Fluid dynamics - flow profile and velocity

1. Sep 19, 2004

kkkkKen

Hi!
Great website I just wished I had discovered it earlier.

Here is my simple question and I would appreciate any tips or links where I can find more information.

Problem: I would like to calculate the mean flow or velocity distribution along the central axis of a narrowing cone as well as the velocity profile changes with increasing depth.
I have looked in the fluid dynamics books that I have at hand and searched the internet without any luck. There is however plenty of information on the opposite problem using a “A-cone” with increasing diameter or a nozzle.

Any ideas where I can this kind of information and the equations needed to simulate this in Matlab?

Cheers!

/Ken

2. Sep 19, 2004

Clausius2

Incompressible flow?

Inviscius Flow?

Viscous flow?

Please, clarify a bit more your question and I will be able to give you the governing equations.

3. Sep 19, 2004

kkkkKen

I'll try to clarify my setup a bit.

The fluid is destilled water (newtonian, viscocity= 0.982[cp]) and the flow is driven by a static pressure from a resarvoar placed above the cone.

For simplicity one could just assume that we have a lamianar flow of water in a long pipe and that the pipe has a section that is tapered. To my understanding the the cone will have an accelerated velocity and the flow profile will change from laminar to plug flow with increasing depth of the cone. I would like to be able to calculate the velocity at different depth of the cone knowing the volume flow after the passage of the cone, and be able to estimate at what depths the flow profile will change from laminar, transient and plug.

Thanks,

/Ken

4. Sep 20, 2004

Clausius2

I have a doubt:
Do you mean turbulent flow when you said "plug flow"?

Let's name coordinate "z" the axis of symmetry of the cone (pointing upwards) and "r" the radial coordinate. We will suppose incompressible, viscous, laminar and steady flow. This is valid at low Strouhal and Reynolds numbers, and very accurate for your problem supposing the rate of discharge and characteristic velocity is not too fast. All this depend on the structural form of the vessel. Surely you will have to derive a condition involving water density, water viscosity, characteristic reservoir lenght and characteristic pipe lenght, in order to set up this problem as viscous. This condition has to be derived in order to check that your simulation really fits with the real event. If I had some picture about the system I could help you to derive it.

Once you have checked the flow as viscous (u=axial velocity, v=radial velocity):

Continuity: $$\frac{\partial u}{\partial z}+\frac{1}{r}\frac{\partial rv}{\partial r}=0$$

x-Momentum: $$\rho u\frac{\partial u}{\partial z}+\rho v\frac{\partial u}{\partial r}= -\frac{\partial (P+\rho g z)}{\partial z} + \frac{\mu}{r}\frac{\partial}{\partial r}(r \frac{\partial u }{\partial r})$$

y-Momentum: $$\frac{\partial P}{\partial r}=0$$

with the boundary conditions: u(z=0)=Uo, P(z=0)=Po, v(z=0)=0. where z=0 is the highest section of the cone.
u(r=r(z))=0, v(r=r(z))=0, and du/dr(r=0)=0.

Please look into the above three equations and see they are "self contained".

Notice that I have employed the Boundary Layer Approximation. I mean, the pressure is constant along any transversal section, and the second derivative (see the viscous term) of the axial velocity respect to z is negligible compared with the second derivative of the axial velocity respect to r. Why is that? I'm supposing the cone has a lenght L and characteristic radius R. In the boundary layer approximation R/L<<<1 and radial variations are the most important. The transverse velocity "v" almost does not vary if the cone shape is narrow enough.

Notice too those equations does not model neither turbulent regimen, nor transient regimen. Why does it not? You have not included any flow of information to rearwards (you will not see any second derivative respect to z). Thus, it is impossible for you to simulate this equations and watch a recirculating flow because of boudary layer separation. You only are enabled to see the point in which the flow changes to turbulent viewing the shear stress at the cone wall. You should know the separation (and so the transition) is produced when the shear stress is 0 at some point (do you know it?). A flow completely turbulent is governed by Reynolds equations (they are very difficult to solve).

The mathematical behaviour of this equations is parabolic. I mean, it only has one characteristic curve (z=constant). Only the past history of the flow influence on the next abcissae. The flow has no information of what is happening in front of it. You will have to writte a program by means of Crank Nicholson or some implicit numerical method, taking steps in each "z".

In order to generate the computational mesh in Matlab, you will have to use a coordinate transformation. You have to build a transformation which transforms the non-rectangular domain into a rectangular one. This coordinate transformation will affect to the original equations. The three equations will be affected by a metric coefficient.

The initial data line (Uo, Po) corresponds to a viscous flow. Please see some literature about that because Uo surely corresponds to a Poiseuille profile and Po is the hidrostatic pressure (I'm not sure). Neglect the entrance effects at the pipe.

Regards.
Javier.

Last edited: Sep 20, 2004
5. Sep 20, 2004

kkkkKen

Wow!

Thank you ever so much Javier!
What a well structured answer, it will take me some time to go through
the details and work out a solution. I also appreciate your offer to help me derive this, but in order to understand it, I really must give it a try myself. As you can tell my background is not in fluid dynamics and I really thought this could be approximated by the use of the Bernoulli equations and the definition of the Raynolds number as in the case of an endless pipe.

And for "plug flow" I was reffering to the velocity profile,which occurs at turbulent flow which you pointed out.

I'll let you know how things turn out.

Thanks a million!

/Ken

6. Sep 20, 2004

Clausius2

Thanks no needed, Ken. I'm very happy of answering Fluid Dynamics questions. Your question about that will be always welcome. Although I have not finished my engineering graduate program yet, Fluid Dynamics is the area of my main interest, and I'll be grateful of answering any doubts.

Although you have to start working on it, no approximation with Bernoulli equation is possible. That is very important. Bernoulli behaviour is the opposite of Viscous behaviour. Elect one of the two behaviours depending on your flow problem, but never mix together both things. If you employ Bernoulli equation, you will never obtain U=0 at the wall, because no boundary condition is possible, except in pressure.

Regards.
Javier.

7. Dec 23, 2009

AzeemA

Dear Javier,
I am impressed from the way you answer this problem, and i really would be so grateful to you if we can cooperate in solving a big fluid dynamics dilemma i have right now,
Send me your contacts details including your phone number in order to communicate.
Best Regards,
Mohamed