Page 302 - Modeling of Chemical Kinetics and Reactor Design

272 Modeling of Chemical Kinetics and Reactor Design

C = C (1 − X ) (5-28)
A
A
AO
Differentiation of A in Equation 5-27 gives

−dC A =C AO dX A (5-29)

Substituting Equations 5-26, 5-28, and 5-29 into Equation 5-24 gives

dX X ){ C )} 2
C AO A = kC (1 − A C BO − C ( 2 AO − A (5-30)
AO
dt
2
= kC − (1 X )(C − C X ) 2 (5-31)
AO A BO AO A


dX   C  2
C AO A = kC (1 − X ) C AO BO −2 X A
AO
A

dt   C AO 

dX 2
2
A = kC (1 − X )(θ − 2 X ) (5-32)
dt AO A B A
where θ = C /C . Rearranging Equation 5-32 and integrating
B BO AO
between the limits t = 0, X = 0, t = t , X = X gives
A holding time A AF=0.9
=
tt holding time
X A 09= . dX
∫ A 2 = kC 2 AO ∫ dt (5-33)
=
A
A
X A 0= ( 1− X )(θ B − 2 X ) t 0
The integral on the left side of Equation 5-33 can be expressed by
partial fraction to give
1 ≡ A + B + C
1− ( X A )( B − 2X A ) 2 1− X A θ B − 2X A ( B − 2X A ) 2 (5-34)
θ
θ
Rearranging Equation 5-34 gives

A θ − 2X ) + ( B 1− X θ − 2X )+ ( 1− X ) (5-35)
2
C
1 ≡ ( B A A )( B A A
2
Putting X = 1 into Equation 5-35 gives 1 = A(θ – 2) or
A
B