Is the pressure at points P1 and P2 the same in surface tension controversy?

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The discussion centers on whether the pressures at points P1 and P2 in a liquid's meniscus are the same, given the hemispherical shape of the meniscus. One participant argues that if P1 and P2 are equal, then the atmospheric pressures P1' and P2' would differ due to varying heights, contradicting their equality to atmospheric pressure. Another contributor points out that the initial pressure calculations neglect surface tension, which is crucial for understanding the meniscus shape and pressure distribution. The conversation highlights the need for a correct equation that incorporates surface tension effects, suggesting that the problem may require advanced analytical or numerical methods to resolve. Ultimately, the participants express confusion about the correct approach to analyze the meniscus curvature and its implications for pressure.
i_island0
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We know about capillary rise. And we say that meniscus is hemispherical in shape.
Now please see this picture.
http://www.flickr.com/photos/63184961@N00/3030969897/

There i am mentioning two points where the pressure is indicated as P1 and P2.
I am also mentioning two other points 1' and 2'. the pressure there is atmospheric (Po).
Are these pressure P1 and P2 same or different.
If same, then pressure P1' and p2' will be different owing to different heights h1 and h2. But we know that P1' = Po and P2' = Po.
If different, then how can the shape of meniscus be hemispherical. It must be flat then.
Where am i going wrong??
 
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i_island0 said:
Are these pressure P1 and P2 same or different.
If same, then pressure P1' and p2' will be different owing to different heights h1 and h2.

The flaw in your reasoning lies here. I suspect you are calculating P1' and P2' by using an equation that neglects surface tension? If so, you're ignoring the effect that you want to study.
 
Mapes said:
The flaw in your reasoning lies here. I suspect you are calculating P1' and P2' by using an equation that neglects surface tension? If so, you're ignoring the effect that you want to study.

Surface tension is a surface phenomenon. So when i am inside the liquid, surface tension has no role to play.
I am using P = Po + dgh (d: density of liquid) to find the pressure. So where is the flaw?
 
i_island0 said:
So when i am inside the liquid, surface tension has no role to play.

Sure it does. Any portion of the liquid will move to reduce the total energy of the system; this is the meaning of equilibrium. Surface energy affects the entire system.

Your equation for pressure only includes potential energy changes related to gravity, not surface tension, so it predicts a flow from 2' to 1' because of a pressure difference. It neglects the fact that this flow would rearrange the surface profile to be flatter, which would result in a net energy penalty.
 
Mapes said:
Your equation for pressure only includes potential energy changes related to gravity, not surface tension, so it predicts a flow from 2' to 1' because of a pressure difference. It neglects the fact that this flow would rearrange the surface profile to be flatter, which would result in a net energy penalty.

So can you please tell me what should be the correct equation then. As i m really confused about all this.
 
Analytically, the meniscus shape looks like a problem for variation calculus: find the curve that minimizes total system energy. There may be some literature out there on this problem, but I don't know. You could also solve it numerically through simulation. Hopefully someone with more fluids experience can weigh in.
 
thx.. i wish some one can reply on this. AS far as i understand, the curvature of meniscus should not be hemispherical.
 
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