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I'm trying to understand how to use Bernoulli's equation in practice. So I will make an example of where is my confusion.

I found this great explanation on Wikipedia about the siphon, but there is a point which is not very clear.

enter link description here As far as I understood, Bernoulli's equation can be use considering two section of the same tube.

Nevertheless in the part where the velocity is obtained the two sections considered are the surface of the upper reservoir and the section of the siphon tube in C for istance. How can this be correct?

These two section do not belong to the same tube, one is of the resevoir, the other of the siphon tube.

At the height of the surface of the upper reservoir, but inside the tube of the siphon, the fluid has non zero velocity, while the velocity is taken to be zero there (and that's because it does not consider the tube of the siphon but the resevoir, as said before).

I do not see how this can be the correct application of Bernoulli equation.

So in general, can Bernoulli's equation be used for two section of different tubes, as in this case?

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  • $\begingroup$ Can you please copy the image into your question? Just in case the picture with the Wikipedia article changes at some point in the future. $\endgroup$ – Han-Kwang Nienhuys May 23 '16 at 14:06
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    $\begingroup$ The confusion comes from the fact that Bernoulli's equation does not refer to "physical" tubes, but to streamtubes: see e.g. the legend of the figure in the derivation here $\endgroup$ – L. Levrel May 25 '16 at 13:11
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In this case, it is allowed. As a thought experiment, you can replace the top bucket with the siphon hose by an S-shaped tube that has a gradual change in diameter, like this:

           _
         /   \
|www|   |     |
\www|   |     |
 \ww|   |     |
  \w|   |     |
   \|__/      |
              |
              |

With this type of analysis, it is more important that the downstream side of the tube makes a gradual transition to a large cross section. If you don't, then the kinetic energy of the high-velocity stream will be converted to heat rather than back to a pressure rise. This is not an issue at the upstream (tube inlet) side.

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Just visualize the upper beaker as a tube connected to the siphon tube(both tubes having different area of cross section)

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