# Applying Bernoulli's Equation to a Siphon

I'm not sure if this is the right place to post this (because it isn't homework), but it could easily be a homework question.

This wiki page, http://en.wikipedia.org/wiki/Siphon#Explanation_using_Bernoulli.27s_equation", shows how to apply Bernoulli's equation to a siphon where both the surface of the water and the exit point of the siphon (point "c" in the diagram) are at the same pressure.

My question is this: how do you use the equation if the pressure at the surface of the water is different than the pressure at point c?

Thanks,
Brent

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gneill
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I'm not sure if this is the right place to post this (because it isn't homework), but it could easily be a homework question.

This wiki page, http://en.wikipedia.org/wiki/Siphon#Explanation_using_Bernoulli.27s_equation", shows how to apply Bernoulli's equation to a siphon where both the surface of the water and the exit point of the siphon (point "c" in the diagram) are at the same pressure.

My question is this: how do you use the equation if the pressure at the surface of the water is different than the pressure at point c?

Thanks,
Brent
Put the appropriate pressures into the Bernoulli equation. The wiki page that you cited had Patm as the pressure on both sides of the equation. It doesn't have to be the same pressure at both locales.

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I understand now. Thanks a lot for you help.

Another question. If I solve for VC and I'm ending up with a negative square root, does that just mean the siphon won't flow?

For example, the height of the starting point of my siphon is at, say, x=0. The end of the siphon is at x=-20m (20 meters below the start point). Also, the pressure at the starting point is standard (101 325 pascals) and the end point is at a pressure of 302 225 pascals. Let's just assume that $\rho$=1.

So (02)/2 + (9.8)(0) + 101.325pa/1 = (VC2)/2 + (9.8m/s)(20m) + 302225pa

Solving for VC, I get $\sqrt{}-402 192$

Is that right? Thanks again for all of your help.