• 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