Vibroacoustic Simulation. Alexander Peiffer
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Название: Vibroacoustic Simulation

Автор: Alexander Peiffer

Издательство: John Wiley & Sons Limited

Жанр: Отраслевые издания

Серия:

isbn: 9781119849865

isbn:

СКАЧАТЬ the direction of the unit vector n and with F=pAn we get:

       bold upper I left-parenthesis t right-parenthesis equals p bold v bold n (2.44)

      As in Equation (1.48) the time average is given by:

      Using the harmonic plane wave solutions for pressure (2.34) and velocity (2.36)

StartLayout 1st Row 1st Column bold-italic p left-parenthesis x comma t right-parenthesis 2nd Column equals 3rd Column bold-italic upper A e Superscript j left-parenthesis omega t minus k x right-parenthesis 2nd Row 1st Column bold-italic v Subscript x Superscript asterisk Baseline left-parenthesis x comma t right-parenthesis 2nd Column equals 3rd Column StartFraction bold-italic upper A Superscript asterisk Baseline Over rho 0 c 0 EndFraction e Superscript j left-parenthesis omega t minus k x right-parenthesis EndLayout

      the time averaged mean intensity yields:

      and finally:

      We see that the specific impedance z0=ρ0c0 relates the intensity to the squared pressure.

      The kinetic energy density ekin in a control volume V0 is written as

       e Subscript normal k normal i normal n Baseline equals StartFraction upper E Subscript normal k normal i normal n Baseline Over upper V 0 EndFraction equals one-half rho 0 v Subscript normal x Superscript 2 Baseline equals StartFraction p squared Over 2 rho 0 c 0 squared EndFraction (2.48)

      The potential energy density epot follows from the adiabatic work integral as in equation (2.9)

       e Subscript normal p normal o normal t Baseline equals StartFraction upper E Subscript normal p normal o normal t Baseline Over upper V 0 EndFraction equals minus StartFraction 1 Over upper V 0 EndFraction integral Subscript upper V Baseline 0 Superscript upper V Baseline upper P d upper V (2.49)

      If we use Equation (2.15) we get the change in density as a start for the change in volume

d rho equals StartFraction 1 Over c 0 squared EndFraction d upper P

      With unit mass M in the control volume V0 it follows from ρ=M/V0 that

d upper V equals minus StartFraction upper V Over rho 0 EndFraction d rho equals minus StartFraction upper V Over rho 0 c 0 squared EndFraction d upper P

      Finally we get:

       e Subscript normal p normal o normal t Baseline equals StartFraction upper E Subscript normal p normal o normal t Baseline Over upper V 0 EndFraction equals integral Subscript upper P 0 Superscript upper P 0 plus p Baseline StartFraction upper P Over rho 0 c 0 squared EndFraction d upper P equals StartFraction p squared Over 2 rho 0 c 0 squared EndFraction (2.50)

       e left-parenthesis x comma t right-parenthesis equals e Subscript normal k normal i normal n Baseline plus e Subscript normal p normal o normal t Baseline equals StartFraction p left-parenthesis x comma t right-parenthesis squared Over rho 0 c 0 squared EndFraction (2.51)

      Using Equation (2.34) the time average over one period leads to:

      Finally, we can see that the speed of sound relates energy density to the sound intensity.

       e equals StartFraction upper I Over c 0 EndFraction (2.53)

      All those above expressions are useful for the description and evaluation of sound fields. Especially in case of statistical methods that are based on the energy density of acoustic subsystems they link the wave fields to the energy in the systems and the power irradiating at the system boundaries.

      If the intensity can be determined over a certain surface the source power is calculated by integrating the intensity component perpendicular to the surface

      2.3.4 Damping in Waves

      There is no motion without damping, and a sound wave propagating over a long distance will vanish. This is considered by adding a damping component to the one-dimensional solution of the wave equation similar to the decay rate in (1.22)

       bold-italic p equals bold-italic upper A e Superscript minus alpha x Baseline e Superscript j left-parenthesis omega t minus k x right-parenthesis (2.55)

      Here СКАЧАТЬ