A Selection of Chemical Engineering Problems
Solved using Mathematica
Housam BINOUS
National Institute of Applied Sciences and Technology
binoushousam@yahoo.com
1- Chemical Kinetics and Catalysis
2- Applied Thermodynamics
Successive First-Order Reversible Reactions
We consider successive first-order reversible reactions :
k1 2
k 23
k n 1 , n
k 21
k 32
k n , n 1
A1  A 2  A3 ...  A n
Governing equations are :
dA 1

  k 12 A1  k 21 A 2

dt
 dA
2

 k 12 A1   k 21  k 23  A 2  k 32 A3
dt

 dA 3  k A   k  k  A  k A
23
2
32
34
3
43
4
 dt

...

dA n

 k n 1 , n A n 1  k n , n 1 A n

dt
Steady state solution
[Ai]
0.001
1. ´ 10 - 12
1. ´ 10
- 21
1. ´ 10 - 30
1. ´ 10
- 39
20
50
100
200
500
1000
i
Forward and Backward rate constants:
ki,i+1=1 and ki+1,i=0.9
0.01
Transient solution
C oncentration
0.008
0.006
0.004
A100, A200, A300, A400, A500,
A600, A700, A800 and A900
0.002
2000 4000 6000 8000 10000 12000 14000
t
1
C oncentration
0.8
0.6
A1
0.4
0.2
50
100
150
200
250
300
350
400
t
0.14
C oncentration
0.12
0.1
0.08
0.06
A1000
0.04
0.02
2000
4000
6000
8000 10000 12000 14000
t
Eley-Rideal Mechanism
Rate expressions for Reaction A + B  C
 A  S  As

 As  B  C s
C  C  S
 s
1/ equilibrium constant KA and rate constants are k1 and k2
2/ rate limiting step, rate constant is kp, equilibrium constant KAB
3/ equilibrium constant 1/KC and rate constants are k5 and k6
A
C
B
As
Cs
catalyst
catalyst
rate 
ABk p K A S tot
1  AK
A
 ABK
A
K AB
Rate expressions for Reaction A + B  C
Adsorption competition with an inert component
 A  S  As

B  S  Bs


D  S  Ds
A  B  C
s
s
 s
 C s  C  S
A
1/ equilibrium constant KA and rate constants are k1 and k2
2/ equilibrium constant KB and rate constants are k3 and k4
3/ equilibrium constant KD and rate constants are k7 and k8
4/ rate limiting step, rate constant is kp, equilibrium constant KAB
5/ equilibrium constant 1/KC and rate constants are k5 and k6
B
C
D
2
rate 
As
Bs Ds
catalyst
Cs
catalyst
ABk p K A K B S tot
1 
AK
A
 BK
B
 ABK
A
K AB K B  DK
D
2
Reaction A + H2  AH2 (for example: hydrogenation reactions)
Rate expression when H2 follows dissociative adsorption
 A  S  As

H 2  2S  2H s

 A s  2 H s  AH 2 s  2 S
 AH  AH  S
2s
2

H2
1/ equilibrium constant KA and rate constants are k1 and k2
2/ equilibrium constant KH2 and rate constants are k3 and k4
3/ rate limiting step, rate constant is kp
4/ equilibrium constant KAH2 and rate constants are k5 and k6
A
AH2
3
rate 
Hs
Hs
catalyst
A
s
AH2 s
catalyst
1  AK
AH 2 K A K H 2 k p S tot
A
 AH 2 K AH 2 

3
H 2K H2
Liquid-liquid Equilibrium of Ternary mixture
Equilibriu m
xi  i  xi  i
1
1
2
i  1, 2 ,3
2
Liquid phase activity coefficients from NRTL model :


G
x
 ji ji j C  x G
j 1
j
ij
ln  i  C
 C

 G ki x k j 1   G kj x k
k 1
 k 1
C
G ji  exp   ji
ji



 
 ij


 ij 

C
  kj G kj x k  
k 1
C
G
kj
k 1
( g ij  g jj )
RT
xk





i3
i 1
x 1
1
i

i3
i 1
xi  1
2
So far we have 5 equations and 6 unknowns,
we need one more equation.
L  L 1
1
2
L x  L x  Xi
1
1
i
2
2
i
i  1, 2
We choose values for X1 and X2 than
we solve the nonlinear system of 8
equations with 8 unknowns.
Liquid-liquid equilibrium
for Water-Benzene-Ethanol at 25 °C
Ethanol
Tie line
1
0.8
0.6
Plait point
0.4
0.2
-0.2
0.2
Water
-0.2
0.4
0.6
0.8
1
Benzene
Liquid Liquid Extraction
Liquid Liquid Equilibrium of ternary system
Isopropyl ether-water-acetic acid at 20 °C and 1 atm :
100
wt % acetic acid
isopropyl ether rich phase
80
tie line
60
water rich phase
40
20
20
40
60
80
wt % water
100
Hunter and Nash Graphical Equilibrium Stage Method
40
Mixing Point
F
30
M = F + S= E1 + RN
20
E1
10
-50
-25
S
RN
M
25
50
75
100
P = Ri-1 - Ei = F - E1 = RN - S
P
E1
E2
1
F
Operating Point
EN
2
R1
N-1
S
N
RN-1
RN
Stepping off Equilibrium Stages
wt % acetic acid
40
F
30
20
E1
RN
10
-50
-25
S
25
50
75
100
wt % water
P
5.35 equilibrium stages are needed to achive raffinate specifications
wt % Acetic Acid in Extract
McCabe and Thiele Diagram
35
Equilibrium Curve
30
25
20
15
10
Operating Line
5
10
20
30
40
wt % Acetic Acid in Raffinate
5.35 equilibrium stages are needed to achive raffinate specifications
Two Feed Extraction Column
Total Feed
FT = F1 + F2
40
F1
F2
20
EN
10
Mixing Point
M = FT + S= R1 + EN
FT
30
M
R1
Operating Points
OP1 = Ri+1 - Ei = R1 - E0
OP2 = Ek - Rk+1= EN RN+1
OP1 + OP2 = F2
OP2
-50
-25
S
25
50
75
100
OP1
F2
S=E0
E1
1
R1
EN-1
2
R2
N-1
EN
N
RN
F1=RN+1
wt % acetic acid
Stepping off Equilibrium Stages
40
F1
FT
30
F2
20
EN
R1
10
OP2
-50
-25
S
25
50
75
100
wt % water
OP1
2.89 equilibrium stages are needed to achive raffinate specifications
Residue Curve Map
vapor
liquid
y
x
dx
i
d
 x  y
i
i
i  1, 2 , 3
Obtaining the boiling temperature :
P  Psat x   Psat x 
1
1
1
2
2
2
 Psat x 
3
3
3
Liquid phase activity coefficients from Wilson model :
ln  k
 C

 1  ln   x j  kj  
 j 1

C

i 1


x i  ik

k

x 
  j ij
 j 1






 ij 
v jL
v iL
  ij   ii
exp  
RT

Obtaining equilibrium vapor phase mole fractions :
Psat x 
i
y 
i
i
P
i
i  1, 2 ,3




Residue curve map for the ternary system
acetone-methanol-chloroform at P=760 mmHg
Chloroform
1
SP
0.8
Azeotrope
SN
UN
0.6
Residue Curve
0.4
0.2
SP
SP
SN
-0.2
0.2
Methanol
-0.2
0.4
0.6
0.8
UN
1
Acetone
Simple reactive distillation
C
Chemical reaction

i
Ai  0
i 1
y
Py i  Pi  i x i
sat
Phase equilibrium
x
C

i
Ai  0
i 1
C
Reaction equilibrium
K eq 

 x  
i
i 1
i
i
Transformed compositions
xi 
Xi 
1
i
k
T
k
yi 
xk
i  1 ..C
i k
Yi 
1
xk
C

i
k
T
k
yk
i  1 ..C
i k
yk
C
Xi 
i 1
i k
Y
i
1
i 1
i k
Equation for simple distillation with reaction equilibrium
dX
d
i
 X i  Yi
i  1 ..C  1
i k
Residue curve map for the isopropyl acetate chemistry at P=1 atm
CH 3 COOH  CH 3 CH ( OH ) CH
3
 CH 3 COOCH ( CH 3 ) 2  H 2 O
Isopropyl acetate
Isopropanol
1
Reactive azeotrope
0.8
XB
0.6
0.4
0.2
0.2
0.4
0.6
water
0.8
1
Acetic Acid
XA
Need to take into account acetic acid dimerization
Residue curve map for the methyl acetate chemistry at P=1 atm
CH 3 COOH  CH 3 OH  CH 3 COOCH
Methanol
3
 H 2O
Methyl acetate
1
0.8
XB
0.6
0.4
0.2
0.2
0.4
0.6
water
0.8
1
Acetic Acid
XA
Need to take into account acetic acid dimerization
Flash Distillation
Vapor
V
yi
Feed
F
zi
P and T
Liquid
L
xi
C
Rachford and Rice :
z i  K i  1
 1   K
i 1
i
 1
0
 
V
F
Equilibrium constants
Equilibrium constants using the equations that fit the DePriester Charts :
ln K 
A
T
2

B
 C  D ln P 
T
E
P
2

F
P
Equilibrium constants using virial equation of state:
K 
Phase Equilibrium :
 v L  B  P  P sat  
exp 

P
RT


P
sat
y i  K i xi
Hydrocarbon Mixture
P=3.5 bars and T=300 K
C om ponent/Equilibrium C onstant
Propane
n-B utane
Isobutane
n-Pentane
Isopentane
Feed
z1=0.2
z2=0.3
z3=0.4
z4=0.05
z5=0.05
D ePriester
2.51561
0.784838
1.09598
0.243987
0.310074
V irial
2.48655
0.750883
1.04149
0.23883
0.312025
  0 . 5770
Vapor and Liquid Compositions
Mass Balance equations give vapor and liquid compositions :
Feed
F=1
z1=0.2
z2=0.3
z3=0.4
z4=0.05
z5=0.05
Vapor
V=0.5770
y1=0.2683
y2=0.2688
y3=0.4153
y4=0.0216
y5=0.0257
P=3.5 bars
T=300 K
Liquid
L=0.4330
x1=0.1066
x2=0.3425
x3=0.3790
x4=0.0886
x5=0.0830
McCabe and Thiele Method for Distillation of Binary Ideal Mixture
distillate
xd=0.9
Binary ideal mixture with
constant relative volatility =
feed
Zf=0.5
y
bottom
xb=0.05
 x
1  1    x

Pinch Point and Minimum Reflux Ratio
1
Feed line :
0.8
y
y
0.6
qx
q 1

zf
q 1
q  0 . 85
0.4
Rectifying operating line :
0.2
0.2
0.4
0.6
x
0.8
1
y
Rx
R 1

xd
R 1
McCabe and Thiele Diagram
1
0.8
y
0.6
R=1.5 Rmin
0.4
0.2
0.2
0.4
0.6
0.8
1
x
9 equilibrium stages are needed to achieve the separation
Murphree Liquid Stage Efficiency
1
0.8
y
EML=0.5
0.6
0.4
E ML 
0.2
0.2
0.4
0.6
0.8
x j  x j 1
x j  x j 1
eq
1
x
19 equilibrium stages are needed to achieve the separation
Multicomponent Distillation
distillate
xd1=0.95
xd2=0.049
xd3=0.001
Ternary ideal mixture of Pentane,
Hexane and Heptane with constant
relative volatilities = 6.35, 2.47 and 1
feed
Z1=0.3
Z2=0.3
Z3=0.4
yi 
 i xi
C

j 1
bottom
xb1=0.05
j
xj
Liquid Compositions
1
Rectifying operating line :
Stripping Section
0.8
y i , n 1 
Rectifying Section
0.6
Rx i , n
R 1

xd i
R 1
Stripping operating line :
0.4
B
y i ,n 
F
0.2
( S  1) x i , n 1

S
D
0.2
0.4
0.6
0.8
1
R=2.5 and S=1.35
Pentane mole fraction
8.3 equilibrium stages are needed to achieve the separation
xb i
S
Enthalpy-Composition Diagram
for hexane-octane system at 760 mmHg
17500
H(y)
15000
Conjugate line
tie line
12500
10000
7500
5000
h(x)
2500
0.2
0.4
0.6
0.8
Hexane mole fraction
1
Ponchon and Savarit method
distillate
xd1=0.95
P1V
P1
 1 . 548
VL
20000
15000
V
feed
q=0.41
Z1=0.5
10000
D
F
B
5000
L
bottom
xb1=0.05
0.2
P2
0.4
0.6
0.8
1
Hexane mole fraction
7 stages are needed to achieve the separation
Conclusion
Mathematica’s
graphical
advantage
algebraic,
capabilities
to
solve
numerical
can
be
several
put
and
into
chemical
engineering and chemistry problems including
equilibrium-staged separations with McCabe-
Thiele,
Methods.
Hunter-Nash
and
Ponchon-savarit
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