Governing equations on integral form: Difference between revisions

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==== The Continuity Equation ====
==== The Continuity Equation ====


{{quote|Mass can be neither created nor destroyed, which implies that mass is conserved}}
{{QuoteBox|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>
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.
Line 57: Line 65:
==== The Momentum Equation ====
==== The Momentum Equation ====


{{quote|The time rate of change of momentum of a body equals the net force exerted on it}}
{{QuoteBox|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 82:
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.
Line 124: Line 132:
==== The Energy Equation ====
==== The Energy Equation ====


{{quote|Energy can be neither created nor destroyed; it can only change in form}}
{{QuoteBox|Energy can be neither created nor destroyed; it can only change in form}}


<math display="block">
<math display="block">
Line 136: Line 144:
;<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
\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=</math><br><br><math>
</math>
\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV
</math>|align=center}}


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


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=</math><br><br><math>\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV
</math>
</math>}}


==== Summary ====
==== Summary ====
Line 222: Line 231:
The integral form of the governing equations for inviscid compressible flow has been derived
The integral form of the governing equations for inviscid compressible flow has been derived


Continuity:
<div style="border: solid 1px;">
 
{{OpenInfoBox|<math>
<math display="block">
\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>|description=Continuity:}}
 
Momentum:


<math display="block">
{{OpenInfoBox|<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>|description=Momentum:}}
 
Energy:


<math display="block">
{{OpenInfoBox|<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=</math><br><br><math>\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV
</math>
</math>|description=Energy:}}
</div>

Latest revision as of 08:03, 2 April 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 Ω in Fig. \ref{fig:generic:cv} is obtained by integrating mass flux over the control volume surface Ω

Ωρ𝐯𝐧dS(Eq. 2.1)

Now, let's consider a small infinitesimal volume dV inside Ω. The mass of dV is ρdV. Thus, the mass enclosed within Ω can be calculated as

ΩρdV(Eq. 2.2)

The rate of change of mass within Ω is obtained as

ddtΩρdV(Eq. 2.3)

Mass is conserved, which means that the rate of change of mass within Ω must equal the net flux over the control volume surface.

ddtΩρdV=Ωρ𝐯𝐧dS(Eq. 2.4)

or

ddtΩρdV+Ωρ𝐯𝐧dS=0(Eq. 2.5)

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"
ddt(m𝐯)=𝐅(Eq. 2.6)

What type of forces do we have?


  • Body forces acting on the fluid inside Ω
    • gravitation
    • electromagnetic forces
    • Coriolis forces
  • Surface forces: pressure forces and shear forces

Body forces inside Ω:

Ωρ𝐟dV(Eq. 2.7)

Surface force on Ω:

Ωp𝐧dS(Eq. 2.8)

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

𝐅=Ωρ𝐟dVΩp𝐧dS(Eq. 2.9)

The fluid flowing through Ω will carry momentum and the net flow of momentum out from Ω is calculated as

Ω(ρ𝐯𝐧dS)𝐯=Ω(ρ𝐯𝐧)𝐯dS(Eq. 2.10)

Integrated momentum inside Ω

Ωρ𝐯dV(Eq. 2.11)

Rate of change of momentum due to unsteady effects inside Ω

ddtΩρ𝐯dV(Eq. 2.12)

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

ddtΩρ𝐯dV+Ω(ρ𝐯𝐧)𝐯dS=Ωρ𝐟dVΩp𝐧dS(Eq. 2.13)

combining the surface integrals, we get

ddtΩρ𝐯dV+Ω[(ρ𝐯𝐧)𝐯+p𝐧]dS=Ωρ𝐟dV(Eq. 2.14)

which is the momentum equation on integral form.

The Energy Equation

"Energy can be neither created nor destroyed; it can only change in form"

E1+E2=E3

E1 Rate of heat added to the fluid in Ω from the surroundings
heat transfer
radiation
E2 Rate of work done on the fluid in Ω
E3 Rate of change of energy of the fluid as it flows through Ω
E1=Ωq˙ρdV(Eq. 2.15)

where q˙ is the rate of heat added per unit mass

The rate of work done on the fluid in Ω due to pressure forces is obtained from the pressure force term in the momentum equation.

E2pressure=Ω(p𝐧dS)𝐯=Ωp𝐯𝐧dS(Eq. 2.16)

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

E2body forces=Ω(ρ𝐟dV)𝐯=Ωρ𝐟𝐯dV(Eq. 2.17)
E2=E2pressure+E2body forces=Ωp𝐯𝐧dS+Ωρ𝐟𝐯dV(Eq. 2.18)

The energy of the fluid per unit mass is the sum of internal energy e (molecular energy) and the kinetic energy V2/2 and the net energy flux over the control volume surface is calculated by the following integral

Ω(ρ𝐯𝐧dS)(e+V22)(Eq. 2.19)

Analogous to mass and momentum, the total amount of energy of the fluid in Ω is calculated as

Ωρ(e+V22)dV(Eq. 2.20)

The time rate of change of the energy of the fluid in Ω is obtained as

ddtΩρ(e+V22)dV(Eq. 2.21)

Now, E3 is obtained as the sum of the time rate of change of energy of the fluid in Ω and the net flux of energy carried by fluid passing the control volume surface.

E3=ddtΩρ(e+V22)dV+Ω(ρ𝐯𝐧dS)(e+V22)(Eq. 2.22)

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

ddtΩρ(e+V22)dV+Ω[ρ(e+V22)(𝐯𝐧)+p𝐯𝐧]dS=

Ωρ𝐟𝐯dV+Ωq˙ρdV
(Eq. 2.23)

The surface integral in the energy equation may be rewritten as

Ω[ρ(e+V22)(𝐯𝐧)+p𝐯𝐧]dS=

Ωρ[e+pρ+V22](𝐯𝐧)dS
(Eq. 2.24)

and with the definition of enthalpy h=e+p/ρ, we get

Ωρ[h+V22](𝐯𝐧)dS(Eq. 2.25)

Furthermore, introducing total internal energy eo and total enthalpy ho defined as

eo=e+12V2(Eq. 2.26)

and

ho=h+12V2(Eq. 2.27)

the energy equation is written as

ddtΩρeodV+Ωρho(𝐯𝐧)dS=

Ωρ𝐟𝐯dV+Ωq˙ρdV
(Eq. 2.28)

Summary

The integral form of the governing equations for inviscid compressible flow has been derived

Continuity:ddtΩρdV+Ωρ𝐯𝐧dS=0
Momentum:ddtΩρ𝐯dV+Ω[(ρ𝐯𝐧)𝐯+p𝐧]dS=Ωρ𝐟dV
Energy:ddtΩρeodV+Ωρho(𝐯𝐧)dS=

Ωρ𝐟𝐯dV+Ωq˙ρdV