One-dimensional flow with heat addition: Difference between revisions

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


\noindent The equations presented in the previous section gives us the flow state after heat addition but since the heat addition, unlike the normal shock, is a continuous process, it is of interest to study the the heat addition from start to end. In order to do so we will now derive differential relations starting from the governing equations on differential form. We will start with converting the integral equation for conservation of mass for one-dimensional flows to differential form.\\
The equations presented in the previous section gives us the flow state after heat addition but since the heat addition, unlike the normal shock, is a continuous process, it is of interest to study the the heat addition from start to end. In order to do so we will now derive differential relations starting from the governing equations on differential form. We will start with converting the integral equation for conservation of mass for one-dimensional flows to differential form.


\begin{equation}
<math display="block">
\rho_1 u_1 = \rho_2 u_2 = const \Rightarrow d(\rho u)=0
\rho_1 u_1 = \rho_2 u_2 = const \Rightarrow d(\rho u)=0
\label{eq:governing:cont:diff}
</math>
\end{equation}\\


\[d(\rho u)=\rho du+ud\rho=0\]
<math display="block">
d(\rho u)=\rho du+ud\rho=0
</math>


\noindent Divide by $\rho u$ gives
Divide by <math>\rho u</math> gives


\begin{equation}
<math display="block">
\dfrac{d\rho}{\rho}=\dfrac{du}{u}
\dfrac{d\rho}{\rho}=\dfrac{du}{u}
\label{eq:governing:cont:diff:b}
</math>
\end{equation}\\


\noindent The integral form of the conservation of momentum equation for one-dimensional flows is converted to differential form as follows.\\
The integral form of the conservation of momentum equation for one-dimensional flows is converted to differential form as follows.


\begin{equation}
<math display="block">
p_1+\rho_1u_1^2=p_2+\rho_2u_2^2=const \Rightarrow d(p+\rho u^2)=0
p_1+\rho_1u_1^2=p_2+\rho_2u_2^2=const \Rightarrow d(p+\rho u^2)=0
\label{eq:governing:mom:diff}
</math>
\end{equation}\\


\begin{equation}
<math display="block">
dp+\rho udu+u\underbrace{d(\rho u)}_{=0}=0\Rightarrow dp=-\rho udu
dp+\rho udu+u\underbrace{d(\rho u)}_{=0}=0\Rightarrow dp=-\rho udu
\label{eq:governing:mom:diff:b}
</math>
\end{equation}\\


\noindent with $\rho=\dfrac{p}{RT}$ and $u^2=M^2a^2=M^2\gamma RT$ in Eqn.~\ref{eq:governing:mom:diff:b}, we get\\
with <math>\rho=\dfrac{p}{RT}</math> and <math>u^2=M^2a^2=M^2\gamma RT</math> in Eqn.~\ref{eq:governing:mom:diff:b}, we get\\


\[dp=-\dfrac{p}{RT}u^2\dfrac{du}{u}=-\dfrac{p}{RT}M^2\gamma RT\dfrac{du}{u}\Rightarrow \]
<math display="block">
dp=-\dfrac{p}{RT}u^2\dfrac{du}{u}=-\dfrac{p}{RT}M^2\gamma RT\dfrac{du}{u}\Rightarrow
</math>


\begin{equation}
<math display="block">
\dfrac{dp}{p}=-\gamma M^2\dfrac{du}{u}
\dfrac{dp}{p}=-\gamma M^2\dfrac{du}{u}
\label{eq:governing:mom:diff:c}
</math>
\end{equation}\\


\noindent which gives the relative change in pressure, $dp/p$, as a function of the relative change in flow velocity, $du/u$. The next equation to derive is an equation that describes the relative change in temperature, $dT/T$, as a function of the relative change in flow velocity, $du/u$. The starting point is the equation of state (the gas law).\\
which gives the relative change in pressure, <math>dp/p</math>, as a function of the relative change in flow velocity, <math>du/u</math>. The next equation to derive is an equation that describes the relative change in temperature, <math>dT/T</math>, as a function of the relative change in flow velocity, <math>du/u</math>. The starting point is the equation of state (the gas law).


\begin{equation}
<math display="block">
p=\rho RT\Rightarrow dp = R(\rho dT+ Td\rho)\Rightarrow dT=\dfrac{1}{R\rho}dp-\dfrac{T}{\rho}d\rho
p=\rho RT\Rightarrow dp = R(\rho dT+ Td\rho)\Rightarrow dT=\dfrac{1}{R\rho}dp-\dfrac{T}{\rho}d\rho
\label{eq:governing:temp:diff:a}
</math>
\end{equation}\\


\noindent Divide by $T$\\
Divide by <math>T</math>


\begin{equation}
<math display="block">
\dfrac{dT}{T}=\dfrac{1}{\rho RT}dp-\dfrac{1}{\rho}d\rho
\dfrac{dT}{T}=\dfrac{1}{\rho RT}dp-\dfrac{1}{\rho}d\rho
\label{eq:governing:temp:diff:b}
</math>
\end{equation}\\


\noindent substitute $dp$ from Eqn.~\ref{eq:governing:mom:diff:c} and $d\rho$ from Eqn.~\ref{eq:governing:cont:diff:b} gives\\
substitute <math>dp</math> from Eqn.~\ref{eq:governing:mom:diff:c} and <math>d\rho</math> from Eqn.~\ref{eq:governing:cont:diff:b} gives


\begin{equation}
<math display="block">
\dfrac{dT}{T}=(1-\gamma M^2)\dfrac{du}{u}
\dfrac{dT}{T}=(1-\gamma M^2)\dfrac{du}{u}
\label{eq:governing:temp:diff:c}
</math>
\end{equation}\\


\noindent The entropy equation reads\\
The entropy equation reads


\begin{equation}
<math display="block">
ds=C_v\dfrac{dp}{p}-C_p\dfrac{d\rho}{\rho}
ds=C_v\dfrac{dp}{p}-C_p\dfrac{d\rho}{\rho}
\label{eq:governing:entropy:diff:a}
</math>
\end{equation}\\


\noindent which after substituting $dp$ from Eqn.~\ref{eq:governing:mom:diff:c} and $d\rho$ from Eqn.~\ref{eq:governing:cont:diff:b} becomes\\
which after substituting <math>dp</math> from Eqn.~\ref{eq:governing:mom:diff:c} and <math>d\rho</math> from Eqn.~\ref{eq:governing:cont:diff:b} becomes


\begin{equation}
<math display="block">
ds=C_v\gamma(1-M^2)\dfrac{du}{u}
ds=C_v\gamma(1-M^2)\dfrac{du}{u}
\label{eq:governing:entropy:diff:b}
</math>
\end{equation}\\


\noindent From the definition of total temperature $T_o$ we get\\
From the definition of total temperature <math>T_o</math> we get




\[T_o=T+\dfrac{u^2}{2C_p}\Rightarrow dT_o=dT+\dfrac{1}{C_p}udu=dT+\dfrac{\gamma-1}{\gamma RT}T u^2\dfrac{du}{u}\Rightarrow\]
<math display="block">
T_o=T+\dfrac{u^2}{2C_p}\Rightarrow dT_o=dT+\dfrac{1}{C_p}udu=dT+\dfrac{\gamma-1}{\gamma RT}T u^2\dfrac{du}{u}\Rightarrow
</math>


\begin{equation}
<math display="block">
dT_o=dT+(\gamma-1)M^2 T\dfrac{du}{u}
dT_o=dT+(\gamma-1)M^2 T\dfrac{du}{u}
\label{eq:governing:To:diff:a}
</math>
\end{equation}\\


\noindent Inserting $dT$ from Eqn~\ref{eq:governing:temp:diff:c} in Eqn~\ref{eq:governing:To:diff:a} we get\\
Inserting <math>dT</math> from Eqn~\ref{eq:governing:temp:diff:c} in Eqn~\ref{eq:governing:To:diff:a} we get


\[dT_o=(1-\gamma M^2)T\dfrac{du}{u}+(\gamma-1)M^2 T\dfrac{du}{u}\]
<math display="block">
dT_o=(1-\gamma M^2)T\dfrac{du}{u}+(\gamma-1)M^2 T\dfrac{du}{u}
</math>


\noindent or
or


\begin{equation}
<math display="block">
dT_o=(1-M^2)T\dfrac{du}{u}
dT_o=(1-M^2)T\dfrac{du}{u}
\label{eq:governing:To:diff:b}
</math>
\end{equation}\\


\noindent Dividing Eqn.~\ref{eq:governing:To:diff:b} by $T_o$ and using\\
Dividing Eqn.~\ref{eq:governing:To:diff:b} by <math>T_o</math> and using


\[T_o=T\left(1+\dfrac{\gamma-1}{2}M^2\right)\]
<math display="block">
T_o=T\left(1+\dfrac{\gamma-1}{2}M^2\right)
</math>


\noindent we get\\
we get


\begin{equation}
<math display="block">
\dfrac{dT_o}{T_o}=\dfrac{1-M^2}{1+\dfrac{\gamma-1}{2}M^2}\dfrac{du}{u}
\dfrac{dT_o}{T_o}=\dfrac{1-M^2}{1+\dfrac{\gamma-1}{2}M^2}\dfrac{du}{u}
\label{eq:governing:To:diff:c}
</math>
\end{equation}\\


Finally, we will derive a differential relation that describes the change in Mach number.


\noindent Finally, we will derive a differential relation that describes the change in Mach number.\\
<math display="block">
M=\dfrac{u}{\sqrt{\gamma RT}}\Rightarrow dM=\dfrac{1}{\sqrt{\gamma R}}(T^{1/2}du+ud(T^{-1/2}))=\dfrac{du}{\sqrt{\gamma RT}}-\dfrac{u}{2\sqrt{\gamma R}}T^{-3/2}dT\Rightarrow
</math>


\[M=\dfrac{u}{\sqrt{\gamma RT}}\Rightarrow dM=\dfrac{1}{\sqrt{\gamma R}}(T^{1/2}du+ud(T^{-1/2}))=\dfrac{du}{\sqrt{\gamma RT}}-\dfrac{u}{2\sqrt{\gamma R}}T^{-3/2}dT\Rightarrow\]
<math display="block">
dM=\dfrac{1}{\sqrt{\gamma RT}}\dfrac{du}{u}-\dfrac{1}{2}\dfrac{u}{\sqrt{\gamma RT}}\dfrac{dT}{T}=M\dfrac{du}{u}-\dfrac{M}{2}\dfrac{dT}{T}
</math>


\[dM=\dfrac{1}{\sqrt{\gamma RT}}\dfrac{du}{u}-\dfrac{1}{2}\dfrac{u}{\sqrt{\gamma RT}}\dfrac{dT}{T}=M\dfrac{du}{u}-\dfrac{M}{2}\dfrac{dT}{T}\]\\
Inserting <math>dT/T</math> from Eqn.~\ref{eq:governing:temp:diff:c}, we get


\noindent Inserting $dT/T$ from Eqn.~\ref{eq:governing:temp:diff:c}, we get\\
<math display="block">
 
\begin{equation}
\dfrac{dM}{M}=\dfrac{1+\gamma M^2}{2}\dfrac{du}{u}
\dfrac{dM}{M}=\dfrac{1+\gamma M^2}{2}\dfrac{du}{u}
\label{eq:governing:M:diff:a}
</math>
\end{equation}\\


\noindent All the derived differential relations are expressed as functions of $du/u$ but it would be more convenient to relate the changes in flow properties to the added heat or the change in total temperature, which can be related to the added heat through the energy equation.\\
All the derived differential relations are expressed as functions of <math<du/u</math> but it would be more convenient to relate the changes in flow properties to the added heat or the change in total temperature, which can be related to the added heat through the energy equation.


\[dT_o=\dfrac{\delta q}{C_p}\]
<math display="block">
dT_o=\dfrac{\delta q}{C_p}
</math>


\noindent From Eqn.~\ref{eq:governing:To:diff:c}, we get\\
From Eqn.~\ref{eq:governing:To:diff:c}, we get


\begin{equation}
<math display="block">
\dfrac{du}{u}=\left(\dfrac{1+\dfrac{\gamma-1}{2}M^2}{1-M^2}\right)\dfrac{dT_o}{T_o}
\dfrac{du}{u}=\left(\dfrac{1+\dfrac{\gamma-1}{2}M^2}{1-M^2}\right)\dfrac{dT_o}{T_o}
\label{eq:governing:du:diff:final}
</math>
\end{equation}\\


\noindent Now, we can substitute $du/u$ in all the above relations using Eqn.~\ref{eq:governing:du:diff:final}, we get the following relations\\
Now, we can substitute $du/u$ in all the above relations using Eqn.~\ref{eq:governing:du:diff:final}, we get the following relations


\begin{equation}
<math display="block">
\dfrac{d\rho}{\rho}=-\left(\dfrac{1+\dfrac{\gamma-1}{2}M^2}{1-M^2}\right)\dfrac{dT_o}{T_o}
\dfrac{d\rho}{\rho}=-\left(\dfrac{1+\dfrac{\gamma-1}{2}M^2}{1-M^2}\right)\dfrac{dT_o}{T_o}
\label{eq:governing:dro:diff:final}
</math>
\end{equation}\\


\begin{equation}
<math display="block">
\dfrac{dp}{p}=\gamma M^2\left(\dfrac{1+\dfrac{\gamma-1}{2}M^2}{1-M^2}\right)\dfrac{dT_o}{T_o}
\dfrac{dp}{p}=\gamma M^2\left(\dfrac{1+\dfrac{\gamma-1}{2}M^2}{1-M^2}\right)\dfrac{dT_o}{T_o}
\label{eq:governing:dp:diff:final}
</math>
\end{equation}\\


\begin{equation}
<math display="block">
\dfrac{dT}{T}=(1-\gamma M^2)\left(\dfrac{1+\dfrac{\gamma-1}{2}M^2}{1-M^2}\right)\dfrac{dT_o}{T_o}
\dfrac{dT}{T}=(1-\gamma M^2)\left(\dfrac{1+\dfrac{\gamma-1}{2}M^2}{1-M^2}\right)\dfrac{dT_o}{T_o}
\label{eq:governing:dT:diff:final}
</math>
\end{equation}\\


\begin{equation}
<math display="block">
\dfrac{dT}{T}=\left(\dfrac{1+\gamma M^2}{2}\right)\left(\dfrac{1+\dfrac{\gamma-1}{2}M^2}{1-M^2}\right)\dfrac{dT_o}{T_o}
\dfrac{dT}{T}=\left(\dfrac{1+\gamma M^2}{2}\right)\left(\dfrac{1+\dfrac{\gamma-1}{2}M^2}{1-M^2}\right)\dfrac{dT_o}{T_o}
\label{eq:governing:dM:diff:final}
</math>
\end{equation}\\


\begin{equation}
<math display="block">
\dfrac{dT}{T}=C_v\gamma \left(1+\dfrac{\gamma-1}{2}M^2\right)\dfrac{dT_o}{T_o}
\dfrac{dT}{T}=C_v\gamma \left(1+\dfrac{\gamma-1}{2}M^2\right)\dfrac{dT_o}{T_o}
\label{eq:governing:ds:diff:final}
</math>
\end{equation}\\


==== Heat Addition Process ====
==== Heat Addition Process ====
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