Multi Component 55o6o

Multi-Component Distillation

Introduction to Multicomponent Distillation

• Most of the distillation processes deal with multicomponent mixtures.

• Multicomponent phase behaviour is much more complex than that for the binary mixtures

• Rigorous design requires computers

• Short cut methods exist to outline the scope and limitations of a particular process.

Continued… • Objective of any distillation process is to recover pure products.

• In case of multi component mixtures, we may be interested in one, two or more than two components in pure form. D, xD,1

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

F zF,1 zF,2

B, xB,1

How many distillation columns are required ?

1

Alternative?

1 1, 2

F

II

2 2,3,4

F zF,1 zF,2 zF,3 zF,4

3 3,4

2

I

4 3

3, 4

III Separation of C components requires (C1) distillation columns. 4

• Unlike in binary distillation, fixing mole fraction of one of the components in a product does not fix the mole fraction of other components.

• On the other hand fixing compositions of all the components in the distillate and the bottoms product, makes almost impossible to meet specifications exactly. • 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.

Key components In practice we usually choose two components separation of which serves as an good indication that a desired degree of separation is achieved. These two components are called key components:

- light key - heavy key

There are different strategies to select these key components: • Choosing two components that are next to each other on the relative volatility. • Based on distribution of component.

Distributed and undistributed components • Components that are present in both the distillate and the bottoms product are called distributed components. • The key components are always distributed components. • Components with negligible concentration (<10-6) in one of the products are called undistributed.

A

B

light non-distributed components (will end up in the overhead product)

C

D

key components

E

G heavy non-distributed components (will end up in bottoms product)

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

design for separation

component

a

1

1.5

Light non-key (LNK)

2

1.4

Light key (LK)

assume exclusively in distillate specify recovery in distillate

3

1.3

Heavy key (HK)

specify recovery in bottoms

4

1.2

Heavy non-key (HNK)

5

1.0

Heavy non-key (HNK)

designation

assume exclusively in bottoms

• 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.

Overall mass balance with non-distribution of non-keys If non-keys do not distribute, then LNK: 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. 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. 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

Hengstebeck and Geddes Equation: Exact distribution of non - key components in top product and bottom product is not possible At the starting of design. But the same can be determined approximately by Hengstebeck and Geddes equation,

 di  log    A  C log a i  bi  Where, di = moles of ith component in distillate bi = moles of ith component in residue αi = relative volatility of ith component with respect to heavy key component A, C = constants

Ex.6. A bubble cap fractionating column consisting of 12 plates working at an average efficiency of 75% is being used to distill 1000 kg/hr of aqueous methanol entering the tower. The feed consists of 3/4th vapor and 1/4th liquid. The feed, overhead product and bottom product are 50 mole %, 90 mole % and 10 mole % methanol respectively. A total condenser is provided. The reflux is sent at its saturation temperature. If the reflux ratio is 1.7 times the minimum check whether the column available is satisfactory. VLE DATA is given below:

X

8

10

20

30

40

50

70

80

y

36.5

41.8

57.9

66.5

72.9

77.9

87

95.8

The calculation of the number of equilibrium stages in multi component distillation: Method Valid for Multi component Distillation (MD): Methods available for MD can be broadly divided into two groups: 1. Short Cut Methods: FUG method, Smith – Brinkley Method etc. 2. Rigorous Method: •Lewis Matheson Method •Thiele –Geddes Method •Relaxation Methods •Tridiagonal Matrix Method •Linear Algebra Method etc. Short Cut Methods: • an estimation of the minimum number of trays • the estimation of minimum reflux rate and • number of stages at finite reflux

A widely used approximate method is commonly referred to as the Fenske-Underwood-Gilliland method

FUG 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 to estimate Rmin (with N = ∞) 3. Use Gilliland correlation to estimate Nactual And for Nfeed one 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:

yi,1 1 xi,0

é yA ù é xA ù ê ú = a AB ê ú ë y B ûN+1 ë xB ûN+1 • at total reflux, the operating line is y = x yi,j+1 = xi,j so yi,N+1 = xi,N

yi,N+1

N xi,N N+1

é xA ù é xA ù ê ú = a AB ê ú ë xB ûN ë xB ûN+1 • move up to stage N: é yA ù é xA ù é xA ù 2 ê ú = a AB ê ú = a AB ê ú ë y B ûN ë xB ûN ë xB ûN+1 • continue to the top of the column

Estimating Nmin • at the top of the column

Fenske equation

• solve for Nmin

é xA ù é y A ù Nmin é xA ù ê ú = ê ú = a AB ê ú ë xB û0 ë y B û1 ë xB ûN+1

Nmin

éæ x ö æ xA ö ù A ln êç ÷ ú ç ÷ êëè xB ø 0 è xB ø N+1 úû = lna AB

• can also write in of fractional recoveries (FR)

Nmin

éæ FR ö ù æ 1- FRB ö A ln êç ú ÷ ç FR ÷ 1FR è êëè A ø distillate B ø bottoms ú û = ln a AB

Other uses of the Fenske equation Once Nmin is known, • check non-distribution of non-keys, or estimate their distribution:

FRC,distillate =

N a CB

min

æ FRB ö Nmin + a CB ç 1- FR ÷ è ø B bottoms

• estimate optimum feed stage at total reflux:

NF,min

éæ x ö æ zA ö ù A ln êç ÷ çz ÷ ú x êëè B ø 0 è B ø F úû = lna AB

Minimum reflux ratio, using Underwood’s method:

Where, αA, αB, αC, etc., are the volatilities with respect to the least volatile component, and θ is the root of equation 1, which lies between the values of αA and αB.

Gilliland Correlation We have Nmin (Fenske) and Rmin (Underwood). R is often specified as a multiple of Rmin, e.g., R = 1.5 Rmin

y=

x= Later Molokanov et al. (1972) represented the Gilliland correlation as,

y = 1- e

éæ 1+54.4 x ö æ x-1ö ù êç ÷ ç 0.5 ÷ ú ëè 11+117.2x ø è x ø û

Optimum Feed Stage

• Obtain Nmin and NF,min from Fenske eqn • Obtain N from Gilliland correlation D

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

NF,min Nmin

=

NF N

D F

F

NF

NF,min

! Nmin and N include partial reboiler (and partial condensor). Works well for symmetrical columns (feed stage close to middle).

B

B

Kirkbride equation Better feed stage estimation for unsymmetrical columns:

Nrectifying Nstripping

éæ z ö æ x ö 2 æ ö ù B = êç HK ÷ ç LK ,B ÷ ç ÷ ú êè zLK ø è xHK ,D ø è D ø ú F ë û

F

D

0.206

D

where Nrect and Nstrip do not include the feed stage:

B

F

N = Nrect + Nstrip + 1

B

Example 7: A feed 100 kmoles /h of saturated liquid containing 10 mole % LNK, 55 mole % LK, and 35 mole % HK and is to be separated in a distillation column. The reflux ratio is 1.2 times of the minimum. It is desired to have 99.5 % recovery of the light key in the distillate. The mole fraction of the light key in the distillate should be 0.75. Equilibrium data: αLNK = 4.0, αLK = 1.0,αHK = 0.75. Find (i) Minimum number of stages required by Fenske method. (ii) Minimum reflux ratio by Underwood method. (iii) Number of ideal stages at R = 1.2 Rmin by Gilliland method. (iv) Also find the number of ideal stages at rectifying section and the stripping section at the operating reflux ratio and location of feed stage.

SOLUTION:

K

Using following equation,

Nmin

éæ x ö æ xA ö ù A ln êç ÷ çx ÷ ú x êëè B ø 0 è B ø N+1 úû = lna AB

Nmin =…………………

(ii) To find the minimum reflux at the condition of saturated liquid, q = 1, Using following equation:

Gives θ = 0.83

Rmin= 2.47 (iii) Now using the Gilliland correlation to determine number of ideal stages at R = 1.2 Rmin =2.97

We can get N = 48.89

(iv) Using Kirkbride Equation

NR =…………………… NS =……………………

Lewis – Matheson Method, Rigorous Method for Multicomponent Distillation Method assumes the constant molar overflow

To use this method following information must be specified. 1. 2. 3. 4. 5. 6.

Feed flow rate, conditions and compositions Product compositions or distribution of key components. Reflux ratio One product flow Column pressure Assumed values for the distribution of the non-key components

Method involves: 1. If the composition of the liquid on any plate is known, then the composition of the vapour in equilibrium is calculated using relative volatilities of the individual components.

2. The composition of the liquid on the plate above is then found by using an operating equation.

If a mixture of components A, B, C, D, has mole fractions xA, xB, xC, xD, in liquid and yA, yB, yC, yD, in the vapour, then,

…………………(1) Alternatively,

n

xi a ihk xi   yi i 1 (1) =>

Multi Component

a1 x1 y1  a i xi

Binary System

ax y 1  a - 1x

Ex.8. A mixture of ortho, meta, and para-mono nitro toluenes containing 60, 4, and 36 mole per cent respectively of the three isomers is to be continuously distilled to give a top product of 98 moleper cent ortho, and the bottom is to contain 12.5 mole per cent ortho. The mixture is to be distilled at a temperature of 410 K requiring a pressure in the boiler of about 6.0 kN/m2. If a reflux ratio of 5 is used, how many ideal plates will be required and what will be the approximate compositions of the product streams? The volatility of ortho relative to the para isomer may be taken as 1.70 and of the meta as 1.16 over the temperature range of 380–415 K. Solution: For 100 kmol of feed with D and W kmol of product and bottoms, respectively and xdo and xwo the mole fraction of the ortho in the distillate and bottoms, then an overall material balance gives: 100 = D + W

As a first estimate, it is supposed that the distillate contains 0.6 mole per cent meta and 1.4 mole per cent para:

Ortho:

And above the feed plate:

Now, starting the calculation from bottom: The composition of the vapour in the still is found from the relation yso = αoxso/

αxs .

Therefore,

α 1.7 1.16 1

yso = 0.191

Ex. 9.

Pinch points in binary distillation Recall for binary distillation: Rmin

Binary composition profile •xD

xD • •

•

•

pinch point

• x

zF•

•zF

• • • • • plateau of • constant • composition • •

• xB

Using Rmin generates a pinch point at the feed stage

xB• • condensor

•

• feed

• reboiler

stage number

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

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 profile LK-HK-HNK

1

stripping

rectifying

Consider LK • LK is the most volatile component in the system on every stage • there is almost no LK at the reboiler stage

xi

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

0

C

F

stage number

R

Behavior of HK

Consider HK:

Ternary composition profile LK-HK-HNK

1

pinch point

• HK behavior is most complex

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

• near reboiler, distillation is almost binary

~binary

~binary

• above feed stage, 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

0

C

F stage number

R

Pinch points in multicomponent distillation pinch F LK, HK, HNK

pinch

F LNK, LK, HK

pinch F LNK, LK, HK, HNK

pinch

• LK-HK-HNK system

• LNK-LK-HK system

• LNK-LK-HK-HNK

has a pinch point in

has a pinch point in

has pinch points in

the stripping section

the rectifying section

both sections

Graphical methods are not helpful in finding these pinch points.

Second Underwood equation CMB in rectifying section at minimum reflux: at the pinch point:

xi,j = xi,j +1

VLE:

y i,j +1 = K i xi,j +1

relative volatility:

a i = K i Kref

Lmin Vmin y i,j +1 = y i,j +1 + Dxi,D a i Kref Vmin y i,j +1 =

a i Dxi,D Lmin ai VminK ref

Vmin y i,j +1 = Lmin xi,j + Dxi,D

y i,j +1 = a i Kref xi,j +1 æ Lmin ö Vmin y i,j +1 ç 1= Dxi,D ÷ è a iVminK ref ø define: f =

a i Dxi,D 2nd Underwood equation: Vmin = åVmin y i,j +1 = å ai - f i i

Lmin VminK ref

First Underwood equation in the stripping section: assume

a i Bxi,B -Vmin = åVmin y i,k +1 = å i i ai - f

f =f DVfeed

é a i Dxi,D a i Bxi,B ù = Vmin -Vmin = å ê + ú a f a f i ê úû i ë i é a i Fzi,F ù = åê ú i ê ë a i - f úû

1st Underwood equation:

DVfeed

feed quality definition:

DVfeed = F(1- q)

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.

a i Dxi,D Vmin = åVmin y i,j +1 = å ai - f i i To get xi,D values for non-keys:

• assume non-distribution OR • assume distribution is the same at minimum reflux and total reflux i.e., use Fenske equation

Lmin = Vmin – D Rmin = Lmin/D

Sandwich components At 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. component

a

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)

designation

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.

D F S

B