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Rainfall drop velocity from a given height (not terminal!)

  1. Apr 4, 2012 #1
    hi everybody,

    I posted this in an engineering forum but I think it's more relevant here, because it's really just a question of fluid and newtonian mechanics.

    I'm working on a project where I'm trying to measure raindrop parameters, and one thing I'm looking at is the sub-terminal speed of drops released from a certain height. The equation that I'm using is from http://staff.science.uva.nl/~jboxel/Publications/PDFs/Gent_98.pdf

    The gist of the equation that I was considering is:

    F = g*ρw*∏*d^3/6 - 3*∏*d*μ*V*Ct*Cd

    where Ct = 1+0.16*Re^(2/3)

    and Re = ρVD/μ;

    and Cd = 1+a(We+b)^c - ab^c

    where a,b,c are empirically derived constants and We = ρ*V^2*d/σ

    Basically, when I put everything together and try to calculate fall velocity, I get stuck with a disgusting integral, because I use

    V(t)=∫a(t) = (1/m)*∫F(t)

    Does anybody have suggestions for how to approach this? I just want to make a model in matlab.. it seems like I could do some kind of step approach, because I looked at the integral and it's really nasty, but I don't know what to do, because I have V(t) on both sides...

    Or if anybody knows of a simpler model presented in a paper, I could use that too. I just want to compare my data with a preexisting model; it's not critical to my project, but I think it's important.
  2. jcsd
  3. Apr 11, 2012 #2
    As I see it you need to write it in the from of a differential equation and Matlab will solve it (numericaly or otherwise):
    Sum( Force( V(t) ) ) = mV'(t)
    i.e. the sum of all the forces acting on the drop (drag ect.) - you need to write the those forces as a function of the drops velocity, equals the mass of the drop times acceleration (derivative of velocity with time).


  4. Apr 12, 2012 #3


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    The most obvious numerical solutions might accumulate significant errors.
    If you're interested in the way it approaches terminal velocity, you should try working with the dependent variable being the difference between V and Vt. You might then be able to make suitable approximations to obtain an analytic solution for the asymptotic behaviour.
    But it would require knowing the values of the constants and figuring out what terms can be ignored.
  5. Apr 12, 2012 #4
    Sweet, i think I figured it oUt. Thanks
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