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So how does a siphon really work?
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11th May 2010 | 18:11
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Pugilistic Animus
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Quite simply:
w= ∫ρgAhdh} h2,h1
Last edited by Pugilistic Animus; 24th May 2010 at
05:30
. Reason: add an h otherwise it's pressure not a work integral
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