# Find equation for self similar solution to the thin shear layer equations

1. Jul 7, 2017

### fahraynk

1. The problem statement, all variables and given/known data
An alternative to the usual no-slip wall boundary condition $u = 0, v = 0$, is the porous-wall condition $u = 0$, $v = v_w(x)$. A porous wall with negative $v_w$ (suction) is known to delay transition to turbulence, for example.
1a) Assuming a uniform $u_e(x)$, determine what differential equation needs to be satisfied by a self-similar boundary layer solution for a nonzero vw. Assume that $v_w≪u_e$ so that the TSL equations remain valid.

1b) Determine the restrictions that must be placed on $v_w(x)$ so that the similar boundary layer solution you derived in Part a) can occur.
1c) Now assuming a non-uniform $u_e(x)$ determine what differential equation needs to be satisfied by a self-similar boundary layer solution for a nonzero $v_w$
1d) Determine the restrictions that must be placed on $v_w(x)$ and $u_e(x)$ so that the similar boundary
layer solution you derived in Part c) can occur.

2. Relevant equations
density, pressure, velocity in x, velocity in y, velocity external, viscosity = $\rho$,$P$,$u$,$v$,$u_e$,$\vartheta$
$X_y=\frac{\partial X}{\partial Y}$
stream function : $-\phi_x=v$, $\phi_y=u$
stream function form of thin shear equations :
$$\phi_y * \phi_{xy} - \phi_x*\phi_{yy}=U_e*\frac{\partial U_e}{\partial x}+\vartheta * \phi_{yyy}$$

3. The attempt at a solution

So the answer for "non self similar solution" would be the regular thin shear layer equations with a boundary condition at the wall of $v(y=0)=v_w$
$$u_x+u_y=0\\\\ u_t+u*u_x+v*u_y=-\frac{1}{\rho}P_x+\vartheta*u{yy}$$
with a solution $u=u_e[1-e^\frac{v_wy}{\vartheta}]$

But the question wants to know the equation that needs to be solved by a self similar solution.

The professor gave a derivation of a self similar solution with no velocity at the wall, but I think there is a mistake in it. Please take a look :
Self similar solution condition :
$$BU(x,y)=U(x+dx,\frac{y}{A})\\\\ B=(1+\epsilon')\\\\ \frac{y}{A}=y+\epsilon y$$
through a taylor approximation, he gets a similar solution condition :
$$\frac{\partial U}{\partial x}+\frac{\epsilon}{dx}Y\frac{\partial U}{\partial y} = \frac{\epsilon'}{dx}U$$
with $\frac{\epsilon}{dx}=\theta$ and $\frac{\epsilon'}{dx}=\alpha$, put in stream function form :
$$\phi_{xy}+\theta Y \phi_{yy}= \alpha \phi_y$$
now take the integral of both sides in order to get a function for $\phi_x$ that can be used to make the stream function as a function of y alone. :
$$\phi_x=v_w-\theta y \phi_y + \phi(\alpha + \theta)-\alpha\phi(y=0)$$
So here is my first question/point of confusion. The professor explains that $\phi(y=0)=0$. He derived this for a situation with no suction, I am trying to modify it for suction. I think $\phi(y=0)=-v_w*x$ because $-\phi_x=v$. Is this correct?

Then, we plug I plug it into the stream equation I get :
$$(\alpha+\theta)(\phi_y)^2-[(\alpha+\theta) \phi+v_w-\alpha \phi (y=0)] \phi_{yy}=u_e \frac{\partial u_e}{\partial x} + \vartheta \phi_{yyy}$$
The professor leaves off a $\theta$ from the first term, with $\phi=0$ and $v_w=0$ he gets :
$$\alpha(\phi_y)^2-(\alpha+\theta) \phi \phi_{yy}=u_e \frac{\partial u_e}{\partial x} + \vartheta \phi_{yyy}$$

So... I think he just substituted wrong which is why the first term with $(\phi_y)^2$ does not equal $(\alpha + \theta)$

Is my equation the correct answer? Also should $\phi=-v_wx$?

2. Jul 12, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

3. Jul 17, 2017

### fahraynk

I basically just need someone who knows about similar solutions to the thin shear layer approximation of navier stokes equations on a flat plate (blasius equation). I don't get how adding suction at the wall affects the self similar stream equation.