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An incompressible viscous fluid of constant densite and kinematic viscosity occupies the space between porous walls at $y=0$ and $y=d$. The steady two dimensional flow is subject to a constant pressure gradient $$\frac{dp}{dx}=-G$$ Fluid enters through the wall at $y=d$ and leaves the wall at $y=0$ at a consant normal speed $V$.

Assume $\textbf u = (u(x,y_, v(x,y),0)$ and $p=p(x)$. Determine the pdes of $u$ and $v$.


$$u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=-\frac{1}{\rho} \frac{\partial p}{\partial x} + ν \bigg( \frac{\partial^2 u}{\partial x^2} +\frac{\partial^2 u}{\partial y^2} \bigg)$$

$$u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}=ν \bigg( \frac{\partial^2 v}{\partial x^2} +\frac{\partial^2 v}{\partial y^2} \bigg)$$ That is what I got for the $x$ and $y$ components.

Are the conditions $u=0$ when $y=0$ or $y=d$?

The next part says, show that $v(x,y)=-V$ is a solution to the $y$ component. Do you just sub it in and then get zero on both sides since the derivatives of a constant are zero?

The next part is to find a second order pde for $u$. So we have: $$\frac{\partial^2 u}{\partial y^2} +\frac{V}{ν} \frac{\partial u}{\partial y} +\frac{G}{\rho ν} =0$$ Is this correct? I used the continuity equation which $\partial u / \partial x = \partial V / \partial y = 0$

And to solve this, can we just turn it into a first order by letting $\eta ' = u''$ and $\eta = u'$?

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  • $\begingroup$ I think the only thing that you know is zero is $$\frac{\mathrm dp}{\mathrm dy}=0.$$ $\endgroup$ – David May 18 '16 at 1:49
  • $\begingroup$ And the differential with t terms are zero right? $\endgroup$ – snowman May 18 '16 at 13:46
  • $\begingroup$ This looks correct. There is a solution with $v(x,y) = -V$. You just show the constant satisfies the $v$-momentum equation trivially by subbing in as you say. We still have the no-slip condition $u = 0$ at $y = 0,d$. Also $\frac{\partial u}{\partial x} = 0$ by continuity since $v$ is constant. $\endgroup$ – RRL May 18 '16 at 15:16
  • $\begingroup$ You are also correct about solving a first-order differential equation for $\eta = u'$. $\endgroup$ – RRL May 18 '16 at 15:19
  • $\begingroup$ @RRL Thank god, I don't understand how to find some simplified solution when the Reynolds number is <<1. In my notes, they do some type of expansion but it doesn't seem like binomial... $\endgroup$ – snowman May 18 '16 at 15:47
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The differential equation you have seems to be correct, but you are missing terms, why did you set ∂V/∂y=0? It is correct that ∂u/∂x=∂V/∂y from the continuity equation but not that they are equal to 0.

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  • $\begingroup$ Oh ok, are my other stuff correct though? because I am concerned about them since that is where you get the differential equation from $\endgroup$ – snowman May 18 '16 at 15:14
  • $\begingroup$ Actually, you may be right, as ∂V/∂y should be 0 as V is just a constant, is this what you thought too? Your other stuff is correct, but when moving the second order derivative over you forgot to put a minus in front of it. $\endgroup$ – Arky May 18 '16 at 15:25
  • $\begingroup$ Did you consider V to be constant too?? $\endgroup$ – Arky May 18 '16 at 17:14
  • $\begingroup$ yeah, and I checked my equation, I cant find any sign error... :/ $\endgroup$ – snowman May 18 '16 at 17:20

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