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Thermodynamics of a floating cylinder

by private_donkey
Tags: cylinder, floating, thermodynamics
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private_donkey
#1
Aug13-14, 01:44 PM
P: 2
I am wondering if my reasoning is correct for determining the energy due to the buoyancy of a deformed water surface.

Essentially, one has a floating cylinder that depresses the surface of a liquid in an infinite tank as seen in the figure. I want to compare the energy of a flat surface with the energy of the deformed surface due to the displacement of the liquid by air (I know there are other energies involved but for the time being I am only looking at the energy due to the buoyancy of the displace liquid due to the deformed surface), not including the liquid displaced by the cylinder.

My reasoning is
[tex]
\begin{align}
E_{deformed} - E_0 &= \int_0^{-h} -F_{buoyancy} dz \\
&= \int_0^{-h} -P A dz \\
&= \rho g \int_0^{h} z \pi f(z)^2 dz
\end{align}
[/tex]

where [itex] \rho [/itex] is the density of the liquid. P is pressue, given by [itex] P = \rho g z [/itex], A is area, and thus A dz is volume of the displace liquid. [itex] A [/itex] can be given by [itex] A = \pi f(z)^2 [/itex], where [itex] f(z) [/itex] describes the radius from the centre of the cylinder to the edge of the meniscus to make infinitesimal disks as a function of z. g is gravity accel.

Is this reasoning correct?
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Greg Bernhardt
#2
Aug19-14, 09:08 PM
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I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?
private_donkey
#3
Aug21-14, 11:47 AM
P: 2
Not sure...Would a depressed liquid surface have a buoyancy? Because you are displacing liquid against gravity right...?


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