4-61
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca1)/d(t) = (Cao/2 -Ca1)/tau -k*Ca1^2
[2] d(Ca2)/d(t) = (Ca1 – Ca2)/tau -k*Ca2^2
Explicit equations as entered by the user
[1] k = 0.025
[2] Cao = 2
[3] tau = 10
P4-33 (b)
99% of the steady state concentration of A (the concentration of A leaving the third reactor) is:
P4-33 (c)
The plot was generated from the Polymath program given above.
P4-33 (d)
We must reexamine the mole balance used in parts a-c. The flow rates have changed and so the mole
balance on species A will change slightly. Because species B is added to two different reactors we will
Mole balance on reactor 1 species A:
1
0 0 1 1
A
A A A A
dN
C v C v r V
dt
! ! =
with
0 0
2
3
A
v v=
and
0
200
15
v=
01
0 1 1
2
3
AA
A A
CdN
v C v r V
dt
! ! =
Liquid phase reaction so V and v are constant.
01 1
1
2
AA A
A
CC dC
r
dt
! !
“ ” =
4-62
Mole balance on reactor 1 species B:
1
0 0 1 1
B
B B B B
dN
C v C v r V
dt
! ! =
and
0 0
1
3
B
v v=
Stoichiometry has not changed so that –rAi = -rBi and it is a liquid phase reaction with V and v constant.
Mole balance on reactor 2 species A:
We are adding more of the feed of species B into this reactor such that v2 = v0 + vB0 = 20
Mole balance on reactor 2 species B:
2
1 0 0 0 2 2
B
B B B B B
dN
C v C v C v r V
dt
+! ! =
0 0
1 2 2
2
1 2
B B
B B B
A
C v
C C dC
r
V dt
! !
+“ ” =
Mole balance for reactor 3 species A:
Mole balance for reactor 3 species B:
3 3
2
3
2 2
B B
B
A
C dC
Cr
dt
! !
“ ” =
Rate law:
Ai Ai Bi
r kC C!=
4-63
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 100 100
Ca1 0 0 1.1484734 1.1484734
Ca2 0 0 0.7281523 0.7281523
Ca3 0 0 0.6278144 0.6278144
k 0.025 0.025 0.025 0.025
Cao 2 2 2 2
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca1)/d(t) = (2*Cao/3 -Ca1)/tau -k*Ca1*Cb1
[2] d(Ca2)/d(t) = Ca1/tau – Ca2/tau2 -k*Ca2*Cb2
[3] d(Ca3)/d(t) = (Ca2 – Ca3)/tau2 -k*Ca3*Cb3
Explicit equations as entered by the user
[1] k = 0.025
[2] Cao = 2
[3] tau = 200/15
[4] X = 1 – 2*Ca3/Cao
Equilibrium conversion is 0.372.
This conversion is reached at t = 85.3 minutes.
P4-33 (e) Individualized solution
4-64
CDP4-A
CH3I + AgClO4 CH3ClO4 + AgI
CDP4-B
a)
4-65
Polymath solution(Ans CDP4-B-a)
4-66
Polymath solution(Ans CDP4-B-b)
c)
Polymath solution(Ans CDP4-B-c)
d) This part is almost same as part(b) with minor changes:
V = 15000000 – 10000t
The reason the graph looks so different from(a) is that pure water is evaporated, but water with atrazine is
coming in.
Polymath code:
d(ca)/dt=cao*vo/v)-
(ca(kv+vout)/v)
ca(0)=4.5 #
vo = 80000 #
CDP4-C
4-67
4-68
CDP4-D
Batch reaction: 2A + B 2C
Mole Balance:
dt =“r
AV
Rate law: -rA = k1CACB + k2CACB
2
CA0
dX
dt =CA0
2(1“X)k1(0.9 “0.5X)+k2CA0(1“X)(0.9 “0.5X)2
[ ]
Integrating between X=0 to X=0.65 for t = 0 to t = t gives
CDP4-E
4-69
Because of the low concentration of B in the feed, such properties as SG, the specific gravity,
MW, the molecular weight, and the solubility of H2 are essentially that of component A. assume
4-70
CDP4-F
4-71
CDP4-G
Develop a design equation
Mole balance: FA(r) – FA(r + Δr) + rA(2πrΔrh) = 0
=>
FA(r) FA(r +“r)
“r=r
A(2
#
rh)
=>
dFA
dr =r
A(2
“
rh)
Rate law:
“r
A=k1CACB
Combine:
dX
dr =2
“
rhkCA0
2(1#X)2
FA0
=>
dX
(1“X)2
0
X
#=(2
$
kCA0
2)rdr
FA0
R0
r
#
4-72
By Integration we get:
X
1“X=k
#
hCA0
2
FA0
(r2“Ro
2)
1+k
FA0
(r2#Ro
b) Now, with the pressure drop,
CA = CB = CA0(1-X)y
Hence,
“rA=kCA0
2(1“X)2y2
=>
d(X)/d(r) = 2*3.1416*h*k*Cao^2/Fao*(1-X)^2*y^2*r
y = (1-alfa*W)^0.5
Cao = 0.1
initial value : 0.1
c) Increasing the value of k increases conversion while decreasing it decreases the conversion.
Increasing FA0 will decrease the conversion and decreasing it will increase the conversion.
Increasing CA0 causes a dramatic increase of conversion. Similarly, decreasing CA0 results in a
large decrease in conversion. Increasing the height will only slightly increase the conversion and
4-73
CDP4-H