Fluid mechanics : Two dimensional converging flow

AI Thread Summary
The discussion focuses on modeling two-dimensional converging flow of a non-compressible Newtonian viscous fluid using polar coordinates. The user is attempting to derive an ordinary differential equation (ODE) for a function f(A) related to the radial velocity V_r, but encounters difficulties with boundary conditions and the angular momentum equation. The user finds that including the convective term p/u * F^2 is necessary to satisfy the boundary conditions, leading to a solution that behaves differently than expected. Despite initial confusion and numerical challenges, the user concludes that a solution exists when the convection term is retained, with the initial value of F(0) influencing the roots of the solution. The discussion highlights the complexities of fluid dynamics modeling and the importance of boundary conditions in deriving accurate equations.
Chuckstabler
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So, here's the problem I've come up with that I wanted to solve.

We're going to 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?
 
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Suppose ##P=\frac{2\mu f}{r^2}##
 
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
 
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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.
 
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