• This method is based on the Lewis modification of the Sorel method,

• It assumes equimolal overflow in the rectifying section, in the stripping section, and equimolal latent heats,

• L0 is a saturated liquid

• Column pressure and reflux ratio are fixed,

F

Lm

L0

DxD

Vm+1

qD

m

B

qB

p

Overall mass balance:

F = D + B

pL

1pV

ENVELOPE A

Vm+1 = Lm + D

Vm+1 ym+1 = Lm xm + D xD

D1m

m1m

m1m x

VD

xVL

y

(1)

(2)

(3)

This is an equation of a straight line on a plot of vapor composition versus liquid composition, where (Lm/Vm+1) is the slope and (DxD/Vm+1) is the intercept which passes through the point (xD, xD) and (xm, ym+1),

v1

v2 L1

L0

DxD

A

F

Lm

Vm+1

v1

qD

m

Since all L values are equal and all V values are equal (due to constant molal overflow assumption:

Dm

mm

m1m x

VD

xVL

y (4)

Equation (4) is the operating line or material balance line for the rectifying section,

Since:

DL

R m

Vm = Lm + D 1RR

1DLDL

DLL

VL

m

m

m

m

m

m

1R1

1DL1

DLD

VD

mmm

In term of R, equation (4) can be written as:

1Rx

x1R

Ry D

m1m

(5)

x xD

1Rx

intercept D

1RR

slope

ENVELOPE B

B1p

p1p

p1p x

VB

xVL

y

(6)

(7)

(8)

B

qB

1pV

pL

1NV

NL

p

p+1

BLV p1p

Bpp1p1p xBxLyV

Since all L values are equal and all V values are equal (due to constant molal overflow assumption:

Bp

pp

p1p x

VB

xVL

y (9)

Equation (9) is the operating line or material balance line for the stripping section,

This is an equation of a straight line with slope and intercept passing through (xB, xB) and (xp, yp+1),

This line can be drawn from point (xB, yB) to point

or with slope

pp VL

pB VBx

pB VBx,0 pp VL

qFLL mp

FLL

q mp

Fq1VV pm

(10)

(11)

(12)

and is calculated by material and enthalpy balancerelationship around the feed plate,

qFDBVqFFVFq1VV mmmp

BqFLV mp (13)

The problem is, how to calculate and ?

FLmVm

pV pL

pV pL

pV pL

q is the number of moles of saturated liquid formed on the feed plate by the introduction of 1 mole of feed:

• q = 1 : saturated liquid feed, xF = xi

• q = 0 : saturated vapor feed, xF = yi

• q > 1 : cold liquid feed, xF < xi

• q < 1 : superheated vapor, xF > xi

• 0 < q < 1 : two-phase feed, xF xi

BqFLxB

xBqFL

qFLy

m

Bp

m

m1p

Substituting eqs, (10) and (13) to eq, (9) yields:

(14)

This equation gives the slope of the operating line in the stripping section as

There is an easier way to draw the operating line in the stripping section, i,e, by using the q-line, which started from point (xF, yF = xF),

BqFLqFL mm

Component material balance of the feed:

ipmimpF yVVxLLxF

i

pmi

mpF y

FVV

xF

LLx

iiF y1qxqx

iiF y1qxqx

1qx

x1q

qy F

ii

(15)

Eq, (12) is the equation of the q line having a slope of q/(q – 1) and terminating at xF on the 45 line and at point (xi, yi),

• Saturated liquid feed : q = 1 : slope = • Saturated vapor feed : q = 0 : slope = 0• Cold liquid feed : q > 1 : slope = +• Superheated vapor feed : q < 1 : slope = – • Two-phase feed : 0 < q < 1 : slope = –

xF xDxB

q = 1

q > 1

0 < q < 1

q = 0

xF xDxB

1Rx

intercept

D

1qq

slope

xF xDxB

x1, y1

x2, y2

x3, y3

x4, y4

x1, y2

x2, y3

x3, y4

MINIMUM REFLUX

xF xDxB

1Rx

intercept

min

D

MINIMUM REFLUX

xF xDxB

1Rx

intercept

min

D

TOTAL REFLUX

xF xDxB

EXAMPLE 2

Using the data of EXAMPLE 1, determine:

a. The number of equilibrium stages needed for saturated-liquid feed and bubble-point reflux with R = 2,5 using McCabe-Thiele graphical method

b. Rmin

c. Minimum number of equilibrium stages at total reflux.

SOLUTION(a) The slope of the operating line in the rectifying section:

715.015.2

5.21R

Rslope

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.00.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

y

x

N = 11

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

x

y

(b)

Intercept = 45.01R

x

min

D

Rmin = 1,18

(c)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

x

y

N = 8

SIDE PRODUCT

• If a product of intermediate composition is required, a vapor or a liquid side stream can be withdrawn,

• This kind of column configuration is typical of the petrochemical plants, where the most common running unit operation is the fractional distillation,

• This consists in splitting a mixture of various components, the crude oil, into its components, Because of their different boiling temperatures, the components (or so-called fractions) of the crude oil are separated at different level (i,e, plate) of the column, where different boiling temperatures are present,

• The fractions are then withdrawn from the plate where they form, therefore the column presents numerous side streams,

Lm Vm

D, xD

B, xB

F, xF

L0

S, xS

Rectifying section

Middle section

Stripping section

nL nV

pL pV

D, xD L0

S, xSVm+1

Lmm

MATERIAL BALANCE IN RECTIFYING SECTION

Assuming constant molar overflow, then for the rectifying section the operating line is given by:

m

Dm

m

m1m V

xDx

VL

y (16)

D, xD

F, xF

L0

S, xS

Rectifying section

Middle section nL1nV

MATERIAL BALANCE IN MIDDLE SECTION

Overall: DSLV n1n

DSnn1n1n xDxSxLyV

1n

DSn

1n

n1n V

xDxSx

VL

y

(19)

(18)

(17)

Component:

Since the side stream is normally removed as a liquid:

SLL mn mn VV

For constant molal overflow:

n

DSn

n

n1n V

xDxSx

VL

y

(20)

Vm Lm

nV nL

SLL mn mn VV

S

DSxDxS

xy DS

which is the mean molar composition of the overhead product and side streams,

Since xS < xD and , this additional operating line cuts the line y = x at a lower value than the operating line though it has a smaller slope,

Equation (20) represents a line of slope , which passes through the point

nn VL

mn LL

MATERIAL BALANCE IN STRIPPING SECTION

F, xF

B, xB

pL1pV

Overall: BLV p1p

Bnp1p1p xBxLyV

1p

Bp

1p

p1p V

xBx

VL

y

(23)

(22)

(21)

Component:

For constant molal overflow:

(24)p

Bp

p

p1p V

xBx

VL

y

Equation (24) represents a line of slope , which passes through the point (xB, xB)

pp VL

nV nL

qFLL np FqVV pn 1

F

pV pL

FqVV np 1

m

m

VL

slope

n

n

VL

slope

p

p

VL

slope

xB xD

MULTIPLE-FEED

Lm Vm

D, xD

B, xB

F2, xF2

L0

F1, xF1

Rectifying section

Middle section

Stripping section

nL nV

pL pV

D, xD L0

F1, xF1

Vm+1

Lmm

MATERIAL BALANCE IN RECTIFYING SECTION

Assuming constant molar overflow, then for the rectifying section the operating line is given by:

m

Dm

m

m1m V

xDx

VL

y (16)

D, xD

F2, xF2

L0

F1, xF1

Rectifying section

Middle section nL1nV

MATERIAL BALANCE IN MIDDLE SECTION

Overall: DLFV nn 11

DnnFnn xDxLxFyV 1111

1

11

11

n

FDn

n

nn V

xFxDx

VL

y(27)

(26)

(25)

Component:

For constant molal overflow:

n

FDn

n

nn V

xFxDx

VL

y 111

(29)

Equation (29) represents a line of slope nn VL

Material balance around feed plate F1 :

11FqLL mn

111 FqVV mn

F1

Vm Lm

nLnV

(30)

(31)

111 FqVV nm

MATERIAL BALANCE IN STRIPPING SECTION

F2, xF2

B, xB

pL1pV

Overall: BLV pp 1

Bnppp xBxLyV 11

111

p

Bp

p

pp V

xBx

VL

y(34)

(33)

(32)

Component:

For constant molal overflow:

(35)p

Bp

p

pp V

xBx

VL

y 1

Equation (24) represents a line of slope , which passes through the point (xB, xB)

pp VL

Material balance around feed plate F2 :

22FqLL np

221 FqVV pn

F2

pLpV

(36)

(37)

nV nL

221 FqVV np

EXAMPLE 3

Benzene (1) is to be separated from toluene (2) in a distillation column. Three feed streams are available:

No. of stream 1 2 3

Lb-mol/h 200 500 300

Benzene mole fraction 0.7 0.5 0.2

Fraction vaporized 0.25 0.0 0.5

q 0.75 1.0 0.5

A high-purity product containing 98-mol% benzene is required. Toluene is to be recovered at 95 mol % purity. A total overhead condenser is to be used. The column is designed to operate at R = 1.25 Rmin, how many stages is required, and where should be the feed streams?

T (K) x1 y1

385 0.000 0.000

380 0.100 0.210

375 0.230 0.410

370 0.365 0.575

365 0.525 0.725

360 0.700 0.850

355 0.910 0.965353 1.000 1.000

SOLUTION

D, xD

B, xB

F2, z2

L0 F1, z1

Rectifying section

Middle section I

Stripping section

F3, z3

Middle section II

Slope of feed lines:

1. Feed 1: 3175,0

75,011

1

q

q

2. Feed 2:

11

112

2

qq

3. Feed 3: 115,0

5,013

3

q

q

OVERALL MATERIAL BALANCE

300500200321 FFFBD

1000 DB

COMPONENT MATERIAL BALANCE

332211 zFzFzFDxBx DB

2.03005.05007.020098.005.0 DB

45098.005.0 DB

(a)

(b)

Simultaneous solution of eqs. (a) and (b) results in:

D = 430.1 and B = 569.9

5.01

98.01

interceptminmin

RR

xD

Minimum Reflux:

96.0min R

Column is operated at R = 1.25 Rmin = 1.2

(see the following graph)

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.00.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

x

y

F1F2

F3

Inte

rcep

t =

0.5

1. Rectifying section:

5455.012.1

2.11

slope

R

R

hmol-lb1.5161.4302.10 RDL

hmol-lb2.9461.4301.51601 DLVV

2. Middle section I:

7433.02.8961.666

slope VL

hmol-lb1.66620075.01.51611 FqLL

hmol-lb2.89620025.02.9461 11 FqVV

3. Middle section II:

3.12.8961.1166

slope VL

hmol-lb1.11665000.11.66622 FqLL

hmollb2.8962000.02.8961 22 FqVV

4. Rectifying section:

76.12.7461.1316

slope VL

hmol-lb1.13163005.01.116633 FqLL

hmollb2.7463005.02.8961 33 FqVV

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.00.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

x

yy

x

N = 25

12

43

65

978

1011