Commit cac61be3 authored by Olivier's avatar Olivier

Some re-organization, some pruning, some renaming

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......@@ -24,9 +24,12 @@
{\LARGE Syllabus}\par
Fluid Mechanics for Master Students\\
(draft version for 2019)\\
\href{https://fluidmech.ninja/}{https://fluidmech.ninja/}\\
Summer semester 2018\\
Summer semester 2019\\
Olivier Cleynen
\mecafluboxtmp
\vspace{0em}
\end{center}
......@@ -82,23 +85,23 @@ Welcome to the Fluid Mechanics course of the \textit{Chemical and Energy Enginee
\clearpage
\section*{Time plan}
Our time plan, which is affected by the holidays in Saxony-Anhalt, should be as follows:
Week 14: Ch. 0 (Important concepts)\\
Week 15: Ch. 1 (Effects of pressure), \textcolor{vocabcolor}{No exercise session}\\
Week 16: Ch. 2 (Effects of shear)\\
Week 17: Ch. 3 (Analysis of existing flows) part 1/2\\
Week 18: Ch. 3 (Analysis of existing flows) part 2/2\\
Week 19: \textcolor{vocabcolor}{No lecture}\\
Week 20: Ch. 4 (Predicting fluid flow)\\
Week 21: Ch. 5 (Flow in pipes), \textcolor{vocabcolor}{Exam briefing}\\
Week 22: Ch. 6 (Scale effects)\\
Week 23: Ch. 7 (Flow near walls)\\
Week 24: Ch. 8 (Large- and small-scale flows), \textcolor{vocabcolor}{Feedback}\\
Week 25: Ch. 9 (Introduction to compressible flow)\\
Week 26: revision \& practice\\
Our time plan, taking into account the holidays in Saxony-Anhalt, should be as follows:
Week 14: Ch. 1 (Basic flow quantities)\\
Week 15: Ch. 2 (Analysis of existing flows, 1D)\\
Week 16: Ch. 3 (Analysis of existing flows, 3D), \textcolor{vocabcolor}{No exercise session}\\
Week 17: Ch. 3 (Analysis of existing flows, 3D)\\
Week 18: Ch. 4 (Effects of pressure) \\
Week 19: Ch. 5 (Effects of shear), \textcolor{vocabcolor}{Exam briefing}\\
Week 20: Ch. 6 (Predictin of fluid flows)\\
Week 21: Ch. 7 (Pipe flows)\\
Week 22: \textcolor{vocabcolor}{No lecture}\\
Week 23: Ch. 8 (Dealing with turbulence), \textcolor{vocabcolor}{Feedback}\\
Week 24: Ch. 9 (Engineering models)\\
Week 25: Ch. 10 (Flow near walls)\\
Week 26: Ch. 11 (Large- and small-scale flows)\\
Week 27: revision \& practice\\
Week 28: Examination: July 12\up{th}, from 15:00 to 17:00 in room G26-HS1
Week 28: Examination: July 11\up{th}
\section*{Copyright and remixing}
......@@ -113,7 +116,7 @@ It’s a pleasure to join you for this course this semester! Fluid mechanics is
\begin{flushright}
Olivier Cleynen\\
April 2018
%April 2018
\end{flushright}
\restoregeometry\restoredefaultfootoffset\cleardoublepage
......@@ -16,8 +16,9 @@
\vspace{0.5em}
{A one-semester course for students in\\
the \textit{Chemical and Energy Engineering} program\\
at the Otto-von-Guericke-Universität Magdeburg\\}%\\
at the University Otto von Guericke of Magdeburg\\}%\\
%\ccLogo\ \ccAttribution\ \ccShareAlike}
\mecafluboxtmp
\end{center}
\vspace{\stretch{1}}
......
\renewcommand{\lastedityear}{2018}
\renewcommand{\lasteditmonth}{04}
\renewcommand{\lasteditday}{02}
\renewcommand{\numberofthischapter}{0}
\renewcommand{\titleofthischapter}{Important concepts}
\renewcommand{\numberofthischapter}{1}
\renewcommand{\titleofthischapter}{Basic flow quantities}
\setcounter{chapter}{-1}
\setcounter{chapter}{0}
\fluidmechchaptertitle
\mecafluboxen
\mecafluboxtmp
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Concept of a fluid}
......
\renewcommand{\lastedityear}{2018}
\renewcommand{\lasteditmonth}{04}
\renewcommand{\lasteditday}{01}
\renewcommand{\numberofthischapter}{0}
\renewcommand{\titleofthischapter}{Important concepts}
\renewcommand{\numberofthischapter}{1}
\renewcommand{\titleofthischapter}{Basic flow quantities}
\atstartofexercises
\fluidmechexercisestitle
......
......@@ -7,6 +7,7 @@
\fluidmechchaptertitle
\mecafluboxen
\mecafluboxtmp
\section{Motivation}
......@@ -20,10 +21,10 @@
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Describing the boundary layer}
\section{The concept of boundary layer}
%%%%%%%%%%%%%%%%%
\subsection{Concept}
\subsection{Rationale}
At the very beginning of the 20\up{th} century, \we{Ludwig Prandtl} observed that for most ordinary fluid flows, viscous effects played almost no role outside of a very small layer of fluid along solid surfaces. In this area, shear between the zero-velocity solid wall and the outer flow dominates the flow structure. He named this zone the \vocab{boundary layer}.
We indeed observe that around any solid object within a fluid flow, there exists a thin zone which is significantly slowed down because of the object’s presence. This deceleration can be visualized by measuring the velocity profile (\cref{fig_boundarylayer}).
......@@ -137,7 +138,7 @@
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The laminar boundary layer}
\section{Laminar boundary layers}
%%%%%%%%%%%%%%%%%
\subsection{Governing equations}
......@@ -205,6 +206,7 @@
c_{f_{(x)}} &=& \frac{\num{0,664}}{\sqrt{\rex}} \IEEEyessubnumber\label{eq_cf_lam}
\end{IEEEeqnarray}
\begin{comment}
%%%%%%%%%%%%%%%%%
\subsection{Pohlhausen’s model}
......@@ -249,11 +251,11 @@
It can be seen now that neither Blasius nor Pohlhausen’s models are fully satisfactory. The first is drawn from a handful of realistic, known boundary conditions, but we fail to find an analytical solution to match them. The second is drawn from the hypothesis that the analytical solution is simple, but it is not very accurate. And we have not even considered yet the case for turbulent flow!\\
Here, it is the process which matters to us. Both of those methods are frequently-attempted, sensible approaches to solving problems in modern fluid dynamics research, where, just as here, one often has to contend with approximate solutions.
\end{comment}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Transition}
\section{Boundary layer transition}
\label{ch_bl_transition}
After it has traveled a certain length along the wall, the boundary layer becomes very unstable and it transits rapidly from a laminar to a turbulent regime (\cref{fig_bl_transition}). We have already briefly described the characteristics of a turbulent flow in chapter~5; the same apply to turbulence within the boundary layer. Again, we stress that the boundary layer may be turbulent in a globally laminar flow (it may conversely be laminar in a globally turbulent flow): that is commonly the case around aircraft in flight, for example.
......@@ -275,7 +277,7 @@
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The turbulent boundary layer}
\section{Turbulent boundary layers}
The extensive description of turbulent flows remains an unsolved problem. As we have seen in chapter~5 when studying duct flows, by contrast with laminar counterparts turbulent flows result in
\begin{itemize}
......@@ -313,7 +315,7 @@
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Separation}
\section{Flow separation}
%%%%%%%%%%%%%%%%%
......@@ -371,7 +373,7 @@
Predicting in practice the position of a separation point is difficult, because an intimate knowledge of the boundary layer profile and of the (external-flow-generated) pressure field are required —\ and as the flow separates, these are no longer independent. Resorting to experimental measurements, in this case, is often a wise idea!
\begin{comment}
%%%%%%%%%%%%%%%%%
\clearfloats %handmade
\subsection{Separation according to Pohlhausen}
......@@ -402,5 +404,6 @@
~\\
is what should be remembered here, for it is typical of the research process in fluid dynamics.
\end{comment}
\atendofchapternotes
\renewcommand{\lastedityear}{2018}
\renewcommand{\lasteditmonth}{06}
\renewcommand{\lasteditday}{18}
\renewcommand{\numberofthischapter}{7}
\renewcommand{\numberofthischapter}{10}
\renewcommand{\titleofthischapter}{Flow near walls}
\atstartofexercises
......
\renewcommand{\lastedityear}{2018}
\renewcommand{\lasteditmonth}{06}
\renewcommand{\lasteditday}{18}
\renewcommand{\numberofthischapter}{8}
\renewcommand{\titleofthischapter}{Modeling large- and small-scale flows}
\renewcommand{\numberofthischapter}{11}
\renewcommand{\titleofthischapter}{Large- and small-scale flows}
\fluidmechchaptertitle
\mecafluboxen
\mecafluboxtmp
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Motivation}
......
\renewcommand{\lastedityear}{2018}
\renewcommand{\lasteditmonth}{06}
\renewcommand{\lasteditday}{18}
\renewcommand{\numberofthischapter}{8}
\renewcommand{\titleofthischapter}{Modeling large- and small-scale flows}
\renewcommand{\numberofthischapter}{11}
\renewcommand{\titleofthischapter}{Large- and small-scale flows}
\atstartofexercises
\fluidmechexercisestitle
......
\renewcommand{\lastedityear}{2017}
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\renewcommand{\lasteditday}{23}
\renewcommand{\numberofthischapter}{2}
\renewcommand{\titleofthischapter}{Analysis of existing flows (1D)}
\fluidmechchaptertitle
\mecafluboxen
\mecafluboxtmp
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Motivation}
todo
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{One-dimensional flow problems}
todo
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Balance of mass}
todo
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Balance of momentum}
todo
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Balance of energy}
todo
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The Bernoulli equation}
\label{ch_bernoulli}
The Bernoulli equation has very little practical use for us; nevertheless it is so widely used that we have to dedicate a brief section to examining it. We will start from equation~\ref{eq_rtt_energy_simple} and add five constraints:
\begin{enumerate}
\item Steady flow.\\
Thus $\timederivative{} \iiint_\cv \rho\ e \diff \vol = 0$.\\
In addition, $\dot m$ has the same value at inlet and outlet;
\item Incompressible flow.\\
Thus, $\rho$ stays constant;
\item No heat or work transfer.\\
Thus, both $\dot Q_{\net\ \inn}$ and $\dot W_\text{shaft, net in}$ are zero;
\item No friction.\\
Thus, the fluid energy $i$ cannot increase due to an input from the control volume;
\item One-dimensional flow.\\
Thus, our control volume has only one known entry and one known exit, all fluid particles move together with the same transit time, and the overall trajectory is already known.
\end{enumerate}
With these five restrictions, equation~\ref{eq_rtt_energy_simple} simply becomes:
\begin{IEEEeqnarray}{rCl}
0 & = & \sum_\out \left\{ \dot m (i + \frac{p}{\rho} + e_k + e_p)\right\} - \sum_\inn \left\{ \dot m (i + \frac{p}{\rho} + e_k + e_p)\right\} \nonumber\\
& = & \dot m \left[(i + \frac{p_2}{\rho} + \frac{1}{2} V_2^2 + g z_2) - (i + \frac{p_1}{\rho} + \frac{1}{2} V_1^2 + g z_1)\right] \nonumber
\end{IEEEeqnarray}
and we here obtain the \vocab{Bernoulli equation}:
\begin{IEEEeqnarray}{rCl}
\frac{p_1}{\rho} + \frac{1}{2} V_1^2 + g z_1 & = & \frac{p_2}{\rho} + \frac{1}{2} V_2^2 + g z_2 \label{eq_bernoulli}
\end{IEEEeqnarray}
This equation describes the properties of a fluid particle in a steady, incompressible, friction-less flow with no energy transfer.
The Bernoulli equation can also be obtained starting from the linear momentum equation (eq.~\ref{eq_rtt_linearmom} p.\pageref{eq_rtt_linearmom}):
\begin{IEEEeqnarray*}{rCl}
\vec F_\net & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol + \iint_\cs \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A
\end{IEEEeqnarray*}
When considering a fixed, infinitely short control volume along a known streamline $s$ of the flow, this equation becomes:
\begin{IEEEeqnarray*}{rCl}
\diff \vec F_\text{pressure} + \diff \vec F_\text{shear} + \diff \vec F_\text{gravity} & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol + \rho V A \diff \vec V
\end{IEEEeqnarray*}
\begin{equationterms}
\item along a streamline, where the velocity $\vec V$ is aligned (by definition) with the streamline.
\end{equationterms}
Now, adding the restrictions of steady flow ($\diff/\diff{t}=0$) and no friction ($\diff \vec F_\text{shear} = \vec 0$), we already obtain:
\begin{IEEEeqnarray*}{rCl}
\diff \vec F_\text{pressure} + \diff \vec F_\text{gravity} & = & \rho V A \diff \vec V
\end{IEEEeqnarray*}
The projection of the net force due to gravity $\diff \vec F_\text{gravity}$ on the streamline segment $\diff s$ has norm $\diff \vec F_\text{gravity} \cdot \diff \vec{s} = -g \rho A \diff z$, while the net force due to pressure is aligned with the streamline and has norm $\diff F_{\text{pressure}, s} = - A \diff p$. Along this streamline, we thus have the following scalar equation, which we integrate from points~1 to~2:
\begin{IEEEeqnarray*}{rCl}
- A \diff p - \rho g A \diff z & = & \rho V A \diff V \nonumber\\
-\frac{1}{\rho} \diff p - g \diff z & = & V \diff V \nonumber\\
-\int_1^2 \frac{1}{\rho} \diff p - \int_1^2 g \diff z & = & \int_1^2 V \diff V
\end{IEEEeqnarray*}
The last obstacle is removed when we consider flows without heat or work transfer, where, therefore, the density $\rho$ is constant. In this way, we arrive to equation.~\ref{eq_bernoulli} again:
\begin{IEEEeqnarray*}{rCl}
\frac{p_1}{\rho} + \frac{1}{2} V_1^2 + g z_1 & = & \frac{p_2}{\rho} + \frac{1}{2} V_2^2 + g z_2
\end{IEEEeqnarray*}
Let us insist on the incredibly frustrating restrictions brought by the five conditions above:
\begin{enumerate}
\item Steady flow.\\
This constrains us to continuous flows with no transition effects, which is a reasonable limit;
\item Incompressible flow.\\
We cannot use this equation to describe flow in compressors, turbines, diffusers, nozzles, nor in flows where $M > \num{0,6}$.
\item No heat or work transfer.\\
We cannot use this equation in a machine (e.g. in pumps, turbines, combustion chambers, coolers).
\item No friction.\\
This is a tragic restriction! We cannot use this equation to describe a turbulent or viscous flow, e.g. near a wall or in a wake.
\item One-dimensional flow.\\
This equation is only valid if we know precisely the trajectory of the fluid whose properties are being calculated.
\end{enumerate}
Among these, the last is the most severe (and the most often forgotten):\\ \textbf{the Bernoulli equation does not allow us to predict the trajectory of fluid particles}. Just like all of the other equations in this chapter, it requires a control volume with a known inlet and a known outlet.
\atendofchapternotes
\renewcommand{\lastedityear}{2018}
\renewcommand{\lasteditmonth}{06}
\renewcommand{\lasteditday}{04}
\renewcommand{\numberofthischapter}{2}
\renewcommand{\titleofthischapter}{Analysis of existing flows (1D)}
\atstartofexercises
\fluidmechexercisestitle
\mecafluboxen
\mecafluexboxen
\begin{boiboiboite}
todo
\end{boiboiboite}
todo
\atendofexercises
......@@ -2,11 +2,12 @@
\renewcommand{\lasteditmonth}{09}
\renewcommand{\lasteditday}{23}
\renewcommand{\numberofthischapter}{3}
\renewcommand{\titleofthischapter}{Analysis of existing flows}
\renewcommand{\titleofthischapter}{Analysis of existing flows (3D)}
\fluidmechchaptertitle
\mecafluboxen
\mecafluboxtmp
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Motivation}
......@@ -103,7 +104,7 @@
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Mass conservation}
\section{Balance of mass}
\commonsvideo{Niccolò Paganini - Caprice No.5 - David Hernando.ogv}{https://frama.link/yVesHaSk}{with sufficient skills (and lots of practice!), it is possible for a musician to produce an uninterrupted stream of air into an instrument while still continuing to breathe, a technique called \vocab{circular breathing}. Can you identify the different terms of eq.~\ref{eq_rtt_mass} as they apply to the clarinetist’s mouth?}{David Hernando Vitores (\ccbysa)}
In this section, we focus on simply asserting that mass is conserved (eq.~\ref{eq_massconservation} p.\pageref{eq_massconservation}). Our study of the fluid’s properties at the borders of the control volume is made by replacing variable $B$ by mass $m$. Thus $\timederivative{B}$ becomes $\timederivative{m_\sys}$, which by definition is~zero.\\
......@@ -137,7 +138,7 @@
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Change of linear momentum}
\section{Balance of linear momentum}
In this section, we apply Newton’s second law: we assert that the variation of system’s linear momentum is equal to the net force being applied to it (eq.~\ref{eq_secondlaw} p.\pageref{eq_secondlaw}). Our study of the fluid’s properties at the control surface is carried out by replacing variable $B$ by the quantity $m \vec V$: momentum. Thus, $\timederivative{B_\sys}$ becomes $\timederivative{(m \vec V_\sys)}$, which is equal to $\vec F_\net$, the vector sum of forces applied on the system as it transits the control volume.\\
In a similar fashion, $b \equiv B/m = \vec V$ and equation~\ref{eq_rtt}, the Reynolds transport theorem, becomes:
......@@ -177,7 +178,7 @@
\clearpage %handmade
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Change of angular momentum}
\section{Balance of angular momentum}
\youtubethumb{nmEe7Dq01AU}{pre-lecture briefing for this chapter, part 2/2}{\oc (\ccby)}
In this third spin on the Reynolds transport theorem, we assert that the change of the angular momentum of a system about a point X is equal to the net moment applied on the system about this point (eq.~\ref{eq_secondlawmom} p.\pageref{eq_secondlawmom}). Our study of the fluid’s properties at the borders of the control volume is made by replacing the variable $B$ by the angular momentum $\vec r_{\X m} \wedge m \vec V$. Thus, $\timederivative{B_\sys}$ becomes $\timederivative{(\vec r_{\X m} \wedge m \vec V_\sys)}$, which is equal to~$\vec M_\net$, the vector sum of moments applied on the system about point X as it transits through the control volume.
......@@ -210,7 +211,7 @@
\dontbreakpage %handmade
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Energy conservation}
\section{Balance of energy}
We conclude our frantic exploration of control volume analysis with the first principle of thermodynamics. We now simply assert that the change in the energy of a system can only be due to well-identified transfers (eq.~\ref{eq_firstprinciple} p.\pageref{eq_firstprinciple}). Our study of the fluid’s properties at the borders of the control volume is made by replacing variable $B$ by an amount of energy~$E_\sys$. Now, $\timederivative{E_\sys}$ can be attributed to three contributors:
\begin{equation}
......@@ -260,82 +261,6 @@
\end{adjustwidth}
This equation~\ref{eq_rtt_energy_simple} is particularly attractive, but it necessitates the input of a large amount of experimental data to provide useful results. It is indeed very difficult to predict how the terms $i$ and $p /\rho$ will change for a given flow process. For example, a pump with given power $\dot Q_{\net \ \inn}$ and $\dot W_\text{shaft, net in}$ will generate large increases of terms $p$, $V$ and $z$ if it is efficient, or a large increase of terms $i$ and $1/\rho$ if it is inefficient. This equation~\ref{eq_rtt_energy_simple}, sadly, does not allow us to quantify the net effect of shear and the extent of irreversibilities in a fluid flow.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The Bernoulli equation}
\label{ch_bernoulli}
The Bernoulli equation has very little practical use for us; nevertheless it is so widely used that we have to dedicate a brief section to examining it. We will start from equation~\ref{eq_rtt_energy_simple} and add five constraints:
\begin{enumerate}
\item Steady flow.\\
Thus $\timederivative{} \iiint_\cv \rho\ e \diff \vol = 0$.\\
In addition, $\dot m$ has the same value at inlet and outlet;
\item Incompressible flow.\\
Thus, $\rho$ stays constant;
\item No heat or work transfer.\\
Thus, both $\dot Q_{\net\ \inn}$ and $\dot W_\text{shaft, net in}$ are zero;
\item No friction.\\
Thus, the fluid energy $i$ cannot increase due to an input from the control volume;
\item One-dimensional flow.\\
Thus, our control volume has only one known entry and one known exit, all fluid particles move together with the same transit time, and the overall trajectory is already known.
\end{enumerate}
With these five restrictions, equation~\ref{eq_rtt_energy_simple} simply becomes:
\begin{IEEEeqnarray}{rCl}
0 & = & \sum_\out \left\{ \dot m (i + \frac{p}{\rho} + e_k + e_p)\right\} - \sum_\inn \left\{ \dot m (i + \frac{p}{\rho} + e_k + e_p)\right\} \nonumber\\
& = & \dot m \left[(i + \frac{p_2}{\rho} + \frac{1}{2} V_2^2 + g z_2) - (i + \frac{p_1}{\rho} + \frac{1}{2} V_1^2 + g z_1)\right] \nonumber
\end{IEEEeqnarray}
and we here obtain the \vocab{Bernoulli equation}:
\begin{IEEEeqnarray}{rCl}
\frac{p_1}{\rho} + \frac{1}{2} V_1^2 + g z_1 & = & \frac{p_2}{\rho} + \frac{1}{2} V_2^2 + g z_2 \label{eq_bernoulli}
\end{IEEEeqnarray}
This equation describes the properties of a fluid particle in a steady, incompressible, friction-less flow with no energy transfer.
The Bernoulli equation can also be obtained starting from the linear momentum equation (eq.~\ref{eq_rtt_linearmom} p.\pageref{eq_rtt_linearmom}):
\begin{IEEEeqnarray*}{rCl}
\vec F_\net & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol + \iint_\cs \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A
\end{IEEEeqnarray*}
When considering a fixed, infinitely short control volume along a known streamline $s$ of the flow, this equation becomes:
\begin{IEEEeqnarray*}{rCl}
\diff \vec F_\text{pressure} + \diff \vec F_\text{shear} + \diff \vec F_\text{gravity} & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol + \rho V A \diff \vec V
\end{IEEEeqnarray*}
\begin{equationterms}
\item along a streamline, where the velocity $\vec V$ is aligned (by definition) with the streamline.
\end{equationterms}
Now, adding the restrictions of steady flow ($\diff/\diff{t}=0$) and no friction ($\diff \vec F_\text{shear} = \vec 0$), we already obtain:
\begin{IEEEeqnarray*}{rCl}
\diff \vec F_\text{pressure} + \diff \vec F_\text{gravity} & = & \rho V A \diff \vec V
\end{IEEEeqnarray*}
The projection of the net force due to gravity $\diff \vec F_\text{gravity}$ on the streamline segment $\diff s$ has norm $\diff \vec F_\text{gravity} \cdot \diff \vec{s} = -g \rho A \diff z$, while the net force due to pressure is aligned with the streamline and has norm $\diff F_{\text{pressure}, s} = - A \diff p$. Along this streamline, we thus have the following scalar equation, which we integrate from points~1 to~2:
\begin{IEEEeqnarray*}{rCl}
- A \diff p - \rho g A \diff z & = & \rho V A \diff V \nonumber\\
-\frac{1}{\rho} \diff p - g \diff z & = & V \diff V \nonumber\\
-\int_1^2 \frac{1}{\rho} \diff p - \int_1^2 g \diff z & = & \int_1^2 V \diff V
\end{IEEEeqnarray*}
The last obstacle is removed when we consider flows without heat or work transfer, where, therefore, the density $\rho$ is constant. In this way, we arrive to equation.~\ref{eq_bernoulli} again:
\begin{IEEEeqnarray*}{rCl}
\frac{p_1}{\rho} + \frac{1}{2} V_1^2 + g z_1 & = & \frac{p_2}{\rho} + \frac{1}{2} V_2^2 + g z_2
\end{IEEEeqnarray*}
Let us insist on the incredibly frustrating restrictions brought by the five conditions above:
\begin{enumerate}
\item Steady flow.\\
This constrains us to continuous flows with no transition effects, which is a reasonable limit;
\item Incompressible flow.\\
We cannot use this equation to describe flow in compressors, turbines, diffusers, nozzles, nor in flows where $M > \num{0,6}$.
\item No heat or work transfer.\\
We cannot use this equation in a machine (e.g. in pumps, turbines, combustion chambers, coolers).
\item No friction.\\
This is a tragic restriction! We cannot use this equation to describe a turbulent or viscous flow, e.g. near a wall or in a wake.
\item One-dimensional flow.\\
This equation is only valid if we know precisely the trajectory of the fluid whose properties are being calculated.
\end{enumerate}
Among these, the last is the most severe (and the most often forgotten):\\ \textbf{the Bernoulli equation does not allow us to predict the trajectory of fluid particles}. Just like all of the other equations in this chapter, it requires a control volume with a known inlet and a known outlet.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
......
......@@ -2,7 +2,7 @@
\renewcommand{\lasteditmonth}{06}
\renewcommand{\lasteditday}{04}
\renewcommand{\numberofthischapter}{3}
\renewcommand{\titleofthischapter}{Analysis of existing flows}
\renewcommand{\titleofthischapter}{Analysis of existing flows (3D)}
\atstartofexercises
\fluidmechexercisestitle
......
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\renewcommand{\titleofthischapter}{Effects of pressure}
\atstartofexercises
......@@ -193,7 +193,7 @@ A system of lock doors is set up to allow boats to travel up the side of a hill
If the atmospheric density is considered uniform, estimate the buoyancy force exerted on an Airbus A380, both on the ground and in cruise flight ($\rho_\text{cruise} = \SI{0,4}{\kilogram\per\metre\cubed}, T_\text{cruise} = \SI{-40}{\degreeCelsius}$).
\begin{figure}[hbc!]
\begin{figure}[hb!]
\begin{center}
\includegraphics[width=\textwidth]{380}
\end{center}
......@@ -202,7 +202,7 @@ A system of lock doors is set up to allow boats to travel up the side of a hill
\label{fig_athreeeighty}
\end{figure}
\begin{table}[hbc!]
\begin{table}[hb!]
\begin{center}
\renewcommand{\tabcolsep}{0.8em}
\renewcommand{\arraystretch}{1.3}
......
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\renewcommand{\numberofthischapter}{5}
\renewcommand{\titleofthischapter}{Effects of shear}
\fluidmechchaptertitle
\mecafluboxen
\mecafluboxtmp
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Motivation}
......@@ -20,7 +21,63 @@
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Concept of shear}
\section{Shear forces on walls}
todo
The calculation of forces due to shear is very similar to that which we used in the previous chapter with pressure.
When the shear $\vec \tau$ exerted on a flat surface of area~$S$ is uniform, the resulting force $F$ is easily easily calculated ($F = \tau_\cst S$). In the more general case of a three-dimensional object immersed in a fluid with non-uniform shear, the force must be expressed as a vector and obtained by integration:
\begin{IEEEeqnarray}{rCcCl}
\vec F_\text{shear} & = & \int_S \diff \vec F & = & \int_S \vec \tau_\text{surface} \diff S \label{eq_shear_force_vector}
\end{IEEEeqnarray}
Much like equation~\ref{eq_pressure_force_vector} in the previous chapter, eq.~\ref{eq_shear_force_vector} is easily implemented in software algorithms to obtain numerically, for example, the force resulting from shear due to fluid flow around a body such as an aircraft wing. In our academic study of fluid mechanics, however, we will restrict ourselves to the simpler cases where the surface is perfectly flat and the shear has uniform direction. The force in eq.~\ref{eq_shear_force_vector}, exterting in the $i$-direction due to shear on a surface perpendicular to the $j$-direction, then becomes:
\begin{IEEEeqnarray}{rCcCl}
F_{\text{shear}\ ji} & = & \int_S \diff F & = & \int_S ||\vec \tau_{ji}|| \diff S \label{eq_shear_force_vector_flat}
\end{IEEEeqnarray}
What is required to calculate the scalar $F$ in eq.~\ref{eq_shear_force_vector_flat} is an expression of $\tau$ as a function of $S$. We proceed like we did in the previous chapter, splitting $\diff S$ as $\diff S = \diff i \diff k$ before proceeding with the calculation starting from;
\begin{IEEEeqnarray}{rCl}
F_{\text{shear} ji} &=& \iint \tau_{ji \ (i, k)} \diff i \diff k \label{eq_shear_force_twod_integration_general}\\
&=& \mu \iint \partialderivative{u_i}{j} \diff i \diff k \label{eq_shear_force_twod_integration}
\end{IEEEeqnarray}
\begin{equationterms}
\item where \tab $i$ \tab is the direction of the force;
\item \tab $j$ \tab is the direction perpendicular to the flat surface;
\item and \tab $k$ \tab is the third (orthonormal) direction.
\end{equationterms}
The above expression \ref{eq_shear_force_twod_integration} is perhaps easier to read when it is developed. For example, the shear force $\vec F_{\text{shear} yi}$ exerting on a plate perpendicular to the $y$-direction is:
\begin{IEEEeqnarray*}{rCl}
\vec F_{\text{shear} yi} &=& \left(\begin{array}{c}%
F_{\text{shear} yx}\\
0\\
F_{\text{shear} yz}%
\end{array}\right)
\end{IEEEeqnarray*}
with
\begin{IEEEeqnarray*}{rCl}
F_{\text{shear} yx} &=& \mu \iint \partialderivative{u}{y} \diff x \diff z\\
F_{\text{shear} yz} &=& \mu \iint \partialderivative{w}{y} \diff x \diff z
\end{IEEEeqnarray*}
Just like for pressure, the moment $\vec M_{\text{X} j}$ generated about a point X by the shear forces in the $i$-direction on a plane surface perpendicular to the $j$-direction can be calculated as
\begin{IEEEeqnarray}{rCcCl}
\vec M_{\text{X} j} & = & \int_S \diff \vec M_\text{X} = \int_S \vec r_{\text{X}F} \wedge \diff \vec F & = & \int_S \vec r_{\text{X}F} \wedge \vec \tau_{ij} \diff S
\end{IEEEeqnarray}
If the point $X$ is in the plane of the surface of interest, this can be expressed as:
\begin{IEEEeqnarray}{rCl}
M_{\text{X} j} &=& \mu \iint r_{\text{X}F} \partialderivative{u_i}{j} \diff i \diff k\\
M_{\text{X} y} &=& \mu \iint r_{\text{X}F} \partialderivative{u_x}{y} \diff x \diff z
\end{IEEEeqnarray}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Shear fields in fluids}
We approached the concept of shear in the introductory chapter with the notion that it represented force parallel to a given flat surface (eq.~\ref{eq_first_def_shear}), for example a flat plate of area $A$:
\begin{IEEEeqnarray}{rCl}
......@@ -152,8 +209,6 @@
\end{itemize}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Slip and viscosity}
\subsection{The no-slip condition}
\label{ch_no_slip_condition}
......@@ -220,54 +275,8 @@
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Wall shear forces}
\section{Special case: shear in laminar flows}
The calculation of forces due to shear is very similar to that which we used in the previous chapter with pressure.
When the shear $\vec \tau$ exerted on a flat surface of area~$S$ is uniform, the resulting force $F$ is easily easily calculated ($F = \tau_\cst S$). In the more general case of a three-dimensional object immersed in a fluid with non-uniform shear, the force must be expressed as a vector and obtained by integration:
\begin{IEEEeqnarray}{rCcCl}
\vec F_\text{shear} & = & \int_S \diff \vec F & = & \int_S \vec \tau_\text{surface} \diff S \label{eq_shear_force_vector}
\end{IEEEeqnarray}
Much like equation~\ref{eq_pressure_force_vector} in the previous chapter, eq.~\ref{eq_shear_force_vector} is easily implemented in software algorithms to obtain numerically, for example, the force resulting from shear due to fluid flow around a body such as an aircraft wing. In our academic study of fluid mechanics, however, we will restrict ourselves to the simpler cases where the surface is perfectly flat and the shear has uniform direction. The force in eq.~\ref{eq_shear_force_vector}, exterting in the $i$-direction due to shear on a surface perpendicular to the $j$-direction, then becomes:
\begin{IEEEeqnarray}{rCcCl}
F_{\text{shear}\ ji} & = & \int_S \diff F & = & \int_S ||\vec \tau_{ji}|| \diff S \label{eq_shear_force_vector_flat}
\end{IEEEeqnarray}
What is required to calculate the scalar $F$ in eq.~\ref{eq_shear_force_vector_flat} is an expression of $\tau$ as a function of $S$. We proceed like we did in the previous chapter, splitting $\diff S$ as $\diff S = \diff i \diff k$ before proceeding with the calculation starting from;
\begin{IEEEeqnarray}{rCl}
F_{\text{shear} ji} &=& \iint \tau_{ji \ (i, k)} \diff i \diff k \label{eq_shear_force_twod_integration_general}\\
&=& \mu \iint \partialderivative{u_i}{j} \diff i \diff k \label{eq_shear_force_twod_integration}
\end{IEEEeqnarray}
\begin{equationterms}
\item where \tab $i$ \tab is the direction of the force;
\item \tab $j$ \tab is the direction perpendicular to the flat surface;
\item and \tab $k$ \tab is the third (orthonormal) direction.
\end{equationterms}
The above expression \ref{eq_shear_force_twod_integration} is perhaps easier to read when it is developed. For example, the shear force $\vec F_{\text{shear} yi}$ exerting on a plate perpendicular to the $y$-direction is:
\begin{IEEEeqnarray*}{rCl}
\vec F_{\text{shear} yi} &=& \left(\begin{array}{c}%
F_{\text{shear} yx}\\
0\\
F_{\text{shear} yz}%
\end{array}\right)
\end{IEEEeqnarray*}
with
\begin{IEEEeqnarray*}{rCl}
F_{\text{shear} yx} &=& \mu \iint \partialderivative{u}{y} \diff x \diff z\\
F_{\text{shear} yz} &=& \mu \iint \partialderivative{w}{y} \diff x \diff z
\end{IEEEeqnarray*}
Just like for pressure, the moment $\vec M_{\text{X} j}$ generated about a point X by the shear forces in the $i$-direction on a plane surface perpendicular to the $j$-direction can be calculated as
\begin{IEEEeqnarray}{rCcCl}
\vec M_{\text{X} j} & = & \int_S \diff \vec M_\text{X} = \int_S \vec r_{\text{X}F} \wedge \diff \vec F & = & \int_S \vec r_{\text{X}F} \wedge \vec \tau_{ij} \diff S
\end{IEEEeqnarray}
If the point $X$ is in the plane of the surface of interest, this can be expressed as:
\begin{IEEEeqnarray}{rCl}
M_{\text{X} j} &=& \mu \iint r_{\text{X}F} \partialderivative{u_i}{j} \diff i \diff k\\
M_{\text{X} y} &=& \mu \iint r_{\text{X}F} \partialderivative{u_x}{y} \diff x \diff z
\end{IEEEeqnarray}
todo
\atendofchapternotes
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\section{Computing turbulent flow}
(content extracted from chapter 6: prediction of fluid flows)
The second flaw appears once we consider the effect of grid coarseness. Every decrease in the size of the grid cell and in the length of the time step increases the total number of equations to be solved by the algorithm. Halving each of $\delta x$, $\delta y$, $\delta z$ and $\delta t$ multiplies the total number of equations by \num{16}, so that soon enough the experimenter will wish to know what maximum (coarsest) grid size is appropriate or tolerable. Furthermore, in many practical cases, we may not even be interested in an exhaustive description of the velocity field, and just wish to obtain a general, coarse description of the fluid flow.
Using a coarse grid and large time step, however, prevents us from resolving small variations in velocity: movements which are so small they fit in between grid points, or so short they occur in between time steps. Let us decompose the velocity field into two components: one is the \vocab{average} flow $\left(\overline u, \overline v, \overline w\right)$, and the other the \vocab{instantaneous fluctuation} flow $\left(u', v', w'\right)$, too fine to be captured by our grid:
\begin{IEEEeqnarray}{rCl}
u_i &\equiv& \overline u_i + u_i'\label{eq_def_average_u}\\
\overline u_i' &\equiv& 0
\end{IEEEeqnarray}
With the use of definition~\ref{eq_def_average_u}, we re-formulate equation~\ref{eq_ns_cartone} as follows:
\begin{multline}
\rho \left[ \partialtimederivative{(\overline u + u')} + (\overline u + u') \partialderivative{(\overline u + u')}{x} + (\overline v + v') \partialderivative{(\overline u + u')}{y} + (\overline w + w') \partialderivative{(\overline u + u')}{z} \right] \nonumber\\
= \rho g_x - \partialderivative{(\overline p + p')}{x} + \mu \left[ \secondpartialderivative{(\overline u + u')}{x} + \secondpartialderivative{(\overline u + u')}{y} + \secondpartialderivative{(\overline u + u')}{z} \right]
\end{multline}
Taking the \emph{average} of this equation —thus expressing the dynamics of the flow as we calculate them with a finite, coarse grid— yields, after some intimidating but easily conquerable algebra:
\begin{IEEEeqnarray}{lr}