VI. Electrokinetics
Lecture 30: Linear Electrokinetic Phenomena
Notes by MIT Student (and MZB)
1
Linear Electrokinetic Resonse of a Nanochannel
y
Fixed surface charge
--------------------Apply
∆P, ∆V
x
Area
A
z
Observe
Q,I
--------------------
We start with the system in equilibrium
Q=0
I=0
ε'2 ψ = ρ(ψ)
'pE = − ρeq 'ψ
Where pe is the electrostatic pressure.
Now, consider applyig a small perturbation ΔP, ΔV and calculate linear
response, and assume diffuse charge does not change:
φ(x, y, z) =
−
ψ(x, y)
o
equilibrium potential profile
p(x, y, z) = pE (x, y) − G0 z ẑ
'p = −'⊥ pE + G0 ẑ
o
small
1
E0 zẑ
o
small perturbation: axial electric field
Lecture 30
10.626 Electrochemical Energy Systems (2010)
The transverse gradient '⊥ =
∂
∂x x̂
+
∂
∂y ŷ.
Bazant
The Poisson equation is now:
ρ = −ε'2 φ = −ε'2⊥ ψ = ρeq (ψ)
The full PDES:
ρ = −ε'2 φ
u
'p = η'2 u + ρE
reduce to
ρeq (ψ) = −ε'2⊥ ψ
−G0 = η'2⊥ u − ε('2⊥ ψ)E0
Let u = uE + up ; the velocity has electrosmotic and pressure driven compo­
nents, where:
−G0 = η'2⊥ uup
(1)
2
ε('⊥
ψ)E0 = η'2⊥ uuE
(2)
To solve this, we have
uuE =
ε(ψ − ϕ)
E0
η
where we introduce an another harmonic function, '2⊥ ϕ = 0 which satisfies
ϕ = ψ on the boundary (no slip). For a symmetric cross section (e.g.
parallel plates or a cylindrical pore), the potential of the surface is constant
by symmetry, so ϕ = ζ = constant (since the unique solution of Laplace’s
equation with constant Dirichlet boundary condition is a constant function).
1.1
Pressure driven flow
up dxdy ≡ Akp G0
Qp =
A
where kp is hydrodanmic permeability, eg kp =
ternatively,
Qp = Kp ΔP
where
ΔP
L
Akp
Kp =
L
G0 = −
2
h2
12η
for parallel plates. Al­
Lecture 30
1.2
10.626 Electrochemical Energy Systems (2010)
Bazant
Electrical current
Z
IE =
σE0 dxdy
A
= AkE E0 = KE ΔV
where
σ(ψ) = axial conductivity =
�
e2 � 2
2
z+ D+ c+ (ψ) + z−
D− c− (ψ)
kB T
for a binary electrolyte, and c± = equilibrium ion profiles.
1.3
Electro-osmotic flow
Z
QE =
uE dxdy
Z
ε
= E0 (ψ − ϕ)dxdy
η
A
≡ AkEO E0
A
= KEO ΔV
where
Z
ε
(ψ − ϕ)dxdy
KEO =
ηL A
−ΔV
E0 =
L
AkEO
KEO =
L
3
Lecture 30
1.4
10.626 Electrochemical Energy Systems (2010)
Bazant
Streaming current
Z
Ip =
ρup dxdy
Z
= −ε ('2⊥ ψ)up dxdy
ZA
2
= −ε ('⊥
(ψ − ϕ))up dxdy
A
Z
= −ε (ψ − ϕ)'2⊥ up dxdy
ZA
ε
=−
(ψ − ϕ)G0 dxdy
η A
≡ AkSC G0
A
= KSC ΔP
In the fourth line, we make use of the identify, ('2 f )gdxdy = f ('2 g)dxdy
if f and g vanish on the boundary 1 , which is the case for the pressure driven
flow up and the electro-osmotic flow ue ∼ ψ − ϕ.
Thus we have that
KSC = KEO
(Onsager relation)
This result is very general, for any charge distribution ρe (ψ) and any crosssectional geometry.
2
General Linear Electrokinetics
For any small disturbance (linear), the driving forces and resulting fluxes
can be expressed as:
⎛
⎞
⎛
⎞⎛
⎞
symmetrix
gradients
⎝f luxes⎠ = ⎝ matrix ⎠ ⎝thermodynamic⎠
K = KT
f orces
1
j
j
Proof: For volume
V and surface S, V (v2 f )g dV = V (v · (gvf ) − vfj · vg) dV =
j
n̂·(gvf ) dS− V vf ·vg dV (divergence theorem) = S n̂·(gvf −f vg) dS+ V f v2 g dV .
S
The surface integral vanishes if f and g vanish on the boundary. This is a generalization
of integration by parts.
4
Lecture 30
10.626 Electrochemical Energy Systems (2010)
Bazant
Specifically, for a nanochannel,
Q
I
=
Kp KEO
KEO KE
ΔP
ΔV
With the Onsager relations K = KT . Onsager (1931) derived this rela­
tion for linear response of a general system near thermal equilibrium, assum­
ing local, microscopic time reversibility of the equaitons of motion. Here we
see it emerge explicitly for linear electrokinetic response in a nanochannel.
5
MIT OpenCourseWare
http://ocw.mit.edu
10.626 Electrochemical Energy Systems
Spring 2014
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.