I Fluid separation in boundary layer

[dRic2]

Hi PF,

I'm trying to derive the Prandlt condition (not sure if it is Prandlt's work tough) for a fluid to separate from a surface in the boundary layer. The equation should be:

$\frac {\partial^2 v_x} {\partial y^2} = 0$

which is quite "intuitive" to me.

To derive it let's start from the simplified version of NS equations: the boundary layer equation derived under Prandlt's assumpions

$v_x \frac {\partial v_x} {\partial x} + v_y \frac {\partial v_x} {\partial y} = \nu \frac {\partial^2 v_x} {\partial y^2} + \frac 1 {\rho} \frac {dp} {dx}$

and the continuity equation

$\frac {\partial v_x} {\partial x} + \frac {\partial v_y} {\partial y} = 0$

Now, I found on my professor's notes that (hope my translation will be good enough):

The point of separation indicates the transition (near the surface) between different currents flowing in opposite directions, thus $\frac {\partial v_x} {\partial y} |_{y=0} = 0$

I can't understand why $\frac {\partial v_x} {\partial y} |_{y=0} = 0$. Does someone have a clue?If a sketch is need I found this picture online:

 

[mfig]

Flow separation occurs where shear stress vanishes on the surface. Look at the definition of shear stress.

 

[dRic2]

occurs where shear stress vanishes on the surface

Why?

 

[mfig]

Where separation occurs, the flow along the surface reverses direction, as your image shows. So the shear stress acts in the $x^+$ direction on one side and in the $x^-$ direction on the other side of the separation point, leaving a zero value right at the separation point.

 

[dRic2]

I have two questions now:

1) Since, as you said, shear stress vanishes when separation occurs, $\frac {\partial v_x} {\partial y} |_{y=0} = 0$ is wrong. I don't need to evaluate the derivate in $y=0$, but I need, instead, $\frac {\partial v_x} {\partial y} |_{y=p} = 0$ (where $p$ is the point of separation), right? But then $p$ should be a max or a min for $v_x$ and this doesn't seem right compared to the image.

2) So (let's take question one for answered) where separation takes place I have both $\frac {\partial v_x} {\partial y} |_{y=p} = 0$ and $\frac {\partial^2 v_x} {\partial y^2} |_{y=p} = 0$. This means $p$ is a stationary point. But it doesn't look that way in the figure...

 

[boneh3ad]

1) Since, as you said, shear stress vanishes when separation occurs, $\frac {\partial v_x} {\partial y} |_{y=0} = 0$ is wrong. I don't need to evaluate the derivate in $y=0$, but I need, instead, $\frac {\partial v_x} {\partial y} |_{y=p} = 0$ (where $p$ is the point of separation), right? But then $p$ should be a max or a min for $v_x$ and this doesn't seem right compared to the image.

It's not wrong. For a Newtonian fluid, shear stress at the wall is
\tau_{w} = \mu\left(\dfrac{\partial u}{\partial y}\right)_{y=0}.
If $\tau_w$ goes to zero, then the boundary layer separates. This is only at the wall. The separation point is a specific coordinate in $x$ where separation occurs, not a coordinate in $y$. So the point $x=p$ would be defined as the location where $\tau_w = 0$.

2) So (let's take question one for answered) where separation takes place I have both $\frac {\partial v_x} {\partial y} |_{y=p} = 0$ and $\frac {\partial^2 v_x} {\partial y^2} |_{y=p} = 0$. This means $p$ is a stationary point. But it doesn't look that way in the figure...

The condition of
\frac{\partial^2 u}{\partial y^2} = 0
is not taken at the wall, necessarily. That only has to be true at some point in the boundary layer for separation to occur. Usually it is used to discuss the conditions under which separation can occur. For example, if you look at the boundary-layer equation and simplify it near the wall, you get
\mu\left(\dfrac{\partial^2 u}{\partial y^2}\right)_{y=0} = \dfrac{dp}{dx}.
In other words, the curvature of the velocity profile near the wall is proportional to the streamwise pressure gradient. If the pressure gradient is favorable ($dp/dx < 0$) then the curvature is negative everywhere in the boundary layer. If the pressure gradient is adverse ($dp/dx > 0$) then the curvature is positive near the wall (a requirement for $\tau_w = 0$) but still negative near the edge, so there must be an inflection point at some $y$ location above the wall where $\partial^2 u/\partial y = 0$.

 

[dRic2]

Okay I finally got it, thank you! I think I got confused because my professor didn't specify where the derivate is evaluated...