Be it in cooking or an experiment, when transferring liquid from one container to another using a funnel should you:

(a) pour it all straight into the funnel, filling it up and waiting for the liquid to drain through

(b) pour it such that the end of the funnel is still always filled but the volume of 'blocking' liquid in the funnel is small - so you pour it slowly

Does one of the above options achieve a faster rate flow of liquid through the funnel? I'm thinking bernoulli's principle is going to come into this? Intuitively the blocking liquid in (a) is going to increase the pressure at the bottom of the funnel and hence 'force' more liquid through...

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    $\begingroup$ I assume there is adequate airspace for the gas to escape the containers without using the hole in the funnel? $\endgroup$ – JMac Nov 19 '18 at 20:45
  • $\begingroup$ Yes exactly, but point is both scenarios allow for gas to escape the containers at the same rate $\endgroup$ – peegs77 Nov 19 '18 at 20:53
  • $\begingroup$ I once heard about a bartender "trick" to whirl a bottle when draining it to get the liquid out faster. It works. Probably by creating a vortex that allows air to enter the bottle to balance the air pressure. So in your case your might want to ensure air can leave the destination container and pour in a way to create a vortex (pour near to the rim). The shape of the funnel might play a role. $\endgroup$ – try-catch-finally Dec 29 '18 at 8:05
  • $\begingroup$ Hmmm interesting!! $\endgroup$ – peegs77 Dec 31 '18 at 15:22

The Bernoulli equation for a non-turbulent fluid with density $\rho$ is

$$ P_\text{fluid} + \frac12 \rho v^2 + \rho g h = \text{constant}. $$

At the top and the bottom of the funnel, where the fluid is exposed to the atmosphere, the fluid surface will move until the fluid pressure and the atmospheric pressure are the same. For an ordinary funnel, that effectively means that the inlet and outlet pressures are identical. If you are continuously filling the funnel, so that the fluid speed $v$ at the top is zero --- or alternatively, if the top of the funnel is so much larger than the outlet that the flow speed at the top is negligible --- then the outlet speed is

$$ v = \sqrt{2g\Delta h}. $$

So doubling the distance $\Delta h$ between the exit of your funnel and the surface of your fluid will make your exit velocity about 40% larger.

An engineer friend of mine once solved a problem with a water reservoir that was draining too slowly by adding a longer hose to dangle under the outlet. It's the same trick: increase the flow rate by increasing $\Delta h$.

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  • $\begingroup$ Another trick is to pour the liquid from high enough that it shoots directly into the narrow part of the funnel. $\endgroup$ – S. McGrew Dec 29 '18 at 5:23
  • $\begingroup$ Yeah, sounds possible. However surely it becomes an optimisation problem given the constraints of funnel width and liquid density, because the higher you pour the liquid the more it will 'disperse' as it falls and eventually be too wide to fall directly into the narrow part of the funnel? $\endgroup$ – peegs77 Dec 31 '18 at 15:24
  • $\begingroup$ @S.McGrew I would think that if you could do that you wouldn't bother with the funnel. $\endgroup$ – rob Dec 31 '18 at 16:18
  • $\begingroup$ Good point. I've noticed, however, that the momentum of the water being poured into the top of the narrow part of the funnel can increase the pressure at the top of the narrow part enough to make the liquid flow faster than it flows when the wide part of the funnel is full. When the poured water hits the top of water that fills the wide part, it doesn't change the pressure at the top of the narrow part much. $\endgroup$ – S. McGrew Dec 31 '18 at 16:24
  • $\begingroup$ @S.McGrew That sounds like a non-laminar flow, for which my simple analysis probably doesn't work. $\endgroup$ – rob Dec 31 '18 at 16:28

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