Distillation IV
McCabe thiele method (2)
Mass Transfer for 4th Year
Chemical Engineering Department
Faculty of Engineering
Cairo University
Today isA
Special cases of binary systems distillation
using McCabe Thiele method.
The Simple Case
•Only one feed
•Consisting of two sections
•Condenser and Reboiler
•No side products
x
R
1
Steps:
x
1. q-line
R 1
2. Rmin
3. Rop and Top section oper.
Line
4. Bottom section oper. line
(XD,XD)
D
min
D
(XW,XW)
XW
XF
XD
Other Cases
1.
2.
3.
4.
5.
6.
Enriching Section
Stripping Section
Complex Feed or Multiple feeds
Open Steam
Top Side Product
Bottom Side Product
NOTE: q-line NOT ONLY represents feed, but it
represents any stream that changes the
flow rates inside the column.
1- Enriching Section
• Used in case when it’s needed to
recover light component from feed
containing little amount of it.
• Feed composition is near that of the
bottom product (xF is very small)
• Feed is usually saturated vapour
• No reboiler is used.
V
yo
L
xo
F
xF
W
xW
D
xD
1- Enriching Section
Steps:
• Feed is saturated vapour
• No. of stages in bottom
section =0
• Get point of intersection
of bottom section and
q-line directly
• Draw the top section
operating line.
(XD,XD)
XW
XF
XD
2- Stripping Section
• Used in case when it’s needed to
recover heavy component from feed
containing little amount of it.
• Feed composition is near that of the
top product (xF is very big)
• Feed is usually saturated liquid
• No reflux is needed.
D
xD
F
xF
V’
yr’
L’
xo’
W
xW
2- Stripping Section
Steps:
• Feed is saturated liqiud
• No. of stages in top
section =0
• Get point of intersection
of top section and
q-line directly
• Draw the bottom section
operating line.
XW
XF
XD
3- Multiple Feeds
Generally F1 is saturated liquid, and F2
is saturated vapour.
The column can now be divided into
THREE section: Enriching, Stripping and
Middle section.
Both operating lines of the top and
bottom sections will not be changed.
Only we will need to get the middle
section line.
V
yo
L
xo
D
xD
F1
xF1
F2
xF2
V”
yr
L’’
xo
W
xW
3- Multiple Feeds
Since we assumed constant molar flow
rates in the column, we can say that:
L’=L+F1
V’=V
And
L”=L’
V’=V”+F2
To draw the middle section operating
line we have to calculate its slope from
the above relations
V
yo
L
xo
L
V
L’
V’
L”
V”
D
xD
F1
xF1
F2
xF2
V”
yr
L’’
xo
W
xW
3- Multiple Feeds
Given information:
F1, xf1, F2, xf2, xD, xW, R
F1 is saturated liqiud
F2 is saturated vapour
(XD,XD)
xD
Steps:
R 1
Top section operating line
as it is
(XW,XW)
XW
XF2
XF1
XD
3- Multiple Feeds
Middle line starts from the
end of the top section and
ends at the beginning of
the bottom section.
So point of intersection of
top section and first q-line
is on the middle section
line.
We need another point or
a slope to draw the middle
line.
(XD,XD)
xD
R 1
(XW,XW)
XW
XF2
XF1
XD
3- Multiple Feeds
The slope of any operating
line is L/V in the section it
represents.
So in the middle section
slope is L’/V’
As we know
L’=F1+L and V’=V
Where
L=D*R and V=L+D=(R+1)D
So we can now get L’/V’
and draw its line
(XD,XD)
L’/V’
xD
R 1
(XW,XW)
XW
XF2
XF1
XD
4- Upper Side Product
• Any side product is withdrawn as
saturated liquid.
• As this side product will change the flow
rates inside the column, there will be a
q-line representing the side product.
F
• The column will be divided into
x
3 section: Top, middle and bottom
• Still the top section not affected, and we
want to draw the middle section line.
V
yo
L
xo
D
xD
S
xS
F
V’
yr
L’
xo
W
xW
4- Upper Side Product
V
yo
To get Operating line of the middle section do
MB on the loop:
V=L+S+D
V.yn+1=L.xn+S.xS+D.xD
(equation of st. line as S,D,xS,xD are constants)
L
xo
S
xS
V
yn+1
This line starts at the intersection of the top
section with the q-line of side product
D
xD
L
xn
F
xF
V’
yr
L’
xo
W
xW
4- Upper Side Product
Steps:
1. Side product is saturated
liquid.
2. Feed q-line is drawn
whatever its state
3. Draw the top section line
We need to use the x D
equation of the middle R  1
section to draw its operating
line.
(XD,XD)
(XW,XW)
XW
XF
XS
XD
4- Upper Side Product
Back to equations:
V=L+S+D
V-L=S+D
V.yn+1=L.xn+S.xS+D.xD
We know a point on the line
and need to get another
point to draw it
The easy point is on 45 line
V.x=L.x+S.xS+D.xD
xD
(V-L).x=S.xS+D.xD
R 1
(S+D).x=S.xS+D.xD
x y
S .x S  D .x D
SD
(XD,XD)
x=y
(XW,XW)
XW
XF
XS
XD
5- Bottom Side Product
• The column will be divided into 3
section: Top, middle and bottom
• Still the top section not affected, and
we want to draw the middle section
line.
• We will derive the middle section
operating line equation (as in
previous case)
V
yo
L
xo
D
xD
F
xF
S
xS
V’
yr
L’
xo
W
xW
5- Bottom Side Product
To get Operating line of the middle section do
MB on the loop:
L’=V’+S+W
L’.x’m+1=V’.y’m+S.xS+W.xW
(equation of st. line as S,W,xS,xw are constants)
V
yo
L
xo
D
xD
F
xF
This line ends at the intersection of the
bottom section with the q-line of side product
V’
ym
L’
Xm+1
S
xS
V’
yr
L’
xo
W
xW
5- Bottom Side Product
Steps:
1. Side product is saturated
liquid.
2. Feed q-line is drawn
whatever its state
3. Draw the top section line
(XD,XD)
xD
We need to use theR  1
equation of the middle
section to draw its operating
line.
(XW,XW)
XW
XS
XF
XD
5- Bottom Side Product
Back to equations:
L’=V’+S+W
L’-V’=S+W
L’.x’m+1=V’.y’m+S.xS+W.xw
We know a point on the line
and need to get another
point to draw it
The easy point is on 45 line x D
L’.x=V’.x+S.xS+W.xW
R 1
(L’-V’).x=S.xS+W.xW
(S+W).x=S.xS+W.xW
x y
S . x S  W . xW
S W
(XD,XD)
x=y
(XW,XW)
XW
XS
XF
XD
6- Open Steam
• Used in cases when the feed
contains water, so heat is added to
column in the form of direct heating
by steam instead of reboiler.
• The bottom section operating line
will change as the equations will be
changed.
• Still the intersection of the top
section line and the q-line is on the
bottom section line.
V
yo
L
xo
F
xF
S
W
xW
D
xD
6- Open Steam
Operating line of bottom section:
L’+S=V’+W
L’-V’=W-S
L’.x’m+1+S.yS=V’.y’m+W.xW
To draw this line we need another point,
try 45o line
L’.x=V’.x+W.xW
(L’-V’)x=W.xW
(W-S)x=W.xW
x y
V
yo
L
xo
F
xF
L’
Xm+1
V’
ym
S
W . xW
W S
W
xW
D
xD
6- Open Steam
Steps:
1. Draw the top section line
2. Draw the q-line
3. Locate the point derived
on the 45o line
4. Draw the bottom section
line.
xD
Here the bottom section willR  1
end at a point (xw,0)
(XD,XD)
W . xW
W S
Usually S needed to be calculated.
XW(XW,ys)
XF
XD
4- 100 Kgmol/hr of saturated liquid containing 70 mol%
benzene enters a stripping tower at 1 atm. The bottom
product flow rate is 15 Kgmol/hr containing only 10%
benzene. Saturated steam at 4 atmospheres is available for
the reboiler duty. (Hv=2740 KJ/Kg, hL=610 KJ/Kg).
Calculate:
a- The overhead product flow rate and its composition.
b- Number of theoretical plates required.
c- Steam consumption in reboiler.
Equilibrium data:
T, C
PBo, mmHg
PTo, mmHg
80
760
205
85
877
345
90
1016
405
95
1068
475
100
1344
557
105
1532
645
lBenzene=7360 Cal/gmol, lToluene=7960 Cal/gmol
110
1800
760
Stripping tower
F=100 Kgmol/hr
(saturated liquid)
xF=0.7
P=1 atm
W=15 Kgmol/hr
xW=0.1
Steam: Hv=2740 KJ/Kg, hL=610 KJ/Kg
lst=2130 KJ/Kg
a)
D=F-W=100-15=85 Kmol/hr
xD=(F.xF-W.xW)/D=(100*0.7-15*0.1)/85=0.806
T, C
PBo, mmHg
PTo, mmHg
X
Y
80
760
205
1
1
85
877
345
0.78
0.90
90
1016
405
0.58
0.78
95
1068
475
0.48
0.678
100
1344
557
0.26
0.46
105
1532
645
0.13
0.26
110
1800
760
0
0
1
b)
NTS= Reboiler + 4.6
0.9
0.8
c)
Qr=mst.lst=V’.lr
lr is calculated at yr
lr =0.2*7360+0.8*7960
lr =7840 Cal/gmol
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
V’=???
Slope of operating line is L’/V’=1.1667
And
1
W+V’=L’
0.9
Get L’ and V’
0.8
V’=90 kmol/hr
0.7
L’=105 Kmol/hr
0.6
L’/V’
0.5
V’
y’m
L’
x’m+1
0.4
V’
y’m
L’
x’m
0.3
0.2
V’
yr
L’
X’1
0.1
W
xw
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Finally
Qr=90*7840=705600 Cal/hr= 2949.408 KJ/hr
lst=2130 KJ/Kg
2949.408 =2130*mst
mst= 1384.7 Kg/hr