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July 3, 2016, 11:42 |
Mystery of the two models
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#1 |
Senior Member
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Dear Friends,
I am doing a case of incompressible flow in a pipe and utilizing kepsilon and kw SST models. I am modelling flow inside a grooved tube with Re range 3000-10000. The grooves are in the form of helical shape. I am comparing Nusselt and friction factor with experiment. The mesh size is in accordance to y+~1. when I use kwSST model I get a very good comparison for friction factor with an error overall less than 8 %. For Nusselt my error is more than 22 %. Specially at high Reynolds numbers. I am comparing komega SST and experiment too with k-epsilon realizable, where the case is 180 deg opposite. Nu is in good comparison (for kepsilon) while friction has more than 22% error. Could you please help me in judging my results or am I doing something wrong? Please help me, i am in serious trouble. |
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July 4, 2016, 15:31 |
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#2 |
Senior Member
Lucky
Join Date: Apr 2011
Location: Orlando, FL USA
Posts: 5,764
Rep Power: 66 |
I think the result is reasonable. The are too many reasons for why each model (even similar models) produce different trends for f and Nu.
See my 2015 AIAA paper where I did a comparison of different models on Ralph Webb's circular tube with orthogonal ribs and compared f and Nu. DOI: 10.2514/6.2015-3927. Interestingly, RSM model did pretty well. I did some further analysis and calculated the reynolds stress anisotropy tensor based on the RSM results and compared that to the two-equation models. The two-equation models predicted values very far off compared to the RSM model, which means that the Boussinesq hypothesis is not 100% accurate (on average it was only 80%). When the fundamental assumption of your RANS model (assuming that Boussinesq hypothesis is true) breaks down, is it any surprise that your models don't all produce accurate results, or the same results? Rather than say one model is better than another, it's much more accurate to say that they're all equally bad. You can do a lot of design and engineering with the existing models, but CFD is far from an exact science. You could go even further and do the same to the temperature fluctuations, which is an even bigger beast. The closure model for the energy equation has been mostly neglected by turbulence modelers that are fixated with the fluid dynamics. Although some people are looking into it now, development isn't widely accepted and results haven't made its way into commercial CFD codes yet. |
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July 4, 2016, 15:59 |
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#3 | |
Senior Member
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Quote:
When I changed Energy Prandtl number, however, miraculously I got very close agreement to experiment. however, in one of your previous posts you did not recommend it. I am also using intensity and hydraulic diameter as reference parameters in boundary condition. as suggested byy Fluent that TI=0.16 Re^(-1/8). I used fully develped (power law) profile as velocity input via UDF. |
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July 5, 2016, 04:09 |
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#4 |
Senior Member
Lucky
Join Date: Apr 2011
Location: Orlando, FL USA
Posts: 5,764
Rep Power: 66 |
The energy Prandtl number is a crucial part of the closure model for the energy equation! A model that actually models the turbulent Prandtl number is what is needed to advance the state-of-the-art. Presently, everyone uses a simple constant.
I don't recommend blindly tweaking model parameters for the sake of matching data. I also don't recommend lowering the Prandtl number any lower than it already is. If you want to experiment with tweaking them, certainly it is a fun exercise. But regardless of how well you match the data, I would not consider that you have discovered any new physics. We can always tune constants to better match any given set of data. There are certainly a lot of comparisons you can do with results from different turbulence models. But the outcome is usually a naive interpretation of the results (i.e. this model does better than this model) without any regard to what is actually being modeled or how the model is implemented. You already have the experimental data and you already know the outcome that eventually CFD must predict! You just want to understand why one model performs poorly and another not so poorly. But unfortunately, this is very hard to do because of how complicated turbulence modeling has become. If you just want to use CFD as a tool, then leave the model comparisons to people that are actually working on turbulence modeling. If not, then you should consider becoming a turbulence model developer. |
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July 6, 2016, 00:23 |
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#5 | |
Senior Member
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Quote:
By the way , I read your paper and found interesting, one thing I want to ask that for thermal performance you put friction factor ratio in cube root. While in most of the papers I saw it without cube root. May I ask any significance of cube root power. |
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July 6, 2016, 01:04 |
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#6 | |
Senior Member
Lucky
Join Date: Apr 2011
Location: Orlando, FL USA
Posts: 5,764
Rep Power: 66 |
Quote:
But if I had to guess. The flowfield doesn't change appreciably with Reynolds number except close to the steps. Most of the differences in predictions are right around the corners of the ribs. The corners is where the Boussinesq hypothesis is most invalid (although the model doesn't know this). If you wanted to investigate why the models produce difference trends w.r.t. Reynolds number, that's where I would look. For any models that uses Boussinesq, you should eventually be able to track everything down to the shear stress tensor because that's ultimately how the models interact with the mean-flow. So maybe a comparison of the strain rates in these areas will give some hints to what is going on. This was the line of thinking that I started on and eventually found that the strain rate tensor and Reynolds stress anisotropy tensor was ideal for this comparison. Pumping power ratio goes like cube root of friction factor. Classically, pumping power is defined based on the cube root. That is, the ratio of heat transfer enhancement over the ratio of pumping power (Q*dP). When you are designing a large industrial heat exchanger where space isn't an issue, you have many scaling options such that ultimately the total thermodynamic loss (the pumping power) is what you care about. I've seen the second pumping power without the cube root mostly in gas turbine literature. I'm not sure what the physical interpretation of this thermal performance is. The design constraint for gas turbine engineering design problems and probably others is a fixed driving pressure ratio. So for these design engineers, they are more interested in how much heat transfer they get per pressure ratio. I don't know what the utility of it is since it's more straightforward to just plot heat transfer rate versus pressure ratio if that's what you really want to look at. Interestingly, thermal performance with a cube root is always smaller than thermal performance without a cube root. So not taking a cube root makes your data look better than it is! I think there's some psychology to it. Data looks much better without the cube root! A lot of rib turbulated cooling configurations have such a high pressure drop that if you take the cube root performance, it's essentially unity and not very appealing (also makes funding hard to obtain). |
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July 10, 2016, 11:21 |
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#7 | ||
Senior Member
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2. What is still pinging me is that what makes the k-w SST behave so differently in terms of heat transfer while the model is very well tested. What thing from velocity or momentum or continuity is going to drop down the Nusselt number? I wonder if something from these is being used in energy equation that is wrongly calculating the delta T . By the way RSM is behaving very accurately in Nu and very far in case of friction. Sorry I said it other way in previous posts. havg=q/A(delta T) and Nu=hD/k, so every thing is fixed. |
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July 11, 2016, 14:12 |
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#8 |
Senior Member
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Any response yet?
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July 18, 2016, 16:26 |
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#9 |
Senior Member
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Guys! please help me in this regard, I am in a fix really!
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July 20, 2016, 13:54 |
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#10 |
Senior Member
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Lucky Tran can you spare some time pls>?
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