Proving Vorticity of Flow in Rotating Cylinder

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SUMMARY

The discussion centers on proving the vorticity of flow in a rotating cylinder, specifically using the equation of continuity and boundary conditions. The velocity field is defined as u = ΩxR, where Ω is the constant angular velocity. The calculated vorticity is initially stated as zero but is later corrected to 2Ω. Boundary conditions are also discussed, including the requirement that the fluid's velocity at the wall matches the wall's velocity.

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Raees
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Homework Statement
If liquid contained within a finite closed circular cylinder rotates about the axis k of the cylinder prove that the equation of continuity and boundary conditions are satisfied by u = ΩxR where Ω = Ωk is the constant angular velocity of the cylinder. What is the vorticity of the flow? Here R=xi+yj+zk.
Relevant Equations
Can someone check if my answer is correct please?
Can someone check if my answer is correct please?

Question:
If liquid contained within a finite closed circular cylinder rotates about the axis k of the cylinder prove that the equation of continuity and boundary conditions are satisfied by u = ΩxR where Ω = Ωk is the constant angular velocity of the cylinder. What is the vorticity of the flow? Here R=xi+yj+zk.

My answer:

u = (-Ωy) i -(-Ωx) j
Therefore: ·u = 0

vorticity: ω = x u = (0) i + (0) j + (-Ω + Ω) k = 0
 
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Raees said:
My answer:

u = (-Ωy) i -(-Ωx) j
Therefore: ·u = 0
OK. Often, the continuity equation is taken to be ·u) + ∂ρ/∂t = 0. Are you required to show that ·u = 0 implies that ∇·u) + ∂ρ/∂t = 0 for this problem?

vorticity: ω = x u = (0) i + (0) j + (-Ω + Ω) k = 0
Check this. I don't think the vorticity is zero.

What about the boundary conditions at the wall of the container?
 
TSny said:
OK. Often, the continuity equation is taken to be ·u) + ∂ρ/∂t = 0. Are you required to show that ·u = 0 implies that ∇·u) + ∂ρ/∂t = 0 for this problem?

Check this. I don't think the vorticity is zero.

What about the boundary conditions at the wall of the container?

Thanks for the reply.

Are you required to show that ·u = 0 implies that ∇·u) + ∂ρ/∂t = 0 for this problem?
No we are not, I can't find that equation in the notes nor has it been taught to us.

Check this. I don't think the vorticity is zero.
I recalculated the vorticity to be 2Ω.

What about the boundary conditions at the wall of the container?
Im not too sure about this part.
 
Has the topic of boundary conditions come up in your course?
 
Last edited:
TSny said:
Has the topic of boundary conditions come up in your course?

Not yet
 
Your result for the vorticity looks good.

It's kind of strange that you are asked about the boundary conditions if you haven't covered the topic yet. But you can check your textbook or search the net. For example here.
 
TSny said:
Your result for the vorticity looks good.

It's kind of strange that you are asked about the boundary conditions if you haven't covered the topic yet. But you can check your textbook or search the net. For example here.

Thanks, would the boundary condition be u·n = 0, where n is the vector perpendicular to the cylinder? Therefore, n = (1,1,0)?
 
Raees said:
Thanks, would the boundary condition be u·n = 0, where n is the vector perpendicular to the cylinder? Therefore, n = (1,1,0)?
Yes, that is one of the boundary conditions. Another condition is that, at the wall, the component of u that is parallel to the wall should equal the velocity of the wall. In other words, at any point of the wall, the fluid and the wall should have the same velocity so that the relative velocity is zero.
 
TSny said:
Yes, that is one of the boundary conditions. Another condition is that, at the wall, the component of u that is parallel to the wall should equal the velocity of the wall. In other words, at any point of the wall, the fluid and the wall should have the same velocity so that the relative velocity is zero.

Thanks you!

So, for u·n = 0, we have:

u = (-Ωy) i -(-Ωx) j and n = (1,1,0)

which gives: u·n = Ω(x-y)

This is not equal to zero. Did I do something wrong here?
 
  • #10
Raees said:
So, for u·n = 0, we have:

u = (-Ωy) i -(-Ωx) j and n = (1,1,0)
The expression for n is not correct. Sorry I didn't catch that in your post #7. If (x, y, z) are the coordinates of a point on the wall, then what is the expression for n at that point?
 
  • #11
TSny said:
The expression for n is not correct. Sorry I didn't catch that in your post #7. If (x, y, z) are the coordinates of a point on the wall, then what is the expression for n at that point?

If it is a cylinder, then the perpendicular vector to the curved surface is (rcosθ,rsinθ,0) and to the top/bottom surfaces is (0,0,±1). Thus, we both these vectors we get 0.
 
  • #12
Raees said:
If it is a cylinder, then the perpendicular vector to the curved surface is (rcosθ,rsinθ,0) and to the top/bottom surfaces is (0,0,±1). Thus, we both these vectors we get 0.
OK. If you want n to represent a unit vector, then (rcosθ,rsinθ,0) needs to be normalized so that it has a magnitude of 1. But it won't change the result.

What about the other boundary condition which states that the velocity of the fluid at the wall should equal the velocity of the wall?
 

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