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 $\eta = u'$?

  • $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.