Post on 16-Dec-2015
• 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
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