The Q1D equations
\section{Governing Equations}
\begin{figure}[ht!] \begin{center} \includegraphics[]{figures/standalone-figures/Chapter06/pdf/control-volume.pdf} \caption{Quasi-one-dimensional flow - control volume} \label{fig:cv} \end{center} \end{figure}
\noindent In the following quasi-one-dimensional flow will be assumed. That means that the cross-section is allowed to vary smoothly but flow quantities varies in one direction only. The equations that are derived will thus describe one-dimensional flow in axisymmetric tubes. Let's assume flow in the $x$-direction, which means that all flow quantities and the cross-section area will vary with the axial coordinate $x$.
\[A=A(x),\ \rho=\rho(x),\ u=u(x),\ p=p(x),\ ...\]\\
\noindent We will further assume steady-state flow, which means that unsteady terms will be zero.\\
\noindent The equations are derived with the starting point in the governing flow equations on integral form\\
%\newpage
\subsection{Continuity Equation}
\noindent Applying the integral form of the continuity equation on the quasi-one-dimensional flow control volume (Fig. \ref{fig:cv}) gives\\
\begin{equation} \underbrace{\frac{d}{dt}\iiint_{\Omega}\rho d{\mathscr{V}}}_{=0}+\varoiint_{\partial \Omega}\rho {\mathbf{v}}\cdot {\mathbf{n}}dS=0 \label{eq:governing:cont:a} \end{equation}\\
\[\varoiint_{\partial \Omega}\rho {\mathbf{v}}\cdot {\mathbf{n}}dS=-\rho_1 u_1 A_1+\rho_2 u_2 A_2\]\\
\begin{equation} \rho_1 u_1 A_1=\rho_2 u_2 A_2 \label{eq:governing:cont:b} \end{equation}\\
\subsection{Momentum Equation}
\noindent Applying the integral form of the momentum equation on the quasi-one-dimensional flow control volume (Fig. \ref{fig:cv}) gives\\
\begin{equation} \underbrace{\frac{d}{dt}\iiint_{\Omega}\rho{\mathbf{v}} d{\mathscr{V}}}_{=0}+\varoiint_{\partial \Omega}\left[\rho ({\mathbf{v}}\cdot {\mathbf{n}}){\mathbf{v}}+p{\mathbf{n}}\right]dS=0 \label{eq:governing:mom:a} \end{equation}\\
\[\varoiint_{\partial \Omega} \rho ({\mathbf{v}}\cdot {\mathbf{n}}){\mathbf{v}}dS=-\rho_1u_1^2A_1+\rho_2u_2^2A_2\]\\
\[\varoiint_{\partial \Omega} p{\mathbf{n}}dS=-p_1A_1+p_2A_2-\int_{A_1}^{A_2}pdA\]\\
\noindent collecting terms\\
\begin{equation} \left(\rho_1u_1^2+p_1\right)A_1+\int_{A_1}^{A_2}pdA=\left(\rho_2u_2^2+p_2\right)A_2 \label{eq:governing:mom:b} \end{equation}\\
\subsection{Energy Equation}
\noindent Applying the integral form of the energy equation on the quasi-one-dimensional flow control volume (Fig. \ref{fig:cv}) gives\\
\begin{equation} \underbrace{\frac{d}{dt}\iiint_{\Omega}\rho e_o d{\mathscr{V}}}_{=0}+\varoiint_{\partial \Omega}\left[\rho h_o ({\mathbf{v}}\cdot{\mathbf{n}})\right]dS=0 \label{eq:governing:energy:a} \end{equation}\\
\[\varoiint_{\partial \Omega}\left[\rho h_o ({\mathbf{v}}\cdot{\mathbf{n}})\right]dS=-\rho_1u_1h_{o_1}A_1+\rho_2u_2h_{o_2}A_2\]\\
\[\rho_1u_1h_{o_1}A_1=\rho_2u_2h_{o_2}A_2\]\\
\noindent Now, using the continuity equation $\rho_1u_1A_1=\rho_2u_2A_2$ gives\\
\begin{equation} h_{o_1}=h_{o_2} \label{eq:governing:energy:b} \end{equation}\\
\subsection{Differential Form}
\noindent The integral term appearing the momentum equation is undesired and therefore the governing equations are converted to differential form.\\
\noindent The continuity equation (Eqn. \ref{eq:governing:cont:b}) is rewritten in differential form as\\
\[\rho_1u_1A_1=\rho_2u_2A_2=const\]\\
\begin{equation} d(\rho uA)=0 \label{eq:governing:cont:c} \end{equation}\\
\noindent The momentum equation (Eqn. \ref{eq:governing:mom:b}) is rewritten in differential form as\\
\[\left(\rho_1u_1^2+p_1\right)A_1+\int_{A_1}^{A_2}pdA=\left(\rho_2u_2^2+p_2\right)A_2\Rightarrow d\left[(\rho u^2+p)A\right]=pdA\]\\
\[d(\rho u^2A)+d(pA)=pdA\]\\
\[ud(\rho uA)+\rho uAdu+Adp+\cancel{pdA}=\cancel{pdA}\]\\
\noindent From the continuity equation we have $d(\rho uA)$ and thus\\
\[\rho u\cancel{A}du+\cancel{A}dp=0\Rightarrow\]\\
\begin{equation} dp=-\rho udu \label{eq:governing:mom:c} \end{equation}\\
\noindent which is the momentum equation on differential form. Also referred to as Euler's equation. Finally, the energy equation (Eqn. \ref{eq:governing:cont:b}) is rewritten in differential form as\\
\[h_{o_1}=h_{o_2}=const\Rightarrow dh_o=0\]\\
\[h_o=h+\frac{1}{2}u^2\Rightarrow dh+\frac{1}{2}d(u^2)=0\]\\
\begin{equation} dh+udu=0 \label{eq:governing:energy:c} \end{equation}\\
%\newpage
\subsection{Summary}
\noindent Continuity:
\[d(\rho uA)=0\]
\noindent Momentum:
\[dp=-\rho udu\]
\noindent Energy:
\[dh+udu=0\]
\noindent The equations are valid for:\\
\begin{itemize} \item quasi-one-dimensional flow \item steady state \item all gas models (no gas model assumptions made) \item inviscid flow \end{itemize}
\noindent It should be noted that equations are exact but they are applied to a physical model that is approximate, i.e., the approximation that flow quantities varies in one dimension with a varying cross-section area. In reality, a variation of cross-section area would imply flow in three dimensions.
