Multicomponent distillation and the FUG method

Distillation separates a liquid mixture by boiling point. Heat the mixture and the lighter components (the ones that boil at a lower temperature) turn to vapor more readily than the heavier ones. Collect and condense that vapor and you have a product richer in the light components; what stays behind is richer in the heavy ones. That is the whole idea, repeated many times over.

One stage separates a little. Many stages separate a lot.

A single equilibrium contact, one flash, only partly splits the mixture: the vapor it makes is a bit richer in the light component, the liquid a bit richer in the heavy one. To reach a nearly pure product you stack many stages and feed some product back (reflux at the top, boilup at the bottom) so the enrichment compounds stage after stage.

Two components versus three or more

In a mixture at vapor-liquid equilibrium, each component distributes itself between the two phases. The distribution coefficient (K value, or K factor) measures that split:

A large K (K > 1) means the component prefers the vapor, so it is light (volatile). A small K (K < 1) means it stays in the liquid, so it is heavy. For an ideal mixture, combine Raoult's law (p_i = x_iP_i^sat) with Dalton's law (y_i = p_i/P_total) and the K value becomes simply:

Vapor pressure rises quickly with temperature, so K values are strongly temperature dependent. That single fact is the reason bubble point and dew point calculations need iteration: change T and every K in the table changes with it.

For light hydrocarbon systems the classic source of K values is the DePriester chart: a nomograph where you lay a straight line from the pressure axis to the temperature axis and read K for each compound where the line crosses its curve. For example, at 5 atm and 40 °C the chart gives K ≈ 7.2 for ethane (C₂).

Charts are awkward inside a spreadsheet, so we use the McWilliams (1973) correlation, which is a curve fit of the DePriester chart. The full form has six constants, but for the paraffins we use in this course only three are nonzero, which leaves:

The fit is valid from about 460 to 760 °R (0 to 300 °F) and 14.7 to 120 psia, with mean errors of roughly 4 to 6 percent for these compounds. That accuracy is fine for shortcut design work.

The DePriester chart from the lecture (SI version: P in atm, T in °C). One straight edge, one reading per compound.

How to read it

1. Mark the operating pressure on the left axis.
2. Mark the temperature on the right axis.
3. Join the two marks with a straight line (a ruler on paper).
4. Read K for each compound where the line crosses its curve.

The lecture example: at 5 atm and 40 °C the line crosses the ethane (C₂) curve at K ≈ 7.2. Try the same line on butane and you will read a value below 1: at these conditions ethane wants to be vapor and butane wants to stay liquid.

The chart and the McWilliams equation are the same information in two formats. In the exam you may be given either one, so practice both.

The relative volatility compares how easily two components separate. It is the ratio of their K values, and under Raoult's law it reduces to a ratio of vapor pressures:

In multicomponent work we reference every component to one chosen compound, almost always the heavy key, and write α_iH = K_i/K_HK. The heavy key then has α = 1 by definition, lighter components have α > 1, and heavier ones have α < 1. Because K values move together as temperature changes, the ratio α changes much more slowly with T than K itself. That is what makes the shortcut method possible.

The bubble point is the temperature where a liquid of known composition x_i forms its first bubble of vapor. Each x_i is fixed, the vapor formed obeys y_i = K_ix_i, and the vapor fractions must add up to 1:

The dew point is the mirror image: a vapor of known composition y_i forms its first drop of liquid. The liquid obeys x_i = y_i/K_i, and those must also add up to 1:

Neither equation can be solved for T directly, because T hides inside every K_i through the McWilliams correlation. So we solve them the practical way:

1. Assume a temperature T.
2. Compute every K_i at that T and P.
3. Form the sum (ΣK_ix_i for bubble, Σy_i/K_i for dew).
4. If the sum is not 1, change T and repeat until it is.

In Excel this trial-and-error loop is one click: Goal Seek adjusts the temperature cell until the sum cell equals 1. That is exactly what the workbook from class does, and the Goal Seek Lab in this app reproduces it so you can watch the iterations happen.

In binary distillation two purity specs (x_D and x_B) pin down the whole separation, and the material balance gives D/F and B/F directly. With three or more components that no longer works: fixing the concentration of one component in each product does not tell you where everything else goes.

So the designer picks two components whose recoveries define the quality of the separation. They are the key components:

• LK  The light key is the component we want to keep out of the bottoms. It leaves mainly in the distillate.
• HK  The heavy key is the component we want to keep out of the distillate. It leaves mainly in the bottoms.

Both keys appear in finite amounts in both products; we say they are distributed. Everything lighter than the light key goes almost entirely to the distillate, and everything heavier than the heavy key goes almost entirely to the bottoms. Those non-key components are essentially non-distributed. As a working rule from the lecture, a component whose K value (or relative volatility) is 10 percent or more above the light key's will nearly always be non-distributed.

A worked illustration: the depropanizer from the lecture

The lecture (slides 18 to 19) runs these ideas through a depropanizer: a column whose job is to take propane overhead. The product spec is about propane against butane, so C₃H₈ is the light key and n-C₄H₁₀ is the heavy key. Watch how the feed splits (basis 100 mol of feed):

Only the two keys show up in both products (24.6/0.4 and 0.3/16.7): they are distributed. Methane and ethane never reach the bottoms, pentane and hexane never reach the top. This is exactly the pattern the column diagram above predicts, and it is why estimating the keys well is most of the job.

The split you are asked for usually arrives as a fractional recovery, for example "95% of the n-pentane in the feed must reach the distillate." Recovery times feed moles gives the product moles of that key, and the material balance gives the rest. Note that a recovery is not a mole fraction; 95% recovery of n-pentane says nothing directly about x_D of n-pentane.

Selecting keys is not guesswork. It is a short procedure driven by what the column must produce:

1. Rank the components by volatility (K values at feed conditions, or boiling points).
2. Draw the cut line from the product spec. Everything the spec sends overhead sits above the line, everything it sends down sits below.
3. The two components that straddle the cut are the keys. The lighter one is the LK, the heavier one is the HK.
4. Prefer adjacent keys. If a component sits between your keys in volatility (a sandwich component), it distributes heavily to both products and the shortcut method loses accuracy.

Columns in gas processing are literally named after this choice. A "de-something-izer" is a column whose light key is that something: a depropanizer takes propane overhead, so its LK is propane and its HK is the next heavier component in the feed. The classic NGL fractionation train applies the rule three times in a row:

Our Example 5.3 fits the same logic. The spec pushes n-pentane up and n-hexane down, so the cut falls between them: n-C5 is the LK, n-C6 is the HK, and the column is, in industry language, a depentanizer.

Same feed, different columns, different keys

Key choice belongs to the column, not to the feed. The textbook makes this point with an alkylation-reactor effluent that must yield three products (Seader, Figure 9.2). Two column sequences are possible, and each first column has its own keys:

Seader, Henley and Roper, Fig. 9.2 (3rd ed., p. 360): one feed, three specified products.

• Sequence 1: deisobutanizer first. The spec limits n-C4 in the isobutane recycle and i-C4 in the n-butane product, so LK = isobutane, HK = n-butane. A hard split: these two differ by only about 20 °F in boiling point.
• Sequence 2: debutanizer first. Now n-butane must go overhead, so LK = n-C4. No spec names the HK, so you choose it: i-C5 is picked simply because it keeps the keys adjacent.

Two lessons. The keys come from each column's job, and when the spec does not fix the HK, the engineer picks the neighbor. This is also why exam problems always tell you the recovery of exactly two components: those two are the keys.

Solving a multicomponent column rigorously means stage-by-stage balances on every component with temperature-dependent K values, which is simulator work (you will do it in Aspen Plus in Lecture 6). Before that, engineers want quick, reliable estimates for preliminary design, cost estimation, studies of how R and N trade off, and optimization of simple distillation. That is the job of the Fenske-Underwood-Gilliland (FUG) shortcut method.

The idea rests on two conceptual operating limits that bracket every real column:

• Total reflux: everything condensed is returned, no products leave. The column needs the fewest stages possible, N_min. The Fenske equation computes it.
• Minimum reflux: the least liquid that can be returned and still (barely) make the split, requiring an infinite number of stages. The Underwood equations compute R_min.

A real column runs between the limits, typically at R = 1.1 to 1.5 times R_min (this course usually uses 1.3). The Gilliland correlation is an empirical bridge: given N_min, R_min and the actual R, it returns the actual number of theoretical stages N. Two helper correlations complete the toolkit: the Hengstebeck-Geddes line estimates how non-key components split between the products, and the Kirkbride equation places the feed stage.

To run the FUG method you must specify: the distribution (recoveries) of the two keys, the feed composition and thermal condition q, and the operating reflux as a multiple of R_min.

The official roadmap: Seader, Henley and Roper, Fig. 9.1 (3rd ed., p. 360). The textbook algorithm for the FUG method.

How the lecture maps onto it

• Specify key splits, estimate non-key splits: our Steps 1 to 3 (recoveries, then the Hengstebeck-Geddes line instead of a first Fenske pass).
• Bubble point / dew point calculations: our Goal Seek runs for the feed, top and bottom temperatures.
• Fenske, Underwood, Gilliland, Kirkbride: our Steps 4 to 7, identical.
• The loop on the left is the honest part: if the non-key split you assumed disagrees badly with what the equations return, update and repeat. In our example one pass is enough.
• The last box (condenser and reboiler duties) is an energy balance, which this course picks up in the equipment design lectures.

Assumptions to keep in mind: constant molar overflow, relative volatility roughly constant over the column, adjacent keys, and a simple column. The output is a preliminary design, good for sizing and screening, to be confirmed by rigorous simulation.

• At the class first guess of 150 °F for the feed liquid, the sum is about 0.47. Well short of 1, so 150 °F is far below the bubble point. Slide T up and watch every K grow.
• The converged bubble point of the feed is 214.89 °F, and the dew point of the same composition taken as vapor is 261.40 °F. Between them the mixture is partly vapor, partly liquid.
• Switch to the distillate preset in dew mode and the bottoms preset in bubble mode: those two runs are exactly how Example 5.3 finds the column top and bottom temperatures (170.8 °F and 273.1 °F).
• The vapor made at the bubble point is much richer in n-C4 than the liquid (y = 0.339 from x = 0.1). The first bit of vapor is always enriched in the light components. Distillation is this effect, stacked stage upon stage.

• Raise both recoveries from 95% to 99%. N_min and N climb steeply: purity is paid for in stages.
• Drop R/R_min from 1.3 toward 1.05 and watch N grow fast; raise it toward 2 and watch N flatten out near N_min. That is the Gilliland curve in action, and it is the capital versus energy trade-off.
• Change the feed so the keys are scarce (for example 0.4 / 0.1 / 0.1 / 0.4). The split gets easier to state but the temperatures move, and so do all three α sets.
• Raise the pressure. K values fall, both end temperatures rise, α shrinks toward 1, and the column needs more stages. Distillation prefers the lowest pressure the condenser can afford.

A feed of 100 kmol/h has the composition and relative volatilities below (α already given at average column conditions, so no K correlation is needed). The desired recovery of the light key O in the distillate is 94.84%. The recovery of the heavy key P in the bottoms is 95.39%. The feed is at its bubble point. Determine the number of ideal stages at R = 1.2 R_min, and the feed stage.

Compute d and b for both keys, then C and C₁. Hint: with 100 kmol/h of feed, f_O = 25 and f_P = 23.

Apply the line to M, N, Q, R, S. For each: r = 10^(C log α + C₁), then b = f/(r + 1), d = f − b.

Use the same α set throughout (they are already averages). Remember q = 1, and R = 1.2 R_min.

• Units in the K correlation. T must be in °R (add 460 to °F) and P in psia (kPa × 0.14498). A K value of 10⁸ or 10^-9 means the units slipped.
• Swapping the two summation tests. Bubble point sums K_ix_i; dew point sums y_i/K_i. If your "bubble point" lands above your dew point, this is what happened.
• One α used everywhere. Hengstebeck-Geddes wants feed-T α, Fenske wants the geometric mean of the end values, Underwood wants α at the arithmetic mean temperature.
• Recovery treated as mole fraction. "95% recovery of n-C5 in D" means d = 0.95 f. It does not mean x_D = 0.95.
• θ out of range. The useful Underwood root always lies strictly between α_HK = 1 and α_LK. If your solver wandered elsewhere, restart the guess at the midpoint.
• Forgetting q. Feed at its bubble point means q = 1, so the first Underwood equation sums to zero. A partially vaporized feed changes the right side to 1 − q, and θ moves.
• Gilliland algebra. From Y, solve N = (N_min + Y)/(1 − Y). N includes the reboiler as a stage; a total condenser is not a stage.
• Kirkbride direction. m counts stages above the feed and p counts stages below it. A feed rich in heavies (like Example 5.3, where m/p = 0.735) needs the larger stripping section, so p > m and the feed sits in the upper half of the column. If your m and p come out the other way around for such a feed, they are probably swapped.

Lecture 5, ChE 335, Dr. Jatupon Chaiwasu, KMUTT · Seader, Henley and Roper, Separation Process Principles, 3rd ed., Chapter 9 (Fenske eq. 9-11/9-12, Underwood eq. 9-28/9-29, Gilliland/Molokanov eq. 9-34, Kirkbride eq. 9-36) · McWilliams, "An Equation to Relate K-factors to Pressure and Temperature", Chemical Engineering, 80(25), 138 (1973) · Practice problem adapted from Coker, Ludwig's Applied Process Design for Chemical and Petrochemical Plants, Vol. 2, 4th ed. · Figures: DePriester chart and column photo from the Lecture 5 slides; FUG algorithm and alkylation example reproduced from Seader Figs. 9.1 and 9.2; NGL train operating ranges and product uses cross-checked against published gas-processing practice; Hengstebeck-Geddes and Gilliland plots computed directly from the class numbers and the Molokanov equation.