7
$\begingroup$

I am trying to plot the propagation of sound from a fixed source in a 2D environment using Finite-Difference Time-Domain (FDTD) Method taking into account the effects of the wind velocity. I came across the following set of equations for solving the equation for pressure with respect to time at different grid points in the 2-D environment.

$$ \frac{∂p}{∂t} = -( v_x\frac{∂}{∂x}+v_y\frac{∂}{∂y})p - \kappa(\frac{∂w_x}{∂x}+\frac{∂w_y}{∂y}) +\kappa Q $$ $$ \frac{∂w_x}{∂t} = -(w_x\frac{∂}{∂x}+w_y\frac{∂}{∂y})v_x- (v_x\frac{∂}{∂x}+v_y\frac{∂}{∂y})w_x-b\frac{∂p}{∂x}+bF_x $$ $$ \frac{∂w_y}{∂t} = -(w_x\frac{∂}{∂x}+w_y\frac{∂}{∂y})v_y- (v_x\frac{∂}{∂x}+v_y\frac{∂}{∂y})w_y-b\frac{∂p}{∂y}+bF_y $$ where, $ b = \frac{1}{\rho} $ is the mass bouyancy.

But, I am unable to figure out what $\vec{F}$ and $Q$ are? They are mentioned somewhere as dipole and monopole pressure sources respectively. What does they actually mean?

|cite|improve this question
$\endgroup$
1
$\begingroup$

A monopole pressure source is one that radiates the pressure source in all directions. That is why it is a source term directly on the pressure term -- it is "injecting" pressure at a point. Think of dropping a pebble in the water, the waves that ripple outwards are a monopole source.

The dipole pressure source terms are ones that do not radiate equally but sort of oscillate back and forth. A piston moving is a good example -- on the leading edge, it generates high pressure while on the trailing edge, it generates low pressure.

A simple search for "monopole pressure source" turned up the following site in the top few results that shows animations for monopole, dipole and quadrupole sources as well as descriptions that make it clear:

http://www.acs.psu.edu/drussell/demos/rad2/mdq.html

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Can you guide me how can I calculate $ \vec{F} $ and $ Q $ in the above equation? $\endgroup$ – Divanshu Jun 27 '13 at 13:29
  • $\begingroup$ Not really. It is based on the problem you are trying to solve. If you don't have any pressure sources, then they are 0. If you do, then it depends on the strength of the source(s). $\endgroup$ – tpg2114♦ Jun 27 '13 at 13:37
  • $\begingroup$ What is actually the "strength" of the source? $\endgroup$ – Divanshu Jun 27 '13 at 13:43
  • $\begingroup$ If you defined your variables and constants than I could be more help, but as it is I have no idea what the units are on those terms. But the strength of the source will depend on what is causing it. A loud speaker turned up to 11 is a stronger source than a person whispering. But without additional information on both the equations and the conditions you are solving, there is no way to better define those terms. $\endgroup$ – tpg2114♦ Jun 27 '13 at 13:49
  • $\begingroup$ My source is a Gunshot within an open field, which I am using for a project. Is this any help to you? $\endgroup$ – Divanshu Jun 27 '13 at 13:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.