Now I've been playing with this for some time but I can't seem to put things into perspective.

I wan't to find out the curvature of the bowl so that the vertical drain speed of the liquid is constant.

Imagine that the bowl is tube shaped. Then the vertical drain speed would be higher when the container is more full. If the container was shaped like an hourglass then the vertical speed would be something thats definitely unlinear with time.

I know that $$v = \sqrt{2gh}$$ and that $$v_1 \times A_1 = v_2 \times A_2$$

My guess would be that the bowl should be conus shaped but don't know how to prove it.

Any tips?

  • 1
    $\begingroup$ I am not sure what do you mean by keeping the draining a constant (do you mean the flowrate?). Do you mean for a change in h? (and you also want the diameter of the draining hole constant?) $\endgroup$ – Wolphram jonny Nov 21 '14 at 2:22
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    $\begingroup$ I don't understand the question. $\endgroup$ – Mike Dunlavey Nov 22 '14 at 20:45
  • $\begingroup$ I think that the goal here is to make the height of the water in the bowl decrease at a constant rate by varying the cross-sectional area of the bowl to compensate for changes in pressure at the outflow. Is that right? $\endgroup$ – user3823992 Nov 23 '14 at 8:16
  • $\begingroup$ Yes, user, you are right. Its just that im not sure how to formalize this $\endgroup$ – ditoslav Nov 23 '14 at 8:18

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