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diff --git a/source/know/concept/blasius-boundary-layer/index.md b/source/know/concept/blasius-boundary-layer/index.md index de80f96..bca933f 100644 --- a/source/know/concept/blasius-boundary-layer/index.md +++ b/source/know/concept/blasius-boundary-layer/index.md @@ -12,88 +12,76 @@ 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. +at large [Reynolds number](/know/concept/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 $$v_x(x, y)$$: +Let the half-plane lie in the $$(x,z)$$-plane (i.e. at $$y = 0$$) +and exist for all $$x \ge 0$$, such that its edge lies on the $$z$$-axis. +A fluid with velocity field $$\va{v} = U \vu{e}_x$$ +approaches the half-plane's edge head-on. +To describe the fluid's movements around the plane, +we make an ansatz for the so-called **slip-flow region**'s $$x$$-velocity $$v_x(x, y)$$: $$\begin{aligned} v_x = U f'(s) - \qquad \quad + \qquad \qquad s \equiv \frac{y}{\delta(x)} \end{aligned}$$ +Where $$\delta(x) \equiv \sqrt{\nu x / U}$$ is the boundary layer thickness +estimate that was used to derive the Prandtl equations. 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 $$s$$: +and that it obeys the boundary conditions $$f'(0) = 0$$ and $$f'(\infty) = 1$$, +i.e. the fluid is stationary at the half-plane's surface $$s = 0$$, +and has velocity $$U$$ far away at $$s \to \infty$$. -$$\begin{aligned} - \pdv{s}{x} - = - y \frac{\delta'}{\delta^2} - = - s \frac{\delta'}{\delta} - \qquad \quad - \pdv{s}{y} - = \frac{1}{\delta} -\end{aligned}$$ - -Inserting the ansatz for $$v_x$$ into the incompressibility condition then yields: +Inserting the ansatz into the incompressibility condition $$\nabla \cdot \va{v} = 0$$ yields: $$\begin{aligned} \pdv{v_y}{y} = - \pdv{v_x}{x} - = U s f'' \frac{\delta'}{\delta} + = - \pdv{v_x}{s} \pdv{s}{x} + = U \frac{\delta'}{\delta} s f'' \end{aligned}$$ -Which we integrate to get an expression for the $$y$$-velocity $$v_y$$, namely: +Which we integrate by parts to get an expression for the $$y$$-velocity $$v_y$$, namely: $$\begin{aligned} v_y = U \frac{\delta'}{\delta} \int s f'' \dd{y} + = U \delta' \bigg( s f' - \int f' \dd{s} \bigg) = U \delta' \: (s f' - f) \end{aligned}$$ 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: +assuming the attack velocity $$U$$ is constant. +Inserting our expressions for $$v_x$$ and $$v_y$$ into it gives: $$\begin{aligned} - \nu f''' + U \delta' \delta f'' f - = 0 + \nu \pdvn{2}{v_x}{y} + &= v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y} + \\ + \nu \pdvn{2}{v_x}{s} \pdvn{2}{s}{y} + &= v_x \pdv{v_x}{s} \pdv{s}{x} + v_y \pdv{v_x}{s} \pdv{s}{y} + \\ + \nu U \frac{1}{\delta^2} f''' + &= - U^2 \frac{\delta'}{\delta} s f'' f' + U^2 \frac{\delta'}{\delta} f'' (s f' - f) \end{aligned}$$ -Then, substituting $$\delta(x) = \sqrt{\nu x / U}$$ and $$\delta'(x) = (1/2) \sqrt{\nu / (U x)}$$ yields: +We multiply by $$\delta^2 / U$$, cancel out some terms, +and substitute $$\delta(x) \equiv \sqrt{\nu x / U}$$, leaving: $$\begin{aligned} - \nu f''' + U \frac{\nu}{2 U} f'' f - = 0 + \nu f''' + &= - U \delta' \delta f'' f + = - U \frac{\nu}{2 U} f'' f \end{aligned}$$ -Simplifying this leads us to the **Blasius equation**, +This leads us to the **Blasius equation**, which is a nonlinear ODE for $$f(s)$$: $$\begin{aligned} |