One-dimensional flow with heat addition: Difference between revisions

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


\noindent When the heat addition reaches $q^\ast$ the flow becomes sonic and the flow is said to thermally choked. Thermal choking is illustrated in Figure~\ref{fig:TSPV:d}, where the curve representing the energy equation (the blue line in the $p\nu$-diagram) is tangent to the Rayleigh line and if more heat is added the blue line will move to the right of the Rayleigh line and thus there are no solutions for $q>q^\ast$. So what happens if more heat is added to the flow after thermal choking is reached. The answer is different if the flow is subsonic or supersonic. For a subsonic flow, the upstream flow will be adjusted such that the slope of the Rayleigh line changes and the energy equation curve becomes tangent to the Rayleigh line. This means that the massflow per unit area ($C$) is reduced and $q^\ast$ is increased such that $q^\ast$ equals the heat added to the flow. Note that the upstream total conditions will not be changed in this process (see Figure~\ref{fig:thermal:choking:sub}).
When the heat addition reaches $q^\ast$ the flow becomes sonic and the flow is said to thermally choked. Thermal choking is illustrated in Figure~\ref{fig:TSPV:d}, where the curve representing the energy equation (the blue line in the <math>p\nu</math>-diagram) is tangent to the Rayleigh line and if more heat is added the blue line will move to the right of the Rayleigh line and thus there are no solutions for <math>q>q^\ast</math>. So what happens if more heat is added to the flow after thermal choking is reached. The answer is different if the flow is subsonic or supersonic. For a subsonic flow, the upstream flow will be adjusted such that the slope of the Rayleigh line changes and the energy equation curve becomes tangent to the Rayleigh line. This means that the massflow per unit area (<math>C</math>) is reduced and <math>q^\ast</math> is increased such that <math>q^\ast</math> equals the heat added to the flow. Note that the upstream total conditions will not be changed in this process (see Figure~\ref{fig:thermal:choking:sub}).


\[
<math display="block">
\begin{aligned}
M_{1^\prime} = f(q^\ast)
M_{1'} & = f(q^\ast)\\
</math>
T_{1'} & = f(T_o, M_{1'})\\
p_{1'} & = f(p_o, M_{1'})\\
\rho_{1'} & = f(p_{1'}, T_{1'})\\
a_{1'} & = f(T_{1'})\\
u_{1'} & = M_{1'}a_{1'}\\
\end{aligned}
\]


<math display="block">
T_{1^\prime} = f(T_o,\ M_{1^\prime})
</math>
<math display="block">
p_{1^\prime} = f(p_o,\ M_{1^\prime})
</math>
<math display="block">
\rho_{1^\prime} = f(p_{1^\prime},\ T_{1^\prime})
</math>
<math display="block">
a_{1^\prime} = f(T_{1^\prime})
</math>
<math display="block">
u_{1^\prime} = M_{1^\prime}a_{1^\prime}
</math>
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\begin{figure}[ht!]
\begin{figure}[ht!]
\begin{subfigure}[b]{0.5\textwidth}
\begin{subfigure}[b]{0.5\textwidth}
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\label{fig:thermal:choking:sub}
\label{fig:thermal:choking:sub}
\end{figure}
\end{figure}
-->


\noindent In a choked supersonic flow, there is no possibility for pressure waves to travel upstream in the flow and thus the upstream flow conditions can not be changed as in the subsonic case. Moreover, since a normal shock is an adiabatic process (a jump between two points on the same Rayleigh line), the total temperature is not changed over a chock. From before we have
In a choked supersonic flow, there is no possibility for pressure waves to travel upstream in the flow and thus the upstream flow conditions can not be changed as in the subsonic case. Moreover, since a normal shock is an adiabatic process (a jump between two points on the same Rayleigh line), the total temperature is not changed over a chock. From before we have


\[\dfrac{T_o}{T_o^\ast}=\dfrac{(\gamma+1)M_1^2}{(1+\gamma M_1^2)^2}(2+(\gamma-1)M_1^2)\]
<math display="block">
\dfrac{T_o}{T_o^\ast}=\dfrac{(\gamma+1)M_1^2}{(1+\gamma M_1^2)^2}(2+(\gamma-1)M_1^2)
</math>


\noindent Inserting the normal shock relation
Inserting the normal shock relation


\[M_2^2=\dfrac{2+(\gamma-1)M_1^2}{2\gamma M_1^2-(\gamma-1)}\]
<math display="block">
M_2^2=\dfrac{2+(\gamma-1)M_1^2}{2\gamma M_1^2-(\gamma-1)}
</math>


\noindent one can show that  
one can show that  


\[\dfrac{T_o}{T_o^\ast}=\dfrac{(\gamma+1)M_2^2}{(1+\gamma M_2^2)^2}(2+(\gamma-1)M_2^2)\]
<math display="block">
\dfrac{T_o}{T_o^\ast}=\dfrac{(\gamma+1)M_2^2}{(1+\gamma M_2^2)^2}(2+(\gamma-1)M_2^2)
</math>


\noindent and thus $T_o^\ast$ is not changed by the normal shock and consequently $q^\ast$ is unchanged if there is a normal shock between station 1 and 2. So, it is not possible to change the upstream static flow conditions and a normal shock will not make it possible to add more heat. The only possible solution is a normal shock upstream of station 1 and thus subsonic flow through the heat addition process.
and thus <math>T_o^\ast</math> is not changed by the normal shock and consequently <math>q^\ast</math> is unchanged if there is a normal shock between station 1 and 2. So, it is not possible to change the upstream static flow conditions and a normal shock will not make it possible to add more heat. The only possible solution is a normal shock upstream of station 1 and thus subsonic flow through the heat addition process.
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