Area-velocity relation - Revision history

Nian at 13:38, 1 April 2026

2026-04-01T13:38:21Z

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Nian at 11:06, 1 April 2026

2026-04-01T11:06:28Z

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	=== The Area-Velocity Relation ===		=== The Area-Velocity Relation ===

	Starting point - the continuity equation (Eqn. \ref{eq:governing:cont}):		Starting point - the continuity equation (Eqn. \ref{eq:governing:cont}):

	<math ~~display="block"~~>		{{NumEqn\|<math>
	d(\rho uA)=0 \Rightarrow \rho u dA+\rho Adu +uAd\rho=0		d(\rho uA)=0 \Rightarrow \rho u dA+\rho Adu +uAd\rho=0
	</math>		</math>}}

	divide by <math>\rho uA</math> gives		divide by <math>\rho uA</math> gives

	<math ~~display="block"~~>		{{NumEqn\|<math>
	\frac{d\rho}{\rho}+\frac{du}{u}+\frac{dA}{A}=0		\frac{d\rho}{\rho}+\frac{du}{u}+\frac{dA}{A}=0
	</math>		</math>}}

	As the name suggests, the area-velocity relation is a relation including the area and the flow velocity. Therefore, the next step is to replace the density terms.		As the name suggests, the area-velocity relation is a relation including the area and the flow velocity. Therefore, the next step is to replace the density terms.
Line 26:		Line 34:
	This can be achieved using the momentum equation (Eqn. \ref{eq:governing:mom})		This can be achieved using the momentum equation (Eqn. \ref{eq:governing:mom})

	<math ~~display="block"~~>		{{NumEqn\|<math>
	dp=-\rho udu\Leftrightarrow \frac{dp}{\rho}=-udu		dp=-\rho udu\Leftrightarrow \frac{dp}{\rho}=-udu
	</math>		</math>}}

	<math ~~display="block"~~>		{{NumEqn\|<math>
	\frac{dp}{\rho}=\frac{dp}{d\rho}\frac{d\rho}{\rho}=-udu		\frac{dp}{\rho}=\frac{dp}{d\rho}\frac{d\rho}{\rho}=-udu
	</math>		</math>}}

	If we assume adiabatic and reversible flow processes, i.e., isentropic flow		If we assume adiabatic and reversible flow processes, i.e., isentropic flow

	<math ~~display="block"~~>		{{NumEqn\|<math>
	\frac{dp}{d\rho}=\left(\frac{dp}{d\rho}\right)_s=a^2\Rightarrow a^2\frac{d\rho}{\rho}=-udu		\frac{dp}{d\rho}=\left(\frac{dp}{d\rho}\right)_s=a^2\Rightarrow a^2\frac{d\rho}{\rho}=-udu
	</math>		</math>}}

	<math ~~display="block"~~>		{{NumEqn\|<math>
	a^2\frac{d\rho}{\rho}=-udu=-u^2\frac{du}{u}		a^2\frac{d\rho}{\rho}=-udu=-u^2\frac{du}{u}
	</math>		</math>}}

	<math ~~display="block"~~>		{{NumEqn\|<math>
	\frac{d\rho}{\rho}=-M^2\frac{du}{u}		\frac{d\rho}{\rho}=-M^2\frac{du}{u}
	</math>		</math>}}

	Eqn. \ref{eq:governing:mom:b} inserted in Eqn. \ref{eq:governing:cont:b} gives		Eqn. \ref{eq:governing:mom:b} inserted in Eqn. \ref{eq:governing:cont:b} gives

	<math ~~display="block"~~>		{{NumEqn\|<math>
	-M^2\frac{du}{u}+\frac{du}{u}+\frac{dA}{A}=0		-M^2\frac{du}{u}+\frac{du}{u}+\frac{dA}{A}=0
	</math>		</math>}}

	or		or

	<math ~~display="block"~~>		{{NumEqn\|<math>
	\frac{dA}{A}=(M^2-1)\frac{du}{u}		\frac{dA}{A}=(M^2-1)\frac{du}{u}
	</math>		</math>}}

	which is the area-velocity relation.		which is the area-velocity relation.

Nian at 20:59, 21 March 2026

2026-03-21T20:59:57Z

← Older revision		Revision as of 20:59, 21 March 2026
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	~~\section{~~The Area-Velocity Relation}		=== The Area-Velocity Relation ===

	~~\noindent~~ Starting point - the continuity equation (Eqn. \ref{eq:governing:cont}):\\		Starting point - the continuity equation (Eqn. \ref{eq:governing:cont}):

	\[d(\rho uA)=0 \Rightarrow \rho u dA+\rho Adu +uAd\rho=0~~\]\\~~		<math display="block">
			d(\rho uA)=0 \Rightarrow \rho u dA+\rho Adu +uAd\rho=0
			</math>

	~~\noindent~~ divide by $\rho uA$ gives\\		divide by <math>\rho uA</math> gives

	~~\begin{equation}~~		<math display="block">
	\frac{d\rho}{\rho}+\frac{du}{u}+\frac{dA}{A}=0		\frac{d\rho}{\rho}+\frac{du}{u}+\frac{dA}{A}=0
	~~\label{eq:governing:cont:b}~~		</math>
	~~\end{equation}\\~~

	~~\noindent~~ As the name suggests, the area-velocity relation is a relation including the area and the flow velocity. Therefore, the next step is to replace the density terms.\\		As the name suggests, the area-velocity relation is a relation including the area and the flow velocity. Therefore, the next step is to replace the density terms.

	~~\noindent~~ This can be achieved using the momentum equation (Eqn. \ref{eq:governing:mom})\\		This can be achieved using the momentum equation (Eqn. \ref{eq:governing:mom})

	\[dp=-\rho udu\Leftrightarrow \frac{dp}{\rho}=-udu~~\]\\~~		<math display="block">
			dp=-\rho udu\Leftrightarrow \frac{dp}{\rho}=-udu
			</math>

	\[\frac{dp}{\rho}=\frac{dp}{d\rho}\frac{d\rho}{\rho}=-udu~~\]\\~~		<math display="block">
			\frac{dp}{\rho}=\frac{dp}{d\rho}\frac{d\rho}{\rho}=-udu
			</math>

	~~\noindent~~ If we assume adiabatic and reversible flow processes, i.e., isentropic flow\\		If we assume adiabatic and reversible flow processes, i.e., isentropic flow

	\[\frac{dp}{d\rho}=\left(\frac{dp}{d\rho}\right)_s=a^2\Rightarrow a^2\frac{d\rho}{\rho}=-udu~~\]\\~~		<math display="block">
			\frac{dp}{d\rho}=\left(\frac{dp}{d\rho}\right)_s=a^2\Rightarrow a^2\frac{d\rho}{\rho}=-udu
			</math>

	\[a^2\frac{d\rho}{\rho}=-udu=-u^2\frac{du}{u}~~\]\\~~		<math display="block">
			a^2\frac{d\rho}{\rho}=-udu=-u^2\frac{du}{u}
			</math>

	~~\begin{equation}~~		<math display="block">
	\frac{d\rho}{\rho}=-M^2\frac{du}{u}		\frac{d\rho}{\rho}=-M^2\frac{du}{u}
	~~\label{eq:governing:mom:b}~~		</math>
	~~\end{equation}\\~~

	~~\noindent~~ Eqn. \ref{eq:governing:mom:b} inserted in Eqn. \ref{eq:governing:cont:b} gives\\		Eqn. \ref{eq:governing:mom:b} inserted in Eqn. \ref{eq:governing:cont:b} gives

	\[-M^2\frac{du}{u}+\frac{du}{u}+\frac{dA}{A}=0~~\]\\~~		<math display="block">
			-M^2\frac{du}{u}+\frac{du}{u}+\frac{dA}{A}=0
			</math>

	or		or

	~~\begin{equation}~~		<math display="block">
	\frac{dA}{A}=(M^2-1)\frac{du}{u}		\frac{dA}{A}=(M^2-1)\frac{du}{u}
	~~\label{eq:governing:av}~~		</math>
	~~\end{equation}\\~~

	~~\noindent~~ which is the area-velocity relation.\\		which is the area-velocity relation.

	~~%\newpage~~		From the area-velocity relation (Eqn. \ref{eq:governing:av}), we can learn that in a subsonic flow, the flow will accelerate if the cross-section area is decreased and decelerate if the cross-section area is increased. It can also be seen that for supersonic flow, the relation between flow velocity and cross-section area will be the opposite of that for subsonic flows, see Fig. \ref{fig:areavelocity}. For sonic flow, <math>M=1</math>, the relation shows that <math>dA=0</math>, which means that sonic flow can only occur at a cross-section area maximum or minimum. From the subsonic versus supersonic flow discussion, it can be understood that sonic flow at the minimum cross section area is the only valid option (see Fig. \ref{fig:sonic}).

	~~\noindent~~ From the area-velocity relation (Eqn. \ref{eq:governing:av}), we can learn that in a subsonic flow, the flow will accelerate if the cross-section area is decreased and decelerate if the cross-section area is increased. It can also be seen that for supersonic flow, the relation between flow velocity and cross-section area will be the opposite of that for subsonic flows, see Fig. \ref{fig:areavelocity}. For sonic flow, $M=1$, the relation shows that $dA=0$, which means that sonic flow can only occur at a cross-section area maximum or minimum. From the subsonic versus supersonic flow discussion, it can be understood that sonic flow at the minimum cross section area is the only valid option (see Fig. \ref{fig:sonic}).\\

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Nian at 19:53, 20 March 2026

2026-03-20T19:53:11Z

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			[[Category:Compressible flow:Topic]]
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	__TOC__		__TOC__

Nian: Created page with "Category:Compressible flow Category:Quasi-one-dimensional flow Category:Inviscid flow TOC \section{The Area-Velocity Relation} \noindent Starting point - the continuity equation (Eqn. \ref{eq:governing:cont}):\\ \[d(\rho uA)=0 \Rightarrow \rho u dA+\rho Adu +uAd\rho=0\]\\ \noindent divide by $\rho uA$ gives\\ \begin{equation} \frac{d\rho}{\rho}+\frac{du}{u}+\frac{dA}{A}=0 \label{eq:governing:cont:b} \end{equation}\\ \noindent As the name suggests,..."

2026-03-16T15:00:10Z

Created page with "Category:Compressible flow Category:Quasi-one-dimensional flow Category:Inviscid flow __TOC__ \section{The Area-Velocity Relation} \noindent Starting point - the continuity equation (Eqn. \ref{eq:governing:cont}):\\ \[d(\rho uA)=0 \Rightarrow \rho u dA+\rho Adu +uAd\rho=0\]\\ \noindent divide by $\rho uA$ gives\\ \begin{equation} \frac{d\rho}{\rho}+\frac{du}{u}+\frac{dA}{A}=0 \label{eq:governing:cont:b} \end{equation}\\ \noindent As the name suggests,..."

New page

[[Category:Compressible flow]]
[[Category:Quasi-one-dimensional flow]]
[[Category:Inviscid flow]]

__TOC__

\section{The Area-Velocity Relation}

\noindent Starting point - the continuity equation (Eqn. \ref{eq:governing:cont}):\\

\[d(\rho uA)=0 \Rightarrow \rho u dA+\rho Adu +uAd\rho=0\]\\

\noindent divide by $\rho uA$ gives\\

\begin{equation}
\frac{d\rho}{\rho}+\frac{du}{u}+\frac{dA}{A}=0
\label{eq:governing:cont:b}
\end{equation}\\

\noindent As the name suggests, the area-velocity relation is a relation including the area and the flow velocity. Therefore, the next step is to replace the density terms.\\

\noindent This can be achieved using the momentum equation (Eqn. \ref{eq:governing:mom})\\

\[dp=-\rho udu\Leftrightarrow \frac{dp}{\rho}=-udu\]\\

\[\frac{dp}{\rho}=\frac{dp}{d\rho}\frac{d\rho}{\rho}=-udu\]\\

\noindent If we assume adiabatic and reversible flow processes, i.e., isentropic flow\\

\[\frac{dp}{d\rho}=\left(\frac{dp}{d\rho}\right)_s=a^2\Rightarrow a^2\frac{d\rho}{\rho}=-udu\]\\

\[a^2\frac{d\rho}{\rho}=-udu=-u^2\frac{du}{u}\]\\

\begin{equation}
\frac{d\rho}{\rho}=-M^2\frac{du}{u}
\label{eq:governing:mom:b}
\end{equation}\\

\noindent Eqn. \ref{eq:governing:mom:b} inserted in Eqn. \ref{eq:governing:cont:b} gives\\

\[-M^2\frac{du}{u}+\frac{du}{u}+\frac{dA}{A}=0\]\\

or

\begin{equation}
\frac{dA}{A}=(M^2-1)\frac{du}{u}
\label{eq:governing:av}
\end{equation}\\

\noindent which is the area-velocity relation.\\

%\newpage

\noindent From the area-velocity relation (Eqn. \ref{eq:governing:av}), we can learn that in a subsonic flow, the flow will accelerate if the cross-section area is decreased and decelerate if the cross-section area is increased. It can also be seen that for supersonic flow, the relation between flow velocity and cross-section area will be the opposite of that for subsonic flows, see Fig. \ref{fig:areavelocity}. For sonic flow, $M=1$, the relation shows that $dA=0$, which means that sonic flow can only occur at a cross-section area maximum or minimum. From the subsonic versus supersonic flow discussion, it can be understood that sonic flow at the minimum cross section area is the only valid option (see Fig. \ref{fig:sonic}).\\

\begin{figure}[ht!]
\begin{center}
\includegraphics[]{figures/standalone-figures/Chapter06/pdf/area-velocity.pdf}
\caption{Area-velocity relation - subsonic flow vs. supersonic flow}
\label{fig:areavelocity}
\end{center}
\end{figure}

\begin{figure}[ht!]
\begin{center}
\includegraphics[]{figures/standalone-figures/Chapter06/pdf/area-velocity-sonic-flow.pdf}
\caption{Area-velocity relation - sonic flow}
\label{fig:sonic}
\end{center}
\end{figure}