Commit 15b70349 authored by Olivier's avatar Olivier

Chapter 6: re-organization + proof-read

Basically re-wrote the chapter, even if core content remains same.
* Change in structure: first, general equations, then,
  incompressible equations
* (even) more realistic commentary about usefulness + suitability
  of the equations. Text now explains (but does not show…) that
  other terms and equations can be used in practice, according to
   needs
* Derivation of Bernoulli from Navier-Stokes moved to appendix.
parent 107458d0
This diff is collapsed.
......@@ -116,7 +116,7 @@
When applied to a vector field, it is equal to the gradient of the divergent of the field, and produces a vector field:
\begin{IEEEeqnarray}{rCl}
\laplacian{\vec A} &\equiv& %
\laplacian{\vec A} = \left(\divergent{\gradient{}}\right) \vec A &\equiv& %
\left(\begin{array}{c}%
\laplacian{A_x}\\
\laplacian{A_y}\\
......@@ -219,7 +219,47 @@
\left(p + \frac{1}{2} \rho V^2 + \rho g z \right)_1 &=& \left(p + \frac{1}{2} \rho V^2 + \rho g z \right)_2
\end{IEEEeqnarray*}
\subsection{The Bernoulli equation from the Navier-Stokes equation}
\label{appendix_benoulli_navier_stokes}
We start by following a particle along its path in an arbitrary flow, as displayed in \cref{fig_integration_navierstokes_bernoulli}. The particle path is known (condition 5 in \S\ref{ch_bernoulli} p.\pageref{ch_bernoulli}), but its speed $V$ is not.
\begin{figure}
\begin{center}
\includegraphics[width=8cm]{integration_navierstokes_bernoulli}
\end{center}
\supercaption{Different pathlines in an arbitrary flow. We follow one particle as it travels from point 1 to point 2. An infinitesimal path segment is named $\diff \vec s$.}{\wcfile{Integration from Navier-Stokes to Bernoulli along path.svg}{Figure} \cczero \oc}
\label{fig_integration_navierstokes_bernoulli}
\end{figure}
We are now going to project every component of the Navier-Stokes equation (eq.~\ref{eq_navierstokes} p.\pageref{eq_navierstokes}) onto an infinitesimal portion of trajectory $\diff \vec s$. Once all terms have been projected, the Navier-Stokes equation becomes a scalar equation:
\begin{IEEEeqnarray}{rCl}
\rho \partialtimederivative{\vec V} + \rho \advective{\vec V} & = & \rho \vec g - \gradient{p} + \mu \laplacian{\vec V}\nonumber\\
\rho \partialtimederivative{\vec V} \cdot \diff \vec s + \rho \advective{\vec V} \cdot \diff \vec s & = & \rho \vec g \cdot \diff \vec s - \gradient{p} \cdot \diff \vec s + \mu \laplacian{\vec V} \cdot \diff \vec s \nonumber
\end{IEEEeqnarray}
Because the velocity vector $\vec V$ of the particle, by definition, is always aligned with the path, its projection is always equal to its norm: $\vec V \cdot \diff \vec s = V \diff s$. Also, the downward gravity $g$ and the upward altitude $z$ have opposite signs, so that $\vec g \cdot \diff \vec s = - g \diff z$; we thus obtain:
\begin{IEEEeqnarray*}{rCl}
\rho \partialtimederivative{V} \diff s + \rho \derivative{V}{s} V \diff s & = & -\rho g \diff z - \derivative{p}{s} \diff s + \mu \laplacian{\vec V} \cdot \diff \vec s
\end{IEEEeqnarray*}
When we restrict ourselves to steady flow (condition 1 in \S\ref{ch_bernoulli}), the first left-hand term vanishes. Neglecting losses to friction (condition 4) alleviates us from the last right-hand term, and we obtain:
\begin{IEEEeqnarray*}{rCl}
\rho \derivative{V}{s} V \diff s & = & - \rho g \diff z - \derivative{p}{s} \diff s\\
\rho V \diff V & = & - \rho g \diff z - \diff p
\end{IEEEeqnarray*}
This equation can then be integrated from point 1 to point 2 along the pathline:
\begin{IEEEeqnarray}{rCl}
\rho \int_1^2 V \diff V & = & - \int_1^2 \rho g \diff z - \int_1^2 \diff p\nonumber
\end{IEEEeqnarray}
When no work or heat transfer occurs (condition 3) and the flow remains incompressible (condition 2), the density $\rho$ remains constant, so that we indeed have returned to eq.~\ref{eq_bernoulli} p.\pageref{eq_bernoulli}:
\begin{IEEEeqnarray}{rCl}
\Delta \left(\frac{1}{2} V^2\right) + g \Delta z + \frac{1}{\rho} \Delta p &=& 0 \\
\left(p + \frac{1}{2} \rho V^2 + \rho g z \right)_1 &=& \left(p + \frac{1}{2} \rho V^2 + \rho g z \right)_2
\end{IEEEeqnarray}
Thus, we can see that if we follow a particle along its path, in a steady, incompressible, frictionless flow with no heat or work transfer, its change in kinetic energy is due only to the result of gravity and pressure, in accordance with the Navier-Stokes equation.
......
......@@ -294,3 +294,11 @@
publisher= {McGraw-Hill Education},
isbn= {9781259002380},
}
@book{batchelor1967,
title= {An Introduction to Fluid Dynamics},
author= {Batchelor, George Keith},
year= {1967},
publisher= {Cambridge University Press},
isbn= {9780521663960},
}
../../a/copyright_images
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