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13-1$
Solutions)for)Chapter)13)–)Unsteady)State)Non-isothermal)
Reactor)Design)
)
P13-1)(a))Example)13-1)
(i)$The$reaction$runs$away$at$T0$=$282.2$K$
$
(ii)$$
(1)$Runaway$occurs$at$Ta1$=$292$K$
(2)$As$Ta1$increases,$the$maxima$of$the$Qr$versus$t,$and$Qg$versus$T$curves$occur$at$shorter$times.$This$is$
$
(iv)$$
If$the$heat$of$mixing$had$been$neglected,$the$shape$of$the$graphs$would$have$been$as$follows:$
)
(v)$$
The$new$T0$of$20$˚F$(497$˚R)$gives$a$new$δHRn$and$T.$With$T=497+89.8X$the$polymath$program$of$
13-2$
P13-1)(a)$Continued$
(vi)$
Calculated values of DEQ variables$$
Variable
Initial value
Minimal value
Maximal value
Final value
1
t
0
0
9000.
9000.
2
X
0
0
1.
1.
3
T
480.
480.
894.193
894.193
4
k
8.307E-06
8.307E-06
56.65781
56.65781
5
Ta
896.4
896.4
896.4
896.4
6
Ua
0.22
0.22
0.22
0.22
7
Qr
91.608
0.4855311
91.608
0.4855311
8
V
1.2
1.2
1.2
1.2
9
Nao
1.
1.
1.
1.
10
NiCpi
403.
403.
403.
403.
11
Ca
0.8333333
-5.468E-07
0.8333333
-1.392E-09
12
ra
-6.923E-06
-0.0038571
8.639E-06
7.887E-08
13
Qg
-0.3016323
-168.055
0.3763917
0.0034364
Differential equations$$
1
d(X)/d(t) = k * (1 - X)
2
d(T)/d(t) = (Qr - (-36309) * (-ra) * V) / NiCpi
$
Explicit equations$$
1
k = 0.000273 * exp(16306 * ((1 / 535) - (1 / T)))
2
Ta = 896.4
3
Ua = .22
4
Qr = Ua * (Ta - T)
5
V = 1.2
6
Nao = 1
7
NiCpi = 403
8
Ca = (Nao/V) * (1 - X)
9
ra = -k * Ca
10
Qg = (-36309) * (-ra) * V
13-3$
P13-1)(a)$Continued$
$
P13-1)(b))Example)13-2)
(i)$No$explosion$would$occur$for$NA0$<$8.9$kmol.$Similarly,$no$explosion$would$have$occurred$for$NB0$<$
32.2$kmol$
$
13-4$
P13-1)(b)$Continued$
(v)$
$
P13-1)(c))Example)13-3)
(i)$1650$lbmol/hr$
$
(ii)$No,$it$is$not$possible.$
13-5$
P13-1)(c)$Continued$
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
Ca
0.1
0.0074672
0.1
0.0309593
2
Ca0
0.1812152
0.1812152
0.1812152
0.1812152
3
Cb
3.45
2.114934
3.45
2.114934
4
Cb0
2.26519
2.26519
2.26519
2.26519
5
Cc
0
0
0.1502559
0.1502559
6
Cm
0
0
0.226519
0.226519
7
Cm0
0.226519
0.226519
0.226519
0.226519
8
dh
-3.6E+04
-3.6E+04
-3.6E+04
-3.6E+04
9
Fa0
80.
80.
80.
80.
10
Fb0
1000.
1000.
1000.
1000.
11
Fm0
100.
100.
100.
100.
12
k
41.71035
31.59807
144.4831
32.06998
13
mc
1000.
1000.
1000.
1000.
14
Na
6.680919
0.4988795
6.680919
2.068369
15
Nb
230.4917
141.2971
230.4917
141.2971
16
Nc
0
0
10.03847
10.03847
17
NCp
4382.683
3372.614
4382.683
3372.614
18
Nm
0
0
15.13355
15.13355
19
Qg
1.003E+07
2.339E+06
1.003E+07
2.388E+06
20
Qr
2.586E+06
2.377E+06
3.578E+06
2.388E+06
21
Qr1
1.706E+06
1.564E+06
2.383E+06
1.571E+06
22
Qr2
8.798E+05
8.133E+05
1.195E+06
8.168E+05
23
ra
-4.171035
-4.171035
-0.9725584
-0.9928658
24
rb
-4.171035
-4.171035
-0.9725584
-0.9928658
25
rc
4.171035
0.9725584
4.171035
0.9928658
26
T
150.
143.7291
179.7336
144.0607
27
t
0
0
4.
4.
28
T0
75.
75.
75.
75.
29
Ta1
67.
67.
67.
67.
30
Ta2
115.8777
112.1848
133.3874
112.3801
31
tau
0.1513355
0.1513355
0.1513355
0.1513355
32
ThetaCp
284.375
284.375
284.375
284.375
33
UA
1.6E+04
1.6E+04
1.6E+04
1.6E+04
34
V
66.80919
66.80919
66.80919
66.80919
35
v0
441.464
441.464
441.464
441.464
13-6$
P13-1)(c)$Continued$
Differential equations
1
d(Ca)/d(t) = 1/tau*(Ca0-Ca)+ra
2
d(Cb)/d(t) = 1/tau*(Cb0-Cb)+rb
3
d(Cc)/d(t) = 1/tau*(0-Cc)+rc
4
d(Cm)/d(t) = 1/tau*(Cm0-Cm)
5
d(T)/d(t) = (Qg-Qr)/NCp
Explicit equations
1
Fa0 = 80
2
T0 = 75
3
V = (1/7.484)*500
4
UA = 16000
5
dh = -36000
6
Ta1 = 67
7
k = 16.96e12*exp(-32400/1.987/(T+460))
8
Fb0 = 1000
9
Fm0 = 100
10
mc = 1000
11
ra = -k*Ca
12
rb = -k*Ca
13
rc = k*Ca
14
Nm = Cm*V
15
Na = Ca*V
16
Nb = Cb*V
17
Nc = Cc*V
18
ThetaCp = 35+Fb0/Fa0*18+Fm0/Fa0*19.5
19
v0 = Fa0/0.923+Fb0/3.45+Fm0/1.54
20
Ta2 = T-(T-Ta1)*exp(-UA/(18*mc))
21
Ca0 = Fa0/v0
22
Cb0 = Fb0/v0
23
Cm0 = Fm0/v0
24
Qr2 = mc*18*(Ta2-Ta1)
25
tau = V/v0
26
NCp = Na*35+Nb*18+Nc*46+Nm*19.5
27
Qr1 = Fa0*ThetaCp*(T-T0)
28
Qr = Qr1+Qr2
29
Qg = ra*V*dh
$
13-7$
P13-1)(c)$Continued$
(v)$See$polymath$program$p13-1cv.pol$
$
$
$
(vi)$
$
13-8$
P13-1)(c)$Continued$
$
$
$
$
$
(vii)$No$solution$will$be$provided$
13-9$
P13-1)(c)$Continued$
(ix)$Time$to$reach$steady$state$increases$with$increase$in$tank$volume.$When$the$tank$volume$is$at$its$
minimum,$practical$stability$limit$is$not$exceeded$for$any$of$the$three$sets$of$initial$conditions.$However,$
as$we$increase$the$volume,$it$is$observed$that$for$Ti$=$340$K$and$CAi$=$1.4$M,$the$practical$stability$is$
P13-1)(d))Example)13-4$
(i)$The$maximum$in$CC$increases$with$an$increase$in$CB0.$
$
P13-1)(e))Example)13-5$
(i)$One$of$the$possible$combinations$is$CA0$=$8$mol/dm3$and$𝑣!=$240$dm3/h.$
$
13-10$
P13-1)(e)$Continued$
$
We$can$see$from$the$graph$that$NA$and$NB$become$constant$at$large$times.$
(vi)$See$polymath$program$P13-1evi.pol$
$
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
Ca
1.
0.0786053
2.036072
0.0786053
2
Cao
4.
4.
4.
4.
3
Cb
0
0
0.8803026
0.247476
4
Cc
0
0
5.018042
5.018042
5
k1a
0.2664781
0.2664781
27.26262
3.288066
6
k2b
0.08
0.08
2.421438
0.5095541
7
Na
100.
29.21648
413.9672
290.8397
8
Nb
0
0
915.6613
915.6613
9
ra
-0.2664781
-21.60838
-0.2584595
-0.2584595
10
rb
0.133239
-0.7139195
9.023013
0.0031273
11
rc
0
0
6.394794
0.3783073
12
t
0
0
15.
15.
13
T
290.
290.
403.1775
342.148
14
V
100.
100.
3700.
3700.
15
vo
240.
240.
240.
240.
Differential equations
1
d(Ca)/d(t) = ra+(Cao-Ca)*vo/V
2
d(Cb)/d(t) = rb-Cb*vo/V
3
d(Cc)/d(t) = rc-Cc*vo/V
4
d(T)/d(t) = (35000*(298-T)-Cao*vo*30*(T-305)+((-6500)*(-k1a*Ca)+(8000)*(-
k2b*Cb))*V)/((Ca*30+Cb*60+Cc*20)*V+100*35)
13-11$
P13-1)(e)$Continued$
Explicit equations
1
Cao = 4
2
vo = 240
3
k1a = 1.25*exp((9500/1.987)*(1/320-1/T))
4
k2b = 0.08*exp((7000/1.987)*(1/290-1/T))
5
ra = -k1a*Ca
6
V = 100+vo*t
7
rc = 3*k2b*Cb
8
rb = k1a*Ca/2-k2b*Cb
9
Na = Ca*V
10
Nb = Cb*V
$
We$find$that$after$running$the$code$for$a$final$time$of$10$hours,$we$get$NA$=$290.8397$mol$and$NB$=$
915.6613$mol.$$
𝐶
!!𝑣!
𝑘!!
=4∗
240
3.2881
=291.962 𝑚𝑜𝑙$
$
P13-1)(f))Example)13-6)
(i)$The$pressure$shoots$up$when$the$volumetric$flow$rate$is$3976$dm3.$
(v)$Individualized$solution$
(vi)$Individualized$solution$
(vii)$The$transition$to$runaway$occurs$over$a$very$small$range$of$Ua.$If$U$a$value$is$changed$such$that$Ua$is$
13-12$
P13-1)(f)$Continued$
But$the$cooling$starts$only$for$T>455K$
Thus$Ua$=$2.77$×$106$J/hr/K,$T>$455K$
$$$$$$$$$$$$$$$=$0$for$T<$455K$
$
Calculated values of DEQ variables$$
Variable
Initial value
Minimal value
Maximal value
Final value
1
A1A
4.0E+14
4.0E+14
4.0E+14
4.0E+14
2
A2S
1.0E+84
1.0E+84
1.0E+84
1.0E+84
3
CA
4.3
0.0457129
4.3
0.0457129
4
CB
5.1
0.8457129
5.1
0.8457129
5
CS
3.
2.999997
3.
2.999997
6
Cv1
3360.
3360.
3360.
3360.
7
Cv2
5.36E+04
5.36E+04
5.36E+04
5.36E+04
8
DHRx1A
-4.54E+04
-4.54E+04
-4.54E+04
-4.54E+04
9
DHRx2S
-3.2E+05
-3.2E+05
-3.2E+05
-3.2E+05
10
E1A
1.28E+05
1.28E+05
1.28E+05
1.28E+05
11
E2S
8.0E+05
8.0E+05
8.0E+05
8.0E+05
12
FD
2467.445
61.27745
8874.034
61.27745
13
Fvent
2467.445
61.27745
8874.034
61.27745
14
k1A
0.0562573
0.0562573
1.82787
0.7925113
15
k2S
8.428E-16
8.428E-16
2.367E-06
1.276E-08
16
P
4.4
4.4
4.4
4.4
17
r1A
-1.233723
-4.437017
-0.0306385
-0.0306385
18
r2S
-2.529E-15
-7.102E-06
-2.529E-15
-3.829E-08
19
SumNCp
1.26E+07
1.26E+07
1.26E+07
1.26E+07
20
SW1
1.
1.
1.
1.
21
Sw2
0
0
0
0
22
T
422.
422.
466.4882
454.9731
23
t
0
0
4.
4.
24
UA
0
0
2.77E+06
0
25
V0
4000.
4000.
4000.
4000.
26
VH
5000.
5000.
5000.
5000.
$
Differential equations
1$$
d(CA)/d(t)$=$SW1*r1A$$
$$$
mol/dm3/hr$$
2$$
d(CB)/d(t)$=$SW1*r1A$$
$$$
change in concentration of cyclomethylpentadiene$$
3$$
d(CS)/d(t)$=$SW1*r2S$$
$$$
change in concentration of diglyme$$
4$$
d(P)/d(t)$=$SW1*((FD-Fvent)*0.082*T/VH)$$
5$$
d(T)/d(t)$=$SW1*((V0*(r1A*DHRx1A+r2S*DHRx2S)-SW1*UA*(T-373.15))/SumNCp)$+$SW1*Sw2*4/60$$
13-13$
P13-1)(f)$Continued$
Explicit equations
1$$
V0$=$4000$$
$$$
dm3$$
2$$
VH$=$5000$$
$$$
dm3$$
3$$
DHRx1A$=$-45400$$
$$$
J/mol Na$$
4$$
DHRx2S$=$-3.2E5$$
$$$
J/mol of Diglyme$$
5$$
SumNCp$=$1.26E7$$
$$$
J/K$$
6$$
A1A$=$4E14$$
$$$
per hour$$
7$$
E1A$=$128000$$
$$$
J/kmol/K$$
8$$
k1A$=$A1A*exp(-E1A/(8.31*T))$$
$$$
rate constant reaction 1$$
9$$
A2S$=$1E84$$
$$$
per hour$$
10$$
E2S$=$800000$$
$$$
J/kmol/K$$
11$$
k2S$=$A2S*exp(-E2S/(8.31*T))$$
$$$
rate constant reaction 2$$
12$$
SW1$=$if$(T>600$or$P>45)$then$(0)$else$(1)$$
13$$
r1A$=$-k1A*CA*CB$$
$$$
mol/dm3/hour (first order in sodium and cyclomethylpentadiene)$$
14$$
r2S$=$-k2S*CS$$
$$$
mol/dm3/hour (first order in diglyme)$$
15$$
FD$=$(-0.5*r1A-3*r2S)*V0$$
16$$
Cv2$=$53600$$
17$$
Cv1$=$3360$$
18$$
Fvent$=$if$(FD<11400)$then$(FD)$else(if$(P<28.2)$then$((P-1)*Cv1)$else$($(P-1)*(Cv1$+Cv2)))$$
19$$
UA$=$if$T>$455$then$(2.77e6)$else$(0)$$
$$$
no cooling$$
20$$
Sw2$=$if$((T>300)$and$(T<422))$then$(1)$else$(0)$$
$
$
13-14$
P13-1)(f)$Continued$
$ $
$
(ix)$Solving$the$equations$with$Ua$=$0$for$the$entirety$of$the$process,$we$get$that$the$temperature$
reaches$at$0.94$hours$while$the$system$reaches$runaway$at$3.649$hours.$This$implies$the$maximum$time$
$
$
(x)$Individualized$solution$
)
P13-1)(g)$Example)RE13-1)
Using$the$code$from$Example$PRS$13.5,$we$can$determine$the$value$of$kc$for$which$the$reactor$will$fall$to$
the$lower$steady-state$and$when$it$becomes$unstable.$The$following$two$graphs$show$those$points$
13-15$
P13-1)(g)$Continued$
$
P13-1)(h))Example)RE13-2$
Using$the$code$from$Example$PRS$13.5,$we$can$change$the$value$of$kc$and$τ1$and$find$values$that$
$
$
P13-1)(i))No$solution$will$be$given$
$
P13-1)(j))Individualized$solution$
$
)
13-16$
P13-2)
After$the$feed$was$shut$off,$$$
)
$
)
See$Polymath$program$P13-2.pol$
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
Cpa
0.38
0.38
0.38
0.38
2
dH
-336.
-336.
-336.
-336.
3
k
0.0081711
0.0081711
4.285E+17
4.285E+17
4
T
980.
980.
3.302E+20
3.302E+20
5
t
0
0
4.
4.
Differential equations
1
d(T)/d(t) = -dH * k / Cpa
Explicit equations
1
dH = -336
2
Cpa = .38
3
k = (0.307/60)* exp(44498 * (1 / 970 - 1 / T))
13-17$
P13-2$Continued$
As$can$be$seen$from$the$above$plots,$temperature$shoots$at$time$t)=)3.16)mins)
Since$NA$doesn’t$come$in$the$temperature$equation,$there$will$not$be$any$effect$of$feed$present$in$the$
$
)
13-18$
P13-3)(a))
$
$
$
$
$
$
13-19$
P13-3)(a)$Continued$
See$Polymath$program$P13-3-a.pol$
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
Cbo
10.
10.
10.
10.
2
Fbo
10.
0
10.
0
3
k
0.0089352
0.0089352
10.08511
10.08511
4
Na
500.
0.1396707
500.
0.1396707
5
Nao
500.
500.
500.
500.
6
Nb
0
0
42.43357
0.1397745
7
Nc
0
0
499.8603
499.8603
8
Qg
0
0
1.24E+05
11.81314
9
Qr
0
0
0
0
10
ra
0
-0.3460808
0
-1.969E-05
11
T
298.
298.
510.4441
510.4441
12
t
0
0
120.
120.
13
V
50.
50.
100.
100.
14
vo
1.
0
1.
0
15
X
0
0
0.9997207
0.9997207
Differential equations
1
d(Na)/d(t) = ra*V
2
d(Nb)/d(t) = ra*V+Fbo
3
d(Nc)/d(t) = -ra*V
4
d(T)/d(t) = ((6000*(-ra*V))-(Fbo*15*(T-323)))/(15*Na+15*Nb+30*Nc)
5
d(X)/d(t) = -ra*V/Nao
Explicit equations
1
Fbo = if(t<50)then(10)else(0)
2
Nao = 500
3
Cbo = 10
4
k = .01*exp((10000/1.987)*(1/300-1/T))
5
vo = Fbo/Cbo
6
V = if(t<50)then(50+(vo*t))else(100)
7
ra = -k*Na*Nb/(V^2)
8
Qg = -6000*ra*V
9
Qr = 0
