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Say I have a cylindrical container of some volume $V$, that I fill with water or some other fluid with dynamic viscosity $\mu$. Here, the bottom of the container is milled such that some fraction, $p$, of its surface area, $A$, is removed to allow passage of fluid. One could, for example, drill pores until this criterion is met.

After positioning a plate at the top of the container, I press down on the plate with some force $F$ to compress the fluid. As a function of $F$, how quickly will I drain the fluid through the bottom of the container?

To calculate this, what additional measurements/parameters, such as surface hydrophobicity/etc., will I require, and how does this change with the minimum cross-sectional area of the pores I mill in the bottom of the container?

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You may find some useful material here:… I don't think the relevant variables are $p$ and $A$; changing $A$ while keeping $pA$ constant should have no effect. Relying on gravity is not the same as pressing down on the piston; if it's only gravity, then the result depends on the depths at which the holes are drilled. The force $F$ is irrelevant; what matters is the pressure. If the walls are thick compared to the diameters of the holes, then the thickness matters. – Ben Crowell Aug 7 '11 at 2:15
@Ben Crowell, I did a poor job writing this question, and I'll fix it when I have a moment. To quickly respond though, I meant gravity as in the lid being heavy and pressing down on the fluid, I agree about pA and didn't mean to suggest otherwise, the walls of my container should be arbitrarily thin, and of course the force of the lid pressing down is relevant, right? – TheSheepMan Aug 7 '11 at 3:19
@Ben, I suggested the variable 'A' for the surface area at the bottom of the container since, with the volume, this properly specifies what my container looks like. – TheSheepMan Aug 7 '11 at 3:27
up vote 0 down vote accepted

Your answer will depend greatly upon the size of the holes you drill. If I remember correctly, fluid flow through an orifice is proportional to the fourth power of the orifice diameter. So, for a given force, you would get twice as much fluid out of a single hole of diameter 1 mm as you would out of eight holes, each of 1/2 mm diameter.

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great, your answer + google was very helpful. – TheSheepMan Aug 22 '11 at 1:15
The fourth power is correct for laminar flow. It gets worse with turbulence. – dmckee Jul 30 '12 at 0:56

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