Chemical Reaction Engineering Calculations for PFR and CSTR

a) For PFR,
V =F AO∫
0
x
dX
−r A
= F AO
kC AO
∫
0
x (1+εX )
(1−X ) dX where ε=Y A 1 δ=1 ( 2+1−1 )=2
¿ F AO
k CAO
¿
¿( 2.5
0.44∗0.3 )¿
¿ 189.4 d m3∗5.11
¿ 967 d m3
b) For CSTR, our equation is of the form A → B+2 C
V = F AO X
−t A
Where −t A=k CA
C A=C AO
1−X
1+ εX
C A 0= PAO
RT = 10
0.082∗400 0.3
V = F AO X (1−εX )
CAO k (1−X )
k =10− 4 e
85000
8.31 ( 1
323 − 1
400 )
=0.044
Hence V = 2.5∗0.9(1+2∗0.9)
0.044∗0.3∗(1−0.9)=4772 d m3
c) dV
dX = F AO
−r A
where −r A = k C AO
( 1+ εX ) ( ( 1−X ) − 4 CAO
2 X2
( 1+εX ) 2 KC
)
Solving for this equation givesV =290 dm3

We want to obtain a plot of x and y against v. we shall have our equations as
dX
dV =−r A
F AO
and dY
dV =−∝
2 y (1+ εX ). Solving for the differential equations on polymath
gives the following output.
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 alfa 0.001 0.001 0.001 0.001
2 C 0.3 0.00767 0.3 0.00767
3 Co 0.3 0.3 0.3 0.3
4 esp 2. 2. 2. 2.
5 fo 2.5 2.5 2.5 2.5
6 k 0.044 0.044 0.044 0.044
7 r -0.0132 -0.0132 -0.0003375 -0.0003375
8 v 0 0 500. 500.
9 x 0 0 0.656431 0.656431
10 y 1. 0.1721111 1. 0.1721111
To get the values of x and y at v=500 as x=0.67 and y=0.17
d) For PFR and 1st order, KT =−ln ( 1−X A )