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Peeter
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Homework Statement
In [1], problem 2.3 (ii), we are asked to show that for a pipe with circular cross section [itex]r = a[/itex] and constant pressure gradient [itex]P = -dp/dz[/itex] one has
[tex]u_z = \frac{P}{4 \mu}\left( a^2 - r^2 \right)[/tex]
[tex]u_r = 0[/tex]
[tex]u_\theta = 0[/tex]
References:
[1] D.J. Acheson. <em>Elementary fluid dynamics</em>. Oxford University Press, USA, 1990.
Homework Equations
If one does assume that [itex]u_r = u_\theta = 0[/itex], then the Navier-stokes equations for an incompressible steady state flow takes the form
[tex]\begin{align}u_z \frac{\partial {u_z}}{\partial {z}} = \frac{P}{\rho} + \nu \left(\frac{1}{{r}} \frac{\partial {}}{\partial {r}}\left(r \frac{\partial {u_z}}{\partial {r}} \right) + \frac{\partial^2 {{u_z}}}{\partial {{\theta}}^2} + \frac{\partial^2 {{u_z}}}{\partial {{z}}^2}\right)\end{align} [/tex]
[tex]\begin{align}0 = \frac{\partial {p}}{\partial {r}}\end{align} [/tex]
[tex]\begin{align}0 = \frac{\partial {p}}{\partial {\theta}}\end{align} [/tex]
The Attempt at a Solution
I can actually solve this problem completely, but have some trouble justifying some of the steps.
a) One of them is the assumption that [itex]u_r = u_\theta = 0[/itex], which I use as a starting point to express NS above. This intuitively makes sense, but are there other physical principles (other than intuition) that lead to this conclusion?
b) In NS above we can kill of the [itex]\partial_{\theta\theta}[/itex] portion of the Laplacian, again by intuition. However, it's not too hard to imagine that you could have a flow where one could have an angular dependence in the flow (ie. imagine a really big pipe where fluid is being injected harder at the base of the pipe than at the top). Something like that would probably have a different pressure dependence. It isn't clear to me what sort of mathematical argument that one could use to show that for this constant pressure gradient scenerio one must have no [itex]\theta[/itex] dependence in the velocity.
c) Finally, once we make assumption (b) so that [itex]\partial_\theta u_z = 0[/itex], NS takes the form (with [itex]w = u_z[/itex])
[tex]\begin{align}w \frac{\partial {w}}{\partial {z}} = \frac{P}{\rho} + \nu \left(\frac{1}{{r}} \frac{\partial {}}{\partial {r}}\left(r \frac{\partial {w}}{\partial {r}} \right) + \frac{\partial^2 {{w}}}{\partial {{z}}^2}\right).\end{align} [/tex]
Attempting separation of variables with [itex]w = Z(z) R(r)[/itex], this appears to be inseparable due to the non-linearity of the [itex](\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u}[/itex] term on the LHS. The fact that this is inseparable doesn't seem like it's a good justificaion for requiring that [itex]w = w(r)[/itex] only and not [itex]w = w(r,z)[/itex]. I could imagine that it is still possible to find solutions of the form [itex]w = w(r,z)[/itex]. Is there some other argument that can be made to show that there is no [itex]z[/itex] dependence in this remaining component of the fluid's velocity?
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