Commit 15b70349 by 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 \ No newline at end of file
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