8-1$
Solutions)for)Chapter)8)–)Multiple)Reactions)
)
P8-1)(a))Example)8-1)
(i)$k1$affects$the$selectivity$and$conversion$the$most.$
(ii)$No$solution$will$be$given.$
(iii)$For$PFR$(gas$phase$with$no$pressure$drop$or$liquid$phase),$
also$at$τ$=$350,$SB/XY$(instantaneous)$$is$at$its$maximum$value$of$0.84$
See$polymath$problem$P8-1-a.pol$
$
Calculated)values)of)the)DEQ)variables
Variable initial value minimal value maximal value final value
tau 0 0 783 783
Ca 0.4 0.0166165 0.4 0.0166165
ODE)Report)(RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(tau) = –k1–k2*Ca–k3*Ca^2
8-2$
P8–1)(a))Example)8–1)Continued$
Explicit equations as entered by the user
[1] Cao = 0.4
[2] X = 1–Ca/Cao
[3] k1 = 0.0001
$
P8-1)(b))Example)8-2)
(a)$CSTR:$intense$agitation$is$needed,$good$temperature$control.$
(b)$PFR:$High$conversion$attainable,$temperature$control$is$hard$–$non–exothermic$reactions,$selectivity$
not$an$issue$
(c)$Batch:$High$conversion$required,$expensive$products$$
P8-1)(c))Example)8-3
(i)-(iii)$Individualized$Solution$
(iv)$$
We$know$
𝐶
!=𝐶
!!𝑒!!!!$and$$
$
P8-1)(d))Example)8-4)
(i)–(ii)$Individualized$Solution$
$
(iii)$According$to$the$plot$of$CB$versus$T$the$$maximum$temperature$is$T$=$310.52$0C.)
8-4$
P8–1)(d))Example)8–4)Continued$
POLYMATH)Results
See$polymath$problem$P8-1-d.pol$
Calculated)values)of)the)DEQ)variables
Variable initial value minimal value maximal value final value
t 0 0 100 100
T 300 300 400 400
ODE)Report)(RKF45)
Differential equations as entered by the user
[1] d(T)/d(t) = 1
Explicit equations as entered by the user
[1] Cao = 5
[2] tau = .5
)
P8-1)(f))Individualized$solution$
)
P8-1)(g))Individualized$solution$
)
P8-1)(h))Example)8-8)
(i)$Individualized$Solution$
$
(ii)$$
SD/U Original Problem
SD/U P8–2 h
2.58
1.01
0.666
0.208
$
Doubling$the$incoming$flow$rate$of$species$B$lowers$the$selectivity.$
$
The$selectivity$becomes$6.52$when$the$first$reaction$is$changed$to$A+2B$!$D$
)
8-5$
P8-1)(i))CDROM)Example))
$Original$Case$–$CDROM$example$ $ $ $
$
The$reaction$does$not$go$as$far$to$completion$when$the$changes$are$made.$The$exiting$concentration$of$
P8-1)(j))
For$equal$molar$feed$in$hydrogen$and$mesitylene.$
CHO$=$yHOCTO$=$(0.5)(0.032)lbmol/ft3$=0.016$lbmol/ft3$$
38.0=
opt
τ
$hr.$$At
5.0=
τ
hr$all$of$the$H2$is$reacted$and$only$the$decomposition$of$X$takes$place.$
CD–ROM
example
This question
XH
0.50
0.99
CH
0.0105
0.00016
CM
0.0027
0.0042
CX
0.00507
0.0077
τ
0.2hr
0.38hr
~
/X T
S
0.596
1.865
See$Polymath$program$P8-1-j.pol.$
POLYMATH)Results
Calculated)values)of)the)DEQ)variables
Variable initial value minimal value maximal value final value
tau 0 0 0.43 0.43
CH 0.016 1.64E–06 0.016 1.64E–06
8-6$
P8–1)(j))Continued$
ODE)Report)(RKF45)
Differential equations as entered by the user
[1] d(CH)/d(tau) = r1H+r2H
Explicit equations as entered by the user
[1] k1 = 55.2
Increasing$θH$decreases$τopt$and$
S
X/T.$
P8-1)(k)))
From$CD–ROM$example$
NLES)Solution)
Variable Value f(x) Ini Guess
CH 4.783E–05 –4.889E–11 1.0E–04
CM 0.0134353 –1.047E–11 0.013
NLES)Report)(safenewt)
Nonlinear equations
[1] f(CH) = CH–CHo+K1*(CM*CH^.5+K2*CX*CH^.5)*tau = 0
Explicit equations
[1] tau = 0.5
[2] K1 = 55.2
P8–1)(k))Continued$
The$yield$of$xylene$from$mesitylene$based$on$molar$flow$rates$exiting$the$CSTR$for$
τ
=0.5:$
producedxylenemole
C
F
Y
X
X
MX ⋅⋅
⋅⋅
=
=
=
=89.0
00232.0
$
8-8$
P8-3)(a)))
Plot$of$CA,$CD$and$CU$as$a$function$of$time$(t):$$
See$Polymath$program$P8-3-a.pol.$
POLYMATH)Results
Calculated)values)of)the)DEQ)variables
Variable initial value minimal value maximal value final value
t 0 0 15 15
Ca 1 0.0801802 1 0.0801802
X 0 0 0.9198198 0.9198198
ODE)Report)(RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = –(k1*(Ca–Cd/K1a)–k2*(Ca–Cu/K2a))
Explicit equations as entered by the user
[1] k1 = 1.0
[2] k2 = 100
[3] K1a = 10
To$maximize$CD$stop$the$reaction$after$a$long$time.$The$concentration$of$D$only$increases$with$time$
)
P8-3)(b)))
Conc.$Of$U$is$maximum$at$t$=$0.31$min.(CA$=$0.53)$
)
P8-3)(c))
Equilibrium$concentrations:$
8-9$
P8-3)(d))
See$Polymath$program$P8-3-d.pol.$
POLYMATH)Results
NLES)Solution)
Variable Value f(x) Ini Guess
Ca 0.0862762 –3.844E–14 1
t 100
NLES)Report)(safenewt)
Nonlinear equations
[1] f(Ca) = Ca0–t*(k1*(Ca–Cd/K1a)+k2*(Ca–Cu/K2a))–Ca = 0
100min
Explicit equations
[1] Ca0 = 1
[2] k1 = 1
P8-4)(a))
A→X
$ $
rX=k1
CA
1/2
$ $
k1=0.004 mol
dm3
!
“
#$
%
&
1/2
min
$
A→B
$ $
r
B=k2CA
$ $
k2=0.3min−1
$
A→Y
$ $
r
Y=k3CA
2
$ $
k3=0.25 dm3
mol. min
$
Sketch$SBX,$SBY$and$SB/XY$as$a$function$of$CA$
See$Polymath$program$P8-4-a.pol.$$
It$shows$the$table$of$SBX,$SBY$and$SB/XY$in$correspondence$with$values$of$$CA.$
(1)$ SB/X$=$
2/1
1
2
2/1
1
2
A
A
A
X
BC
k
k
Ck
Ck
r
r==
$
CAexit
0.295
0.133
0.0862
CDexit
0.2684
0.666
0.784
CUexit
0.436
0.199
0.129
P8–4)(a))Continued$
$
(2)$ SB/Y$=$
A
B
k
Ck
r
2
2
2==
$
P8–4)(b))Continued$
And$
)( 2
32
2/1
1AAAYBXA CkCkCkrrrr ++=++=−
$
=>$
3
2
*
3
*
2
2/1
*
1
*
00
*
00 4.92
)()( dm
CCv
CCv
V
AAA
AA
AA =
−
=
−
=
)
)
P8-4)(c)))
Effluent$concentrations:$
We$know,$τ$=$9.24$min$=>$
3
*
11.0
mol
C
C
C
B
B
B=⇒==
τ
$
P8–4)(f))Continued$
Now$we$use$a$mole$balance$on$species$A$
0A A
A
F F
V
r
−
=−
$
( )
0A A
A
v C C
V
r
−
=−
$
0 0
0.5 2
1 2 3
A A A A
A A A A
C C C C
r k C k C k C
τ
− −
= =
−+ +
$
A$mole$balance$on$the$other$species$gives$us:$
i i i
F vC rV= =
$
i i
C r
τ
=
$
$
$
$
Using$these$equations$we$can$make$a$Polymath$program$and$by$varying$the$temperature,$we$can$find$a$
maximum$value$for$CB$at$T$=$306$K.$At$this$temperature$the$selectivity$is$only$5.9.$$This$may$result$in$too$
See$Polymath$program$P8-4-f.pol.$
POLYMATH)Results
NLE)Solution)
Variable Value f(x) Ini Guess
Ca 0.0170239 3.663E–10 0.05
T 306
R 1.987
k1 0.0077215
Cb)versus)temp) 290$
292$
294$
296$
298$
300$
302$
304$
8-13$
P8–4)(f))Continued$
NLE)Report)(safenewt)
Nonlinear equations
[1] f(Ca) = (Cao–Ca)/(k1*Ca^.5+k2*Ca+k3*Ca^2)–10 = 0
Explicit equations
[1] T = 306
[2] R = 1.987
P8-4)(g)))
Concentration$is$proportional$to$pressure$in$a$gas–phase$system.$Therefore:$
SB/XY ~P
A
P
A+P
A
2
which$would$suggest$that$a$low$pressure$would$be$ideal.$But$as$before$the$tradeoff$is$
lower$production$of$B.$A$moderate$pressure$would$probably$be$best.$
8-14$
P8–4)(g))Continued$
$
$
$
P8-5)
US$legal$limit:$0.8$g/l$
Sweden$legal$limit:$0.5$g/l$
1 2k k
A B C⎯⎯→ ⎯⎯→
$
Where$A$is$alcohol$in$the$gastrointestinal$tract$and$B$is$alcohol$in$the$blood$stream$
1
A
A
dC k C
dt =−
$
1 2
B
A
dC k C k
dt =−
$
1
110k hr −
=
$
20.192 g
kL hr
=
$
Two$tall$martinis$=$80$g$of$ethanol$
Body$fluid$=$40$L$
0
80 2
40
A
g g
CL L
= =
$
Now$we$can$put$the$equations$into$Polymath.$
See$Polymath$program$P8-5.pol.$
8-15$
P8-5)Continued$
POLYMATH)Results
Calculated)values)of)the)DEQ)variables
ODE)Report)(RKF45)
Differential equations as entered by the user
P8-5)(a))
In$the$US$the$legal$limit$it$0.8$g/L.$
P8-5)(b)$$
In$Sweden$CB$=$0.5$g/l$,$t$=$7.8$hrs.$
P8-5)(c))In$Russia$CB$=$0.0$g/l,$t$=$10.5$hrs$
)
P8-5)(d)$$
For$this$situation$we$will$use$the$original$Polymath$code$and$change$the$initial$concentration$of$A$to$1$
8-16$
P8-5)(d))Continued$
for$the$US$t$=$6.2$hours$
$
P8-5)(e)))
The$mole$balance$on$A$changes$if$the$drinks$are$consumed$at$a$continuous$rate$for$the$first$hour.$$80$g$
of$ethanol$are$consumed$in$an$hour$so$the$mass$flow$rate$in$is$80$g/hr.$Since$volume$is$not$changing$the$
rate$of$change$in$concentration$due$to$the$incoming$ethanol$is$2$g/L/hr.$
POLYMATH)Results
Calculated)values)of)the)DEQ)variables
Variable initial value minimal value maximal value final value
t 0 0 11 11
ODE)Report)(RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = if(t<1)then(–k1*Ca+2*t)else(–k1*Ca)
P8-5)(f))
60$g$of$ethanol$immediately$!$CA$=$1.5$g/L$
$
P8-5)(g))
A$heavy$person$will$have$more$body$fluid$and$so$the$initial$concentration$of$CA$would$be$lower.$This$
)
8-17$
P8-6)(a))
$
$
See$Polymath$program$P8-6-a.pol.$
POLYMATH)Results
Calculated)values)of)the)DEQ)variables
Variable initial value minimal value maximal value final value
t 0 0 4 4
Ca 6.25 0.3111692 6.25 0.3111692
ODE)Report)(RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = –k1*Ca–k2*Ca
8-18$
P8–6)(a))Continued$
P8-6)(b))
$
$
$
P8-6)(c))
If$one$takes$initially$two$doses$of$Tarzlon,$it$is$not$recommended$to$take$another$dose$within$the$first$six$
P8-6)(d))
If$the$drug$is$taken$on$a$full$stomach$most$of$it$will$not$reach$the$wall$at$all.$The$processed$food$can$also$
drag$the$drug$to$the$intestines$and$may$limit$its$effectiveness.$This$effect$can$be$seen$in$the$adsorption$
$
)
P8-7)(a))
Reactor$selection$
))))
DBA →+
)
AD rr 1
−=
) ) –
BAA CCTKr )/8000exp(10
1−=
)
))))
UBA →+
)
AU rr 2
−=
) ) –
2/32/1
2)/1000exp(100 BAA CCTKr −=
)
k1$=$2.62$x$10–11$$&$ k2$=$3.57$ SD/U$
2/1
1035.7
B
A
C
C
×
=
$
At$T$=$1000K$
k1$=$3.35$x$10-3$$&$ k2$=36.78$$ SD/U$
2/1
102.9
B
A
C
C
×
=
$
Hence$In$order$to$maximize$SDU,$use$higher$concentrations$of$A$and$lower$concentrations$of$B.$This$can$
be$achieved$using:$
8-20$
P8-7)(a))Continued$
1)$A$semibatch$reactor$in$which$B$is$fed$slowly$into$a$large$amount$of$A$
2)$A$tubular$reactor$with$side$streams$of$B$continually$fed$into$the$reactor$
3)$A$series$of$small$CSTR’s$with$A$fed$only$to$the$first$reactor$and$small$amounts$of$B$fed$to$each$reactor.$
$
Also,$since$ED$>$EU,$so$the$specific$reaction$rate$for$D$increases$much$more$rapidly$with$temperature.$
Consequently,$the$reaction$system$should$be$operated$at$highest$possible$temperature$to$maximize$SDU.$$
Note$that$the$selectivity$is$extremely$low,$and$the$only$way$to$increase$it$is$to$keep$
1
2
6
10
B
C
−
⎛ ⎞ <
⎜ ⎟
$and$
P8-7)(b))
)))))
DBA →+
)and) –
BAA CCTKr )/1000exp(100
1−=
)
))))
UBA →+
)and$$$$$-$
BAA CCTKr )/8000exp(106
2−=
)
P8-7)(c))$
DBA →+
)) and) –
110exp( 8000 / )
A A B
r K T C C=−
)
B D U+→
)and$$$$$$–
9
210 exp( 10,000 / )
A B D
r K T C C=−
)
( )
( )
1
9
2
10exp 8000 /
10 exp 10000 /
A B
A
DU
A B D
K T C C
r
S
r K T C C
−
= = −
$
( )
( )
8
exp 8000 /
10 exp 10000 /
A
DU
D
T C
S
T C
−
=−
$