13-58
3
/998.3 dmmoleC A!=
004.0=
CD
S
3
/0065.0 dmmolCB=
368.3=
DE
S
3
/000277.0 dmmolCC=
286.4=
EF
S
3
/068.0 dmmolCD=
3
/0202.0 dmmolCE=
3
/0047.0 dmmolCF=
Segregation Maximum Mixedness CSTR PFR
00196.0=
CD
S
0009.0=
CD
S
068.0=
CD
S
004.0=
CD
S
009.3=
DE
S
004.3=
DE
S
342.3=
DE
S
368.3=
DE
S
28433=
EF
S
1343110=
EF
S
815844=
EF
S
286.4=
EF
S
EF
S
For the PFR is very much smaller than for the others, because CF is not so small at the exit
of the PFR in turn due to exit CC is not so small either. The conversion of CA in the PFR is
virtually complete at the exit of the PFR, hence greater CC.
P13-16 (d)
See Polymath program P13-16-d.pol
E(t)=IF (t<=10) THEN (0.01) ELSE ( IF(t>=20) THEN ( 0 ) ELSE ( 0.2-0.01 t))
Segregation Maximum Mixedness CSTR PFR
004.0=
CD
S
0035.0=
CD
S
068.0=
CD
S
004.0=
CD
S
109.3=
DE
S
106.3=
DE
S
342.3=
DE
S
368.3=
DE
S
24078=
EF
S
41503=
EF
S
815844=
EF
S
286.4=
EF
S
For the Segregation and Maximum Mixedness models,
EF
S
is much lower than
for the CSTR but still far greater than for the PFR. The CSTR and PFR values are unchanged as
they do not depend on E(t).
P13-17
Multiple parallel reactions, isothermal
Asymmetric RTD: E=IF (t <=1.26) THEN (E1) ELSE (E2)
E1=-2.104t4+4.167t3-1.596t2-0.353t-0.004
P13-17 (a)
Segregation model
DAEBAD
ACCkCCk
dt
dC
2
2
13!!=
2
3
2
1
2CBFBAD
BCCkCCk
dt
dC !!=
2
32
2
12CBFDAEBAD
CCCkCCkCCk
dt
dC !+=
2
32
2
12CBFDAEBAD
DCCkCCkCCk
dt
dC +!=
DAE
ECCk
dt
dC
2
=
2
3CBF
FCCk
dt
dC =
( )
tEC
dt
Cd
A
A=
( )
tEC
dt
Cd
B
B=
( )
tEC
dt
Cd
C
C=
( )
tEC
dt
Cd
D
D=
( )
tEC
dt
Cd
E
E=
( )
tEC
dt
Cd
F
F=
D
C
CD C
C
S=
E
D
DE C
C
S=
F
E
EF C
C
S=
See Polymath program P13-17-a.pol
Exit concentrations: Selectivities:
3
/819.0 dmmolC A=
272.0=
CD
S
3
/767.0 dmmolCB=
330.11=
DE
S
3
/163.0 dmmolCC=
267.0=
EF
S
3
/600.0 dmmolCD=
3
/053.0 dmmolCE=
3
/199.0 dmmolC F=
P13-17 (b)
Maximum Mixedness model
As the RTD is asymmetric we can use the same equations for E(λ) as we did for E(t), with :
E(λ)=IF (λ<=1.26) THEN (E1) ELSE (E2)
13-60
( ) ( )
( )
!
!
!
F
E
CCr
d
dC
AoAiA
A
“
“+“=#1
( ) ( ) ( )
( )
!
!
!
F
E
CCCCkCCk
d
dC
AoADAEBAD
A
“
“+“““=1
32
2
1
The same applies to the equations for the other species as in Part (a).
See Polymath program P13-17-b.pol
Exit concentrations: Selectivities:
3
/847.0 dmmolC A=
281.0=
CD
S
3
/824.0 dmmolCB=
7.10=
DE
S
3
/162.0 dmmolCC=
280.0=
EF
S
3
/576.0 dmmolCD=
3
/054.0 dmmolCE=
3
/192.0 dmmolC F=
P13-17 (c)
Ideal CSTR
tm=τ and
( )
!
!
/t
e
tE
“
=
( ) AAAo
ArCCv
dt
dC +!=
Where
( )
vvo
CCkCCkr DAEBADA
=
!!=2
2
12
and so on for the other species.
See Polymath program P13-17-c-1.pol
Exit concentrations: Selectivities:
3
/386.1 dmmolC A=
832.0=
CD
S
3
/774.1 dmmolCB=
100.71=
DE
S
3
/095.0 dmmolCC=
198.0=
EF
S
3
/114.0 dmmolCD=
3
/002.0 dmmolCE=
3
/008.0 dmmolC F=
13-61
Ideal PFR
tm=τ and RTD function
( ) ( )
!“
#=ttE
!
“
#
$
%
&
=v
r
dV
dC AA
Where
( )
vv
CCkCCkr
o
DAEBADA
=
!!=2
2
12
and so on for the other species.
See Polymath program P13-17-c-2.pol
Exit concentrations: Selectivities:
3
/310.0 dmmolC A=
162.0=
CD
S
3
/338.0 dmmolCB=
618.3=
DE
S
3
/106.0 dmmolCC=
497.0=
EF
S
3
/652.0 dmmolCD=
3
/180.0 dmmolCE=
3
/362.0 dmmolC F=
Segregation Maximum Mixedness CSTR PFR
272.0=
CD
S
281.0=
CD
S
832.0=
CD
S
162.0=
CD
S
330.11=
DE
S
7.10=
DE
S
100.71=
DE
S
618.3=
DE
S
237.0=
EF
S
280.0=
EF
S
198.0=
EF
S
497.0=
EF
S
CD
S
is significantly greater than for the others, because exit CD is lower.
Similarly
DE
S
is much greater in the CSTR than for the others, because exit CE is so low. This is
because the achievable conversion in a CSTR is not so high.
P13-18 (a)
P13-18 (b)
P13-18 (c)
P13-18 (d)
13-62
P13-19 (a)
External age distribution E(t)
By plotting C 105 as a function of time, the curve shown is obtained
C(t)
0
100
200
300
400
500
600
700
800
900
0 10 20 30 40 50 60 70
t[min]
C
To obtain the E(t) curve from the C(t) curve, we just divide C(t) by the integral
!!! !! +++=
“60
50
50
10
6
0
10
60 )()()()()( dttCdttCdttCdttCdttC
( ) ][ 55
6
0104.417310)650()720(3)785(3)831(2)812(3)622(3)0(11
8
3
)( !! “=++++++=
#dttC
( )
[ ] 55
10
6107.210610)418()523(4)650(2
3
1
)( !! “=++=
#dttC
( ) ][ 55
50
10 107.3671105)8(4)14(2)25(4)44(2)77(4)136(2)238(44185
3
1
)( !! “=++++++++=
#dttC
( )
[ ] 55
60
50 103010)1(1)5(110
2
1
)( !! “=+=
#dttC
1.0107.9981)( 5
0!“=#
$
%dttC
We now calculate:
1.0
)(
)(
)(
tC
tC
tE ==
E(t)
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64
t[min]
E(t) [1/min]
Using Excel we fit E(t) to a polynomial:
!
$
%+
%&%+%&=‘‘
&
&&&
109505.9
107492.4101355.1101675.1)(30
2
223243
1
t
ttttEtfor
0
0.2
0.4
0.6
0.8
1
1.2
0 10 20 30 40 50 60
t[min]
F(t)
P13-19 (c)
Mean residence time and variance
!“
=0)( tdttEtm
The area under the curve of a plot tE(t) as a function of t will yield tm.
t C(t) E(t) tE(t) t-tm (t-tm)2E(t)
0
0
0
0
-10
0
0.4
329
0.0329
0.01316
-9.6
3.032064
1
622
0.0622
0.0622
-9
5.0382
2
812
0.0812
0.1624
-8
5.1968
6
650
0.065
0.39
-4
1.04
8
523
0.0523
0.4184
-2
0.2092
10
418
0.0418
0.418
0
0
15
238
0.0238
0.357
5
0.595
35
25
0.0025
0.0875
25
1.5625
40
14
0.0014
0.056
30
1.26
45
8
0.0008
0.036
35
0.98
50
5
0.0005
0.025
40
0.8
min10min88.9 !==
“
m
t
We can calculate the variance by calculating the area under the curve of a plot of: (t-tm)2E(t)
( ) ( ) ( )
( ) ( )
!!
!!!
“+“
+“+“=“=#
50
2
10
2
10
6
2
6
0
2
0
2
2
)()(
)()()(
dttEttdttEtt
dttEttdttEtttdttEtt
mm
mmm
$
σ2=74min2
13-66
P13-19 (d)
Fraction of the material that spends between 2 and 4min in the reactor
[ ] 16.0)0785.0(1)0831.0(4)0812.0(1
3
1
)(
4
2=++==
!areashadeddttE
P13-19 (e)
Fraction of the material that spends longer than 6min
13-67
!! !! ++==
“60
50
10
6
50
106 )()()()( dttEdttEdttEareashadeddttE
( ) 210.0)0418.0(1)0523.0(4)065.0(1
3
2
)(
10
6=++=
!dttE
(
)367.0)0005.0(1)0008.0(4)0014.0(2
)0025.0(4)0044.0(2)0077.0(4)0136.0(2)0238.0(4)0418.0(1
3
5
)(
50
10
=+++
+++++=
!dttE
( ) 003.00001.00005.0
2
10
)(
60
50 =+==
!taildttE
581.0)(
6==
!“areashadeddttE
P13-19 (f)
Fraction of the material that spends less than 3min
[ ] 192.0)0831.0(1)0812.0(3)0622.0(3)0(1
8
3
)(
3
0=+++==
!areashadeddttE
P13-19 (g)
Normalized distributions
Normalized RTD
( ) ( )
tEE
t
!
!
=“=“
13-68
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0 1 2 3 4 5 6 7
Normalized cumulative RTD
( ) ( ) ( )
!! “” =““=“00
dttEdEF
0.2
0.4
0.6
0.8
1
1.2
0123456
P13-19 (h)
Reactor Volume
F=10 dm3/min
3
Θ
E(Θ)
Θ
F(Θ)
P13-19 (i)
Internal age distribution
( ) ( )
[ ]
tFtI !=1
1
“
13-70
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 70 70
amean 0 0 0.5778625 0.5778625
kd 0.1 0.1 0.1 0.1
E1 0 -2.436E+04 0.0838847 -2.436E+04
E2 0.092343 -21.261016 0.092343 -21.261016
E 0 0 0.0838847 0
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(amean)/d(t) = a*E
Explicit equations as entered by the user
[1] kd = 0.1
[2] E1 = -0.0011675*t^4+0.011355*t^3-0.047492*t^2+0.0995005*t
[3] E2 = -1.8950*10^(-6)*t^4+8.7202*10^(-5)*t^3-1.1739*10^(-3)*t^2-1.7979*10^(-4)*t+0.092343
[4] E4 = 0
[5] E3 = 1.2618*10^(-8)*t^4-2.4995*10^(-6)*t^3+1.8715*10^(-4)*t^2-6.3512*10^(-3)*t+0.083717
[7] a = 1/(1+kd*t)
Ideal PFR
2nd order, liquid phase, kCAo =0.1min-1, CAo=1mol/dm3
τ=10 min (from P13.19 (c))
5.0
1=
+
=
Ao
Ao
Ck
Ck
X
!
!
P13-19 (n)
LFR
Laminar Flow Reactor
2nd order, liq. phase, irreversible reaction kCAo =0.1 min-1,τ=10 min.
We apply the segregation model, using Polymath:
Ao
Ao
ktC
ktC
X
+
=1
and
!
“
#
$
<
=%min5min)2/()10(
min00.50
)( 132 tfort
tfor
tE
See Polymath program P13-19-n.pol
13-71
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 0.4504221 0.4504221
cao 1 1 1 1
k 0.1 0.1 0.1 0.1
tau 10 10 10 10
E1 5.0E+05 1.852E-06 5.0E+05 1.852E-06
t1 5 5 5 5
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 = 1.0
[2] k = .1
[3] tau = 10
[4] E1 = tau^2/2/(t^3+0.0001)
[5] t1 = tau/2
[6] E = if (t<t1) then (0) else (E1)
[7] x = k*cao*t/(1+k*cao*t)
451.0)/21ln(
2
1=
!
“
$
%
‘
(
*
+
–=Da
DaX
with Da= kCAoτ=1
P13-19 (o)
Ideal CSTR
2nd order, liquid phase, kCAo =0.1 min–1, CAo=1mol/dm3
( ) Ao
Ck
X
X
!
=
“2
1
→
382.0=X
P13-19 (p)
Segregation Model
2nd order, liquid phase, kCAo =0.1 min-1, CAo=1mol/dm3
( ) ( )
!
“
=
0
dttEtXX
Where
( ) tkC
tkC
tX
Ao
Ao
+
=1
See Polymath program P13-19-p.pol
13-72
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 70 70
Xbar 0 0 0.4224876 0.4224876
kCao 0.1 0.1 0.1 0.1
E1 0 -2.436E+04 0.0836855 -2.436E+04
E2 0.092343 -21.261016 0.092343 -21.261016
E4 0 0 0 0
E3 0.083717 1.949E-05 0.083717 0.0017977
E 0 0 0.0836855 0
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Xbar)/d(t) = E*X
Explicit equations as entered by the user
[1] kCao = 0.1
[2] E1 = -0.0011675*t^4+0.011355*t^3-0.047492*t^2+0.0995005*t
[3] E2 = -1.8950*10^(-6)*t^4+8.7202*10^(-5)*t^3-1.1739*10^(-3)*t^2-1.7979*10^(-4)*t+0.092343
[4] E4 = 0
[7] X = kCao*t/(1+kCao*t)
Maximum Mixedness Model
2nd order, liquid phase, kCAo =0.1 min-1, CAo=1mol/dm3
( )
( )X
F
E
C
r
d
dX
Ao
A
!
!
!
“
+= 1
Rate Law :
2
AA kCr =!
( )
XCC AoA !=1
( )22 1XkCr AoA !=
where k=0.1 dm3 /mol min
C
Ao
( )
zE
dz
dF !=
where z=60-λ
13-73
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 60 60
X 0 0 0.4773052 0.4047103
F 0.99 -0.010344 0.99 -0.010344
cao 1 1 1 1
lam 60 0 60 0
E2 -9.8680524 -9.8680524 0.092343 0.092343
E3 2.228E-05 1.806E-05 0.083717 0.083717
ra -0.1 -0.1 -0.0273247 -0.035437
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
[2] lam = 60-z
[3] E1 = -0.0011675*lam^4+0.011355*lam^3-0.047492*lam^2+0.0995005*lam
[4] E2 = -1.8950*10^(-6)*lam^4+8.7202*10^(-5)*lam^3-1.1739*10^(-3)*lam^2-1.7979*10^(-
4)*lam+0.092343
[5] E3 = 1.2618*10^(-8)*lam^4-2.4995*10^(-6)*lam^3+1.8715*10^(-4)*lam^2-6.3512*10^(-
3)*lam+0.083717
[7] Ca = cao*(1-X)
[8] k = 0.1
[9] ra = -k*Ca^2
[10] E = if(lam<=3)then(E1)else(if(lam<=20)then(E2)else(if(lam<60)then(E3)else(E4)))
XLFR
XMM
Xseg
XPFR
XCSTR