# crosswind modelling on a vehicle

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 April 25, 2012, 07:52 crosswind modelling on a vehicle #1 New Member   ma3ou2 Join Date: Apr 2012 Posts: 11 Rep Power: 6 hi i am a student from IRAN and interested to (( modelling crosswind )) on a vehicle. but i have problem with simulate in fluent. i need your needs for simulate crosswind. please help me with regards

April 25, 2012, 13:46
#2
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Lucky Tran
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Quote:
 Originally Posted by ma3ou2 hi i am a student from IRAN and interested to (( modelling crosswind )) on a vehicle. but i have problem with simulate in fluent. i need your needs for simulate crosswind. please help me with regards
what's giving you trouble?
cross-wind is just a wind (velocity) coming from the side of the vehicle.

you just need to set up the boundary conditions properly.

 April 26, 2012, 00:54 #3 New Member   ma3ou2 Join Date: Apr 2012 Posts: 11 Rep Power: 6 yes , my problem is to crosswind boundary conditions. i want to know if vehicle has long length, what's B.C. for side vehicle?

April 26, 2012, 01:01
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Lucky Tran
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Quote:
 Originally Posted by ma3ou2 yes , my problem is to crosswind boundary conditions. i want to know if vehicle has long length, what's B.C. for side vehicle?

There are a number of ways of setting up this problem.

I think the easiest to visualize is to have on one side, the inlet (velocity/pressure), and opposite side the outlet. Then the front and rear and on top of the vehicle can be treated with the appropriate "farfield" boundary condition (this can be symmetry or farfield pressure or periodic, many ways). Below the vehicle should be a wall to simulate ground effects? Make sense?

I don't want to say exactly, this is how to do it since there are a number of ways and it also depends on how the domain is setup (computational domain, mesh extent, etc).

 April 26, 2012, 01:22 #5 New Member   ma3ou2 Join Date: Apr 2012 Posts: 11 Rep Power: 6 tnx, but my B.C. model is: front=inlet velocity rear=pressure outlet road=wall top=symmetry and because my model is symmetric, only 1/2 car is modelling.and so right side=symmetry left side=? this is my B.C. but it's not correct

April 26, 2012, 01:25
#6
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Lucky Tran
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Quote:
 Originally Posted by ma3ou2 tnx, but my B.C. model is: front=inlet velocity rear=pressure outlet road=wall top=symmetry and because my model is symmetric, only 1/2 car is modelling.and so right side=symmetry left side=? this is my B.C. but it's not correct
right that will no longer work. You will need to correct it. It is the fluid flow that needs to be symmetric, not the car.

You will have to rebuild the geometry and mesh. You will not be able to use any symmetry unless the front and rear of the car are symmetric. And even then this would only work if the crosswind is exactly from the side of the car. It is probably better to model the whole car.

 April 26, 2012, 01:27 #7 New Member   ma3ou2 Join Date: Apr 2012 Posts: 11 Rep Power: 6 my purpose is modelling a moving long car with crosswind condition. and i want to simulate moving car = airflow coming forward to front and crosswind flow to side. is it correct?

 April 26, 2012, 01:31 #8 New Member   ma3ou2 Join Date: Apr 2012 Posts: 11 Rep Power: 6 my model is a train that is symmetric,therefor i used a symmetric model car(train)

April 26, 2012, 02:19
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Quote:
 Originally Posted by ma3ou2 my model is a train that is symmetric,therefor i used a symmetric model car(train)
As I said before, the train model may be symmetric, but the flow may not be! When you apply a symmetry boundary condition, you are applying it to the flow and not the train. The flow could care less whether there is a train there or not.

Furthermore, consider that you have cross wind on one side, if symmetry condition is used, then you have an equivalent cross wind on the other side of the train. Then you have crosswind from two sides. It should be obvious that this is difficult to occur physically. It also means that you do not have crosswind from only one side.

If you want to simulate the crosswind, then you must throw away symmetry boundary along centerline. If you want to simulate oncoming wind and cross wind effects then you need to throw away mid-plane or meridional symmetry.

Quote:
 Originally Posted by ma3ou2 my purpose is modelling a moving long car with crosswind condition. and i want to simulate moving car = airflow coming forward to front and crosswind flow to side. is it correct?
The case of oncoming wind and then a purely crosswind condition is highly non-physical. Even if you ignored the realizability, it still can't be done with symmetric condition as described. It is really one wind.

For example, oncoming wind @ 4 m/s + crosswind @ 3 m/s is really a wind at an angle relative to the centerline of the train @ 5 m/s. This should not be too difficult to understand, it is grade school algebra.

It is highly suggested that you combine both the oncoming wind and crosswind into a single wind (since that is also what would occur naturally). You can apply velocity at an angle relative to the inlet and achieve this type of velocity. You will need two inlets and make sure that your velocities at each inlet are consistent with one another.

 April 26, 2012, 04:15 #10 Senior Member     Alex Pasic Join Date: Aug 2011 Location: Croatia Posts: 184 Rep Power: 8 Or, just rotate your model and keep the rectangular domain with the wind just coming directly at the vehicle? I think that's an even easier case to pull off. P.S. No centerline symmetries, as LuckyTran said do a whole vehicle model and rotate it by an angle which will depend on your headwind/velocity and crosswind magnitude (atan 3/4 etc)..

April 26, 2012, 04:24
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You can do it like the attached picture .... as the others mentioned you can't apply any symmetric BC
Attached Images
 corsswind.JPG (14.0 KB, 47 views)

 April 26, 2012, 04:47 #12 New Member   ma3ou2 Join Date: Apr 2012 Posts: 11 Rep Power: 6 tanx very much of all friends, and a question. with this method(rotate model) for calculate Drag,Lift,Side force&Moments what's Area&Lengh???

April 26, 2012, 04:51
#13
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Alex Pasic
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Quote:
 Originally Posted by ma3ou2 tanx very much of all friends, and a question. with this method(rotate model) for calculate Drag,Lift,Side force&Moments what's Area&Lengh???
To calculate drag/lift/side force you do not need reference values. You only need those for the coefficients of lift/drag etc. I don't really know what the general agreement is on projected surfaces when doing yaw cases, but you can either use the frontal area that you use in 0° yaw case and the length of the vehicle, or you can find the new projected area in Fluent via Reports -> Projected areas and there pick the walls that represent the vehicle and pick the direction X/Y/Z and decrease the Min feature size to 0.0001 m or something to get an accurate surface area projection.

April 26, 2012, 08:44
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Lucky Tran
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Quote:
 Originally Posted by scipy Or, just rotate your model and keep the rectangular domain with the wind just coming directly at the vehicle? I think that's an even easier case to pull off. P.S. No centerline symmetries, as LuckyTran said do a whole vehicle model and rotate it by an angle which will depend on your headwind/velocity and crosswind magnitude (atan 3/4 etc)..
If you have the time, I highly suggest you do it this way.

Quote:
 Originally Posted by scipy To calculate drag/lift/side force you do not need reference values. You only need those for the coefficients of lift/drag etc.
One more thing.

For coefficients you need the reference quantities for forces you do not need reference quantities. In both cases however, you do need a reference vector. With the rotated car model your reference vectors may not be aligned with the rectangular domain (depending on how you decide which direction lift and drag is).

I'm not familiar with what convention for reference length/area is appropriate in this case either for a long train. I would take projected frontal area or total frontal area. Length could be either the physical length or width of the train depending on interpretation.

April 26, 2012, 08:44
#15
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ma3ou2
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if i have use this B.C.
of course velocity inlet1=velocity inlet2=v relative
(side+oncoming flow air)
is it correct???
noticed that domain is not square
Attached Images
 333.jpg (24.9 KB, 31 views)

April 26, 2012, 08:48
#16
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Lucky Tran
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Quote:
 Originally Posted by ma3ou2 if i have use this B.C. of course velocity inlet1=velocity inlet2=v relative (side+oncoming flow air) is it correct??? noticed that domain is not square
both should have the complete relative velocity. this is again referring to the "it is really one wind" statement I made earlier.

Try to imagine the wind coming from two ways, how can wind do that?

Also if you try to run the case as is, your domain extends are probably restricting the flow and you may encounter convergence difficulties or inaccurate solutions. You need to move your inlets and outlets further away from the vehicle.

April 26, 2012, 08:57
#17
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ma3ou2
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Quote:
 Originally Posted by LuckyTran both should have the complete relative velocity. this is again referring to the "it is really one wind" statement I made earlier. Try to imagine the wind coming from two ways, how can wind do that? Also if you try to run the case as is, your domain extends are probably restricting the flow and you may encounter convergence difficulties or inaccurate solutions. You need to move your inlets and outlets further away from the vehicle.
therefore the best choice is a rectangular domain with rotating model train?

April 26, 2012, 09:07
#18
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Quote:
 Originally Posted by ma3ou2 therefore the best choice is a rectangular domain with rotating model train?
If you define both inlets to have the same oncoming + crosswind velocity correctly then they are essentially the same whether or not you rotate the model train. However, now your velocity inlet is too close to the train and may produce wrong results. If you are desperately short on time and need it done really fast this is what I would recommend as you do not need to redo the meshing.

However I would still recommend the setup that Zigainer showed. It is much better. Also note, even with this approach, I still would recommend that the width of the domain be extended slightly. With the purely oncoming wind the lateral spreading of the influence of the car is symmetrical and therefore small. With the crosswind there is a chance for more lateral effects as the flow may have a tendency to turn towards the train.

Last edited by LuckyTran; April 26, 2012 at 16:52.

April 26, 2012, 16:15
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Quote:
 Originally Posted by LuckyTran Also note, even with this approach, I still would recommend that the width of the domain be extended slightly.
My draft is not scaled correctly, itīs rather to show what I mean with rotated car.
As LuckyTran mentioned be sure that there is enough space/mesh for your wind to develop. If itīs to small, the wall will affect your flow over your car. So make it wide enough.
Your "approach" with two inlets and two outlets is not really good. Could be hard to get good results. Just imagine that at the corner your inlet is just next to the outlet and thats a thing you do not want! So go for the rotated car/train with a wide enough surrounding. It would be better if itīs too wide than too mall (of course you can use a coarser grid far away of the car and therefore you cell number should not be increased to much).

 April 27, 2012, 01:43 #20 New Member   ma3ou2 Join Date: Apr 2012 Posts: 11 Rep Power: 6 can u tell me which one of 2 method is the best? 1-moving train body + crosswind 2-stationary train body + relative wind(cross+oncoming wind) if 2 method give 1conclusion ?

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