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Old   June 13, 2016, 03:52
Default number density size distribution
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Alaska1964
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Dear All,

I have a question regarding particle size distribution, I would be grateful if you could suggest me some ideas about solving it.

I have obtained a plot of frequency distribution (q3(xi)) versus particle size (xi) which was measured by laser diffraction method. I want to get number density distribution from this plot. How can I get this?
I want to know how how many particles per volume are in each particle size interval and plot it versus the diameter.

If you could suggest me some sources regarding this problem to study, I would be so thankful.

Kind regards,
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Old   June 13, 2016, 11:52
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Christopher
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When you say frequency distribution, I hear 'number distribution'. The frequency with which particle diameters are distributed is a number based distribution. I do not understand the exact definition of the output data from this laser diffraction method, so it may be an issue of semantics(word choice). Can you provide more detail about the data that you have?


For reference, here is an article that discusses the difference between number weighted distributions and volume weighted distributions.

http://www.horiba.com/us/en/scientif...distributions/
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Old   June 14, 2016, 02:39
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The data I get is from HELOS from Sympatec GmbH. The data are volume based since they are presented by Q3(xi) and q3*.

I am showing an example here which is very close to my data. I would like to get the number density distribution of these data. I want to learn the number of the particles in each size interval.

My idea is this:
from this eqn ,ΔQ3(xi)=xiłNi ∕ΣxiNi, for each size interval by using the midpoint data I can extract a linear equation. at the end to the number of the intervals I have equations and the same amount unknowns. Then I solve asystem of n linear equations in n unknowns.
Ni is the number of the particles in i th size interval.
Now if I plot Ni-dp graph I have the number density distribution.


Is it correct?

Last edited by Alaska1964; June 14, 2016 at 02:46. Reason: insert image
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Old   June 14, 2016, 02:47
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My Example:
http://images.google.de/imgres?imgur...=1049&biw=2144

this link shows the table and plot of my example. I couldn't put it here. But in the link you can see the table includes ΔQ3(xi). and xi is the midpoint of the interval.
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Old   June 14, 2016, 13:34
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Christopher
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I think you are on the right track. From that webpage it looks like Q3(xi) is the CDF and the q3*(xi) is the PDF distribution. If you have the data for the PDF histogram(q3*(xi)), then I would imagine that you can just take the value of the vertical histogram quantity and multiply it by the cube-root of the x-axis coordinate of the histogram.


Going forward from number density to volume would be done as follows:

q_vol(xi) = xi^3*q_num(xi) . Then properly normalize the new PDF to have an area of 1 by dividing each point by ΣΔxi*q_vol(xi).

I'm currently working on a very similar problem for a simulation, so this is how I went about going from a number distribution to a volume distribution that I needed. The only difference that I see between your method and mine is that the denominator used to normalize Q3(xi) to have an area of 1 is different.
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Old   June 15, 2016, 10:15
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The data I get is volume based with the index of 3!

q3*(xi)=ΔQ3(xi) / (log xi−log xi-1) which is
q3(xi)=ΔQ3(xi) / (xi-xi-1),

ΔQ3(xi)= volume fraction=mass fraction in the interval of [xi-1, xi]
ΔQ3(xi)=mi/mtotal-injection ; mtotal-injection is known. so I can have mi which is the mass in the ith size interval.

massi = (ρparticles)(volumeith-size-interval)(Nith)=ρVithNith=ρ(π/6)[(xi+xi-1)/2]łNith

where Nith is the number of the particles in the ith size interval. So now we have the number of particles in different intervals. Therefore, we can get the Q0(xi) and q0(xi) easily.
q0(xi) is called number density distribution what I was looking for. q0(xi) shows the number of the particles in unite volume in ith size range.

Please let me know your idea if this approach is correct.

Cheers
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