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Siphon acceleration or equilibrium?

  1. Jun 8, 2005 #1
    If I had a barrel of water on a pedistool, and at the bottom of the barrel there is a siphon, that would pump out, my question is, whether the rate of water that would normally flow from the barrel following down the path of least resistance, would the syphon be able to increase the volume and thus the velocity from the barrel ,or would it equal out to a natural drain rate?
     
  2. jcsd
  3. Jun 8, 2005 #2

    OlderDan

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    I interpret this to mean you are comparing the drain rate from a hole or short pipe at the base of the barrel to a siphon tube of the same diameter. You need to apply Bernoulli's equation to both situations. Here is a link to the considerations for a siphon.

    http://www.answers.com/topic/siphon
     
  4. Jun 11, 2005 #3
    First, thanks for the link, it is an excellent site for info, after I reread my post , I concur it could have been phrased alot better.
    A follow up question would be, is this right?, no matter how much you force the water through the siphon, by manually pumping, it is theoreticly impossible to increase the discharge more than what I would have if it just drained out of the barrel?
    Same size hole as diameter of siphon tube.
     
  5. Jun 13, 2005 #4

    OlderDan

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    I don't know what you mean by manually pumping it. With a mechanical pump, you could drain the barrel as fast as the pump can pump. The result for the siphon shows that velocity depends on how far the drain point is below the upper surface of the liquid, as ling as the intermediate height is not excessive. That means you can increase the flow rate by lowering the drain point. If you just had a hole at the bottom of the barrel, what would the velocity be? Think about what happens as the upper surface gets near the bottom of the barrel.
     
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