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Homework Help: Rotating, submerged cylinder problem.

  1. Dec 2, 2007 #1
    [SOLVED] Rotating, submerged cylinder problem.

    1. The problem statement, all variables and given/known data
    I need to develop the equations for describing a submerged cylinder which is rotating at an angular velocity of [tex] \omega [/tex], and there is a constant fluid flow of [itex] \overrightarrow{v} = v_0\overrightarrow{i} [/itex] acting on the cylinder (assume that the position of the cylinder is fixed, so that it doesn't get swept away, but that it rotates about its centre). I am to assume that the fluid is incompressible (at least for the first part. I'm suppose to drop that assumption later on).

    The cylinder is considered to be solid, and so there is no fluid flow through the cylinder.

    The cylinder is considered to be infinitely long in the z axis (cylindrical coordinates are used through out). Because of this, the z term in the laplacian can be dropped.

    2. Relevant equations
    I need to solve:
    [tex] \nabla^2 \phi = 0 [/tex]

    Where [tex] \phi [/tex] is the velocity potential of the fluid, i.e. [tex] \overrightarrow{v} = \nabla \phi [/tex]

    From what I can tell, these are the boundary Conditions of the problem:
    1) For [tex] r >> a [/tex] (a being the radius of the cylinder), [tex] \overrightarrow{v} \rightarrow v_0\overrightarrow{i} [/tex].
    2) [tex] \frac{\partial \phi}{\partial n} = 0 [/tex] for r = a, where n represents the vector normal to the surface of the cylinder (since the centre of the cylinder is at the coordinate system origin, this can be replaced with r).
    3) at r = a, [tex] v = \omega a [/tex]. I am very unsure of this boundary condition.

    3. The attempt at a solution

    At the moment, I have little clue how to do this, but I've tried two methods so far:
    Method 1:
    I took the case for a non-rotating cylinder, and assumed that there was another function of r, theta and omega which was added to it, and then substituted this into the laplacian equation, but I didn't get very far with that.
    Method 2:
    I started from the general solution for a non-rotating sphere, and then attempted to see how the new boundary conditions affected the constants given, but again, I was met with failure.
    Any help at all would be appreciated.
  2. jcsd
  3. Dec 2, 2007 #2


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    Laplace's equation is linear. So, I think a superposition of easier solutions might work.

    Have you tried looking at this problem as a combination of a uniform flow around a stationary cylinder and rotation of the cylinder in stagnant fluid, and adding the solutions?
  4. Dec 3, 2007 #3
    That is a very good idea. Thank you very much Siddharth. After a little bit of thought, this is what I have come up with (there is a problem still, but I'm much more optimistic about this problem now :D)

    This is what I've got now:
    Solution for a non-rotating cylinder:
    [tex] \phi_1 = v_0\left\{r + \frac{a^2}{r} \right\} cos \theta[/tex]

    Now, I know that the general solution to the problem should be something along the lines of:
    [tex] \phi = a_1\phi_1 + a_2\phi_2 [/tex]

    [itex]\phi_1 [/itex] is the solution for the non-rotating cylinder in uniform flow, and [itex] \phi_2 [/itex] is the solution for a rotating cylinder in a stationary flow.

    To get [itex] \phi_2 [/itex], I looked at boundary Conditions. I thought that for a stationary flow (but rotating cylinder), the velocity at large r would approach zero, and so the velocity potential would be constant in terms of r (i.e. it would be independant of r).

    I think thought, if the cylinder is rotating with an angular speed of [itex] \omega [/itex] in the theta-hat direction, then the velocity of the fluid in the theta-hat direction at the boundary between the cylinder and the fluid (i.e. r = a), would be [itex] v = \omega a [/itex].

    They are the conclusions that I came to after some thought. Tell me if the boundary conditions I have decided upon are incorrect.

    Putting all that into some mathematics:
    [tex] \frac{\partial \phi_2}{\partial r} =0 [/tex]
    [tex] \frac{1}{r}\frac{\partial \phi_2}{\partial \theta} = \omega a [/tex]
    The equation above represents the boundary condition of the problem, and so the LHS should be evaluated at r = a, but I didn't know how to put that into latex.

    From the second BC, I found that [itex] \phi_2 = \omega a^2 \theta [/itex]. This solution seems to fit with the problem, and so, it implies that the solution to the original problem is:
    [tex] \phi = a_1v_0\left\{r + \frac{a^2}{r} \right\} cos \theta+ a_2 \omega a^2 \theta [/tex]

    Where a1 and a2 are constants. It seems to be right because when I put it back into the laplace equation, it appears to be a solution. Also, when I grad the function, I get a function for the velocity which looks correct. The stagnation points are where they are suppose to be for if the angular speed of rotation is zero. Although, from the result of this equation, there is no stagnation point if the cylinder is rotating. That seems to make sense as a result, but I must ask... is my solution right?
    Last edited: Dec 3, 2007
  5. Dec 4, 2007 #4


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    That's right.

    I think a_1 and a_2 should both be 1. Only then will it reduce to the individual cases. For example, when [itex]\omega = 0 [/itex] in the general case, the solution should reduce to [itex]\phi_1[/itex] and this won't happen unless a_1 is 1.

    Yeah, looks right to me. In fact, since [itex]v_r = \frac{\partial \phi}{\partial r}[/itex] is 0, you can straight away say this.

    I think there's a typo in the last equation. It should be

    [tex] \frac{1}{r}\frac{\partial \phi_2}{\partial \theta} = \frac{\omega a^2}{r} [/tex]

    Why is there no stagnation point? When v is 0, can't you solve for theta to find the new stagnation points? ie, when

    [tex] -2v_0 \sin \theta + a \omega = 0[/tex]
    Last edited: Dec 4, 2007
  6. Dec 4, 2007 #5
    oh yes, you are correct. Sorry for the typos I made in my last post. I don't think they helped! You are correct though about the stagnation point. I assumed that there was no stagnation point because all fluid in contact with the surface of the cylinder had the velocity of the edge of the sphere, and so (at first) it appeared that there was nowhere on the surface of the cylinder where the speed of the fluid became zero, but on reflection, that's wrong.

    Thanks for all the help. It is greatly appreciated.
  7. Dec 4, 2007 #6


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    No problem, glad to help.
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