Engineering Acoustics. Malcolm J. Crocker

3.8 Sound intensity I_n, being measured on (a) segment dS of an imaginary hemispherical enclosure surface and (b) an elemental area dS of a rectangular enclosure surface surrounding a source having a sound power W.

Figure 3.9 Sound intensity probe microphone arrangement commonly used.

      Example 3.8

By making measurements around a source (an engine exhaust pipe) it is found that it is largely omnidirectional at low frequency (in the range of 50–200 Hz). If the measured sound pressure level on a spherical surface 10 m from the source is 60 dB at 100 Hz, which is equivalent to a mean‐square sound pressure p²_rms of (20 × 10⁻³)² (Pa)², what is the sound power in watts at 100 Hz frequency?

      Solution

      Assuming that ρ = 1.21 kg/m³ and c = 343 m/s, so ρc = 415 ≈ 400 rayls:

      then from Eq. (3.47):

      Example 3.9

      If the sound intensity level, measured using a sound intensity probe at the same frequency, as in Example 3.8, but at 1 m from the exhaust exit, is 80 dB (which is equivalent to 0.0001 W/m²), what is the sound power of the exhaust source at this frequency?

      Solution

      From Eq. (3.41) (for an omnidirectional source). Then W = 1.26 × 10⁻³ watts (the same result as Example 3.8).

      Sound intensity measurements do and should give the same result as sound pressure measurements made in a free field.

      Far away from omnidirectional sound sources, provided there is no background noise and reflections can be ignored:

(3.48)

      (3.49)

      and by taking 10 log throughout this equation

(3.50)

      where L_p is the sound pressure level, L_W is the source sound power level, and r is the distance, in metres, from the source center. (Note we have assumed here that ρc = 415 ≅400 rayls.) If ρc ≅ 400 rayls (kg/m²s), then since I = p²_rms/ρc

      So,

      (3.51)

3.8 Sound Sources Above a Rigid Hard Surface

In practice many real engineering sources (such as machines and vehicles) are mounted or situated on hard reflecting ground and concrete surfaces. If we can assume that the source of sound power W radiates only to a half‐space solid angle 2π, and no power is absorbed by the hard surface (Figure 3.10), then

      (3.52)

      where L_W is the sound power level of the source and r is the distance in metres.

Figure 3.10 Source above a rigid surface.

In this discussion we have assumed that the sound source radiates the same sound intensity in all directions; that is, it is omnidirectional. If the source of sound power W becomes directional, the mean square sound pressure in Eqs. (3.48) and (3.46) will vary with direction, and the sound power W can only be obtained from Eqs. (3.41) and (3.47) by measuring either the mean‐square pressure (p²_rms) all over a surface enclosing the source (in the far acoustic field, the far field) and integrating Eq. (3.47) over the surface, or by measuring the intensity all over the surface in the near or far acoustic field and integrating over the surface (Eq. (3.41)). We shall discuss source directivity in Section 3.9.

      Example 3.10

      If the sound power level of a source is 120 dB (which is equivalent to 1 acoustical watt), what is the sound pressure level at 50 m (a) for sound radiation to whole space and (b) for radiation to half space?

      Solution

      1 For whole space: I = 1/4π(50)2 = 1/104 π (W/m2), then

       Since we may assume r = 50 m is in the far acoustic field, Lp ≅ LI = 75 dB as well (we have also assumed ρc ≅ 400 rayls).

       For half space: I = 1/2π(50)2 = 2/104 π (W/m2), then

      and L_p ≅ L_I = 78 dB also.

       It is important to note that the sound power radiated by a source can be significantly affected by its environment. For example, if a simple constant‐volume velocity source (whose strength Q will be unaffected by the environment) is placed on a floor, its sound power will be doubled (and its sound power level increased by 3 dB). If it is placed at a floor–wall intersection, its sound power will be increased