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Commits (5)
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......@@ -32,12 +32,13 @@
F_\text{pressure} & = & p_\text{uniform} \ S_\text{flat wall}
\end{IEEEeqnarray}
When the fluid pressure $p$ is not uniform (for example, as depicted on the right side of the wall in figure~\ref{fig_pressure_distribution_plate_two}), the situation is more complex: the force must be obtained by integration. The surface is split in infinitesimal portions $\diff S$, and the corresponding forces are summed up as:
When the fluid pressure $p$ is not uniform (for example, as depicted on the right side of the wall in figure~\ref{fig_pressure_distribution_plate_two}), the situation is more complex: the force must be obtained by integration. The surface is split in infinitesimal portions of area $\diff S$, and the corresponding forces are summed up as:
\begin{IEEEeqnarray}{rCcCl}
F_\text{pressure} & = & \int_S \diff F & = & \int_S p \diff S \label{eq_pressure_force_scalar}
F_\text{pressure} & = & \int_S \diff F_\text{pressure} & = & \int_S p \diff S \label{eq_pressure_force_scalar}
\end{IEEEeqnarray}
\begin{equationterms}
\item for a flat surface.
\item where the $S$-integral denotes an integration over the entire surface.
\end{equationterms}
\begin{figure}[ht!]
\begin{center}
......@@ -66,6 +67,7 @@
%%%%
\subsection{Position of the pressure force}
\label{ch_position_pressure_force}
We are often interested not only in the \emph{magnitude} of the pressure force, but also its position. This position can be evaluated by calculating the magnitude of moment generated by the pressure forces about any chosen point X. This moment $\vec M_\text{X}$, using notation shown in \cref{fig_moment_forces_pressure_statics}, is expressed as:
\begin{IEEEeqnarray}{rCcCl}
......
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\atstartofexercises
\fluidmechexercisestitle
......@@ -9,6 +9,17 @@
\mecafluexboxen
\begin{boiboiboite}
Shear force on a flat surface in the plane $x,y$:
\begin{IEEEeqnarray}{rCcCl}
F_\text{{shear, direction } i} & = & \iint \tau_{\text{direction } i (x, y)} \diff x \diff y \ztag{\ref{eq_shear_force_twod_integration}}
\end{IEEEeqnarray}
Shear in the direction $j$, on a plane perpendicular to direction $i$:
\begin{IEEEeqnarray}{rCl}
||\vec \tau_{ij}|| &=& \mu \partialderivative{V_j}{i} \ztag{\ref{eq_shear_velocity_gradient}}
\end{IEEEeqnarray}
\end{boiboiboite}
\subsubsection{Flow in between two plates}
\wherefrom{Munson \& al. \smallcite{munsonetal2013} Ex1.5}
......@@ -26,8 +37,8 @@
\begin{figure}[ht!]
\begin{center}
\includegraphics[width=0.8\textwidth]{velocity_distribution_couette_flow.png}
\end{center}
\supercaption{Velocity distribution for laminar flow in between two plates, also known as \vocab{Couette flow}.}{\wcfile{Couette flow flat plate laminar velocity distributions.svg}{Figure} \cczero \oc}
\end{center}\vspace{-0.5cm}%handmade
\supercaption{Velocity distribution for laminar flow in between two plates, also known as \vocab{Couette flow}.}{\wcfile{Couette flow flat plate laminar velocity distributions.svg}{Figure} \cczero \oc}\vspace{-0.5cm}%handmade
\label{fig_plates}
\end{figure}
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\atstartofexercises
\fluidmechexercisestitle
......@@ -10,15 +10,15 @@
\mecafluexboxen
\begin{boiboite}
Continuity equation:
\begin{equation}
\frac{1}{\rho} \totaltimederivative{\rho} + \divergent{\vec V} = 0 \tag{\ref{eq_continuity_der_alt}}
\end{equation}
Continuity equation for incompressible flow:
\begin{IEEEeqnarray}{rCl}
\divergent{\vec V} & = & 0\ztag{\ref{eq_continuity_der_inc}}
\end{IEEEeqnarray}
Navier-Stokes equation for incompressible flow:
\begin{equation}
\rho \totaltimederivative{\vec V} = \rho \vec g - \gradient{p} + \mu \laplacian{\vec V} \tag{\ref{eq_navierstokes}}
\end{equation}
\begin{IEEEeqnarray}{rCl}
\rho \totaltimederivative{\vec V} &=& \rho \vec g - \gradient{p} + \mu \laplacian{\vec V} \ztag{\ref{eq_navierstokes}}
\end{IEEEeqnarray}
\end{boiboite}
......@@ -26,7 +26,7 @@ Navier-Stokes equation for incompressible flow:
\label{exo_ns_revision}
%homemade
For the continuity equation (eq.~\ref{eq_continuity_der_alt}), and then for the incompressible Navier-Stokes equation (eq.~\ref{eq_navierstokes}),
For the continuity equation (eq.~\ref{eq_continuity_der_inc}), and then for the incompressible Navier-Stokes equation (eq.~\ref{eq_navierstokes}),
\begin{enumerate}
\item Write out the equation in its fully-developed form in three Cartesian coordinates;
\item State in which flow conditions the equation applies.
......@@ -35,8 +35,8 @@ Navier-Stokes equation for incompressible flow:
Also, in order to revise the notion of total (or substantial) derivative:
\begin{enumerate}
\shift{2}
\item Describe a situation in which the total time derivative $\text{D}/\text{D}t = 0$ of a property is non-zero, even though the flow is entirely steady ($\partial{}/\partial{t} \neq 0$).
\item Describe a situation in which the the flow is unsteady, although some property of the fluid, when measured from the point of view of the particle, is not changing with time.
\item Describe a situation in which the total time derivative $\text{D}/\text{D}t$ of a property is non-zero, even though the flow is entirely completely ($\partial{}/\partial{t} = 0$).
\item Describe a situation in which the the flow is unsteady, although some property of the fluid, when measured from the point of view of a fluid particle, is not changing with time.
\end{enumerate}
......@@ -131,7 +131,7 @@ Navier-Stokes equation for incompressible flow:
\NumTabs{2}
\begin{description}
\item [\ref{exo_ns_revision}]%
\tab 1) For continuity, use eqs.~\ref{eq_totaltimederivativeheavy} and~\ref{eq_def_gradient} in~equation~\ref{eq_continuity_der_alt}. For Navier-Stokes, see eqs.~\ref{eq_ns_cartone}, \ref{eq_ns_carttwo} and~\ref{eq_ns_cartthree} p.~\pageref{eq_ns_cartone};
\tab 1) Continuity: eq.~\ref{eq_continuity_der_inc_dev} p.\pageref{eq_continuity_der_inc_dev}. Navier-Stokes: see eqs.~\ref{eq_ns_cartone}, \ref{eq_ns_carttwo} and~\ref{eq_ns_cartthree} p.~\pageref{eq_ns_cartone};
\tab 2) Read \S\ref{ch_continuity_der} p.~\pageref{ch_continuity_der} for continuity, and \S\ref{ch_navier-stokes} p.\pageref{ch_navier-stokes} for Navier-Stokes;
\tab 3) and 4) see \S\ref{ch_substantial_derivative} p.~\pageref{ch_substantial_derivative}.
\item [\ref{exo_acceleration_field}]%
......@@ -139,20 +139,20 @@ Navier-Stokes equation for incompressible flow:
\item [\ref{exo_volumetric_dilatation_rate}]%
\tab $\divergent{\vec V} = -x^2 + x - z$; thus at the probe it takes the value $\left(\divergent{\vec V}\right)_\text{probe} = \SI{-4}{\per\second}$.
\item [\ref{exo_incompressiblity}]%
\tab Apply equation~\ref{eq_conti_inc_twod} p.\pageref{eq_conti_inc_twod} to $\vec V$: the answer is yes.
\tab Apply equation~\ref{eq_continuity_der_inc_dev} p.\pageref{eq_continuity_der_inc_dev} to $\vec V$: the answer is yes.
\item [\ref{exo_missing_components}]%
\tab 1) Applying equation~\ref{eq_conti_inc_twod}: $w_1 = \num{-3}xz - \frac{1}{2}z^2 + f_{(x,y,t)}$;
\tab 1) Applying equation~\ref{eq_continuity_der_inc_dev}: $w_1 = \num{-3}xz - \frac{1}{2}z^2 + f_{(x,y,t)}$;
\tab 2) idem, $v_2 = \num{-3}axy - bzy^2 + f_{(x,z,t)}$.
\item [\ref{exo_another_acceleration_field}]%
\tab $\totaltimederivative{\vec V} = (3)\vec i + (3z + y^2x)t\vec j + (y^2 + 2xyzt) \vec k$. At the probe it takes the value $\num{3}\vec{i} + \num{250}\vec{j} + \num{496}\vec{k}$.
\item [\ref{exo_vortex_continuity}]%
\tab Apply equation~\ref{eq_conti_inc_twod} to $\vec V$ to verify incompressibility.
\item [\ref{exo_pressure_fields}]%
\tab Note: the constant (initial) value~$p_0$ is sometimes implicitly written in the unknown functions $f$.
\tab\tab 1) $p = -\rho \left[abx + \frac{1}{2}a^2 x^2 + bcy + \frac{1}{2}a^2 y^2 \right] + p_0 + f_{(t)}$;
\tab 2) $p = -\rho \left(8 x^2 + 8 y^2\right) + p_0 + f_{(t)}$;
\tab Note: the constant (initial) value~$p_{ini}$ is sometimes implicitly written in the unknown functions $f$.
\tab\tab 1) $p = -\rho \left[abx + \frac{1}{2}a^2 x^2 + bcy + \frac{1}{2}a^2 y^2 \right] + p_{ini} + f_{(t)}$;
\tab 2) $p = -\rho \left(8 x^2 + 8 y^2\right) + p_{ini} + f_{(t)}$;
\tab 3) $\partialderivative{}{x}\left(\partialderivative{p}{y}\right) \neq \partialderivative{}{y}\left(\partialderivative{p}{x}\right)$, thus we cannot describe the pressure with a mathematical function;
\tab 4) $p = -\rho \left[\frac{U_0^2}{L} \left(x + \frac{x^2}{2L} + \frac{y^2}{2L}\right) - g_x x - g_y y \right] + p_0 + f_{(t)}$.
\tab 4) $p = -\rho \left[\frac{U_0^2}{L} \left(x + \frac{x^2}{2L} + \frac{y^2}{2L}\right) - g_x x - g_y y \right] + p_{ini} + f_{(t)}$.
\end{description}
\atendofexercises
......@@ -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},
}
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