Governing equations on differential form

The Differential Equations on Conservation Form

Conservation of Mass

Apply Gauss's divergence theorem on the surface integral in Eqn. \ref{eq:governing:cont:int} gives

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

(Eq. 2.32)

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. 2.33)

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. 2.34)

$\Omega$ is an arbitrary control volume and thus

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

(Eq. 2.35)

which is the continuity equation on partial differential form.

Conservation of Momentum

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. 2.36)

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

(Eq. 2.37)

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. 2.38)

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. 2.39)

$\Omega$ is an arbitrary control volume and thus

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

(Eq. 2.40)

which is the momentum equation on partial differential form

Conservation of Energy

Gauss's divergence theorem applied to the surface integral term in the energy equation (Eqn. \ref{eq:governing:energy:int}) gives

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

(Eq. 2.41)

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. 2.42)

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. 2.43)

$\Omega$ 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. 2.44)

which is the energy equation on partial differential form

Summary

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

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

(Eq. 2.45)

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

(Eq. 2.46)

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

(Eq. 2.47)

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. 2.48)

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. 2.49)

From before we have the continuity equation on differential form as

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

(Eq. 2.50)

which can be rewritten as

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

(Eq. 2.51)

and thus

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

(Eq. 2.52)

Eqn. \ref{eq:governing:cont:non} 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. 2.53)

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. 2.54)

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. 2.55)

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

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

(Eq. 2.56)

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 (Eqn. \ref{eq:governing:energy:pde}), repeated here for convenience

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

(Eq. 2.57)

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

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

(Eq. 2.58)

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. 2.59)

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. 2.60)

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. 2.61)

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. 2.62)

Summary

Continuity:

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

(Eq. 2.63)

Momentum:

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

(Eq. 2.64)

Energy:

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

(Eq. 2.65)

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. 2.66)

Inserted in Eqn. \ref{eq:governing:energy:non}, this gives

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

(Eq. 2.67)

Now, let's replace the substantial derivative $D𝐯/Dt$ using the momentum equation on non-conservation form (Eqn. \ref{eq:governing:mom:non}).

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

(Eq. 2.68)

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

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

(Eq. 2.69)

Divide by $\rho$

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

(Eq. 2.70)

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. 2.71)

Insert in Eqn. \ref{eq:governing:energy:non:b}

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

(Eq. 2.72)

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

(Eq. 2.73)

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. 2.74)

with $De/Dt$ from Eqn. \ref{eq:governing:energy:non:b}

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

(Eq. 2.75)

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

(Eq. 2.76)

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. 2.77)

From the momentum equation (Eqn. \ref{eq:governing:mom:non})

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

(Eq. 2.78)

which gives

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

(Eq. 2.79)

Inserting $Dh/Dt$ from Eqn. \ref{eq:governing:energy:non:c} 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. 2.80)

The substantial derivative operator applied to pressure

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

(Eq. 2.81)

and thus

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

(Eq. 2.82)

which gives

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

(Eq. 2.83)

If we assume adiabatic flow without body forces

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

(Eq. 2.84)

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

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

(Eq. 2.85)

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