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July 17, 2013, 04:08 
Droplet Incident Angle

#1 
New Member
David
Join Date: Oct 2012
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Good morning,
I'm working in the BAI's impingement model on droplets and I'm trying to create my own rebound model but I met one problem. I would like to define a restitution coefficient for my rebound model but I really don't see how can I obtain the incident angle of my droplet in the subroutine DROWBC. If I understand well the code sample in DROWBC, there is a parameter called S1 which represent the cosinus of the angle between cells directions and the global x,y and z directions. But it's not my droplet velocity incident angle. How could I obtain the incident angle of my droplet velocity? Can someone shed any light on this? 

July 17, 2013, 11:16 

#2 
Senior Member
Pauli
Join Date: Mar 2009
Posts: 185
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It sounds like you have enough information to compute the boundary face normal. Combine that with the droplet velocity vector components & you should be able to compute incident angle. Or am I missing something?


July 18, 2013, 05:25 

#3 
New Member
David
Join Date: Oct 2012
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Thank you. In fact, it's like if the programm makes believe that the cell is oriented by an angle even if it's not. So that angle defines effectively the incident angle of the droplet velocity vector. It's just a play on points of view.
I don't know if I'm clear but I have the solution of my problem thank you. 

July 24, 2013, 08:46 

#4 
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David
Join Date: Oct 2012
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Me again !
I'm sorry Pauli I thought I found my solution alone but after few trials I realized that the solution I found doesn't represent my incident angle... So I'll try to explain my approach : In the subroutine DROWBC.f, there is an example to how compute the droplet velocity after rebound. In this example, there are 3 parameters called S1,S2 and S3 which represent the outwardpointing area components of the cell face in the global x, y and z directions (m2) and the AREA ratio with AREA = SQRT(SBX²+SBY²+SBZ²). The mistake I did is simply that to understand well this, I made a drawing in 2D of this. In 2D, in a basic drawing of a cell, S1 and S2 correspond to my sinus and cosinus of the angle between my global x,y and z directions and my cell's local directions. But I'm computing a rebound model in 3D so in that case, S1 and S2 don't represent a sinus or a cosinus. Here is the approach I used (but it's not correct obviously): I calculate the incident droplet velocity magnitude as VDMAG = SQRT(UD²+VD²+WD²) Then I use the variable used in DROWBC which represents (I think) the incident droplet velocity magnitude on the normal direction of my cell: VNMAG = UD.S1 + VD.S2 + WD.S3 As I'm considering a plane rebound, I just need to calculate the angle between VDMAG and VNMAG using the the function ACOS(). I did it but to have a proof of this, I imposed a 10°incidentangle on my wall and my solver finds an incident angle of 22.7°. I don't understand what I did wrong... Any suggestions to help me ? Thank you in advance. 

July 24, 2013, 10:19 

#5 
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Pauli
Join Date: Mar 2009
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I believe you need to compute the angle between the vector defining the cell face normal (you report this as S1, S2, and S3) and the droplet velocity vector.
Do not use the plane geometry formulation, use the "long" formula which computes the angle between two vectors in 3D (http://www.wikihow.com/FindtheAngl...enTwoVectors). The result should be the angle between the cell face normal and the droplet velocity vector. Depending upon how you define incident angle, you may need to subtract the computed value from 90 degrees. 

July 24, 2013, 10:20 

#6 
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David
Join Date: Oct 2012
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After few researches, I maybe found my mistake. Indeed, when I look at my incident droplet velocity components (UD, VD and WD), I should fall on my incident angle that I imposed at the pipe exit (10°). Whereas, when I do this check, I don't find it so the only explaination possible is that my incident angle is not well defined.
I find : UD = 1.8459, VD = 1.1095e4, WD = 4.3926 The angle between UD and WD is 22.79° and not 10°. When I defined it in proSTAR, I fill in in the Injection Points pannel :  Injection Points panel  Selected option Single Point from the Set Type menu  Entered 1 in the Point Coordinate System and Injection Direction Coordinate System boxes  Entered 1 in the Number of Points box  Selected option Specify for Point and entered its coordinates (30, 0, 2674) in the X, Y, Z boxes  Specified the entrance velocity direction in terms of unit vector components by entering (0.173648, 0, 0.984808) in the DX, DY, DZ boxes to have an incident angle of 10°.  Ensured Add New Set is selected from the Action menu and clicked Apply. Thank you in advance for your help, 

July 24, 2013, 10:30 

#7 
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David
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Hello Pauli,
Thank you for responding. The solution that you propose is the approach I used. Indeed, the variable VNMAG is exactly the normal vector of my cell (I'm quite sure when I look my drawing). Then considering that 2 vectors are in the same plane, I can use directly the function cosinus to find my angle. I already tried to use the "long" definition in 3D, scalarproduct of 2 vectors but I don't have the components of the normal vector of my cell. Deadend for me with this way. Did you see my latest post? Do you think I didn't define very well my incident angle ? 

July 25, 2013, 09:37 

#8 
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Pauli
Join Date: Mar 2009
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Looks like you defined it correctly. Therefore your problem is likely correctly interpreting the output. Have you made a droplet plot & graphically viewed what is happening?
Didn't you say S1 is the cell face direction cosines? What is the normal vector for the surface you are spraying against? 

July 25, 2013, 11:07 

#9 
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David
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Hi Pauli,
I made a droplet plot and also a particle tracking to see if there is an angle of 10° at the exit of my pipe. This check gave me an incident angle of 10° so the injection is correctly modelised. You are right, S1 is the cell face direction cosinus but only in one of my drawing which is unfortunately a particular case of my problem so I can't say that S1 is always the cell face direction cosinus. I've got news about VNMAG, it's obviously my boudary normal direction. I found something intersting. I forgot that my case in running with the gravity effect and as I am shouting out droplets on a ceiling, gravity plays a crucial part in the droplet velocity direction and by consequence in the incident angle. I decided to put my pipe close too my ceiling to check that and ... as if by magic, I found an incident angle of 10° between my boundary normal direction and my droplet velocity vector. So I think my problem is now solved... Thank you very much Pauli for your interest and your patience. 

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