Flash calculations

Why it matters

A flash drum is the simplest separator: a feed is brought to a new temperature or pressure, splits into vapour and liquid, and the two are drawn off separately. Flash calculations size these drums, are the building block of every distillation stage in a simulator, and appear whenever a hot pressurised stream is let down through a valve.

Key ideas

The problem. A feed of F mol/s with composition zᵢ reaches equilibrium at a specified T and P. Find the vapour fraction V/F and the compositions xᵢ (liquid) and yᵢ (vapour). By Duhem's theorem, fixing T and P for a given feed fixes everything.

Equations. Component balance zᵢ = xᵢ(1 − V) + yᵢV (with F = 1), equilibrium yᵢ = Kᵢxᵢ, and the sums Σxᵢ = Σyᵢ = 1. Eliminating gives xᵢ = zᵢ/[1 + V(Kᵢ − 1)] and yᵢ = Kᵢxᵢ. Using Σyᵢ − Σxᵢ = 0 gives the Rachford-Rice equation in the single unknown V.

K-values. With Raoult's law Kᵢ = Pᵢ^sat/P, depending only on T and P, so V comes from one equation. With modified Raoult's law Kᵢ = γᵢPᵢ^sat/P, and γ depends on the unknown x, so an outer loop updates γ. For high-pressure hydrocarbon systems K-values come from an equation of state (φ-φ method) or from DePriester-type charts in a data book.

First check whether the feed really splits. At the flash T, compute the feed's bubble pressure (ΣzᵢPᵢ^sat for Raoult) and dew pressure (1/Σzᵢ/Pᵢ^sat). A two-phase result exists only if P_dew < P < P_bubble. Equivalently, with K-values: Σzᵢ Kᵢ > 1 and Σzᵢ/Kᵢ > 1. If P is above the bubble pressure the feed is all liquid (V = 0); below the dew pressure it is all vapour (V = 1).

Solving Rachford-Rice. The function f(V) = Σzᵢ(Kᵢ − 1)/[1 + V(Kᵢ − 1)] decreases monotonically with V between 0 and 1, with f(0) = ΣzᵢKᵢ − 1 > 0 and f(1) = 1 − Σzᵢ/Kᵢ < 0 for a two-phase feed. So bisection always works and Newton's method converges quickly from V = 0.5. For a binary there is a closed form: x₁ = (1 − K₂)/(K₁ − K₂), y₁ = K₁x₁, and V from the lever rule.

Lever rule. On a T-x-y or P-x-y diagram, V/F = (z₁ − x₁)/(y₁ − x₁): the vapour fraction is the length from the liquid line to the feed point divided by the tie-line length.

Adiabatic flash. In a let-down valve and drum without heat input, T is unknown and an energy balance h_F = V·h_V + (1 − V)·h_L is added. The usual approach iterates on T until the energy balance closes, with an isothermal flash inside each step.

Formulas

zᵢ = (1 − V)·xᵢ + V·yᵢ (per mole of feed) yᵢ = Kᵢ·xᵢ Kᵢ = Pᵢ^sat/P (Raoult); Kᵢ = γᵢ·Pᵢ^sat/P (modified Raoult) xᵢ = zᵢ / [1 + V·(Kᵢ − 1)], yᵢ = Kᵢ·xᵢ Σ zᵢ·(Kᵢ − 1) / [1 + V·(Kᵢ − 1)] = 0 (Rachford-Rice) P_dew < P < P_bubble, i.e. Σ zᵢ·Kᵢ > 1 and Σ zᵢ/Kᵢ > 1 (two-phase test) x₁ = (1 − K₂)/(K₁ − K₂), y₁ = K₁·x₁ (binary) V/F = (z₁ − x₁)/(y₁ − x₁) (lever rule) h_F = V·h_V + (1 − V)·h_L (adiabatic flash, per mole of feed)

Symbols: F feed rate (mol/s); V vapour fraction of the feed (V/F, dimensionless); zᵢ feed, xᵢ liquid and yᵢ vapour mole fractions; Kᵢ K-value; Pᵢ^sat vapour pressure (kPa); P drum pressure (kPa); γᵢ activity coefficient; h molar enthalpies (J/mol).

Worked examples

Example 1 (standard): binary flash with Raoult's law. Given: an equimolar benzene (1) and toluene (2) feed is flashed at 95 °C and 101.33 kPa. At 95 °C, P₁^sat = 157.29 kPa and P₂^sat = 63.67 kPa (from Antoine constants in a data book). Find V, x₁ and y₁.

1. Two-phase test: P_bubble = 0.5 × 157.29 + 0.5 × 63.67 = 110.48 kPa; P_dew = 1/(0.5/157.29 + 0.5/63.67) = 90.64 kPa. Since 90.64 < 101.33 < 110.48, the feed splits.
2. K₁ = 157.29/101.33 = 1.5523; K₂ = 63.67/101.33 = 0.6283.
3. x₁ = (1 − K₂)/(K₁ − K₂) = 0.3717/0.9240 = 0.4023.
4. y₁ = K₁x₁ = 1.5523 × 0.4023 = 0.6245.
5. V = (z₁ − x₁)/(y₁ − x₁) = (0.5 − 0.4023)/(0.6245 − 0.4023) = 0.0977/0.2222 = 0.440. Answer: V/F = 0.440, x₁ = 0.402, y₁ = 0.624

Example 2 (GATE level): ternary Rachford-Rice. Given: feed z = (0.3, 0.3, 0.4) with K = (3.0, 1.0, 0.4) at the drum T and P. Find V and both phase compositions.

1. Two-phase test: ΣzᵢKᵢ = 0.9 + 0.3 + 0.16 = 1.36 > 1; Σzᵢ/Kᵢ = 0.1 + 0.3 + 1.0 = 1.4 > 1. Two phases.
2. f(V) = 0.3 × 2/(1 + 2V) + 0 + 0.4 × (−0.6)/(1 − 0.6V) = 0.6/(1 + 2V) − 0.24/(1 − 0.6V).
3. Set f = 0: 0.6(1 − 0.6V) = 0.24(1 + 2V), so 0.6 − 0.36V = 0.24 + 0.48V, giving V = 0.36/0.84 = 0.4286. (Newton from V = 0.5 gives 0.4278, then 0.4286 in two steps.)
4. xᵢ = zᵢ/[1 + V(Kᵢ − 1)]: x₁ = 0.3/1.8571 = 0.1615; x₂ = 0.3/1 = 0.3000; x₃ = 0.4/0.7429 = 0.5385. Sum = 1.000.
5. yᵢ = Kᵢxᵢ: y₁ = 0.4846, y₂ = 0.3000, y₃ = 0.2154. Sum = 1.000. Answer: V/F = 0.429; x = (0.162, 0.300, 0.538); y = (0.485, 0.300, 0.215) Note: a component with K = 1 has the same mole fraction in both phases and in the feed.

Common mistakes

• Solving Rachford-Rice without first checking that the feed is two-phase, and accepting a V outside 0 to 1.
• Using Σxᵢ = 1 alone as the equation to solve; it has a spurious root and converges badly. Use Σ(yᵢ − xᵢ) = 0.
• Treating K as constant when T or P changes; K depends on both (and on composition if γ ≠ 1).
• Confusing vapour fraction V/F with vapour mole fraction yᵢ.
• Writing the lever rule upside down: V/F = (z − x)/(y − x), not (y − z)/(y − x) (that is L/F).

For GATE CH

• NAT problems: vapour fraction and phase compositions of a binary flash with vapour pressures or K-values given; ternary Rachford-Rice with simple K-values.
• Deciding whether a feed at given T and P is subcooled, two-phase or superheated using bubble and dew pressures.
• Conceptual MCQs on K-values, the lever rule and the effect of T and P on vaporisation.
• Practise solving Rachford-Rice by hand (bisection or Newton) in three or four steps.

Quick check

1. Write the Rachford-Rice equation.
2. If ΣzᵢKᵢ < 1, what is the state of the feed?
3. For a binary with K₁ = 2 and K₂ = 0.5, find x₁ and y₁.
4. An equimolar binary feed gives x₁ = 0.4 and y₁ = 0.6. What is V/F?
5. What extra equation does an adiabatic flash need?

Answers: 1. Σzᵢ(Kᵢ − 1)/[1 + V(Kᵢ − 1)] = 0. 2. Subcooled liquid (below its bubble point, V = 0). 3. x₁ = 1/3, y₁ = 2/3. 4. 0.5. 5. An energy balance, h_F = V·h_V + (1 − V)·h_L.