6-15$
P6-6$(a))Continued$
Rename$
Transport$out$the$sides$of$the$reactor:$
RA$=$kACA$=
$
–rA$=$rB$=1/2$rC$
Combine$and$solve$in$Polymath$code:$
See$Polymath$program$P6-6-a.pol.$
POLYMATH)Results$
Calculated)values)of)the)DEQ)variables$
Variable initial value minimal value maximal value final value$
v 0 0 20 20 $
Fa 100 57.210025 100 57.210025$
Fb 0 0 9.0599877 1.935926 $
Fc 0 0 61.916043 61.916043$
Rb 0 0 2.9904791 0.6396478$
Fao 100 100 100 100 $
X 0 0 0.4278998 0.4278998$
$
ODE)Report)(RKF45)$
Differential equations as entered by the user$
[1] d(Fa)/d(v) = ra – Ra$
$
Explicit equations as entered by the user$
[1] Kc = 0.01$
[2] Ft = Fa+ Fb+ Fc$
[3] Co = 1$
6-16$
P6-6$(a))Continued$
$
P6-6$(b)$
The$setup$is$the$same$as$in$part$(a)$except$there$is$no$transport$out$the$sides$of$the$reactor.$
See$Polymath$program$P6-6-b.pol.$
POLYMATH)Results$
Calculated)values)of)the)DEQ)variables$
Variable initial value minimal value maximal value final value$
v 0 0 20 20 $
Fa 100 84.652698 100 84.652698$
Fb 0 0 15.347302 15.347302$
)
ODE)Report)(RKF45)$
Differential equations as entered by the user$
[1] d(Fa)/d(v) = ra$
$
Explicit equations as entered by the user$
[1] Kc = 0.01$
6-17$
P6-6$(b))Continued$
$
$
P6-6$(c))Conversion$would$be$greater$if$C$were$diffusing$out.$
P6-6$(d))Individualized$solution$
$
)
P6-)7)(a)$
$
Assuming$catalyst$distributed$uniformly$over$the$whole$volume$
Mole$balance:$$
$ $ $
$$
P6-)7)(a)$
Rate$law:$
$
6-18$
P6-7$(a))Continued$
POLYMATH)Results$
Calculated)values)of)the)DEQ)variables$
Variable initial value minimal value maximal value final value$
W 0 0 100 100 $
Fa 2 0.7750721 2 0.7750721$
Fb 2 0.7750721 2 0.7750721$
$
ODE)Report)(RKF45)$
Differential equations as entered by the user$
[1] d(Fa)/d(W) = r$
[1] Keq = 1.44$
[2] Ft = Fa+Fb+Fc+Fd$
P6–7)(b)$
In$a$PFR$no$hydrogen$escapes$and$the$equilibrium$conversion$is$reached.$
$
P6–7)(c)$
If$feed$rate$is$doubled,$then$the$initial$values$of$Fa$and$Fb$are$doubled.$This$results$in$a$conversion$of$
.459$
$
$
)
P6-8$
Suppose$the$volumetric$flow$rate$could$be$increased$to$as$much$as$6,000$dm3/h$(9,000$mol/h)$and$the$
total$time$to$fill,$heat,$empty$and$clean$is$4.5$hours.$What$is$the$maximum$number$of$moles$of$ethylene$
glycol$(CH2OH)2$you$can$make$in$one$24$hour$period?$The$feed$rate$of$ethylene$chlorohydrin$will$be$
adjusted$so$that$the$volume$of$fluid$at$the$end$of$the$reaction$time$will$be$2500$dm3.$Now$suppose$CO2$
leaves$the$reactor$as$fast$as$it$is$formed.$
A+B→C+D+CO2
$
Mole)Balance$
$
dNA
dt =rAV
dNB
dt =F
B0+r
BV
dNC
dt =r
CV
ND=NC
$
Overall)Mass)Balance$
$Accumulation$=$In$–$Out$
P6-8)Continued$
Rate)Law)and)Relative)Rates$
$
−rA=kCACB
r
B=rA
$
6-21$
P6–8)continued$
$
P6–8(c)$
FA0$–$0.15$mol/min$=$9$mol/hr$
vo$=$Fao/cao$=$(9$mol/hr)$/$(1.5$mol/dm3)$=$6$dm3/hr$
)
P6-9$
$ $ $
V
T=
π
D2L
4=
π
1.5m
( )
22.5m
( )
4
V
T=4.42m3=4,420)dm3
$$
P6-9)continued$
$ $
NA=NB
CA0V0=CB0ΔV=CB0V
T−V0
#
$%
&
V0=
V
T
CA0
CB0
+1
=4420
0.8 +1.0 =2456
ΔV=1964
υ
0=ΔV
tR
k0=0.000052*dm3mol s =0.187*dm3mol h
$
For$1$batch$tR$=$24$–$3$=$21$
$ $
υ
0=
V
T−V0
21
$
(1)$For$one$batch$we$see$that$only$198$moles$of$C$are$made$so$one$batch$will$not$work.$
(2)$For$two$batches,$we$have$a$down$time$of$2$×$3$=$6$h$and$therefore$each$batch$has$a$reaction$time$of$
18h/2$=$9$h.$We$see$that$106$moles$of$C$are$made$in$one$batch$therefore$2$×$106$=$212$moles/day$
6-23$
P6-9)continued$
P6–9)Two)Batches$
$
P6–9)Two)Batches$
$
)
P6–10)Individualized$solution$
$
)
6-24$
P6–11)(a)$
At$equilibrium,$r$=$0$$=>$
$
CACB=
CCCD
KC
$
V$=$V0$+$vot$
$
#
&
$
See$Polymath$program$P6–11–a.pol.$
Solving$in$Polymath$
POLYMATH Report$
Nonlinear Equation$
Calculated values of NLE variables $
$
Variable $
Value $
f(x) $
Initial Guess $
1 $
x $
0 $
0 $
0.495 ( 0 < x < 0.99 ) $
Nonlinear equations $
1 $
f(x) = 2825.25*(x^2/1.08/(1–x)+x) = 0 $
6-25$
P6–11)(a))continued$
$
If$we$solve$in$Excel$
0$
0$
0.05$
148.1466374$
0.1$
311.591358$
0.15$
493.0338235$
0.2$
695.8486111$
0.25$
924.3101852$
0.3$
1183.914286$
0.35$
1481.84765$
0.4$
1827.692593$
0.45$
2234.515909$
0.5$
2720.611111$
0.55$
3312.40216$
0.6$
4049.525$
0.65$
4994.264683$
0.7$
6250.42963$
0.75$
8004.875$
0.8$
10631.31111$
0.85$
15001.7287$
0.9$
23732.1$
0.95$
49902.28611$
0.99$
259188.435$
$
6-26$
P6–11)(a))continued$
X$$$$$$$$$$$$t$(sec)$
$
P6–11)(b)$
See$Polymath$program$P6–11–b.pol.$
POLYMATH)Results$
Calculated)values)of)the)DEQ)variables$
Variable initial value minimal value maximal value final value$
t 0 0 1.5E+04 1.5E+04 $
Ca 7.72 0.2074331 7.72 0.2074331$
$
ODE)Report)(RKF45)$
Differential equations as entered by the user$
[1] d(Ca)/d(t) = ra – Ca*vo/V$
$
Explicit equations as entered by the user$
[1] Kc = 1.08$
[2] k = 0.00009$
$
6-27$
P6–11)(b))continued$
$
Polymath$solution$
P6–11)(c)$
As$ethanol$evaporates$as$fast$as$it$forms:$ CD$=$0$
POLYMATH)Results$
Calculated)values)of)the)DEQ)variables$
Variable initial value minimal value maximal value final value$
t 0 0 6000 6000 $
Ca 7.72 0.0519348 7.72 0.0519348$
$
ODE)Report)(RKF45)$
$
6-28$
P6–11)(c))continued$
$
$
P6–11)(d)$
Change$the$value$of$vo,$CAO,$etc.$in$the$Polymath$program$to$see$the$changes.$
P6–11)(e)$Individualized$solution$
P6–11)(f)$Individualized$solution$
)
$
P6–12)(a))No$solution$will$be$given$
P6–12)(b)$No$solution$will$be$given$
$
)
P6–13)
Reaction:$𝐴⇄𝐵$ $with$𝑘!=𝑘=0.4!
!$ $ 𝐾!=4$
$
Reactor$-$mass$balance$(1):$𝑚!,!” =𝐶
!!∙𝑣!+𝐶
!!⋅𝑣!=100 ⋅(12 +𝑣!)$
Reactor$-$component$A$balance$(2):$input$=$output$+$reaction;$𝑚!,!” =𝑚!,!”# +(−𝑟
!𝑉)$
6-29$
P6–13)Continued$
Combine$(3)$and$(4),$substitute$for$numerical$values$of$CA0,$k$and$KC$to$get$(5):$−𝑟
!𝑉=30𝐶
!−600$
Combine$(5)$and$(1)$with$(2)$and$evaluate$numerically$to$get:$𝑐!=!”##!!””!!
!”!!!
;$𝑐!=100 −𝑐!=!”##
!”!!!
$