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August 23, 2011, 04:57 
Bug in the PilchErdman Breakup Model

#1 
Member
chenkai
Join Date: May 2010
Location: munich
Posts: 44
Rep Power: 8 
Hi everyone,
I have notice, there is a bug in the PilchErdman Breakup Model in OpenFOAM. In the source code of the Model, the following two equ., scalar Vd = Urmag*rho12*(B1_*taubBar * B2_*taubBar*taubBar); ...... scalar Ds = 2.0*Wec*sigma*Vd1/(Vd1*rhoc*sqr(Urmag)); which are not the same as in the paper: Use of breakup time data and velocity history date to predict the maximun size of stable fragments for acceleration induced breakup of a liquid drop. I think it should be rewritten like following: scalar Vd = Urmag*rho12*(B1_*taubBar + B2_*taubBar*taubBar); ...... scalar Ds = 2.0*Wec*sigma/(Vd1*rhoc*sqr(Urmag)); Does anyone also notice this error. Thanks Chenkai 

August 30, 2011, 04:59 

#2 
Member
chenkai
Join Date: May 2010
Location: munich
Posts: 44
Rep Power: 8 
Nobody has ever used this model?
The problem is that, when I change the code of this model as in the paper. My Simulation results are even worse than before. I don't if here is really a bug or not. Hope someone can help me? Thanks a lot! 

January 16, 2013, 08:08 

#3 
New Member
Join Date: Jun 2012
Posts: 14
Rep Power: 6 
A year late but this post already exits so im updating it here.
Im looking at this model currently for my own purposes. You are correct, the model should be what you suggest as a fix i.e. scalar Vd = Urmag*rho12*(B1_*taubBar + B2_*taubBar*taubBar); ...... scalar Ds = 2.0*Wec*sigma/(Vd1*rhoc*sqr(Urmag)); Also, // update the droplet diameter according to the rate eq. (implicitly) d = (d + frac*Ds)/(1.0 + frac); makes no sense whatsoever. It should be d = d*(1.0frac)+Ds*frac when im less busy ill update the git repository if someone else don't get to it first. 

January 16, 2013, 14:28 

#4  
Super Moderator
Niklas Nordin
Join Date: Mar 2009
Location: Stockholm, Sweden
Posts: 693
Rep Power: 22 
Quote:
I cant belieave I've missed this one. I will talk to the guys and correct it. About the 'makes no sense whatsoever' part. You could do it like that. Its not wrong, but neither is the implemented update. (as shown below) f = dt/tau we're trying to solve this eq: dD/dt = (DDs)/tau 1) explicit, approximate D with D0 in the RHS (D1D0)/dt = (D0Ds)/tau D1 – D0 = (D0 – Ds)*f D1 = D0  D0*f + Ds*f D1 = (1f)*D0 + Ds*f 2) implicit, approximate D with D1 in the RHS (D1D0)/dt = (D1 – Ds)/tau (D1D0)= D1*f + Ds*f (1+f)*D1  D0 = Ds*f (1+f)*D1 = D0 + Ds*f D1 = (D0 + f*Ds)/(1+f) 

January 16, 2013, 15:29 

#5 
Super Moderator
Niklas Nordin
Join Date: Mar 2009
Location: Stockholm, Sweden
Posts: 693
Rep Power: 22 
now...you may wonder 'why is method 2 to prefer'?
think of large timesteps (f >> 1) first method yields D1 > f*(DsD0) (makes no physical sense) second method yields D1 > f*Ds/f = Ds (makes perfect sense) 

January 16, 2013, 18:07 

#6 
New Member
Join Date: Jun 2012
Posts: 14
Rep Power: 6 
Ahhh, I see, thanks for reminding me of the two methods!


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