Governing equations on integral form: Difference between revisions

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[[Category:Compressible flow]]
[[Category:Compressible flow]]
[[Category:Governing equations]]
[[Category:Governing equations]]
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\begin{figure}[ht!]
\begin{figure}[ht!]
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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>
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">
{{NumEqn|<math>
-\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS
-\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS
</math>
</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
Now, let's consider a small infinitesimal volume <math>dV</math> 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">
{{NumEqn|<math>
\iiint_{\Omega} \rho dV
\iiint_{\Omega} \rho dV
</math>
</math>}}


The rate of change of mass within <math>\Omega</math> is obtained as
The rate of change of mass within <math>\Omega</math> is obtained as


<math display="block">
{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega} \rho dV
\frac{d}{dt}\iiint_{\Omega} \rho dV
</math>
</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.
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">
{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega} \rho dV=-\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS
\frac{d}{dt}\iiint_{\Omega} \rho dV=-\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS
</math>
</math>}}


or
or


<math display="block">
{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0
\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0
</math>
</math>}}


which is the integral form of the continuity equation.
which is the integral form of the continuity equation.
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{{quote|The time rate of change of momentum of a body equals the net force exerted on it}}
{{quote|The time rate of change of momentum of a body equals the net force exerted on it}}


<math display="block">
{{NumEqn|<math>
\frac{d}{dt}(m\mathbf{v})=\mathbf{F}
\frac{d}{dt}(m\mathbf{v})=\mathbf{F}
</math>
</math>}}


What type of forces do we have?
What type of forces do we have?




* Body forces acting on the fluid inside $\Omega$
* Body forces acting on the fluid inside <math>\Omega</math>
** gravitation
** gravitation
** electromagnetic forces
** electromagnetic forces
Line 74: Line 81:
Body forces inside <math>\Omega</math>:
Body forces inside <math>\Omega</math>:


<math display="block">
{{NumEqn|<math>
\iiint_{\Omega}\rho \mathbf{f}dV
\iiint_{\Omega}\rho \mathbf{f}dV
</math>
</math>}}


Surface force on <math>\partial \Omega</math>:
Surface force on <math>\partial \Omega</math>:


<math display="block">
{{NumEqn|<math>
-\iint_{\partial \Omega} p\mathbf{n}dS
-\iint_{\partial \Omega} p\mathbf{n}dS
</math>
</math>}}


Since we are considering inviscid flow, there are no shear forces and thus we have the net force as
Since we are considering inviscid flow, there are no shear forces and thus we have the net force as


<math display="block">
{{NumEqn|<math>
\mathbf{F}=\iiint_{\Omega}\rho \mathbf{f}dV-\iint_{\partial \Omega} p\mathbf{n}dS
\mathbf{F}=\iiint_{\Omega}\rho \mathbf{f}dV-\iint_{\partial \Omega} p\mathbf{n}dS
</math>
</math>}}


The fluid flowing through $\Omega$ will carry momentum and the net flow of momentum out from <math>\Omega</math> is calculated as
The fluid flowing through <math>\Omega</math> will carry momentum and the net flow of momentum out from <math>\Omega</math> is calculated as


<math display="block">
{{NumEqn|<math>
\iint_{\partial \Omega} (\rho\mathbf{v}\cdot\mathbf{n}dS)\mathbf{v}=\iint_{\partial \Omega} (\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}dS
\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>
</math>}}


Integrated momentum inside <math>\Omega</math>
Integrated momentum inside <math>\Omega</math>


<math display="block">
{{NumEqn|<math>
\iiint_{\Omega} \rho \mathbf{v} dV
\iiint_{\Omega} \rho \mathbf{v} dV
</math>
</math>}}


Rate of change of momentum due to unsteady effects inside <math>\Omega</math>
Rate of change of momentum due to unsteady effects inside <math>\Omega</math>


<math display="block">
{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV
\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV
</math>
</math>}}


Combining the rate of change of momentum, the net momentum flux and the net forces we get
Combining the rate of change of momentum, the net momentum flux and the net forces we get


<math display="block">
{{NumEqn|<math>
\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
\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>
</math>}}


combining the surface integrals, we get
combining the surface integrals, we get


<math display="block">
{{NumEqn|<math>
\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
\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>
</math>}}


which is the momentum equation on integral form.
which is the momentum equation on integral form.
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;<math>E_3</math> Rate of change of energy of the fluid as it flows through <math>\Omega</math>
;<math>E_3</math> Rate of change of energy of the fluid as it flows through <math>\Omega</math>


<math display="block">
{{NumEqn|<math>
E_1=\iiint_{\Omega} \dot{q}\rho dV
E_1=\iiint_{\Omega} \dot{q}\rho dV
</math>
</math>}}


where <math>\dot{q}</math> is the rate of heat added per unit mass
where <math>\dot{q}</math> is the rate of heat added per unit mass


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.
The rate of work done on the fluid in <math>\Omega</math> due to pressure forces is obtained from the pressure force term in the momentum equation.


<math display="block">
{{NumEqn|<math>
E_{2_{pressure}}=-\iint_{\partial \Omega}(p\mathbf{n}dS)\cdot\mathbf{v}=-\iint_{\partial \Omega} p\mathbf{v}\cdot\mathbf{n}dS
E_{2_{pressure}}=-\iint_{\partial \Omega}(p\mathbf{n}dS)\cdot\mathbf{v}=-\iint_{\partial \Omega} p\mathbf{v}\cdot\mathbf{n}dS
</math>
</math>}}


The rate of work done on the fluid in $\Omega$ due to body forces is
The rate of work done on the fluid in $\Omega$ due to body forces is


<math display="block">
{{NumEqn|<math>
E_{2_{body\ forces}}=\iiint_{\Omega}(\rho\mathbf{f}dV)\cdot\mathbf{v}=\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV
E_{2_{body\ forces}}=\iiint_{\Omega}(\rho\mathbf{f}dV)\cdot\mathbf{v}=\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV
</math>
</math>}}


<math display="block">
{{NumEqn|<math>
E_2=E_{2_{pressure}}+E_{2_{body\ forces}}=-\iint_{\partial \Omega} p\mathbf{v}\cdot\mathbf{n}dS+\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV
E_2=E_{2_{pressure}}+E_{2_{body\ forces}}=-\iint_{\partial \Omega} p\mathbf{v}\cdot\mathbf{n}dS+\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV
</math>
</math>}}


The energy of the fluid per unit mass is the sum of internal energy <math>e</math> (molecular energy) and the kinetic energy <math>V^2/2</math> and the net energy flux over the control volume surface is calculated by the following integral
The energy of the fluid per unit mass is the sum of internal energy <math>e</math> (molecular energy) and the kinetic energy <math>V^2/2</math> and the net energy flux over the control volume surface is calculated by the following integral


<math display="block">
{{NumEqn|<math>
\iint_{\partial \Omega}(\rho \mathbf{v}\cdot\mathbf{n}dS)\left(e+\frac{V^2}{2}\right)
\iint_{\partial \Omega}(\rho \mathbf{v}\cdot\mathbf{n}dS)\left(e+\frac{V^2}{2}\right)
</math>
</math>}}


Analogous to mass and momentum, the total amount of energy of the fluid in <math>\Omega</math> is calculated as
Analogous to mass and momentum, the total amount of energy of the fluid in <math>\Omega</math> is calculated as


<math display="block">
{{NumEqn|<math>
\iiint_{\Omega}\rho\left(e+\frac{V^2}{2}\right)dV
\iiint_{\Omega}\rho\left(e+\frac{V^2}{2}\right)dV
</math>
</math>}}


The time rate of change of the energy of the fluid in <math>\Omega</math> is obtained as
The time rate of change of the energy of the fluid in <math>\Omega</math> is obtained as


<math display="block">
{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega}\rho\left(e+\frac{V^2}{2}\right)dV
\frac{d}{dt}\iiint_{\Omega}\rho\left(e+\frac{V^2}{2}\right)dV
</math>
</math>}}


Now, <math>E_3</math> 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.
Now, <math>E_3</math> is obtained as the sum of the time rate of change of energy of the fluid in <math>\Omega</math> and the net flux of energy carried by fluid passing the control volume surface.


<math display="block">
{{NumEqn|<math>
E_3=\frac{d}{dt}\iiint_{\Omega}\rho\left(e+\frac{V^2}{2}\right)dV+\iint_{\partial \Omega}(\rho \mathbf{v}\cdot\mathbf{n}dS)\left(e+\frac{V^2}{2}\right)
E_3=\frac{d}{dt}\iiint_{\Omega}\rho\left(e+\frac{V^2}{2}\right)dV+\iint_{\partial \Omega}(\rho \mathbf{v}\cdot\mathbf{n}dS)\left(e+\frac{V^2}{2}\right)
</math>
</math>}}


With all elements of the energy equation defined, we are now ready to finally compile the full equation
With all elements of the energy equation defined, we are now ready to finally compile the full equation


<math display="block">
{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega}\rho\left(e+\frac{V^2}{2}\right)dV+\iint_{\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}dV+\iiint_{\Omega} \dot{q}\rho dV
\begin{aligned}
</math>
&\dfrac{d}{dt}\iiint_{\Omega}\rho\left(e+\dfrac{V^2}{2}\right)dV+\iint_{\partial \Omega}\left[\rho\left(e+\dfrac{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}dV+\iiint_{\Omega} \dot{q}\rho dV
\end{aligned}
</math>}}


The surface integral in the energy equation may be rewritten as
The surface integral in the energy equation may be rewritten as


<math display="block">
{{NumEqn|<math>
\iint_{\partial \Omega}\left[\rho\left(e+\frac{V^2}{2}\right)(\mathbf{v}\cdot\mathbf{n}) + p\mathbf{v}\cdot\mathbf{n}\right]dS=\iint_{\partial \Omega}\rho\left[e+\frac{p}{\rho}+\frac{V^2}{2}\right](\mathbf{v}\cdot\mathbf{n})dS
\iint_{\partial \Omega}\left[\rho\left(e+\frac{V^2}{2}\right)(\mathbf{v}\cdot\mathbf{n}) + p\mathbf{v}\cdot\mathbf{n}\right]dS=\iint_{\partial \Omega}\rho\left[e+\frac{p}{\rho}+\frac{V^2}{2}\right](\mathbf{v}\cdot\mathbf{n})dS
</math>
</math>}}


and with the definition of enthalpy <math>h=e+p/\rho</math>, we get
and with the definition of enthalpy <math>h=e+p/\rho</math>, we get


<math display="block">
{{NumEqn|<math>
\iint_{\partial \Omega}\rho\left[h+\frac{V^2}{2}\right](\mathbf{v}\cdot\mathbf{n})dS
\iint_{\partial \Omega}\rho\left[h+\frac{V^2}{2}\right](\mathbf{v}\cdot\mathbf{n})dS
</math>
</math>}}


Furthermore, introducing total internal energy <math>e_o</math> and total enthalpy <math>h_o</math> defined as
Furthermore, introducing total internal energy <math>e_o</math> and total enthalpy <math>h_o</math> defined as


<math display="block">
{{NumEqn|<math>
e_o=e+\frac{1}{2}V^2
e_o=e+\frac{1}{2}V^2
</math>
</math>}}


and
and


<math display="block">
{{NumEqn|<math>
h_o=h+\frac{1}{2}V^2
h_o=h+\frac{1}{2}V^2
</math>
</math>}}


the energy equation is written as
the energy equation is written as


<math display="block">
{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega}\rho e_o dV+\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV
\frac{d}{dt}\iiint_{\Omega}\rho e_o dV+\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV
</math>
</math>}}


==== Summary ====
==== Summary ====
Line 224: Line 234:
Continuity:
Continuity:


<math display="block">
{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0
\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0
</math>
</math>}}


Momentum:
Momentum:


<math display="block">
{{NumEqn|<math>
\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
\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>
</math>}}


Energy:
Energy:


<math display="block">
{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega}\rho e_o dV+\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV
\frac{d}{dt}\iiint_{\Omega}\rho e_o dV+\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV
</math>
</math>}}