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Fluid dynamics - finding pressure for a rotating fluid

by Deadstar
Tags: dynamics, fluid, pressure, rotating
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Deadstar
#1
Feb25-11, 02:47 PM
P: 106
1. The problem statement, all variables and given/known data

An ideal fluid is rotating under gravity g with constant angular velocicy [tex]\Omega[/tex], so that relative to the fixed Cartesian axes [tex]\mathbf{u} = (-\Omega y, \Omega x, 0)[/tex]. We wish to find the surfaces of constant pressure, and hence the surface of a uniformly rotating bucket of water (which will be at atmospheric pressure).
'By Bernoulli,' [tex]p/\rho + \mathbf{u}^2/2 + gz[/tex] is constant so the constant pressure surfaces are
[tex]z = \textrm{constant} - \frac{\Omega^2}{2g}(x^2 + y^2)[/tex].
But thes means that the surface of a rotating bucket of water is at the highest in the middle. What is wrong? (Done, not my question)

Write down the Euler equations in component form, integrate them directly to find the pressure, and hence obtain the correct shape for the free surface.

2. Relevant equations

Euler equations

[tex]\frac{D \mathbf{u}}{Dt} = -\frac{1}{\rho} \nabla p + \mathbf{g}[/tex]

and in component form.

[tex]\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} + w\frac{\partial u}{\partial z} = -\frac{1}{\rho}\frac{\partial p}{\partial x}[/tex]

[tex]\frac{\partial v}{\partial t} + u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} + w\frac{\partial v}{\partial z} = -\frac{1}{\rho}\frac{\partial p}{\partial y}[/tex]

[tex]\frac{\partial w}{\partial t} + u\frac{\partial w}{\partial x} + v\frac{\partial w}{\partial y} + w\frac{\partial w}{\partial z} = -\frac{1}{\rho}\frac{\partial p}{\partial z} - g[/tex]

3. The attempt at a solution

I get that the above component form becomes...

[tex]\Omega x \frac{\partial u}{\partial y} = -\Omega^2 x = -\frac{1}{\rho}\frac{\partial p}{\partial x}[/tex]

[tex]\Omega y \frac{\partial v}{\partial x} = -\Omega^2 y = -\frac{1}{\rho}\frac{\partial p}{\partial y}[/tex]

[tex]g = -\frac{1}{\rho}\frac{\partial p}{\partial z}[/tex]

So now I just sort of... integrate and combine them together. I'm not sure whether I can do this though, integrate each line seperatly and add it all together (as each is a component of the total pressure..?)

So I get...

[tex]-\frac{p}{\rho} = -\frac{\Omega^2}{2}(x^2 + y^2) + gz + constant[/tex]

which is pretty close to what I'm after however I'm guessing something has gone wrong at the integration step. Anyone care to assist?

For finding the shape is it just a case of setting [tex]p/\rho[/tex] equal to zero?
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tiny-tim
#2
Feb25-11, 03:35 PM
Sci Advisor
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Thanks
tiny-tim's Avatar
P: 26,148
Hi Deadstar!


(have a rho: ρ and an omega: Ω and a grad: ∇ and a curly d: ∂ )
Quote Quote by Deadstar View Post
'By Bernoulli,'
Bernoulli's equation is only valid along a streamline

the "uphill" line on the surface is not a streamline

(a streamline would be either a horizontal circle which doesn't help! or any vertical line, on the weird principle that if you made little hole in the bottom, the water would go vertically into it )
So I get...

[tex]-\frac{p}{\rho} = -\frac{\Omega^2}{2}(x^2 + y^2) + gz + constant[/tex]

which is pretty close to what I'm after however I'm guessing something has gone wrong at the integration step. Anyone care to assist?

For finding the shape is it just a case of setting [tex]p/\rho[/tex] equal to zero?
That looks ok

your equal-pressure surfaces would be by putting p = constant.

(btw, you could have got the surface shape just by using centripetal acceleration of a tiny drop )
Deadstar
#3
Feb25-11, 06:35 PM
P: 106
Thanks tiny tim but the 'by Bernoulli' part is part of the actual question and you had to figure out why the statement was wrong (which I had done)


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