% vim: tw=50 % 21/02/2023 09AM \subsection{The Vorticity Equation for Viscous Flow} $\nabla \times (\nu \nabla^2 \bf{u}) = \nu \nabla^2 \bf{w}$! Take curl of Navier-Stokes, and use the previous derivation from Euler (Section 2.5.2) to obtain \[ \boxed{\pfrac{\bf{w}}{t} + (\bf{u} \cdot \nabla) \bf{w} = (\bf{w} \cdot \nabla)\bf{u} + \nu \nabla^2 \bf{w}} \] ($(\bf{u} \cdot \nabla)\bf{w}$ is advection, $(\bf{w} \cdot \nabla) \bf{u}$ is vortex stretching, $\nu\nabla^2 \bf{u}$ is diffusion). \myskip For $R_e \gg 1$: Away from rigid boundaries, vorticity is dominantly advected and amplified by vortex stretching, and the final term is small. At rigid boundaries vorticity is generated by the no-slip boundary condition and diffuses a short distance to form a boundary layer. \begin{center} \includegraphics[width=0.6\linewidth] {images/681d39d8b1d111ed.png} \end{center} \myskip For parallel flow: \[ \bf{u} = (u(y, t), 0, 0) \implies \bf{\omega} = (0, 0, \omega(y, t)) \] where $\omega = -\pfrac{\bf{u}}{t}$ \[ \pfrac{\omega}{t} = \nu \pfrac[2]{\omega}{t} \] diffusion equation. \newpage \section{Inviscid Irrotational Flow} AKA Potential Flow. \subsection{The Velocity Potential} In inviscid flow, if $\bf{\omega} = \nabla \times \bf{u} = \bf{0}$ at $t = 0$ (irrotational flow) then \[ \Dfrac{\bf{\omega}}{t} = (\bf{\omega} \cdot \nabla)\bf{u} \implies \nabla \times \bf{u} = 0 \quad \forall t \ge 0 \] ``irrotational flow remains irrotational'' in inviscid flow. \myskip $\nabla \times \bf{u} = \bf{0}$ implies there exists a \emph{velocity potential} $\phi(\bf{x}, t)$ \[ \boxed{\bf{u} = \nabla \phi} \] (for example $\phi = \int^{\bf{x}} \bf{u} \cdot \dd \bf{x}$). Note: \begin{enumerate}[(1)] \item The $+$ sign (cf $\bf{f} = -\nabla \chi$) \item Can add any $f(t)$ without changing $\bf{u}$. \end{enumerate} Incompressibility $\nabla \cdot \bf{u} = 0$ implies \[ \boxed{\nabla^2 \phi = 0} \] Kinematic boundary condition $\bf{u} \cdot \bf{n} = \bf{U} \cdot \bf{n}$ implies \[ \boxed{\bf{n} \cdot \nabla \phi = \bf{U} \cdot \bf{n}} \] So solving the Euler equation for an irrotational flow reduces to solving the more familiar, and linear, Laplace's equation with Neumann boundary conditions (or $\pfrac{\phi}{n}$) -- the solution is non-zero because of the boundary conditions. Important examples of irrotational flow are flows starting from rest or with uniform flow upstream. \subsection{Examples} For simple boundaries (spheres, cylinders, rectangular channel, half-space). We can build solutions using separable solutions to $\nabla^2 \phi = 0$ in suitable coordinate systems cf Methods. \begin{itemize} \item Cartesians: for example uniform flow, $\bf{u} = \bf{U} \implies \phi = \bf{U} \cdot \bf{x} = U_1 x_1 + U_2 y + U_3 z$ For example \[ \phi = Z(z)X(x) \] with \begin{align*} Z(z) &= \begin{cases} e^{\pm kz} \\ \cosh kz \\ \sinh kz \end{cases} \\ X(x) &= \begin{cases} e^{\pm ikx} \\ \cos kx \\ \sin kx \end{cases} \end{align*} so that $\nabla^2 \phi = 0$. Corresponding flows are periodic in the $x$-direction, for example waves in section 5. \item Spherical coordinates: General axisymmetric solution to $\nabla^2 \phi = 0$ in spherical coordinates is \[ \phi = \sum_{n \ge 0} (A_n r^n + B_n r^{n - 1}) P_n(\cos\theta) \] where $P_n$ is a Legendre polynomial. We will only need the first few simple modes. (for example $\phi = A_0 \implies \bf{u} = \bf{0}$) For example $\phi = \frac{B}{r} \implies \bf{u} = \nabla \phi = -\frac{B}{r^2} \bf{e}_r$ radial flow $\propto \frac{1}{r^2}$. Volume fluw across any sphere $r = a$ is \[ q = \int_{r = a} \bf{u} \cdot \bf{n} \dd A = 4\pi A^2 \left( -\frac{B}{a^2} \right) = -4\pi B \] independent of $a$ (by mass conservation). \[ \boxed{\phi = -\frac{q}{4\pi r}} \] gives a \emph{point source of strength (volume flux) $q$}. ($q < 0$ gives a \emph{point sink}). For example $\phi = U r\cos\theta = U_2$ implies $\bf{u} = \nabla \phi = U \bf{e}_z$ \emph{uniform flow again}. Another example $\phi = \frac{U\cos\theta}{r^2} \implies \bf{u} = \nabla \phi = \cdots \implies$ \begin{center} \includegraphics[width=0.6\linewidth] {images/9d993d20b1d411ed.png} \end{center} Another example: Uniform flow past a stationary sphere: \begin{center} \includegraphics[width=0.6\linewidth] {images/e6e529a2b1d511ed.png} \end{center} \[ \begin{cases} \nabla^2 \phi = 0 \text{ in $r > a$} & \text{irrotational / incompressible} \\ \phi \to Ur\cos\theta \text{ as $r \to \infty$} & \text{far-field} \\ \bf{u} \cdot \bf{n} = \pfrac{\phi}{r} = 0 \text{ on $r = a$} & \text{kinematic bc} \end{cases} \] Linear problem, forcing $\alpha \cos \theta = P_1 (\cos\theta)$. Try $\phi = U\cos\theta \left( r + \frac{B}{r^2} \right)$: \[ \left. \pfrac{\phi}{r} \right|_{r = a} = \left. U\cos\theta \left( 1 - \frac{2B}{r^2} \right) \right|_{r = a} \] \[ \implies B = \frac{a^3}{2} \] \[ \implies \phi = U\cos\theta \left( r + \frac{a^3}{2r^2} \right) \] \[ \implies \bf{u} = \nabla \phi = \left( \pfrac{\phi}{r}, \frac{1}{r} \pfrac{\phi}{\theta}, 0 \right) \] in sphericals \[ = \left( U\cos\theta \left( 1 - \frac{a^3}{r^3} \right) , -U\sin\theta \left( 1 + \frac{a^3}{2r^3} \right), 0 \right) \] \begin{center} \includegraphics[width=0.6\linewidth] {images/a2a296c0b1d611ed.png} \end{center} \item Cylindrical Geometry / 2D Flow: General solution to $\nabla^2 \phi = 0$ in plane polars is \[ \phi = C_0 + A_0 \ln r + B_0\theta + \sum_{n \ge 1} (A_n r^n + B_n r^{-n})(C_n \cos n\theta + D_n \sin n\theta) \] -- again we will only need a few modes. \end{itemize}