Page 66 - Advances in Textile Biotechnology

Developments in enzymatic textile treatments 45

This set of rate equations can be rewritten as:
dC E,S
=− S kC E,S [2.18]
dt
dC 1 −α
E,C = S kC E,S − C k C E,C [2.19]
dt α
Equation [2.19] can be solved easily for the initial conditions: t = 0,
C E,S = C E,C = C E,f,0 and C E,B,0 = 0. This means that the initial enzyme concen-
tration in the fabric is everywhere C E,0 and that the initial enzyme concen-
tration in the bath is zero. The solution of equation [2.19] now becomes:

C E,S = C E,f,0 e − kt S [2.20]
Substitution of this expression for C E,S in equation [2.19] gives:

dC E,C 1 −α −
= kC E,f,0 e kt S − k C E,C [2.21]
C
S
dt α

which is a standard first order differential equation of the type:
dy
kx
ab
=+ e + cy [2.22]
dx
The solution of this differential equation can be found in handbooks
(Kreyszig, 1993). By applying the initial conditions t = 0, C E,C = C E,f,0 , the
solution of the standard equation for the enzyme concentration in the con-
vective region gives:
D
⎡ k S −D k C k S (1 − ) kt ⎤
C E,C = C E,f,0 ⎢ e − kt − e − S ⎥ [2.23]
C
⎣ D ( k S − k C ) D ( k S − k C ) ⎦
If we substitute now the expressions for C E,S and C E,C , equations [2.20] and
2.23], into equation [2.15] we finally obtain the expression for the enzyme

concentration in the bath liquid:
α
ρ C E,f,0 ⎡ k S −α k S (1 − ) kt ⎤
C E,B = B ⎢ 1 −(1 −α e ) − kt − k C e − kt + e − S ⎥
S
C
f ρ LCR ⎣ k S − k C k S − k C ⎦
k
[2.24]
For t = 0, this equation results in C E,B = 0, and for t = ∞,
U C E f,,0
C E,B,∞ = B [2.25]
f U LCR

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