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May 5, 2012, 15:59 |
interfoam BC
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#1 |
Senior Member
A_R
Join Date: Jun 2009
Posts: 122
Rep Power: 17 |
Dear Foamers
I want to simulate a single cylinder in 2 phase flow with interface i have tested many conditions but i cannot give regular reason please help me my answer and 0/ folder is attached |
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May 5, 2012, 17:51 |
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#2 |
Senior Member
Nima Samkhaniani
Join Date: Sep 2009
Location: Tehran, Iran
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your problem definition is vague
put the whole case! at least users can see each boundary condition name where it is! however it seems you want to model a cylinder one phase is stagnant and the other phase has the velocity of 1 m/s. first as you use p-rgh, i guess there is no difference between bouyant pressure and zeroGradient second the answer is some how reasonable, as you put inlet1 fixedvalue zero! there is no input mass for phase 1 so gradually, the phase 1 will leave the domain and the level of phase 1 will be reduced |
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May 6, 2012, 00:57 |
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#3 |
Senior Member
A_R
Join Date: Jun 2009
Posts: 122
Rep Power: 17 |
Dear Nima
thanks for your reply. yes you guess truly. I send address of my test case. I want to have stable atmosphere on the top of my domain so I use zero as value for inlet. But if you look more carefully, it seems that the second phase mass want to leave the domain like a dam that is broken. I want to save two phases in my domain. http://amirms81.persiangig.com/damBreak.zip |
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June 15, 2012, 06:39 |
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#4 |
Senior Member
Albrecht vBoetticher
Join Date: Aug 2010
Location: Zürich, Swizerland
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Hello A_R, have you tried to switch your outlet and up boundary conditions in the p_rgh file, e. g. buoyantPressure at the outlet and atmosphere type totalPressure at up, together with type inletOutlet; inletValue uniform 0; value uniform 0; for alpha1 at up?
Or, try the following: Move the mesh from positive to the negative x-coordinate quadrant, that has funny influence on the outlet BC! |
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June 15, 2012, 16:01 |
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#5 | |
Senior Member
A_R
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Quote:
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July 2, 2012, 20:07 |
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#6 |
Member
Amin
Join Date: Mar 2012
Posts: 60
Rep Power: 14 |
Dear A_R
I'm working in free surface like you, and i have problems with your simulate yet. i attached my physical domain picture.for example in my case: Re=180, Fr=.6, D=2m if you can please help 1. what's the BC for alpha, U, P in each boundary? (inlet_air, inlet_water, outlet air, outlet water, up, down, cylinder) 2. you mentioned your simulation became fine, can i know your physical conditions? 3. if you validate your results(Cd, Cl, wave pattern) with any article please inform us. Regards Amin |
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July 3, 2012, 05:13 |
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#7 |
Senior Member
Albrecht vBoetticher
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Location: Zürich, Swizerland
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Hi Amin, will you run only 2D cases? I assume you only model laminar flow with Re=180, so you will not have to account for turbulence with free surface flow and complex geometry. A nice discussion about BC can be found here: http://www.cfd-online.com/Forums/ope...interfoam.html
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July 3, 2012, 09:15 |
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#8 |
Member
Amin
Join Date: Mar 2012
Posts: 60
Rep Power: 14 |
Hi Albercht
tnx for your answer. yes, for now i want to simulate 2D,laminar. after obtain good results i have to change my model to turbulent and 3D. i prefer to use default properties for water and air. with D=2m, rho=1000, nu=10^-6 for Re=180, we should set U=0.00009 m/s. can we use this very low velocity? does it give us reasonable results in OF? also with this velocity Fr will be 2.03*10^-5 so i can't obtain Fr=.6 !!! How can i balance default properties, Re, Fr? my mean is how can we have favorite Re,Fr and default properties all together? which is more important to set first? regards Last edited by amin66; July 3, 2012 at 10:01. |
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July 3, 2012, 11:30 |
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#9 |
Senior Member
Albrecht vBoetticher
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Hi Amin, I expect you define your Froude number using a characteristic length. How did you define your characteristic length that Fr equals 2.03*10^-5?
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July 3, 2012, 15:30 |
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#10 |
Member
Amin
Join Date: Mar 2012
Posts: 60
Rep Power: 14 |
Hi Albercht
characteristic length is equal to cylinder diameter=2m . |
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July 4, 2012, 04:42 |
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#11 |
Senior Member
Albrecht vBoetticher
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Location: Zürich, Swizerland
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Well, if you use Fr = v/(gL)^0.5 and Re = vL/nu the ratio between Re and Fr will always be the same independent of the velocity. How were the values of your case defined, at the undisturbed channel or at the cylinder? I would work with a 3D model in a size, defining Re using the hydraulic radius of the channel and definig Fr using the flow depth as characteristic length. You will not be able to increase Fr, but at least you can have higher velocities, for example with depth and width 4m you can get about 0.00016 m/s with Re 187 in the channel. However, is your focus on the discharge of the channel or on the forces at the cilynder? With the mentioned setup, waves will propagate against the flow much faster than with Fr = 0.6 but both are far from critical discharge. If Fr and Re are determined at the cylinder, with flow depth = water depth at cylinder - diameter, the situation is different and you can get higher velocities and Fr for Re 180.
Last edited by vonboett; July 4, 2012 at 06:40. |
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July 4, 2012, 08:02 |
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#12 |
Member
Amin
Join Date: Mar 2012
Posts: 60
Rep Power: 14 |
ok tnx, let me to clear my problem:
it's about wave making resistance(wave drag) of a submerged body(2D,3D) as shown in post #6. this bodies for example are under the free surface of sea, but for simulation i think i have to use open channel flow and i think InterFoam is the best for my problem. am i right? i want to obtain drag force and Cd of this bodies for different submergence depth and Fr numbers. so the Cd is more important to achieve. i start with laminar for now and then will shift to turbulent model. to calculate Re(Re = vd/nu) and Fr( Fr = v/(gd)^0.5) using the d(cylinder diameter) and properties of water. |
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July 5, 2012, 04:36 |
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#13 |
Senior Member
Albrecht vBoetticher
Join Date: Aug 2010
Location: Zürich, Swizerland
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Hi Amin,
I see. InterFoam is good if your body is fixed in its position. But if your body will have the possibility to move due to drag, I strongly recommend interDyMFoam to account for the interaction. If you look at the floatingObject Tutorial, yust be aware what is discussed here. http://www.cfd-online.com/Forums/ope...-tutorial.html Since it is about wave drag, I would give higher priority to Fr than to Re, and introduce turbulence as very next step. I think Re can go up to 10^5 allowing the neglection of turbulence in a first step. |
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July 5, 2012, 11:45 |
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#14 |
Member
Amin
Join Date: Mar 2012
Posts: 60
Rep Power: 14 |
Hi Albercht
yes my case is fixed. i used other BC and my simulation is better now. i say it's better because the water level don't decrease at outlet as dear A_R said before. now i can see the wave on free surface and vortex shedding behind the cylinder. but unfortunately i dont have any source yet to validate Cd with my results. guys that work on free surface can look at this article: "Flow past a cylinder close to a free surface" By P. REICHL, K. HOURIGAN AND M.C. THOMPSON it's helpful but discussed just about wakes. the BCs taht i used, are similar the new solver (LTSInterfoam) in OF 2. i use OF 1.6 yet!!! thanks very much for your attentions. |
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July 5, 2012, 12:32 |
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#15 |
Senior Member
Albrecht vBoetticher
Join Date: Aug 2010
Location: Zürich, Swizerland
Posts: 240
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Hi Amin, cool. What Fr and Re did you use?
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July 5, 2012, 17:47 |
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#16 |
Member
Amin
Join Date: Mar 2012
Posts: 60
Rep Power: 14 |
Albercht, I used u=2 m/s and d=2m ,so fr=0.45 and Re=4*10^6 !!! and it's funny that i used laminar model.
yes i know it's far from reality, but i just wanted to obtain BC for now. after that i wanted to simulate with Fr=0.6 and Re=180 so i scaled my mesh with gambit to mm, for that mesh diameter became 2mm and i had to use u=.09 m/s to reach Fr=0.6 and Re=180. but unfortunately for every 0.05 pass time my laptop was working about 20min. so i left that till i run with a powerful PC. i will inform you my result after that. |
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October 6, 2024, 14:01 |
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#17 |
New Member
John Philip
Join Date: Mar 2024
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Did you get the answer for your question. I am also stuck with the similar question
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October 6, 2024, 14:36 |
dount
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#18 | |
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John Philip
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October 6, 2024, 14:38 |
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#19 | ||
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John Philip
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