Hi there,

I'm busy modelling flow within a flat container which is only 0.004" deep. When creating the mesh in gambit, do i have to choose a node spacing of less than 0.004 in order to see an accurate 3D image of the depth? Or does the node spacing not work in this way? I have used a spacing of 0.01 which 'looks' fine but i'm concerned that the software is treating the 0.004" depth as a 2D surface. Any advice or suggestions would be appreciated.

Thanks in advance, gary.

Concerning the grid... In such a small space, the fluid (if it has any viscosity at all) would very likely behave like laminar flow, and if that is the case, no refinement at the walls is needed to capture energy production and dissipation. However, there will be development of a parabolic profile as the fluid travels, so that may come into play in deciding the density of the mesh. Many fluid dynmamics codes are always 3D, so to approximate 2D flows, they use a domain that is one-cell thick, as you have done. A cell spacing of 0.01, I believe, means that you are breaking the domain grid into 100 pieces for every unit length. Assuming you are using GAMBIT, then I think the default unit length is 1 meter, but if you have changed it then it is what you have defined it as. If the default is 1 meter, then you're space is 1 cm, which is actually 100 times greater than 0.004 inches. Perhaps your default unit length is 1 cm, in which case the cell size equals the true size of your physical geometry of interest. Whatever the case, I believe it is best to keep the computational geometry at the same size and scale as the real-world geometry.

Considering the nodes, i.e. grid points... in GAMBIT you have the option of a 2 or 3-node setting along the edges, 4 or 8 or 9 nodes on a quadrilateral face, and 8 or 20 or 27 nodes on/in a hexahedral volume. The reason for this is so you can have control over how many grid points are used to define the cells and subsequently the mesh. Since FLUENT calculates values for cell centers, then the actual values at the nodes are interpolated (or averaged) between neighboring cells. If you have a thickness of one cell, it follows that the nodes would have the average of the two cells on either side of it.

Considering the flow... if you have reason to believe the flow has turbulence, then you will absolutely need more than one cell thick (because you have 4 walls and need refinement on each), otherwise the solution will be only a generalization of reality. Are there photographs of the flow? Do you have measurements? Published information? These will tell you whether you have turbulent or laminar flow.

It would probably be beneficial to take out a very small portion of the domain of interest and get a good understanding of it first in 3 dimensions, if possible. Then make it more complicated as each section is understood. The first step, however, is going to be deciding what flow regime you are in and how you wish to simulate it.

Aaron,

Very many thanks for taking the time to post. Your description was very informative. I'm almost certain the flow is laminar, and so i will do some more reading regarding the energy production and dissapation which you mentioned.

Thanks again, Gary.

Most welcome. Actually what I meant by the energy production and dissipation statement is that if your flow is laminar you won't have turbulence to worry about which is how energy is moved about in the system. So, since you're in a laminar flow you won't need to worry about that in order to create a model that simulates such a flow. What is more important, I believe, is determining whether a 2D flow is useful enough for you to describe your flow and that may depend upon formation of a parabolic profile(s) as the fluid cruises downstream.

BTW, what is the viscosity of your fluid? As you described the problem, I got the image of a capillary flow, and if that is the case, then a 1D approximation of the Darcy's equation can be applied to solve it.

hello again Aaron, i see what you mean about turbulence causing the energy to 'upset' fluid flow.

the viscosity of the fluid is 782 kg/m3. What i was worried about was that (correct me if i'm wrong)there will be a point where the gap (which at the moment is 0.004") becomes so small that the fluid will either flow very unconventionally, or will not flow at all? I am trying to design a device which uses expensive chemicals and so the volume is an issue. the device we use at the moment has a depth of 0.008" (and works fine) and i am looking to modify this. it is not a tube, but more of a miniature, very flat oval shape disc with a circular opening at each end.

thanks, gary.

Gary, The gap 0.004" isn't really that small (it's at the very upper limit of what is considered "micro"). Happily, one thing you don't have to worry about is the breakdown of the continuum theory (I am assuming that the number you gave as "viscosity" is actually the density since those are the units it has). In that case, it has a density of the order of water's, and the continuum assumption holds down to length scales of 10 nanometers (your smallest length scale, 0.004", is 100,000 nanometers!) However, it is possible in small gaps to have large velocity gradients and hence large shear streses. Therefore you should be concerned about possible non-Newtonian behavior of the fluid. Whether these can effects can be neglected can be determined by calculating the Weisenberg number, which can be estimated using We~(V/h)*mu*M/(rho*Ru*T) where V is the maximum velocity in your gap, h is the gap depth, mu is the dynamic viscosity, M is the molecular weight, rho is the density, Ru is the unversal gas constant and T is the temperature. Just make sure that you match the units so the result is dimensionless, and if We<<1 then Newtonian behavior can be expected. Hope that resolves your concern.

Peter, thanks very much for the time and information. I'll have a look into the things you suggested! gary.

Hi there,

the viscosity of the fluid is 0.0004 kg/m-s. If you have the time could you possibly explain in the simplest terms what you mean by the continuum theory. i've tried doing some research of my own but can't really understand the difference between newtonian and non-newtonian behaviour? Does it mean that as a general rule some fluids won't flow through a gap of 10 nanometers due to the shear forces??

thanks for any explanation! gary.

If continuum theory doesn't hold then you can't use Fluent (at least, not in any straight-forward manner). You would only have to worry about that for a 0.004" gap if you had a very low density gas. If the fluid behavior is non-Newtonian then the the stress is not proportional to the strain rate (the Newtonian constitutive relationship doesn't apply). Were you able to calculate the We?

Hi peter,

I'm interested in calculating the weisenberg number which you mentioned although i have a question regarding units: i am using a liquid, so does this mean that i still have to include the universal 'gas' constant, Ru? Is this one number for all liquids and gases?

thanks in advance, gary.

Gary, The universal gas constant Ru is equal to the product of Avagadros number NA (the number of something in one mole) and Boltzman's constant, kb. So in the approximate expression for the We the product Ru*T is NA*kb*T. kb*T is a scale of thermal energy leading to increased entropy, and it shows up in this context due to its role in predicting the relaxation time for large molecules. NA shows up because of the units of molecular weight- we are working in moles not single molecules. One way of thinking about the cause of many non-Newtonian effects is that the molecules in the fluid (usually a grease, oil, liquid polymer, or something else with high molecular weight and hence large molecules) become aligned and stretched by the flow, and do not have time to recover. If Ru*T is larger, the molecules have more thermal energy and are less likely, all other things being equal, to be "organized" by being aligned and stretched. So that's a long way of saying: yes, Ru should be included, and yes, it is the "universal" gas constant, which is "universal" regardless of fluid, Ru= 8.315 J/(mol*K). If your fluid is composed of small molecules (molecular weight less than 100, for instance, as the viscosity&density suggest) then you will probably find your We is pretty small.

Hi Peter, the molecular weight was, as you suggested less than 100, it was 41.04. The velocity 0.008 m/s, (slow!), height 0.0001m, and room temperature of about 293K. This gave me a weisenberg number of 0.000000689! So, to clarify, this means that my fluid definitely follows 'newtonian behaviour' as the number is far less than 1? thanks very much, gary.

Yes, you are safe using a Newtonian model.

thanks very much for your time and help!