## 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:

$v = \frac{2}{9}\frac{r^2 g (\rho_p - \rho_f)}{\mu},$

where:
$v$ is the particles' terminal velocity velocity (m/s),
$r$ is the radius of the sphere,
$g$ is the gravitational acceleration,
$\rho_p$ is the density of the falling sphere,
$\rho_f$ 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)?

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 Quote by Polyamorph 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: $v = \frac{2}{9}\frac{r^2 g (\rho_p - \rho_f)}{\mu},$ where: $v$ is the particles' terminal velocity velocity (m/s), $r$ is the radius of the sphere, $g$ is the gravitational acceleration, $\rho_p$ is the density of the falling sphere, $\rho_f$ 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.
Yes, but it's an approximation. From your equation above, you can show that the drag force on the sphere is given by F = 6πrμv. This applies to a sphere moving at a constant velocity. But, if you assume that, to a first approximation, it also describes the drag force during the final stages of the deceleration, then you can write a force balance on the sphere, and use Newton's second law to solve for the velocity as a function of time. This velocity variation will depend on the viscosity. You can plot the velocity as a function of time for various values of the viscosity, and find the curve that best matches your data.

 Quote by Chestermiller Yes, but it's an approximation. From your equation above, you can show that the drag force on the sphere is given by F = 6πrμv. This applies to a sphere moving at a constant velocity. But, if you assume that, to a first approximation, it also describes the drag force during the final stages of the deceleration, then you can write a force balance on the sphere, and use Newton's second law to solve for the velocity as a function of time. This velocity variation will depend on the viscosity. You can plot the velocity as a function of time for various values of the viscosity, and find the curve that best matches your data.
Thanks. I've done this, solving Newton's 2nd law to find v as a function of time, taking into account gravitational force, bouyancy and drag force. The resulting function v(t) can then be fitted to the experimental data using a least square fit to find the optimum η. My remaining question the is regarding Stokes law F = 6πrμv, is this really only valid at constant velocity? Because the only time a falling sphere is at constant velocity is when it is at terminal velocity. During its fall it is subject to the velocity dependent drag force. Is this not F = 6πrμv ?

## 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 $\text{Re}_D \ll 1$. That is absolutely a problem here since your Reynolds numbers are almost certainly not low enough.

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 Quote by boneh3ad Whoa whoa whoa. Keep in mind this is the equation for Stokes flow, meaning that it only applies to flows with $\text{Re}_D \ll 1$. That is absolutely a problem here since your Reynolds numbers are almost certainly not low enough.
Whoa, whoa, whoa, whoa. A standard viscometer for very viscous fluids is the falling ball viscometer. Yes, it applies when the Reynolds number is very low. Typically, the viscosity of the fluid will be on the order of 100's to 1000's of poise, and the ball will only be on the order of a few mm.

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 Quote by Polyamorph Thanks. I've done this, solving Newton's 2nd law to find v as a function of time, taking into account gravitational force, bouyancy and drag force. The resulting function v(t) can then be fitted to the experimental data using a least square fit to find the optimum η. My remaining question the is regarding Stokes law F = 6πrμv, is this really only valid at constant velocity? Because the only time a falling sphere is at constant velocity is when it is at terminal velocity. During its fall it is subject to the velocity dependent drag force. Is this not F = 6πrμv ?
No. Stokes law is only strictly valid at constant velocity, and, in the falling ball viscometer, this is when the ball is at its terminal velocity. But, even when the ball is still accelerating to approach the terminal velocity, Stokes law should still provide an excellent approximation to the drag force. Can you visualize why? It is because, at any time, the far field velocity of the fluid relative to the ball is essentially constant over the length scale of the ball.

 Quote by Chestermiller Whoa, whoa, whoa, whoa. A standard viscometer for very viscous fluids is the falling ball viscometer. Yes, it applies when the Reynolds number is very low. Typically, the viscosity of the fluid will be on the order of 100's to 1000's of poise, and the ball will only be on the order of a few mm.
Yes, a falling ball viscometer is great, but it can't simply use Stokes Law indiscriminately.

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 Quote by boneh3ad Yes, a falling ball viscometer is great, but it can't simply use Stokes Law indiscriminately.
Of course not. It is important to verify a postiori that the Reynolds number is sufficiently low.

 Quote by Chestermiller Of course not. It is important to verify a postiori that the Reynolds number is sufficiently low.
The thing is, it takes a quite viscous fluid, a very buoyant ball or else an incredibly tiny sphere for Stokes Law to apply due to the Reynolds number restriction. Without the OP mentioning nothing about the types of velocities he is measuring or size/type of sphere being used, there is basically no information about the Reynolds number and I feel it important to point out the limits of the technique. In my experience, many inexperienced experimenters aren't aware of that restriction and don't even think to consider it.

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 Quote by boneh3ad The thing is, it takes a quite viscous fluid, a very buoyant ball or else an incredibly tiny sphere for Stokes Law to apply due to the Reynolds number restriction. Without the OP mentioning nothing about the types of velocities he is measuring or size/type of sphere being used, there is basically no information about the Reynolds number and I feel it important to point out the limits of the technique. In my experience, many inexperienced experimenters aren't aware of that restriction and don't even think to consider it.
Your caveat is appropriate. Maybe the OP can provide more details of the experiments.

 Quote by Chestermiller Your caveat is appropriate. Maybe the OP can provide more details of the experiments.
I used Rhenium spheres with diameters of 100-150 μm in a silicate melt, which has a viscosity of about 0.05 Pa s using white beam x-ray radiography with a very fast CCD camera. I just wanted to know the special case when the sphere didn't reach terminal velocity, which happened a couple of times due to the process used to make the spheres means they don't always have uniform density - instead of throwing that data away it would be nice to use it. I think the method described by Chestermiller does the trick, it just means the error bars might be a bit larger in those cases. Cheers.
 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.