fraction. A volume fraction additionally depends on phase densities.
Do not relabel β as gas holdup in a flowing pipe. Holdup also depends on in-situ occupancy and potentially slip, beyond an equilibrium phase split.
Explore the example
Fixed snapshot from the site's CoolProp 8 PR engine, independently compared with the thermo reference fixture. Selecting a row does not run a new flash.
Select one of 6 recorded scenarios. Intermediate states are not interpolated.
Component number (C1 to nC6) = 1. Feed component (mol): 50; Liquid component (mol): 3.32662902; Vapor component (mol): 46.673371.
- Feed component (mol)
- Liquid component (mol)
- Vapor component (mol)
Worked example
Component allocation in one recorded flash
Fixed PR case: 280 K, 2 MPa(a), total feed 100 mol, vapor mole fraction 0.657564115965. Component number selects methane through n-hexane.
Relation and unit basis
feed_i = 100zi; liquid_i = 100(1 − β)xi; vapor_i = 100βyi
Rows 1–6 are C1, C2, C3, nC4, nC5 and nC6. The equilibrium state does not change with component selection.
1. Define the baseline
Component number (C1 to nC6) = 1
Fixed PR case: 280 K, 2 MPa(a), total feed 100 mol, vapor mole fraction 0.657564115965. Component number selects methane through n-hexane.
2. Select the relation
feed_i = 100zi; liquid_i = 100(1 − β)xi; vapor_i = 100βyi
Rows 1–6 are C1, C2, C3, nC4, nC5 and nC6. The equilibrium state does not change with component selection.
3. Substitute the values
Methane feed = 100 × 0.5 = 50 mol; liquid = 100 × (1 − 0.657564116) × 0.0971460404; vapor = 100 × 0.657564116 × 0.70979194; liquid + vapor = 50 mol
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
Feed component (mol) = 50; Liquid component (mol) = 3.32662902; Vapor component (mol) = 46.673371
Phase composition and phase amount have different denominators. Every component must close independently.
Result and interpretation
Feed component (mol) = 50; Liquid component (mol) = 3.32662902; Vapor component (mol) = 46.673371
Phase composition and phase amount have different denominators. Every component must close independently.
The equilibrium state is fixed. Component selection changes the balance row, not the thermodynamic solve.
Exact scenario data
Fixed snapshot from the site's CoolProp 8 PR engine, independently compared with the thermo reference fixture. Selecting a row does not run a new flash.
| Component number (C1 to nC6) | Feed component (mol) | Liquid component (mol) | Vapor component (mol) |
|---|---|---|---|
| 1 | 50 | 3.326629 | 46.67337 |
| 2 | 15 | 4.264388 | 10.73561 |
| 3 | 15 | 8.743734 | 6.256266 |
| 4 | 10 | 8.296928 | 1.703072 |
| 5 | 5 | 4.707416 | 0.2925842 |
| 6 | 5 | 4.904494 | 0.09550565 |
Scroll the table to see all columns. The highlighted row is the worked baseline. Downloads retain full numerical precision.
Test your understanding
Work through the question before revealing its answer. Use the stated assumptions and include units in every numerical step.
1. With β = 0.6 and yi = 0.8 in a 100 mol feed, how many vapor moles of component i exist?
Show the worked answer
0.6 × 100 × 0.8 = 48 mol. The vapor phase contains 60 mol in total.
2. Reproduce the baseline Feed component (mol) without rounding intermediate values.
Show the worked answer
Methane feed = 100 × 0.5 = 50 mol; liquid = 100 × (1 − 0.657564116) × 0.0971460404; vapor = 100 × 0.657564116 × 0.70979194; liquid + vapor = 50 mol. Feed component (mol) = 50. Phase composition and phase amount have different denominators. Every component must close independently.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Component number (C1 to nC6) changes from 1 to 6. Other controls remain as stated in the premise. Feed component (mol) changes from 50 to 5.
4. What would be needed to apply this example to a different physical system?
Show the worked answer
Replace the synthetic controls with compatible documented inputs and check the relation's assumptions. Rows 1–6 are C1, C2, C3, nC4, nC5 and nC6. The equilibrium state does not change with component selection. Phase composition and phase amount have different denominators. Every component must close independently.
Before using the result
- Name every fraction's denominator.
- Keep mole, mass and volume fractions distinct.
- Check each component balance.
Phase composition and phase amount have different denominators. Every component must close independently.
Continue learning
Start with the broader workflow guide
Sources and method
The explanation and synthetic examples are original PetroCalcHub teaching material. The references below support the definitions and model context. They do not supply field validation for the synthetic scenarios.