Tdrum
—°C
Chapter 3
Flash Distillation
The single equilibrium stage — the simplest separation there is, and the building block for everything that follows.
By the end of this chapter you should be able to
- Count degrees of freedom for a flash drum and choose a legitimate set of specifications.
- Solve a binary flash graphically on a or diagram for any of the common specifications.
- Set up and solve the Rachford–Rice equation for a multicomponent flash.
- Explain why a single stage is limited, and how cascading stages overcomes that limit.
- Draw external balances around a distillation column and compute , , and .
Why start with a single stage?
Almost every separation you will meet in this course is built from one repeating idea: bring two phases into contact, let them approach equilibrium, then pull them apart. If you can analyse one such contact, you can analyse a hundred of them stacked in a column.
Everything below is drawn on the phase diagrams of Chapter 2a2a §2.6 — if the and constructions are not yet second nature, go back first.
Flash distillation is the simplest realisation of phase creation1 §1.6 — heat the feed until a second phase appears, and let the two phases share out the components. It is that one contact in its purest form. A liquid feed is heated, its pressure is dropped across a valve, and it partially vaporises into a drum. Vapour leaves the top, liquid leaves the bottom, and — this is the whole point — the two streams that leave are in equilibrium with each other.
The one rule to carry through the whole course
The streams leaving an equilibrium stage are in equilibrium with one another. The streams entering are not. Every graphical construction from here to Chapter 7 is a bookkeeping device for exactly this statement plus a mass balance2a §2.9.
3.1 The flash drum
Write the four families of equations that describe the drum. This same list — material, component, equilibrium, energy — is the skeleton of every stage calculation in this course.
(3.1)
(3.2)
(3.3)
(3.4)
Degrees of freedom
| Specified (5) | Unspecified (7) |
|---|---|
| Liquid phase: [2] Vapour phase: [2] Common: [3] | |
| Degrees of freedom — one is almost always . | |
So after fixing the drum pressure you have exactly one further specification to make — which is the same conclusion the Gibbs phase rule2a §2.2 reaches by a shorter route, once you remember that it counts only intensive variables. Which one you choose is an engineering decision, not a mathematical one:
- Liquid-phase mole fraction,
- Vapour-phase mole fraction,
- Drum temperature,
- Fraction of feed vaporised,
- Fraction of feed remaining liquid,
Why the arithmetic is easy
The energy balance (3.4) contains , which appears in no other equation. That decouples it from the material balances, so you can solve sequentially: (1) solve the material balances and equilibrium for ; (2) go back and use the energy balance to find . You never have to solve them simultaneously.
3.2 Binary flash — the sequential solution
When , or is fixed
- Draw the plot at the chosen pressure.
- If is given, read off the bubble curve and off the dew curve at that temperature. If or is given instead, locate it on the appropriate curve and read the temperature.
- Solve the overall and component balances for and .
- Solve the energy balance to obtain (or check) .
When or is given
Combining (3.1) and (3.2) and eliminating the flow rates gives a straight line on the plane:
(3.5)
which, written in terms of , becomes
(3.6)
or in terms of ,
(3.7)
Three observations make this line easy to draw:
- Equations (3.6) and (3.7) are straight lines on an plot — the operating line for the flash.
- Both and lie between 0 and 1 (all liquid to all vapour), and their slopes lie in .
- Setting in either equation gives . The operating line always crosses the diagonal at the feed composition. That single fixed point plus the slope is all you need.
- Equation (3.6) is the lever-arm rule2a §2.9 rearranged — the same statement, written as a line rather than a ratio.
So the recipe becomes: plot on the diagonal, compute the slope, draw the line, read and where it cuts the equilibrium curve, then get , , and finally .
Binary flash explorer
Drag the slider — or drag directly on either plot — to move the flash condition. The tie line and the operating line update together, and every readout is computed from the same VLE data. diagram
Bubble and dew curves at constant pressure; the horizontal tie line is the flash.
Bubble (liquid)
Dew (vapour)
Tie line ––
diagram
Equilibrium curve, diagonal, and the operating line through .
Equilibrium
Operating line
x (liquid)
—
y (vapour)
—
f = V/F
—
V
—mol/h
L
—mol/h
Slope
—
Recovery in V
—%
✓Two-phase region.
Step 4 — the energy balance
Constant and constant , saturated liquid at 0 °C as the reference state. , , .
hF
—kJ/mol
hL
—kJ/mol
HV
—kJ/mol
Heater duty Q
—kJ/h
Same, as power
—kW
✓—
Show the numbers as a worked balance
Do it now · 10 min
If moles of a mixture at condition H is boiled at constant pressure until it reaches point E, what is the relative amount of liquid and vapour that would be obtained?
Reveal the lever-arm rule
Component balance: . Overall balance: . Eliminating ,
This is the lever-arm rule2a App 5: the ratio of the two phases is the inverse ratio of the two segments of the tie line on either side of the feed point. The phase you are closer to on the tie line is the one you have more of. Watch the –– markers in App 1 while you drag — the feed point slides along the tie line, and is exactly the lever ratio.
3.3 Multicomponent flash
Nothing conceptual changes with components — only the bookkeeping. The variable and equation counts grow, but the degrees of freedom stay at two.
| Variables | Count | Equations | Count |
|---|---|---|---|
| Specified: | Overall balance | 1 | |
| Liquid: | Component balances | ||
| Vapour: | Equilibrium | ||
| Common: | 3 | Energy balance | 1 |
| Stoichiometry , | 2 | ||
| Total unspecified | Total equations · DOF = 2 |
The Rachford–Rice equation
Combining the overall balance, the component balances and the equilibrium relation and solving for the phase compositions gives
(3.8)
Each of these alone can be summed to one, but subtracting the two stoichiometric conditions,
(3.9)
and substituting (3.8) gives a single equation in a single unknown — the Rachford–Rice equation:
(3.10)
Why this form and not another
You could just as well have solved or directly. All three forms share exactly the same poles, at — the poles are a property of the flash equations, not of how you arrange them. What singles out the difference form is that it is strictly monotonically decreasing between consecutive poles, because
always. So once the two-phase test passes, the root in is unique and it is bracketed — bisection on will find it every time. The individual sums and are not monotone, so neither guarantee is available.
Monotone does not mean Newton converges from anywhere. When the feed contains both and , is neither convex nor concave on , and a single Newton step can jump clean over a pole and never return. Try with in App 2: it is genuinely two-phase with a root at , yet Newton from steps to — past the pole at — and runs away. Production flash routines therefore bracket first and use Newton only inside the bracket.
Solution procedure, for a given and . The come from a DePriester chart or from the Antoine equation2a §2.4:
- Assume a value of with and evaluate .
- Update by Newton's method until : Safeguard the step: if it lands outside , discard it and bisect instead. The bracket is guaranteed by the two-phase test; the Newton step is not.
- With known, get from (3.1), then and from (3.8), and finally from the energy balance.
Rachford–Rice solver
Enter and directly, or let the app compute from Antoine constants at your chosen and . The iteration table shows every Newton step, not just the answer. The reported always comes from bisection on the bracket ; the Newton path is shown alongside it, and is drawn in red if it escapes.| Component | zi | Ki | (VP)i, kPa | xi | yi | |
|---|---|---|---|---|---|---|
| Σ | — | — | — |
✓—
The Rachford–Rice function
over the physical range, with the bracketed root and the Newton path marked.
Newton iterates
Root
Newton iterations
Starting from ; converged when . If a step leaves the iteration is abandoned — the answer below still comes from the bracketed solve.
| k | (V/F)k | fk | f′k |
|---|
V/F
—
L/F
—
Iterations
—
f(0)
—
f(1)
—
Check yourself
Before you touch App 2: for a given feed, what must be true about and for a two-phase solution to exist at all? What physically is happening when ?
Reveal
At , . At , . Because is monotonically decreasing, a root in requires and , i.e.
These are exactly the bubble-point and dew-point checks. If then : the feed is below its bubble point at this and , so nothing vaporises — subcooled liquid. If , the feed is above its dew point — superheated vapour. Try setting to 40 °C in Antoine mode and watch the status line.
3.4 Real systems — where the hand calculation stops
Everything above works cleanly for systems that obey Raoult's law2a §2.4. For real systems you must return to the rigorous statement of equilibrium2a §2.3, . Activity coefficients2b §2b.2 and fugacity coefficients are themselves functions of composition and temperature, so the equilibrium relations (3.3) become coupled to the material balances and the sequential solution collapses. These problems need simultaneous numerical schemes.
That is what process simulators are for. ASPEN Plus, HYSYS, Symmetry, VMGSim and PROSIM carry extensive property databases and robust solvers. Your job as the engineer is not to out-compute them — it is to know what answer to expect, and to recognise when the simulator has given you a wrong one.
Summary of §3.1–3.4
- Single equilibrium staged separation is the simplest form of separation.
- Binary ideal systems — graphical solution on or .
- Multicomponent ideal systems — the Rachford–Rice scheme.
- Real systems — coupled and numerical; use a simulator.
- To solve any single-stage problem you need: overall balance, component balance, equilibrium relationship, energy balance.
- And again: the streams leaving the stage are in equilibrium with each other.
3.5 Flash cascades
A single stage buys you one equilibrium step, and that is usually nowhere near enough. Look at what App 1 gives you: for a 30 % ethanol feed a single flash might lift the vapour to 55 % — useful, but far from a product. To raise purity and recovery at the same time, stages are combined into cascades.
Flash cascade builder
Chain up to six flash stages and watch purity and recovery trade against each other. Stripping the liquid drives one product pure; enriching the vapour drives the other. Neither does both — which is exactly why countercurrent columns exist.Stage-by-stage path on the diagram
Each orange segment is one stage: it runs from the feed point on the diagonal out to the equilibrium point (green) that the flash produces.
Equilibrium
Stage operating lines
Stage products
Purity vs. recovery, stage by stage
Product purity and cumulative recovery, both on a 0–1 scale.
Product purity
Cumulative recovery
| Stage | Feed to stage | x | y | T, °C | Product flow, kmol/h | Cum. recovery |
|---|
✓—
Think!
Run App 3 with six stages enriching the vapour. The purity climbs nicely — but look at the recovery column. Where did all the light component go? Now switch to "strip the liquid" and ask the same question about purity. What single structural change to the cascade would fix both at once?
Reveal
In the crosscurrent arrangement each stage throws away the phase it is not chasing. Enriching the vapour six times gives a beautifully pure vapour containing a small fraction of the light component you started with; the rest left in six separate liquid streams. Stripping is the mirror image.
The fix is to stop discarding: feed each rejected stream back into the neighbouring stage rather than out of the system. That is a countercurrent cascade — and once you add a condenser at the top to generate reflux and a reboiler at the bottom to generate boil-up, you have built a distillation column. That is the subject of Chapter 4.
One caveat the app will show you if you push it: on ethanol–water the cascade stalls at however many stages you add. No amount of countercurrent contacting fixes that, because it is an azeotrope2b §2b.6 rather than a stage-count limitation.
3.6 From a cascade of flashes to a distillation column
The sequence in Figure 3.4 is worth dwelling on, because it explains the geometry of every column you will ever design. The vapour rising from stage is hot and needs cooling; the liquid falling from stage is cold and needs heating. Rather than run each through a separate exchanger, let them contact each other directly on a tray. All the intermediate duty cancels, and only the two ends — the condenser and the reboiler — still need external heat exchange.
Design problems and simulation problems
Two quite different questions get asked of the same column, and it is worth naming them explicitly because the solution strategies differ.
| Design problem | Simulation problem | |
|---|---|---|
| Objective | Design a distillation column for a new separation. | Analyse the performance of an existing column. |
| Typical statement | "You have a feed with composition …, and want a top product of composition … and a bottom product of composition …" | "You have a column of the following specifications …. You wish to separate a mixture of composition …. What is the best separation achievable?" |
| What you produce | Column height and diameter, number of stages, feed location. | Top and bottom flow rates and compositions, reboiler and condenser duties. |
3.7 External column balances
Before any tray-by-tray work, draw a box around the whole column. Three balances follow immediately:
(3.11)
(3.12)
(3.13)
Equations (3.11) and (3.12) give and directly — two equations, two unknowns, no iteration and no knowledge whatsoever of what happens inside the column. This is why external balances are always step one.
Getting the condenser duty
(3.14)
(3.15)
Combining (3.11), (3.12) and (3.15) gives the condenser duty in terms of quantities you already know:
(3.16)
With known, follows from the overall energy balance (3.13).
Sign convention — get this wrong and nothing balances
is negative (heat is removed from the system at the condenser) and is positive (heat is added at the reboiler). Since , equation (3.16) returns a negative number automatically — if yours comes out positive, you have swapped an enthalpy.
External column balances
Set the specifications a design problem would give you and read off , , and . The energy waterfall shows every term in equation (3.13) so you can see it close.D (distillate)
—kmol/h
B (bottoms)
—kmol/h
L0
—kmol/h
V1
—kmol/h
QC
—kW
QR
—kW
Recovery in D
—%
✓—
Energy balance, term by term
Equation (3.13) as a waterfall. Blue bars add energy to the column, red bars remove it; the running total closes at zero.
Where the feed goes
Total molar split and the split of the light component.
To distillate
To bottoms
3.8 Worked examples
Three problems in the style of the course question bank, worked in full. Each one is set up so you can reproduce it in the apps above and check yourself.
Worked example 3.1Binary flash · constant relative volatility
We are flashing a mixture of ethanol and propanol. The mixture is ideal with . The feed to the flash still is 100 kmol/h of mole fraction 0.60 ethanol. Your supervisor wants to know the effect of varying .
- Find the composition of the liquid and vapour leaving the still for , analytically.
- We want an outlet liquid composition of 0.50. What achieves this?
Solution
Set up. Two equations in two unknowns for each . The equilibrium relation for constant ,
and the operating line written with , which we rearrange from :
Substituting the first into the second gives one equation in alone:
which rearranges to the quadratic . Take the root in .
The two end points are free. At nothing vaporises, so and the infinitesimal vapour is in equilibrium with it: . At everything vaporises, so and the last drop of liquid is . These bracket every other answer.
| , kmol/h | , kmol/h | Slope | Ethanol recovered in | |||
|---|---|---|---|---|---|---|
| 0.0 | 0.6000 | 0.7590 | 100.0 | 0.0 | 0.0 % | |
| 0.2 | 0.5667 | 0.7331 | 80.0 | 20.0 | −4.000 | 24.4 % |
| 0.4 | 0.5308 | 0.7038 | 60.0 | 40.0 | −1.500 | 46.9 % |
| 0.6 | 0.4930 | 0.6713 | 40.0 | 60.0 | −0.667 | 67.1 % |
| 0.8 | 0.4545 | 0.6364 | 20.0 | 80.0 | −0.250 | 84.9 % |
| 1.0 | 0.4167 | 0.6000 | 0.0 | 100.0 | 0 | 100.0 % |
Read the trend. As more of the feed is vaporised, both product streams get poorer in ethanol: falls from 0.600 to 0.417 and falls from 0.759 to 0.600. The last two columns are the point of the exercise. The purest vapour the still can ever make is 0.759 — but only in the limit of producing none of it. Take 85 % of the ethanol into the vapour and its purity has slipped to 0.636. Purity and recovery move in opposite directions, and a single stage gives you no way to have both. That is the limitation §3.5 sets out to break.
(b) If is required, first find the vapour in equilibrium with it:
then use the lever arm on the operating line, :
So kmol/h and kmol/h.
Check it in App 1. Choose Constant α, set α = 2.10 and , then specify . The readout should give and .
Worked example 3.2Multicomponent flash · Rachford–Rice
A reactor effluent is cooled and flashed to separate light gases from heavier hydrocarbons. At the drum conditions of 500 psia and 100 °F the feed and values are as follows. Calculate the flow rate and composition of the vapour leaving the drum.
| Component | Flow, lbmol/h | ||
|---|---|---|---|
| H₂ | 2000 | 0.43478 | 80 |
| CH₄ | 2000 | 0.43478 | 10 |
| Benzene | 500 | 0.10870 | 0.010 |
| Toluene | 100 | 0.02174 | 0.004 |
| Total | 4600 | 1.00000 |
Solution
Step 1 — confirm two phases exist before solving anything, using the bubble- and dew-point tests:
Both exceed unity, so the drum is in the two-phase region and Rachford–Rice has a root in .
Step 2 — solve Rachford–Rice. With ,
Newton's method from converges in a handful of steps to
Step 3 — back out the compositions from and :
| Component | , lbmol/h | , lbmol/h | ||
|---|---|---|---|---|
| H₂ | 0.00624 | 0.49899 | 1996.3 | 3.7 |
| CH₄ | 0.04925 | 0.49254 | 1970.5 | 29.5 |
| Benzene | 0.78201 | 0.00782 | 31.3 | 468.7 |
| Toluene | 0.16250 | 0.00065 | 2.6 | 97.4 |
| Σ | 1.00000 | 1.00000 | 4000.6 | 599.4 |
Sanity check. Both columns sum to 1 — that is not automatic, it is the condition Rachford–Rice enforced. Physically, 99.8 % of the hydrogen and 98.5 % of the methane leave in the vapour while 94 % of the benzene and 97 % of the toluene stay in the liquid. With values spanning four orders of magnitude, one equilibrium stage does an excellent job — which is exactly when a single flash is the right unit operation.
Check it in App 2. Enter the four components with these and in "Enter K directly" mode. The iteration table will show the same Newton path.
Worked example 3.3Two drums in series · cascade arithmetic
A binary mixture with constant relative volatility is separated in two flash drums in series. The feed is an equimolar mixture at 100 kmol/h. In the first drum, 50 % of the feed entering it is vaporised. The liquid from drum 1 is re-heated and flashed in drum 2, whose vapour product contains 50 mol % of the more volatile component. Find all stream flows and compositions.
Solution
Drum 1. With and , the operating line becomes , i.e. simply . Combine with equilibrium :
kmol/h and kmol/h.
Drum 2. Its feed is : kmol/h at . We are told , so invert the equilibrium relation for the liquid in equilibrium with it:
Now the lever arm gives the vaporised fraction of drum 2:
| Stream | Flow, kmol/h | Mole fraction MVC | MVC carried, kmol/h |
|---|---|---|---|
| Feed | 100.00 | 0.5000 | 50.00 |
| — vapour product | 50.00 | 0.5858 | 29.29 |
| — vapour product | 24.26 | 0.5000 | 12.13 |
| — liquid product | 25.74 | 0.3333 | 8.58 |
| Σ products | 100.00 | 50.00 ✓ |
What the numbers tell you. Overall mass and component balances close, which is the check to always run last. But look at the products: the best stream is only 58.6 % MVC from a 50 % feed, and it took two drums with two heaters to get there. Adding a third drum in this crosscurrent arrangement would improve purity again while cutting recovery again. Countercurrent contacting — Chapter 4 — is what breaks that trade-off.
Related to App 3. App 3 automates exactly this pattern, with one difference: it fixes on every stage, whereas drum 2 here is specified by its vapour composition instead. Either specification closes the degrees of freedom — that is §3.1 in action.
3.9 Check your understanding
Five multiple-choice questions, two short problems and two long problems. Everything is graded in your browser — nothing is submitted or recorded anywhere, and your answers are kept on this device only so a refresh does not lose them. Work them on paper first; the point is the method, not the number.
Multiple choice
Short problems
Enter numbers only. Answers are marked correct within the tolerance shown after you check them; give at least four significant figures where the tolerance is tight.
Long problems
Multi-part. Each part is marked separately, so a slip in part (a) does not cost you the rest — but do check your own carry-through.
Think about it
Think!
- The energy balance decoupled cleanly from the material balance for an ideal binary flash. Name a situation where it would not decouple.
- In App 1, drag towards 1. What happens to the operating line slope, and what does that mean physically?
- For a fixed feed and fixed , does a larger relative volatility give you a better or worse single-stage separation? Verify with the constant- system in App 1.
- Equation (3.16) contains , , , and but no stage count. Why can you compute the condenser duty without knowing how many trays the column has?
Reveal discussion notes
- Whenever appears in more than one equation, or whenever the equilibrium depends on a variable the energy balance sets. Non-isothermal, non-ideal, or reactive flashes all couple. So does an adiabatic flash where is itself an unknown fixed by .
- Slope : the operating line goes horizontal and . All of the feed vaporises, so the vapour must have the feed composition — no separation at all. The same happens at , where the slope goes to and .
- Better. Larger bows the equilibrium curve further from the diagonal, so the same operating line intersects it at a more widely separated pair. This is the sense in which measures "difficulty of separation" — recall the table where for xylene isomers needs 1000 plates2a §2.7.
- Because external balances are a black-box statement. The number of stages determines whether the specified and are achievable at that reflux ratio — but if you assert them as achieved, conservation of mass and energy fixes the duties regardless of the internal arrangement. A column with too few trays simply will not meet the specification you assumed.
Where this chapter connects
- Back to: the and diagrams every construction here is drawn on come from Chapter 2a2a §2.6, and the phase split is the lever-arm rule2a §2.9 in disguise.
- values and : defined in Chapter 2a2a §2.7 — the Antoine mode in the Rachford–Rice solver3 App 2 computes them the same way.
- The limit you cannot cross: the cascade builder stalls on ethanol–water because of an azeotrope2b §2b.6, not because it needs more stages.
- Ahead: Figure 3.4 turns this cascade into a column; the external balances of §3.7 are the first step of every design, and the McCabe–Thiele construction4a §4.6 replaces the stage-by-stage bookkeeping with a single diagram.
Summary & key equations
Single stage
Overall / component balance ·
Equilibrium, for ideal systems
Energy balance
Degrees of freedom2 (usually plus one of )
Binary flash operating line
In terms of
In terms of
Always passes through on the diagonal
Lever-arm rule
Multicomponent flash
Phase compositions,
Rachford–Rice
Two-phase test and
External column balances
Mass,
Energy
Condenser duty
Signs,