Tom's Electrical Problem

Tom has come up with a new problem that involves solving the electric field. Here are the equations at the electrode interface ($1$):

$ i_T^1 = -\kappa \vec{n} \cdot \nabla V_{IR}$

$ i_T^1 = -c_{DL} \frac{\partial V_{DL}}{\partial t} + i_{F} $

$ i_F = \frac{c_m^i}{c_m^{\infty}} \sum_j \theta_j n i_j^{\infty} F \left( \alpha_j, V_{DL} \right) $

$ V_{CELL} = V_{IR} + V_{DL} $

$ D_m \vec{n} \cdot \nabla c_m = \frac{i_F}{n F} $

and in the bulk:

$ \frac{\partial c_m}{\partial t} = \nabla \cdot D_m \nabla c_m + \nabla \cdot \left[\frac{z F}{R T} D_m c_m \nabla V_{IR} \right] $

$ \nabla^2 V_{IR} = 0$

on the top electrode ($2$):

$ V_{IR} =0 $

$ c_m = c_m^{\infty} $

At first glance, I didn't see that this system is closed. Since the $V_{IR}$ equation is steady state, and surrounded by Neumann boundary conditions it only needs one boundary to fix the value (it is an initial value problem for the slope of $V_{IR}$ rather than a boundary value problem for $V_{IR}$.

Solving for the 1D problem

We have (assuming the normal points out of the electrolyte)

$ i_T = -\kappa V_{IR} / \delta$

where $\delta$ is the distance between the electrodes

$ i_T = i_F - C_{DL} \frac{\partial V_{DL}}{\partial t} $

$ i_F = i_0 \exp{\left(-\frac{\alpha F V_{DL}}{R T} \right)} $

This leads to an ODE for $V_{IR}$

$\frac{\partial V_{IR}}{\partial t} = -\frac{1}{C_{DL}} \left[ i_0 \exp{\left(-\frac{\alpha F \left(V_{CELL} - V_{IR} \right)}{R T} \right)} +\kappa V_{IR} / \delta \right] $

This can be solved with scipy's ODE integrator. To do this I took an example <http://docs.scipy.org/doc/scipy/reference/generated/scipy.integrate.ode.html> and used that. The code is source:trunk/moffat/electrical/1D.py.

The results are

source:trunk/moffat/electrical/voltageVersusTime.png

and

source:trunk/moffat/electrical/currentVersusTime.png

With the following numbers:

$i_0 = 40\;\text{A / m}^2$

$\alpha = 0.4$

$F = 9.6485 \times 10^{-4}\;\text{J / V / mol}$

$V_{CELL} = -0.275\;\text{V}$

$R = 8.314\;\text{J / K / mol}$

$T = 298.0\;\text{K}$

$\kappa = 15.26\;\text{A / V / m}$

$\delta = 100 \times10^{-6}\;\text{m}$

$C_{DL} = 0.3\;\text{A s / V / m}^2$

We get the following steady state values:

$i_F = 2294.7\;\text{A / m}^2$

$V_{IR} = -0.01503\;\text{V}$

$V_{DL} = -0.259\;\text{V}$

  • Posted: 2012-01-12 14:25 (Updated: 2012-02-09 09:54)
  • Author: wd15
  • Categories: (none)

Comments

1. wd15 -- 2012-02-01 17:38

The comment at the end about $V_{IR}$ is bogus. Tom really means $\Delta V_{IR}$ so we need to fix $V_{IR}$ at some point. The question in 2D is which point or should it be the top surface.

2. wd15 -- 2012-02-08 11:33

Having chatted with Tom $V_{IR}=0$ is fixed on somewhere. In 1D, we can just assume it's the top surface that is fixed. In 2D, I'm not so sure whether it is the entire top surface or just a single location.

Add New Comment