By Christopher K. W. Tam
Computational Aeroacoustics (CAA) is a comparatively new study quarter. CAA algorithms have constructed speedily and the equipment were utilized in lots of components of aeroacoustics. the target of CAA isn't really just to increase computational equipment but additionally to exploit those how you can clear up sensible aeroacoustics difficulties and to accomplish numerical simulation of aeroacoustic phenomena. by means of studying the simulation info, an investigator can verify noise new release mechanisms and sound propagation approaches. this is often either a textbook for graduate scholars and a reference for researchers in CAA and as such is self-contained. No earlier wisdom of numerical equipment for fixing PDE's is required, besides the fact that, a normal knowing of partial differential equations and easy numerical research is believed. routines are incorporated and are designed to be an essential component of the bankruptcy content material. furthermore, pattern machine courses are incorporated to demonstrate the implementation of the numerical algorithms.
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Additional info for Computational Aeroacoustics: A Wave Number Approach
3. Solve the initial value problem: ∂u ∂u + =0 at t = 0, u = H(x + 50) − H(x − 50), ∂t ∂x computationally on a grid with x = 1 , where H(x) is the unit step function, Plot u(x, t ) at t = 300. Determine from your numerical solution (a) the most negative wave velocity (waves propagating to the left) (b) the wave number of the wave with zero group velocity Compare the wave curve. 1 ). 37 3 Time Discretization There is a fundamental difference between space and time. Space extends from negative infinity to positive infinity, whereas time only increases in one direction.
T 2 ∂x4 45 46 Finite Difference Scheme as Dispersive Waves c. Linearized Korteweg de Vries equation for water waves ∂φ ∂ 3φ + μ 3 = 0. ∂t ∂x d. Linearized Boussinesq equation for water and elastic waves 2 4 ∂ 2φ 2∂ φ 2 ∂ φ − a − μ = 0. ∂t 2 ∂x2 ∂x2 ∂t 2 Because these equations are linear with constant coefficients, they can be readily solved by Fourier-Laplace transform. The Fourier-Laplace transforms of these equations are (a) (ω2 − c2 α 2 − μ2 )φ˜ = H1 (α, ω). (b) (ω2 − μ2 α 4 )φ˜ = H2 (α, ω).
I) The angle of reflection is equal to the angle of incidence. (ii) The reflected wave has the same intensity as the incident wave. (iii) There is a pressure doubling at the wall, that is, p2wall = 2p2incident = 2p2reflected , where the overbar denotes a time average. Suppose the problem is solved by finite difference approximation with a mesh size of x and y. On using a second-order central difference approximation, Eqs. (A) and (C) become pℓ+1,m − 2pℓ,m + pℓ−1,m pℓ,m+1 − 2pℓ,m + pℓ,m−1 ∂ 2 pℓ,m = c2 + 2 2 ∂t ( x) ( y)2 m = 0, incident reflected pincident + preflected = 0, ℓ,m+1 − pℓ,m−1 ℓ,m+1 − pℓ,m−1 (E) (F) where ℓ, m are the spatial indices in the x and y directions and the wall is located at m = 0.
Computational Aeroacoustics: A Wave Number Approach by Christopher K. W. Tam