Transcript Slide 1

Diffusion Effects in Enzymes
Immobilized in a Porous Matrix
At steady state, the intraparticle diffusion rate of
substrate equals to the reaction rate in a
spherical shell:
d[ S ]
d[ S ]
2
De
4r
 De
4r 2
 v4r 2r
r r
r
dr
dr
R
v
is the reaction rate
per unit volume of support (mg/cm3-s).
De is the effective diffusivity (cm2/s).
r
r+Δr
Diffusion Effects in Enzymes
Immobilized in a Porous Matrix
Dividing the two sides of the equation by
yields
When
4r
d[ S ] 2
d[ S ] 2
De
r
 De
r
r


r
r
dr
dr
 vr2
r
r
→0
d
d[S ] 2
(De
r )  vr2
dr
dr
Re-arrange this equation
d 2[S ] 2
d[S ]
De (
r  2r
)  vr 2
2
dr
dr
Diffusion Effects in Enzymes
Immobilized in a Porous Matrix
Dividing the two sides of the equation by r2, yields,
d 2[S ] 2 d[S ]
De (

)v
r dr
dr 2
Then
"
Vm
" [S ]
Vm
v
K m  [S ]
" [S ]
Vm
d 2[S ] 2 d[S ]
De (

)
r dr
K m  [S ]
dr 2
is the maximum reaction rate per unit volume of support
(mg/cm3-s).
De is the effective diffusivity (cm2/s).
The above equation can be written in dimensionless form
by defining the following dimensionless variables:
Km
[S ]
r
S
,r  , 
[S s ]
R
[S s ]
" S
d 2 S 2 d S R 2Vm


2 r dr
S s De S  
dr
d2S 2 dS
S
2


2 r dr
S
dr
"
Vm
=Thiele modules
 R
S s De
d2S 2 dS
S
2


2 r dr
S
dr
With boundary conditions of
S  1, at r  1
d S / d r  0, at r  0
This differential equation can be solved numerically.
Refer to H. Fogler, Elements of Chemical Reaction Engineering
1999, p746 for analytical solution for first order reaction.
At steady state, the rate of substrate consumption is equal to
the rate of substrate transfer through the external surface
of the support particle into the sphere.
d[ S ]
2
rs  Ns  4R De
dr
r R
Under diffusion limitations, the rate per unit volume is usually
expressed in terms of the effectiveness factor as follows:
"
Vm [S s ]
rs  
K m  [S s ]