Fluid Dynamics - Small velocity field

[Deadstar]

Suppose an inviscid, incompressible fluid is rotating uniformly with angular velocity \Omega. Take Cartesian axes fixed in a frame rotating with that angular velocity.

Show that the evolution of a SMALL velocity field, u_1 = (u_1, v_1, w_1), relative to the rotating axes and starting from rest is governed by...
\frac{\partial u_1}{\partial t} + 2 \Omega \times u_1 = -\frac{1}{\rho} \nabla p_1
\nabla \cdot u_1 = 0

By eliminating u1, v1 and w1 , show that
(\frac{\partial^2}{\partial t^2}(\frac{\partial^2}{\partial x^2} +\frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}) + 4 \Omega^2 \frac{\partial^2}{\partial z^2})p_1 = 0First of I have to show that...

2\Omega ( \frac{\partial u_1}{\partial y} - \frac{\partial v_1}{\partial x} ) = -\frac{1}{\rho} \nabla^2 p_1
and
2\Omega ( \frac{\partial u_1}{\partial x} + \frac{\partial v_1}{\partial y} ) = \frac{\partial}{\partial t} ( \frac{\partial u_1}{\partial y} - \frac{\partial v_1}{\partial x})

Which I can sort of do by taking dot product/cross product with respect to nabla except I can't justify why...
(u_1 \cdot \nabla)2\Omega = 0
and I don't see how the left hand side of...
2\Omega ( \frac{\partial u_1}{\partial x} + \frac{\partial v_1}{\partial y} ) = \frac{\partial}{\partial t} ( \frac{\partial u_1}{\partial y} - \frac{\partial v_1}{\partial x})
is not zero as it should be given the incompressible condition.

 

[hunt_mat]

does it not say that the angular velocity is uniform?

The final equation comes from taking the curl of Eulers equation.

Mat

 

[Deadstar]

does it not say that the angular velocity is uniform?
Mat

Do you mean for this equation?

(u_1 \cdot \nabla)2\Omega = 0

If so then yeah I know it should've been something else there but I've sorted that part anyway now (latex wasn't showing earlier so probably added wrong equation).

The final equation comes from taking the curl of Eulers equation.
Mat

Ok I can derive that last equation (the very last one in the OP) but the left hand side should be zero due to incompressibility should it not. Even with that sorted I still don't see how to get the equation I'm after. (The one lacking u1,v1 and w1)...

 

[hunt_mat]

What do you get?

 

[Deadstar]

What do you get?

For what? After getting these equations...

2\Omega ( \frac{\partial u_1}{\partial y} - \frac{\partial v_1}{\partial x} ) = -\frac{1}{\rho} \nabla^2 p_1
and
2\Omega ( \frac{\partial u_1}{\partial x} + \frac{\partial v_1}{\partial y} ) = \frac{\partial}{\partial t} ( \frac{\partial u_1}{\partial y} - \frac{\partial v_1}{\partial x})

I can't see any way to use them to get the final equation.

 

[hunt_mat]

I am confused slightly, at one point you're referring to \Omega as a vector and then as a scalar. can you tell me what the vector \Omega is please.
Just to make thing clear, you're writing the Euler equation as:
<br /> \frac{\partial\mathbf{u}}{\partial t}=-\frac{1}{\rho}\nabla p<br />
As the velocity field is small, you're ignoring the advection term?

 

[Deadstar]

Ok I think I'm just going to have to leave this question alone.

This thread has only confused me more.

 

[hunt_mat]

Lighthill did some work on this, you might want to check his stuff.

 

[Deadstar]

Lighthill did some work on this, you might want to check his stuff.

Don't really have time to be honest. Fluid dynamics exam today and this was only question out of 50 from our book or so that I can't do. I suppose I just need to know one thing which is why is the left hand side of 2\Omega ( \frac{\partial u_1}{\partial x} + \frac{\partial v_1}{\partial y} ) = \frac{\partial}{\partial t} ( \frac{\partial u_1}{\partial y} - \frac{\partial v_1}{\partial x}) not zero? (Or is it and the author just chose to be awkward and write it like that?)

Omega is the angular velocity so is just (0,0,\omega) as standard.

 

[hunt_mat]

I think it is because the continuity equations is:
<br /> \frac{\partial u_{1}}{\partial x}+\frac{\partial v_{1}}{\partial y}+\frac{\partial w_{1}}{\partial z}=0<br />
and so:
<br /> \frac{\partial u_{1}}{\partial x}+\frac{\partial v_{1}}{\partial y}=-\frac{\partial w_{1}}{\partial z}<br />

 

[Deadstar]

I think it is because the continuity equations is:
<br /> \frac{\partial u_{1}}{\partial x}+\frac{\partial v_{1}}{\partial y}+\frac{\partial w_{1}}{\partial z}=0<br />
and so:
<br /> \frac{\partial u_{1}}{\partial x}+\frac{\partial v_{1}}{\partial y}=-\frac{\partial w_{1}}{\partial z}<br />

Oh **** sakes we're in 3 dimensions with this question aren't we...

 

[hunt_mat]

That was you're problem?

 

[Deadstar]

That was you're problem?

Just for that small part. 49 questions all in 2D so was sort of used to that.

But no I still can't get the final solution. But doesn't matter tbh.