User:EMBaero/Sandbox

Insert a picture of continuum versus free-molecular flow perhaps

Gases are composed of molecules which collide with one another and solid objects. If density and velocity are taken to be well-defined at infinitely small points, and are assumed to vary continuously from one point to another, the discrete molecular nature of a gas is ignored.

The continuity assumption becomes less valid as a gas becomes more rarefied. In these cases, statistical mechanics is a more valid method of solving the problem than continuous aerodynamics. The Knudsen number can be used to guide the choice between statistical mechanics and the continuous formulation of aerodynamics.

Give an intro to N-S equations. The derivation of the Navier–Stokes equations begins with an application of Newton's second law: conservation of momentum (often alongside mass and energy conservation) being written for an arbitrary control volume. In an inertial frame of reference, the general form of the equations of fluid motion is:^[7]

${\displaystyle \rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot \nabla \mathbf {u} \right)=-\nabla p+\nabla \cdot \mathbb {T} +\mathbf {f} ,}$

where ${\displaystyle \mathbf {v} }$ is the flow velocity, ${\displaystyle \rho }$ is the fluid density, p is the pressure, ${\displaystyle \mathbb {T} }$ is the (deviatoric) stress tensor, and ${\displaystyle \mathbf {f} }$ represents body forces (per unit volume) acting on the fluid and ${\displaystyle \nabla }$ is the del operator. This is a statement of the conservation of momentum in a fluid and it is an application of Newton's second law to a Continuum; in fact this equation is applicable to any non-relativistic continuum and is known as the Cauchy momentum equation.

Incompressible flow, can assume density, viscosity are constant, neglect body forces and get a simplificiation:

${\displaystyle {\frac {D\mathbf {u} }{Dt}}=-{1 \over \rho }\nabla p+{\mu \over \rho }{\nabla }^{2}\mathbf {u} }$

Compressible flow is more complicated because the flow divergence is not zero so we have the second viscosity coefficient appear (write in aerodynamics nomenclature):

${\displaystyle \rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot \nabla \mathbf {u} \right)=-\nabla p+\mu \nabla ^{2}\mathbf {u} +\left({\tfrac {1}{3}}\mu +\mu ^{v})\nabla (\nabla \cdot \mathbf {u} \right)+\mathbf {f} }$

Then, neglecting viscosity, the Euler equations can work for compressible and incompressible flow.

${\displaystyle \rho \left({\frac {\partial }{\partial t}}+{\mathbf {u} }\cdot \nabla \right){\mathbf {u} }+\nabla p=0}$