# Water Moving Through Coffee Grinds

I was wondering how long it would take water to seep through the void space of a volume of coffee. It can be highly idealized, i.e. the coffee doesn't compress under the weight of water on top of it, the grinds are distributed uniformly, and the water is poured perfectly evenly.

My instincts tell me water wants to convert its gravitational potential energy to kinetic, but gets lost to frictional forces when flowing through the coffee. The simplest equation to model this is $F_D=\frac{1}{2}\rho_w v^2 C_D A$, where $A$ is the effective projected surface area of the projectile. Here are my results so far for the terminal velocity:

\begin{align*} \int_{S} P(x,y,z)dA&=\int_0^y\int_0^x mgh\,\mathrm{d}x\,\mathrm{d}y\\ &=\rho_w g xyh\\ &=\rho_w g V\\ &=\frac{1}{2}\rho_w v^2 C_D A \end{align*}

This implies that the speed at which a fluid must travel to reach equilibrium with frictional forces is $v=\sqrt{\frac{2gV}{C_D A}}$. Now, given porosity $\phi=\frac{V_V}{V_T}$, where $V_V$ is the void space, and $V_T$ is the volume of the bulk material, the effective surface area is $A_T(1-\phi^{\frac{2}{3}})$, which implies \begin{align*} v_{\text{ter}}&=\sqrt{\frac{2gV}{C_D A_T(1-\phi^{\frac{2}{3}})}}\\ &=\sqrt{\frac{2gh}{C_D (1-\phi^{\frac{2}{3}})}}\\ \end{align*} When the porosity is high, then the velocity is unbounded (makes sense), but when the porosity is low, it has a non-zero value. This is nonsense, since a fluid can't move through a perfectly solid object. Any suggestions, corrections, etc?

• I would have assumed that surface tension played the largest role, although one could probably cast that as a friction force. But that does alter how to interpret your 'effective projected surface area'. May 4, 2016 at 13:39
• Surface tension effects are not significant in the bed unless there is air in the bed forming interfaces with the fluid flowing through the bed (i.e., two phase flow). May 5, 2016 at 0:06
• Agree with Mr Miller, not sure I follow the logic behind surface tension being a significant factor, could you clarify a bit? These types of things may be obfuscated in the equations though, as is common in mechanics. Thank you regardless. May 5, 2016 at 11:19