Jon,
Thanks, I understand the 'convection coefficient' issue now...
Regarding your statement:
"However you determine Mc, yva, and d2Gmducdphi, remember that
they must be determined on faces to be consistent with
phi.getFaceGrad()."
How do I find out if I am determining these values on the face? If I am
not, is it just as easy to use one of the other 'getGrad' variations? Ie
getArithmeticFaceValue(), and not have to change how I am determining
the values of Mc, yva, and d2Gmducdphi?
I'll look in the manual too!
Buddy
-----Original Message-----
From: [email protected] [mailto:[EMAIL PROTECTED] On Behalf Of Jonathan Guyer
Sent: Thursday, April 27, 2006 2:14 PM
To: Multiple recipients of list
Subject: Re: Convection solutions
On Apr 27, 2006, at 1:34 PM, Damm, Edward F. (E. Buddy) wrote:
> Sorry, Hope the attached clarifies it. The second to the last
> equation is what I am talking about. The stuff before leads up to it.
> The term that is not in the original reference, but that I added per
> our discussion was the 1/uc.
It's not that you divide by uc; it's that you must factor it out of the
term to determine the coefficient. You have correctly identified your
convection term as
\begin{equation}
\nabla\cdot \left[ M_{c}\cdot u_{c}y_{va}\cdot \frac{\partial
^{2}G_{m}}{\partial u_{c}\partial \phi }\nabla \phi \right]
\end{equation}
A convection \nabla\cdot\left[ u_c \vec{v} \right] represents the field
$u_c$ being transported by a velocity field $\vec{v}$. For FiPy's
purposes, the coefficient of the convection term is the velocity field
which, in your expression, is everything *but* $u_c$.
It's not that you need to explicitly divide by $u_c$; you simply need to
factor it out.
Your convection *coefficient* (not equation!!!) is then:
Mc * yva * d2Gmducdphi * phi.getFaceGrad()
However you determine Mc, yva, and d2Gmducdphi, remember that they must
be determined on faces to be consistent with phi.getFaceGrad().
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