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Diffstat (limited to 'source/know/concept/blasius-boundary-layer')
| -rw-r--r-- | source/know/concept/blasius-boundary-layer/index.md | 34 |
1 files changed, 17 insertions, 17 deletions
diff --git a/source/know/concept/blasius-boundary-layer/index.md b/source/know/concept/blasius-boundary-layer/index.md index fd4af57..de80f96 100644 --- a/source/know/concept/blasius-boundary-layer/index.md +++ b/source/know/concept/blasius-boundary-layer/index.md @@ -12,15 +12,15 @@ 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 $\mathrm{Re} \gg 1$ +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 $x \ge 0$. +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 $v_x(x, y)$: +for the *slip-flow* region's $$x$$-velocity $$v_x(x, y)$$: $$\begin{aligned} v_x @@ -30,13 +30,13 @@ $$\begin{aligned} \equiv \frac{y}{\delta(x)} \end{aligned}$$ -Note that $f'(s)$ is the derivative of an unknown $f(s)$, -and that it obeys the boundary conditions $f'(0) = 0$ and $f'(\infty) = 1$. -Furthermore, $\delta(x)$ is the thickness of the stationary boundary layer at the surface. +Note that $$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)$$ is the thickness of the stationary boundary layer at the surface. To derive the Prandtl equations, -the estimate $\delta(x) = \sqrt{\nu x / U}$ was used, +the estimate $$\delta(x) = \sqrt{\nu x / U}$$ was used, which we will stick with. -For later use, it is worth writing the derivatives of $s$: +For later use, it is worth writing the derivatives of $$s$$: $$\begin{aligned} \pdv{s}{x} @@ -47,7 +47,7 @@ $$\begin{aligned} = \frac{1}{\delta} \end{aligned}$$ -Inserting the ansatz for $v_x$ into the incompressibility condition then yields: +Inserting the ansatz for $$v_x$$ into the incompressibility condition then yields: $$\begin{aligned} \pdv{v_y}{y} @@ -55,7 +55,7 @@ $$\begin{aligned} = U s f'' \frac{\delta'}{\delta} \end{aligned}$$ -Which we integrate to get an expression for the $y$-velocity $v_y$, namely: +Which we integrate to get an expression for the $$y$$-velocity $$v_y$$, namely: $$\begin{aligned} v_y @@ -64,21 +64,21 @@ $$\begin{aligned} \end{aligned}$$ Now, consider the main Prandtl equation, -assuming that the attack velocity $U$ is constant: +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: +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, +After multiplying it by $$\delta^2 / U$$ and cancelling out some terms, it reduces to: $$\begin{aligned} @@ -86,7 +86,7 @@ $$\begin{aligned} = 0 \end{aligned}$$ -Then, substituting $\delta(x) = \sqrt{\nu x / U}$ and $\delta'(x) = (1/2) \sqrt{\nu / (U x)}$ yields: +Then, substituting $$\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 @@ -94,7 +94,7 @@ $$\begin{aligned} \end{aligned}$$ Simplifying this leads us to the **Blasius equation**, -which is a nonlinear ODE for $f(s)$: +which is a nonlinear ODE for $$f(s)$$: $$\begin{aligned} \boxed{ @@ -103,7 +103,7 @@ $$\begin{aligned} \end{aligned}$$ Unfortunately, this cannot be solved analytically, only numerically. -Nevertheless, the result shows a boundary layer $\delta(x)$ +Nevertheless, the result shows a boundary layer $$\delta(x)$$ exhibiting the expected downstream thickening. |