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I Fluid mechanics : Two dimensional converging flow

  1. Oct 28, 2016 #1
    So, here's the problem I've come up with that I wanted to solve.

    We're gonna be using a polar coordinate system for this one. A will represent our angle theta, and r will represent our radial coordinate. We are going to be looking at a non-compressible newtonian viscous fluids. I'll now put the boundary conditions for the flow and explain what I'm trying to model.

    $$ V_r = V_r(r,A)$$
    $$ V_A = 0 $$
    $$ V_r(r, +A_0) = 0 $$
    $$ V_r(r, -A_0) = 0 $$

    As we can see, this try's to model a two dimensional flow through some sort of nozzle with viscosity. With the assumptions above, the continuity equation reduces to

    $$ \frac {1} {r} \frac {\partial (rV_r(r,A))} {\partial r} = 0$$

    Which can only be true if

    $$ V_r(r,A) = \frac {f(A)} {r} $$

    Using the fact that ##V_r(r,A) = \frac {f(A)} {r}## and using our assumptions we can simplify the radial momentum equation to the following form
    $$ \frac {d^2f(A)} {dA^2} \frac {1} {r^3} - \frac {\partial P} {\partial r} \frac {1} {u} = \frac {p} {u} \frac {f(A)^2} {r^3} $$

    Multiplying both sides by ## r^3 ## and rearranging terms we arrive at the following form
    $$\frac {\partial P} {\partial r} \frac {r^3} {u} = \frac {d^2f(A)} {dA^2} + \frac {p} {u} f(A)^2 $$

    This would be all fine and dandy except that the angular momentum equation gives me this

    $$ \frac {\partial P} {\partial A} = \frac {2u} {r^2} \frac {df(A)} {dA} $$

    So now I'm stuck. What exactly should I be doing next? My ultimate goal is to get some ODE for ##f(A)## but I don't know if that's even possible.

    Edit ; I could just assume that r^3/u * dP/dr = constant, but can I really just do that?
    Last edited: Oct 28, 2016
  2. jcsd
  3. Oct 28, 2016 #2
    Suppose ##P=\frac{2\mu f}{r^2}##
  4. Oct 28, 2016 #3
    Oh, that should help. I'll work on it and see what I find. I'll post an update soon. Thanks Chestermill (seriously you're fantastic; you've helped me on two problems so far and I really appreciate it).

    UPDATE :

    So, if we take your ansatz we find that the angular momentum equation is automatically satisfied. That's a good start. We then find that the radial momentum equation reduces to something. I'll figure that out and update
    Last edited: Oct 28, 2016
  5. Nov 6, 2016 #4
    As an update ; now I'm having problems with boundary conditions. I can't actually enforce the damn boundary conditions, which is problematic. I'll post an update soon. I tried numerically using maple and no matter what I try it always ends up with f being 0 at theta = pi/2.

    Weeeeeeellll then. Why is nothing as easy as I ever hope it would be.

    So I end up with this : F" = -4*F + p/u * F^2. I wanted to just ignore the convective term p/u*F^2, but if I do that I end up with something that cannot satisfy the boundary condition F(-Theta) = 0 F(theta) = 0. That's because the solution ends up being c1*cos(2theta) + c2*sin(2theta). Numerically however by keeping the p/u*F^2 term the boundary conditions can be satisfied. This is REALLY weird. I can't even express how weird this is to me at the moment. I don't know why.

    Have I just had too much caffeine? Am I insane? Is there actually no solution unless we include convection?

    Or maybe not, because now I'm having trouble numerically as well.

    Okay well a solution definitely exists given the convection term is retained. The initial value of F(0) determines the roots of the solution given the convection term is retained.
    Last edited: Nov 6, 2016
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