Chemical Reaction Engineering Calculations for PFR and CSTR

This article discusses the calculations involved in chemical reaction engineering for PFR and CSTR reactors. It covers topics such as conversion, volume, and rate of reaction for different types of reactions. The equations and solutions are provided in detail.

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 )

Hence 0.001∗T =−ln ( 1−0.9 ) =23025.85 min which is equivalent to
23025.85
24∗7 =137.057 days
e) For the reaction A → B+2 C, we have −r A =k f CA −k B CB CC
2=0
KC= kf
k b
=0.025
With an initial concentration C AO=0.305 mol /l we have
X3∗4∗0.3052=0.205 ( 1+2 X ) 2 (1−X ) . solving this nonlinear equation gives X =0.512.
For 90% of the equilibrium conversion , X =0.9∗0.512=0.4608
V = F AO X
−r A
where −r A =k f CAO
1− X
1+2 X − kf
KC
C AO
X
1+2 X (CAO )2 X2
( 1+ 2 X )2
−r A = 0.443∗0.305
1+2∗0.4608 [ ( 1−0.4608 ) −
1
0.025∗4∗0.3052∗0.46083
( 1+ 2∗0.4608 ) 2 ]=1.08828∗10−2
For CSTR,V = F AO X (1+ ϵX )
k C AO ( ( 1− X )− 4 C AO
2 X2
( 1+ ϵX )2 KC
)
= 2.5∗0.47∗(1+2 ( 0.4608 ))
0.044∗0.3 ¿ ¿
For PFR, dX
dV =−r A
F AO
and −r A = k C AO
( 1+ ϵX ) ( ( 1− X )− 4 CAO
2 X2
( 1+ϵX )2 KC
). Solving using polymath to
obtain our volume as 290dm3 as indicated by the output below
Calculated values of DEQ variables
Variabl
e
Initial
value
Minimal
value
Maximal
value
Final
value
1 cao 0.3 0.3 0.3 0.3
2 e 2. 2. 2. 2.
3 fao 2.5 2.5 2.5 2.5
4 k 0.044 0.044 0.044 0.044
5 kc 0.025 0.025 0.025 0.025
6 ra -0.0132 -0.0132 -0.0009033 -
0.0009033

7 v 0 0 290.2388 290.2388
8 x 0 0 0.47 0.47
f) For the reaction A → B+2 C, we have −r A =k t CA − kt CB CC
2
KC
=0
Isothermal, no pressure drop C A= C AO (1−X )
1+ ϵX , CB= C AO X
1+ ϵX , CC =2 C AO X
1+ ϵX
where k c=0.08∧K c=0.025 , X =0.47 , ϵ=2, C AO=0.3
−r A = k C AO
( 1+ ϵX ) ( ( 1− X )− 4 CAO
2 X2
( 1+ϵX )2 KC
). Solving the differential equations using polymath
gives:
Calculated values of DEQ variables
Variabl
e
Initial
value
Minimal
value
Maximal
value
Final
value
1 cao 0.3 0.3 0.3 0.3
2 e 2. 2. 2. 2.
3 fao 2.5 2.5 2.5 2.5
4 k 0.044 0.044 0.044 0.044
5 kc 0.025 0.025 0.025 0.025
6 ra -0.0132 -0.0132 -0.0009033 -
0.0009033
7 v 0 0 290.2388 290.2388
8 x 0 0 0.47 0.47
Obtaining the volume as 290.2388
References
[1] H. S. Fogler, Elements of Chemical Reactions Engineering, 3rd ed., New Delhi: Prentice-Hall of
India Dum Ma8Rd, 2004.
[2] E. B. Noman, Chemical Reactor Design, Optimization, and Scaleup, 2nd ed., WILEY, 1987.