Governing equations on differential form: Difference between revisions

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

==== Conservation of Mass ====

\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.

==== Conservation of Momentum ====

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

==== Conservation of Energy ====

\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

==== Conservation of Momentum ====

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 ====

\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

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 ====

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

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

\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}

\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

\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

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}

\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)

\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}

\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

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.