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Developments in enzymatic textile treatments 51
2 αψ + β
Γ ES =ψ −
⎛ β ⎞ −( αψ β [2.45]
⎜ +α ⎟ e +α
2
+ )t
⎝ψ ⎠
where
−+ β − 4 αγ
β
2
ψ = [2.46]
2 α
If the amount of adsorbed enzyme is small compared with the amount
available in the liquid bulk, the model becomes simpler. In that case, it may
be assumed that the enzyme concentration in the bulk is constant and equal
to C E,0 so that equation [2.37] becomes:
d*
A C ES = kV C E,0 A C * E k A C * − des ES [2.47]
ads
C
t d
Substitution of equation [2.39] and simplifi cation gives:
d*
des *
ES = kV C C E,0 * ES,max −( k ads V C E,0 + k ) ES [2.48]
ads
C
dt
Equation [2.48] describes the dynamics of the adsorption process when
the enzyme concentration in the capillary liquid does not change as a result
of the adsorption process, which is the case when there is an excess
of enzymes in the solution. This differential equation can be solved by
separation of the variables. Its solution with the initial conditions t = 0,
Γ ES = 0 reads:
ads
* ES = kV C C E,0 * ES,max ⎣ 1 ⎡ ⎡ − e −(k ads C E,0 +V C k des t ) ⎤ ⎦ [2.49]
kV C C E,0 + k des
ads
For equations [2.45] and [2.49] we have performed some calculations to
show clearly the effect of the decreasing enzyme concentration in the cap-
illary liquid during adsorption of enzymes at the capillary wall. For the
example we have used values calculated earlier:
−10
• A C = 6 × 10 m 2
−16
• V C = 3 × 10 m 3
• C E,0 = 0.1 mol m −3
−8
• Γ ES,max = 6.7 × 10 mol m −2
The values for the adsorption and desorption coefficients have been
−1
−1 −1
15
estimated to be: k ads = 10 mol s and k des = 0.5 s . The results are expressed
as the fraction of the adsorption sites that are occupied by enzyme mole-
cules Γ ES /Γ ES,max as a function of time. Figure 2.13 shows the results
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