Jon,
I think what you are saying is that I do not have a convection term
here at all.
My equation is $\frac{\partial u_{c}}{\partial t}=\nabla \left[ D\cdot
\nabla u_{c}+u\cdot \nabla \phi \right] $
OR, $\frac{\partial u_{c}}{\partial t}=\nabla \left[ D\cdot \nabla
u_{c}+%
\vec{u}\phi \right] $, right?
Your started your last note in the context of solving for $u_{c}$ as a
TransientTerm $\frac{\partial u_{c}}{\partial t}.$ I think you stayed
with
that context, and therefore your examples never reverted to solving the
TransientTerm $\frac{\partial \phi }{\partial t}.$
> Jon Said...
> If you're solving for u_{c}, then:.....\qquad
> $\nabla \left[ D\cdot \nabla u_{c}\right] $ is a
> DiffusionTerm for u_{c}with diffusion coefficient D.
Great, that is precisely my first term. I should factor out a conanical
$%
u_{c}$ in $D$ and solve it like a diffusionTerm, Right?
> $\nabla \cdot \left[ \vec{v}u_{c}\right] $ is a
> ConvectionTerm for u_{c}with velocity coefficient v.
That is NOT what I have in my second term!
> $\nabla \cdot \left[ \vec{v}\phi \right] $ isn't a
> ConvectionTerm for u_{c}at all.
Ah, that \emph{IS} what I have, and I've been trying to solve it as a
convection term. \ I should stop trying to solve it as a convection term
and
instead solve it as ???? \ what, a source term perhaps? \ If so, I do
not
factor out the $u_{c}$ in $\vec{v}$, right?
Buddy
-----Original Message-----
From: [email protected] [mailto:[EMAIL PROTECTED] On Behalf Of Jonathan Guyer
Sent: Friday, May 26, 2006 4:19 PM
To: Multiple recipients of list
Subject: Re: convection coefficent a function of concentration
On May 26, 2006, at 2:51 PM, Damm, Edward F. (E. Buddy) wrote:
>
>
> All the examples you gave had the gradient term expressed as the field
> variable you are solving for in the transientTerm. There were no
> mixed gradient terms with respect to the transientTerm, so I'm still a
> little fuzzy. I may be oversimplifying here.
>
> If I have the equation $\frac{\partial u_{c}}{\partial t}=\nabla
> \left[ D\cdot \nabla u_{c}+u\cdot \nabla \phi \right] $
>
> Here transient term is solving for the Field variable $u$_{c}.
>
> The first gradient term is in $u_{c}$, so I do Not factor out $u_{c}$
>
> The second gradient term is in $\phi $, so now I do need to factor out
> $% u_{c}.$ With that done, I then have...
It has nothing to do with where the gradient is. Each term implicitly
contains the solution variable in one particular location, so you should
never include this *implicit* solution variable in the coefficient for
that term. That would be redundant. You have to figure out which
canonical Term is represented by each part of your equation. Once you
know that, then look where the solution variable appears in that Term
and everything *else* is the coefficient.
If you're solving for u_c, then:
$\frac{\partial u_{c}}{\partial t}$ is a TransientTerm for u_c with
coefficient 1.
$\frac{\partial \rho_0 u_{c}}{\partial t} is a TransientTerm for u_c
with coefficient rho0.
$\frac{\partial \rho_0 u_{c} u_{c}}{\partial t} is a TransientTerm for
u_c with coefficient rho0 * uC.
$\frac{\partial \phi}{\partial t} isn't a TransientTerm for u_c at all.
$\nabla^2 u_c$ is a DiffusionTerm for u_c with diffusion coefficient 1.
$\nabla \left[ D \cdot \nabla u_c \right]$ is a DiffusionTerm for u_c
with diffusion coefficient D.
$\nabla \left[ D u_c \cdot \nabla u_c \right]$ is a DiffusionTerm for
u_c with diffusion coefficient D * uC.
$\nabla \left[ D \cdot \nabla \phi \right]$ isn't a DiffusionTerm for
u_c at all.
$\nabla\cdot\left[ (1,0,0) * u_c \right]$ is a DiffusionTerm for u_c
with velocity coefficient (1,0,0) (in 3D).
$\nabla\cdot\left[ \vec{v} u_c \right]$ is a ConvectionTerm for u_c with
velocity coefficient v.
$\nabla\cdot\left[ \vec{v} * (1 - u_c) u_c \right]$ is a ConvectionTerm
for u_c with velocity coefficient v * (1 - uC).
$\nabla\cdot\left[ \vec{v} \phi \right]$ isn't a ConvectionTerm for u_c
at all.
A convection term is a convection term because it looks like the
divergence of u_c times *some* velocity vector. It doesn't matter if
that velocity vector has a gradient in it. It doesn't matter if the
velocity vector is *also* a function of u_c.
A diffusion term is a diffusion term because it looks like the
divergence of some diffusivity times the gradient of u_c. It doesn't
matter if the diffusivity also has u_c in it.
Maybe this is the source of your confusion: Arguably, if you had a term
$\nabla\cdot\left[ u_c \nabla u_c \right]$
you could think of this as a DiffusionTerm with a diffusion coefficient
uC *or* as a ConvectionTerm with a velocity coefficient
uC.getFaceGrad(). While this may be technically true, I would never
treat it as a ConvectionTerm, and I don't even know if it would work at
all. Maybe Daniel can confirm.
-----------------------------------------
This message and any attachments are intended for the individual or
entity named above. If you are not the intended recipient, please
do not forward, copy, print, use or disclose this communication to
others; also please notify the sender by replying to this message,
and then delete it from your system. The Timken Company / The
Timken Corporation
