Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

PHY454H1S Continuum Mechanics. Lecture 10: Navier-Stokes equation. Taught by Prof. K. Das.

Posted by peeterjoot on February 11, 2012

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

Review. Newtonian fluid.

Reading: section 6.* from [1].

We stated the model for a newtonian fluid

\begin{aligned}\sigma_{ij} = -p \delta_{ij} + 2 \mu e_{ij}\end{aligned} \hspace{\stretch{1}}(2.1)

and started considering conservation of mass with a volume dV through an area element d\mathbf{s}. For the rate of change of mass flowing out of the volume V is

\begin{aligned}\oint \rho \mathbf{u} \cdot d\mathbf{s} = - \frac{\partial {}}{\partial {t}} \int_V \rho dV.\end{aligned} \hspace{\stretch{1}}(2.2)

Application of Green’s theorem, for a fixed (in time) volume V produces

\begin{aligned}0 = \int_V \left( \boldsymbol{\nabla} \cdot (\rho \mathbf{u}) + \frac{\partial {\rho}}{\partial {t}} \right) dV,\end{aligned} \hspace{\stretch{1}}(2.3)

or in differential form for an infinitesimal volume

\begin{aligned}0 = \frac{\partial {\rho}}{\partial {t}} + \boldsymbol{\nabla} \cdot (\rho \mathbf{u}).\end{aligned} \hspace{\stretch{1}}(2.4)

Expanding out the divergence term using

\begin{aligned}\boldsymbol{\nabla} \cdot (a \mathbf{b})&=\partial_i (a b_i) \\ &=b_i \partial_i a +a \partial_i b_i \\ &=\mathbf{b} \cdot \boldsymbol{\nabla} a+ a \boldsymbol{\nabla} \cdot \mathbf{b}\end{aligned}

\begin{aligned}0 = \frac{\partial {\rho}}{\partial {t}} + \rho \boldsymbol{\nabla} \cdot \mathbf{u}+ \mathbf{u} \cdot \boldsymbol{\nabla} \rho.\end{aligned} \hspace{\stretch{1}}(2.7)

For an incompressible fluid

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{u} = 0\end{aligned} \hspace{\stretch{1}}(2.6)

so the conservation of mass equality relation takes the form

\begin{aligned}0 = \frac{\partial {\rho}}{\partial {t}} + \mathbf{u} \cdot \boldsymbol{\nabla} \rho.\end{aligned} \hspace{\stretch{1}}(2.7)

Conservation of momentum.

In classical mechanics we have

\begin{aligned}\mathbf{f} = m \mathbf{a},\end{aligned} \hspace{\stretch{1}}(3.8)

our analogue here is found in terms of the stress tensor

\begin{aligned}\int_V F_i dV = \int_V \frac{\partial {\sigma_{ij}}}{\partial {x_j}} dV\end{aligned} \hspace{\stretch{1}}(3.9)

Here F_i is the force per unit volume. With body forces we have

\begin{aligned}F_i = \rho \frac{du_i}{dt} = \frac{\partial {\sigma_{ij}}}{\partial {x_j}} + \rho f_i\end{aligned} \hspace{\stretch{1}}(3.10)

where f_i is an external force per unit volume. Observe that \sigma_{ij}, through the constituative relation, includes both contributions of linear displacement and the vorticity component.

From the constitutive relation 2.1, we have

\begin{aligned}\frac{\partial {\sigma_{ij}}}{\partial {x_j}} &= - \frac{\partial {p}}{\partial {x_j}} \delta_{ij} + 2 \mu \frac{\partial {e_{ij}}}{\partial {x_j}} \\ &= - \frac{\partial {p}}{\partial {x_i}} + 2 \mu \frac{\partial {}}{\partial {x_j}} \left( \frac{1}{{2}} \left( \frac{\partial {u_i}}{\partial {x_j}}+ \frac{\partial {u_j}}{\partial {x_i}}\right)\right) \\ &= - \frac{\partial {p}}{\partial {x_i}} + \mu \left(\frac{\partial^2 u_i}{\partial x_j \partial x_j}+\frac{\partial^2 u_j}{\partial x_i \partial x_j}\right) \end{aligned}

Observe that the term

\begin{aligned}\frac{\partial^2 u_i}{\partial x_j \partial x_j}\end{aligned} \hspace{\stretch{1}}(3.11)

is the i^{\text{th}} component of \boldsymbol{\nabla}^2 \mathbf{u}, whereas

\begin{aligned}\frac{\partial^2 u_j}{\partial x_i \partial x_j} &= \frac{\partial {}}{\partial {x_i}} \left( \frac{\partial {u_j}}{\partial {x_j}} \right) \\ &= \frac{\partial {}}{\partial {x_i}} (\boldsymbol{\nabla} \cdot \mathbf{u})\end{aligned}

is the i^{\text{th}} component of \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{u}).

We have therefore that

\begin{aligned}\rho \frac{du_i}{dt} = \left( -\boldsymbol{\nabla} p + \mu \boldsymbol{\nabla}^2 \mathbf{u} + \mu \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{u}) + \rho \mathbf{f}\right)_i,\end{aligned} \hspace{\stretch{1}}(3.12)

or in vector notation

\begin{aligned}\rho \frac{d\mathbf{u}}{dt} = -\boldsymbol{\nabla} p + \mu \boldsymbol{\nabla}^2 \mathbf{u} + \mu \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{u}) + \rho \mathbf{f}.\end{aligned} \hspace{\stretch{1}}(3.13)

We can expand this a bit more writing our velocity \mathbf{u} = \mathbf{u}(x, y, z, t) differential

\begin{aligned}du_i = \frac{\partial {u_i}}{\partial {x_j}} \delta x_j + \frac{\partial {u_i}}{\partial {t}} \delta t.\end{aligned} \hspace{\stretch{1}}(3.14)

Considering rates

\begin{aligned}\frac{du_i}{dt} = \frac{\partial {u_i}}{\partial {x_j}} \frac{dx_j}{dt} + \frac{\partial {u_i}}{\partial {t}} .\end{aligned} \hspace{\stretch{1}}(3.15)

In vector notation we have

\begin{aligned}\frac{d\mathbf{u}}{dt} = (\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u} + \frac{\partial {\mathbf{u}}}{\partial {t}}.\end{aligned} \hspace{\stretch{1}}(3.16)

Newton’s second law 3.13 now becomes

\begin{aligned}\boxed{\rho  (\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u} + \rho \frac{\partial {\mathbf{u}}}{\partial {t}} = -\boldsymbol{\nabla} p + \mu \boldsymbol{\nabla}^2 \mathbf{u} + \mu \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{u}) + \rho \mathbf{f}.}\end{aligned} \hspace{\stretch{1}}(3.20)

This is the Navier-Stokes equation. Observe that we have an explicitly non-linear term

\begin{aligned}(\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u} ,\end{aligned} \hspace{\stretch{1}}(3.18)

something we don’t encounter in most classical mechanics. The impacts of this non-linear term are very significant and produce some interesting effects.

Incompressible fluids.

Incompressibility was the condition

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{u} = 0,\end{aligned} \hspace{\stretch{1}}(3.19)

so the Navier-Stokes equation takes the form

\begin{aligned}\begin{aligned}\rho  (\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u} + \rho \frac{\partial {\mathbf{u}}}{\partial {t}} &= -\boldsymbol{\nabla} p + \mu \boldsymbol{\nabla}^2 \mathbf{u} + \rho \mathbf{f} \\ \boldsymbol{\nabla} \cdot \mathbf{u} &= 0\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.20)

Boundary value conditions.

In order to solve any sort of PDE we need to consider the boundary value conditions. Consider the interface between two layers of liquids as in figure (\ref{fig:continuumL9:continuumL10fig1})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL10fig1}
\caption{Rocker tank with two viscosity fluids.}
\end{figure}

Also found an illustration of this in fig 1.13 of white’s text online.

We see the fluids sticking together at the boundary. This is due to matching of the tangential velocities at the interface.

References

[1] D.J. Acheson. Elementary fluid dynamics. Oxford University Press, USA, 1990.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

 
%d bloggers like this: