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I'm trying to understand the Blasius boundary layer solution, but I'm having some difficulties. Using wikipedia, I wonder how they get the first formula: http://en.wikipedia.org/wiki/Blasius_boundary_layer. $$\frac{U^2}{L}\approx\nu \frac{U^2}{\delta^2}$$ And how they get to the fourth formula: $$\delta(x) \approx \sqrt{\frac{\nu x}{U}}$$ I feel that those formulas are correlated somehow, but I don't really see how they derive those. I hope someone can help.

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up vote 2 down vote accepted

The first formula (scaling argument)

$$ \frac{U^2}{L} \sim \nu \frac{U}{\delta^2}, \tag{*} $$

comes directly from equations for boundary layer equations (and not specifically for Blasius boundary layer). We have continuity equation: $$\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}=0 $$ and x-component of momentum equation: $$u\dfrac{\partial u}{\partial x}+v\dfrac{\partial u}{\partial y}=-\dfrac{1}{\rho}\dfrac{d p}{dx}+{\nu}\dfrac{\partial^2 u}{\partial y^2}.$$ The general assumptions behind this equations is that along the x-axis quantities vary slower than along y-axis. Let us denote $L$ the characteristic length scale along the x- axis, $\delta$ the scale along the y-axis and $U$ is the characteristic velocity of the fluid. ($L$ could be, for instance, the length of the body and $\delta$ the boundary layer thickness.) We thus have $L \gg \delta$ and $u \sim U$. Various derivatives could be estimated by $$\dfrac{\partial }{\partial x} \sim \frac{1}{L},\qquad \dfrac{\partial }{\partial y} \sim \frac{1}{\delta}.$$

Now applying this approximations to continuity equation we get for instance the scale of y-component of the velocity: $$v \sim \frac \delta L U.$$

Now for the x-momentum equation both terms of the left side and the term with pressure on the right have all the following estimate: $$ u\dfrac{\partial u}{\partial x}\sim v\dfrac{\partial u}{\partial y} \sim \dfrac{1}{\rho}\dfrac{d p}{dx} \sim \frac{U^2}L ,$$ (the term with pressure could be rewritten using Bernoulli's equation as $u_\infty \dfrac{ d u_\infty}{dx}$ which would lead to the same estimate).

The single remaining term of x-momentum equation will have the following approximation: $$ {\nu}\dfrac{\partial^2 u}{\partial y^2} \sim \nu \frac {U}{\delta^2}.$$

Combining this estimates we get $(*)$. Note, that I wrote it using the $\sim$ and not $\approx$ as in wiki page.

Now we further assume that the problem at hand is the Blasius boundary layer, that is semi-infinite plate in uniform flow parallel to it. In this problem there is no intrinsic length scale (because the plate is infinite), so role of scale along the x-axis would be played by the current value of the x-coordinate $L = L (x) = x$, and the traversal scale $\delta$ also has to be dependent on $x$. Then we simply find from $(*)$ $$\delta (x) \sim \sqrt{\frac{x \nu}{U}},$$ which is yours second equation.

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Thank you for the answer! –  Rayman Aug 22 '13 at 16:10
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