Transcript Lecture 14

“Chemical Engineering Equilibrium
Separations”
Lectures 14
17 Oct 2012
1
Overview
•AspenPlus:
o Shortcut methods: DSTWU
o Rigorous method: RADFRAC
• Efficiencies
• Introduction to multicomponent distillation (Chapter 9)
2
Multicomponent Distillation (Introduction)
• In binary distillation we could specify xD and xB ….
• Now: define “Key Components”
Decreasing
Relative
Volatility
1
2
3
4
Keys
Light
Heavy
Most of LK obtained in distillate product
Most of HK obtained in bottoms product
Today, most multicomponent systems are solved rigorous simulation.
But need to do shortcut methods to get good starting point (FUG-Kirkbride).
3
Multicomponent Shortcut Methods
Minimum number of stages: Fenske Equation
N min
 xLK   xHK
 
ln 
 xHK  Dist  xLK

ln  LK  HK
N min
 
 
 Bot 
 FRLK
ln 
 1  FRLK

  FRHK
 
 Dist  1  FRHK
ln 
 
 
 Bot 
   Dist  Bot or   Dist  Feed Bot 
1
3
4
Multicomponent Shortcut Methods
Minimum reflux ratio: Underwood Equation
Case A:
 DxLK , D 
 DxHK , D
NKs don’t

   LK  HK 

distribute
 Fz

 Fz
Lmin 

F
LK , F

 LK  HK

1
HK , F




Rmin
Lmin

D
Case B & C: NKs distribute, or there is a “sandwich” NK: see Wankat;
numerical iterative procedures can be involved.
R = factor * Rmin
Approximate number of equilibrium stages (N): Gilliland correlation
5
Gilliland Correlation
61 Data points over ranges:
1.
2.
3.
4.
5.
6.
No. components: 2 to 11
q : 0.28 to 1.42
P : vacuum to 42.4 bar
 : 1.11 to 4.05
Rmin : 0.53 to 9.09
Nmin : 3.4 to 60.3
Molokanov Eqn:
 1  54.4 X  X  1 
Y  1  exp
 0.5 
 11 117.2 X  X 
Seader & Henley, 2006
6
Multicomponent Shortcut Methods
Optimum feed stage location (NF): Kirkbride Equation
N R  z HK
 
N S  z LK

  xLK , B  B 


 
z
 D
 Feed  HK , D 

2
0.206
DSTWU (AspenPlus)
• Uses Winn, Underwood, and Gilliland methods to find Nmin, Rmin, & N.
• Specify LK and HK recoveries in the distillate product stream
• If input -1.2 for reflux ratio; it finds N at R = 1.2 * Rmin.
• N given by DSTWU is number of equilibrium stages (includes partial
condensers and/or partial reboilers)
7
In-Class Problem
benzene (17 mol%)
toluene (66 mol%)
m-xylene (17 mol%)
D
S HO RT Y
DSTWU
FEE D
F = 100 kmol/s
sat’d liquid
1 atm
B
1.0135 bar
Component
Benzene
Toluene
m-Xylene
Tbp oC
80.1
110.7
139.1
Pi star [bar]
80.1 oC 123 oC
1.0135 3.217
0.387
1.42
0.151
0.645
8
In-Class Problem
9
In-Class Problem
10
In-Class Problem
11
In-Class Problem
By hand calculations first.
Then use to verify AspenPlus results…
12
In-Class Problem
13
In-Class Problem
14
Homework Problem
benzene (17 mol%)
toluene (66 mol%)
m-xylene (17 mol%)
Xbz = 99 mol%
D
S HO RT Y
DSTWU
FEE D
F = 100 kmol/s
sat’d liquid
1 atm
B
Xbz = 0.1 mol%
1.0135 bar
Component
Benzene
Toluene
m-Xylene
Tbp oC
80.1
110.7
139.1
Pi star [bar]
80.1 oC 123 oC
1.0135 3.217
0.387
1.42
0.151
0.645
15
Overview
•AspenPlus:
o Shortcut methods: DSTWU
o Rigorous method: RADFRAC
• Efficiencies
• Introduction to multicomponent distillation (Chapter 9)
16