Since we are here interested only in steady flows, and we have clearly-identified inlets and outlets, this becomes:

\begin{IEEEeqnarray}{CCCCCCC}

\vec F_\net& = &\Sigma\left(\dot m \vec V\right)_\text{incoming}&+&\Sigma\left(\dot m \vec V\right)_\text{outgoing}\nonumber\\\label{eq_linearmom_oned_two}\\

\begin{array}{c}{\scriptstyle\text{the net force}}\\

{\scriptstyle\text{applying on the fluid}}

\begin{array}{c}

{\scriptstyle\text{the vector sum}}\\

{\scriptstyle\text{of forces}}\\

{\scriptstyle\text{on the fluid}}

\end{array}&=&

\begin{array}{c}{\scriptstyle\text{the sum of incoming}}\\

{\scriptstyle\text{momentum flows}}\\

...

...

@@ -206,13 +208,12 @@

\vec F_\net& = & -\left(\rho |V_\perp| A \vec V \right)_\inn&+&\left(\rho |V_\perp| A \vec V \right)_\out

\end{IEEEeqnarray*}

\youtubethumb{AOcGNeY9ad0}{The net force thing in the momentum balance equation}{\oc (\ccby)}

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

\youtubetopthumb{AOcGNeY9ad0}{The net force thing in the momentum balance equation}{\oc (\ccby)}

Three remarks can be made about this equation.

First, we need to be aware that this is not one, but \emph{three} equations, one for each dimension. In order to express $\vec F_\net$, we need to calculate its three components:

\begin{IEEEeqnarray}{rCl}

...

...

@@ -246,7 +247,7 @@

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Balance of energy}

At which rate does the energy of the system vary when it transits through the control volume? We answer this question by writing an energy balance equation. It compares the rate of change of the system’s energy, to the flow of energy through the borders of the control volume.

What power is applied to the fluid for it to travel through the control volume? We answer this question by writing an energy balance equation. It compares the rate of change of the system’s energy, to the flow of energy through the borders of the control volume.

For this we prefer to express the energy $E$ as the specific energy $e$ (in \si{\joule\per\kilogram}) multiplied by the mass $m$ (\si{\kilogram}). The time rate change of $m e$ is measured in \si{watts} ($\SI{1}{\watt}\equiv\SI{1}{\joule\per\second}$). The energy balance equation is then: