Based on information from (Bird, et. al. 2007). Continuing with the
derivation of the conservation equations, again we use the GBE to derive
a conservation equation that defines the conservation of momentum.
Momentum Equation
Applying the GBE with \(\psi = \rho \vec{v}\),
\(\mathbb{J} = \mathbb{T} = p\mathbb{I} + \tau\), and
\(\dot{\psi}_{g} = \rho\vec{f}_{b} = \rho\vec{g}\), which is to say:
the conservation quantity we will use is momentum, which when normalized
per unit volume, becomes \(\rho \vec{v}\). Note that this is a
vector quantity. The current of momentum across a boundary is a tensor
which involves a symmetric tensor for the pressure in a system (which is
isotropic), and the torque applied to the system. And finally, that the
only generation of momentum is due to body forces, of which gravity is
the only concern (unless we’re working with magnetohydrodynamics). With
these, we get
\[\begin{split}\underbrace{\frac{\partial\left(\rho\vec{v}\right)}{\partial t}}_{\substack{\text{rate of change}\\
\text{of momentum}\\
\text{per unit volume}
}
}+\underbrace{\nabla\cdot\left(\rho\vec{v}\vec{v}\right)}_{\substack{\text{momentum change}\\
\text{by convection}\\
\text{per unit volume}
}
}=-\underbrace{\nabla p}_{\substack{\text{pressure}\\
\text{force}
}
}-\underbrace{\nabla\cdot\boldsymbol{\tau}}_{\substack{\text{viscous}\\
\text{force}
}
}+\underbrace{\rho\vec{v}}_{\substack{\text{gravity}\\
\text{force}
}
}\end{split}\]
By again using the definition of the substantial derivative, we can
create Cauchy’s Equation of motion as
\[\rho\frac{D\vec{v}}{Dt}=-\nabla p-\nabla\cdot\boldsymbol{\tau}+\rho\vec{g}\]
and by inspection, we can see that this matches Newton’s Second Law of
Motion:
\[\rho\vec{a}=\sum\vec{F}\]
Boussinesq Assumption
When gravity is important, we use the Boussinesq assumption(s):
Thermal expansion (\(\beta\)) causes density change.
This happens by the thermal expansion coefficient, \(\beta\),
such that
\[\beta\equiv\frac{1}{\nu}\left.\frac{\partial\nu}{\partial T}\right)_{p}=\frac{1}{\rho}\left.\frac{\partial\rho}{\partial T}\right)_{p}\]
This density change is only important in the gravity term.
Applying those assumptions, we get
\[dp=-\rho\beta dT\;\;\therefore\;\;\rho-\overline{\rho}=-\overline{\rho}\beta\left(T-\overline{T}\right)\]
and putting that into the momentum equations, we get
\[\overline{\rho}\frac{D\vec{v}}{Dt}=-\cancelto{\nabla p_{dyn}}{\nabla p}-\nabla\cdot\boldsymbol{\tau}+\left[\cancelto{0}{\overline{\rho}}-\overline{\rho}\beta\left(T-\overline{T}\right)\right]\vec{g}\]
.
Bibliography
