Governing equations on integral form: Difference between revisions
Created page with "Category:Compressible flow Category:Governing equations __TOC__ \section{Governing Equations on Integral Form} \begin{figure}[ht!] \begin{center} \includegraphics[]{figures/standalone-figures/Chapter02/pdf/control-volume.pdf} \caption{Generic control volume} \label{fig:generic:cv} \end{center} \end{figure} \noindent The governing equations stems from mass conservation, conservation of momentum and conservation of energy \subsection{The Continuity Equation}..." |
No edit summary |
||
| Line 4: | Line 4: | ||
__TOC__ | __TOC__ | ||
<!-- | |||
\begin{figure}[ht!] | \begin{figure}[ht!] | ||
\begin{center} | \begin{center} | ||
| Line 13: | Line 12: | ||
\end{center} | \end{center} | ||
\end{figure} | \end{figure} | ||
--> | |||
The governing equations stems from mass conservation, conservation of momentum and conservation of energy | |||
==== The Continuity Equation ==== | |||
{{quote|Mass can be neither created nor destroyed, which implies that mass is conserved}} | |||
\ | The net massflow into the control volume <math>\Omega</math> in Fig. \ref{fig:generic:cv} is obtained by integrating mass flux over the control volume surface <math>\partial \Omega</math> | ||
\ | <math display="block"> | ||
-\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS | |||
</math> | |||
Now, let's consider a small infinitesimal volume $d\mathscr{V}$ inside <math>\Omega</math>. The mass of <math>dV</math> is <math>\rho dV</math>. Thus, the mass enclosed within <math>\Omega</math> can be calculated as | |||
\ | <math display="block"> | ||
\iiint_{\Omega} \rho dV | |||
</math> | |||
The rate of change of mass within <math>\Omega</math> is obtained as | |||
\ | <math display="block"> | ||
\frac{d}{dt}\iiint_{\Omega} \rho dV | |||
</math> | |||
Mass is conserved, which means that the rate of change of mass within <math>\Omega</math> must equal the net flux over the control volume surface. | |||
<math display="block"> | |||
\frac{d}{dt}\iiint_{\Omega} \rho dV=-\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS | |||
</math> | |||
or | |||
\ | <math display="block"> | ||
\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0 | |||
</math> | |||
which is the integral form of the continuity equation. | |||
==== The Momentum Equation ==== | |||
{{quote|The time rate of change of momentum of a body equals the net force exerted on it}} | |||
<math display="block"> | |||
\frac{d}{dt}(m\mathbf{v})=\mathbf{F} | |||
</math> | |||
What type of forces do we have? | |||
* Body forces acting on the fluid inside $\Omega$ | |||
** gravitation | |||
** electromagnetic forces | |||
** Coriolis forces | |||
* Surface forces: pressure forces and shear forces | |||
Body forces inside <math>\Omega</math>: | |||
\ | <math display="block"> | ||
\iiint_{\Omega}\rho \mathbf{f}dV | |||
</math> | |||
Surface force on <math>\partial \Omega</math>: | |||
\ | <math display="block"> | ||
-\iint_{\partial \Omega} p\mathbf{n}dS | |||
</math> | |||
Since we are considering inviscid flow, there are no shear forces and thus we have the net force as | |||
\ | <math display="block"> | ||
\mathbf{F}=\iiint_{\Omega}\rho \mathbf{f}dV-\iint_{\partial \Omega} p\mathbf{n}dS | |||
</math> | |||
The fluid flowing through $\Omega$ will carry momentum and the net flow of momentum out from <math>\Omega</math> is calculated as | |||
\ | <math display="block"> | ||
\iint_{\partial \Omega} (\rho\mathbf{v}\cdot\mathbf{n}dS)\mathbf{v}=\iint_{\partial \Omega} (\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}dS | |||
</math> | |||
Integrated momentum inside <math>\Omega</math> | |||
\ | <math display="block"> | ||
\iiint_{\Omega} \rho \mathbf{v} dV | |||
</math> | |||
Rate of change of momentum due to unsteady effects inside <math>\Omega</math> | |||
\ | <math display="block"> | ||
\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV | |||
</math> | |||
Combining the rate of change of momentum, the net momentum flux and the net forces we get | |||
\ | <math display="block"> | ||
\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV+\iint_{\partial \Omega} (\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}dS=\iiint_{\Omega}\rho \mathbf{f}dV-\iint_{\partial \Omega} p\mathbf{n}dS | |||
</math> | |||
combining the surface integrals, we get | |||
\ | <math display="block"> | ||
\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV+\iint_{\partial \Omega} \left[(\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}+p\mathbf{n}\right]dS=\iiint_{\Omega}\rho \mathbf{f}dV | |||
</math> | |||
which is the momentum equation on integral form. | |||
==== The Energy Equation ==== | |||
{{quote|Energy can be neither created nor destroyed; it can only change in form}} | |||
\[E_1+E_2=E_3\]\\ | \[E_1+E_2=E_3\]\\ | ||
Revision as of 08:24, 17 March 2026
The governing equations stems from mass conservation, conservation of momentum and conservation of energy
The Continuity Equation
Mass can be neither created nor destroyed, which implies that mass is conserved
The net massflow into the control volume in Fig. \ref{fig:generic:cv} is obtained by integrating mass flux over the control volume surface
Now, let's consider a small infinitesimal volume $d\mathscr{V}$ inside . The mass of is . Thus, the mass enclosed within can be calculated as
The rate of change of mass within is obtained as
Mass is conserved, which means that the rate of change of mass within must equal the net flux over the control volume surface.
or
which is the integral form of the continuity equation.
The Momentum Equation
The time rate of change of momentum of a body equals the net force exerted on it
What type of forces do we have?
- Body forces acting on the fluid inside $\Omega$
- gravitation
- electromagnetic forces
- Coriolis forces
- Surface forces: pressure forces and shear forces
Body forces inside :
Surface force on :
Since we are considering inviscid flow, there are no shear forces and thus we have the net force as
The fluid flowing through $\Omega$ will carry momentum and the net flow of momentum out from is calculated as
Integrated momentum inside
Rate of change of momentum due to unsteady effects inside
Combining the rate of change of momentum, the net momentum flux and the net forces we get
combining the surface integrals, we get
which is the momentum equation on integral form.
The Energy Equation
Energy can be neither created nor destroyed; it can only change in form
\[E_1+E_2=E_3\]\\
\begin{itemize} \item[$E_1$:] Rate of heat added to the fluid in $\Omega$ from the surroundings \begin{itemize} \item heat transfer \item radiation \end{itemize} \item[$E_2$:] Rate of work done on the fluid in $\Omega$ \item[$E_3$:] Rate of change of energy of the fluid as it flows through $\Omega$ \end{itemize}
\[E_1=\iiint_{\Omega} \dot{q}\rho d\mathscr{V}\]\\
\noindent where $\dot{q}$ is the rate of heat added per unit mass\\
\noindent The rate of work done on the fluid in $\Omega$ due to pressure forces is obtained from the pressure force term in the momentum equation.\\
\[E_{2_{pressure}}=-\oiint_{\partial \Omega}(p\mathbf{n}dS)\cdot\mathbf{v}=-\oiint_{\partial \Omega} p\mathbf{v}\cdot\mathbf{n}dS\]\\
\noindent The rate of work done on the fluid in $\Omega$ due to body forces is\\
\[E_{2_{body\ forces}}=\iiint_{\Omega}(\rho\mathbf{f}d\mathscr{V})\cdot\mathbf{v}=\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}d\mathscr{V}\]\\
\[E_2=E_{2_{pressure}}+E_{2_{body\ forces}}=-\oiint_{\partial \Omega} p\mathbf{v}\cdot\mathbf{n}dS+\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}d\mathscr{V}\]\\
\noindent The energy of the fluid per unit mass is the sum of internal energy $e$ (molecular energy) and the kinetic energy $V^2/2$ and the net energy flux over the control volume surface is calculated by the following integral\\
\[\oiint_{\partial \Omega}(\rho \mathbf{v}\cdot\mathbf{n}dS)\left(e+\frac{V^2}{2}\right)\]\\
\noindent Analogous to mass and momentum, the total amount of energy of the fluid in $\Omega$ is calculated as\\
\[\iiint_{\Omega}\rho\left(e+\frac{V^2}{2}\right)d\mathscr{V}\]\\
\noindent The time rate of change of the energy of the fluid in $\Omega$ is obtained as\\
\[\frac{d}{dt}\iiint_{\Omega}\rho\left(e+\frac{V^2}{2}\right)d\mathscr{V}\]\\
\noindent Now, $E_3$ is obtained as the sum of the time rate of change of energy of the fluid in $\Omega$ and the net flux of energy carried by fluid passing the control volume surface.\\
\[E_3=\frac{d}{dt}\iiint_{\Omega}\rho\left(e+\frac{V^2}{2}\right)d\mathscr{V}+\oiint_{\partial \Omega}(\rho \mathbf{v}\cdot\mathbf{n}dS)\left(e+\frac{V^2}{2}\right)\]\\
\noindent With all elements of the energy equation defined, we are now ready to finally compile the full equation\\
\begin{equation} \frac{d}{dt}\iiint_{\Omega}\rho\left(e+\frac{V^2}{2}\right)d\mathscr{V}+\oiint_{\partial \Omega}\left[\rho\left(e+\frac{V^2}{2}\right)(\mathbf{v}\cdot\mathbf{n}) + p\mathbf{v}\cdot\mathbf{n}\right]dS=\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}d\mathscr{V}+\iiint_{\Omega} \dot{q}\rho d\mathscr{V} \label{eq:governing:integral:energy} \end{equation}\\
\noindent The surface integral in the energy equation may be rewritten as\\
\[\oiint_{\partial \Omega}\left[\rho\left(e+\frac{V^2}{2}\right)(\mathbf{v}\cdot\mathbf{n}) + p\mathbf{v}\cdot\mathbf{n}\right]dS=\oiint_{\partial \Omega}\rho\left[e+\frac{p}{\rho}+\frac{V^2}{2}\right](\mathbf{v}\cdot\mathbf{n})dS\]\\
\noindent and with the definition of enthalpy $h=e+p/\rho$, we get\\
\[\oiint_{\partial \Omega}\rho\left[h+\frac{V^2}{2}\right](\mathbf{v}\cdot\mathbf{n})dS\]\\
\noindent Furthermore, introducing total internal energy $e_o$ and total enthalpy $h_o$ defined as\\
\[e_o=e+\frac{1}{2}V^2\]\\
and\\
\[h_o=h+\frac{1}{2}V^2\]\\
\noindent the energy equation is written as\\
\begin{equation} \frac{d}{dt}\iiint_{\Omega}\rho e_o d\mathscr{V}+\oiint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}d\mathscr{V}+\iiint_{\Omega} \dot{q}\rho d\mathscr{V} \label{eq:governing:integral:energy:b} \end{equation}\\
\subsection{Summary}
\noindent The integral form of the governing equations for inviscid compressible flow has been derived\\
\noindent Continuity:\\
\[\frac{d}{dt}\iiint_{\Omega} \rho d\mathscr{V}+\oiint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0\]\\
\noindent Momentum:\\
\[\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} d\mathscr{V}+\oiint_{\partial \Omega} \left[(\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}+p\mathbf{n}\right]dS=\iiint_{\Omega}\rho \mathbf{f}d\mathscr{V}\]\\
\noindent Energy:\\
\[\frac{d}{dt}\iiint_{\Omega}\rho e_o d\mathscr{V}+\oiint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}d\mathscr{V}+\iiint_{\Omega} \dot{q}\rho d\mathscr{V}\]