- #1
guitarstorm
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Homework Statement
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]
Homework Equations
[itex]m=\rho _{p}\frac{\pi }{6}D_{p}^{3}n_{n}(D_{p})[/itex]
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...