This material is from (Bird, et. al. 2007). We will finish up our
derivation of the conservation equations with the conservation of
energy. This equation is significantly more complicated, and is given in
many forms without the assumptions required to simplify it.
Energy equation
Applying the GBE with
\(\psi=\rho\left(u+\frac{\vec{v}^{2}}{2}\right)\),
\(\mathbb{J}=\vec{q}+\mathbb{T}\vec{v}\) and
\(\dot{\psi_{g}}=\rho\vec{v}\cdot\vec{g}+\dot{q}\). We can recognize
that these simply the assumptions for the momentum equation multiplied
by velocity, with an extra term \(\dot{q}\) in the generation term,
which corresponds to volumetric heat generation in the flux. This is
possible through chemical reactions, radiant heat absorption, or
fission/nuclear reactions. This gives us the equation
\[\begin{split}\underbrace{\frac{\partial\left(\rho\left(u+\frac{\vec{v}^{2}}{2}\right)\right)}{\partial t}}_{\substack{\text{rate of energy change}\\
\text{per unit volume}
}
}+\underbrace{\nabla\cdot\rho\vec{v}\left(u+\frac{\vec{v}^{2}}{2}\right)}_{\substack{\text{rate of energy change}\\
\text{by convection per unit}\\
\text{volume}
}
}\\=-\underbrace{\nabla\cdot\vec{q}}_{\substack{\text{conduction}}
}-\underbrace{\nabla\cdot\left(p\vec{v}\right)}_{\substack{\text{work done by}\\
\text{pressure}
}
}-\underbrace{\nabla\cdot\left(\boldsymbol{\tau}\cdot\vec{v}\right)}_{\substack{\text{work done by}\\
\text{viscous force}
}
}+\underbrace{\rho\left(\vec{v}\cdot\vec{g}\right)}_{\substack{\text{work done}\\
\text{by gravity}
}
}+\underbrace{\dot{q}}_{\substack{\text{heat}\\
\text{generation}
}
}\end{split}\]
and we want to put this in a more usable form. We split these into
kinetic energy and internal energy forms, by FOIL-ing the flux terms:
\[\begin{split}-\nabla\cdot\left(p\vec{v}\right)=-\underbrace{p\nabla\cdot\vec{v}}_{\substack{\text{reversible}\\
\text{\ensuremath{pdV} work}
}
}-\underbrace{\vec{v}\cdot\nabla p}_{\substack{\text{kinetic}\\
\text{energy}
}
}\end{split}\]
\[\begin{split}-\nabla\cdot\left(\boldsymbol{\tau}\cdot\vec{v}\right)=-\underbrace{\boldsymbol{\tau}:\nabla\vec{v}}_{\substack{\text{irreversible}\\
\text{viscous work}
}
}-\underbrace{\vec{v}\cdot\nabla\cdot\boldsymbol{\tau}}_{\substack{\text{kinetic}\\
\text{energy}
}
}\end{split}\]
. Then our forms are:
Internal Energy:
\[\frac{Du}{Dt}=-\nabla\cdot\vec{q}-p\nabla\cdot\vec{v}-\boldsymbol{\tau}:\nabla\vec{v}+\dot{q}\]
Kinetic Energy:
\[\frac{D\frac{\vec{v}^{2}}{2}}{Dt}=-\vec{v}\cdot\nabla p-\vec{v}\cdot\nabla\cdot\boldsymbol{\tau}+\rho\vec{v}\cdot\vec{g}\]
These are not basic balance equations, though, because they don’t
follow the form \(\nabla \left(\;\right)\).
Some other equations to remember, which are created though assumptions
and simplification of the energy equations are:
Poisson’s Heat Equation:
\[\rho c_{p}\frac{\partial T}{\partial t}=\nabla\cdot\left(k\nabla T\right)+\dot{q}\]
Temperature Equation:
\[\rho c_{v}\frac{DT}{Dt}=\rho c_{v}\frac{\partial T}{\partial t}+\rho c_{v}\vec{v}\nabla\cdot T=k\nabla^{2}T-\boldsymbol{\tau}:\nabla\vec{v}+T\left.\frac{\partial p}{\partial T}\right)_{\rho}\nabla\cdot\vec{v}+\dot{q}\]
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