11-15#
P11-6)(b))
Mole#balance:#
dX
dV =−rA
FA0
#
Rate#law:####
−rA=kCA
#
Stoichiometry:###
CA=CA01
1−X
1+
ε
X
T0
T
#
ε
=yA0
δ
#
δ
=1+1−1=1
#
yA0=FA0
F
T0
=FA0
FA0+Fi0
=1
1+
θ
i
#
ε
=1
1+
θ
i
#
#
Enter#these#equations#into#Polymath#
See#Polymath#program#P11-6-b.pol.#
POLYMATH)Results
Calculated)values)of)the)DEQ)variables
Variable initial value minimal value maximal value final value
V 0 0 500 500
X 0 0 0.417064 0.417064
)
11-16#
P11–6)(b))Continued#
Explicit equations as entered by the user
[1] Cao = 2/(.082*1100)
[2] Cio = Cao
[3] theta = 100
#
#
#
#
P11-6)(c))
There#is#a#maximum#at#θ#=#8.#This#is#because#when#θ#is#small,#adding#inerts#keeps#the#temperature#low#
P11-6)(d))
The#only#change#to#the#Polymath#code#from#part#(b)#is#that#the#heat#of#reaction#changes#sign.#The#new#
11-17#
P11–6)(d))Continued#
See#Polymath#program#P11-6-d.pol.#
#
#
The#maximum#conversion#occurs#at#low#values#of#theta#(θ#<#8)#because#the#reaction#is#now#exothermic.#
P11-6)(e))
We#need#to#alter#the#equations#from#part#(c)#such#that#
2
A A
r kC−=
#and#CA0#=#1#
A#plot#of#conversion#versus#theta#shows#a#maximum#at#about#θ#=#5.#
#
See#Polymath#program#P11-6-e.pol.#
#
P11-6)(f))
We#need#to#alter#the#equations#from#part#(c)#such#that#
−rA=k CA−CBCC
KC
“
#
$
$
%
&
‘
‘
#
We#already#know#that#
CA=CA0
1−X
1+
ε
X
T0
T
.#Now#we#need#expressions#for#CB#and#CC.#From#stoichiometry#
P11–7)(a))
Adiabatic#
Mole#Balance#(1)#
dX
dW =−ʹ
rAFA0
#
# # #
W=
ρ
bV
#
# # #
dX
dV =−ʹ
rA
ρ
B
FA0
=−
rA
FA0
#
Rate#Law#(2)#
rA=−k CA−CB
KC
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
#
# # (3)#
k=k1exp E
R
1
T
1
−1
T
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
#
# # (4)#
KC=KC2exp ΔHRx
R
1
T
2
−1
T
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
#
Stoichiometry#(5)#
CA=CA01−X
( )
y T0T
( )
#
# # (6)#
CB=CA0Xy T0T
( )
#
# # #
dy
dW =
α
y
F
T
F
T0
T
T0
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟=−
α
2y
T
T0
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
W=
ρ
V
dy
dV =−
αρ
b
2y
T
T0
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
#
Parameters#(7)#–#(15)#
FA0,#k1,#E,#R,#T
1,#KC2,#ΔHRx ,#T
2,#CA0
,#
T0
,#a,#rb#
11-20#
P11–7)(a)#continued#
#Adiabatic#
Adiabatic#
11-21#
P11-7)(c))
FA0
dX
dW =−rA‘
−rA‘=k(1−
α
W)0.5CA0[(1−X)−X
KC
]
(1−
α
W)0.5 dW
∫=FA0
1
1−(1+1
KC
)X
∫dX
2
3
α
[1−(1−
α
W)1.5]=−FA0
(1+1
KC
)
ln[1 −(1+1
KC
)X]
ln[1 −(1+1
KC
)X]=−
2(1+1
KC
)
3
α
FA0
[1−(1−
α
W)1.5]
1−(1+1
KC
)X=exp(−
2(1+1
KC
)
3
α
FA0
[1−(1−
α
W)1.5])
)
X=
1−exp(−
2(1+1
KC
)
3
α
FA0
[1−(1−
α
W)1.5])
1+1
KC
)
Qr=FA0X(−ΔHR(T0)) =
FA0(−ΔHR(T0))(1 −exp(−
2(1+1
KC
)
3
α
FA0
[1−(1−
α
W)1.5]))
1+1
KC
Qr=FA0X(−ΔHR(T0)) =
FA0(−ΔHR(T0))(1 −exp(−
2(1+1
KC
)
3
α
FA0
[1−(1−
αρ
bV)1.5]))
1+1
KC
)
)
Qr#vs#V#can#be#plotted#according#to#the#above#equation.#
11-22#
P11-7)(c))Continued#
#
#
)
P11–8)(a))
#(1)#
dX
dW =−“
rA
FA0
#
#(2)#
−rA=k CACB−CC
2
KC
“
#
$
$
%
&
‘
‘
#
#(3)#
k=k1exp E
R
1
T
−1
T
“
$
$
%
‘
‘
(
*
*
+
–
–
#
#(13)# #
CA=CA01−X
( )
pT0
T
#
#(14)# # #
ΘB=2
#
#(15)# #
CB=CA02−X
( )
T0
Tp
#
0
20
40
60
80
100
120
0 10 20 30 40
Qr*(kcal/min)
V*(dm3)
Qr*vs.*V
P11-8)(a))Continued#
#(16)# #
CC=2CA0XT0
Tp
#
)
P11–8)(b))
−rA=k CA01−X
( )
pT0
T
“
#
$
$
%
&
‘
‘CA02−X
( )
pT0
T
“
#
$
$
%
&
‘
‘−
2CA0XpT0
T
“
#
$
$
%
&
‘
‘
2
KC
(
)
*
*
*
*
*
*
+
,
–
–
–
–
–
–
#
−rA=kCA0
21−X
( )
2−X
( )
−4X2
KC
“
#
$
$
%
&
‘
‘pT0
T
(
)
*
*
+
,
–
–
2
#
P11–8)(c))
Equilibrium#Conversion# #
KC=4CA0
2Xe
2
CA0
21−Xe
( )
2−Xe
( )
#
# # #
KC
4=Xe
2
1−Xe
( )
2−Xe
( )
2
KC
4−3KC
4
Xe+KCXe
2
4=Xe
2
KC
4−1
“
#
$
$
%
&
‘
‘Xe
2−3KC
4
Xe+2
KC
4=0
#
Equilibrium#Conversion#(19)#
Xe=
3KC
4−3KC
4
“
#
$
$
%
&
‘
‘
2
−2KC
KC
4−1
“
#
$
$
%
&
‘
‘
2
KC
4−1
“
#
$
$
%
&
‘
‘
#
# #
Section)1.01 Case)1)Adiabatic)
XeT
WW
X
Xe
P11–8)(d))
Energy#Balance#(16)# #
T=T0+
−ΔHRx
( )
X
∑ΘiCP
i
#
Energy#Balance#(17)# # T0#=#325#
# # #
ΔHRx =ΔHR
+ΔCPT−TR
( )
#
P11–8)(e))
# # #
T=T0+−ΔHRx
( )
X
∑ΘiCP
=400 +20,000X
100 =400 +200X
P11–8)(f))Continued#
Gas#Phase#Adiabatic#
#
#
P11–8)(g)))
For#isothermal#solution,#set#dT/dW#to#be#zero#in#the#polymath#code#provided#in#P11–8(f)#and#add#an#
P11-9)
( )
2
2
1
1
C D e
C
A B e
C
e
C
C C X
K
C C X
K
X
K
= =
−
⇒=+
#
( )
( )
0
30000
300 300 600
25 25
A B
R
P P
H X
T T X X
C C
−
Δ
=−=−= +
+ +
#
See#Polymath#program#P11-9.pol.#
11-26#
P11-9)Continued#
Calculated values of NLE variables #
Variable
Value
f(x)
Initial Guess
1
Xe
0.9997644
3.518E–11
0.5 ( 0 < Xe < 1. )
#
Variable
Value
1
T
300.
2
Kc
1.8E+07
#
Nonlinear equations##
1
f(Xe) = Xe – (1 – Xe) * Kc ^ 0.5 = 0
#
Explicit equations##
1
T = 300
2
Kc = 500000 * exp(–30000 / 1.987 * (1 / 323 – 1 / T))
T#
X#
300#
1#
320#
0.999#
340#
0.995#
360#
0.984#
380#
0.935#
400#
0.887#
420#
0.76#
440#
0.585#
460#
0.4#
480#
0.26#
500#
0.1529#
520#
0.091#
540#
0.057#
560#
0.035#
Xe
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
300 350 400 450 500 550
Temp (K)
Xe (Eqm Conversion)
Xe
P11–10)(a))
For#first#reactor,#
1
1
1
or
1 1
e C
C e
e C
X K
K X
X K
= =
−+
#
For#second#reactor#
( ) ( )
( )( )
01
2
0
02 0 02
0
.2 .12
1.7
0
i
A
B
A
A P i A R
pA B pB
R
F
F
F C T T F X H
C C T T
XH
θ
θ
θ
= =
− Σ − +−Δ =
+−
=−Δ
#
Slope#is#now#negative#
11-28#
P11–10)(a))continued#
)
P11–10)(b))
The#same#setup#and#equations#can#be#used#as#in#part#(a).#The#entering#temperature#for#reactor#1#is#now#
450#K#and#the#outlet#is#520#K.#When#the#two#streams#are#joined#prior#to#entering#reactor#2#the#
temperature#is#(520+450)/2#=#485#K#
Say#that#the#outlet#temperature#for#reactor#2#is#510#K.#Then#the#entering#temperature#for#reactor#3#
)