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General Question: Inlet- Outlet Mesh - have to be simular?

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Old   July 5, 2011, 06:09
Default General Question: Inlet- Outlet Mesh - have to be simular?
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Hi guys,
iam kinda new in the whole cfd business, so here my question:

i was wondering how important it is, that inlet and outlet have the same meshing (and by meshing in mean: simular number of cells).
Is it at all important? My outlet got way more cells as my inlet (because i want to incert some particles and im really interested in where the particles "land" on the outlet screen) - was just wondering what and if that means anything for my results?!
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Old   July 5, 2011, 07:06
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Hi Andi,
The only important thing is that you have to achieve grid independent results.

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Old   July 5, 2011, 09:57
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And by that u mean, that i have to change the mesh-type (for example from size function on curvature, on proximity) to another method of meshing? or can i show grid independence also by changes the number of cells.

another general question that comes to mind (its not connected with the question above).
when i use symmetry on my tube-system, do i have to take only the half flow rate? that would also mean a lower velocity because of the half surface - so i guess not?!? i marked all symmetry-faces, so i thought (until now), that fluent is aware of that and handles it like "full". is that correct?
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Old   July 5, 2011, 15:32
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Quote:
Originally Posted by thisisit View Post
And by that u mean, that i have to change the mesh-type (for example from size function on curvature, on proximity) to another method of meshing? or can i show grid independence also by changes the number of cells.
it doesn't need to change grid topologies unless your grid's quality is not satisfactory. i.e. you can compare any 2 grids which have good qualities with different cells for grid study checking.

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another general question that comes to mind (its not connected with the question above).
when i use symmetry on my tube-system, do i have to take only the half flow rate? that would also mean a lower velocity because of the half surface - so i guess not?!? i marked all symmetry-faces, so i thought (until now), that fluent is aware of that and handles it like "full". is that correct?
when you implement symmetric condition over a surface, fluent just forces zero gradient normal to that boundary and nothing else, so you have to correct flow rates, but it doesn't lead to lower velocity because your surface is reduced! fluent is not very genius!
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Old   July 10, 2011, 07:07
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Yes that makes sense. velocity will not change.
when i calculate the turbulent intensity, do i have to use half-surface as a factor to calculate the Renolds-number as well? Re goes directly into the formular
I = 0.16(RE)^(-1/8)

its not topic related but i have another general question - and because you allready helped me that good, i ll try to post it in this thread :-)

Imagine a CLOSED simple tube-system where i have 1 inlet and 1 outlet.
how can i simulate the system beeing closed? i can asign 1 inlet (with my known velocity) and 1 pressure-outlet. but that doesnt fit my situation. i tryed to make the pressure-outlet a inlet with negative velocity, but that doesnt work and continuity should allready been considerated by fluent. i thought about making the pressure-outlet kind of a "pressure drop" - but i honestly dont know, how huge the drop is, because its not exactly a tube, its a little more complicated in the geometry. any suggestions?

Last edited by thisisit; July 10, 2011 at 07:55.
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Old   July 10, 2011, 08:18
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Quote:
Originally Posted by thisisit View Post
Yes that makes sense. velocity will not change.
when i calculate the turbulent intensity, do i have to use half-surface as a factor to calculate the Renolds-number as well? Re goes directly into the formular
I = 0.16(RE)^(-1/8)
you have to use full-surface for Re calculation, because this formula is derived for this purpose and it's dimensionless.

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Originally Posted by thisisit View Post
its not topic related but i have another general question - and because you allready helped me that good, i ll try to post it in this thread :-)

Imagine a CLOSED simple tube-system where i have 1 inlet and 1 outlet.
how can i simulate the system beeing closed? i can asign 1 inlet (with my known velocity) and 1 pressure-outlet. but that doesnt fit my situation. i tryed to make the pressure-outlet a inlet with negative velocity, but that doesnt work and continuity should allready been considerated by fluent. i thought about making the pressure-outlet kind of a "pressure drop" - but i honestly dont know, how huge the drop is, because its not exactly a tube, its a little more complicated in the geometry. any suggestions?
Honestly, I didn't understand what you mean?!! you have closed system with some open boundaries! put a image here.
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Old   July 10, 2011, 12:44
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just imagine a simple tube as a part of a closed system (by that i mean circular flow). i take 1 part of it and select inlet and outlet as named surfaces. then i put in the velocity at the inlet. the outlet i select as pressure-outlet (with no other options seleceted).
i believe that this system is simplified (by my settings). so from inlet there is fluid that flows out at the outlet (by in reality, it doenst flow out, its still a part of the circular flow). i dont know, if i loose information by just selection the outlet as simple pressure-outlet.

sorry i cant post a picture.
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Old   July 10, 2011, 14:54
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Quote:
Originally Posted by thisisit View Post
just imagine a simple tube as a part of a closed system (by that i mean circular flow). i take 1 part of it and select inlet and outlet as named surfaces. then i put in the velocity at the inlet. the outlet i select as pressure-outlet (with no other options seleceted).
i believe that this system is simplified (by my settings). so from inlet there is fluid that flows out at the outlet (by in reality, it doenst flow out, its still a part of the circular flow). i dont know, if i loose information by just selection the outlet as simple pressure-outlet.

sorry i cant post a picture.
I think that periodic boundary condition is what you need.
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Old   July 11, 2011, 11:26
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Sorry for my poor discription, let me try it again.
I want to simulate suction at the outlet. What i think is , that a simple pressure-outlet isnt enough - i tryed to make the outlet a inlet with negative velocity, but that doesnt really work, i guess.
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