# I Fluid mechanics : Two dimensional converging flow

1. Oct 28, 2016

### Chuckstabler

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. Oct 28, 2016

### Staff: Mentor

Suppose $P=\frac{2\mu f}{r^2}$

3. Oct 28, 2016

### Chuckstabler

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
4. Nov 6, 2016

### Chuckstabler

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