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Normalisation of Experimental Data

  1. Oct 25, 2015 #1
    Hi all,

    I am currently operating a piece of equipment that essentially collects particles and separates them based on their size. Essentially you have 8 stages, and each stage has a differing size of particles it collects. For example:

    Stage----Size of Particles (D) (um)-----Mass Collected (M) (mg)
    1-------------0.1 - 0.2------------------------------1
    3-------------0.5 - 1--------------------------------1
    4-------------1 - 2----------------------------------0.5
    5-------------2 - 4-----------------------------------1
    6-------------4 - 5----------------------------------0.5
    7-------------5 - 9------------------------------------3
    8-------------9 - 10----------------------------------2

    Now, as you can see each stage has a different range of sizes collected. Stage 1 has a "width" of 0.1 um, while stage 7 has a width of 4 um. Because of these different stage widths it is common to normalise the plot to make the results indepedent of stage width, like so:

    dM/dlogDp = dM/log(Du)-log(Dl)

    where Du is the upper stage width and Dl is the lower stage width (ex, for stage one Du is 0.2 and Dl is 0.1). If you then plot this data:


    Now, my question is, what exactly is the physical meaning of these new results. The initial results show me the mass of particles between 0.1 and 10 um collected after a certain time of experiment or some other factor. The total mass is 9.1 mg. But, these normalised results have much higher numbers. Do they have any inherent meaning other than being independent of stage width. Am I missing some fundamental calculus principle that imparts meaning to dM/dlogDp?

  2. jcsd
  3. Oct 25, 2015 #2

    Simon Bridge

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    Science Advisor
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    You can imagine that the dimensions of the catcher will affect the number of things caught ... imagine you are catching raindrops in a bucket, the bigger bucket tends to catch more rain. If you want to compare the amount of rain, then, you need to account for the size of the bucket
    In this case, it is sensible to divide the number of drops caught by the area of the bucket, getting rain per unit area ... and compare that: so we'd do ##dM/d\pi r^2## ... though not happy with that notation since we are actually doing ##M/\pi r^2##

    Your normalization procedure does something similar.
    To get what exactly ti is doing, you need to go into detail about what exactly is being counted and how.
  4. Oct 26, 2015 #3
    Hi Simon,

    Just to give you a bit of background info, I essentially draw air through a device that separates out particulates based on their inertia and therefore size. There are 8 stages, each separates out the particles in the size range given in my initial post. Now, we know the mass of air that passes through the device (let's say 10 grams), and we know the mass of particles collected on each stage, so what I then do is calculated the amount of particles in a given air of mass (mgparticles/gair), for example:

    Stage 1 mass collected = 1mg. 10 grams air drawn through. Therefore, 0.1mgparticles/1gramair. (Particles between 0.1-0.2 um)
    Stage 2 mass collected = 0.1mg. 10 grams air drawn through. Therefore, 0.01mgparticles/1gramair. (Particles between 0.2-0.5 um)
    etc etc.

    Now, typically the 8 stages are summed and we get total particles (mg) per gram of air. This doesn't require normalisation. However, if we want to to see which size range of particles are predominant, then we need to take the size range of each stage into account. That's where we do the normalisation outlined previously.

    I understand the principle of the normalisation. Indeed, how can you compare the mass collected in each stage when some stages collect a bigger range of particles than others?

    What I can't get my head around is how the two are directly comparable. If you look at my original post the total mass collected is 9.1 mg, or 0.91mgparticles/gair if we assume 10g of air again. dM/dlogDp has the same units as mg or mgparticles/gair (you can do normalisation for either), but the numbers are far higher (72.5mg in fact).

    So, ultimately, which one is correct? Are they directly comparable? Does the normalised have any physical meaning. Both each individual stage and the total mass of particles is much higher. If someone asked me "What mass of particles are in 10g of air?", would I tell them 9.1mg or 72.5mg?

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