One-dimensional flow with heat addition: Difference between revisions

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==== Rayleigh Line ====
==== Rayleigh Line ====


\noindent The continuity equation for steady-state, one-dimensional flow reads
The continuity equation for steady-state, one-dimensional flow reads


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


\noindent where $C$ is the massflow per square meter (massflow divided by area). Inserted in the momentum equation we get\\
where <math>C</math> is the massflow per square meter (massflow divided by area). Inserted in the momentum equation we get


\begin{equation}
<math display="block">
p_1+\dfrac{C^2}{\rho_1}=p_2+\dfrac{C^2}{\rho_2} \Leftrightarrow p_1+\nu_1 C^2=p_2+\nu_2 C^2\Rightarrow \dfrac{p_2-p_1}{\nu_2-\nu_1}=-C^2
p_1+\dfrac{C^2}{\rho_1}=p_2+\dfrac{C^2}{\rho_2} \Leftrightarrow p_1+\nu_1 C^2=p_2+\nu_2 C^2\Rightarrow \dfrac{p_2-p_1}{\nu_2-\nu_1}=-C^2
\label{eq:governing:mom:b}
</math>
\end{equation}\\


\noindent Eqn.~\ref{eq:governing:mom:b} tells us that any solution to the governing flow equations must lie along a line (a so-called Rayleigh line) in a $p\nu$-diagram. In Figure~\ref{fig:PV}, 1 corresponds to the flow state before heat addition and states 2 and 3 corresponds to the flow state after heat is added. If the flow in state 1 is subsonic, adding heat will change the flow state following the Rayleigh line to the right, i.e. towards flow state 2. If the initial flow state instead is supersonic, heat addition will move the flow state towards state 3.\\
Eqn.~\ref{eq:governing:mom:b} tells us that any solution to the governing flow equations must lie along a line (a so-called Rayleigh line) in a <math>p\nu</math>-diagram. In Figure~\ref{fig:PV}, 1 corresponds to the flow state before heat addition and states 2 and 3 corresponds to the flow state after heat is added. If the flow in state 1 is subsonic, adding heat will change the flow state following the Rayleigh line to the right, i.e. towards flow state 2. If the initial flow state instead is supersonic, heat addition will move the flow state towards state 3.


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\label{fig:PV}
\label{fig:PV}
\end{figure}
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\noindent Now we know in which direction we will move along the Rayleigh curve when heat is added but in order to find the flow state after heat addition we need to add the energy equation to the problem. If we draw a curve corresponding to the energy equation including the heat addition in the same $p\nu$-diagram, the intersection of this curve and the Rayleigh line corresponds to the downstream flow state (the flow state that fulfils the continuity, momentum, and energy equations). To be able to do this we will rewrite the energy equation such that it can be represented by a line in the $p\nu$-diagram.\\
Now we know in which direction we will move along the Rayleigh curve when heat is added but in order to find the flow state after heat addition we need to add the energy equation to the problem. If we draw a curve corresponding to the energy equation including the heat addition in the same <math>p\nu</math>-diagram, the intersection of this curve and the Rayleigh line corresponds to the downstream flow state (the flow state that fulfils the continuity, momentum, and energy equations). To be able to do this we will rewrite the energy equation such that it can be represented by a line in the <math>p\nu</math>-diagram.


\noindent The energy equation for one-dimensional flow with heat addition reads\\


\begin{equation}
The energy equation for one-dimensional flow with heat addition reads
 
<math display="block">
h_1+\dfrac{1}{2}u_1^2+q=h_2+\dfrac{1}{2}u_2^2
h_1+\dfrac{1}{2}u_1^2+q=h_2+\dfrac{1}{2}u_2^2
\label{eq:governing:energy}
</math>
\end{equation}\\


\noindent Inserting the constant $C$ from above (the massflow per $m^2$) and and and $h=C_pT$, we get\\
Inserting the constant <math>C</math> from above (the massflow per <math>m^2</math>) and and and <math>h=C_pT</math>, we get


\begin{equation}
<math display="block">
\dfrac{\gamma R}{\gamma - 1}T_1+\dfrac{1}{2}C^2\nu_1^2+q=\dfrac{\gamma R}{\gamma-1}+\dfrac{1}{2}C^2\nu_2^2
\dfrac{\gamma R}{\gamma - 1}T_1+\dfrac{1}{2}C^2\nu_1^2+q=\dfrac{\gamma R}{\gamma-1}+\dfrac{1}{2}C^2\nu_2^2
\label{eq:governing:energy:b}
</math>
\end{equation}\\


\noindent which may be rewritten as\\
which may be rewritten as


\begin{equation}
<math display="block">
\dfrac{p_2}{p_1}=\left(\dfrac{\nu_2}{\nu_1}-\dfrac{\gamma+1}{\gamma-1}-\dfrac{2q}{RT_1}\right)\left(1-\dfrac{\gamma+1}{\gamma-1}\dfrac{\nu_2}{\nu_1}\right)^{-1}
\dfrac{p_2}{p_1}=\left(\dfrac{\nu_2}{\nu_1}-\dfrac{\gamma+1}{\gamma-1}-\dfrac{2q}{RT_1}\right)\left(1-\dfrac{\gamma+1}{\gamma-1}\dfrac{\nu_2}{\nu_1}\right)^{-1}
\label{eq:governing:energy:c}
</math>
\end{equation}\\


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\label{fig:TSPV:a}
\label{fig:TSPV:a}
\end{figure}
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\label{fig:TSPV:b}
\label{fig:TSPV:b}
\end{figure}
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\label{fig:TSPV:c}
\label{fig:TSPV:c}
\end{figure}
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\label{fig:TSPV:d}
\label{fig:TSPV:d}
\end{figure}
\end{figure}
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\noindent As you can see in the examples above (Figures~\ref{fig:TSPV:b} and \ref{fig:TSPV:d}), sonic conditions are reached when the Rayleigh line is tangent to the curve representing the energy equation in the $p\nu$-diagram. Adding more heat would move the energy equation line upwards and thus there can not be any solution after reaching this state unless the upstream conditions are changed such that the energy line intersects the Rayleigh line after further heat addition. Let's have a second look at the equations and see if it is possible to verify that the case where the Rayleigh line is a tangent to the energy-equation curve is in fact the sonic state.\\
As you can see in the examples above (Figures~\ref{fig:TSPV:b} and \ref{fig:TSPV:d}), sonic conditions are reached when the Rayleigh line is tangent to the curve representing the energy equation in the <math>p\nu</math>-diagram. Adding more heat would move the energy equation line upwards and thus there can not be any solution after reaching this state unless the upstream conditions are changed such that the energy line intersects the Rayleigh line after further heat addition. Let's have a second look at the equations and see if it is possible to verify that the case where the Rayleigh line is a tangent to the energy-equation curve is in fact the sonic state.


\noindent Starting from Eqn.~\ref{eq:governing:energy:b}, it is easy to see that for any point along the energy equation curve the flow state may be expressed as a function of the initial flow state and the added heat $q$ as\\
Starting from Eqn.~\ref{eq:governing:energy:b}, it is easy to see that for any point along the energy equation curve the flow state may be expressed as a function of the initial flow state and the added heat <math>q</math> as


\begin{equation}
<math display="block">
\dfrac{\gamma}{\gamma-1}p\nu+\dfrac{1}{2}C^2\nu^2=\dfrac{\gamma}{\gamma-1}p_1\nu_1+\dfrac{1}{2}C^2\nu_1^2+q=D
\dfrac{\gamma}{\gamma-1}p\nu+\dfrac{1}{2}C^2\nu^2=\dfrac{\gamma}{\gamma-1}p_1\nu_1+\dfrac{1}{2}C^2\nu_1^2+q=D
\label{eq:governing:energy:d}
</math>
\end{equation}\\


\noindent where $D$ is a constant.\\
where <math>D</math> is a constant.


\noindent Now, let's different the Eqn.\ref{eq:governing:energy:d} with respect to $\nu$\\
Now, let's different the Eqn.\ref{eq:governing:energy:d} with respect to <math>\nu</math>


\begin{equation}
<math display="block">
\dfrac{\gamma}{\gamma-1}\left(\nu\dfrac{dp}{d\nu}+p\right)+C^2\nu=0\Rightarrow \dfrac{dp}{d\nu}=-\dfrac{\gamma}{\gamma-1}C^2-\dfrac{p}{\nu}
\dfrac{\gamma}{\gamma-1}\left(\nu\dfrac{dp}{d\nu}+p\right)+C^2\nu=0\Rightarrow \dfrac{dp}{d\nu}=-\dfrac{\gamma}{\gamma-1}C^2-\dfrac{p}{\nu}
\label{eq:governing:energy:e}
</math>
\end{equation}\\


\noindent The Rayleigh line is a tangent to the energy equation curve when $dp/d\nu=-C^2$ and thus\\
The Rayleigh line is a tangent to the energy equation curve when <math>dp/d\nu=-C^2</math> and thus


\begin{equation}
<math display="block">
\dfrac{C^2}{\gamma}=\dfrac{p}{\nu}
\dfrac{C^2}{\gamma}=\dfrac{p}{\nu}
\label{eq:governing:energy:f}
</math>
\end{equation}\\


\noindent By definition $C=\rho u$ and $\nu=1/\rho$, which inserted in Eqn.~\ref{eq:governing:energy:f} gives\\
By definition <math>C=\rho u</math> and <math>\nu=1/\rho</math>, which inserted in Eqn.~\ref{eq:governing:energy:f} gives


\begin{equation}
<math display="block">
u=\sqrt{\dfrac{\gamma p}{\rho}}=a
u=\sqrt{\dfrac{\gamma p}{\rho}}=a
\label{eq:governing:energy:g}
</math>
\end{equation}\\


==== Thermal Choking ====
==== Thermal Choking ====
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