THESE de DOCTORAT - cerfacs
THESE de DOCTORAT - cerfacs THESE de DOCTORAT - cerfacs
60 Chapter 4: Validation of the acoustic code AVSP-f Ŝ(y, ω) = Ŝ 0 δ(y − y 0 )e −iφ (4.4) where Ŝ 0 = 1E10 W/m 3 s, y 0 = [0.0, 0.0] m, φ = 0 rad, and ω = 2π · 5000 rad/s. The same case is computed with the numerical tool AVSP-f. The outputs of this code are the fields of both modulus | ˆp|and argument arg( ˆp) of the acoustic pressure. They are illustrated in Fig. 4.2. The reconstruction of the acoustic field in the time domain for t 0 = 0 is computed by p ′ = | ˆp| cos(arg( ˆp)) and is shown in Fig. 4.3. Fig. 4.4 shows the isotropic pattern in the sound radiation produced by a monopolar source. It is also observed from this figure that a good agreement is reached between analytical an numerical results. a) Modulus of Pressure | ˆp| b) Argument of pressure arg( ˆp) Figure 4.2: Outputs of AVSP-f. Monopole case Figure 4.3: Acoustic field produced by a monopole. p ′ = | ˆp| cos(arg( ˆp))
4.1 Fundamental validation cases 61 150 210 120 240 90 270 0.008 60 0.006 0.004 0.002 180 0 300 30 330 r = 0.05 r = 0.15 Pressure Fluctuation (Pa) 0.03 0.02 0.01 0 −0.01 −0.02 Analytics AVSP−f −0.03 −0.2 −0.1 0 0.1 0.2 X axis (m) a) | ˆp| Directivity of a monopole b) Pressure fluctuation for two different radii. along the horizontal e 1 axis [ —— Analytics, (◦△) AVSP-f ] [ —— Analytics, (◦) AVSP-f ] Figure 4.4: Analytical solution vs Numerical solution. 4.1.2 A Dipole in free space Once pure radiation is tested in section (4.1.1), the influence on the acoustic field of another monopole is tested. If this monopole radiates sound out of phase with respect to the first monopole, a substraction of acoustic fluctuations will occur leading to a diminution of the acoustic amplitude in some regions. If this phase shift is equal to π a perfect cancelation should occur in the plane of antisymmetry between the two monopoles. Two monopoles with such a shift phase conform a dipole. The source is defined as Ŝ(y, ω) = Ŝ 0 δ(y − y 1 )e −iφ 1 + Ŝ 0 δ(y − y 2 )e −iφ 2 (4.5) where φ 1 = φ 2 + π. From Eq. (4.3) ˆp(x, ω) = i ∫ 4c 2 0 V 0 (y) (Ŝ0 δ(y − y 1 )e −iφ 1 + Ŝ 0 δ(y − y 2 )e 2) −iφ H0(kr)dV 2 (4.6) Finally ˆp(x, ω) = i (Ŝ0 δ(y − y 1 )e 1) −iφ H0(kr 2 1 )dV 1 + 4c 2 0 i 4c 2 0 (Ŝ0 δ(y − y 2 )e 2) −iφ H0(kr 2 2 )dV 2 (4.7)
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60 Chapter 4: Validation of the acoustic co<strong>de</strong> AVSP-f<br />
Ŝ(y, ω) = Ŝ 0 δ(y − y 0 )e −iφ (4.4)<br />
where Ŝ 0 = 1E10 W/m 3 s, y 0 = [0.0, 0.0] m, φ = 0 rad, and ω = 2π · 5000 rad/s. The same case is<br />
computed with the numerical tool AVSP-f. The outputs of this co<strong>de</strong> are the fields of both modulus<br />
| ˆp|and argument arg( ˆp) of the acoustic pressure. They are illustrated in Fig. 4.2. The reconstruction<br />
of the acoustic field in the time domain for t 0 = 0 is computed by p ′ = | ˆp| cos(arg( ˆp))<br />
and is shown in Fig. 4.3. Fig. 4.4 shows the isotropic pattern in the sound radiation produced<br />
by a monopolar source. It is also observed from this figure that a good agreement is reached<br />
between analytical an numerical results.<br />
a) Modulus of Pressure | ˆp| b) Argument of pressure arg( ˆp)<br />
Figure 4.2: Outputs of AVSP-f. Monopole case<br />
Figure 4.3: Acoustic field produced by a monopole. p ′ = | ˆp| cos(arg( ˆp))
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