T&K §5.1.
Fick’s Second Law
CHEN 6603
Thursday, March 8, 12 1
Fick’s 2nd Law
Assume no reaction
Mass Form Molar Form
⇥( ) (c)
= ⇤ · (n) = ⇤ · v( ) ⇤ · (j) = ⇥ · (N) = ⇥ · u(c) ⇥ · (J)
⇥t t
Assume ρt is constant Assume ct is constant
⇤(⇥) 1 (x) 1
= ⇤ · v(⇥) ⇤ · (j) = ⇥ · u(x) ⇥ · (J)
⇤t t t ct
o Fick’s Law
(j) = t [D ](r⇥) (J) = ct [D](rx)
(simplified)
⇥( ) (x)
+ ⇥ · v( ) = ⇥ · [Do ](⇥ ) + ⇥ · u(x) = ⇥ · [D](⇥x)
⇥t t
Assume [Do] is constant Assume [D] is constant
⇥( ) (x)
+ ⇥ · v( ) = [Do ](⇥2 ) + ⇥ · u(x) = [D](⇥2 x)
⇥t t
• Does this describe multicomponent effects?
• When is it reasonable to assume that [D] is constant?
Thursday, March 8, 12 2
Scaling & Fick’s Second Law
(x)
+ ⇥ · u(x) = [D](⇥2 x)
t
Non-dimensionalization: we have length scale, time scale, and D.
⇤t
t = Dimensionless time
⌧
x 2
⇤ Dimensionless space
`
x = If D*=1 then ⌧ =
` D
⇤ ⌧
D = D 2 Dimensionless diffusivity
`
• Given D, we can estimate how long it will require for a species to diffuse distance ℓ.
• Given D, we can estimate how far the diffusion boundary layer will reach in time τ.
• If we measure the time that it takes to detect a species that diffuses some distance ℓ
through a pure fluid, we can estimate the binary diffusion coefficient for that species in
that fluid at the given temperature and pressure.
Thursday, March 8, 12 3
Diffusion Equations
⇥
“Generic” simplified form: + ⇥ · v = D ⇥2
⇥t
Temperature (Fourier’s Law of Conduction) ⇥T k 2
• constant properties + ⇥ · vT = ⇥ T
⇥t cp
• No heat released by chemical reactions ⇥T k 2
For systems
• low Mach numbers (negligible viscous heating) r T
{ =
with flow
• No species diffusion ⇥t cp
• Pressure is steady.
Velocity (Newton’s Law of Viscosity) ⇥v µ 2
• constant properties + · (vv) = v
⇥t
• pressure & density are constant
Note: the form without the convective term
• no body forces is meaningless in the case of momentum.
Species (Fick’s 2nd Law)
• Constant properties (x)
+ ⇥ · u(x) = [D](⇥2 x)
• Pressure is constant t
• Body forces act equally (x)
• No reaction. = [D](r2 x)
• No thermal diffusion (Soret effect) t
Thursday, March 8, 12 4
T&K §5.4
Example: 2 Bulb Problem
(Equimolar Counterdiffusion)
Assume:
• T, p are constant
xi = xi0 z = 0,
• No reaction xi = xiL z = L,
• 1-D domain z=[0,L]
• Compositions are constant at z=0 z=L
domain boundaries z = 0, z = L.
• Ideal gas behavior.
n
Molar reference frame is most convenient, ct si P
because uz=constant for an ideal gas mixture = ⇥ · ct u + ct = V̄ 1
=
t Mi RT
in 1D. (we can also deduce that uz = 0) i=1
At “steady state,” [D]( 2 x) = 0,
d2 (x)
= 0,
Equations are decoupled dz 2
because diffusive fluxes d(x)
are constant. = (C1 ), get C1, C2 using the BCs,
dz (xL ) (x0 )
(x) = (C1 )z + (C2 ), (x) = z + (x0 )
L
⇥
⇤x
Diffusive fluxes can (J) = ct [D] ,
⇤z
be calculated using: ct
= [D](xL x0 ).
L
Thursday, March 8, 12 5
T&K §5.4
Example - Balance on “Bulbs”
Assume:
• T, p are constant
• No reaction
• 1-D domain z=[0,L] z=0 z=L
• Each bulb is well-mixed (no spatial gradients) xi = xi0 z = 0,
• Ideal gas behavior xi = xiL z = L,
• @ t=0, each bulb has a known composition.
• @ t=∞, we can determine the composition (equilibrium).
• Can we determine the composition in each bulb as a function of
time?
xi
ct = ⇥ · Ni What assumptions
Z t Z have been made?
xi
Mole balance on a bulb: ct dV = Ni · a dS
V t S ?
xi
ct V = Ni A = Ji A (because u = 0) .
t
Need to get Ji...
Thursday, March 8, 12 6
For the bulb at z=0:
Recall from our previous discussion
dx0i (Fick’s second law):
ct V0 = Ji A
dt (xL ) (x0 )
(x) = z + (x0 )
n
L
dx0i ⇤1 xL
j x0j ct
V0 = A Dij (J) = [D](xL x0 ).
dt j=1
L L
d(x0 ) A ⇥ What were the assumptions?
= [D] (xL ) (x ) 0
dt LV0
Use equilibrium balance to eliminate (xL): V0 x0i + VL xL
i = (V0 + VL )xi ,
(assume tube has “negligible” volume) xL
i = xi (1 + V0 /VL ) x0i V0 /VL
0
✓ ◆
d(x ) A V0 1 0 Must solve this (coupled)
= 1+ [D] (x ) (x ) system of ODEs for the
dt LV0 VL change in the composition
= [D] (x1 ) (x0 ) , in bulb 0 in time.
✓ ◆
A V0 Constant for a
⌘ 1+
LV0 VL given geometry.
Thursday, March 8, 12 7
Solution Strategy
d(x0 ) ⇥
= [D] (x ) (x )
0
dt
Solution Options:
• Solve this as a system of coupled ODEs.
• Make some assumptions to decouple the
system.
• note: we have already made some
assumptions to get the ODEs and the
expression for (J).
Thursday, March 8, 12 8
Heat Transfer Analogy
⌅T k 2
From before: + ⇤ · vT = ⇤ T,
⌅t cp
d2 T
Steady state, v=0: = 0,
dz 2
TL T0
Temperature solution: T = z + T0 ,
L
dT k
Heat flux: q = k = (TL T0 ).
dz L
(xL ) (x0 )
(x) = z + (x0 )
L
ct
(J) = [D](xL x0 )
L
k If we know Le, ρ, k, cp, we
Leij =
cp Dij can find Dij. More later.
Thursday, March 8, 12 9
