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The Stokes law equation for the drag on the oil droplet is:

$$ F_d = 6 \pi \eta r v $$

wher $\eta$ is the viscosity of the air, $r$ is the radius of the oil drop and $v$ is the velocity of the oil drop. The trouble is that when the oil drop is very small its radius is comparable to the mean free path of the air molecules. That means the air no longer behaves as a completely homogeneous fluid, and while this is a very small effect it is detectable in the oil drop experiment.

The correction that you refer, which is to due to Cunningham, is to write the drag as:

$$ F_d = 6 \pi \left(\frac{\eta}{1 + \frac{b}{p\,r}}\right) r v $$

where the quantity in brackets can be treated as a corrected viscosity. I assume this is what your document refers to, so the $\eta_\text{corr}$ is given by:

$$ \eta_\text{corr} = \frac{\eta}{1 + \frac{b}{P\,r}} $$

In this equation $P$ is the pressure and $b$ is an empirical constant. Some Googling finds that the constant $b$ is approximately $0.00822$ Pa m.

Response to comment:

If we observe an oil drop falling at constant velocity $v$ we know that the drag force is equal to the gravitational force $F_d = mg$, and the mass is just the volume of the drop multiplied by the density of the oil. If we feed this into the Cunningham equation above we get:

$$ \tfrac{4}{3}\pi r^3 \rho g = 6 \pi \left(\frac{\eta}{1 + \frac{b}{p\,r}}\right) r v $$

We know everything in this equation except for $r$, so we can solve the equation to get $r$. It's going to turn into a quarticquadratic equation in $r$, so give it to your computerwhich is easy to solve numerically.

The error due to the Cunningham correction isn't that large. For a one micron sized oil drop it's about 10%, so as a first attempt I would just ignore it and use the uncorrected air vicosity.

The Stokes law equation for the drag on the oil droplet is:

$$ F_d = 6 \pi \eta r v $$

wher $\eta$ is the viscosity of the air, $r$ is the radius of the oil drop and $v$ is the velocity of the oil drop. The trouble is that when the oil drop is very small its radius is comparable to the mean free path of the air molecules. That means the air no longer behaves as a completely homogeneous fluid, and while this is a very small effect it is detectable in the oil drop experiment.

The correction that you refer, which is to due to Cunningham, is to write the drag as:

$$ F_d = 6 \pi \left(\frac{\eta}{1 + \frac{b}{p\,r}}\right) r v $$

where the quantity in brackets can be treated as a corrected viscosity. I assume this is what your document refers to, so the $\eta_\text{corr}$ is given by:

$$ \eta_\text{corr} = \frac{\eta}{1 + \frac{b}{P\,r}} $$

In this equation $P$ is the pressure and $b$ is an empirical constant. Some Googling finds that the constant $b$ is approximately $0.00822$ Pa m.

Response to comment:

If we observe an oil drop falling at constant velocity $v$ we know that the drag force is equal to the gravitational force $F_d = mg$, and the mass is just the volume of the drop multiplied by the density of the oil. If we feed this into the Cunningham equation above we get:

$$ \tfrac{4}{3}\pi r^3 \rho g = 6 \pi \left(\frac{\eta}{1 + \frac{b}{p\,r}}\right) r v $$

We know everything in this equation except for $r$, so we can solve the equation to get $r$. It's going to turn into a quartic equation, so give it to your computer to solve numerically.

The error due to the Cunningham correction isn't that large. For a one micron sized oil drop it's about 10%, so as a first attempt I would just ignore it and use the uncorrected air vicosity.

The Stokes law equation for the drag on the oil droplet is:

$$ F_d = 6 \pi \eta r v $$

wher $\eta$ is the viscosity of the air, $r$ is the radius of the oil drop and $v$ is the velocity of the oil drop. The trouble is that when the oil drop is very small its radius is comparable to the mean free path of the air molecules. That means the air no longer behaves as a completely homogeneous fluid, and while this is a very small effect it is detectable in the oil drop experiment.

The correction that you refer, which is to due to Cunningham, is to write the drag as:

$$ F_d = 6 \pi \left(\frac{\eta}{1 + \frac{b}{p\,r}}\right) r v $$

where the quantity in brackets can be treated as a corrected viscosity. I assume this is what your document refers to, so the $\eta_\text{corr}$ is given by:

$$ \eta_\text{corr} = \frac{\eta}{1 + \frac{b}{P\,r}} $$

In this equation $P$ is the pressure and $b$ is an empirical constant. Some Googling finds that the constant $b$ is approximately $0.00822$ Pa m.

Response to comment:

If we observe an oil drop falling at constant velocity $v$ we know that the drag force is equal to the gravitational force $F_d = mg$, and the mass is just the volume of the drop multiplied by the density of the oil. If we feed this into the Cunningham equation above we get:

$$ \tfrac{4}{3}\pi r^3 \rho g = 6 \pi \left(\frac{\eta}{1 + \frac{b}{p\,r}}\right) r v $$

We know everything in this equation except for $r$, so we can solve the equation to get $r$. It's going to turn into a quadratic equation in $r$, which is easy to solve.

The error due to the Cunningham correction isn't that large. For a one micron sized oil drop it's about 10%, so as a first attempt I would just ignore it and use the uncorrected air vicosity.

2 Respond to comment
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The Stokes law equation for the drag on the oil droplet is:

$$ F_d = 6 \pi \eta r v $$

wher $\eta$ is the viscosity of the air, $r$ is the radius of the oil drop and $v$ is the velocity of the oil drop. The trouble is that when the oil drop is very small its radius is comparable to the mean free path of the air molecules. That means the air no longer behaves as a completely homogeneous fluid, and while this is a very small effect it is detectable in the oil drop experiment.

The correction that you refer, which is to due to Cunningham, is to write the drag as:

$$ F_d = 6 \pi \left(\frac{\eta}{1 + \frac{b}{p\,r}}\right) r v $$

where the quantity in brackets can be treated as a corrected viscosity. I assume this is what your document refers to, so the $\eta_\text{corr}$ is given by:

$$ \eta_\text{corr} = \frac{\eta}{1 + \frac{b}{P\,r}} $$

In this equation $P$ is the pressure and $b$ is an empirical constant. Some Googling finds that the constant $b$ is approximately $0.00822$ Pa m.

Response to comment:

If we observe an oil drop falling at constant velocity $v$ we know that the drag force is equal to the gravitational force $F_d = mg$, and the mass is just the volume of the drop multiplied by the density of the oil. If we feed this into the Cunningham equation above we get:

$$ \tfrac{4}{3}\pi r^3 \rho g = 6 \pi \left(\frac{\eta}{1 + \frac{b}{p\,r}}\right) r v $$

We know everything in this equation except for $r$, so we can solve the equation to get $r$. It's going to turn into a quartic equation, so give it to your computer to solve numerically.

The error due to the Cunningham correction isn't that large. For a one micron sized oil drop it's about 10%, so as a first attempt I would just ignore it and use the uncorrected air vicosity.

The Stokes law equation for the drag on the oil droplet is:

$$ F_d = 6 \pi \eta r v $$

wher $\eta$ is the viscosity of the air, $r$ is the radius of the oil drop and $v$ is the velocity of the oil drop. The trouble is that when the oil drop is very small its radius is comparable to the mean free path of the air molecules. That means the air no longer behaves as a completely homogeneous fluid, and while this is a very small effect it is detectable in the oil drop experiment.

The correction that you refer, which is to due to Cunningham, is to write the drag as:

$$ F_d = 6 \pi \left(\frac{\eta}{1 + \frac{b}{p\,r}}\right) r v $$

where the quantity in brackets can be treated as a corrected viscosity. I assume this is what your document refers to, so the $\eta_\text{corr}$ is given by:

$$ \eta_\text{corr} = \frac{\eta}{1 + \frac{b}{P\,r}} $$

In this equation $P$ is the pressure and $b$ is an empirical constant. Some Googling finds that the constant $b$ is approximately $0.00822$ Pa m.

The Stokes law equation for the drag on the oil droplet is:

$$ F_d = 6 \pi \eta r v $$

wher $\eta$ is the viscosity of the air, $r$ is the radius of the oil drop and $v$ is the velocity of the oil drop. The trouble is that when the oil drop is very small its radius is comparable to the mean free path of the air molecules. That means the air no longer behaves as a completely homogeneous fluid, and while this is a very small effect it is detectable in the oil drop experiment.

The correction that you refer, which is to due to Cunningham, is to write the drag as:

$$ F_d = 6 \pi \left(\frac{\eta}{1 + \frac{b}{p\,r}}\right) r v $$

where the quantity in brackets can be treated as a corrected viscosity. I assume this is what your document refers to, so the $\eta_\text{corr}$ is given by:

$$ \eta_\text{corr} = \frac{\eta}{1 + \frac{b}{P\,r}} $$

In this equation $P$ is the pressure and $b$ is an empirical constant. Some Googling finds that the constant $b$ is approximately $0.00822$ Pa m.

Response to comment:

If we observe an oil drop falling at constant velocity $v$ we know that the drag force is equal to the gravitational force $F_d = mg$, and the mass is just the volume of the drop multiplied by the density of the oil. If we feed this into the Cunningham equation above we get:

$$ \tfrac{4}{3}\pi r^3 \rho g = 6 \pi \left(\frac{\eta}{1 + \frac{b}{p\,r}}\right) r v $$

We know everything in this equation except for $r$, so we can solve the equation to get $r$. It's going to turn into a quartic equation, so give it to your computer to solve numerically.

The error due to the Cunningham correction isn't that large. For a one micron sized oil drop it's about 10%, so as a first attempt I would just ignore it and use the uncorrected air vicosity.

1
source | link

The Stokes law equation for the drag on the oil droplet is:

$$ F_d = 6 \pi \eta r v $$

wher $\eta$ is the viscosity of the air, $r$ is the radius of the oil drop and $v$ is the velocity of the oil drop. The trouble is that when the oil drop is very small its radius is comparable to the mean free path of the air molecules. That means the air no longer behaves as a completely homogeneous fluid, and while this is a very small effect it is detectable in the oil drop experiment.

The correction that you refer, which is to due to Cunningham, is to write the drag as:

$$ F_d = 6 \pi \left(\frac{\eta}{1 + \frac{b}{p\,r}}\right) r v $$

where the quantity in brackets can be treated as a corrected viscosity. I assume this is what your document refers to, so the $\eta_\text{corr}$ is given by:

$$ \eta_\text{corr} = \frac{\eta}{1 + \frac{b}{P\,r}} $$

In this equation $P$ is the pressure and $b$ is an empirical constant. Some Googling finds that the constant $b$ is approximately $0.00822$ Pa m.