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4-61
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca1)/d(t) = (Cao/2 -Ca1)/tau -k*Ca1^2
P4-33 (b)
99% of the steady state concentration of A (the concentration of A leaving the third reactor) is:
(0.99)(0.611) = 0.605
This occurs at t =
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
also need a mole balance for species B.
Mole balance on reactor 1 species A:
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Mole balance on reactor 1 species B:
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:
Mole balance for reactor 3 species B:
See Polymath program P4-33-d.pol.
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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
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca1)/d(t) = (2*Cao/3 -Ca1)/tau -k*Ca1*Cb1
Explicit equations as entered by the user
[1] k = 0.025
Equilibrium conversion is 0.372.
This conversion is reached at t = 85.3 minutes.
P4-33 (e) Individualized solution
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CDP4-A
CH3I + AgClO4 CH3ClO4 + AgI
CDP4-B
a)
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Polymath solution(Ans CDP4-B-a)
b)
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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:
The reason the graph looks so different from(a) is that pure water is evaporated, but water with atrazine is
coming in.
Polymath code:
Polymath solution(Ans CDP4-B-d)
CDP4-C
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CDP4-D
Batch reaction: 2A + B 2C
k1 = 1.98 ft3/lbmol.min, k2 = 9.2 x 103 (ft3/lbmol)2/min
V = 5 gal =0.67 ft3, X = 0.65
CDP4-E
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Liquid feed is a mixture of A and B yAo = 1 – yBo = 0.999
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CDP4-F
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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)
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By Integration we get:
X
1"X=k
#
hCA0
2
FA0
(r2"Ro
2)
b) Now, with the pressure drop,
CA = CB = CA0(1-X)y
Differential equations
d(X)/d(r) = 2*3.1416*h*k*Cao^2/Fao*(1-X)^2*y^2*r
y = (1-alfa*W)^0.5
W = density*3.1416*h*(r^2- ro^2)
c) Increasing the value of k increases conversion while decreasing it decreases the conversion.
CDP4-H
