13-1
Solutions for Chapter 13 – Distributions of Residence
Times for Chemical Reactors
P13-1 No solution will be given.
P13-2 (a)
The area of a triangle (h=0.044, b=5) can approximate the area of the tail :0.11
P13-2 (b)
~0.11
P13-2 (c)
For a PFR/CSTR Series
E t
( )
=
0t<
“
p
e#t#
“
p
( )
/
“
s
“
s
t$
“
p
%
&
‘
‘
(
‘
‘
13-3
P13-2 (d)
X=0.75
For a PFR first order reaction:
( ) 39.14ln
1
1
ln ==
!
“
#
$
%
&
‘
=X
Da
where
!
kDa =
Solving iteratively Hilder approximate formula with an initial value
Dao (i.e. DaPFR<Dao< DaCSTR).
Da
Da
Da
Da
Da
Da
+
!
“
#
$
%
&
+
+
!
“
#
$
%
&
=
2
exp4
2
exp
75.0
where
!
kDa =
Da=2.58
The ratio of the Damköhler numbers is equal to the ratios of the sizes.
Relative sizes:
0.46
V
V
CSTR
PFR =
,
0.86
V
V
CSTR
LFR =
;
For a PFR, τ=5.15min, first order, liquid phase, irreversible reaction with k=0.1min-1.
402.01 =!=!
“
k
eX
For a CSTR, τ=5.15min, first order, liquid phase, irreversible reaction with k=0.1min–1.
402.0
1=
+
=
!
!
k
k
X
Xseg
XPFR
XCSTR
0.385
0.402
0.340
Page 851, only the RTD is necessary to calculate the conversion for a first-order reaction in any
type of reactor. Not good when the RTD has a long tail that is difficult to interpret or interpolate.
Decrease of 10
°
C in temperature
See Polymath program P13-2-f.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 2.0E+04 2.0E+04
Xbar 0 0 0.6023837 0.6023837
k 0.0025446 0.0025446 0.0025446 0.0025446
Cao 0.75 0.75 0.75 0.75
X 0 0 0.9744692 0.9744692
E2 5.0E+10 6.25E-08 5.0E+10 6.25E-08
E 0 0 0.0023564 6.25E-08
[1] d(Xbar)/d(t) = X*E
Explicit equations as entered by the user
[1] k = .00493*exp(13300/1.9872*(1/323.15-1/313.15))
[2] Cao = .75
[3] X = k*Cao*t/(1+k*Cao*t)
[4] tau = 1000
[7] E = if (t<t1) then (0) else (E2)
13-5
Decrease in reaction order from 2nd to pseudo 1st
See Polymath program P13-2-f-2.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 2.0E+04 2.0E+04
Xbar 0 0 0.9391084 0.9391084
k 0.004 0.004 0.004 0.004
Cao 0.75 0.75 0.75 0.75
X 0 0 1 1
E2 5.0E+10 6.25E-08 5.0E+10 6.25E-08
E 0 0 0.0024915 6.25E-08
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 = 0.004
[2] Cao = .75
[3] X = 1-exp(-k*t)
[4] tau = 1000
[5] t1 = tau/2
[6] E2 = tau^2/2/(t^3+.00001)
mol/dm3 then the rate of consumption of A is larger and hence resulting in a larger conversion.
13-6
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 2.0E+04 2.0E+04
Xbar 0 0 0.999375 0.999375
X 0 0 1 1
T 323.15 323.15 823.15 823.15
Cao 0.75 0.75 0.75 0.75
E 0 0 0.0022142 6.299E-08
k 0.01 0.01 19.337202 19.337202
Differential equations as entered by the user
[1] d(Xbar)/d(t) = X*E
[2] d(X)/d(t) = k*(1-X)
Explicit equations as entered by the user
[2] Cao = .75
[3] tau = 1000
[4] t1 = tau/2
[5] E2 = tau^2/2/(t^3+.00001)
[6] E = if (t<t1) then (0) else (E2)
The mean conversion Xbar, the integral, is estimated to be 99.9%. The reaction is adiabatic and
exothermic as the temperature increases to a maximum of 1373.15 K once the batch conversion
within the globules has reached 100% which occurs after only – 4 seconds. Hence, the adiabatic
increase in temperature considerably increases the rate at which conversions increases with time
and hence also the final value.
For a PFR, τ=40min, second order, liquid phase, irreversible reaction with k=0.01 dm3 /mol⋅min-1.
76.0
1=
+
=
Ao
Ao
C
Ck
X
!”
“
13-7
For a CSTR, τ=40min, second order, liquid phase, irreversible reaction with k=0.01 dm3
/mol⋅min-1.
( ) 58.0
12=!=
“
XCk
X
X
Ao
#
Maximum Mixedness Model and Segregation model are given in E13-7
Maximum Mixedness Model
See Polymath program P13-2-g.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 200 200
x 0 0 0.5938635 0.5632738
cao 8 8 8 8
k 0.01 0.01 0.01 0.01
lam 200 0 200 0
E1 0.1635984 0.0028734 0.1635984 0.028004
E2 2.25E-04 2.25E-04 0.015011 0.015011
E 2.25E-04 2.25E-04 0.028004 0.028004
EF 0.075005 0.0220689 0.075005 0.028004
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(ra/cao+E/(1-F)*x)
Explicit equations as entered by the user
[1] cao = 8
[2] k = .01
[3] lam = 200-z
[4] ca = cao*(1-x)
[5] E1 = 4.44658e-10*lam^4-1.1802e-7*lam^3+1.35358e-5*lam^2-.000865652*lam+.028004
[6] E2 = -2.64e-9*lam^3+1.3618e-6*lam^2-.00024069*lam+.015011
[7] F1 = 4.44658e-10/5*lam^5-1.1802e-7/4*lam^4+1.35358e-5/3*lam^3-
.000865652/2*lam^2+.028004*lam
[10] E = if (lam<=70) then (E1) else (E2)
[11] F = if (lam<=70) then (F1) else (F2)
[12] EF = E/(1-F)
13-8
XMM
Xseg
XPFR
XCSTR
56%
61%
76%
58%
P13-2 (h)
Liquid phase, first order, Maximum Mixedness model
Rate Law:
A1A Ckr =!
where
-1
10.08minkk == Ao
C
CA=CAo 1“X
( )
r
A
CAo
=“k11“X
( )
dX
=r
A
+E
“
( )
X
d
“
1#F
“
( )
dX
dz =k1“X
( )
“E
#
( )
1“F
#
( )
X
See Polymath program P13-2-h-1.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 200 200
x 0 0 0.7829342 0.7463946
cao 8 8 8 8
k 0.08 0.08 0.08 0.08
E1 0.1635984 0.0028731 0.1635984 0.028004
E2 2.25E-04 2.25E-04 0.015011 0.015011
E 2.25E-04 2.25E-04 0.028004 0.028004
EF 0.075005 0.0220691 0.075005 0.028004
13-9
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(ra/cao+E/(1-F)*x)
Explicit equations as entered by the user
[1] cao = 8
[2] k = 0.08
[3] lam = 200-z
[4] ca = cao*(1-x)
[5] E1 = 4.44658e-10*lam^4-1.1802e-7*lam^3+1.35358e-5*lam^2-.000865652*lam+.028004
[6] E2 = -2.64e-9*lam^3+1.3618e-6*lam^2-.00024069*lam+.015011
[7] F1 = 4.44658e-10/5*lam^5-1.1802e-7/4*lam^4+1.35358e-5/3*lam^3-
.000865652/2*lam^2+.028004*lam
[8] F2 = -(-9.30769e-8*lam^3+5.02846e-5*lam^2-.00941*lam+.618231-1)
[11] F = if (lam<=70) then (F1) else (F2)
[12] EF = E/(1-F)
At z = 200, i.e. λ = 0 (exit), conversion X = 75 %.
The decrease in reaction order from 2nd to 1st has the effect of increasing the exit conversion by
19%. Once the concentration of A drops below 1 mol/dm3 then the rate of consumption of A does
not fall as rapidly (as the 2nd order reaction) and hence resulting in a larger conversion.
Liquid phase, third order, Maximum Mixedness model
( )
X1CC AoA !=
( )32
Ao
Ao
AX1Ck’
C
r!!=
Where
1
2min08.0‘ !
== kCk Ao
( )
( )X
F
E
C
r
d
dX
Ao
A
!
!
!
“
+= 1
( ) ( )
( )X
F
E
XCk
dz
dX
Ao
!
!
“
““=1
1‘ 32
See Polymath program P13-2-h-2.pol
13-10
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 200 200
x 0 0 0.4867311 0.4614308
cao 8 8 8 8
k 0.08 0.08 0.08 0.08
lam 200 0 200 0
E2 2.25E-04 2.25E-04 0.015011 0.015011
F 0.9970002 0 0.9970002 0
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(-k*(1-x)^3+E/(1-F)*x)
Explicit equations as entered by the user
[1] cao = 8
[2] k = 0.08
[3] lam = 200-z
[4] ca = cao*(1-x)
[5] E1 = 4.44658e-10*lam^4-1.1802e-7*lam^3+1.35358e-5*lam^2-.000865652*lam+.028004
[6] E2 = -2.64e-9*lam^3+1.3618e-6*lam^2-.00024069*lam+.015011
[7] F1 = 4.44658e-10/5*lam^5-1.1802e-7/4*lam^4+1.35358e-5/3*lam^3-
.000865652/2*lam^2+.028004*lam
[9] E = if (lam<=70) then (E1) else (E2)
[10] F = if (lam<=70) then (F1) else (F2)
[11] EF = E/(1-F)
10%. Once the concentration of A drops below 1 mol/dm3 then the rate falls rapidly and CA is not
consumed so quickly, resulting in a smaller conversion.
Rate Law:
2/1
AA Ck’r =!
( )
X1CC AoA !=
( ) 2/12/1
Ao
Ao
AX1Ck’
C
r!!=!
Where
1
2/1 min08.0‘ !
!== Ao
Ckk
( )
( )X
F
E
C
r
d
dX
Ao
A
!
!
!
“
+= 1
( ) ( )
( )X
F
E
XkC
dz
dX
Ao
!
!
“
““=“
1
12/12/1
13-11
See Polymath program P13-2-h-3.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 200 200
x 0 0 0.9334778 0.9038179
cao 8 8 8 8
k 0.08 0.08 0.08 0.08
lam 200 0 200 0
E2 2.25E-04 2.25E-04 0.015011 0.015011
F 0.9970002 0 0.9970002 0
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(-k*(1-x)^(.5)+E/(1-F)*x)
Explicit equations as entered by the user
[1] cao = 8
[2] k = 0.08
[3] lam = 200-z
[4] ca = cao*(1-x)
[5] E1 = 4.44658e-10*lam^4-1.1802e-7*lam^3+1.35358e-5*lam^2-.000865652*lam+.028004
[6] E2 = -2.64e-9*lam^3+1.3618e-6*lam^2-.00024069*lam+.015011
[7] F1 = 4.44658e-10/5*lam^5-1.1802e-7/4*lam^4+1.35358e-5/3*lam^3-
[9] E = if (lam<=70) then (E1) else (E2)
[10] F = if (lam<=70) then (F1) else (F2)
[11] EF = E/(1-F)
At z = 200, i.e. λ = 0 (exit), conversion X = 90 %.
The decrease in reaction order from 2nd to ½ has the effect of increasing the exit conversion by
34%. The smaller the dependency of the rate on CA means that when CA falls below 1 mol/dm3
13-12
then the rate of consumption of A does not fall as rapidly ( as the 2nd order reaction) and hence
resulting in a larger conversion.
See Polymath program P13-2-i-1.pol
POLYMATH Results
Variable initial value minimal value maximal value final value
t 0 0 2.52 2.52
ca 1 0.0228578 1 0.0228578
cb 1 0.2840909 1 0.2840909
cc 0 0 0.3992785 0.3992785
cabar 0 0 0.1513598 0.1513306
cbbar 0 0 0.4543234 0.4539723
ccbar 0 0 0.3570959 0.3566073
cebar 0 0 0.1782569 0.1778722
T 350 350 350 350
k1 1 1 1 1
rc 1 0.0064937 1 0.0064937
k3 1 1 1 1
ra -2 -2 -0.0293515 -0.0293515
re 0 0 0.1762951 0.0742139
E -0.004 -0.0272502 0.958793 -0.0272502
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(ca)/d(t) = ra
[2] d(cb)/d(t) = rb
[3] d(cc)/d(t) = rc
[4] d(cabar)/d(t) = ca*E
[5] d(cbbar)/d(t) = cb*E
[6] d(ccbar)/d(t) = cc*E
[7] d(cd)/d(t) = rd
[10] d(cebar)/d(t) = ce*E
13-13
Explicit equations as entered by the user
[1] T = 350
[2] k1 = exp((5000/1.987)*(1/350-1/T))
[3] k2 = exp((1000/1.987)*(1/350-1/T))
[4] E1 = -2.104*t^4+4.167*t^3-1.596*t^2+0.353*t-0.004
[5] E2 = -2.104*t^4+17.037*t^3-50.247*t^2+62.964*t-27.402
[6] rc = k1*ca*cb
[7] k3 = exp((9000/1.987)*(1/350-1/T))
[9] re = k3*cb*cd
[10] E = if(t<=1.26)then(E1)else(E2)
[11] rb = -k1*ca*cb-k3*cb*cd
[12] Scd = cc/(cd+.000000001)
[13] Sde = cd/(ce+.000000000001)
If the temperature is raised, the conversion of A increases. The selectivity Sc/d increases with
temperature and Sd/e decreases with increasing temperature
Bimodal RTD
See Polymath program P13-2-i-2.pol
Variable initial value minimal value maximal value final value
z 0 0 6 6
ca 1 0.2660482 1 0.2660482
cb 1 0.5350642 1 0.5352659
cc 0 0 0.2872257 0.2745726
F 0.99 -0.0033987 0.99 -0.0033987
cbo 1 1 1 1
cao 1 1 1 1
cco 0 0 0 0
k2 1 1 1 1
k1 1 1 1 1
rc 1 0.1424065 1 0.1424065
E2 6737.4446 0.0742397 6737.4446 925.46463
E3 0.00911 0.0061156 1.84445 1.84445
re 0 0 0.1694414 0.144103
E 0.00911 0.0061156 0.6288984 0.20909
EF 0.911 0.2083818 1.8694436 0.2083818
Scd 0 0 1.0856272 1.0198908
13-14
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(ca)/d(z) = –(-ra+(ca-cao)*EF)
[2] d(cb)/d(z) = –(-rb+(cb-cbo)*EF)
[3] d(cc)/d(z) = –(-rc+(cc-cco)*EF)
[4] d(F)/d(z) = -E
[5] d(cd)/d(z) = –(-rd+(cd-cdo)*EF)
[6] d(ce)/d(z) = –(-re+(ce-ceo)*EF)
[2] cao = 1
[3] cco = 0
[4] cdo = 0
[5] ceo = 0
[6] lam = 6-z
[7] T = 350
[10] rc = k1*ca*cb
[11] k3 = exp((9000/1.987)*(1/350-1/T))
[12] E1 = 0.47219*lam^4-1.30733*lam^3+0.31723*lam^2+0.85688*lam+0.20909
[13] E2 = 3.83999*lam^6-58.16185*lam^5+366.2097*lam^4-1224.66963*lam^3+2289.84857*lam^2-
2265.62125*lam+925.46463
[17] rb = -k1*ca*cb-k3*cb*cd
[18] E = if(lam<=1.82)then(E1)else(if(lam<=2.8)then(E2)else(E3))
[19] EF = E/(1-F)
[20] rd = k2*ca-k3*cb*cd
[21] Scd = cc/(cd+.0000000001)
P13-2 (j)
Exothermic Reaction: E=45Kj/mol
See Polymath program P13-2-j-1.pol
13-15
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 200 200
x 0 0 0.9628524 0.9579239
cao 8 8 8 8
T 320 320 464.4279 463.68858
lam 200 0 200 0
F1 5.6333387 0 5.6333387 0
F2 0.9970002 0.381769 0.9970002 0.381769
k 0.01 0.01 1.924805 1.8893687
F 0.9970002 0 0.9970002 0
ra -0.64 -0.8227063 -0.1699892 -0.2140759
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(ra/cao+E/(1-F)*x)
Explicit equations as entered by the user
[1] cao = 8
[2] T = 320+150*x
[3] lam = 200-z
[4] ca = cao*(1-x)
[5] E1 = 4.44658e-10*lam^4-1.1802e-7*lam^3+1.35358e-5*lam^2-.000865652*lam+.028004
[6] E2 = -2.64e-9*lam^3+1.3618e-6*lam^2-.00024069*lam+.015011
[7] F1 = 4.44658e-10/5*lam^5-1.1802e-7/4*lam^4+1.35358e-5/3*lam^3-
.000865652/2*lam^2+.028004*lam
[9] k = .01*exp(45000/8.314*(1/320-1/T))
[10] E = if (lam<=70) then (E1) else (E2)
[11] F = if (lam<=70) then (F1) else (F2)
[13] ra = -k*ca^2
Endothermic Reaction: E=45Kj/mol
13-16
See Polymath program P13-2-j-1.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 200 200
x 0 0 0.3017158 0.2860515
cao 8 8 8 8
T 320 289.82835 320 291.39485
F1 5.6333387 0 5.6333387 0
F2 0.9970002 0.381769 0.9970002 0.381769
k 0.01 0.0017191 0.01 0.0019006
F 0.9970002 0 0.9970002 0
ra -0.64 -0.64 -0.0536459 -0.0620023
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(ra/cao+E/(1-F)*x)
Explicit equations as entered by the user
[1] cao = 8
[2] T = 320-100*x
[3] lam = 200-z
[4] ca = cao*(1-x)
[5] E1 = 4.44658e-10*lam^4-1.1802e-7*lam^3+1.35358e-5*lam^2-.000865652*lam+.028004
[6] E2 = -2.64e-9*lam^3+1.3618e-6*lam^2-.00024069*lam+.015011
[7] F1 = 4.44658e-10/5*lam^5-1.1802e-7/4*lam^4+1.35358e-5/3*lam^3-
.000865652/2*lam^2+.028004*lam
[9] k = .01*exp(45000/8.314*(1/320-1/T))
[10] E = if (lam<=70) then (E1) else (E2)
[11] F = if (lam<=70) then (F1) else (F2)
[12] EF = E/(1-F)
13-17
P13-2 (k)
Base case:
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = ra/vo
[2] d(Cb)/d(t) = rb/vo
[3] d(Cc)/d(t) = rc/vo
Explicit equations as entered by the user
[2] k1 = 1
[3] k2 = 1
[4] tau = 1.26
[5] ra = -k1*Ca
[6] rc = k2*Cb
PFR
K1/K2=1
K1/K2=2
K1/K2=0.5
CA
0.284
0.080
0.284
CB
0.357
0.406
0.203
CC
0.359
0.513
0.513
See Polymath program P13-2-k-2.pol
POLYMATH Results
NLES Solution
Variable Value f(x) Ini Guess
ca 0.4424779 4.704E-10 1
cb 0.2466912 -3.531E-10 0
cc 0.3108309 0 0
cao 1
tau 1.26
cco 0
k1 1
k2 1
ra -0.4424779
NLES Report (safenewt)
Nonlinear equations
[1] f(ca) = cao+ra*tau-ca = 0
[2] f(cb) = cbo+rb*tau-cb = 0
[3] f(cc) = cco+rc*tau-cc = 0
13-18
Explicit equations
[1] cao = 1
[2] tau = 1.26
[3] cbo = 0
[4] cco = 0
[5] k1 = 1
[8] rc = k2*cb
[9] rb = k1*ca-k2*cb
CA
0.443
0.284
0.443
CB
0.247
0.317
0.158
CC
0.311
0.399
0.399
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 2.52 2.52
ca 1 0.0804596 1 0.0804596
cb 0 0 0.3678466 0.2027582
cc 0 0 0.7167822 0.7167822
cabar 0 0 0.3050655 0.304964
cbbar 0 0 0.3350218 0.3347693
k2 1 1 1 1
cao 1 1 1 1
E1 -0.004 -27.462382 0.9523809 -27.462382
x 0 0 0.9195404 0.9195404
rb 1 -0.1353314 1 -0.1222986
E -0.004 -0.0272502 0.9568359 -0.0272502
Differential equations as entered by the user
[1] d(ca)/d(t) = ra
[2] d(cb)/d(t) = rb
[3] d(cc)/d(t) = rc
[4] d(cabar)/d(t) = ca*E
[5] d(cbbar)/d(t) = cb*E
[6] d(ccbar)/d(t) = cc*E
13-19
Explicit equations as entered by the user
[1] k1 = 1
[2] k2 = 1
[3] cao = 1
[4] E1 = -2.104*t^4+4.164*t^3-1.596*t^2+0.353*t-0.004
[5] E2 = -2.104*t^4+17.037*t^3-50.247*t^2+62.964*t-27.402
[6] ra = -k1*ca
[10] E = if(t<=1.26)then(E1)else(E2)
Asymmetric RTD
Model
CA
0.306
0.110
0.306
CB
0.335
0.390
0.195
CC
0.347
0.486
0.486
X
0.694
0.89
0.694
See Polymath program P13-2-k-4.pol
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 7 7
ca 1 9.119E-04 1 9.119E-04
cb 0 0 0.3678325 0.0063832
cc 0 0 0.9927049 0.9927049
cabar 0 0 0.3879174 0.3879174
cbbar 0 0 0.2782572 0.2782572
cao 1 1 1 1
ra -1 -1 -9.119E-04 -9.119E-04
rc 0 0 0.3675057 0.0063832
E4 0 0 0 0
E3 1.84445 0.0061369 1.84445 0.09781
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(ca)/d(t) = ra
[2] d(cb)/d(t) = rb
[3] d(cc)/d(t) = rc
[4] d(cabar)/d(t) = ca*E
[5] d(cbbar)/d(t) = cb*E
[6] d(ccbar)/d(t) = cc*E
13-20
Explicit equations as entered by the user
[1] k1 = 1
[2] k2 = 1
[3] cao = 1
[4] E1 = 0.47219*t^4-1.30733*t^3+0.31723*t^2+0.85688*t+0.20909
[5] E2 = 3.83999*t^6-58.16185*t^5+366.20970*t^4-1224.66963*t^3+2289.84857*t^2-
2265.62125*t+925.46463
[9] rb = k1*ca-k2*cb
[10] E4 = 0
[11] E3 = 0.00410*t^4-0.07593*t^3+0.52276*t^2-1.59457*t+1.84445
[12] E = if(t<=1.82)then(E1)else(if(t<=2.8)then(E2)else(if(t<6)then(E3)else(E4)))
Bimodal RTD
Segregation
Model
K1/K2=1
K1/K2=2
K1/K2=0.5
CA
0.388
0.213
0.388
CB
0.278
0.350
0.175
CC
0.327
0.430
0.430
Asymmetric RTD
Maximum
Mixedness
K1/K2=1
K1/K2=2
K1/K2=0.5
CA
0.306
0.110
0.306
CB
0.335
0.390
0.195
CC
0.347
0.486
0.486
X
0.694
0.89
0.694
P13-2 (l-r) No solution will be given at this time.
P13-3
Equivalency Maximum Mixedness and Segregation model for first order reaction:
Maximum Mixedness Model:
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