equilibrium constant K AH2 and rate constants are k 5 and k 6
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Transcript equilibrium constant K AH2 and rate constants are k 5 and k 6
A Selection of Chemical Engineering Problems
Solved using Mathematica
Housam BINOUS
National Institute of Applied Sciences and Technology
[email protected]
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
i3
i 1
x 1
1
i
i3
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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