Distillation IV McCabe thiele method (2)

Report
Distillation V
Multicomponent Distillation
Mass Transfer for 4th Year
Chemical Engineering Department
Faculty of Engineering
Cairo University
Distillation
operations
Single Stage
Simple
Differential
Distillation
√
Steam
Distillation
√
Multistage
Flash
vaporization
Distillation
√
Binary
system
√
Multicomponent
systems
Introduction
Multicomponent
Distillation
Lewis-Matheson
Method
Shortcut
Methods
Hengstebeck’s
method
Gilliland, Fenske ,
Underwood
Method
Constant relative
volatility method
KEY COMPONENTS
The solution in some methods is based on choice of two
“key” components between which it’s desired to make
the separation: “Light Key (LK)” and “Heavy Key (HK)”.
• Light Key: is the component that is desired to be
kept out of the bottom product.
• Heavy Key: is the component that is desired to be
kept out of the top product.
Concentrations of the key components in the top and
bottom products must be specified.
All other components except light and heavy keys are
called the non-keys components.
a
Logically ...
In the case shown:
Heavy Key (HK):C7
Light Key (LK):C6
The solution in some methods is
based on the fact that if the light key is
eliminated (nearly or completely) from
bottom product then OF COURSE the
heavier components will be also
eliminated from bottom product.
And the same for the heavy key.
V
L
C4
C5
C6
C7
C4
C5
C6
C7
C8
C9
V’ L’
C6
C7
C8
C9
1- Lewis Matheson Method
• Similar to Lewis method we used in
binary system.
• Tray to tray calculations are done with
the assumption of constant molar
flow rates of liquid and vapour in
each section.
• First Calculate L, V, L’, V’ (How?)
• Top section tray to tray calculations
are done till xi≤xFi
• Bottom section tray to tray
calculations are done till y ≥xFi
V
y1
L
Xoi
V
D
xDi
L
F
xFi
V’ L’
V’
yri
L’
x’1i
W
xWi
1- Lewis Matheson Method
For Top Section:
If total condenser
y1i=xoi=xDi
Equilibrium relation:
x1i=y1i/Ki
Operating line:
V.y2i=L.x1i + D.xDi
y 2i 
L
V
x 1i 
D
V
x Di
V1
y1
V
y1
V
y2
L
x1
L
x2
L
xo
D
xD
1- Lewis Matheson Method
For Bottom Section:
First: Reboiler (m=0):
Equilibrium relation:
yri=KixWi
Operating line:
L’.x’1i=V’.y’ri +W.xWi
y' ri 
L'
V'
x' 1i 
W
V'
x Wi
V’
y’2
L’
x’2
L’
x’1
L’
X’1
V’
y’1
V’
yr
W
xw
1- Lewis Matheson Method
For Bottom Section:
For m=1:
Equilibrium relation:
y1i=Kix1i
Operating line:
L’.x’2i=V’.y’1i +W.xWi
y' 1i 
L'
V'
x' 2i 
W
V'
x Wi
V’
y’2
L’
x’2
L’
x’1
L’
X’1
V’
y’1
V’
yr
W
xw
1- Lewis Matheson Method
NOTE:
To do these calculations we must know the
value of “K”
K=f(T,P)
Operating pressure is known
BUT operating temperature varies from tray to
another, so each tray calculation will be done by
assuming T and checking it from Sx or Sy (as if
it’s a normal flashing problem)
1) A distillation column is designed for light gases fractionation operates at
a reflux ratio of 2.5, if the condenser temperature is 60oC and it’s required
to get a top product of the following specs:
Component
C3
i-C4
n-C4
i-C5
n-C5
Distillate flow rate
(mol/sec)
5
15
24
1
0.1
xD
0.111
0.333
0.532
0.022
0.002
The temperature of the first plate is 65oC and that of the second plate is
70oC, the data for K values for the components are as follows:
Component
K at 65oC
K @70oC
C3
i-C4
n-C4
i-C5
n-C5
2.36
1.19
0.86
0.42
0.32
2.58
1.37
0.9
0.51
0.4
Estimate the composition of vapour and liquid leaving the second stage.
V=(R+1)D=3.5*45.1=157.85 mol/sec
L=RD=2.5*45.1=112.75 mol/sec
Since x1 and y1 left the first stage and
total condenser (y1i=xDi=xoi )
x1i=y1i/Ki
xC3)1=0.111/2.36=0.047
xi-C4)1=0.333/1.19=0.279
xn-C4)1=0.532/0.86=0.619
xi-C5)1=0.022/0.42=0.053
xn-C5)1=0.002/0.32=0.007
C3
i-C4
n-C4
i-C5
n-C5
D
5
15
24
1
0.1
K at 65oC
2.36
1.19
0.86
0.42
0.32
C3
i-C4
n-C4
i-C5
n-C5
xD
0.111
0.333
0.532
0.022
0.002
K @70oC
2.58
1.37
0.9
0.51
0.4
V1
y1
V
y1
V
y2
L
x1
L
x2
L
xo
D
xD
Do material balance for the loop:
VyC3)2=LxC3)1+DxC3)D
Vyi-C4)2=Lxi-C4)1+Dxi-C4)D
Vyn-C4)2=Lxn-C4)1+Dxn-C4)D
Vyi-C5)2=Lxi-C5)1+Dxi-C5)D
Vyn-C5)2=Lxn-C5)1+Dxn-C5)D
Substitue:
157.85*yC3)2=112.75*0.047+5
157.85*yi-C4)2=112.75*0.279+15
157.85*yn-C4)2=112.75*0.619+24
157.85*yi-C5)2=112.75*0.053+1
157.85*yn-C5)2=112.75*0.007+0.1
yC3)2=0.0652
yn-C4)2=0.5942
yn-C5)2=0.0056
yi-C4)2=0.2943
yi-C5)2=0.0442
C3
i-C4
n-C4
i-C5
n-C5
D
5
15
24
1
0.1
K at 65oC
2.36
1.19
0.86
0.42
0.32
C3
i-C4
n-C4
i-C5
n-C5
xD
0.111
0.333
0.532
0.022
0.002
K @70oC
2.58
1.37
0.9
0.51
0.4
V1
y1
V
y1
V
y2
L
x1
L
x2
L
xo
D
xD
Since x2and y1left the second stage
x2i=y2i/Ki
xC3)2=0.0652/2.58=0.0253
xi-C4)2=0.2943/1.37=0.2148
xn-C4)2=0.5942/0.9=0.6602
xi-C5)2=0.0442/0.51=0.0867
xn-C5)2=0.0056/0.4=0.0140
C3
i-C4
n-C4
i-C5
n-C5
D
5
15
24
1
0.1
K at 65oC
2.36
1.19
0.86
0.42
0.32
C3
i-C4
n-C4
i-C5
n-C5
xD
0.111
0.333
0.532
0.022
0.002
K @70oC
2.58
1.37
0.9
0.51
0.4
V1
y1
V
y1
V
y2
L
x1
L
x2
L
xo
D
xD
2- Constant Relative Volatility Method
• Before in Lewis, during calculations you need to
apply equilibrium relation but to do so you need
temperature of that plate which is dependent on
composition so trial and error needed.
• In this method to get red of this difficulty will use
relative volatility instead of k-values in relating the
vapor and liquid composition (which are in eqm).
• So, no more trial and error will be done on T.
How?
2- Constant Relative Volatility Method
• What happens here is that for each component get average
relative volatility (constant) and work with it a long the tower.
So, Temperature will not be included in calculations.
• Finally, this method will be like Lewis-Matheson in steps but
without trials on T only the difference that equilibrium
relation will be.
 =
 
 
 =






 = (


) =

Where  is constant for each component along the
tower.


3- Shortcut Methods:
a) Hengstebeck’s
• It’s also called Pseudo binary system method.
• System is reduced to an equivalent binary system and
is then solved by McCabe-Thiele method graphically.
• The only method that solves the multi-component
systems graphically (in our course of course).
• Used as for preliminary design work.
Logically ...
• Using the concept of light and heavy key components
• So, we can consider the system as a binary system where
it’s desired to separate the light key from the heavy key.
According to that the molar flow rate of the non-key
components can be considered constant. ****
Also the total flow rates of vapour and liquid are
considered constant.
The method used for these calculations was
developed by R.J.Hengstebeck, that’s why it’s called
Hengstebeck’s method.
Hengstebeck’s Method
Let’s say that:
V=total molar vapour flow rate in the top section
L=total molar liquidflow rate in the top section
V’=total molar vapour flow rate in the bottom section
L’=total molar liquidflow rate in the bottom section
yni=mole fraction of component “i” in vapour phase on tray “n”
xni=mole fraction of component “i” in liquid phase on tray “n”
uni=molar vapour flow rate of component “i” from stage “n”
lni=molar liquid flow rate of component “i” from stage “n”
di=molar liquid flow rate of component “i” in top product
wi=molar liquid flow rate of component “i” in bottom product
a
Hengstebeck’s Method
This means that:
uni=yni*V
u'ni=y’ni*V’
lni=xni*L
l’ni=x’ni*L’
yni=uni/V
y'ni=u'ni/V’
xni=lni/L
x’ni=l’ni/L’
di=xDi*D
wi=xWi*W
a
Hengstebeck’s Method
To reduce the system to an equivalent binary system we have to
calculate the flow rates of the key components through the
column (operating line slope is always L/V)
The total flow rates (L and V) are constant, and the molar flow
rates of non key components are constant, then we can
calculate the molar flow rates of key components in terms of
them.
NOTE: The total flow rate is constant and the molar flow rates
of non key components are constant, this does not mean
that the molar flow rates of key components are
constant as the mass transfer is equimolar.
a
Hengstebeck’s Method
If Le and Ve are the estimated flow rates of the combined keys.
And li and ui are flow rates of the non-key components lighter
than the keys in the top section.
And l’i and u’i are flow rates of the non-key components heavier
than the keys in the bottom section.
Then slope of top section operating line will be Le/Ve
And slope of bottom section operating line will be L’e/V’e
SO:
Le=L-Sli
Ve=V-Sui
L’e=L’-Sl’i
V’e=V’-S’ui
a
Hengstebeck’s Method
The final shape of the x-y diagram will be as shown
We need to calculate:
Le (or li’s)
Ve (or ui’s)
L’e
V’e
xD 
Le/Ve
L’e/V’e
xW 
 LK
 LK W
W   HK W
xF 
 LK
 LK  F
 F   HK  F
 LK
 LK  D
 D   HK  D
Hengstebeck’s Method
To calculate them we will need to calculate “a” where “a”
is the relative volatility WITH RESPECT TO THE HEAVY KEY
FOR ALL COMPONENTS.
So for components heavier than the heavy key a<1
And for components lighter than the heavy key a>1
o
Pi
o
αi 
Pi
PT

o
o
P HK
P HK
Ki

K HK
PT
o
P LK
o
α LK 
P LK
P
o
HK
PT

P
o
HK

K LK
K HK
PT
Hengstebeck’s Method
For TOP SECTION
Equilibrium Relation:
yi=Ki.xi
ui
V
 Ki
li
ui 
L
V
L
Material balance:
ui=li+di
K ili
V
L
K ili  li  d i
For heavy key:
V
L
V
L
K HK l i  l i  d i
K HK  1 
di
li
And di/li=0
K HK 
L
V
Hengstebeck’s Method
For TOP SECTION
Equilibrium Relation:
yi=Ki.xi
ui
V
 Ki
li
L
ui 
V
L
Material balance:
ui=li+di
K ili
V
L
K ili  li  d i
For any light non-key component:
V
L
K ili  li  d i
1
K HK
li 
K ili  li  d i
di
α i -1
u i  d i  li
Hengstebeck’s Method
For BOTTOM SECTION
Equilibrium Relation:
yi=Ki.xi
ui
V
 Ki
li
li 
L
L'
V' K i '
Material balance:
l’i=u’i+wi
L'
ui
V' K' i
u 'i  u 'i  w i
For Light key:
L'
V' K' LK
u 'i  u 'i  w i


L'

 1 u ' i  w i
 V' K'
LK


And


≈0
K LK 
L'
V'
Hengstebeck’s Method
For BOTTOM SECTION
Equilibrium Relation:
yi=Ki.xi
ui
V
 Ki
li
L
l' i 
L'
V' K i '
Material balance:
l’i=u’i+wi
υ' i
L'
V' K' i
u 'i  u 'i  w i
For any heavy non-key component:
L'
V' K' i
K LK
Ki
u 'i  u 'i  w i
u 'i  u 'i  w i
u 'i 
a LK u ' i
ai
 u 'i  w i
αiw i
α LK - α i
l' i  υ' i  w i
2) Estimate the number of ideal stages needed in the butanepentane splitter defined by the compositions given in the table
below. The column will operate at a pressure of 8.3 bar, with a
reflux ratio of 2.5. The feed is at its boiling point.
Compositions of feed, top and bottom products are shown in
table below:
Propane, C4
i-Butane, i-C4
n- Butane, n-C4
i-Pentane, i-C5
n-Pentane, n-C5
Total, kmol
Feed (F)
5
15
25
20
35
100
Tops (d)
5
I5
24
1
0
45
Bottoms (w)
0
0
1
19
35
55
Equilibrium constants were calculated and found to be:
Component
C3
iC4
nC4
iC5
nC5
Average value of K
5.0
2.6
2.0
1.0
0.85
Propane, C4
i-Butane, i-C4
n- Butane, n-C4
i-Pentane, i-C5
n-Pentane, n-C5
Total, kmol
Light key will be:
n-C4
Heavy key will be:
i-C5
SO
aC3=5/1=5
aiC4=2.6/1=2.6
anC4=2/1=2
aiC5=1/1=1
anC5=0.85/1=0.85
xD 
d nC 4
d nC 4  d iC 5

F
5
15
25
20
35
100
Component
C3
iC4
nC4
iC5
nC5
24
24  1
 0 . 96
xw 
xF 
w nC4
w nC4  w iC5

1
1  19
d
5
I5
24
1
0
45
w
0
0
1
19
35
55
Average value of K
5.0
2.6
2.0
1.0
0.85
f nC4
f nC4  f iC5
 0.05

25
25  2 0
 0.56
For Top section:
Le=L-Sli
L=R*D=2.5*45=112.5 Kmoles
 =

5
15
=
+
= 10.625
 − 1 5 − 1 2.6 − 1
C3
i-C4
n-C4
i-C5
n-C5
Total
F
5
15
25
20
35
100
d
5
I5
24
1
0
45
Le=112.5-10.625=101.875 Kmoles
And
Ve=V-Sui
V=(R+1)*D=3.5*45=157.5 Kmoles
 =
5
15
  +  =
+5 +
+ 15 = 30.625
5−1
2.6 − 1
Ve=157.5-30.625=126.875
SO
Le/Ve=0.8
w
0
0
1
19
35
55
For Bottom section:
L’=L+F
(feed is at its boiling point)
L’=112.5+100=212.5 Kmoles
V’=V=157.5 Kmoles
V’e=V-Su’i
′ =
 
0.85 ∗ 35
=
= 25.87
 − 
2 − 0.85
Ve=157.5-25.87=131.63 Kmoles
And
L’e=L’-Sl’I
′ =
′
0.85 ∗ 35
+ 35) = 60.87
 +  = (
2 − 0.85
L’e=212.5-60.87=151.63 Kmoles
So
C4
i-C4
n-C4
i-C5
n-C5
Total
L’e/V’e=1.15
F
5
15
25
20
35
100
d
5
I5
24
1
0
45
w
0
0
1
19
35
55
Equilibrium curve:
y
a LK x
1  a LK  1  x
0.2

0.3
2x
1 x
x
0
0.1
0.4
0.5
0.6
0.7
0.8
0.9
1
y
0
0.18 0.33 0.46 0.57 0.66 0.75 0.82 0.88 0.94
1
1
0.9
NTS in top section
6
NTS in bottom section
Reboiler+5.5
0.8
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
3- Shortcut Methods:
b) Gilliland, Fenske , Underwood Method
• It is an empirical method for calculating number of stages
in Multi-component distillation.
• I t is composed of 3 equations to work with.
1- Gilliland equation (Has graph.): Used to calculate N stages
(In this equation min. reflux ratio and min. number of
stages is need)
2- Fenske equation: Used to calculate (N stages )min
3- Underwood equation: Used to calculate minimum reflux
ratio
3- Shortcut Methods:
b) Gilliland, Fenske , Underwood Method
1- Gilliland equation:
 − 
 − 
= 0.75[1 −
+1
+1
0.567
]
2- Fenske equation:
 =


ln(  )
 
ln(, )
3- Underwood equation:
1− =
 
 −
 + 1 =
 
 −
q: (Hv-hf)/ (Hv-hL)
Where θ : is a relative volatility lies between the relative volatility of light
and heavy components.
3- Shortcut Methods:
b) Gilliland, Fenske , Underwood Method
(Gilliland Chart to use instead of equation)
3) A mixture of Hexane, Heptane, and Octane is to be
separated to give the following products. Use the shortcut
method to:
a) Calculate the approximate minimum number of stages
b) Calculate the approximate minimum reflux ratio
c) Show how to get the approximate number of stages
Note: The feed is liquid at its bubble point.
compon
ent
Feed
F
(kmole)
Hexane
40
Heptane
35
Octane
25
Distillate
xf
0.4
0.35
0.25
D
(kmole)
40
34
1
xD
0.534
0.453
0.013
Relative
Volatility
Residue
W
(kmole)
0
1
24
xW
0
0.04
0.96
2.7
2.22
1
- Calculate  from Underwood Equation as follows:
1− =
 
 − 
Feed is saturated liquid so, (q=1)
2.7 ∗ 0.4 2.22 ∗ 0.35 1 ∗ 0.25
+
+
= 1−1
2.7 − 
2.22 − 
1−
The above is one equation in one unknown; however, it will be
solved using trial and error
Note: the value of  will lie between the relative volatilities of the
light and heavy keys (Heptane and octane respectively)
1 <  < 2.22
Get  = 1.1725
- Calculate Rmin from Underwood Equation as follows:
 + 1 =
 
 − 
2.7 ∗ 0.534
2.22 ∗ 0.453
1 ∗ 0.013
+
+
=  + 1
2.7 − 1.1725 2.22 − 1.1725 1 − 1.1725
Rmin = 0.8287
- Calculate Nmin from Fenske Equation:

 
ln(
)
 
=
ln(, )
Nmin = 14.13
- Finally to get approximate number of stage
1-ass. R=1.5Rmin=1.243
2-Get (R-R min)/(R+1)=0.185
3- Go to Gilliland chart and get the ratio (N-Nmin)/(N+1)=0.462
4- Knowing Nmin you can get N=27.13
KOL SANA W ENTO TAYEBEN

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