Commit 0816fdac authored by Olivier's avatar Olivier

Chapters 2 & 3: typo and language fixes reported by student

Language and typo fixes for errors reported by a sharp-eyed student
(who wished to remain anonymous) during the last exercise session,
implemented with gratitude.
parent eb41f737
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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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