Attention ! Gitlab fournissant maintenant nativement des certificats Let’s Encrypt aux domaines personnalisés des Gitlab Pages, nous avons coupé notre service qui le faisait automatiquement pour vous.

Il est impératif, pour que votre domaine personnalisé continue à avoir un certificat Let’s Encrypt à jour, d’activer la fonctionnalité native dans les paramètres de votre projet. Cette activation remplacera votre certificat actuel par un nouveau certificat Let’s Encrypt géré par Gitlab.

Voir les détails sur https://docs.framasoft.org/fr/gitlab/gitlab-pages-le.html

Commit fcf7c864 authored by Olivier's avatar Olivier

Chapter 8: always v instead of u for cylindrical components of velocity

parent 1833b09c
\renewcommand{\lastedityear}{2017}
\renewcommand{\lasteditmonth}{10}
\renewcommand{\lasteditday}{30}
\renewcommand{\lastedityear}{2018}
\renewcommand{\lasteditmonth}{06}
\renewcommand{\lasteditday}{18}
\renewcommand{\numberofthischapter}{8}
\renewcommand{\titleofthischapter}{Modeling large- and small-scale flows}
......@@ -214,15 +214,15 @@
This stream function allows us to describe the velocity everywhere:
\begin{IEEEeqnarray}{rCl}
u_r &=& \frac{1}{r} \partialderivative{\psi}{\theta} = U_\infty \cos \theta \left(1 - \frac{R^2}{r^2}\right)\label{eq_ur_cylinder_nolift}\\
u_\theta &=& - \partialderivative{\psi}{r} = - U_\infty \sin \theta \left(1 + \frac{R^2}{r^2}\right)\label{eq_utheta_cylinder_nolift}
v_r &=& \frac{1}{r} \partialderivative{\psi}{\theta} = U_\infty \cos \theta \left(1 - \frac{R^2}{r^2}\right)\label{eq_ur_cylinder_nolift}\\
v_\theta &=& - \partialderivative{\psi}{r} = - U_\infty \sin \theta \left(1 + \frac{R^2}{r^2}\right)\label{eq_utheta_cylinder_nolift}
\end{IEEEeqnarray}
We can even calculate the lift and drag applying on the cylinder surface. Indeed, along the cylinder wall, $r=R$ and
\begin{IEEEeqnarray}{rCl}
\left.u_r\right|_{r=R} &=& 0\\
\left.u_\theta\right|_{r=R} &=& - 2 U_\infty \sin \theta
\left.v_r\right|_{r=R} &=& 0\\
\left.v_\theta\right|_{r=R} &=& - 2 U_\infty \sin \theta
\end{IEEEeqnarray}
Since the Bernoulli equation can be applied along any streamline in this (steady, constant-energy, inviscid, incompressible) flow, we can express the pressure~$p_\text{s}$ on the cylinder surface as a function of $\theta$:
\begin{IEEEeqnarray}{rCl}
......@@ -260,8 +260,8 @@
With this function, several key characteristics of the flow field can be obtained. The first is the velocity field at the cylinder surface:
\begin{IEEEeqnarray}{rCl}
\left.u_r\right|_{r=R} &=& 0\\
\left.u_\theta\right|_{r=R} &=& - 2 U_\infty \sin \theta + \frac{\Gamma}{2 \pi R}\label{eq_utheta_rorating_cylinder}
\left.v_r\right|_{r=R} &=& 0\\
\left.v_\theta\right|_{r=R} &=& - 2 U_\infty \sin \theta + \frac{\Gamma}{2 \pi R}\label{eq_utheta_rorating_cylinder}
\end{IEEEeqnarray}
and we immediately notice that the velocity distribution is no longer symmetrical with respect to the horizontal axis (\cref{fig_lifting_cylinder}): the fluid is deflected, and so there will be a net force on the cylinder.
\begin{figure}
......@@ -273,7 +273,7 @@
\end{figure}
\begin{comment}
The position of the stagnation points can be determined by setting $u_\theta = 0$ in eq.~\ref{eq_utheta_rorating_cylinder}:
The position of the stagnation points can be determined by setting $v_\theta = 0$ in eq.~\ref{eq_utheta_rorating_cylinder}:
\begin{IEEEeqnarray}{rCl}
\theta_{\text{stagnation}} &=& \sin^{-1} \left(\frac{\Gamma}{4 \pi R \ U}\right)
\end{IEEEeqnarray}
......
\renewcommand{\lastedityear}{2017}
\renewcommand{\lastedityear}{2018}
\renewcommand{\lasteditmonth}{06}
\renewcommand{\lasteditday}{18}
\renewcommand{\numberofthischapter}{8}
......@@ -60,8 +60,8 @@
Certain flows in which both compressibility and viscosity effects are negligible can be described using the potential flow assumption (the hypothesis that the flow is everywhere irrotational). If we compute the two-dimensional laminar steady fluid flow around a cylinder profile, we obtain the velocities in polar coordinates as:
\begin{IEEEeqnarray}{rCl}
u_r &=& \frac{1}{r} \partialderivative{\psi}{\theta} = U_\infty \cos \theta \left(1 - \frac{R^2}{r^2}\right)\ztag{\ref{eq_ur_cylinder_nolift}}\\
u_\theta &=& - \partialderivative{\psi}{r} = - U_\infty \sin \theta \left(1 + \frac{R^2}{r^2}\right)\ztag{\ref{eq_utheta_cylinder_nolift}}
v_r &=& \frac{1}{r} \partialderivative{\psi}{\theta} = U_\infty \cos \theta \left(1 - \frac{R^2}{r^2}\right)\ztag{\ref{eq_ur_cylinder_nolift}}\\
v_\theta &=& - \partialderivative{\psi}{r} = - U_\infty \sin \theta \left(1 + \frac{R^2}{r^2}\right)\ztag{\ref{eq_utheta_cylinder_nolift}}
\end{IEEEeqnarray}
\begin{equationterms}
\item where \tab the origin ($r = 0$) is at the center of the cylinder profile;
......@@ -199,26 +199,26 @@
\rho \left[ \partialtimederivative{v_z} + v_r \partialderivative{v_z}{r} + \frac{v_\theta}{r} \partialderivative{v_r}{\theta} + v_z \partialderivative{v_z}{z} \right] &\nonumber\\
\ \ \ \ \ = \ \rho g_z - \partialderivative{p}{z} + \mu \left[ \frac{1}{r} \partialderivative{}{r} \left(r \partialderivative{v_z}{r} \right) + \frac{1}{r^2} \secondpartialderivative{v_z}{\theta} + \secondpartialderivative{v_z}{z} \right] \nonumber \\
\\
\frac{1}{r} \partialderivative{r u_r}{r} + \frac{1}{r} \partialderivative{u_\theta}{\theta} + \partialderivative{u_z}{z} = 0
\frac{1}{r} \partialderivative{r v_r}{r} + \frac{1}{r} \partialderivative{v_\theta}{\theta} + \partialderivative{v_z}{z} = 0
\end{IEEEeqnarray}
In this exercise, we are interested in solving the pressure field in a simplified description of a tornado. For this, we consider only a horizontal layer of the flow, and we consider that all properties are independent of the altitude $z$ and of the time $t$.% and of the angular position $\theta$.
We start by modeling the tornado as a vortex imparting an angular velocity such that:
\begin{equation}
u_\theta = \frac{\Gamma}{2 \pi r}
v_\theta = \frac{\Gamma}{2 \pi r}
\end{equation}
\begin{equationterms}
\item in which $\Gamma$ is the \vocab{circulation} (measured in \si{\per\second}) and remains constant everywhere.
\end{equationterms}
\begin{enumerate}
\item What form must the radial velocity $u_r$ take in order to satisfy continuity?
\item What form must the radial velocity $v_r$ take in order to satisfy continuity?
\end{enumerate}
Among all the possibilities for $u_r$, we choose the simplest in our study, so that:
Among all the possibilities for $v_r$, we choose the simplest in our study, so that:
\begin{equation}
u_r = 0
v_r = 0
\end{equation}
\begin{equationterms}
\item all throughout the tornado flow field.
......@@ -229,7 +229,7 @@
\item What is the pressure field throughout the horizontal layer of the tornado?
\end{enumerate}
The flow field described above becomes unphysical in the very center of the tornado vortex. Indeed, our model for $u_\theta$ is typical of an \vocab{irrotational vortex}, which, like all irrotational flows, is constructed under the premise that the flow is inviscid. However, in the center of the vortex, we are confronted with high velocity gradients over very small distances: viscous effects can no longer be neglected and our model breaks down.
The flow field described above becomes unphysical in the very center of the tornado vortex. Indeed, our model for $v_\theta$ is typical of an \vocab{irrotational vortex}, which, like all irrotational flows, is constructed under the premise that the flow is inviscid. However, in the center of the vortex, we are confronted with high velocity gradients over very small distances: viscous effects can no longer be neglected and our model breaks down.
It is observed that in most such vortices, a \vocab{vortex core} region forms that rotates just like a solid cylindrical body. This flow is \vocab{rotational} and its governing equations result in a realistic pressure distribution.
......
......@@ -392,8 +392,8 @@
With $\psi$, we get the velocity field:\pause
\begin{IEEEeqnarray*}{rCl}
u_r &=& \frac{1}{r} \partialderivative{\psi}{\theta} = U \cos \theta \left(1 - \frac{R^2}{r^2}\right)\\
u_\theta &=& - \partialderivative{\psi}{r} = - U \sin \theta \left(1 + \frac{R^2}{r^2}\right)
v_r &=& \frac{1}{r} \partialderivative{\psi}{\theta} = U \cos \theta \left(1 - \frac{R^2}{r^2}\right)\\
v_\theta &=& - \partialderivative{\psi}{r} = - U \sin \theta \left(1 + \frac{R^2}{r^2}\right)
\end{IEEEeqnarray*}
\end{frame}
......@@ -402,8 +402,8 @@
\begin{frame}
On the cylinder wall, $r=R$ and
\begin{IEEEeqnarray*}{rCl}
\left.u_r\right|_{r=R} &=& 0\\
\left.u_\theta\right|_{r=R} &=& - 2 U \sin \theta
\left.v_r\right|_{r=R} &=& 0\\
\left.v_\theta\right|_{r=R} &=& - 2 U \sin \theta
\end{IEEEeqnarray*}
\end{frame}
......
Markdown is supported
0% or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment