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the droplet diameter according to the rate eq. (implicitly)

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Old   September 13, 2011, 08:56
Default the droplet diameter according to the rate eq. (implicitly)
Join Date: May 2010
Location: munich
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Hello everyone,

I have a question about the droplet diameter rate equation in the breakup model:

In many paper this rate equation is like following: dr/dt=-(r-rc)/t_brekup

But in the breakup models e.g. PilchErdman and ReitzDiawakar the rate equation is implemented as following:

d = (d + frac*Ds)/(1.0 + frac)

I do not know why it is so implemented. In my opinion it should be like following:

d = d - d*frac + frac*Ds

Does anyone also meet this problem.

Thanks in advance
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