Governing equations on integral form

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 $\Omega$ in Fig. \ref{fig:generic:cv} is obtained by integrating mass flux over the control volume surface $\partial \Omega$

$$-\underset{\partial \Omega }{\iint }\rho 𝐯\cdot 𝐧dS$$

Now, let's consider a small infinitesimal volume $d\mathscr{V}$ inside $\Omega$. The mass of $dV$ is $\rho dV$. Thus, the mass enclosed within $\Omega$ can be calculated as

$$\underset{\Omega }{\iiint }\rho dV$$

The rate of change of mass within $\Omega$ is obtained as

$$\frac{d}{dt}\underset{\Omega }{\iiint }\rho dV$$

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

$$\frac{d}{dt}\underset{\Omega }{\iiint }\rho dV=-\underset{\partial \Omega }{\iint }\rho 𝐯\cdot 𝐧dS$$

or

$$\frac{d}{dt}\underset{\Omega }{\iiint }\rho dV+\underset{\partial \Omega }{\iint }\rho 𝐯\cdot 𝐧dS=0$$

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

$$\frac{d}{dt}(m𝐯)=𝐅$$

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 $\Omega$:

$$\underset{\Omega }{\iiint }\rho 𝐟dV$$

Surface force on $\partial \Omega$:

$$-\underset{\partial \Omega }{\iint }p𝐧dS$$

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

$$𝐅=\underset{\Omega }{\iiint }\rho 𝐟dV-\underset{\partial \Omega }{\iint }p𝐧dS$$

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

$$\underset{\partial \Omega }{\iint }(\rho 𝐯\cdot 𝐧dS)𝐯=\underset{\partial \Omega }{\iint }(\rho 𝐯\cdot 𝐧)𝐯dS$$

Integrated momentum inside $\Omega$

$$\underset{\Omega }{\iiint }\rho 𝐯dV$$

Rate of change of momentum due to unsteady effects inside $\Omega$

$$\frac{d}{dt}\underset{\Omega }{\iiint }\rho 𝐯dV$$

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

$$\frac{d}{dt}\underset{\Omega }{\iiint }\rho 𝐯dV+\underset{\partial \Omega }{\iint }(\rho 𝐯\cdot 𝐧)𝐯dS=\underset{\Omega }{\iiint }\rho 𝐟dV-\underset{\partial \Omega }{\iint }p𝐧dS$$

combining the surface integrals, we get

$$\frac{d}{dt}\underset{\Omega }{\iiint }\rho 𝐯dV+\underset{\partial \Omega }{\iint }[(\rho 𝐯\cdot 𝐧)𝐯+p𝐧]dS=\underset{\Omega }{\iiint }\rho 𝐟dV$$

which is the momentum equation on integral form.

The Energy Equation

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

$$E_1+E_2=E_3$$

$E_1$ Rate of heat added to the fluid in $\Omega$ from the surroundings

heat transfer

radiation

$E_2$ Rate of work done on the fluid in $\Omega$

$E_3$ Rate of change of energy of the fluid as it flows through $\Omega$

$$E_1=\underset{\Omega }{\iiint }\dot{q}\rho dV$$

where $\dot{q}$ 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.

$$E_{2_{pressure}}=-\underset{\partial \Omega }{\iint }(p𝐧dS)\cdot 𝐯=-\underset{\partial \Omega }{\iint }p𝐯\cdot 𝐧dS$$

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

$$E_{2_{bodyforces}}=\underset{\Omega }{\iiint }(\rho 𝐟dV)\cdot 𝐯=\underset{\Omega }{\iiint }\rho 𝐟\cdot 𝐯dV$$

$$E_2=E_{2_{pressure}}+E_{2_{bodyforces}}=-\underset{\partial \Omega }{\iint }p𝐯\cdot 𝐧dS+\underset{\Omega }{\iiint }\rho 𝐟\cdot 𝐯dV$$

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

$$\underset{\partial \Omega }{\iint }(\rho 𝐯\cdot 𝐧dS)(e+\frac{V^2}{2})$$

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

$$\underset{\Omega }{\iiint }\rho (e+\frac{V^2}{2})dV$$

The time rate of change of the energy of the fluid in $\Omega$ is obtained as

$$\frac{d}{dt}\underset{\Omega }{\iiint }\rho (e+\frac{V^2}{2})dV$$

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.

$$E_3=\frac{d}{dt}\underset{\Omega }{\iiint }\rho (e+\frac{V^2}{2})dV+\underset{\partial \Omega }{\iint }(\rho 𝐯\cdot 𝐧dS)(e+\frac{V^2}{2})$$

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

$$\frac{d}{dt}\underset{\Omega }{\iiint }\rho (e+\frac{V^2}{2})dV+\underset{\partial \Omega }{\iint }[\rho (e+\frac{V^2}{2})(𝐯\cdot 𝐧)+p𝐯\cdot 𝐧]dS=\underset{\Omega }{\iiint }\rho 𝐟\cdot 𝐯dV+\underset{\Omega }{\iiint }\dot{q}\rho dV$$

The surface integral in the energy equation may be rewritten as

$$\underset{\partial \Omega }{\iint }[\rho (e+\frac{V^2}{2})(𝐯\cdot 𝐧)+p𝐯\cdot 𝐧]dS=\underset{\partial \Omega }{\iint }\rho [e+\frac{p}{\rho }+\frac{V^2}{2}](𝐯\cdot 𝐧)dS$$

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

$$\underset{\partial \Omega }{\iint }\rho [h+\frac{V^2}{2}](𝐯\cdot 𝐧)dS$$

Furthermore, introducing total internal energy $e_o$ and total enthalpy $h_o$ defined as

$$e_o=e+\frac{1}{2}{V}^2$$

and

$$h_o=h+\frac{1}{2}{V}^2$$

the energy equation is written as

$$\frac{d}{dt}\underset{\Omega }{\iiint }\rho e_{o}dV+\underset{\partial \Omega }{\iint }\rho h_o(𝐯\cdot 𝐧)dS=\underset{\Omega }{\iiint }\rho 𝐟\cdot 𝐯dV+\underset{\Omega }{\iiint }\dot{q}\rho dV$$

Summary

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

Continuity:

$$\frac{d}{dt}\underset{\Omega }{\iiint }\rho dV+\underset{\partial \Omega }{\iint }\rho 𝐯\cdot 𝐧dS=0$$

Momentum:

$$\frac{d}{dt}\underset{\Omega }{\iiint }\rho 𝐯dV+\underset{\partial \Omega }{\iint }[(\rho 𝐯\cdot 𝐧)𝐯+p𝐧]dS=\underset{\Omega }{\iiint }\rho 𝐟dV$$

Energy:

$$\frac{d}{dt}\underset{\Omega }{\iiint }\rho e_{o}dV+\underset{\partial \Omega }{\iint }\rho h_o(𝐯\cdot 𝐧)dS=\underset{\Omega }{\iiint }\rho 𝐟\cdot 𝐯dV+\underset{\Omega }{\iiint }\dot{q}\rho dV$$