classical mechanics - How does the period of an hourglass depend on the grain size?

Suppose I have an hourglass that takes 1 full hour on average to drain. The grains of sand are, say, $1 \pm 0.1\ {\rm mm}$ in diameter.

If I replace this with very finely-grained sand $0.1 \pm 0.01\ {\rm mm}$ in diameter but keep the hourglass otherwise the same, how long a time will the "hourglass" now measure? Does this depend on the size of the funnel, or should all one-hour hourglasses change in roughly the same way?

Bonus: is the new hourglass more or less precise? (precision defined at $\sigma_t /t$, with $t$ the time to drain)

Answer

I have at least an answer :

Does this depend on the size of the funnel ?

Yes, as it was expected (for a very large funnel, all sand falls in the same time, whereas for a very small funnel it doesn't fall at all. More precisely ... the complete answer is probably extremely complex, as one has to take into account the shape of the hourglass, the dynamics of the grains, and so on (see http://arxiv.org/abs/0707.4550 for example). However we can have a rough idea with some dimensional analysis.

Let us consider a cylinder of diameter $D$, with a circular hole punched on the bottom side with radius $a$. We fill the cylinder with a height $H$ of sand. If we look at the speed of sand grains going out the bucket, we observe (experimentally) that it does not depend of the height of sand $H$, if $H$ is big enough (compared to the diameter $D$ - because the constraint saturates). We are left with two parameters : the diameter of the hole $a$ and the gravity field $g$ that makes it fall, so the output speed $v$ has to be proportional to $\sqrt{g a}$. The flow rate is the speed times the section, thus it is $Q \propto v \, a^2$, so it is of order $Q \propto g^{1/2} a^{5/2}$ (this is the Beverloo law).

However we have to take into account the size of grains ; let us suppose they are spherical and let be $d$ their diameter. All grains that are less than half on the hole won't fall, so there is a ring-shaped exclusion region at the border of the hole. The effective diameter of the hole, were grains will actually fall, is thus reduced by $d/2 + d/2 = d$, and is thus $a-d$. We thus replace $a$ with $a-d$ in the Beverloo law and get

$Q \propto g^{1/2} (a-d)^{5/2}$

Now, this becomes invalid when $H$ and $D$ are of the same order, but we can get an idea of the time to drain $T$ by saying that as the flow rate does not depend of $H$, $Q T = V = 4 \pi (D/2)^2 H = \pi D^2 H$ so, dropping the $\pi$ we get

$T \propto g^{1/2} \, \frac{(a-d)^{5/2}}{D^2 \, H}$

and thus for everything but the size of sand grains $d$ constant

$\frac{T(d_1)}{T(d_2)} = \left( \frac{a - d_1}{a - d_2} \right)^{5/2}$