View source for Governing equations on differential form

[[Category:Compressible flow]]
[[Category:Topic]]
[[Category:Governing equations]]

__TOC__

=== 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

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

Also, if <math>\Omega</math> is a fixed control volume

<math display="block">
\frac{d}{dt}\iiint_{\Omega} \rho dV=\iiint_{\Omega} \frac{\partial \rho}{\partial t} dV
</math>

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

<math display="block">
\iiint_{\Omega} \left[\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})\right]dV=0
</math>

<math>\Omega</math> is an arbitrary control volume and thus

<math display="block">
\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0
</math>

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.

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

<math display="block">
\iint_{\partial \Omega} p\mathbf{n}dS=\iiint_{\Omega} \nabla pdV
</math>

Also, if <math>\Omega</math> is a fixed control volume

<math display="block">
\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV=\iiint_{\Omega}  \frac{\partial}{\partial t}(\rho \mathbf{v}) dV
</math>

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

<math display="block">
\iiint_{\Omega}\left[\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p - \rho \mathbf{f}\right]dV=0
</math>

<math>\Omega</math> is an arbitrary control volume and thus

<math display="block">
\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f} 
</math>

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

<math display="block">
\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=\iiint_{\Omega}\nabla\cdot(\rho h_o\mathbf{v})dV
</math>

Fixed control volume

<math display="block">
\frac{d}{dt}\iiint_{\Omega}\rho e_o dV=\iiint_{\Omega}\frac{\partial}{\partial t}(\rho e_o) dV
</math>

The energy equation can now be written as

<math display="block">
\iiint_{\Omega}\left[\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) - \rho\mathbf{f}\cdot\mathbf{v} - \dot{q}\rho \right]dV=0
</math>

<math>\Omega</math> is an arbitrary control volume and thus

<math display="block">
\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>

which is the energy equation on partial differential form

==== Summary ====

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

<math display="block">
\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0
</math>

<math display="block">
\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}
</math>

<math display="block">
\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>

=== The Differential Equations on Non-Conservation Form ===

==== The Substantial Derivative ====

The substantial derivative operator is defined as

<math display="block">
\frac{D}{Dt}=\frac{\partial}{\partial t}+\mathbf{v}\cdot\nabla
</math>

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

<math display="block">
\frac{D\rho}{Dt}=\frac{\partial \rho}{\partial t}+\mathbf{v}\cdot\nabla\rho
</math>

From before we have the continuity equation on differential form as

<math display="block">
\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0
</math>

which can be rewritten as

<math display="block">
\frac{\partial \rho}{\partial t} + \rho(\nabla\cdot\mathbf{v}) + \mathbf{v}\cdot\nabla\rho=0
</math>

and thus

<math display="block">
\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0
</math>

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

<math display="block">
\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}
</math>

Expanding the first and the second terms gives

<math display="block">
\rho\frac{\partial \mathbf{v}}{\partial t} + \mathbf{v}\frac{\partial \rho}{\partial t} + \rho\mathbf{v}\cdot\nabla\mathbf{v} + \mathbf{v}(\nabla\cdot\rho\mathbf{v}) + \nabla p = \rho \mathbf{f}
</math>

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

<math display="block">
\rho\underbrace{\left[\frac{\partial \mathbf{v}}{\partial t}+\mathbf{v}\cdot\nabla\mathbf{v}\right]}_{=\frac{D\mathbf{v}}{Dt}}+\mathbf{v}\underbrace{\left[\frac{\partial \rho}{\partial t}+\nabla\cdot\rho\mathbf{v}\right]}_{=0}+ \nabla p = \rho \mathbf{f}
</math>

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

<math display="block">
\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}
</math>

==== 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

<math display="block">
\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>

Total enthalpy, <math>h_o</math>, is replaced with total energy, <math>e_o</math>

<math display="block">
h_o=e_o+\frac{p}{\rho}
</math>

which gives

<math display="block">
\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho e_o\mathbf{v}) + \nabla\cdot(p\mathbf{v})= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>

Expanding the two first terms as

<math display="block">
\rho\frac{\partial e_o}{\partial t} + e_o\frac{\partial \rho}{\partial t} + \rho\mathbf{v}\cdot\nabla e_o + e_o\nabla\cdot(\rho \mathbf{v}) + \nabla\cdot(p\mathbf{v})= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>

Collecting terms, we can identify the substantial derivative operator applied on total energy, <math>De_o/Dt</math> and the continuity equation

<math display="block">
\rho\underbrace{\left[ \frac{\partial e_o}{\partial t} + \mathbf{v}\cdot\nabla e_o \right]}_{=\frac{De_o}{Dt}}  + e_o\underbrace{\left[\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho \mathbf{v}) \right]}_{=0} + \nabla\cdot(p\mathbf{v})= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>

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

<math display="block">
\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>

==== Summary ====

Continuity:

<math display="block">
\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0
</math>

Momentum:

<math display="block">
\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}
</math>

Energy:

<math display="block">
\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>

=== Alternative Forms of the Energy Equation ===

==== Internal Energy Formulation ====

Total internal energy is defined as

<math display="block">
e_o=e+\frac{1}{2}\mathbf{v}\cdot\mathbf{v}
</math>

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

<math display="block">
\rho\frac{De}{Dt} + \rho\mathbf{v}\cdot\frac{D \mathbf{v}}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>

Now, let's replace the substantial derivative <math>D\mathbf{v}/Dt</math> using the momentum equation on non-conservation form (Eqn. \ref{eq:governing:mom:non}).

<math display="block">
\rho\frac{De}{Dt} -\mathbf{v}\cdot\nabla p + \cancel{\rho\mathbf{f}\cdot\mathbf{v}} + \nabla\cdot(p\mathbf{v}) = \cancel{\rho\mathbf{f}\cdot\mathbf{v}} + \dot{q}\rho
</math>

Now, expand the term <math>\nabla\cdot(p\mathbf{v})</math> gives

<math display="block">
\rho\frac{De}{Dt} \cancel{-\mathbf{v}\cdot\nabla p} + \cancel{\mathbf{v}\cdot\nabla p} +  p(\nabla\cdot\mathbf{v}) = \dot{q}\rho\Rightarrow \rho\frac{De}{Dt} + p(\nabla\cdot\mathbf{v}) = \dot{q}\rho
</math>

Divide by <math>\rho</math>

<math display="block">
\frac{De}{Dt} + \frac{p}{\rho}(\nabla\cdot\mathbf{v}) = \dot{q}
</math>

Conservation of mass gives

<math display="block">
\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0\Rightarrow \nabla\cdot\mathbf{v} = -\frac{1}{\rho}\frac{D\rho}{Dt}
</math>

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

<math display="block">
\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}
</math>

<math display="block">
\frac{De}{Dt} + p\frac{D\nu}{Dt} = \dot{q}
</math>

Compare with the first law of thermodynamics: <math>de=\delta q-\delta w</math>

==== Enthalpy Formulation ====

<math display="block">
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)
</math>

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

<math display="block">
\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)}
</math>

<math display="block">
\frac{Dh}{Dt}=\dot{q} + \frac{1}{\rho}\frac{Dp}{Dt}
</math>

==== Total Enthalpy Formulation ====

<math display="block">
h_o=h+\frac{1}{2}\mathbf{v}\mathbf{v}\Rightarrow\frac{Dh_o}{Dt}=\frac{Dh}{Dt}+\mathbf{v}\cdot\frac{D\mathbf{v}}{Dt}
</math>

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

<math display="block">
\frac{D\mathbf{v}}{Dt}=\mathbf{f}-\frac{1}{\rho}\nabla p
</math>

which gives

<math display="block">
\frac{Dh_o}{Dt}=\frac{Dh}{Dt}+\mathbf{v}\cdot\mathbf{f} -\frac{1}{\rho}\mathbf{v}\cdot\nabla p
</math>

Inserting <math>Dh/Dt</math> from Eqn. \ref{eq:governing:energy:non:c} gives

<math display="block">
\frac{Dh_o}{Dt}=\dot{q} + \frac{1}{\rho}\frac{Dp}{Dt}+\mathbf{v}\cdot\mathbf{f} -\frac{1}{\rho}\mathbf{v}\cdot\nabla p = \frac{1}{\rho}\left[\frac{Dp}{Dt}-\mathbf{v}\cdot\nabla p\right] + \dot{q} + \mathbf{v}\cdot\mathbf{f}
</math>

The substantial derivative operator applied to pressure

<math display="block">
\frac{Dp}{Dt}=\frac{\partial p}{\partial t}+\mathbf{v}\cdot\nabla p
</math>

and thus

<math display="block">
\frac{Dp}{Dt}-\mathbf{v}\cdot\nabla p=\frac{\partial p}{\partial t}
</math>

which gives

<math display="block">
\frac{Dh_o}{Dt}=\frac{1}{\rho}\frac{\partial p}{\partial t} + \dot{q} + \mathbf{v}\cdot\mathbf{f}
</math>

If we assume adiabatic flow without body forces

<math display="block">
\frac{Dh_o}{Dt}=\frac{1}{\rho}\frac{\partial p}{\partial t}
</math>

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

<math display="block">
\frac{Dh_o}{Dt}=0
</math>

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