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Commits (3)
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......@@ -109,7 +109,7 @@
Looking again at an example case where there were two inlets and two outlets, this equation~\ref{eq_mass_oned} would become:
\begin{IEEEeqnarray*}{rCCCCCCCC}
0 & = & \rho_\text{in 1} V_{\perp \text{ in 1}} A_\text{in 1} &+& \rho_\text{in 2} V_{\perp \text{ in 2}} A_\text{in 2} &+& \rho_\text{out 1} V_{\perp \text{ out 1}} A_\text{out 1} &+& \rho_\text{out 1} V_{\perp \text{ out 2}} A_\text{out 1}\nonumber\\
0 & = & \rho_\text{in 1} V_{\perp \text{ in 1}} A_\text{in 1} &+& \rho_\text{in 2} V_{\perp \text{ in 2}} A_\text{in 2} &+& \rho_\text{out 1} V_{\perp \text{ out 1}} A_\text{out 1} &+& \rho_\text{out 2} V_{\perp \text{ out 2}} A_\text{out 2}\nonumber\\
0 & = & \left(\rho V_\perp A \right)_\text{in 1} &+& \left(\rho V_\perp A \right)_\text{in 2} &+& \left(\rho V_\perp A \right)_\text{out 1} &+& \left(\rho V_\perp A \right)_\text{out 2}\nonumber\\
0 & = & -\left(\rho |V_\perp| A \right)_\text{in 1} &-& \left(\rho |V_\perp| A \right)_\text{in 2} &+& \left(\rho |V_\perp| A \right)_\text{out 1} &+& \left(\rho |V_\perp| A \right)_\text{out 2}
\end{IEEEeqnarray*}
......@@ -184,7 +184,7 @@
In the example case where there is one inlet and one outlet, we would write:
\begin{IEEEeqnarray*}{rCCCCCCCC}
\vec F_\net & = & \left(\rho V_\perp A \vec V\right)_\inn &+& \left(\rho V_\perp A \vec V \right)_\out\\
\vec F_\net & = & -\left(\rho |V_\perp| A \vec V \right)_\inn &-& \left(\rho |V_\perp| A \vec V \right)_\out
\vec F_\net & = & -\left(\rho |V_\perp| A \vec V \right)_\inn &+& \left(\rho |V_\perp| A \vec V \right)_\out
\end{IEEEeqnarray*}
To make clear a few things, let us focus on the simple case where a considered volume is traversed by a steady flow with mass flow $\dot m$, with one inlet (point~1) and one outlet (point~2). The net force $\vec F_\net$ applying on the fluid is
......@@ -335,7 +335,7 @@
With these five restrictions, equation~\ref{eq_sfee} simply becomes:
\begin{IEEEeqnarray*}{rCcl}
0 + 0 &=& & \left[\dot m \left(i_\cst + \frac{p}{\rho_\cst} + \frac{1}{2} V^2 + g z \right) \right]_1 \\
&& +& \left[\dot m \left(i\cst + \frac{p}{\rho_\cst} + \frac{1}{2} V^2 + g z \right) \right]_2
&& +& \left[\dot m \left(i_\cst + \frac{p}{\rho_\cst} + \frac{1}{2} V^2 + g z \right) \right]_2
\end{IEEEeqnarray*}
Dividing by $|\dot m|$ and canceling $i_\cst$, as follows,
\begin{IEEEeqnarray*}{rCl}
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......@@ -65,25 +65,28 @@
\begin{IEEEeqnarray}{rCl}
\timederivative{B_\cv} = \timederivative{} \iiint_\cv \frac{B}{\vol} \diff \vol = \timederivative{} \iiint_\cv \rho b \diff \vol \label{eq_secondbit}
\end{IEEEeqnarray}
\begin{equationterms}
\item where \tab CV \tab is the control volume,
\item and \tab $\vol$ \tab\tab\tab is volume (\si{\metre\cubed}).
\end{equationterms}
The term $\dot B_\net$ can be evaluated by quantifying, for each area element $\diff A$ of the control volume’s surface, the surface flow rate $\rho b V_\perp$ of property $B$ that flows through it, as shown in \cref{fig_cv_da}. The integral over the entire control volume surface CS of this term is:
\begin{IEEEeqnarray}{rCl}
\dot B_\net = \iint_\cs \rho b V_\perp \diff A = \iint_\cs \rho b \ (\vec V_\rel \cdot \vec n) \diff A \label{eq_thirdbit}
\end{IEEEeqnarray}
\begin{equationterms}
\item where \tab CV \tab\tab is the control volume,
\item \tab CS \tab\tab\tab is the the control surface (enclosing the control volume),
\item where \tab CS \tab\tab\tab is the control surface (enclosing the control volume CV),
\item \tab $\vec n$ \tab\tab\tab\tab is a unit vector on each surface element $\diff A$ pointing outwards,
\item \tab $\vec V_\rel$ \tab is the local velocity of fluid relative to the control surface,
\item and \tab $V_\perp \equiv \vec V_\rel \cdot \vec n$ is the local cross-surface speed (positive outwards, neg. inwards)
\end{equationterms}
\begin{figure}[ht]
\begin{center}
\begin{center}\vspace{-0.25cm}%handmade
\includegraphics[width=0.7\textwidth]{concept_vrel_vecn.png}
\end{center}
\supercaption{Part of the system may be flowing through an arbitrary piece of the control surface with area $\diff A$. The $\vec n$ vector defines the orientation of $\diff A$ surface, and by convention is always pointed outwards.}{\wcfile{System control volume integral analysis.svg}{Figure} \cczero \oc}
\label{fig_cv_da}
\vspace{-0.8cm}%handmade
\vspace{-1cm}%handmade
\end{figure}
By inserting equations~\ref{eq_secondbit} and~\ref{eq_thirdbit} into equation~\ref{eq_rtt_basic}, we obtain:\dontbreakpage
......@@ -182,7 +185,7 @@
To make clear a few things, let us focus on the simple case where a considered volume has only one inlet (point~1) and one outlet (point~2). From equation~\ref{eq_rtt_linearmom}, the net force $\vec F_\net$ applying on the fluid is:
\begin{IEEEeqnarray}{rCl}
\vec F_\net & = & \timederivative{}\iiint_\cv \rho \vec V \diff \vol + \iint_\out \rho_2 |V_{\perp 2}| \vec V_2 \diff A_2 - \iint_\out \rho_1 |V_{\perp 1}| \vec V_1 \diff A_1 \nonumber\\\label{eq_fnet_twovectors_unsteady}
\vec F_\net & = & \timederivative{}\iiint_\cv \rho \vec V \diff \vol + \iint_\out \rho_2 |V_{\perp 2}| \vec V_2 \diff A_2 - \iint_\inn \rho_1 |V_{\perp 1}| \vec V_1 \diff A_1 \nonumber\\\label{eq_fnet_twovectors_unsteady}
\end{IEEEeqnarray}
In this equation~\ref{eq_fnet_twovectors_unsteady}, what could cause $\vec F_\net$ to be non-zero?
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......@@ -13,7 +13,7 @@
\section{Motivation}
\youtubethumb{4MVn1PoxGqY}{pre-lecture briefing for this chapter (back when it had a different chapter number)}{\oc (\ccby)}
In fluid mechanics, only three types of forces apply to fluid particles: forces due to gravity, pressure, and shear. This chapter focuses on pressure (we will address shear in \chapterfive), and should allow us to answer two questions:
In fluid mechanics, only three types of forces apply to fluid particles: forces due to gravity, pressure, and shear. This chapter focuses on pressure (we will address shear in \chapterfiveshort), and should allow us to answer two questions:
\begin{itemize}
\item How is the effect of pressure described and quantified?
\item What are the pressure forces generated on walls by static fluids?
......@@ -151,7 +151,7 @@
\item where $\diff \vol \equiv \diff x \diff y \diff z$ is the volume of the infinitesimal cube.
\end{equationterms}
Now generalizing eq.\ref{eq_force_pressure_x} for the other two directions, we can write:
Now generalizing eq.~\ref{eq_force_pressure_x} for the other two directions, we can write:
\begin{IEEEeqnarray*}{rCl}
F_{\text{net, pressure}, x} & = & \diff \vol \frac{-\partial p}{\partial x}\\
F_{\text{net, pressure}, y} & = & \diff \vol \frac{-\partial p}{\partial y}\\
......@@ -192,14 +192,14 @@
What are the forces applying on an arbitrary particle in a static fluid?
\begin{itemize}
\item The force due to pressure is related to the pressure gradient: we just quantified this with eq~\ref{eq_pressure_force_in_fluid}.
\item The force due to pressure is related to the pressure gradient: we just quantified this with eq.~\ref{eq_pressure_force_in_fluid}.
\item The force due to shear is zero. We will indeed see in \chapterfive that shear efforts can be expressed as a function of viscosity and velocity. All ordinary fluids are unable to exert shear when static.
\item The force due to gravity is easy to quantify: it is the mass $m$ of the fluid multiplied by the gravity vector $\vec g$.
\end{itemize}
In a moving fluid, the sum of these forces would add up to the mass of the particle times its acceleration. But in a static fluid, the velocity is zero and never changes. We can thus write:
\begin{IEEEeqnarray}{rCCCCCl}
\vec F_\text{net, pressure} &+& \vec F_\text{shear} &=& \vec F_\text{gravity} &=& \vec 0\nonumber\\
\begin{IEEEeqnarray}{cCcCcCc}
\vec F_\text{net, pressure} &+& \vec F_\text{shear} &+& \vec F_\text{gravity} &=& \vec 0\nonumber\\
-\diff \vol \ \gradient{p} &+& \vec 0 &+& m \vec g &=& \vec 0\nonumber\\
-\gradient{p} &+& \vec 0 &+& \rho \vec g &=& \vec 0\nonumber
\end{IEEEeqnarray}
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