diff --git a/notes.tex b/notes.tex index 4a95f76..45462ae 100644 --- a/notes.tex +++ b/notes.tex @@ -2627,6 +2627,8 @@ if you have $\tau_0 = 0$ (the Kelvin-Voigt model of a viscoelastic solid) $\delt \pagebreak \chapter{Continuum Mechanics in 3D} +\section{Navier-Stokes Derivation} + \begin{description} \item[Material Coordinates] (lagrangian) @@ -2674,7 +2676,7 @@ At $t = 0$, $X(A, t= 0) = A \implies F = \mathbb I \implies J = 1$ Since $J(t = 0) = 1$, $J \ne 0$ always implies $J(t) > 0$ \week{} -\lecture +\lecture{} \begin{ex} Uniform dilatation @@ -3050,8 +3052,6 @@ Gives $\vec t = T^T \vec n$ where $T$ is the \emph{Cauchy stress tensor} -\lecture {} - The left hand side of $( * * ) $ can be written as $$\varrho D_t \vec v + \vec v \cdot D_t \varrho + \vec v \varrho(\nabla \cdot @@ -3119,6 +3119,7 @@ So the full system we have so far is What is missing is the constitutive relation for $T$ +\lecture{} \subsection{Material frame indifference} Any constitutive law should not depend on translation or rotation: any rigid body motion. @@ -3134,9 +3135,520 @@ smooth functions, and $Q$ is a rotation matrix. $Q Q^T = \mathbb I \forall t > other coordinate frame. You should be able to compute the stress tensor in either and get the same result. + \item Form invariance: any constitutive law $T = G(T) $ stress tensor in + terms of another quantity $R$ should be the same in the transformed + coordinate frame. +\end{enumerate} + +These conditions give: + +$$G(R^*) = T^* = Q G(R) Q^T \qquad\text{For all } R \text{ and } T$$ + + +\begin{ex} + $X^* = QX + b(t) \quad |_{\p t}$ + + $V^* - Q'(t) X + QV + b'(t) $ this is how the velocity field transforms. + + And a constitutive law such as $T = G(v)$ now in spatial coordinates. (We are + using $R = \vec v$ and $R^* = \vec v^*$). + + MFI in this case is + + $$Q G(v) Q^T = G(Q'(t) x + Q v + b'(t))$$ + + For all rotations $Q(t)$ and shifts $b(t)$, but even if $Q = \mathbb I$ + $C(v) = G(v + b'(t))$ is only true if $b'(t) = 0$ ie $b$ is constant. + + +\end{ex} + +\begin{ex} + $T = \mu (\nabla v + (\nabla v)^T) $ ie $R = \nabla v + \nabla v ^T$ is + materially frame indifferent. + +\end{ex} + +\begin{thm}{}{} Rivilin Ericson Theorem + + Assuming that $T$ and $R$ are symmetric and objective and $T = G(R)$ is form + invariant, then it can be written as + + $$T = \alpha_0 \mathbb I + \alpha_1 R + \alpha_2 R^2$$ + + Where $\alpha_0, \alpha_1,\alpha_2 \in \R$ are functions of + + $I_R = tr(R) = + \lambda_1 + \lambda_2 + \lambda_3$ + + $II_R = \frac 12(tr(R)^2 - tr(R^2)) = \lambda_1 \lambda_2 + \lambda_1 + \lambda_3 + \lambda_2 \lambda_3$ + + $III_R = \det(R) = \lambda_1 \lambda_2 \lambda_3$ + + Which are the principle invariants of $R$, ie, they are conserved under + rotations and $\lambda_1 , \lambda_2 \lambda_3$ are eigenvalues of $R$. + + How to show this: + + Either constrctive by picking a range of potential rotations, or assuming + that $G$ has a (taylor) expansion $G(R) = \sum_{n = 0}^\infty \kappa_n R^n$ + with constants $\kappa_n$, and using the Cayley-Hamilton theorem: + + $R \in \R^{3\times 3} $ satisfies its characteristic equation: $\det(R - X + \mathbb I) = 0 \implies R^3 I_R R^2 - II_R + III_R \mathbb I$. + +$... = \alpha_{2,n} R^2 + \alpha_{1,n} R + \alpha_{0,n} \mathbb I$. +\end{thm} + +\week{} +\lecture{} + +\subsection{Newtonian Fluids} + +The stress depends linearly on the strain rate. (In 1D strain rate was $\p_t +\varepsilon$). In 3D strain rate is pressure. + +\begin{itemize} + \item Pressure is modelled by $T = -p \mathbb I$. Acting through an + arbitrary plane with unit normal $\vec n$, it is given by $-p \vec n$. + Pressure is isotripic. + \item Viscous stress. In 1D this was $\varepsilon_t = \p_t (U_A) = V_A$ + to model relative velocity differences between neighbouring particles. + In 3D spatial coordinates an analog would be $\p_x v, \p_y v, \p_z v$ ie + $\nabla \vec v$. + + Decompose this into its symmetric and anti-symmetric component: $\nabla + \vec v = D + W $ where: + + $D$ is the rate of deformation tensor defined as: +$$ +\begin{aligned} +{D} & \equiv \frac{1}{2}\left(\nabla \vec{v}+\nabla \vec{v}^{T}\right) \\ +&=\left(\begin{array}{ccc} +\frac{\partial v_{1}}{\partial x} & \frac{1}{2}\left(\frac{\partial v_{1}}{\partial y}+\frac{\partial v_{2}}{\partial x}\right) & \frac{1}{2}\left(\frac{\partial v_{1}}{\partial z}+\frac{\partial v_{3}}{\partial x}\right) \\ +\frac{1}{2}\left(\frac{\partial v_{1}}{\partial y}+\frac{\partial v_{2}}{\partial x}\right) & \frac{\partial v_{2}}{\partial y} & \frac{1}{2}\left(\frac{\partial v_{2}}{\partial z}+\frac{\partial v_{3}}{\partial y}\right) \\ +\frac{1}{2}\left(\frac{\partial v_{1}}{\partial z}+\frac{\partial v_{3}}{\partial x}\right) & \frac{1}{2}\left(\frac{\partial v_{2}}{\partial z}+\frac{\partial v_{3}}{\partial y}\right) & \frac{\partial v_{3}}{\partial z} +\end{array}\right) +\end{aligned} +$$ + +And $W$ is the vorticity or spin tensor: +$$ +\begin{aligned} +{W} & \equiv \frac{1}{2}\left(\nabla \vec{v}-\nabla \vec{v}^{T}\right) \\ +&=\left(\begin{array}{ccc} +0 & \frac{1}{2}\left(\frac{\partial v_{1}}{\partial y}-\frac{\partial v_{2}}{\partial x}\right) & \frac{1}{2}\left(\frac{\partial v_{1}}{\partial z}-\frac{\partial v_{3}}{\partial x}\right) \\ +\frac{1}{2}\left(\frac{\partial v_{2}}{\partial x}-\frac{\partial v_{1}}{\partial y}\right) & 0 & \frac{1}{2}\left(\frac{\partial v_{2}}{\partial z}-\frac{\partial v_{3}}{\partial y}\right) \\ +\frac{1}{2}\left(\frac{\partial v_{3}}{\partial x}-\frac{\partial v_{1}}{\partial z}\right) & \frac{1}{2}\left(\frac{\partial v_{3}}{\partial y}-\frac{\partial v_{2}}{\partial z}\right) & 0 +\end{array}\right) +\end{aligned} +$$ +\end{itemize} + +This suggests a constitutive law of the form: + +$$T = -p \mathbb I + G(D,w) \qquad(*)$$ + +However under rigid body motion $X^* = Q(t) X + b(t)$ and $D$ and $w$ transform +according to $D^* = QDQ^T$ $w^* = QwQ^T + +\underbrace{Q'Q^T}_{\text{an antisimmetric +matrix}}$ + +$(*)$ is MFI: $QG(R)Q^T = G(R^*)$ if + +$$QG(D,w) Q^T = G(D^*, w^*) = G(Q D Q^T, QwQ^T, + Q'Q^T)\quad \forall Q(t)$$ + +But if $Q = \mathbb I$ and $Q' = M$ (anti-symmetric), + +$$G(D,w) = G(D, w + M)$$ + +Which is only true if $M = \hat 0 \in \R^{3 \times 3}$ is the zero matrix. +$\implies T = -p \mathbb I + G(D)$. + +The rivilin-erickson theorem now tells us that + +$$G(D) = \alpha_0 \mathbb I + \alpha_q D + \alpha_2 D^2$$ + +And $\alpha_0 = 0$ if $D = \mathbf 0$, there is no viscous stress if $D = \mathbf 0$. + +Finally, since linearity characterises newtonian fluids, we simplify the model +by assuming linear dependence of $T$ on $D$. + +$\implies \alpha_2 = 0$ because $D^2$ is not linear. + +$\alpha_1 = 2\mu$, a constant, since $D$ is linear. + +$\alpha_0 = \lambda I_D = \lambda tr(D) = \lambda \nabla \cdot \vec v$ since +$I_d$ is the only principle invariant that is linear. + + +\begin{itemize} + \item [$\mu$] Dynamic shear velocity + \item [$p$] pressure + \item [$\lambda$] bulk viscosity +\end{itemize} + +$\implies \nabla \cdot T = - \nabla p + \lambda \nabla (\nabla \cdot \vec v ) ++ \mu(\Delta \vec v + \nabla (\nabla \cdot \vec v))$ + +And substitute into the momentum equation to obtain the Navier-Stokes equation. + +$$\varrho D_t \vec v = -\nabla p (\lamba + \mu) \nabla (\nabla \cdot \vec v) + +\mu \Delta \vec v + \varrho \vec f$$ + +Typically couple to the continuity equation: + +$$D_t \varrho + \varrho \nabla \cdot \vec v = 0$$ + +Ideal gas law (with known temperature $T$): + +$$p = \varrho R T$$ + +Usually this system is solved numerically, for example using OpenFoam software +package. + +\lecture{} + +A common simplification is to assume incompressibility: $\p_t \varrho = 0 = +\p_{\vec x_i } \varrho \implies \p_t \varrho + \Nabla \cdot (\varrho \vec v) = 0$ + +$\implies \nabvla \cot \vec v = 0$ + +The stress in this case is $T = -p \I + 2\mu D$ and we obtain the incompressible +navier-stokes equation: + +$$\varrho D_t \vec v = - \nabla p + \mu \Delta \vec v + \varrho \vec f$$ + +Coupled to : $\nabvla \cdot \vec v = 0$ + +And treat $p$ as lagrangian multiplier to enforce $\nabla\cdot \vec v = 0 $ + +Then for an incompressible newtonian fluid in cartesian coordinates, letting +$\vec v = (u,v,w)$ and $\vec f = (f,g,h)$: + +\begin{gathered} +\rho\left(\frac{\partial u}{\partial t}+u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}+w \frac{\partial u}{\partial z}\right)=-\frac{\partial p}{\partial x}+\mu\left(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}+\frac{\partial^{2} u}{\partial z^{2}}\right)+\rho f \\ +\rho\left(\frac{\partial v}{\partial t}+u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}+w \frac{\partial v}{\partial z}\right)=-\frac{\partial p}{\partial y}+\mu\left(\frac{\partial^{2} v}{\partial x^{2}}+\frac{\partial^{2} v}{\partial y^{2}}+\frac{\partial^{2} v}{\partial z^{2}}\right)+\rho g \\ +\rho\left(\frac{\partial w}{\partial t}+u \frac{\partial w}{\partial x}+v \frac{\partial w}{\partial y}+w \frac{\partial w}{\partial z}\right)=-\frac{\partial p}{\partial z}+\mu\left(\frac{\partial^{2} w}{\partial x^{2}}+\frac{\partial^{2} w}{\partial y^{2}}+\frac{\partial^{2} w}{\partial z^{2}}\right)+\rho h \\ +\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0 +\end{gathered} + + +Note: $D_t \vec v = \p_t \vec v + (\vec v \cdot \nabla) \vec v = \p_t +\vec v + (\nabla\vec v) \vec v $ + +$\Delta \vec v = \begin{pmatrix} \Delta u\\ \Delta v \\ \Delta w \end{pmatrix} $ + + +\subsubsection{Remarks} + +\begin{itemize} \item +Proving global existance in time for the incompressible navier stokes +equation is one of the millenium problems. +\item +This model is the basis for +simulations of fluids on all spatial scales: galaxies, earths mantle, oceans, +hydraulics, in biology; blood flow, tissue, cytoplasm, etc. +\item + $\mu$ is the shear viscosity: + + \begin{center} +\begin{array}{l|l|l} +\hline \text { Fluid } & \text { Viscosity (Pa-s) } & \text { Density }\left(\mathrm{kg} / \mathrm{m}^{3}\right) \\ +\hline \text { Air } & 1.8 \times 10^{-5} & 1.18 \\ +\hline \text { Water } & 8.9 \times 10^{-4} & 0.997 \times 10^{3} \\ +\hline \text { Mercury } & 1.5 \times 10^{-3} & 1.3 \times 10^{4} \\ +\hline \text { Olive oil } & 0.08 & 0.92 \times 10^{3} \\ +\hline \text { Honey } & 37 & 1.4 \times 10^{3} \\ +\hline \text { Pitch } & 2.3\times 10^8& \\ +\hline +\end{array} +\end{center} +\end{itemize} + +\section{Solutions to steady flow problems} + +Steady flow is characterised by $\p_t \vec v = 0$, and $\p_t p = 0$ and $\p_t +\varrho = 0$. + +This implies that $D_t \vec v = \p_t \vec v + (\vec v \cdot \nabla) \vec v$ + +So we now the Newtonian Navier--Stokes becomes: + +$$\begin{cases} \varrho(\vec v \cdot \nabla) \vec v = - \nabla p + \mu \Delta +\vec v + \varrho \vec f \\ \nabla \vec v = 0\end{cases}$$ + +Note that $\vec v, p \varrho $ do not change with time but single particles +still move, and their trajectories must satisfy in general: + +$$\begin{cases}\frac {d X} {dt} = V(A,t) = v(X(A,t), t)\\ X(A,t=0) = A +\end{cases}$$ + +So in steady flow: + +$$\begin{cases} \frac {dX} {dt} = V(A) = v(X(A,t)) \\ X(A,t=0) = A\end{cases}$$ + +The trajectory lines do not change with time. + + +\subsection{Coette plane flow} + +Steady flow of fluid between two parallel planes. + +\incfigw{coette}{0.6} + +Lower plate is stationary, upper plate has speed in x direction of $u_0$. +$dt y = h$. + +Use the boundary condition: assume there is a no-slip condition between the fluid +and the boundary with the plate: the fluid travels at the speed of the plate. + +At the upper plate $\vec v (x, y = h, z) = (u_0 , 0 , 0)$ + +At the lower plate $\vec v(x, y= 0, z) = (0,0,0)$ + +Assume laminar flow: $\vec v = (u,0,0)$: flow is only in the $x$ direction, +solution is homogeneous in $z$-direction: $\p_z u = 0, \p_z p = 0$ + +Now our model becomes: + +$$ +(* * *) +\begin{cases} + u_x = 0 \\ + \varrho u_x u = -p_x + \mu (u_{x x} + u_{y y } + u_{z z}) \\ + 0 = -py + 0 \\ + 0 = 0 + 0 \\ +\end{cases} +$$ + +$\p_z u = 0 = \p_x u \implies u = u(y)$ + +$\p_z p = 0 = \p_y p \implies p = p(x)$ + +So $(* * * ) $ becomes $0 = -p_x + \mu u_{y y}$, i.e., $p'(x) = \mu u''(y)$. + +Use method of separation: $p'(x) = const = \mu u''(y) \implies p' = const +\implies$ + +$$p(x) = p_0 + x p_1$$ + +Modelling intuition: $p_1 \ne 0$ would impl;y that the pressure blows up at $x = +\pm \infty \implies $ assume that $p_1 = 0$, with that $p(x) = p_0$. + +\lecture{} + +The second equation becomes $\mu u''(y) = 0 \implies u(y) = ay + b$. The +boundary conditions are $u(y = 0) = 0, u(h = 0) = u_0$, so $a = \frac {u_0} h +\implies u = \gamma y$ where $\gamma = \frac {u_0} h$ + +$\gamma = \frac {u_0} {h} $ is the \emph{shear rate} which has dimension $\frac +1t$. + +So we have a solution: + +$\vec v = \begin{pmatrix} \gamma y \\ 0 \\0 \end{pmatrix}$ and $p = p_0$ + +Remark: The stress tensor is given by + +$$T = -p \mathbb I + 2\mu D = -p_0 \mathbb I + \mu(\nabla v + {\nabla v}^T) = -p_0 + \mathbb I + \mu \begin{pmatrix}0 & \gamma & 0 \\ \gamma & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$ + + So the stress through a horizontal plane, for example $ \{ y = 0\} , + \vec n = (0,1,0)$, is given by + + + $$ + T \vec n = \left(-p_0 \mathbb I + \mu \begin{pmatrix} 0 & \gamma & 0 \\ + \gamma & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \right) \begin{pmatrix} 0 \\ 1 \\ + 0\end{pmatrix} = \underbrace{-p_0 \begin{pmatrix} 0\\1\\0 + \end{pmatrix}}_{\vec t_s} + + \underbrace{\mu + \begin{pmatrix} \gamma \\ 0 \\ 0 \end{pmatrix} }_{\vec t_p} + $$ + + $\vec t_s$ --- the shear component. + + $\vec t_p$ --- the normal component. + +\subsubsection{Shear Test} + +A Shear test can be used to investgate whether a fluid is Newtonian (where +stress is linear with respect to $\gamma$) and what the shear viscosity is. + +Put the fluid between two plates and move the upper plate with differnet speeds, +$\gamma = \frac {u_0}{h}$. Measure the force (stress $\times$ area), to see if +it is linear with respect to the velocity of the top plate. + +Since shear stress $\vec t_s = \mu \gamma \begin{pmatrix} 1 \\ 0 \\ 0 +\end{pmatrix} $, the gradient of your shear stress / shear rate graph is the +shear viscosity $\mu$. + +\subsection{Poiseuille Flow} + +Steady flow through a cylindrical pipe of length $L$ with radius $R$. + +Use the incompressible Navier-Stokes equation written in cylindrical +coordinates. + +Write the problem in polar coordinatnes about the centre of the pipe, with $z$ +running along its length: + +$\vec v = (v_R, v_\theta, v_z) = v_r \vec e_r + v_\theta \vec e_\theta ++ v_z + \vec e_z$, and + +$\vec f = f_r \vec e_r + f_\theta \vec e_\theta + f_z \vec e_z$, + +$x = r\cos \theta$ and $y = r \sin \theta$ and $z = z$. + +The incompressible Navier--Stokes system for newtonian fluid in +cyndrical coordinates is: + +\begin{aligned} +&\rho\left(\frac{\partial v_{r}}{\partial t}+v_{r} \frac{\partial v_{r}}{\partial r}+\frac{v_{\theta}}{r} \frac{\partial v_{r}}{\partial \theta}-\frac{v_{\theta}^{2}}{r}+v_{z} \frac{\partial v_{r}}{\partial z}\right)\\ +&\quad=-\frac{\partial p}{\partial r}+\mu\left[\frac{\partial}{\partial r}\left(\frac{1}{r} \frac{\partial}{\partial r}\left(r v_{r}\right)\right)+\frac{1}{r^{2}} \frac{\partial^{2} v_{r}}{\partial \theta^{2}}-\frac{2}{r^{2}} \frac{\partial v_{\theta}}{\partial \theta}+\frac{\partial^{2} v_{r}}{\partial z^{2}}\right]+\rho f_{r}\\\\ +&\rho\left(\frac{\partial v_{\theta}}{\partial t}+v_{r} \frac{\partial v_{\theta}}{\partial r}+\frac{v_{\theta}}{r} \frac{\partial v_{\theta}}{\partial \theta}+\frac{v_{r} v_{\theta}}{r}+v_{z} \frac{\partial v_{\theta}}{\partial z}\right)\\ +&\quad=-\frac{1}{r} \frac{\partial p}{\partial +\theta}+\mu\left[\frac{\partial}{\partial r}\left(\frac{1}{r} +\frac{\partial}{\partial r}\left(r v_{\theta}\right)\right)+\frac{1}{r^{2}} +\frac{\partial^{2} v_{\theta}}{\partial \theta^{2}}+\frac{2}{r^{2}} +\frac{\partial v_{r}}{\partial \theta}+\frac{\partial^{2} v_{\theta}}{\partial +z^{2}}\right]+\rho f_{\theta}\\ \\ +&\rho\left(\frac{\partial v_{z}}{\partial t}+v_{r} \frac{\partial v_{z}}{\partial r}+\frac{v_{\theta}}{r} \frac{\partial v_{z}}{\partial \theta}+v_{z} \frac{\partial v_{z}}{\partial z}\right)\\ +&\quad =-\frac{\partial p}{\partial z}+\mu\left[\frac{1}{r} +\frac{\partial}{\partial r}\left(r \frac{\partial v_{z}}{\partial +r}\right)+\frac{1}{r^{2}} \frac{\partial^{2} v_{z}}{\partial +\theta^{2}}+\frac{\partial^{2} v_{z}}{\partial z^{2}}\right]+\rho f_{z}\\ \\ +&\frac{1}{r} \frac{\partial\left(r v_{r}\right)}{\partial r}+\frac{1}{r} \frac{\partial v_{\theta}}{\partial \theta}+\frac{\partial v_{z}}{\partial z}=0 +\end{aligned} + +Unlike before there is pressure difference across the length ($z$) of the pipe +and the flow is coming from the pressure difference. + + +The boundary conditions are $\vec v = \begin{pmatrix} 0\\0\\0 \end{pmatrix} $ at +$r = R$ (no-slip boundary condition). $p(z = 0) = p_0$ and $p(z = L) = p_1 < +p_0$. + +$\vec v = \begin{pmatrix} v_r \\ v_\theta \\ v_z \end{pmatrix} $ + +Again assume laminar flow (trajectory not depending on $\theta$ or $r$). + +The velocity only has a $z$ component: + +$v_r(z = 0) = 0$ and $v_\theta(z = 0) = 0$ + +$v_r(z = L) = 0$ and $v_\theta (z = L) = 0$ + +$v_\tehta = v_r = 0 \implies \vec v = \begin{pmatrix} 0 \\ 0 \\ v_z +\end{pmatrix} $ + +And assume the solution satisfies rotational symmetry + +$v_z = v_z(r,z)$ and $p = p(r,z)$. + +And assume body forces are $0$. + +Now we can get rid of most of the Navier Stokes terms because of these +assumptions which leaves the governing equations + +$$ +\begin{cases} + 0 = \p_r p \\ + \varrho v_z \frac {\p v_z} {\p z} = -p_z + \mu (\frac 1r \p_r( r \p_r + v_z ) + \p _{z z} v_z)\\ + 0 = \p_z v_z +\end{cases} +$$ + + +$\implies p = p(z) $ and $v_z = (r)$ and $p'(z) = \mu(v_z'' (r) + \frac 1r v_z' +(r)$ + +Use method of separation: $p'(z) = constant = \mu (v_z '' + \frac 1r v_z')$ + +Use $p (z = 0) = p_0$ and $p(z = L) = p_1$ $\implies$ + +$p = p_0 + \alpha z$ and $l(L) = p_1 = p_0 + \alpha l \implies \alpha = \frac +{p_1 - p_0 } L$ + +$\implies p(z) = p_0 + \frac {p_1 - p_0}L z$ + +$\frac {p_1 - p_0} L = \mu (v_z'' + \frac 1r v_z ' ) $ + +$$\frac r \mu \frac {p_1 - p_0} L = r v_z '' + vz' = (rvz') ' $$ + +Integrate + +$$\frac {r^ 2} 2 \frac {p_1 - p_0} {\mu L} = r v_z ' - C$$ + +Integrate + +$$v_z ' = \frac r2 \frac {p_1 - p_0} {\mu L} + \frac 1r C$$ + +Integrate using $v_z(R) = 0$ + +$$v_z = (\frac {r^2}2 - \frac {R^2} 2) \frac {p_1 - p_0} {2\mu L} + C \log r$$ + +If $C \ne 0$ the velocity would blow up towards the centre of the pipe, so $C = +0$. + + $$\implies v_z = + \begin{pmatrix} 0 \\ 0 \\ R^2 - r^2 \end{pmatrix} + \frac {p_0 - p_1} {4 \mu L} $$ + + So you get a parabolic velocity profile. + + The stress tensor in cylindrical coordinates is given by: + + $$T = \begin{pmatrix} -p & 0 & \mu v_z' \\ 0 & -p & 0 \\ + \mu v_z ' & 0 & -p \end{pmatrix} $$ + + Eg the stress at the surface of a cylinder with radius $r < R$, with normal + to the cylinder of $\vec n = \begin{pmatrix} 1\\0\\0 \end{pmatrix} + \implies$ $T \vec n = \begin{pmatrix} -p \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} + 0 \\ 0 \\ \mu v_z '\end{pmatrix} = \vec t_p + \vec t_s$, shear stress and + normal stress. + +\section{Closing remarks} + +\begin{enumerate} + \item Vorticity: the curl of the velocity at a specific point + + $$\vec \omega = \nabla \times \vec v = \begin{pmatrix} \frac {\p + w} {\p y} - \frac {\p v} {\p t} \\ \frac {\p v} {\p t} - \frac {\p w } + {\p x} \\ \frac {\p v} {\p x} - \frac {\p u} {\p y} \end{pmatrix} $$ + + It is like `how much the particle rotates'. + + For example in plane Couette flow: $\vec v = \begin{pmatrix} \gamma y \\ 0 + \\ 0 \end{pmatrix} $, the vorticity is $\vec \omega = \nabla \times \vec v + = \begin{pmatrix} 0 \\ 0 \\ - \gamma\end{pmatrix} $ + +\item + Helmholtz Representation Theorem + + Let $\vec q (\vex x) : \R^3 \to \R^3$ diffble, then + + $\vec q (\vec x) = \nabla \phi + \nabla\times \vec g $ where + + $\phi:\R^3 \to + R$ and $\vec g: \R^3 \to \R^3, \nabla \cdot \vec g = 0$. + + It can be split into the scalar potential with divergence and no rotational + component (curl) and + the vector potential that is pure rotation. + + \end{enumerate} + +\week{} +\lecture{} + + \pagebreak \appendix