Microchip problem
January 30, 2018
Assuming that the transport boundary layer has a thickness δ L, the transport equation reduces to :
yU H
∂C
∂x =D∂2C
∂y2 (1)
A dimensional analysis of the equation provides a scaling law forδ(x) :
δ(x) =x1/3H2/3P e−1/3H (2) whereP eH is the Peclet number based on the channel thicknessH.
We seek solutions for the concentration field as self-similar profiles : C =C0f(ζ) where ζ =y/δ(x) is the rescaled vertical coordinate.
We have the following relations for the spatial derivatives of C :
∂2C
∂y2 = C0
δ2 d2f
dζ2 (3)
∂C
∂x =C0
df dζ
∂ζ
∂x =−C0df dζ
y δ2
dδ
dx =−C0
3 df dζ
y
δx (4)
Substituting in (1), we get :
−1 3
df dζ
y2 δx
U
H =D1 δ2
d2f
dζ2 (5)
or :
−1 3
df dζ
ζ2δ3
xH2P eH = d2f
dζ2 (6)
Sinceδ3 =xH2/P eH, we get finally an ordinary differential equation for f :
−ζ2 3
df
dζ = d2f
dζ2 (7)
or :
1
∂
∂ζ
lndf dζ
=−ζ2
3 (8)
Integrating once, we have :
df
dζ =Aexp(−ζ3/9) (9)
The boundary conditions are C = 0 at the wall (f = 0 at ζ = 0) and C → C0 when yδ, orf →1 whenζ → ∞.
The solution is : f =
Rζ
0 exp(−ξ3/9)dξ R∞
0 exp(−ξ3/9)dξ ≈0.54 Z ζ
0
exp(−ξ3/9)dξ (10)
The local mass flux at the wall is : JD(x) =−D∂C
∂y =−DC0
δ(x)f0(0)≈ −0.54DC0
δ(x) (11)
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