Area-velocity relation: Difference between revisions

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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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Revision as of 20:59, 21 March 2026


The Area-Velocity Relation

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

d(ρuA)=0ρudA+ρAdu+uAdρ=0

divide by ρuA gives

dρρ+duu+dAA=0

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.

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

dp=ρududpρ=udu

dpρ=dpdρdρρ=udu

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

dpdρ=(dpdρ)s=a2a2dρρ=udu

a2dρρ=udu=u2duu

dρρ=M2duu

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

M2duu+duu+dAA=0

or

dAA=(M21)duu

which is the area-velocity relation.

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}).