One-dimensional flow with friction: Difference between revisions

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Created page with "Category:Compressible flow Category:One-dimensional flow Category:Inviscid flow Category:Continuous solution __TOC__ \section{One-Dimensional Flow with Friction} \noindent The starting point is the governing equations for one-dimensional steady-state flow\\ \subsection{Continuity} \begin{equation} \rho_1 u_1=\rho_2 u_2 \label{eq:governing:cont} \end{equation}\\ \subsection{Momentum} \begin{equation} \rho_1 u_1^2+p_1-\bar{\tau}_w\frac{bL}{A}=\rho_2..."
 
 
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\section{One-Dimensional Flow with Friction}
--><noinclude><!--
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\noindent The starting point is the governing equations for one-dimensional steady-state flow\\
-->
==== Flow-station data ====


\subsection{Continuity}
The starting point is the governing equations for one-dimensional steady-state flow


\begin{equation}
===== Continuity =====
 
<math display="block">
\rho_1 u_1=\rho_2 u_2
\rho_1 u_1=\rho_2 u_2
\label{eq:governing:cont}
</math>
\end{equation}\\


\subsection{Momentum}
===== Momentum =====


\begin{equation}
{{NumEqn|<math>
\rho_1 u_1^2+p_1-\bar{\tau}_w\frac{bL}{A}=\rho_2 u_2^2+p_2
\rho_1 u_1^2+p_1-\bar{\tau}_w\frac{bL}{A}=\rho_2 u_2^2+p_2
\label{eq:governing:mom}
</math>}}
\end{equation}\\


\noindent where $\bar{\tau}_w$ is the average wall-shear stress\\
where <math>\bar{\tau}_w</math> is the average wall-shear stress


\begin{equation}
{{NumEqn|<math>
\bar{\tau}_w=\frac{1}{L}\int_0^L\tau_w dx
\bar{\tau}_w=\frac{1}{L}\int_0^L\tau_w dx
\label{eq:governing:tauw}
</math>}}
\end{equation}\\


\noindent $b$ is the tube perimeter, and $L$ is the tube length. For circular cross sections\\
<math>b</math> is the tube perimeter, and <math>L</math> is the tube length. For circular cross sections


\[\frac{bL}{A}=\left\{A=\frac{\pi D^2}{4}, b=\pi D\right\}=\frac{4L}{D}\]\\
{{NumEqn|<math>
\frac{bL}{A}=\left\{A=\frac{\pi D^2}{4}, b=\pi D\right\}=\frac{4L}{D}
</math>}}


\noindent and thus\\
and thus


\begin{equation}
{{NumEqn|<math>
\rho_1 u_1^2+p_1-\frac{4}{D}\int_0^L\tau_w dx=\rho_2 u_2^2+p_2
\rho_1 u_1^2+p_1-\frac{4}{D}\int_0^L\tau_w dx=\rho_2 u_2^2+p_2
\label{eq:governing:mom:b}
</math>}}
\end{equation}\\


\subsection{Energy}
===== Energy =====


\begin{equation}
{{NumEqn|<math>
h_1 + \frac{1}{2}u_1^2=h_2 + \frac{1}{2}u_2^2
h_1 + \frac{1}{2}u_1^2=h_2 + \frac{1}{2}u_2^2
\label{eq:governing:energy}
</math>}}
\end{equation}\\


\subsection{Differential Form}
==== Differential Form ====


In order to remove the integral term in the momentum equation, the governing equations are written in differential form\\
In order to remove the integral term in the momentum equation, the governing equations are written in differential form


\subsection{Continuity}
===== Continuity =====


\[\rho_1 u_1=\rho_2 u_2=const\Rightarrow\]\\
{{NumEqn|<math>
\rho_1 u_1=\rho_2 u_2=const\Rightarrow
</math>}}


\begin{equation}
{{NumEqn|<math>
\frac{d}{dx}(\rho u)=0
\frac{d}{dx}(\rho u)=0
\label{eq:governing:cont}
</math>}}
\end{equation}\\


\subsection{Momentum}
===== Momentum =====


\[(\rho_2 u_2^2+p_2-\rho_1 u_1^2+p_1)=-\frac{4}{D}\int_0^L\tau_w dx\Rightarrow\]\\
{{NumEqn|<math>
(\rho_2 u_2^2+p_2-\rho_1 u_1^2+p_1)=-\frac{4}{D}\int_0^L\tau_w dx\Rightarrow
</math>}}


\begin{equation}
{{NumEqn|<math>
\frac{d}{dx}(\rho u^2+p)=-\frac{4}{D}\tau_w
\frac{d}{dx}(\rho u^2+p)=-\frac{4}{D}\tau_w
\label{eq:governing:mom:c}
</math>}}
\end{equation}\\


\[\frac{d}{dx}(\rho u^2+p)=\rho u\frac{du}{dx}+u\frac{d}{dx}(\rho u)+\frac{dp}{dx}=\left\{\frac{d}{dx}(\rho u)=0\right\}=\rho u\frac{du}{dx}+\frac{dp}{dx}\]\\
{{NumEqn|<math>
\frac{d}{dx}(\rho u^2+p)=\rho u\frac{du}{dx}+u\frac{d}{dx}(\rho u)+\frac{dp}{dx}=\left\{\frac{d}{dx}(\rho u)=0\right\}=\rho u\frac{du}{dx}+\frac{dp}{dx}
</math>}}


\begin{equation}
{{NumEqn|<math>
\rho u\frac{du}{dx}+\frac{dp}{dx}=-\frac{4}{D}\tau_w
\rho u\frac{du}{dx}+\frac{dp}{dx}=-\frac{4}{D}\tau_w
\label{eq:governing:mom:d}
</math>}}
\end{equation}\\


\noindent The wall shear stress is often approximated using a shear-stress factor, $f$, according to\\
The wall shear stress is often approximated using a shear-stress factor, <math>f</math>, according to


\begin{equation}
{{NumEqn|<math>
\tau_w=f\frac{1}{2}\rho u^2
\tau_w=f\frac{1}{2}\rho u^2
\label{eq:tauw:b}
</math>}}
\end{equation}\\


\noindent and thus\\
and thus


\begin{equation}
{{NumEqn|<math>
\rho u\frac{du}{dx}+\frac{dp}{dx}=-\frac{2}{D}f\rho u^2
\rho u\frac{du}{dx}+\frac{dp}{dx}=-\frac{2}{D}f\rho u^2
\label{eq:governing:mom:d}
</math>}}
\end{equation}\\


\subsection{Energy}
===== Energy =====


\[h_1 + \frac{1}{2}u_1^2=h_2 + \frac{1}{2}u_2^2=const\]\\
{{NumEqn|<math>
h_1 + \frac{1}{2}u_1^2=h_2 + \frac{1}{2}u_2^2=const
</math>}}


\[h_{o_1}=h_{o_2}=const\]\\
{{NumEqn|<math>
h_{o_1}=h_{o_2}=const
</math>}}


\begin{equation}
{{NumEqn|<math>
\frac{d}{dx}h_o=0
\frac{d}{dx}h_o=0
\label{eq:governing:energy}
</math>}}
\end{equation}\\


\subsection{Summary}
==== Summary ====


\noindent continuity:\\
continuity:


\[\frac{d}{dx}(\rho u)=0\]\\
{{NumEqn|<math>
\frac{d}{dx}(\rho u)=0
</math>}}


\noindent momentum:\\
momentum:


\[\rho u\frac{du}{dx}+\frac{dp}{dx}=-\frac{2}{D}f\rho u^2\]\\
{{NumEqn|<math>
\rho u\frac{du}{dx}+\frac{dp}{dx}=-\frac{2}{D}f\rho u^2
</math>}}


\noindent energy:\\
energy:


\[\frac{d}{dx}h_o=0\]
{{NumEqn|<math>
\frac{d}{dx}h_o=0
</math>}}


\noindent From chapter 3.9 we have the following expression for the momentum equation for one-dimensional flow with friction (equation (3.95))\\
From chapter 3.9 we have the following expression for the momentum equation for one-dimensional flow with friction (equation (3.95))


\begin{equation*}
{{NumEqn|<math>
dp+\rho u du=-\frac{1}{2}\rho u^2 \frac{4 f dx}{D}\ (3.95)
dp+\rho u du=-\frac{1}{2}\rho u^2 \frac{4 f dx}{D}
\end{equation*}\\
</math>}}


\noindent For cases dealing with calorically perfect gas, (3.95) can be recast completely in terms of Mach number using the following relations\\
For cases dealing with calorically perfect gas, (3.95) can be recast completely in terms of Mach number using the following relations


\begin{itemize}
* speed of sound: <math>a^2=\gamma p/\rho</math>
\item[] speed of sound: $a^2=\gamma p/\rho$
* the definition of Mach number: <math>M^2=u^2/a^2</math>
\item[] the definition of Mach number: $M^2=u^2/a^2$
* the ideal gas law for thermally perfect gas: <math>p=\rho R T</math>
\item[] the ideal gas law for thermally perfect gas: $p=\rho R T$
* the continuity equation: <math>\rho u=const</math>
\item[] the continuity equation: $\rho u=const$
* the energy equation: <math>C_p T+u^2/2=const</math>
\item[] energy equation: $c_p T+u^2/2=const$
\end{itemize}


\subsection{Continuity equation}
==== Continuity equation ====


\noindent We start with the continuity equation which for one-dimensional steady flows reads\\
We start with the continuity equation which for one-dimensional steady flows reads


\begin{equation}
{{NumEqn|<math>
\rho u=const
\rho u=const
\label{eqn:cont:a}
</math>}}
\end{equation}\\


\noindent Differentiating (\ref{eqn:cont:a}) gives\\
Differentiating (\ref{eqn:cont:a}) gives


\begin{equation}
{{NumEqn|<math>
d(\rho u)=0. \Leftrightarrow \rho du + ud\rho=0.
d(\rho u)=0. \Leftrightarrow \rho du + ud\rho=0.
\label{eqn:cont:b}
</math>}}
\end{equation}\\


\noindent If $u\neq 0.$ we can divide by $\rho u$ which gives us\\
If <math>u\neq 0.</math> we can divide by <math>\rho u</math> which gives us


\begin{equation}
{{NumEqn|<math>
\frac{du}{u}+\frac{d\rho}{\rho}=0.
\frac{du}{u}+\frac{d\rho}{\rho}=0.
\label{eqn:cont:b}
</math>}}
\end{equation}\\


\noindent Now, if we divide and multiply the first term in (\ref{eqn:cont:b}) by $2u$ and use the chain rule for derivatives we get\\
Now, if we divide and multiply the first term in (\ref{eqn:cont:b}) by <math>2u</math> and use the chain rule for derivatives we get


\begin{equation}
{{NumEqn|<math>
\frac{d(u^2)}{2u^2}+\frac{d\rho}{\rho}=0.
\frac{d(u^2)}{2u^2}+\frac{d\rho}{\rho}=0.
\label{eqn:cont}
</math>}}
\end{equation}\\


\subsection{Energy equation}
==== Energy equation ====
\noindent For an adiabatic one-dimensional flow we have that \\


\begin{equation}
For an adiabatic one-dimensional flow we have that
c_p T+\frac{u^2}{2}=const
\label{eqn:ttot:a}
\end{equation}\\


\noindent If we differentiate (\ref{eqn:ttot:a}) we get\\
{{NumEqn|<math>
C_p T+\frac{u^2}{2}=const
</math>}}


\begin{equation}
If we differentiate (\ref{eqn:ttot:a}) we get
c_p dT+\frac{1}{2}d(u^2)=0.
\label{eqn:ttot:b}
\end{equation}\\


\noindent We replace $c_p$ with $\gamma R/(\gamma-1)$ and multiply and divide the first term with $T$ which gives us\\
{{NumEqn|<math>
C_p dT+\frac{1}{2}d(u^2)=0.
</math>}}


\begin{equation}
We replace <math>C_p</math> with <math>\gamma R/(\gamma-1)</math> and multiply and divide the first term with <math>T</math> which gives us
 
{{NumEqn|<math>
\frac{\gamma RT}{(\gamma-1)}\frac{dT}{T}+\frac{1}{2}d(u^2)=0.
\frac{\gamma RT}{(\gamma-1)}\frac{dT}{T}+\frac{1}{2}d(u^2)=0.
\label{eqn:ttot:c}
</math>}}
\end{equation}\\


\noindent Now, divide by $\gamma RT/(\gamma-1)$ and multiply and divide the second term by $u^2$ gives\\
Now, divide by <math>\gamma RT/(\gamma-1)</math> and multiply and divide the second term by <math>u^2</math> gives


\begin{equation}
{{NumEqn|<math>
\frac{dT}{T}+\frac{(\gamma-1)}{2}M^2\frac{d(u^2)}{u^2}=0.
\frac{dT}{T}+\frac{(\gamma-1)}{2}M^2\frac{d(u^2)}{u^2}=0.
\label{eqn:ttot}
</math>}}
\end{equation}\\


\noindent We want to remove the $dT/T$-term in (\ref{eqn:ttot}). From the definition of Mach number we have that\\
We want to remove the <math>dT/T</math>-term in (\ref{eqn:ttot}). From the definition of Mach number we have that


\begin{equation}
{{NumEqn|<math>
a^2M^2=u^2
a^2M^2=u^2
\label{eqn:Mach:a}
</math>}}
\end{equation}\\


\noindent which we can rewrite using the expression for speed of sound $(a^2=\gamma RT)$ according to\\
which we can rewrite using the expression for speed of sound <math>(a^2=\gamma RT)</math> according to


\begin{equation}
{{NumEqn|<math>
\gamma RTM^2=u^2
\gamma RTM^2=u^2
\label{eqn:Mach:b}
</math>}}
\end{equation}\\


\noindent Differentiating (\ref{eqn:Mach:b}) gives us\\
Differentiating (\ref{eqn:Mach:b}) gives us


\begin{equation}
{{NumEqn|<math>
\gamma RM^2 dT+\gamma RT d(M^2)=d(u^2)
\gamma RM^2 dT+\gamma RT d(M^2)=d(u^2)
\label{eqn:Mach:c}
</math>}}
\end{equation}\\


\noindent Now, if we divide (\ref{eqn:Mach:c}) by $\gamma RT M^2$ and use $a^2=\gamma RT$ and $a^2M^2=u^2$ we get\\
Now, if we divide (\ref{eqn:Mach:c}) by <math>\gamma RT M^2</math> and use <math>a^2=\gamma RT</math> and <math>a^2M^2=u^2</math> we get


\begin{equation}
{{NumEqn|<math>
\frac{dT}{T}+\frac{d(M^2)}{M^2}=\frac{d(u^2)}{u^2}
\frac{dT}{T}+\frac{d(M^2)}{M^2}=\frac{d(u^2)}{u^2}
\label{eqn:Mach}
</math>}}
\end{equation}\\


\noindent Equation (\ref{eqn:Mach}) may now be used to replace the $dT/T$-term in equation (\ref{eqn:ttot})\\
Equation (\ref{eqn:Mach}) may now be used to replace the <math>dT/T</math>-term in equation (\ref{eqn:ttot})


\begin{equation}
{{NumEqn|<math>
-\frac{d(M^2)}{M^2}+\frac{d(u^2)}{u^2}+\frac{(\gamma-1)}{2}M^2\frac{d(u^2)}{u^2}=0.
-\frac{d(M^2)}{M^2}+\frac{d(u^2)}{u^2}+\frac{(\gamma-1)}{2}M^2\frac{d(u^2)}{u^2}=0.
\label{eqn:ttot:Mach:a}
</math>}}
\end{equation}\\


\noindent which can be rewritten according to\\
which can be rewritten according to


\begin{equation}
{{NumEqn|<math>
\frac{d(u^2)}{u^2}=\left[1+\frac{(\gamma-1)}{2}M^2\right]^{-1}\frac{d(M^2)}{M^2}
\frac{d(u^2)}{u^2}=\left[1+\frac{(\gamma-1)}{2}M^2\right]^{-1}\frac{d(M^2)}{M^2}
\label{eqn:ttot:Mach:b}
</math>}}
\end{equation}\\


\noindent Using the chain rule for derivatives, the last term may be rewritten according to\\
Using the chain rule for derivatives, the last term may be rewritten according to


\[\frac{d(M^2)}{M^2}=2M\frac{dM}{M^2}=2\frac{dM}{M}\]\\
{{NumEqn|<math>
\frac{d(M^2)}{M^2}=2M\frac{dM}{M^2}=2\frac{dM}{M}
</math>}}


\noindent which gives\\
which gives


\begin{equation}
{{NumEqn|<math>
\frac{d(u^2)}{u^2}=2\left[1+\frac{(\gamma-1)}{2}M^2\right]^{-1}\frac{dM}{M}
\frac{d(u^2)}{u^2}=2\left[1+\frac{(\gamma-1)}{2}M^2\right]^{-1}\frac{dM}{M}
\label{eqn:ttot:Mach}
</math>}}
\end{equation}\\


\subsection{The ideal gas law}
==== The ideal gas law ====
\noindent For a perfect gas the ideal gas law reads\\


\begin{equation}
For a perfect gas the ideal gas law reads
 
{{NumEqn|<math>
p=\rho R T
p=\rho R T
\label{eqn:gaslaw:a}
</math>}}
\end{equation}\\


\noindent Differentiating (\ref{eqn:gaslaw:a}) gives:\\
Differentiating (\ref{eqn:gaslaw:a}) gives:


\begin{equation}
{{NumEqn|<math>
dp=\rho R dT+RT d\rho
dp=\rho R dT+RT d\rho
\label{eqn:gaslaw:b}
</math>}}
\end{equation}\\


\noindent If $p\neq0.$, we can divide (\ref{eqn:gaslaw:b}) by $p$ which gives\\
If <math>p\neq0.</math>, we can divide (\ref{eqn:gaslaw:b}) by <math>p</math> which gives


\begin{equation}
{{NumEqn|<math>
\frac{dp}{p}=\frac{dT}{T}+\frac{d\rho}{\rho}
\frac{dp}{p}=\frac{dT}{T}+\frac{d\rho}{\rho}
\label{eqn:gaslaw:c}
</math>}}
\end{equation}\\


\noindent which can be rearranged according to\\
which can be rearranged according to


\begin{equation}
{{NumEqn|<math>
\left[\frac{dp}{p}-\frac{d\rho}{\rho}\right]=\frac{dT}{T}
\left[\frac{dp}{p}-\frac{d\rho}{\rho}\right]=\frac{dT}{T}
\label{eqn:gaslaw:d}
</math>}}
\end{equation}\\


\noindent Now, inserting $dT/T$ from equation (\ref{eqn:ttot}) gives\\
Now, inserting <math>dT/T</math> from equation (\ref{eqn:ttot}) gives


\begin{equation}
{{NumEqn|<math>
\left[\frac{dp}{p}-\frac{d\rho}{\rho}\right]+\frac{(\gamma-1)}{2}M^2\frac{d(u^2)}{u^2}=0.
\left[\frac{dp}{p}-\frac{d\rho}{\rho}\right]+\frac{(\gamma-1)}{2}M^2\frac{d(u^2)}{u^2}=0.
\label{eqn:gaslaw:b}
</math>}}
\end{equation}\\


\noindent The $d\rho/\rho$-term can be replaced using equation (\ref{eqn:cont})\\
The <math>d\rho/\rho</math>-term can be replaced using equation (\ref{eqn:cont})


\begin{equation}
{{NumEqn|<math>
\frac{dp}{p}+\frac{d(u^2)}{2u^2}+\frac{(\gamma-1)}{2}M^2\frac{d(u^2)}{u^2}=0.
\frac{dp}{p}+\frac{d(u^2)}{2u^2}+\frac{(\gamma-1)}{2}M^2\frac{d(u^2)}{u^2}=0.
\label{eqn:gaslaw:c}
</math>}}
\end{equation}\\


\noindent Collect terms and rewrite gives\\
Collect terms and rewrite gives


\begin{equation}
{{NumEqn|<math>
\frac{dp}{p}+\left[\frac{1+(\gamma-1)M^2}{2}\right]\frac{d(u^2)}{u^2}=0.
\frac{dp}{p}+\left[\frac{1+(\gamma-1)M^2}{2}\right]\frac{d(u^2)}{u^2}=0.
\label{eqn:gaslaw}
</math>}}
\end{equation}\\


\subsection{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


\[dp+\rho u du=-\frac{1}{2}\rho u^2 \frac{4 f dx}{D}\ (3.95)\]\\
{{NumEqn|<math>
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}
{{NumEqn|<math>
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}
{{NumEqn|<math>
\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}
{{NumEqn|<math>
\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}
{{NumEqn|<math>
\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*}
{{NumEqn|<math>
\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>}}


\subsection{Differential Relations}
==== Differential Relations ====


\noindent In analogy with the heat addition process discussed in the previous section, one-dimensional flow with heat addition is a continuous process. We will derive the differential relations for one-dimensional flow with friction, which will lead to trends for supersonic and supersonic flow with friction.
In analogy with the heat addition process discussed in the previous section, one-dimensional flow with heat addition is a continuous process. We will derive the differential relations for one-dimensional flow with friction, which will lead to trends for supersonic and supersonic flow with friction.


<!--
\begin{figure}[ht!]
\begin{figure}[ht!]
\begin{center}
\begin{center}
Line 348: Line 346:
\label{fig:fanno:dx}
\label{fig:fanno:dx}
\end{figure}
\end{figure}
-->


\noindent The continuity equation gives
The continuity equation gives


\[d(\rho u)=ud\rho+\rho du \Rightarrow\]
{{NumEqn|<math>
d(\rho u)=ud\rho+\rho du \Rightarrow
</math>}}


\begin{equation}
{{NumEqn|<math>
\dfrac{d\rho}{\rho}=-\dfrac{du}{u}
\dfrac{d\rho}{\rho}=-\dfrac{du}{u}
\label{eq:fanno:drho}
</math>}}
\end{equation}
 
\noindent The addition of friction does not affect total temperature and thus the total temperature is constant


\[T_o=T+\dfrac{u^2}{2C_p}=const\]
The addition of friction does not affect total temperature and thus the total temperature is constant


\noindent differentiating gives
{{NumEqn|<math>
T_o=T+\dfrac{u^2}{2C_p}=const
</math>}}


\[dT_o=dT+\dfrac{1}{Cp}udu=0\]
differentiating gives


\noindent with $u=M\sqrt{\gamma RT}$, we get
{{NumEqn|<math>
dT_o=dT+\dfrac{1}{Cp}udu=0
</math>}}


with <math>u=M\sqrt{\gamma RT}</math>, we get


\begin{equation}
{{NumEqn|<math>
\dfrac{dT}{T}=-(\gamma-1)M^2\dfrac{du}{u}
\dfrac{dT}{T}=-(\gamma-1)M^2\dfrac{du}{u}
\label{eq:fanno:dT}
</math>}}
\end{equation}\\


\noindent A differential relation for pressure can be obtained from the ideal gas relation
A differential relation for pressure can be obtained from the ideal gas relation


\[p=\rho RT\Rightarrow dp=R(Td\rho+\rho dT)\Rightarrow\]
{{NumEqn|<math>
p=\rho RT\Rightarrow dp=R(Td\rho+\rho dT)\Rightarrow
</math>}}


\begin{equation}
{{NumEqn|<math>
\dfrac{dp}{p}=-\left(1+(\gamma-1)M^2\right)\dfrac{du}{u}
\dfrac{dp}{p}=-\left(1+(\gamma-1)M^2\right)\dfrac{du}{u}
\label{eq:fanno:dp}
</math>}}
\end{equation}\\


\noindent The entropy increase can be obtained from
The entropy increase can be obtained from


\[ds=C_v \dfrac{dp}{p}-C_p\dfrac{d\rho}{\rho}\]
{{NumEqn|<math>
ds=C_v \dfrac{dp}{p}-C_p\dfrac{d\rho}{\rho}
</math>}}


\noindent and thus
and thus


\begin{equation}
{{NumEqn|<math>
ds=-R(1-M^2)\dfrac{du}{u}
ds=-R(1-M^2)\dfrac{du}{u}
\label{eq:fanno:ds}
</math>}}
\end{equation}\\


\noindent Finally, a relation describing the change in Mach number can be obtained from \\
Finally, a relation describing the change in Mach number can be obtained from


\[M=\dfrac{u}{\sqrt{\gamma RT}}\Rightarrow dM=M\dfrac{du}{u}-\dfrac{M}{2}\dfrac{dT}{T}\]
{{NumEqn|<math>
M=\dfrac{u}{\sqrt{\gamma RT}}\Rightarrow dM=M\dfrac{du}{u}-\dfrac{M}{2}\dfrac{dT}{T}
</math>}}


\noindent which can be rewritten as
which can be rewritten as


\begin{equation}
{{NumEqn|<math>
\dfrac{dM}{M}=\left(1+(\gamma-1)M^2\right)\dfrac{du}{u}
\dfrac{dM}{M}=\left(1+(\gamma-1)M^2\right)\dfrac{du}{u}
\label{eq:fanno:dM}
</math>}}
\end{equation}\\


\noindent Eqns.~\ref{eq:fanno:drho} - \ref{eq:fanno:dM} are expressed as functions of $du$ and in order to get a direct relation to the addition of friction caused by the increase in pipe length $dx$, the equations are rewritten so that all variable changes are functions of the entropy increase $ds$.\\
Eqns.~\ref{eq:fanno:drho} - \ref{eq:fanno:dM} are expressed as functions of <math>du</math> and in order to get a direct relation to the addition of friction caused by the increase in pipe length <math>dx</math>, the equations are rewritten so that all variable changes are functions of the entropy increase <math>ds</math>.


\begin{equation}
{{NumEqn|<math>
\dfrac{d\rho}{\rho}=-\dfrac{1}{R(1-M^2)}ds
\dfrac{d\rho}{\rho}=-\dfrac{1}{R(1-M^2)}ds
\label{eq:fanno:drho:b}
</math>}}
\end{equation}


\begin{equation}
{{NumEqn|<math>
\dfrac{dT}{T}=-(\gamma-1)M^2\dfrac{1}{R(1-M^2)}ds
\dfrac{dT}{T}=-(\gamma-1)M^2\dfrac{1}{R(1-M^2)}ds
\label{eq:fanno:dT:b}
</math>}}
\end{equation}


\begin{equation}
{{NumEqn|<math>
\dfrac{dp}{p}=-\left(1+(\gamma-1)M^2\right)\dfrac{1}{R(1-M^2)}ds
\dfrac{dp}{p}=-\left(1+(\gamma-1)M^2\right)\dfrac{1}{R(1-M^2)}ds
\label{eq:fanno:dp:b}
</math>}}
\end{equation}


\begin{equation}
{{NumEqn|<math>
\dfrac{dM}{M}=\left(1+\dfrac{(\gamma-1)}{2}M^2\right)\dfrac{1}{R(1-M^2)}ds
\dfrac{dM}{M}=\left(1+\dfrac{(\gamma-1)}{2}M^2\right)\dfrac{1}{R(1-M^2)}ds
\label{eq:fanno:dp:b}
</math>}}
\end{equation}


\begin{equation}
{{NumEqn|<math>
\dfrac{du}{u}=\dfrac{1}{R(1-M^2)}ds
\dfrac{du}{u}=\dfrac{1}{R(1-M^2)}ds
\label{eq:fanno:drho:b}
</math>}}
\end{equation}\\


\noindent A relation for the change in total pressure can be obtained from  
A relation for the change in total pressure can be obtained from  


\[ds=C_p\dfrac{dT_o}{T_o}-R\dfrac{dp_o}{p_o}\]\\
{{NumEqn|<math>
ds=C_p\dfrac{dT_o}{T_o}-R\dfrac{dp_o}{p_o}
</math>}}


\noindent Since total temperature is constant the relation above gives
Since total temperature is constant the relation above gives


\begin{equation}
{{NumEqn|<math>
\dfrac{dp_o}{p_o}=-\dfrac{ds}{R}
\dfrac{dp_o}{p_o}=-\dfrac{ds}{R}
\label{eq:fanno:dpo:b}
</math>}}
\end{equation}\\


\noindent Using the differential relations above, we can get a good picture of the development of flow variables as friction is continuously added to the flow (see Figure~\ref{fig:fanno:trends}).\\
Using the differential relations above, we can get a good picture of the development of flow variables as friction is continuously added to the flow (see Figure~\ref{fig:fanno:trends}).


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\begin{figure}[ht!]
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\label{fig:fanno:trends}
\label{fig:fanno:trends}
\end{figure}
\end{figure}
-->


\subsection{Friction Choking}
==== Friction Choking ====


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\begin{figure}[ht!]
\begin{figure}[ht!]
\centering
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\label{fig:friction:Ts}
\label{fig:friction:Ts}
\end{figure}
\end{figure}
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\noindent Figure~\ref{fig:friction:Ts} shows the Fanno flow process in a $Ts$-diagram. The dashed line represents the sonic temperature, which means that the flow states along the process line above the dashed line are subsonic flow states and the part of the line below the dashed line represents supersonic flow states. In both subsonic and supersonic flow addition of friction leads to a change in temperature in the direction towards the sonic temperature, i.e. the flow approaches sonic conditions ($M=1$). When the length of the pipe through which the fluid flows is equal to the length at which the flow is sonic, the flow is choked (friction choking) and further pipe length cannot be added without a change in the flow conditions. For an initially subsonic flow, a pipe longer than $L^\ast$, the change in flow conditions is analogous to the what happens for addition of heat to a subsonic flow that has reached sonic state discussed in the previous section. The inlet conditions will change such that the massflow is reduced without changing the inlet total conditions such as the pipe length is equal to $L^\ast$ for the new inlet conditions.
Figure~\ref{fig:friction:Ts} shows the Fanno flow process in a <math>Ts</math>-diagram. The dashed line represents the sonic temperature, which means that the flow states along the process line above the dashed line are subsonic flow states and the part of the line below the dashed line represents supersonic flow states. In both subsonic and supersonic flow addition of friction leads to a change in temperature in the direction towards the sonic temperature, i.e. the flow approaches sonic conditions (<math>M=1</math>). When the length of the pipe through which the fluid flows is equal to the length at which the flow is sonic, the flow is choked (friction choking) and further pipe length cannot be added without a change in the flow conditions. For an initially subsonic flow, a pipe longer than <math>L^\ast</math>, the change in flow conditions is analogous to the what happens for addition of heat to a subsonic flow that has reached sonic state discussed in the previous section. The inlet conditions will change such that the massflow is reduced without changing the inlet total conditions such as the pipe length is equal to <math>L^\ast</math> for the new inlet conditions.


\[
{{InfoBox|<math>
\begin{aligned}
M_{1^\prime} = f(L^\ast)
M_{1'} & = f(L^\ast)\\
</math><br><br><math>
T_{1'} & = f(T_o, M_{1'})\\
T_{1^\prime} = f(T_o,\ M_{1^\prime})
p_{1'} & = f(p_o, M_{1'})\\
</math><br><br><math>
\rho_{1'} & = f(p_{1'}, T_{1'})\\
p_{1^\prime} = f(p_o,\ M_{1^\prime})
a_{1'} & = f(T_{1'})\\
</math><br><br><math>
u_{1'} & = M_{1'}a_{1'}\\
\rho_{1^\prime} = f(p_{1^\prime},\ T_{1^\prime})
\end{aligned}
</math><br><br><math>
\] \\
a_{1^\prime} = f(T_{1^\prime})
</math><br><br><math>
u_{1^\prime} = M_{1^\prime}a_{1^\prime}
</math>}}


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\begin{subfigure}[b]{0.5\textwidth}
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\label{fig:friction:choking:sub}
\label{fig:friction:choking:sub}
\end{figure}
\end{figure}
-->


 
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\begin{figure}[ht!]
\begin{figure}[ht!]
\centering
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\label{fig:friction:choking:sup}
\label{fig:friction:choking:sup}
\end{figure}
\end{figure}
-->


\noindent For a choked supersonic flow, addition of more friction (increasing the length of the pipe such that $L>L^\ast$) may lead to the generation of a shock inside the pipe. In contrast to the one-dimensional flow with heat addition where a shock does not change $q^\ast$, $L^\ast$ is increased over a shock. The internal shock will be generated in an axial location such that $L^\ast$ downstream of the shock equals the remaining pipe length at the shock location (see Figure~\ref{fig:friction:choking:sup}). As more length is added to the pipe, the shock will move further and further upstream in the pipe until it stands at the pipe entrance. If the pipe is longer than $L^\ast$ after o shock standing at the inlet, the shock will move to the upstream system and the pipe flow will be subsonic and the massflow will be adjusted such that $L=L^\ast$ according to the process described for subsonic choking above.  
For a choked supersonic flow, addition of more friction (increasing the length of the pipe such that <math>L>L^\ast</math>) may lead to the generation of a shock inside the pipe. In contrast to the one-dimensional flow with heat addition where a shock does not change <math>q^\ast</math>, <math>L^\ast</math> is increased over a shock. The internal shock will be generated in an axial location such that <math>L^\ast</math> downstream of the shock equals the remaining pipe length at the shock location (see Figure~\ref{fig:friction:choking:sup}). As more length is added to the pipe, the shock will move further and further upstream in the pipe until it stands at the pipe entrance. If the pipe is longer than <math>L^\ast</math> after o shock standing at the inlet, the shock will move to the upstream system and the pipe flow will be subsonic and the massflow will be adjusted such that <math>L=L^\ast</math> according to the process described for subsonic choking above.  


\noindent From prvevious derivations, we know that $L^\ast$ is a function of mach number according to\\
From prvevious derivations, we know that <math>L^\ast</math> is a function of mach number according to


\[\dfrac{4\bar{f}L^\ast}{D}=\dfrac{1-M^2}{\gamma M^2}+\left(\dfrac{\gamma+1}{2\gamma}\right)\ln\left(\dfrac{(\gamma+1)M^2}{2+(\gamma-1)M^2}\right)\]\\
{{NumEqn|<math>
\dfrac{4\bar{f}L^\ast}{D}=\dfrac{1-M^2}{\gamma M^2}+\left(\dfrac{\gamma+1}{2\gamma}\right)\ln\left(\dfrac{(\gamma+1)M^2}{2+(\gamma-1)M^2}\right)
</math>}}


\noindent by dividing both the numerator and denominator in the fractions by $M^2$ it is easy to see that the choking length (Figure~\ref{fig:friction:factor}) approaches a finite length for great Mach numbers and thus the upper limit for the choking length $L^\ast_1$ is given by\\
by dividing both the numerator and denominator in the fractions by <math>M^2</math> it is easy to see that the choking length (Figure~\ref{fig:friction:factor}) approaches a finite length for great Mach numbers and thus the upper limit for the choking length <math>L^\ast_1</math> is given by


\[\left.\dfrac{4\bar{f}L_1^\ast}{D}(M_1)\right|_{M_1\rightarrow \infty}=-\dfrac{1}{\gamma}+\left(\dfrac{\gamma+1}{2\gamma}\right)\ln\left(\dfrac{\gamma+1}{\gamma-1}\right)\]\\
{{NumEqn|<math>
\left.\dfrac{4\bar{f}L_1^\ast}{D}(M_1)\right|_{M_1\rightarrow \infty}=-\dfrac{1}{\gamma}+\left(\dfrac{\gamma+1}{2\gamma}\right)\ln\left(\dfrac{\gamma+1}{\gamma-1}\right)
</math>}}


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\begin{figure}[ht!]
\begin{figure}[ht!]
\centering
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\label{fig:friction:factor}
\label{fig:friction:factor}
\end{figure}
\end{figure}
-->


\noindent From the normal shock relations we know that the downstream Mach number approaches the finite value $\sqrt{(\gamma-1)/2\gamma}$ large Mach numbers and thus the choking length downstream the shock is limited to \\
From the normal shock relations we know that the downstream Mach number approaches the finite value <math>\sqrt{(\gamma-1)/2\gamma}</math> large Mach numbers and thus the choking length downstream the shock is limited to


\[\left.\dfrac{4\bar{f}L_2^\ast}{D}(M_2)\right|_{M_1\rightarrow \infty}=\left(\dfrac{\gamma+1}{\gamma(\gamma-1)}\right)+\left(\dfrac{\gamma+1}{2\gamma}\right)\ln\left(\dfrac{(\gamma+1)(\gamma-1)}{4\gamma+(\gamma-1)^2}\right)\]\\
{{NumEqn|<math>
\left.\dfrac{4\bar{f}L_2^\ast}{D}(M_2)\right|_{M_1\rightarrow \infty}=\left(\dfrac{\gamma+1}{\gamma(\gamma-1)}\right)+\left(\dfrac{\gamma+1}{2\gamma}\right)\ln\left(\dfrac{(\gamma+1)(\gamma-1)}{4\gamma+(\gamma-1)^2}\right)
</math>}}


\noindent From the relations above we get \\
From the relations above we get


\[\left.\left(\dfrac{4\bar{f}L_2^\ast}{D}(M_2)-\dfrac{4\bar{f}L_1^\ast}{D}(M_1)\right)\right|_{M_1\rightarrow \infty}=\left(\dfrac{2}{\gamma-1}\right)+\left(\dfrac{\gamma+1}{2\gamma}\right)\ln\left[\left(\dfrac{(\gamma+1)(\gamma-1)}{4\gamma+(\gamma-1)^2}\right)\left(\dfrac{\gamma-1}{\gamma+1}\right)\right]\]\\
{{NumEqn|<math>
\left.\left(\dfrac{4\bar{f}L_2^\ast}{D}(M_2)-\dfrac{4\bar{f}L_1^\ast}{D}(M_1)\right)\right|_{M_1\rightarrow \infty}=\left(\dfrac{2}{\gamma-1}\right)+\left(\dfrac{\gamma+1}{2\gamma}\right)\ln\left[\left(\dfrac{(\gamma+1)(\gamma-1)}{4\gamma+(\gamma-1)^2}\right)\left(\dfrac{\gamma-1}{\gamma+1}\right)\right]
</math>}}


\noindent Figure~\ref{fig:friction:factor:shock} shows the development of choking length $L_1^\ast$ in a supersonic flow as a function of Mach number in relation to the corresponding choking length $L_2^\ast$ downstream of a normal shock generated at the same Mach number. As can be seen from the figure, a normal shock will always increase the choking length.
Figure~\ref{fig:friction:factor:shock} shows the development of choking length <math>L_1^\ast</math> in a supersonic flow as a function of Mach number in relation to the corresponding choking length <math>L_2^\ast</math> downstream of a normal shock generated at the same Mach number. As can be seen from the figure, a normal shock will always increase the choking length.


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\begin{figure}[ht!]
\centering
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\label{fig:friction:factor:shock}
\label{fig:friction:factor:shock}
\end{figure}
\end{figure}
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