|
[Sponsors] |
May 10, 2002, 09:50 |
skew-symmetric form
|
#1 |
Guest
Posts: n/a
|
Dear all,
I found there are two versions of the so-called skew-symmetric form for the convective term in the incompressible flow, i.e., 1) u \cdot \nabla u = 1/2 ( u \cdot \nabla u ) + 1/2 \nabla (u \cdot u); 2) u \cdot \nabla u = 1/2 ( u \cdot \nabla u ) + 1/2( (\p u_i u_j)/(\p x_j)); The second version is easy to understand but i think the first one never holds, though it is so widely used. Any one can give some explanations ? Thank you so much! |
|
May 16, 2002, 08:38 |
Re: skew-symmetric form
|
#2 |
Guest
Posts: n/a
|
Hi Annie,
this is a simple problem of differential calculus. If i have not made a mistake in understanding the formula, the following should be the answer to your question. The thesis is: u \cdot \nabla u = \nabla (u \cdot u) that is U_j d/dX_j (U_i) = d/dX_i (U_j U_j) (using the summation convention for repeated indices). We have simply: U_j d/dX_j (U_i) = d/dX_j (U_j U_i)- U_i d/dX_j (U_j) and the last term is zero because it is the divergence of the velocity vector. I hope this have cleared your doubts. Best regards, Nicola |
|
Thread Tools | Search this Thread |
Display Modes | |
|
|
Similar Threads | ||||
Thread | Thread Starter | Forum | Replies | Last Post |
Form drag | mnabi | Main CFD Forum | 0 | March 5, 2011 16:58 |
RANS in integral form | siw | Main CFD Forum | 0 | August 17, 2010 03:18 |
Derivation of Momentum Equation in Integral Form | Demonwolf | Main CFD Forum | 2 | October 29, 2009 19:53 |
symmetric flow around fuselage | Christoph | FLUENT | 0 | November 3, 2005 05:50 |
how to reduce Cell Equivolume Skew | santosh | FLUENT | 1 | August 16, 2005 14:38 |