One-dimensional flow with friction: Difference between revisions

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==== Momentum equation ====
==== Momentum equation ====


By combining the above derived relations and the momentum equation on the form given by (3.95), we can get an expression where the friction force is a function of Mach number only\\
By combining the above derived relations and the momentum equation on the form given by (3.95), we can get an expression where the friction force is a function of Mach number only


\noindent For convenience equation (3.95) is written again here\\
For convenience equation (3.95) is written again here


\[dp+\rho u du=-\frac{1}{2}\rho u^2 \frac{4 f dx}{D}\ (3.95)\]\\
<math display="block">
dp+\rho u du=-\frac{1}{2}\rho u^2 \frac{4 f dx}{D}
</math>


\noindent if $u\neq 0.$, we can divide by $0.5\rho u^2$ which gives\\
if <math>u\neq 0.</math>, we can divide by <math>0.5\rho u^2</math> which gives


\begin{equation}
<math display="block">
2\frac{dp}{\rho u^2}+2\frac{\rho u du}{\rho u^2}=-\frac{4 f dx}{D}
2\frac{dp}{\rho u^2}+2\frac{\rho u du}{\rho u^2}=-\frac{4 f dx}{D}
\label{eqn:mom:a}
</math>
\end{equation}\\


\noindent using $M^2=u^2/a^2$, $a^2=\gamma p/\rho$ and the chain rule in (\ref{eqn:mom:a}) gives\\
using <math>M^2=u^2/a^2</math>, <math>a^2=\gamma p/\rho</math> and the chain rule in (\ref{eqn:mom:a}) gives


\begin{equation}
<math display="block">
\frac{2}{\gamma M^2}\frac{dp}{p}+\frac{d(u^2)}{u^2}=-\frac{4 f dx}{D}
\frac{2}{\gamma M^2}\frac{dp}{p}+\frac{d(u^2)}{u^2}=-\frac{4 f dx}{D}
\label{eqn:mom:b}
</math>
\end{equation}\\


\noindent From equation (\ref{eqn:gaslaw}) we can get a relation that expresses the pressure derivative term, $dp/p$, in terms of Mach  number and $d(u^2)/u^2$. Inserting this in (\ref{eqn:mom:b}) gives\\
From equation (\ref{eqn:gaslaw}) we can get a relation that expresses the pressure derivative term, <math>dp/p</math>, in terms of Mach  number and <math>d(u^2)/u^2</math>. Inserting this in (\ref{eqn:mom:b}) gives


\begin{equation}
<math display="block">
\frac{2}{\gamma M^2}\left\{-\left[\frac{1+(\gamma-1)M^2}{2}\right]\frac{d(u^2)}{u^2}\right\}+\frac{d(u^2)}{u^2}=-\frac{4 f dx}{D}
\frac{2}{\gamma M^2}\left\{-\left[\frac{1+(\gamma-1)M^2}{2}\right]\frac{d(u^2)}{u^2}\right\}+\frac{d(u^2)}{u^2}=-\frac{4 f dx}{D}
\label{eqn:mom:c}
</math>
\end{equation}\\


\noindent collecting terms and rearranging gives\\
collecting terms and rearranging gives


\begin{equation}
<math display="block">
\frac{M^2-1}{\gamma M^2}\frac{d(u^2)}{u^2}=\frac{4 f dx}{D}
\frac{M^2-1}{\gamma M^2}\frac{d(u^2)}{u^2}=\frac{4 f dx}{D}
\label{eqn:mom:d}
</math>
\end{equation}\\


\noindent if we now use equation (\ref{eqn:ttot:Mach}) to get rid of the $d(u^2)/u^2$-term we end up with the following expression\\
if we now use equation (\ref{eqn:ttot:Mach}) to get rid of the <math>d(u^2)/u^2</math>-term we end up with the following expression


\begin{equation*}
<math display="block">
\frac{4  f dx}{D}=\frac{2}{\gamma M^2}(1-M^2)\left[1+\frac{(\gamma-1)}{2}M^2\right]^{-1}\frac{dM}{M}
\frac{4  f dx}{D}=\frac{2}{\gamma M^2}(1-M^2)\left[1+\frac{(\gamma-1)}{2}M^2\right]^{-1}\frac{dM}{M}
\end{equation*}
</math>


==== Differential Relations ====
==== Differential Relations ====
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