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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?

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?

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