8. Multi-component distillation

Prof. S. Scott, ChE 128, UC Santa Barbara

How many distillation columns are required?

One distillation column can be optimized to separate one pair of volatile components.

We can specify two fractional recoveries:FR1 = (DxD,1)/(FzF,1) = 0.95

FR2 = (BxB,2)/(FzF,2) = 0.98

If the feed contains more than two volatile components, we cannot specify the recoveries of the additional components.

However, we can add more distillation columns.

FzF,1

zF,2

B, xB,1

D, xD,1

Distillation of multicomponent mixtures

I

3, 4

FzF,1

zF,2

zF,3

zF,4

II

2

1

1, 2

III

4

3

Separation of C components requires (C-1) distillation columns.

F

1

2,3,4

2

3,4

3

4

Alternative?

Key components

• Each column is designed to separate two components of adjacent relative volatility. These components are the keys.• All other components are non-keys.

component a designation

1 1.5 Light non-key (LNK)

2 1.4 Light key (LK)

3 1.3 Heavy key (HK)

4 1.2 Heavy non-key (HNK)

5 1.0 Heavy non-key (HNK)

design for separation

• Distributions in the distillate and bottoms streams are specified for the two key components.

• If we assume that the non-keys do not distribute, the overall mass balance is easily solved.

assume exclusively in distillatespecify recovery in distillate

specify recovery in bottoms

assume exclusively in bottoms

Overall mass balancewith non-distribution of non-keys

If non-keys do not distribute, thenLNK: Dxi,D = Fzi,F

HNK: Bxi,B = Fzi,F

Define fi, di, and bi as molar flow rates of component i in feed, distillate and bottoms, respectively.

component i zF,i fi = F zF,i di xD,i = di/D bi xB,i = bi/B

LNK 0.25 25 25 0.50 0 0

LK 0.25 25 20 0.40 5 0.10

HK 0.25 25 5 0.10 20 0.40

HNK 0.25 25 0 0 25 0.50

Σdi = D = 50 ΣxD,i = 1 Σbi = B = 50 ΣxB,i = 1

Ex.: 100 mol/h of an equimolar mixture of four components. Assume 80 % recovery of LK in D, and 80 % recovery of HK in B.

Stage-by-stage calculationsFor a specified external reflux ratio R = L0/D and feed quality q, and assuming constant molal overflow (CMO), first perform the overall column balance to obtain D. Then,

L = (L0/D) D

V = L + D ⇒ L/V

LH = L + q FVH = V + (1 – q) F ⇒ LH/VH

Algebraic solution, starting at top of column:1. yi,1 = xi,D (total condensor)

2. xi,1 = yi,1 / Ki,1 (VLE)

3. yi,2 = (L/V) xi,1 + (1 – L/V) xD (mass balance, rectifying section)

Repeat steps (2) and (3) down to feed stage. Then, change mass balance equation to yi,k+1 = (LH/VH) xi,k + (1 – LH/VH) xB

We just need to find Ki for each stage.

Obtaining Ki(T) for each stage• Consider a ternary system with components A, B, C• Assume constant relative volatility• Choose B as the reference compound

IF relative volatility is not constant, use geometric average, (a1aN)½. Or, find Tj from dewpoint calculation (starting with xD,i), then find Ki(Tj) using VLE.

Trial-and-error solution

Perform stage-by-stage analysis, from top to bottom of column.

xHK,N+1 ≥ xHK,B

xLK,N+1 ≤ xLK,B

must be satisfied simultaneously

• if not true, choose different starting concentrations at the top of the column, and repeat

• cannot assume xD,HNK = 0, but no way to guess

• works best if there are no HNKs. If there is no LNK, can start at bottom of column and work to the top

• not obvious where the optimum feed stage is

• very tedious, convergence is difficult

Approximate shortcut methods for multicomponent distillation

• Suitable for preliminary designs

• Three-part F-U-G method:

1. Use Fenske equation to estimate Nmin (at total reflux)

2. Use Underwood equations (2) to estimate Rmin (with N = ∞)

3. Use Gilliland correlation to estimate Nactual and Nfeed (can also use Kirkbride equation)

Operating at total reflux

• designate the keys A and B, and assume constant aAB

• according to the definition of equilibrium:1

N

N+1

yi,1

yi,N+1

xi,0

xi,N

(even though there is no liquid stream with composition xi,N+1)

• at total reflux, the operating line is y = xyi,j+1 = xi,j so

yi,N+1 = xi,N

• move up to stage N:

• continue to the top of the column

Estimating Nmin

• at the top of the column

• solve for NminFenske equation

• can also write in terms of fractional recoveries (FR)

Other uses of the Fenske equation

Once Nmin is known,• check non-distribution of non-keys, or estimate their distribution:

• estimate optimum feed stage at total reflux:

Pinch points in binary distillation

• xB

•xD

•zF

Rmin

pinch point

Recall for binary distillation: Binary composition profile

x

stage number

•feed

•condensor

•reboiler

Using Rmin generates a pinch point at the feed stage

plateau of constant

composition

• near the pinch point, composition changes little from stage to stage- passing streams are very close to equilibrium- no change in temperature between stages

•

•

•• •••

••

•

••

••

xD•

zF•

xB•

Behavior in ternary systems

• in the presence of a third component whose composition does change through

the feed stage, the pinch point moves

Ternary composition profileLK-HK-HNK

1

0

xi

stage number

FC R

Consider LK• LK is the most volatile component in the

system on every stage• there is almost no LK at the reboiler stage

Consider HNK• HNK is the least volatile component in the system on every stage• almost all HNK is found at the reboiler stage• finite HNK at feed stage, drops off rapidly above

strippingrectifying

Behavior of HK

Consider HK:• HK behavior is most complex

• above feed stage, distillation is almost binary

(HK-LK), with HK the less volatile component

Ternary composition profileLK-HK-HNK

1

0

xi

stage number

FC R

pinch point

~bin

ary

~bin

ary

• near reboiler, distillation is almost binary

(HNK-HK), but HK is the more volatile

component

• HNK goes through a maximum in the stripping

section, creating a pinch point there

Pinch points in multicomponent distillation

Graphical methods are not helpful in finding these pinch points.

FLK, HK,

HNK pinch

FLNK, LK,

HK

pinchF

LNK, LK,HK, HNK

pinch

pinch

• LK-HK-HNK system

has a pinch point in

the stripping section

• LNK-LK-HK system

has a pinch point in

the rectifying section

• LNK-LK-HK-HNK

has pinch points in

both sections

Second Underwood equationCMB in rectifying section at minimum reflux:

at the pinch point:

VLE:

relative volatility:

define:

2nd Underwood equation:

First Underwood equation

in the stripping section:

assume

1st Underwood equation:

feed quality definition:

searching forφ: with C components, the 1st Underwood equation has C roots, one between each adjacent pair of a-values, and one between 0 and the smallest a-value.

We need the root aHK < φ < aLK

• binary system: use quadratic equation to solve forφ• trial-and-error: use ½(aHK + aLK) as initial guess

Estimating Rmin

Onceφis known, obtain Vmin using the 2nd Underwood equation.

Lmin = Vmin – D

Rmin = Lmin/D

To get xi,D values for non-keys:

• assume non-distributionOR

• assume distribution is the same at minimum reflux and total refluxi.e., use Fenske equation

Gilliland correlationWe have Nmin (Fenske) and Rmin (Underwood).

R is often specified as a multiple of Rmin, e.g., R = 1.5 Rmin

From Separation Process Engineering, Third Edition by Phillip C. Wankat (ISBN: 0131382276) Copyright © 2012 Pearson Education, Inc. All rights reserved.

Figure 7-3 Gilliland correlation as modified by Liddle (1968); Reprinted with permission from Chemical Engineering, 75(23), 137 (1968), copyright 1968, McGraw-Hill.

y =

x =

Optimum feed stage

• obtain Nmin and NF,min from Fenske eqn

• obtain N from Gilliland correlation

Assume relative position of the feed stage is the same as it was at total reflux:

Remember! Nmin and N include partial reboiler (and partial condensor).

Works well for symmetrical columns (feed stage close to middle).

NF,min

F

B

D

NF

F

B

D

Kirkbride equation

Better feed stage estimation for unsymmetrical columns:

where Nrect and Nstrip do not include the feed stage:

N = Nrect + Nstrip + 1F

B

D

F

B

D

Sandwich componentsAt the beginning of this section, we specified that the key components should have adjacent relative volatilities. What if there is a non-key with intermediate volatility (i.e., between the two key components)?

This component is called a “sandwich” component.

Sandwich components tend to concentrate in the middle of the column, and can cause flooding even when present at trace concentrations.

If unavoidable, use side-streams to withdraw them.

component a designation

1 1.5 Light non-key (LNK)

2 1.4 Light key (LK)

3 1.3 Sandwich non-key

4 1.2 Heavy key (HK)

5 1.0 Heavy non-key (HNK)

F

B

D

S