Chemical Engineering Chapter 13 This The Same Expression For The Exit

subject Type Homework Help
subject Pages 14
subject Words 282
subject Authors H. Scott Fogler

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page-pf1
13-21
( )
( ) ( )
( )
( )
( )
!
!
!
!
!
!
!
!
!
!
d
F
eE
e
C
C
d
F
E
k
d
F
E
k
Ao
A"#
$
%
&
'
(
)
*
+*
$
%
&
'
(
)
*
+*
$
$
$
%
&
'
'
'
(
)
*
"
*=
"
1
1
1
changing the variables from λ to t in the RHS integral:
( )( ) ( ) ( )( )
( ) dt
tF
tFetE
Fe
C
Ckt
k
Ao
A!"#
#$
%
&
'
(
)
#
#
=#
*
**
1
1
1
P13-4 (a)
Mean Residence Time
P13-4 (b)
Variance
page-pf2
13-22
Using Polymath:
See Polymath program P13-4-b.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 1.596 1.596
sigma 0 0 0.1593161 0.1593161
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(sigma)/d(t) = (t-tau)^2*E
Explicit equations as entered by the user
[1] tau = (2/3.14)^0.5
P13-4 (c)
Conversion predicted by the Segregation model
page-pf3
13-23
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 1.596 1.596
Xbar 0 0 0.4447565 0.4447565
X 0 0 0.7210716 0.7210716
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Xbar)/d(t) = X*E
Explicit equations as entered by the user
[1] tau = (2/3.14)^0.5
P13-4 (d)
Conversion predicted by the Maximum Mixedness model
page-pf4
13-24
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 1.596 1.596
x 0 0 0.4445289 0.4445289
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(-k*(1-x)+E/(1-F)*x)
[2] d(F)/d(z) = -E
Explicit equations as entered by the user
[1] k = .8
%5.44=X
as for the Segregation Model, but we knew this because for first order
reactions Xseg = XMM
page-pf5
P13-5 (a)
The cumulative distribution function F(t) is given:
The real reactor can be modelled as two parallel PFRs:
min25)75.0min*20()1min*10(
1
0
=+== !tdFtm
P13-5 (b)
Variance
P13-5 (c)
For a PFR, second order, liquid phase, irreversible reaction with k = 0.1 dm3 /molmin-1, τ = 25
min and CAo = 1.25 mol/dm3
PFR
page-pf6
13-26
For two parallel PFRs, τ1 = 10 min and τ2 = 30 min, Fa01 = 1/4Fa0 and Fa02 = 3/4Fa0 , second
order, liquid phase, irreversible reaction with k = 0.1 dm3 /molmin-1 and CAo = 1.25 mol/dm3
P13-5 (d)
1-Conversion predicted by the Segregation Model
2-Conversion predicted by the Maximum Mixedness model
( )
( )X
F
E
C
r
d
dX
Ao
A
!
!
!
"
+= 1
See Polymath program P13-5-d.pol
page-pf7
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 40 40
x 0 0 0.7125177 0.7061611
F 0.9999 -1.081E-04 0.9999 -1.081E-04
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(ra/cao+E/(1-F)*x)
[2] d(F)/d(z) = -E
Explicit equations as entered by the user
[1] cao = 1.25
[2] k = .1
Xreal
XMM
Xseg
XPFR
XCSTR
0.731
0.706
0.731
0.758
0.572
P13-5 (e)
Adiabatic Reaction E=10000cal/mol and
XT 500325 !=
page-pf8
P13-5 (f)
Conversion Predicted by an ideal laminar flow reactor
For a LFR, second order, liquid phase, irreversible reaction with k = 0.1 dm3 /molmin-1, τ = 25
min and CAo = 1.25 mol/dm3
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 300 300
xbar 0 0 0.7077852 0.7077852
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(xbar)/d(t) = x*E
Explicit equations as entered by the user
[1] cao = 1.25
We can compare with the exact analytical formula due to Denbigh.
page-pf9
13-29
P13-6 (a)
1
2
2.02 1== t
A
min5
1=t
.
!
!
$
<
ttif
t
t
1
2
1
0
1
11
1
1
0
t
t
P13-6 (b)
Variance
page-pfa
13-30
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 10 10
sigma 0 0 4.1666667 4.1666667
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(sigma)/d(t) = (t-tau)^2*E
Explicit equations as entered by the user
[1] tau = 5
σ2=4.167min2
P13-6 (c)
For a PFR, second order, liquid phase, irreversible reaction with kCAo =0.2 min-1,τ=5 min
P13-6 (d)
1-Segregation model
See Polymath program P13-6-d-1.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 15 15
page-pfb
13-31
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Xbar)/d(t) = X*E
Explicit equations as entered by the user
[1] k1 = .2
%7.47=X
2-Maximum Mixedness Model
kCAo =k
page-pfc
13-32
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 20 20
x 0 0 0.6642538 0.4669205
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(-k*(1-x)^2+E/(1-F)*x)
[2] d(F)/d(z) = -E
Explicit equations as entered by the user
[1] k = 0.2
P13-6 (e)
Laminar Flow Reactor
For a LFR, 2nd order, liq. phase, irreversible reaction kCAo =0.2 min-1,τ=5 min.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 300 300
page-pfd
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(xbar)/d(t) = x*E
Explicit equations as entered by the user
We can compare with the exact analytical formula due to Denbigh.
P13-7
Irreversible Liquid phase, half order, Segregation model.
Mean conversion
( ) ( )
!"== 0
1.0dttEtXX
(1)
page-pfe
13-34
P13-8 (a)
The E(t) is a square pulse
Third order liquid-phase reaction: rA = kCA
3 with CAo = 2mol/dm3 and k = 0.3 dm6/mol2/min (
Isothermal Operation)
1-Conversions
Segregation Model
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 2 2
Xbar 0 0 0.5296583 0.5296583
page-pff
13-35
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Xbar)/d(t) = X*E
Explicit equations as entered by the user
[1] k = .3
2-Maximum Mixedness Model
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 2 2
x 0 0 0.5215389 0.5215389
F 0.9999 -9.999E-05 0.9999 -9.999E-05
page-pf10
13-36
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(ra/cao+E/(1-F)*x)
[2] d(F)/d(z) = -E
Explicit equations as entered by the user
[1] cao = 2
[2] k = .3
The conversion shows an inflection point in correspondence of z = 1, where start the pulse
0
2
2
=
!
d
Xd
.
P13-8 (b)
Introducing in the Segregated Model and in the MM Model :
The discrepancy is greatest at 300K
P13-8 (c)
Adiabatic Reaction
Introducing the enthalpy balance:
page-pf11
13-37
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 5 5
Xbar 0 0 0.8949502 0.8949502
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Xbar)/d(t) = X*E
Explicit equations as entered by the user
[1] ko = 0.3*exp(20000*(1/300-1/305))
[2] t1 = 1
P13-9 (a)
3rd order, k=175 dm6/(mol2 min), CBo=0.0313 dm3/min
(LFR, PFR, CSTR with τ=100s)
PFR
page-pf12
13-38
( )
!=
"
X
Bo v
V
kC
X
dX
0
2
3
1
( ) 12
1
12
2+=
!v
V
kC
XBo
CSTR
Design equation
( ) VrCCv AAAoo !=!
2
BAA CCkr =!
( )
XCC AoA !=1
( )
XCC BoB !=1
τ=V/v=1000/10=100s=1.67min
( ) ( )
!
"
=
0
dttEtXX
page-pf13
13-39
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 1.0E-05 1.0E-05 100 100
xbar 0 0 0.1827616 0.1827616
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(xbar)/d(t) = x*E
Explicit equations as entered by the user
[1] cbo = 0.0313
P13-9 (b)
(Segregation Model and Maximum Mixedness Model applying RTD of Example 13-1)
Segregation model
Vr
dt
dN
A
A!=!
page-pf14
Similarly
( ) !
!
"
#
$
$
%
&
+
'=tkC
tX
Bo
2
21
1
1
POLYMATH Results
Polynomial Regression Report
Model: C02 = a1*C01 + a2*C01^2 + a3*C01^3 + a4*C01^4
Variable Value 95% confidence
a1 0.0889237 0.0424295
General
Order of polynomial = 4
Number of observations = 13
Statistics
R^2 = 0.8653673
See Polymath program P13-9-b-1.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 14 14
xbar 0 0 0.4106313 0.4106313
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(xbar)/d(t) = E*x
[2] d(F)/d(t) = E
Explicit equations as entered by the user
[1] cbo = 0.0313

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