navier-stokes
 

guys, does anyone know how should the navier-stokes equation for 1D unsteady compressible viscous fluid flow in the variable cross-section duct look like?

I have tried to derive it by myself, but it doesn't look like the one I found on the web... thanks

Re: navier-stokes
 

Re: navier-stokes
 

queram

if you have velocity like (v(x),0,0) there v and x are oriented along with flow you will have

stess tensor with non-zero xx,yy,zz component:

Pxx=-p + m1 dv/dx + 2*m2 dv/dx; Pyy=-p + m1 dv/dx; Pzz=-p + m1 dv/dx;

m1 and m2 - different viscosities but usually m1 = 2/3 m2. Further we have to put the tensor in Navier-Stokes there it turns to vector:

(d(Pxx)/dx,0,0)

Ask more if you need more complicated case. But actually in Cartesian coordinates general case with constant m1,m2 has very convinient form:

(m1+m2)*grad div v + m2*Laplace v (*)

In Cartesian you have:

(m1+m2)*d( div v)/dx + m2*Laplace v (m1+m2)*d( div v)/dy + m2*Laplace u (m1+m2)*d( div v)/dz + m2*Laplace w

But here comes a time for my question. I has to express (*) in cylindical coordinates. I started from Christofell symbols and obtained something very reliable (checked for incompressible) but I want be 100% sure. Does anybody know the source for the Navire-Stokes in cylindrical coordnates for compressible fluid?

Re: navier-stokes
 

The following classics have them:

Bird, Stewart and Lightfoot, Transport Phenomena

Anderson et al, Computational Fluid Mechanics & Heat Transfer

Bachelor, An Intruduction to Fluid Mechanics

Re: navier-stokes
 

Thank you, Rami, but what about URLs?

Re: navier-stokes
 

mmmmmm.... look at my procedure: having NS for compressible 3D unsteady fluid flow in x direction in the form: du/dx=-1/ro*dp/dx+mu(d^2/dx^2+d^2/dy^2+d^2/dz^2)-mu/3*d/dx(du/dx+dv/dy+dw/dz)

I have come to 1D unsteady compressible NS as: du/dt=-1/ro*dp/dx+2/3*mu*d^2u/dx^2

where u, v, w - velocity vectors ro - density mu - kin. viscosity x, y, z - coordinates

Re: navier-stokes
 

Hi, faber

I am afraid you have to change you code before its too late. m1=-2/3*m2 (looks like I've lost minus in the previous message) this done to keep physical sense of the pressure (pxx+pyy+pzz)/2=p. But it's only very common assumption. Originally Navier-Stoke's law contains two viscosities.

For additional check you can look here http://www.efunda.com/formulae/fluids/navier_stokes.cfm Couldn't find any more relevant.

Re: navier-stokes
 

(pxx+pyy+pzz)/2=p -> (pxx+pyy+pzz)/3=p