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---
title: "Blasius boundary layer"
sort_title: "Blasius boundary layer"
date: 2021-05-29
categories:
- Physics
- Fluid mechanics
- Fluid dynamics
layout: "concept"
---
In fluid dynamics, the **Blasius boundary layer** is an application of
the [Prandtl equations](/know/concept/prandtl-equations/),
which govern the flow of a fluid
at large Reynolds number $$\mathrm{Re} \gg 1$$
close to a surface.
Specifically, the Blasius layer is the solution
for a half-plane approached from the edge by a fluid.
A fluid with velocity field $$\va{v} = U \vu{e}_x$$ flows to the plane,
which starts at $$y = 0$$ and exists for $$x \ge 0$$.
To describe this, we make an ansatz
for the *slip-flow* region's $$x$$-velocity $$v_x(x, y)$$:
$$\begin{aligned}
v_x
= U f'(s)
\qquad \quad
s
\equiv \frac{y}{\delta(x)}
\end{aligned}$$
Note that $$f'(s)$$ is the derivative of an unknown $$f(s)$$,
and that it obeys the boundary conditions $$f'(0) = 0$$ and $$f'(\infty) = 1$$.
Furthermore, $$\delta(x)$$ is the thickness of the stationary boundary layer at the surface.
To derive the Prandtl equations,
the estimate $$\delta(x) = \sqrt{\nu x / U}$$ was used,
which we will stick with.
For later use, it is worth writing the derivatives of $$s$$:
$$\begin{aligned}
\pdv{s}{x}
= - y \frac{\delta'}{\delta^2}
= - s \frac{\delta'}{\delta}
\qquad \quad
\pdv{s}{y}
= \frac{1}{\delta}
\end{aligned}$$
Inserting the ansatz for $$v_x$$ into the incompressibility condition then yields:
$$\begin{aligned}
\pdv{v_y}{y}
= - \pdv{v_x}{x}
= U s f'' \frac{\delta'}{\delta}
\end{aligned}$$
Which we integrate to get an expression for the $$y$$-velocity $$v_y$$, namely:
$$\begin{aligned}
v_y
= U \frac{\delta'}{\delta} \int s f'' \dd{y}
= U \delta' \: (s f' - f)
\end{aligned}$$
Now, consider the main Prandtl equation,
assuming that the attack velocity $$U$$ is constant:
$$\begin{aligned}
v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y}
= \nu \pdvn{2}{v_x}{y}
\end{aligned}$$
Inserting our expressions for $$v_x$$ and $$v_y$$ into this leads us to:
$$\begin{aligned}
- U^2 \frac{\delta'}{\delta} s f'' f' + U^2 \frac{\delta'}{\delta} f'' (s f' - f)
= \nu U \frac{1}{\delta^2} f'''
\end{aligned}$$
After multiplying it by $$\delta^2 / U$$ and cancelling out some terms,
it reduces to:
$$\begin{aligned}
\nu f''' + U \delta' \delta f'' f
= 0
\end{aligned}$$
Then, substituting $$\delta(x) = \sqrt{\nu x / U}$$ and $$\delta'(x) = (1/2) \sqrt{\nu / (U x)}$$ yields:
$$\begin{aligned}
\nu f''' + U \frac{\nu}{2 U} f'' f
= 0
\end{aligned}$$
Simplifying this leads us to the **Blasius equation**,
which is a nonlinear ODE for $$f(s)$$:
$$\begin{aligned}
\boxed{
2 f''' + f'' f = 0
}
\end{aligned}$$
Unfortunately, this cannot be solved analytically, only numerically.
Nevertheless, the result shows a boundary layer $$\delta(x)$$
exhibiting the expected downstream thickening.
## References
1. B. Lautrup,
*Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
CRC Press.
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