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Unit problem with calculating the mass concentration of a particle

  1. Nov 11, 2011 #1
    1. The problem statement, all variables and given/known data

    We have to calculate the mass concentration for a particle size distribution. I'm just doing the first size range here, since once I get that the rest is just repetition.

    FYI, the particles are spherical.

    Size range (diameter) = 0.005-0.5 μm
    Average particle size (diameter) = 0.25 μm
    Number concentration = 1 cm[itex]^{-3}[/itex]
    Particle density = 1800 kg m[itex]^{-3}[/itex]

    The answer is supposed to be in units μg m[itex]^{-3}[/itex] μm[itex]^{-1}[/itex]

    2. Relevant equations

    [itex]m=\rho _{p}\frac{\pi }{6}D_{p}^{3}n_{n}(D_{p})[/itex]

    3. The attempt at a solution

    First thing I did was convert the density from kg m[itex]^{-3}[/itex] to μg m[itex]^{-3}[/itex]:

    [itex]\frac{1800 \: kg}{m^{3}} * \frac{1\: \mu g}{1*10^{9\: }kg}=\frac{1.8*10^{12}\: \mu g}{m^{3}}[/itex]

    Then I plugged into the mass concentration formula:

    [itex](1.8*10^{12}\mu g\: m^{-3})(\frac{\pi }{6})(0.25\: \mu m)^{3}(1\: cm^{-3})=1.473*10^{10}\mu g\: m^{-3\: }\mu m^{3}\: cm^{-3}[/itex]

    So now, since there's no cm in the units we're supposed to have, I figured I'd get rid of that:

    [itex](1.473*10^{10}\: \frac{\mu g\: \mu m^{3}}{m^{3}\: cm^{3}})*(\frac{100\: cm}{1\: m})^{3}=1.473*10^{16}\: \frac{\mu g\: \mu m^{3}}{m^{6}}[/itex]

    So now we have this mess, and at this point I give up, because idk how the heck to get this in μg m[itex]^{-3}[/itex] μm[itex]^{-1}[/itex]... Not to mention the number I'm getting seems way too large...
  2. jcsd
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