Fluid dynamics - finding pressure for a rotating fluid

In summary: I'm not sure I've done correctly.In summary, the conversation discusses the rotation of an ideal fluid under gravity and the desire to find the surfaces of constant pressure. This leads to a discussion of Bernoulli's equation and the problem of finding the correct shape for the free surface of a rotating bucket of water. The conversation then shifts to the Euler equations in component form and how to integrate them to find the pressure, ultimately leading to the correct shape for the free surface. The final question is about setting the pressure equal to zero to find the shape, which is confirmed as a valid method.
  • #1
Deadstar
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Homework Statement



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.

Homework 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]

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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  • #2
Hi Deadstar! :smile:


(have a rho: ρ and an omega: Ω and a grad: ∇ and a curly d: ∂ :wink:)
Deadstar said:
'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 :rolleyes:)
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. :smile:

(btw, you could have got the surface shape just by using centripetal acceleration of a tiny drop :wink:)
 
  • #3
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)
 

FAQ: Fluid dynamics - finding pressure for a rotating fluid

What is fluid dynamics?

Fluid dynamics is the study of how fluids (liquids and gases) move and interact with their surroundings. It involves analyzing the forces and pressures that affect the motion of fluids.

What is pressure in fluid dynamics?

In fluid dynamics, pressure refers to the force per unit area that a fluid exerts on its surroundings. It is a fundamental property of fluids and is affected by factors such as density, velocity, and viscosity.

How do you calculate pressure in a rotating fluid?

To calculate pressure in a rotating fluid, you can use the Bernoulli's equation, which relates the fluid's pressure, velocity, and density. This equation takes into account the effects of centrifugal force and can be expressed as P + 1/2ρv^2 + ρgh = constant, where P is pressure, ρ is density, v is velocity, g is the acceleration due to gravity, and h is the height above a reference point.

What is the relationship between pressure and velocity in a rotating fluid?

In a rotating fluid, there is an inverse relationship between pressure and velocity. As the velocity of the fluid increases, the pressure decreases. This is known as the Bernoulli's principle.

How does rotation affect pressure in a fluid?

Rotation can affect pressure in a fluid by causing centrifugal forces, which push the fluid away from the center of rotation. This can result in a decrease in pressure at the center of rotation and an increase in pressure at the outer edges of the rotating fluid.

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