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Stream around a perturbated cylinder

  1. May 16, 2015 #1
    1. The problem statement, all variables and given/known data
    I know that the following problem is really long but bear with me. The questions I have are quite simple.
    The problem: Calculate the speed profile of an inviscid fluid around a perturbated cylinder. The radius of the cylinder is given as $$r=R(1-\varepsilon \sin ^2 \Theta)$$ for ##\varepsilon \ll 1##.Velocity far away from the cylinder is constant.

    2. Relevant equations


    3. The attempt at a solution
    Ok, using continuity equation and assuming the field is irrotational (##\nabla \times \vec v=0##) than as expected one has to solve Laplace equation $$\nabla ^2 v(r,\Theta)=0.$$ The solution to this equation is $$\Phi (r,\Theta)=a ln(r)+b+\sum_{m=1}[A_mr^m+\frac{B_m}{r^m}][c_m\cos(m\Theta)+D_m\sin(m\Theta)]$$ Now let's assume a completely general (orientation of the) velocity far away from the cylinder $$v_\infty=v_0(\cos \varphi,-\sin \varphi)$$ and knowing ##v=-\nabla \Phi## brings me to a boundary condition far away from the cylinder $$\Phi _\infty=-v_0\cos(\Theta +\varphi).$$ Other boundary condition (I seriously doubt this is ok) is that we have a symmetry in $\Theta$ direction, meaning since ##\sin## is an odd function only ##\cos \Theta## can stay in my equation for scalar field. And now my question: What is the next boundary condition? I have a strong feeling that I should say that normal component to the cylinder is 0, written in equation, this should be $$\nabla \Phi=0$$ of course evaluated on the cylinder.
    Since ##\nabla ## is in polar coordinates, my guestion here is probably a stupid one: Would it be enough to say that ##\frac{\partial }{\partial r}\Phi=0## or not?

    I hope nobody gets angry with me if I stop here for a second and continue after "the boundary conditions" part is clear.
     
  2. jcsd
  3. May 16, 2015 #2
    Ok, let me be more specific.

    For the last boundary condition ##\nabla \Phi =0## one gets $$\nabla \Phi=((A-\frac{B}{r^2})\cos\Theta,-(A+\frac{B}{r^2})\sin\Theta)$$ Now looking at the first component the condition is $$A=\frac{B}{r^2}$$ and at the second component $$A=-\frac{B}{r^2}$$.

    Question: Which one makes physical sense and why?
     
  4. May 16, 2015 #3
    This is not how I would approach this problem. I would start out with the solution for the case where ε = 0. Then I would find the velocity (or stream function) that this solution would give on the actual surface. Then I would solve a second problem in which I put negative the velocity (or stream function) from the ε =0 case for the actual surface, and apply it at r = R. I would probably work in terms of the stream function rather than the velocity.

    Chet
     
  5. May 17, 2015 #4
    Yeah I did that also.

    For ##\varepsilon =0## the result is $$\Phi (r, \Theta)=-v_0\cos(\Theta +\varphi)(r+\frac{R^2}{r})$$. And also if I take condition $$A=\frac{B}{r^2}$$ the solution for perturbated cylinder brings me to $$\Phi (r, \Theta)=-v_0\cos(\Theta +\varphi)(r+\frac{R^2(1-2\varepsilon\sin^2\Theta)}{r})=$$ where you can see that for ##\varepsilon =0## the result matches for a perfectly shaped cylinder.
     
  6. May 17, 2015 #5
    By the way,

    What would be a more realistic model of the fluid? What would I have to consider?
     
  7. May 17, 2015 #6
    As I said earlier, your approach using the velocity potential is not a preferable way to solve this problem. It would be better to be working with the stream function. What is the equation for the stream function in the situation where ε =0 (taking the streamline impinging on the leading edge of the cylinder as the zero stream line? After you write that down as a function of r and θ, I can help you with what to do next.

    Chet
     
  8. May 21, 2015 #7
    Hi Chet,

    Hmm.. Ok, I need your help, but I don't understand yours so I might need a bit more help at the beginning. For example: I don't even know what you mean by streamline functions? What are they? In which coordinate system? Are they parametrized functions? How are they defined and stuff like that.

    So if you could give me good intro, maybe I could work it out by myself.
     
  9. May 21, 2015 #8
    The stream function ψ is used to automatically satisfy the continuity equation for a compressible fluid. In cylindrical coordinates, the velocity components are determined from the stream function by:

    $$u = \frac{1}{r}\frac{\partial ψ}{\partial θ}$$
    $$v=-\frac{\partial ψ}{\partial r}$$
    where u is the radial velocity and v is the circumferential velocity.

    These equations automatically satisfy the differential mass balance equation (continuity equation) exactly.
    For potential flow, the stream function satisfies the Laplace equation. For flow past a cylinder, the solution for the stream function is found to be:

    $$ψ=U_∞rsinθ\left[1-\left(\frac{R}{r}\right)^2\right]$$

    Constant values of the steam function correspond to the streamlines of the flow. Note that, at the surface of the cylinder, the stream function is taken to be zero. See what you get if you substitute the equation for the stream function into the equations for the velocity components.

    Chet
     
  10. May 21, 2015 #9
    Ok, I think I understand so far. Just one question. The solution ##\psi ## to Laplace equation you gave, is a solution in case where ##v=v_0 \hat e_y##, right? I can't see how else would one get that ##\sin \Theta## in equation for ##\psi##.

    If ##\psi=v_0r \sin \Theta (1-(\frac R r)^2)## than $$u=v_0\cos \Theta (1-(\frac R r)^2)$$ and $$v=-v_0\sin\Theta (1+(\frac R r)^2).$$ If I understood correctly than these are actually components of speed vector field in polar coordinates for a cylinder. If I plot it, the solution does make sense.
    ffff.PNG
    Now what? Is the idea now to look at ##R\rightarrow R(1-\varepsilon\sin^2\Theta)## or...?
     
  11. May 21, 2015 #10

    pasmith

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    The boundary condition is that the normal component of [itex]\nabla \Phi = \frac{\partial \Phi}{\partial r} \hat r + \frac1r \frac{\partial \Phi}{\partial \theta} \hat \theta[/itex] must vanish on the surface of the cylinder, which I'll call [itex]r = s(\theta) = R(1 - \epsilon \sin^2 \theta)[/itex].

    The tangent to the cylinder is [itex]\frac{d}{d\theta}(s \hat r) = \hat r \frac{ds}{d\theta} + s \hat \theta[/itex], so you can take the normal as [itex]s \hat r - \frac{ds}{d\theta} \hat \theta[/itex]. Thus the boundary condition becomes [tex]
    s\left.\frac{\partial \Phi}{\partial r}\right|_{r = s(\theta)} - \frac1s \frac{ds}{d\theta} \left.\frac{\partial \Phi}{\partial \theta}\right|_{r = s(\theta)} = 0.
    [/tex] Now the tricks:

    - Firstly, expand [itex]\frac1s = R^{-1}(1 - \epsilon\sin^2\theta)^{-1}[/itex] as a binomial series in [itex]\epsilon \sin^2\theta[/itex].
    - Secondly, expand the partial derivatives in Taylor series about [itex]r = R[/itex] so that [tex]
    \left.\frac{\partial \Phi}{\partial r}\right|_{r = s(\theta)} = \left.\frac{\partial \Phi}{\partial r}\right|_{r = R} - R\epsilon \sin^2\theta \left.\frac{\partial^2 \Phi}{\partial r^2}\right|_{r = R} + \dots, \\
    \left.\frac{\partial \Phi}{\partial \theta}\right|_{r = s(\theta)} = \left.\frac{\partial \Phi}{\partial \theta}\right|_{r = R} - R\epsilon \sin^2\theta \left.\frac{\partial^2 \Phi}{\partial r\,\partial \theta}\right|_{r = R} + \dots.
    [/tex]
    - Thirdly, substitute [itex]\Phi = \sum_{n=0} \Phi_n \epsilon^n[/itex].

    Then collect up powers of [itex]\epsilon[/itex] to find the boundary conditions to be satisfied at each order. For example, at leading order we obtain [tex]
    R\left.\frac{\partial \Phi_0}{\partial r}\right|_{r =R} = 0.[/tex]

    You can do a similar thing with the stream function: If you want [tex]
    \Psi(s(\theta),\theta) = 0[/tex] then you can expand as a Taylor series about [itex]r = R[/itex]: [tex]
    \Psi(R,\theta) - R\epsilon\sin^2\theta \left.\frac{\partial\Psi}{\partial r}\right|_{r = R} + \dots = 0.[/tex]
     
    Last edited: May 21, 2015
  12. May 21, 2015 #11
    Well, the first thing I would do would be to write the following:

    $$r=R(1-εsin^2θ)=R\left[\left(1-\frac{ε}{2}\right)+\frac{ε}{2}cos2θ\right]=R\left(1-\frac{ε}{2}\right)\left[1+\frac{ε}{(2-ε)}cos2θ\right]$$
    So,
    $$r=R^*(1-ε^*cos2θ)$$
    where ##R^*=R\left(1-\frac{ε}{2}\right)## and ##ε^*=\frac{ε}{(2-ε)}##

    Then I would write ##ψ=ψ^{0}+ε^*ψ^{1}##, where ψ0 satisfies the differential equation and boundary conditions at r = R*.
     
  13. May 21, 2015 #12
    Would you mind explaining how you got that equation for tangent to the cylinder?
     
  14. May 21, 2015 #13
    Next, what is the value of ψ0 as a function of θ on the actual surface ##r=R^*(1-ε^*cos2θ)##, to linear terms in ε?

    Chet
     
  15. May 25, 2015 #14
    pasmith following you instructions with the method I originally started with, brings me to one problem. I did manage to find out the equation of the tangent vector to the cylinder, so this is solved but there is a new problem.
    Collecting terms with ##\varepsilon ^0## brings me to exactly what you have already written $$\frac{\partial \Phi _0}{\partial r}\left. \right|_{r=R}=0$$ and knowing that $$\Phi (r,\theta)=(Ar+\frac Br)\cos \theta$$ than if I am not mistaken $$\Phi_0(r,\theta)=-v_0(r+\frac{R^2}{r})\cos \theta$$ if far away from the cylinder the velocity of the flow is ##\vec v=v_0\hat e_x##. This solution is good and makes sense. It is exactly what I would expect for a perfect cylinder (order ##\varepsilon ^0##).
    But I have a problem for powers ##\varepsilon ^1##. Because, If I collected up all the powers correctly, than $$ [R\frac{\partial \Phi_1}{\partial r}-R\varepsilon\sin^2\theta\frac{\partial \Phi_0}{\partial r}+ \frac{\varepsilon}{R}\sin(2\theta)\frac{\partial \Phi_0}{\partial \theta}]\left. \right|_{r=R}=0.$$ The second term is 0, because of the condition in power ##\varepsilon ^0## which than brings me to $$\frac{\partial \Phi_1}{\partial r}\left. \right|_{r=R}=-\frac{2}{R}2\sin(2\theta)\sin\theta.$$ And now again saying that $$\Phi _1=(A_1r+\frac{B_1}{r})\cos(\theta)$$ will bring me to a solution where ##B## and ##A## will not be constants but a function of ##\theta##. And that is not good.

    I checked everything three times and I can't find where I got it all wrong. Maybe last assumption saying $$\Phi _1=(A_1r+\frac{B_1}{r})\cos(\theta)$$ is wrong and should be something else. In that case, what? :/
     
  16. May 25, 2015 #15

    pasmith

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    At order [itex]\epsilon^1[/itex] I get [tex]
    R\left.\frac{\partial \Phi_1}{\partial r}\right|_{R} - R^2\sin^2 \theta\left.\frac{\partial^2 \Phi_0}{\partial r^2}\right|_{R} - R\sin^2 \theta \left.\frac{\partial \Phi_0}{\partial r}\right|_{R} + 2\sin \theta \cos \theta \left.\frac{\partial \Phi_0}{\partial \theta}\right|_{R} = 0[/tex] which on making use of the known [itex]\Phi_0[/itex] yields [tex]
    R\left.\frac{\partial \Phi_1}{\partial r}\right|_{R} + 2v_0 \cos \theta \sin^2 \theta + 4v_0R\cos^2 \theta \sin \theta = 0,[/tex] which with use of trigonometric identities becomes [tex]
    R\left.\frac{\partial \Phi_1}{\partial r}\right|_{R} + 2v_0 (\cos \theta - \cos^3 \theta) + 4v_0R(\sin \theta - \sin^3 \theta) = 0.[/tex]
    In these problems you want to eliminate powers of cosines and sines in favour of linear combinations of cosines and sines of integer multiples of [itex]\theta[/itex]. You can do that using de Moivre's Theorem [tex]
    \cos n\theta + i\sin n\theta = (\cos \theta + i\sin \theta)^n[/tex] and the identity [itex]\cos^2 \theta + \sin^2 \theta = 1[/itex]. Doing that with [itex]n = 3[/itex] yields [tex]
    \cos^3\theta = \frac14 \cos 3 \theta + \frac34 \cos \theta[/tex] and [tex]\sin^3 \theta = -\frac14 \sin 3\theta + \frac34 \sin \theta.[/tex]
     
    Last edited: May 25, 2015
  17. May 25, 2015 #16
    I think you have made a couple of small mistakes.
    Firstly I do agree with your ##\varepsilon ^1## terms. I can see missed one. But knowing ##\Phi_ 0## I get $$ R\frac{\partial \Phi_1}{\partial r}\bigg|_R+2v_0R\sin^2\theta\cos\theta +4v_0R\sin^2\theta\cos\theta= 0.$$ I think you are missing one ##R## that comes in the second term and since ##\frac{d}{d\varphi}\cos\varphi =-\sin\varphi## I also think the powers in the last term are misplaced.

    Anway, using identity ##\sin^2x+\cos^2x=1## and ##\cos^3x=\frac 1 4(\cos(3x)+3\cos x)## what I get is $$\frac{\partial \Phi_1}{\partial r}\bigg|_R=\frac 3 2(\cos(3\theta)-\cos \theta).$$

    But stil, How do find ##\Phi _1##?
     
  18. May 25, 2015 #17

    pasmith

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    You are quite correct. I think we can agree that [tex]
    \left.\frac{\partial \Phi_1}{\partial r}\right|_{R} = - 6v_0 \sin^2 \theta \cos \theta = -6v_0(\cos \theta - \cos^3 \theta).[/tex]

    You've lost the factor of [itex]v_0[/itex] on the right, but otherwise I agree.

    It's intuitively obvious that all we require are the [itex](Ar + Br^{-1})\cos \theta[/itex] and [itex](Cr^3 + Dr^{-3})\cos 3\theta[/itex] terms from the general solution in your first post. Applying the other boundary condition ([itex]\Phi_1 \to 0[/itex] as [itex]r \to \infty[/itex]) immediately yields two of the unknowns.

    (In general you would begin by evaluating the general solution (or its derivative with respect to [itex]r[/itex]) on each boundary, which then yields the problem of determining the coefficients of a Fourier series in [itex]\theta[/itex] at each boundary.)
     
  19. May 25, 2015 #18
    Aha, now I understand!

    Ok, doing that brings me to $$ \Phi_1(r,\theta)=\frac{v_0}{2}\left[ 3\frac{R^2}{r}\cos\theta - \frac{R^4}{r^3}\cos(3\theta) \right].$$
    Plotting the vector field $${v}'=-\varepsilon \nabla \Phi_1$$ leaves me with this awesome yet unexpected picture.
    Capturedd.PNG
    Luckily, my potential is ##\Phi =\Phi_ 0 +\varepsilon \Phi_1## and since ##\varepsilon \ll 1## the vector field for ##\varepsilon =0.01## looks much better:
    Captureg.PNG

    Thanky you bigtime, pasmith. I think this is it. I learned a lot.
     
  20. May 26, 2015 #19
    I started solving this problem in terms of the stream function, rather than the velocity potential. For the value of the boundary condition of ψ1 at r = R, I got very simply:
    $$ψ^1=\frac{v_0R}{2}(sin 3θ-sinθ)$$
    ψ1 satisfies Laplace's equation, subject to the other boundary condition ψ1-->0 at r --> ∞.

    Chet
     
  21. May 26, 2015 #20
    Ammm, did you miss out a couple of ##r## or does the result really say that ##\psi ^1## is only a function of polar angle?
     
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