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May 5, 2003, 06:11 
Effects of NOT using mesh inflation

#1 
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How does CFX 5 treat the boundary layer calculation when no mesh inflation is used? Does it use the turbulence model wall functions to calculate a theoretical solution at each boundary node, or is the wall "displaced" to the boundary node?
I have to compare two simulations, one of which did not use inflation, and I need to know if I can attribute the differences to the fact the the boundary layer was not fully resolved in the one case. It seems for instance that separation occurs much later when their is no inflation. In both cases turbulence models and advections schemes were the same. 

May 5, 2003, 08:32 
Re: Effects of NOT using mesh inflation

#2 
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Hi Louwrens If inflation is not used then extremely fine tetrahedral is needed to be used to properly resolve boundary layer. In the case of infalted layers high aspect ratio prism elements are generated which resolve boundary layer properly at relatively less computational cost.


May 5, 2003, 10:51 
Re: Effects of NOT using mesh inflation

#3 
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I know the theory and reason for mesh inflation, I need to know how the code handles the problem when it's not there...


May 5, 2003, 15:44 
Re: Effects of NOT using mesh inflation

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Louwrens,
The only difference is the mesh. There is no fundamental difference between any element type. Robin 

May 6, 2003, 04:59 
Re: Effects of NOT using mesh inflation

#5 
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>The only difference is the mesh. There is no fundamental difference between any element type.
I understand that, specifically for the unstructured solver. On my two sets of simulations, one was done without inflation, and the other (of a slightly different geometry) with inflation. I basically need to know if I should attribute a "lack" of separation (where it is expected) on the mesh without inflation to the fact that the boundary layer was not solved and the flow therefore staying attached, or if I should attribute it to a "better" flow geometry. I know the best way is to rerun the first problem using a grid with mesh inflation, but I unfortunately do not have time, so I have to make postulations from the data I already have. 

May 6, 2003, 06:20 
Re: Effects of NOT using mesh inflation

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What is the geometry like and where is the separation occuring and what are the changes to the geometry that have resulted in the changes seen in the flow field ? You are really shooting into the wind with this one as conclusions on separation using CFD can be tricky at the best of times, even if your model is a good one with a good quality mesh. Sorry for sounding negative. Bob


May 6, 2003, 12:03 
Re: Effects of NOT using mesh inflation

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Hi Louwrens,
I would hesitate to ever compare results for two different geometries with such different meshes. Having no inflation will ceratainly delay the onset of separation. Without resolving the boundary layer you will overpredict the mixing, thus energizing your boundary layer by too much. For this is the same reason, the kepsilon turbulence model will also delay the onset of separation; the wall functions overpredict the momentum transfer. Although this is not nearly as bad as having no boundary layer resolution. In general, for ke wall functions, you should have 5 nodes within the boundary layer. It may be too late for you in this case, but you should consider using the SST turbulence model if boundary layer separation is an issed. The SST model allows you to model more of the boundary layer and capture the instability which leads to separation. For SST, you should have at least 15 nodes in the boundary layer to resolve it accurately. If you have less, it will blend in a wall function, but you should still not go below 5. Best regards, Robin 

May 6, 2003, 13:52 
Re: Effects of NOT using mesh inflation

#8 
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>Having no inflation will ceratainly delay the onset of separation. Without resolving the boundary layer you will overpredict the mixing, thus energizing your boundary layer by too much.
YES, this is what I've been looking for, it would help me explain some of the things I see. I used the ke model for both simulations and am just comparing the results. Correctly predicting the separation point is not as critical as just being able to point to the influence of a nonoptimal grid... >The SST model allows you to model more of the boundary layer and capture the instability which leads to separation. I have read the theory behind the SST and kw models, and did some validation cases using both ke and SST. Interestingly SST seemed somewhat TOO sensitive and predicted separation earlier (and reattachment later, but I can forgive that, I know its damn difficult!) than for certain experimental results... On some others there were very little difference between ke and SST, especially when there's a very big area of separation, eg a sphere, or a diffusing duct with large areas of secondary flow. Only when there was a "gradual" adverse pressure gradient leading to "eventual" separation did the SST show its merits. Thanx for your advice Robin, it is greatly appreciated 

May 6, 2003, 16:58 
Re: Effects of NOT using mesh inflation

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Your welcome Louwrens.
By the way, where SST predicted separation too early, you may want to check that the boundary condition was appropriately specified (right level of turbulence, velocity profile, etc). Best regards, Robin 

May 7, 2003, 03:25 
Re: Effects of NOT using mesh inflation

#10 
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>By the way, where SST predicted separation too early, you may want to check that the boundary condition was appropriately specified (right level of turbulence, velocity profile, etc).
Point noted, but it brings me to another one. Should probably start a new thread for this... How do you decide on the turbulence level if you don't have prior info? For the stuff I've done I've standardised on Default Intensity and Autocompute Length Scale. Is there some rough guide for turbulence levels for certain applications...? 

May 7, 2003, 16:07 
Re: Effects of NOT using mesh inflation

#11 
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Louwrens,
If you do not have experimental data, you area best to look to the literature or textbook references for your application. Robin 

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