2D-flow in a tube: Continuity and Navier-Stokes -- CFD Online Discussion Forums

August 9, 2017, 12:55

2D-flow in a tube: Continuity and Navier-Stokes

Mike Jey

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Hi people,

I want to do some basic stuff with Python 2.7 an the finite differences method. Now I have the following case:

- 2D-flow in a tube, steady state, not compressible
- at the inlet there ist a constant velocity profile so ( $\frac{du}{dy}=0$ )
- the flow is developing into a convex profile

I want to find out the length "L" of the tube, where the velocity profile isn't developing any more.

The basic equation for continuity and Navier-Stokes are:

$\frac{du}{dx} + \frac{dv}{dy}= 0$

$u \frac{du}{dx} + v \frac{du}{dy}= -\frac{1}{\rho} \frac{dp}{dx} +\nu (\frac{d^2u}{dx^2} + \frac{d^2u}{dy^2})$

An assumption is, that I say the velocity "v" is everywhere zero. So every term with "v" in the Navier-Stokes and Continuity equation is zero and I have to solve only the u-Momentum (as shown). In addition to this I say that $\frac{dp}{dx}$ is also zero.

After this assumptions I have these equations left (Continuity and navier-stokes):

$\frac{du}{dx}= 0$

$u \frac{du}{dx} = \nu (\frac{d^2u}{dx^2} + \frac{d^2u}{dy^2})$

If I look to my continuity equation and insert it into the navier-stokes, there is:

$0 = (\frac{d^2u}{dx^2} + \frac{d^2u}{dy^2})$

So in my eyes it doesn't make sense ? It means: The velocity profile is not dependent on the fluid. So it doesn't matter, if there is water or air ? I don't think so.(?)

I tried the case without inserting $\frac{du}{dx}=0$ and I solved the equation $u \frac{du}{dx} = \nu (\frac{d^2u}{dx^2} + \frac{d^2u}{dy^2})$ in Python. It works and there are meaningful results.

But I don't understand. Actually I have to insert the continuity into the Navier-stokes, isn't it ?

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