\youtubethumb{1LXlFVtPoCY}{pre-lecture briefing for this chapter, part 1/2}{\oc (\ccby)}

In this chapter, we use the same tools that we developed in \chaptertwoshort, but we are able to develop and apply them to more complex cases. Specifically, we would like to answer the following questions:

\begin{enumerate}

\item What the mass flows and forces involved when a flow is non-uniform?

\item What are the mass flows and forces involved when a flow has non-uniform velocity?

\item What are the forces and moments involved when a flow changes direction?

\end{enumerate}

...

...

@@ -24,7 +24,7 @@

\subsection{Control volume}

Let us begin, this time, by building a control volume in \emph{any arbitrary flow}: we are no longer limited to one-inlet, one-oulet steady situtations. Instead, we will write equations that work inside any generic velocity field $\vec V =(u, v, w)$ which is a function of space and time ($\vec V = f(x, y, z, t)$).

Let us begin, this time, by building a control volume in \emph{any arbitrary flow}: we are no longer limited to one-inlet, one-oulet steady situtations. Instead, we will write equations that work inside any generic velocity field $\vec V =(u, v, w)$ which is a function of space and time :$\vec V = f(x, y, z, t)$.

Within this flow, we draw an arbitrary volume named \vocab{control volume} (CV) which is free to move and change shape (\cref{fig_cv}). We are going to measure the properties of the fluid at the borders of this volume, which we call the \vocab{control surface} (CS), in order to compute the net effect of the flow through the volume.

\youtubethumb{nmEe7Dq01AU}{pre-lecture briefing for this chapter, part 2/2}{\oc (\ccby)}

What is the moment (the “spinning effort”) applying to a fluid flowing through any arbitrary volume? We answer this question by writing an angular momentum balance in the template provided by the Reynolds transport theorem (eq. \ref{eq_rtt}).

What is the moment (the “twisting effort”) applying to a fluid flowing through any arbitrary volume? We answer this question by writing an angular momentum balance in the template provided by the Reynolds transport theorem (eq. \ref{eq_rtt}).

We now state that the placeholder variable $B$ is angular momentum $\vec r_{\X m}\wedge m \vec V$. It follows that $\inlinetimederivative{B}$ becomes $\inlinetimederivative{\vec r_{\X m}\wedge m \vec V_\sys}$, which by definition is the net moment $\vec M_\net$ applying to the system (see eq.~\ref{eq_secondlawmom} p.\pageref{eq_secondlawmom}). Also, $b \equiv B/m =\vec r_{\X m}\wedge\vec V$ and now the Reynolds transport theorem becomes:\dontbreakpage