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5-21$
P5-14)(c))continued$
$
So,$we$see$the$maximum$rate$in$case$with$pressure$drop$is$at$catalyst$weight$equal$to$around$600$Kg.$
$$
To$achieve$70%$conversion,$catalyst$weight$required$is$932.3$kg$.$
$
In$case$of$(a),$915.5$kg$of$catalyst$is$required$to$achieve$70%$conversion.$
5-22$
P5-14)(d))Individualized$solution$
P5-14)(e))Individualized$solution$
)
$
P5-15)$
Gaseous$reactant$in$a$tubular$reactor:$A$→$B$
$
$
$ $ $
$ $ $
$
At$T2$=$260°F$=$720°R,$with$k1$=$0.0015$min-1$at$T1$=$80°F$=$540°R,$
$
5-23$
P5-16)(a)$
)$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$A$$→$$B/2$
$
$
Combining$
$
(for$the$integration,$refer$to$Appendix$A)$
from$the$Ideal$Gas$assumption.$
$
$
$
5-24$
P5-16)(b))Individualized$solution.$
)
$
P5-17)(a))$
Given:$The$metal$catalyzed$isomerization$$$liquid$phase$reaction$
with$Keq$=$5.8
Case$1:$an$identical$plug$flow$reactor$connected$in$series$with$the$original$reactor.$
$
Since$yA$=$1.0,$∪B$=$0.$For$a$liquid$phase$reaction$ $$and$ $$
$
For$the$first$reactor,$
$
$
Take$advantage$of$the$fact$that$two$PFR’s$in$series$is$the$same$as$one$PFR$with$the$volume$of$the$two$
combined.$$
VF$=$V1$+$V2$=$2V1$and$at$VF,$$$X$=$X2$
$
5-25$
P5-17)(a))Continued$
$
$
P5-17)(b))$
Case$2:$Products$from$1st$reactor$are$separated$and$pure$A$is$fed$to$the$second$reactor,$
$
The$analysis$for$the$first$reactor$is$the$same$as$for$case$1.$
$
By$performing$a$material$balance$on$the$separator,$FA0,2$=$FA0(1-X1)$
Since$pure$A$enters$both$the$first$and$second$reactor$CA0,2$=$CA0,$CB0,2$=$0,$∪B$=$0$$
$$$ $for$the$second$reactor.$
5-26$
P5-17)(b))continued$
$
Overall$conversion$for$this$scheme:$
$
)
P5-17)(c))Individualized$Solution)
)
$
P5-18$
Given:$Meta-$to$ortho-$and$para-$isomerization$of$xylene.$
$
Check$to$see$what$type$of$reactor$is$being$used.$
Case$1:$
$
5-27$
P5-18)Continued$
Case$2:$
$
Assume$plug$flow$reactor$conditions:$
$$$or$
CM0,$k,$and$V$should$be$the$same$for$Case$1$and$Case$2.$
Therefore,$
$
The$reactor$appears$to$be$plug$flow$since$(kV)Case$1$=$(kV)Case$2$
As$a$check,$assume$the$reactor$is$a$CSTR.$
$
)))or$$$ $
Again$kV$should$be$the$same$for$both$Case$1$and$Case$2.$
$
$
kV$is$not$the$same$for$Case$1$and$Case$2$using$the$CSTR$assumption,$therefore$the$reactor$must$be$
modeled$as$a$plug$flow$reactor.$
$
5-28$
P5-18)Continued$
For$the$new$plant,$with$v0$=$5500$gal$/$hr,$XF$=$0.46,$the$required$catalyst$volume$is:$
$
)
$
P5-19$$
A→$B$in$a$tubular$reactor$
$
Tube$dimensions:$L$=$40$ft,$D$=$0.75$in.$
nt$=$50$
$
$or$$ $
Assume$Arrhenius$equation$applies$to$the$rate$constant.$
At$T1$=$600°R,$k1$=$0.00152$ $
5-29$
P5-19)continued$
$
$
From$above$we$have$
$
$
Dividing$both$sides$by$T$gives:$
$
$
Evaluating$and$simplifying$gives:$
$
)
$
P5-20)
$
5-30$
P5-20)Continued$
$
$
$
5-31$
P5-20)(b)$
$
P5-20)(c)$
$
$
P5-21$
Production$of$phosgene$in$a$micro$reactor$
$$$$$$$$$$$CO$$+$$Cl2$$ $$COCl2$$$$(Gas$phase$reaction)$
The$equations$that$need$to$be$solved$are$as$follows:$
d(X)/d(W)$=$-rA/FA0$
P5-21)Continued$
FB0$=$FA0;$
Fb$=$FB0-FA0*X;$
5-32$
P5-21)Continued$
POLYMATH)Results$
Calculated)values)of)the)DEQ)variables$
Variable initial value minimal value maximal value final value$
W 0 0 3.5E-06 3.5E-06 $
$
ODE)Report)(RKF45)$
Differential equations as entered by the user$
[1] d(X)/d(W) = -rA/FA0$
Explicit equations as entered by the user$
[1] e = -.5 [2] FA0 = 2e-5$
P5-21)(a))$
$$ $
P5-21)(b)$
The$outlet$conversion$of$the$reactor$is$0.784$
The$yield$is$then$MW*FA*X$=$99$g/mol$*$2$e-5$mol/s$*$0.784$=$.00155$g/s$=$48.95$g/$year.$
5-33$
P5-21)(c)$
Assuming$laminar$flow,$α$~$Dp
-2,$therefore$
$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $
$$$$ $
$
P5-21)(d)$
A$lower$conversion$is$reached$due$to$equilibrium.$$Also,$the$reverse$reaction$begins$to$overtake$the$
P5-21)(e))Individualized$solution$
$
P5-21)(f))Individualized$solution$
$
P5-21)(g))Individualized$solution$
)
$
P5-22)No$solution$necessary$
)
P5-23)(a)$
$$$
P5-23)(b)$
Elementary$with$respect$to$partial$pressure,$so$the$rate$law$is:$
$$$$$
P5-23)(c)$
Feed$is$pure$A,$so$ .$$
$ $
P5-23)(c))Continued$
$Therefore$ $$
5-35$
P5-23)(g)$Continued$
Stoichiometry:)$
$
Feed$is$pure$A,$so$ $$&$A$is$the$limiting$reactant;$
$
There$is$no$change$in$P$or$T,$so$ $;$$
Therefore:$
$
Combine:$
$$
Separate)and)integrate:)$
$$
Let’s$look$at$the$left$hand$side$first:$
$$ $$
Notice$that$the$equation$is$now$in$quadratic$form,$where$$
$$$$$$ $$$$$$ $$
The$solution$to$the$integral$is,$from$Appendix$A$(equation$A-12):$
$$
Where$p$and$q$are$the$roots$of$the$quadratic$equation:$
p,$q$=$
$
P5-23)(g)$Continued$
So$
$$$$$ $$
P5-23)(h)$
We$want$the$catalyst$weight$at$X$=$0.9$XeXe$=$0.684$
Using$equation$(*),$we$have:$
$$$
P5-23)(i)$
For$a$fluidized$bed$reactor$(CSTR),$when$X$=$0.$9$XeXe$=$0.684:$
$
$
P5-23)(j)$
Since$ $$
It$can$be$easily$seen$that$
$$$$$$$ $
Pressure$fall$below$1$atm:$
$$$$$$$ $
5-37$
P5-23)(k)$
$
)
P5-24$
$ $ $
$ $ $
$ $ $
1.)Mole)Balance$–$Use$Differential$Form$
$ $
dX
dW =−r
A
F
$
2.)Rate)Law$–$Psuedo$Zero$Order$in$B$
$ $
−r
A=kCACB
0=kCA
$
3.)Stoichiometry$–)Gas:$
v=v01+εX
( )
T
T
0
"
#
$
$
%
&
'
'
P
0
P
"
#
$
$
%
&
'
'
$Isothermal,$therefore$T$=$T0$
$
CA=F
A
v=
F
A0 1−X
( )
v01+εX
( )
P
0
P=CA0
1−X
( )
1+εX
( )
p**,**p =P
0
P
ε=yA0δ=1
2
$
&'
)1+1−1−1
( )
=0
$
P5-24)Continued$
$Integrating$
$
p=1− αW
( )
1 2
$
4.)Combine$
$ $
−rA=kCA=kCA0 1−X
( )
p=kCA0 1−X
( )
1− αW
( )
1 2
dX
1 2
$
5-39$
P5-24)Continued$
W$
Y$
1/y$
X$
P$
0$
1$
1$
0$
10$atm$
1$
1$
1$
0.47$
$
2.5$
0.99$
1.01$
0.11$
$
5$
0.975$
1.02$
0.21$
$
15$
0.92$
1.08$
0.5$
$
25$
0.87$
1.16$
0.67$
$
50$
0.71$
1.41$
0.87$
$
75$
0.5$
2.0$
0.94$
$
90$
0.32$
3.16$
0.955$
$
99$
0.1$
10$
0.96$
1$atm$
X$=$0.9$
$ $
W$=$57g$
for$5%$W$=$1.05g$
for$last$5%$(85$to$90%)$W$=$57$–$45$=$12$g$
Ratio$=$ $=$11.4$
5-40$
P5-24)Continued$
$
$
p
p$
p$
