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Imagine I have a small bottle filled with water with a syphon in it which opening at the end is 10 times wider than the opening of the bottle and at the beginning almost as wide as the bottle.

Is it still possible that through the syphon water can flow (after a bit sucking on it). By Bernoullis equation $\frac {1}{2} v^2 = gh$ I think that the height difference should increase but how much, how to calculate?

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  • $\begingroup$ Temeceart. What is that? $\endgroup$
    – Marijn
    Jan 7, 2017 at 9:28

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This will still work. However, it is most unlikely that the output will take up the entire 10x opening.

So long as you have a continuous tube of water and the end of that tube is below the level of water in the bottle, you will get a siphon. The devil is merely in the details. The flow rate will be limited by the narrow part of the tube, due to fluid friction. As you might predict, the water will have to be moving 10x slower near the mouth of your siphon. At these speeds, it is easy for air to move upwards into your siphon.

What happens next depends on a whole slew of factors. If the water flow is fast enough, the water will simply break free from the walls of the siphon. If this occurs below the water level in the bottle, the siphon will flow and the water will dribble out the bottom of the wide part.

If the water flow is not sufficiently fast, the air will be able to sneak up above the level of the water in the bottle. This air can then stop the siphon from functioning.

If we were to modify the setup and your 10x wide mouth was immersed in a bucket of water at a lower elevation, this issue would not occur because air would not be able to get to the mouth of the tube. Instead, you would see the very natural result: the water would be moving 10x slower at the mouth of the tube, and eventually all of the water would be moved from the bottle to this lower bucket.

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