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Physics/calculus/funnel problem?
In my high school calculus class, we are exploring how water flows through a funnel and whether or not the flow is constant (and then writing an equation that depicts this).
Our lab set up was as follows: one graduated cylinder, a small plastic funnel (around 6 inches in diameter), 300mL of water, and a small stopper for the funnel. We filled the funnel with the 300mL of water, which was placed at the top of the graduated cylinder, and as soon as we un-stoppered it, we started a stopwatch. Every three seconds we recorded how much water has collected in the cylinder.
Our data showed that the flow of water is not constant, and that is in fact faster at the beginning of the trial and slower as more water empties into the graduated cylinder.
What we need help on is explaining WHY this is. No one in our group has actually taken physics class, so we don't have much understanding of how this works. We all assumed that the more water that is in the funnel, the more water pressure there is, therefore it is pushed out of the funnel faster, but after a little internet research we're not so sure that that is actually true.
Can we get any insight from anyone? Or perhaps you could point us in the direction of this information..? (Google hasn't been particularly helpful, maybe it's just our keywords...)
Thank you!
2 Answers
- ?Lv 79 years agoFavorite Answer
I don't fully appreciate the final conclusion of your earlier answer. I do follow the first part, agree with it, and used it below. But Bernoulli's equation yields approximately, as the exit velocity of the water:
1. vₑ = √[2⋅g⋅h]
That's actually already enough to know flow rate is ∝ √h, because the instantaneous change in volume is just the exit velocity times the area of the exit times the infinitesimal of time:
2. dV = Aₑ⋅vₑ dt
or,
3. dV ⁄ dt = Aₑ⋅vₑ = Aₑ⋅√[2⋅g⋅h]
So from that, I do NOT get the idea that at half-height that the rate of change in volume (flow rate) is going to be four times slower, but rather that it will be √2 slower.
Your own data should quickly tell which, since the difference in our opinions is so far apart it shouldn't take more than a few data points to know.
...
However, the rate of change in the height would be a different story altogether, since the remaining volume is so much less with lower height due to the shape. Even with a slower rate of change in volume, there is a lot less volume. At the very end of it, I'd expect the rate of change of h with respect to time to rapidly move towards ∞.
I won't bore you with the details (basically did dV ⁄ dt ⨯ dh ⁄ dV to get it), but it works out (I think) to:
4. dh ⁄ dt ∝ 1 ⁄ √[h³]
or,
5. dh ⁄ dt = k(g,θ) ⋅ Aₑ ⁄ √[h³]
where k is a function of 'g' (gravitational acceleration) and 'θ' (half-angle of the cone, where smaller is narrower and taller) and it doesn't vary over time, but only on the geometry of the funnel and the gravity you are experiencing then.
- bskelkarLv 79 years ago
The funnel is conical shaped, right? It has a opening instead of the vertex. Assume that the opening is very small that you can take it as vertex (a point). The the semivetcal angle is the nagle between the radius and height so r = htanA. So volume = (pi/3)(h^3)tanA. Now A is a constant the volume is proportional to CUBE of the height and dV/dt = pi(h^2)tanA9dh/dt) so rate of change of voule is proportional to the squqre of the height. If height is halved, rate od chenage of volume is (1/4)th and so on.
I hope this helps.
