Governing equations

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 $\partial \Omega$

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

(Eq. 2.1)

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

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

(Eq. 2.2)

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

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

(Eq. 2.3)

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

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

(Eq. 2.4)

or

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

(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"

$\frac{d}{dt}(m𝐯)=𝐅$

(Eq. 2.6)

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 $\Omega$:

$\underset{\Omega }{\iiint }\rho 𝐟dV$

(Eq. 2.7)

Surface force on $\partial \Omega$:

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

(Eq. 2.8)

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

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

(Eq. 2.9)

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

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

(Eq. 2.10)

Integrated momentum inside $\Omega$

$\underset{\Omega }{\iiint }\rho 𝐯dV$

(Eq. 2.11)

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

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

(Eq. 2.12)

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

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

(Eq. 2.13)

combining the surface integrals, we get

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

(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"

$$E_1+E_2=E_3$$

$E_1$ 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 $\Omega$

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

(Eq. 2.15)

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.

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

(Eq. 2.16)

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

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

(Eq. 2.17)

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

(Eq. 2.18)

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

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

(Eq. 2.19)

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

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

(Eq. 2.20)

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

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

(Eq. 2.21)

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.

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

(Eq. 2.22)

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

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

(Eq. 2.23)

The surface integral in the energy equation may be rewritten as

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

(Eq. 2.24)

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

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

(Eq. 2.25)

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

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

(Eq. 2.26)

and

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

(Eq. 2.27)

the energy equation is written as

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

(Eq. 2.28)

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:

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

Governing equations on differential form

The Differential Equations on Conservation Form

Conservation of Mass

The continuity equation on integral form reads

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

Apply Gauss's divergence theorem on the surface integral gives

$\underset{\partial \Omega }{\iint }\rho 𝐯\cdot 𝐧dS=\underset{\Omega }{\iiint }\mathrm{\nabla }\cdot (\rho 𝐯)dV$

(Eq. 2.29)

Also, if $\Omega$ is a fixed control volume

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

(Eq. 2.30)

The continuity equation can now be written as a single volume integral.

$\underset{\Omega }{\iiint }[\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\cdot (\rho 𝐯)]dV=0$

(Eq. 2.31)

$\Omega$ is an arbitrary control volume and thus

$\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\cdot (\rho 𝐯)=0$

(Eq. 2.32)

which is the continuity equation on partial differential form.

Conservation of Momentum

The momentum equation on integral form reads

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

As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.

$\underset{\partial \Omega }{\iint }(\rho 𝐯\cdot 𝐧)𝐯dS=\underset{\Omega }{\iiint }\mathrm{\nabla }\cdot (\rho 𝐯𝐯)dV$

(Eq. 2.33)

$\underset{\partial \Omega }{\iint }p𝐧dS=\underset{\Omega }{\iiint }\mathrm{\nabla }pdV$

(Eq. 2.34)

Also, if $\Omega$ is a fixed control volume

$\frac{d}{dt}\underset{\Omega }{\iiint }\rho 𝐯dV=\underset{\Omega }{\iiint }\frac{\partial }{\partial t}(\rho 𝐯)dV$

(Eq. 2.35)

The momentum equation can now be written as one single volume integral

$\underset{\Omega }{\iiint }[\frac{\partial }{\partial t}(\rho 𝐯)+\mathrm{\nabla }\cdot (\rho 𝐯𝐯)+\mathrm{\nabla }p-\rho 𝐟]dV=0$

(Eq. 2.36)

$\Omega$ is an arbitrary control volume and thus

$\frac{\partial }{\partial t}(\rho 𝐯)+\mathrm{\nabla }\cdot (\rho 𝐯𝐯)+\mathrm{\nabla }p=\rho 𝐟$

(Eq. 2.37)

which is the momentum equation on partial differential form

Conservation of Energy

The energy equation on integral form reads

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

Gauss's divergence theorem applied to the surface integral term in the energy equation gives

$\underset{\partial \Omega }{\iint }\rho h_o(𝐯\cdot 𝐧)dS=\underset{\Omega }{\iiint }\mathrm{\nabla }\cdot (\rho h_o𝐯)dV$

(Eq. 2.38)

Fixed control volume

$\frac{d}{dt}\underset{\Omega }{\iiint }\rho e_{o}dV=\underset{\Omega }{\iiint }\frac{\partial }{\partial t}(\rho e_o)dV$

(Eq. 2.39)

The energy equation can now be written as

$\underset{\Omega }{\iiint }[\frac{\partial }{\partial t}(\rho e_o)+\mathrm{\nabla }\cdot (\rho h_o𝐯)-\rho 𝐟\cdot 𝐯-\dot{q}\rho ]dV=0$

(Eq. 2.40)

$\Omega$ is an arbitrary control volume and thus

$\frac{\partial }{\partial t}(\rho e_o)+\mathrm{\nabla }\cdot (\rho h_o𝐯)=\rho 𝐟\cdot 𝐯+\dot{q}\rho$

(Eq. 2.41)

which is the energy equation on partial differential form

Summary

The governing equations for compressible inviscid flow on partial differential form:

Continuity:

$\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\cdot (\rho 𝐯)=0$

Momentum:

$\frac{\partial }{\partial t}(\rho 𝐯)+\mathrm{\nabla }\cdot (\rho 𝐯𝐯)+\mathrm{\nabla }p=\rho 𝐟$

Energy:

$\frac{\partial }{\partial t}(\rho e_o)+\mathrm{\nabla }\cdot (\rho h_o𝐯)=\rho 𝐟\cdot 𝐯+\dot{q}\rho$

The Differential Equations on Non-Conservation Form

The Substantial Derivative

The substantial derivative operator is defined as

$\frac{D}{Dt}=\frac{\partial }{\partial t}+𝐯\cdot \mathrm{\nabla }$

(Eq. 2.42)

where the first term of the right hand side is the local derivative and the second term is the convective derivative.

Conservation of Mass

If we apply the substantial derivative operator to density we get

$\frac{D\rho }{Dt}=\frac{\partial \rho }{\partial t}+𝐯\cdot \mathrm{\nabla }\rho$

(Eq. 2.43)

From before we have the continuity equation on differential form as

$\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\cdot (\rho 𝐯)=0$

(Eq. 2.44)

which can be rewritten as

$\frac{\partial \rho }{\partial t}+\rho (\mathrm{\nabla }\cdot 𝐯)+𝐯\cdot \mathrm{\nabla }\rho =0$

(Eq. 2.45)

and thus

$\frac{D\rho }{Dt}+\rho (\mathrm{\nabla }\cdot 𝐯)=0$

(Eq. 2.46)

Eq. 2.46 says that the mass of a fluid element with a fixed set of fluid particles is constant as the element moves in space.

Conservation of Momentum

We start from the momentum equation on differential form derived above

$\frac{\partial }{\partial t}(\rho 𝐯)+\mathrm{\nabla }\cdot (\rho 𝐯𝐯)+\mathrm{\nabla }p=\rho 𝐟$

(Eq. 2.47)

Expanding the first and the second terms gives

$\rho \frac{\partial 𝐯}{\partial t}+𝐯\frac{\partial \rho }{\partial t}+\rho 𝐯\cdot \mathrm{\nabla }𝐯+𝐯(\mathrm{\nabla }\cdot \rho 𝐯)+\mathrm{\nabla }p=\rho 𝐟$

(Eq. 2.48)

Collecting terms, we can identify the substantial derivative operator applied to the velocity vector and the continuity equation.

$\rho \underset{=\frac{D𝐯}{Dt}}{\underbrace{[\frac{\partial 𝐯}{\partial t}+𝐯\cdot \mathrm{\nabla }𝐯]}}+𝐯\underset{=0}{\underbrace{[\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\cdot \rho 𝐯]}}+\mathrm{\nabla }p=\rho 𝐟$

(Eq. 2.49)

which gives us the non-conservation form of the momentum equation

$\frac{D𝐯}{Dt}+\frac{1}{\rho }\mathrm{\nabla }p=𝐟$

(Eq. 2.50)

Conservation of Energy

The last equation on non-conservation differential form is the energy equation. We start by rewriting the energy equation on differential form (Eq. 2.41), repeated here for convenience

$\frac{\partial }{\partial t}(\rho e_o)+\mathrm{\nabla }\cdot (\rho h_o𝐯)=\rho 𝐟\cdot 𝐯+\dot{q}\rho$

Total enthalpy, $h_o$, is replaced with total energy, $e_o$

$h_o=e_o+\frac{p}{\rho }$

(Eq. 2.51)

which gives

$\frac{\partial }{\partial t}(\rho e_o)+\mathrm{\nabla }\cdot (\rho e_o𝐯)+\mathrm{\nabla }\cdot (p𝐯)=\rho 𝐟\cdot 𝐯+\dot{q}\rho$

(Eq. 2.52)

Expanding the two first terms as

$\rho \frac{\partial e_o}{\partial t}+e_o\frac{\partial \rho }{\partial t}+\rho 𝐯\cdot \mathrm{\nabla }{e}_o+e_o\mathrm{\nabla }\cdot (\rho 𝐯)+\mathrm{\nabla }\cdot (p𝐯)=$ $=\rho 𝐟\cdot 𝐯+\dot{q}\rho$

(Eq. 2.53)

Collecting terms, we can identify the substantial derivative operator applied on total energy, $D{e}_o/Dt$ and the continuity equation

$\rho \underset{=\frac{D{e}_o}{Dt}}{\underbrace{[\frac{\partial e_o}{\partial t}+𝐯\cdot \mathrm{\nabla }{e}_o]}}+e_o\underset{=0}{\underbrace{[\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\cdot (\rho 𝐯)]}}+\mathrm{\nabla }\cdot (p𝐯)=\rho 𝐟\cdot 𝐯+\dot{q}\rho$

(Eq. 2.54)

and thus we end up with the energy equation on non-conservation differential form

$\rho \frac{D{e}_o}{Dt}+\mathrm{\nabla }\cdot (p𝐯)=\rho 𝐟\cdot 𝐯+\dot{q}\rho$

(Eq. 2.55)

Summary

Continuity:

$\frac{D\rho }{Dt}+\rho (\mathrm{\nabla }\cdot 𝐯)=0$

Momentum:

$\frac{D𝐯}{Dt}+\frac{1}{\rho }\mathrm{\nabla }p=𝐟$

Energy:

$\rho \frac{D{e}_o}{Dt}+\mathrm{\nabla }\cdot (p𝐯)=\rho 𝐟\cdot 𝐯+\dot{q}\rho$

Alternative Forms of the Energy Equation

Internal Energy Formulation

Total internal energy is defined as

$e_o=e+\frac{1}{2}𝐯\cdot 𝐯$

(Eq. 2.56)

Inserted in Eq. 2.55, this gives

$\rho \frac{De}{Dt}+\rho 𝐯\cdot \frac{D𝐯}{Dt}+\mathrm{\nabla }\cdot (p𝐯)=\rho 𝐟\cdot 𝐯+\dot{q}\rho$

(Eq. 2.57)

Now, let's replace the substantial derivative $D𝐯/Dt$ using the momentum equation on non-conservation form (Eq. 2.50).

$\rho \frac{De}{Dt}-𝐯\cdot \mathrm{\nabla }p+\cancel{\rho 𝐟\cdot 𝐯}+\mathrm{\nabla }\cdot (p𝐯)=\cancel{\rho 𝐟\cdot 𝐯}+\dot{q}\rho$

(Eq. 2.58)

Now, expand the term $\mathrm{\nabla }\cdot (p𝐯)$ gives

$\rho \frac{De}{Dt}\cancel{-𝐯\cdot \mathrm{\nabla }p}+\cancel{𝐯\cdot \mathrm{\nabla }p}+p(\mathrm{\nabla }\cdot 𝐯)=\dot{q}\rho \Rightarrow$ $\Rightarrow \rho \frac{De}{Dt}+p(\mathrm{\nabla }\cdot 𝐯)=\dot{q}\rho$

(Eq. 2.59)

Divide by $\rho$

$\frac{De}{Dt}+\frac{p}{\rho }(\mathrm{\nabla }\cdot 𝐯)=\dot{q}$

(Eq. 2.60)

Conservation of mass gives

$\frac{D\rho }{Dt}+\rho (\mathrm{\nabla }\cdot 𝐯)=0\Rightarrow \mathrm{\nabla }\cdot 𝐯=-\frac{1}{\rho }\frac{D\rho }{Dt}$

(Eq. 2.61)

Insert in Eq. 2.60

$\frac{De}{Dt}-\frac{p}{{\rho }^2}\frac{D\rho }{Dt}=\dot{q}\Rightarrow \frac{De}{Dt}+p\frac{D}{Dt}\left(\frac{1}{\rho }\right)=\dot{q}$

(Eq. 2.62)

$\frac{De}{Dt}+p\frac{D\nu }{Dt}=\dot{q}$

(Eq. 2.63)

Compare with the first law of thermodynamics: $de=\delta q-\delta w$

Enthalpy Formulation

$h=e+\frac{p}{\rho }\Rightarrow \frac{Dh}{Dt}=\frac{De}{Dt}+\frac{1}{\rho }\frac{Dp}{Dt}+p\frac{D}{Dt}\left(\frac{1}{\rho }\right)$

(Eq. 2.64)

with $De/Dt$ from Eq. 2.60

$\frac{Dh}{Dt}=\dot{q}-\cancel{p\frac{D}{Dt}\left(\frac{1}{\rho }\right)}+\frac{1}{\rho }\frac{Dp}{Dt}+\cancel{p\frac{D}{Dt}\left(\frac{1}{\rho }\right)}$

(Eq. 2.65)

$\frac{Dh}{Dt}=\dot{q}+\frac{1}{\rho }\frac{Dp}{Dt}$

(Eq. 2.66)

Total Enthalpy Formulation

$h_o=h+\frac{1}{2}𝐯𝐯\Rightarrow \frac{D{h}_o}{Dt}=\frac{Dh}{Dt}+𝐯\cdot \frac{D𝐯}{Dt}$

(Eq. 2.67)

From the momentum equation (Eq. 2.50)

$\frac{D𝐯}{Dt}=𝐟-\frac{1}{\rho }\mathrm{\nabla }p$

(Eq. 2.68)

which gives

$\frac{D{h}_o}{Dt}=\frac{Dh}{Dt}+𝐯\cdot 𝐟-\frac{1}{\rho }𝐯\cdot \mathrm{\nabla }p$

(Eq. 2.69)

Inserting $Dh/Dt$ from Eq. 2.66 gives

$\frac{D{h}_o}{Dt}=\dot{q}+\frac{1}{\rho }\frac{Dp}{Dt}+𝐯\cdot 𝐟-\frac{1}{\rho }𝐯\cdot \mathrm{\nabla }p=$ $=\frac{1}{\rho }[\frac{Dp}{Dt}-𝐯\cdot \mathrm{\nabla }p]+\dot{q}+𝐯\cdot 𝐟$

(Eq. 2.70)

The substantial derivative operator applied to pressure

$\frac{Dp}{Dt}=\frac{\partial p}{\partial t}+𝐯\cdot \mathrm{\nabla }p$

(Eq. 2.71)

and thus

$\frac{Dp}{Dt}-𝐯\cdot \mathrm{\nabla }p=\frac{\partial p}{\partial t}$

(Eq. 2.72)

which gives

$\frac{D{h}_o}{Dt}=\frac{1}{\rho }\frac{\partial p}{\partial t}+\dot{q}+𝐯\cdot 𝐟$

(Eq. 2.73)

If we assume adiabatic flow without body forces

$\frac{D{h}_o}{Dt}=\frac{1}{\rho }\frac{\partial p}{\partial t}$

(Eq. 2.74)

If we further assume the flow to be steady state we get

$\frac{D{h}_o}{Dt}=0$

(Eq. 2.75)

This means that in a steady-state adiabatic flow without body forces, total enthalpy is constant along a streamline.

The entropy equation

From the second law of thermodynamics

$\frac{De}{Dt}=T\frac{Ds}{Dt}-p\frac{D}{Dt}\left(\frac{1}{\rho }\right)$

(Eq. 2.76)

From the energy equation on differential non-conservation form internal energy formulation

$\frac{De}{Dt}=\dot{q}-\frac{p}{\rho }(\mathrm{\nabla }\cdot 𝐯)$

(Eq. 2.77)

The continuity equation on differential non-conservation form

$\frac{D\rho }{Dt}+\rho (\mathrm{\nabla }\cdot 𝐯)=0\Rightarrow \mathrm{\nabla }\cdot 𝐯=-\frac{1}{\rho }\frac{D\rho }{Dt}$

(Eq. 2.78)

and thus

$\frac{De}{Dt}=\dot{q}+\frac{p}{{\rho }^2}\frac{D\rho }{Dt}$

(Eq. 2.79)

$\frac{D\rho }{Dt}=-\frac{1}{{\nu }^2}\frac{D\nu }{Dt}$

(Eq. 2.80)

$\rho \frac{De}{Dt}=\rho \dot{q}-\frac{p}{\rho {\nu }^2}\frac{D\nu }{Dt}=\rho \dot{q}-\rho p\frac{D\nu }{Dt}$

(Eq. 2.81)

$\rho [\frac{De}{Dt}+p\frac{D\nu }{Dt}-\dot{q}]=0\Rightarrow \frac{De}{Dt}=\dot{q}-p\frac{D\nu }{Dt}$

(Eq. 2.82)

Insert $De/Dt$ in Eqn. \ref{eq:second:law}

$\dot{q}-p\frac{D}{Dt}\left(\frac{1}{\rho }\right)=T\frac{Ds}{Dt}-p\frac{D}{Dt}\left(\frac{1}{\rho }\right)\Rightarrow$

(Eq. 2.83)

$T\frac{Ds}{Dt}=-\dot{q}$

(Eq. 2.84)

Adiabatic flow:

$T\frac{Ds}{Dt}=0$

(Eq. 2.85)

In an adiabatic, steady-state, inviscid flow, entropy is constant along a streamline.

Crocco's equation

The momentum equation without body forces

$\rho \frac{D𝐯}{Dt}=-\mathrm{\nabla }p$

(Eq. 2.86)

Expanding the substantial derivative

$\rho \frac{\partial 𝐯}{\partial t}+\rho 𝐯\cdot \mathrm{\nabla }𝐯=-\mathrm{\nabla }p$

(Eq. 2.87)

The first and second law of thermodynamics gives

$T\mathrm{\nabla }s=\mathrm{\nabla }h-\frac{\mathrm{\nabla }p}{\rho }$

(Eq. 2.88)

Insert $\mathrm{\nabla }p$ from the momentum equation

$T\mathrm{\nabla }s=\mathrm{\nabla }h+\frac{\partial 𝐯}{\partial t}+𝐯\cdot \mathrm{\nabla }𝐯$

(Eq. 2.89)

Definition of total enthalpy ($h_o$)

$h_o=h+\frac{1}{2}𝐯\cdot 𝐯\Rightarrow \mathrm{\nabla }h=\mathrm{\nabla }{h}_o-\mathrm{\nabla }(\frac{1}{2}𝐯\cdot 𝐯)$

(Eq. 2.90)

The last term can be rewritten as

$\mathrm{\nabla }(\frac{1}{2}𝐯\cdot 𝐯)=𝐯\times (\mathrm{\nabla }\times 𝐯)+𝐯\cdot \mathrm{\nabla }𝐯$

(Eq. 2.91)

which gives

$\mathrm{\nabla }h=\mathrm{\nabla }{h}_o-𝐯\times (\mathrm{\nabla }\times 𝐯)-𝐯\cdot \mathrm{\nabla }𝐯$

(Eq. 2.92)

Insert $\mathrm{\nabla }h$ in the entropy equation gives

$T\mathrm{\nabla }s=\mathrm{\nabla }{h}_o-𝐯\times (\mathrm{\nabla }\times 𝐯)-\cancel{𝐯\cdot \mathrm{\nabla }𝐯}+\frac{\partial 𝐯}{\partial t}+\cancel{𝐯\cdot \mathrm{\nabla }𝐯}$

(Eq. 2.93)

$T\mathrm{\nabla }s=\mathrm{\nabla }{h}_o-𝐯\times (\mathrm{\nabla }\times 𝐯)+\frac{\partial 𝐯}{\partial t}$

(Eq. 2.94)