One-dimensional flow with friction: Difference between revisions

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==== 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.


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\label{fig:fanno:dx}
\label{fig:fanno:dx}
\end{figure}
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\noindent The continuity equation gives
The continuity equation gives


\[d(\rho u)=ud\rho+\rho du \Rightarrow\]
<math display="block">
d(\rho u)=ud\rho+\rho du \Rightarrow
</math>


\begin{equation}
<math display="block">
\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  
The addition of friction does not affect total temperature and thus the total temperature is constant  


\[T_o=T+\dfrac{u^2}{2C_p}=const\]
<math display="block">
T_o=T+\dfrac{u^2}{2C_p}=const
</math>


\noindent differentiating gives
differentiating gives


\[dT_o=dT+\dfrac{1}{Cp}udu=0\]
<math display="block">
dT_o=dT+\dfrac{1}{Cp}udu=0
</math>


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


 
<math display="block">
\begin{equation}
\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


\[p=\rho RT\Rightarrow dp=R(Td\rho+\rho dT)\Rightarrow\]
<math display="block">
p=\rho RT\Rightarrow dp=R(Td\rho+\rho dT)\Rightarrow
</math>


\begin{equation}
<math display="block">
\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


\[ds=C_v \dfrac{dp}{p}-C_p\dfrac{d\rho}{\rho}\]
<math display="block">
ds=C_v \dfrac{dp}{p}-C_p\dfrac{d\rho}{\rho}
</math>


\noindent and thus
and thus


\begin{equation}
<math display="block">
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


\[M=\dfrac{u}{\sqrt{\gamma RT}}\Rightarrow dM=M\dfrac{du}{u}-\dfrac{M}{2}\dfrac{dT}{T}\]
<math display="block">
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}
<math display="block">
\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}
<math display="block">
\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}
<math display="block">
\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}
<math display="block">
\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}
<math display="block">
\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}
<math display="block">
\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  


\[ds=C_p\dfrac{dT_o}{T_o}-R\dfrac{dp_o}{p_o}\]\\
<math display="block">
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}
<math display="block">
\dfrac{dp_o}{p_o}=-\dfrac{ds}{R}
\dfrac{dp_o}{p_o}=-\dfrac{ds}{R}
\label{eq:fanno:dpo:b}
</math>
\end{equation}\\
 
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}).


\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}).\\


\begin{figure}[ht!]
\begin{figure}[ht!]
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