kkkkKen said:
Thanks for your reply Clausius2!
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.
/Ken
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 length 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: [tex]\frac{\partial u}{\partial z}+\frac{1}{r}\frac{\partial rv}{\partial r}=0[/tex]
x-Momentum: [tex]\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})[/tex]
y-Momentum: [tex]\frac{\partial P}{\partial r}=0[/tex]
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 length 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.
I hope it will help you.
Regards.
Javier.
