Governing equations on differential form: Difference between revisions

=== The Differential Equations on Conservation Form ===

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

\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=\iiint_{\Omega}\nabla\cdot(\rho\mathbf{v})dV

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

\frac{d}{dt}\iiint_{\Omega} \rho dV=\iiint_{\Omega} \frac{\partial \rho}{\partial t} dV

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

\iiint_{\Omega} \left[\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})\right]dV=0

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

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

which is the continuity equation on partial differential form.

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

\iint_{\partial \Omega} (\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}dS=\iiint_{\Omega} \nabla\cdot(\rho \mathbf{v}\mathbf{v})dV

\iint_{\partial \Omega} p\mathbf{n}dS=\iiint_{\Omega} \nabla pdV

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

\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV=\iiint_{\Omega}  \frac{\partial}{\partial t}(\rho \mathbf{v}) dV

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

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

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

which is the momentum equation on partial differential form

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

\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=\iiint_{\Omega}\nabla\cdot(\rho h_o\mathbf{v})dV

Fixed control volume

\frac{d}{dt}\iiint_{\Omega}\rho e_o dV=\iiint_{\Omega}\frac{\partial}{\partial t}(\rho e_o) dV

The energy equation can now be written as

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

\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho

which is the energy equation on partial differential form

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

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

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

\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho

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

The substantial derivative operator is defined as

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

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

If we apply the substantial derivative operator to density we get

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

From before we have the continuity equation on differential form as

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

which can be rewritten as

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

and thus

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

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.

We start from the momentum equation on differential form derived above

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

Expanding the first and the second terms gives

\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}

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

\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}

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

\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}

==== 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) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho

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

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

which gives

\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

Expanding the two first terms as

\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

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

\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

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

\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho

==== Summary ====

Continuity:

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

Momentum:

\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}

Energy:

\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho

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

Total internal energy is defined as

e_o=e+\frac{1}{2}\mathbf{v}\cdot\mathbf{v}

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

\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

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}).

\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

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

\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

Divide by <math>\rho</math>

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

Conservation of mass gives

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

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}

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

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

==== 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)

with <math>De/Dt</math> 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)}

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

==== Total Enthalpy Formulation ====

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}

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

\frac{D\mathbf{v}}{Dt}=\mathbf{f}-\frac{1}{\rho}\nabla p

which gives

\frac{Dh_o}{Dt}=\frac{Dh}{Dt}+\mathbf{v}\cdot\mathbf{f} -\frac{1}{\rho}\mathbf{v}\cdot\nabla p

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

\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}

The substantial derivative operator applied to pressure

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

and thus

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

which gives

\frac{Dh_o}{Dt}=\frac{1}{\rho}\frac{\partial p}{\partial t} + \dot{q} + \mathbf{v}\cdot\mathbf{f}

If we assume adiabatic flow without body forces

\frac{Dh_o}{Dt}=\frac{1}{\rho}\frac{\partial p}{\partial t}

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

\frac{Dh_o}{Dt}=0

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