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Viscosity of a falling sphere |
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| Nov16-12, 09:50 AM | #1 |
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Viscosity of a falling sphere
Hello,
I have recently conducted an experiment to measure the viscosity of some liquids using the falling sphere method and a high speed camera. For most of the measurements this is fine since the spheres reach terminal velocity and so it is easy to calculate the viscosity from: [itex]v = \frac{2}{9}\frac{r^2 g (\rho_p - \rho_f)}{\mu},[/itex] where: [itex]v[/itex] is the particles' terminal velocity velocity (m/s), [itex]r[/itex] is the radius of the sphere, [itex]g[/itex] is the gravitational acceleration, [itex]\rho_p[/itex] is the density of the falling sphere, [itex]\rho_f[/itex] is the density of the liquid, and μ is the viscosity. However, in one or two of the measurements the sphere didn't reach terminal velocity, though it is possible to calculate the acceleration. My question is, is it possible to calculate the viscosity from knowing only the acceleration of the sphere (by perhaps comparing it to the gravitational acceleration in a particular medium for example)? Thanks in advance for your help. |
| Nov16-12, 12:53 PM | #2 |
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| Nov19-12, 05:01 PM | #3 |
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| Nov19-12, 05:34 PM | #4 |
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Viscosity of a falling sphere
Whoa whoa whoa. Keep in mind this is the equation for Stokes flow, meaning that it only applies to flows with [itex]\text{Re}_D \ll 1[/itex]. That is absolutely a problem here since your Reynolds numbers are almost certainly not low enough.
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| Nov19-12, 07:19 PM | #5 |
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| Nov19-12, 07:30 PM | #6 |
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| Nov19-12, 08:44 PM | #7 |
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| Nov19-12, 09:02 PM | #8 |
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| Nov19-12, 09:32 PM | #9 |
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| Nov19-12, 10:02 PM | #10 |
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| Nov20-12, 05:34 AM | #11 |
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| Nov20-12, 08:14 AM | #12 |
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Yeah looks like you pretty much have your bases covered on the Reynolds number side. I don't know what the density of your silicate melt is, but it seems like your formula will most likely work.
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