Surface tension of a torus raindrop


by peripatein
Tags: raindrop, surface, tension, torus
peripatein
peripatein is offline
#1
Nov25-12, 07:55 AM
P: 812
1. The problem statement, all variables and given/known data

When calculating the difference in pressure inside a spherical raindrop, the force exerted by the surface tension is calculated to be 2pi*R*gama, where R is the radius of the drop and gamma is dE/dS (dyne/cm).
When the shape of the raindrop is said to be that of a torus, the force exerted by the surface tension is calculated to be 2pi*R*gamma + 2pi(R+2r)*gamma (please see attachment).
My question is simply why does the force in the case of the torus have two components, whereas in the case of a sphere it has only one?

2. Relevant equations



3. The attempt at a solution
Attached Thumbnails
Torus.JPG  
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mfb
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#2
Nov25-12, 08:29 AM
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P: 10,824
In both cases, the expression is the length of the boundary (as seen in a 2D-projection) multiplied by gamma. A circle has one boundary, the torus has 2 (inner+outer).
peripatein
peripatein is offline
#3
Nov25-12, 09:06 AM
P: 812
But isn't the circle's circumference considered a boundary? Or shouldn't it be? Is it not a thin layer of fluid verging on air?

mfb
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#4
Nov25-12, 09:49 AM
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P: 10,824

Surface tension of a torus raindrop


But isn't the circle's circumference considered a boundary?
Of course. That should be the origin of 2pi*R*gamma.
You get a layer of fluid/air in contact there.
peripatein
peripatein is offline
#5
Nov25-12, 10:49 AM
P: 812
Okay. Thank you.


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