Distillation V Multicomponent Distillation
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Transcript of Distillation V Multicomponent Distillation
Distillation VMulticomponent Distillation
Mass Transfer for 4th YearChemical Engineering Department
Faculty of EngineeringCairo University
Distillation operations
Single Stage
Simple Differential Distillation
Steam Distillation
Flash vaporization Distillation
Multistage
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 ...
C4C5C6C7C8C9
C6C7C8C9
C4C5C6C7V L
V’ L’
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.
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
L’x’1i
FxFi
Vy1
V’yri
WxWi
LXoi
DxDi
V L
V’ L’
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V1
y1
Lxo
DxDL
x1
Lx2
Vy2
Vy1
1- Lewis Matheson Method
For Top Section:If total condensery1i=xoi=xDi
Equilibrium relation:x1i=y1i/Ki
Operating line:V.y2i=L.x1i + D.xDi
Di1i2i xVDx
VLy
L’X’1
V’yr
Wxw
V’y’1
V’y’2
L’x’2
L’x’1
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
Wi1iri xV'Wx'
V'L'y'
L’X’1
V’yr
Wxw
V’y’1
V’y’2
L’x’2
L’x’1
1- Lewis Matheson Method
For Bottom Section:For m=1:Equilibrium relation:y1i=Kix1i
Operating line:L’.x’2i=V’.y’1i +W.xWi
Wi2i1i xV'Wx'
V'L'y'
1- Lewis Matheson Method
NOTE:To do these calculations we must know the value of “K”K=f(T,P)Operating pressure is knownBUT 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:
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:
Estimate the composition of vapour and liquid leaving the second stage.
Component Distillate flow rate (mol/sec)
xD
C3 5 0.111i-C4 15 0.333n-C4 24 0.532i-C5 1 0.022n-C5 0.1 0.002
Component K at 65oC K @70oC C3 2.36 2.58
i-C4 1.19 1.37n-C4 0.86 0.9i-C5 0.42 0.51n-C5 0.32 0.4
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=0.047xi-C4)1=0.333/1.19=0.279xn-C4)1=0.532/0.86=0.619xi-C5)1=0.022/0.42=0.053xn-C5)1=0.002/0.32=0.007
D xDC3 5 0.111
i-C4 15 0.333n-C4 24 0.532i-C5 1 0.022n-C5 0.1 0.002
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V1
y1
Lxo
DxDL
x1
Lx2
Vy2
Vy1
K at 65oC K @70oC C3 2.36 2.58
i-C4 1.19 1.37n-C4 0.86 0.9i-C5 0.42 0.51n-C5 0.32 0.4
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+5157.85*yi-C4)2=112.75*0.279+15157.85*yn-C4)2=112.75*0.619+24157.85*yi-C5)2=112.75*0.053+1157.85*yn-C5)2=112.75*0.007+0.1
yC3)2=0.0652 yi-C4)2=0.2943yn-C4)2=0.5942 yi-C5)2=0.0442yn-C5)2=0.0056
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V1
y1
Lxo
DxDL
x1
Lx2
Vy2
Vy1
D xDC3 5 0.111
i-C4 15 0.333n-C4 24 0.532i-C5 1 0.022n-C5 0.1 0.002
K at 65oC K @70oC C3 2.36 2.58
i-C4 1.19 1.37n-C4 0.86 0.9i-C5 0.42 0.51n-C5 0.32 0.4
Since x2and y1left the second stagex2i=y2i/Ki
xC3)2=0.0652/2.58=0.0253xi-C4)2=0.2943/1.37=0.2148xn-C4)2=0.5942/0.9=0.6602xi-C5)2=0.0442/0.51=0.0867xn-C5)2=0.0056/0.4=0.0140
D xDC3 5 0.111
i-C4 15 0.333n-C4 24 0.532i-C5 1 0.022n-C5 0.1 0.002
K at 65oC K @70oC C3 2.36 2.58
i-C4 1.19 1.37n-C4 0.86 0.9i-C5 0.42 0.51n-C5 0.32 0.4
E-9
V1
y1
Lxo
DxDL
x1
Lx2
Vy2
Vy1
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 sectionL=total molar liquidflow rate in the top sectionV’=total molar vapour flow rate in the bottom sectionL’=total molar liquidflow rate in the bottom sectionyni=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 productwi=molar liquid flow rate of component “i” in bottom product a
Hengstebeck’s Method
This means that:uni=yni*Vu'ni=y’ni*V’
lni=xni*Ll’ni=x’ni*L’
di=xDi*Dwi=xWi*W
yni=uni/Vy'ni=u'ni/V’
xni=lni/Lx’ni=l’ni/L’
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
Le/Ve
L’e/V’e FF
FF HKLK
LKx
DD
DD HKLK
LKx
WW
WW HKLK
LKx
We need to calculate:Le (or li’s)Ve (or ui’s)L’e
V’e
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<1And for components lighter than the heavy key a>1
HK
i
T
oHK
T
oi
oHK
oi
i KK
PP
PP
PPα
HK
LK
T
oHK
T
oLK
oHK
oLK
LK KK
PP
PP
PPα
Hengstebeck’s Method
For TOP SECTIONEquilibrium Relation:yi=Ki.xi
For heavy key:
Material balance:ui=li+di
LlK
Vi
ii
uiii lK
LV
uiiii dllK
LV
iiiHK dllKLV
i
iHK l
d1KLV
And di/li=0VLKHK
Hengstebeck’s Method
For TOP SECTIONEquilibrium Relation:yi=Ki.xi
For any light non-key component:
Material balance:ui=li+di
LlK
Vi
ii
uiii lK
LV
uiiii dllK
LV
iiii dllKLV
iiiiHK
dllKK
1 1-α
dli
ii
iii ld u
Hengstebeck’s Method
For BOTTOM SECTIONEquilibrium Relation:yi=Ki.xi
For Light key:
Material balance:l’i=u’i+wi
LlK
Vi
ii
ui
ii 'KV'
L' uliii
i
w''K'V'
L'uu
And V'L'KLK
iiiLK
w''K'V'L'
uu
iiLK
w'1K'V'L'
u
Hengstebeck’s Method
For BOTTOM SECTIONEquilibrium Relation:yi=Ki.xi
For any heavy non-key component:
Material balance:l’i=u’i+wi
LlK
Vi
ii
ui
ii υ'
'KV'L'l' iii
i
w''K'V'
L'uu
iiii
LK w''K
Kuu
iLK
iii α-α
wα' u
iii wυ'l' iii
iLK w''u
aua
iiii
w''K'V'
L'uu
2) Estimate the number of ideal stages needed in the butane-pentane 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:
Equilibrium constants were calculated and found to be:
Feed (F) Tops (d) Bottoms (w)Propane, C4 5 5 0
i-Butane, i-C4 15 I5 0n- Butane, n-C4 25 24 1i-Pentane, i-C5 20 1 19
n-Pentane, n-C5 35 0 35Total, kmol 100 45 55
Component Average value of KC3 5.0iC4 2.6nC4 2.0iC5 1.0nC5 0.85
Light key will be:n-C4
Heavy key will be:i-C5
SOaC3=5/1=5aiC4=2.6/1=2.6anC4=2/1=2aiC5=1/1=1anC5=0.85/1=0.85
F d wPropane, C4 5 5 0
i-Butane, i-C4 15 I5 0n- Butane, n-C4 25 24 1i-Pentane, i-C5 20 1 19
n-Pentane, n-C5 35 0 35Total, kmol 100 45 55
Component Average value of KC3 5.0iC4 2.6nC4 2.0iC5 1.0nC5 0.85
96.0124
24
54
4
iCnC
nCD dd
dx
0.05191
1ww
wxiC5nC4
nC4w
0.560225
25ff
fxiC5nC4
nC4F
For Top section:Le=L-SliL=R*D=2.5*45=112.5 Kmoles
Le=112.5-10.625=101.875 KmolesAndVe=V-SuiV=(R+1)*D=3.5*45=157.5 Kmoles
Ve=157.5-30.625=126.875
SO Le/Ve=0.8
F d wC3 5 5 0
i-C4 15 I5 0n-C4 25 24 1i-C5 20 1 19n-C5 35 0 35Total 100 45 55
For Bottom section:L’=L+F (feed is at its boiling point)L’=112.5+100=212.5 KmolesV’=V=157.5 KmolesV’e=V-Su’i
Ve=157.5-25.87=131.63 KmolesAndL’e=L’-Sl’I
L’e=212.5-60.87=151.63 Kmoles
So L’e/V’e=1.15
F d wC4 5 5 0
i-C4 15 I5 0n-C4 25 24 1i-C5 20 1 19n-C5 35 0 35Total 100 45 55
Equilibrium curve:
NTS in top section6NTS in bottom sectionReboiler+5.5
xx
xxy
LK
LK
12
11 aa
x 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1y 0 0.18 0.33 0.46 0.57 0.66 0.75 0.82 0.88 0.94 1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
• 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:
2- Fenske equation:
3- Underwood equation: 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
3- Shortcut Methods:b) Gilliland, Fenske , Underwood Method
(Gilliland Chart to use instead of equation)
component
Feed Distillate Residue Relative Volatility
F (kmole)
xf D (kmole)
xD W (kmole)
xW
Hexane 40 0.4 40 0.534 0 0 2.7Heptane 35 0.35 34 0.453 1 0.04 2.22Octane 25 0.25 1 0.013 24 0.96 1
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.
- Calculate from Underwood Equation as follows:
Feed is saturated liquid so, (q=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)
Get = 1.1725
2 .7∗0 .42 .7−𝜃 +
2 .22∗0 .352 .22−𝜃 +
1∗0 .251−𝜃 =1−1
1−𝑞=∑ 𝛼 𝑖𝑥 𝑓𝑖
𝛼𝑖−𝜃
- Calculate Rmin from Underwood Equation as follows:
Rmin = 0.8287- Calculate Nmin from Fenske Equation:
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
2 .7∗0 .5342 .7−1.1725 +
2 .22∗0 .4532.22−1 .1725 +
1∗0 .0131−1 .1725 =𝑅𝑚𝑖𝑛+1
KOL SANA W ENTO TAYEBEN