• 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

BqB

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

VDx

VLy

(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

VDx

VLy (4)

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

Since:

DLR 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:

1Rxx

1RRy D

m1m

(5)

x xD

1Rxintercept D

1RRslope

ENVELOPE B

B1p

p1p

p1p x

VBx

VL

y

(6)

(7)

(8)

BqB

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

VBx

VL

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 VLpB 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

BqFLxBx

BqFLqFLy

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

1qxx

1qqy 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 > 10 < q < 1

q = 0

xF xDxB

1Rx

intercept

D

1qqslope

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 2Using 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

xmin

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

xDxVLy (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

xDxSxVLy

(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

xDxSxVLy

(20)

DSxDxSxy 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

xBxVL

y

(23)

(22)

(21)

Component:

For constant molal overflow:

(24)p

Bp

p

p1p V

xBxVL

y

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

pp VL

m

m

VLslope

n

n

VLslope

p

p

VL

slope

xB xD