# Math Help - wouldnt the navier stokes equation would be much easier by using the finite volume me

1. ## wouldnt the navier stokes equation would be much easier by using the finite volume me

wouldn't the naiver stokes equation would be much easier to solve by using the finite volume method ?

2. Originally Posted by janemba
wouldn't the naiver stokes equation would be much easier to solve by using the finite volume method ?
Much easier compared against what?

3. compared to solving naiver stokes equations

4. The Navier-Stokes equations are indeed very difficult to solve analytically. However it is not the fact that they are of a very high level that we shouldn't attempt to solve them. This means that it is important to understand the equations, and live with the fact that very often we need to be satisfied with exact solutions of approximate equations, approximate solutions to the exact equations or even with approximate solutions to approximate equations of NS.

This fact makes numerical methods very attractive. And indeed the NS equations are available in numerical form as Computational Fluid Dynamics. This uses the finite volume method. However this does not mean that everything is solvable now, on the contrary. One needs to make a detailed 3D CAD model and has to create a high quality grid in the domain. This takes up most of the time of the calculation engineer. After this all the boundary conditions and solver strategies must be set-up. Then the model has to be calculated and finally the results can be analyzed (if the calculation converged). One point of attention is the turbulence which is inherent a part of flow and has to be modeled as well because calculating this directly is nowadays for practical cases still impossible.

So in the end it is a matter of the problem difficulty you want to solve. Is it more or less comparable to a known simple geometry and are formula's available or derivable? If so, calculate everything analytically, if not go to the (very expensive) CFD packages.

NS much easier by finite volume? Certainly not.

Hope this clears a few things.