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1 November, 05:22

A chain saw produces a spherical sound wave having a frequency of 214Hz in air at 358C (308.2K or 958F). At a distance of 600mm (23.62 in), the sound pressure level is 100 dB. Determine (a) the acoustic power level and (b) the rms acoustic particle velocity and velocity level at a distance of 600mm from the saw.

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  1. 1 November, 05:34
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    Given Information:

    Frequency = 214 Hz

    Temperature = 358° C = 308.2 K

    Sound Pressure = p = 100 dB = 2 pascal

    Distance = 600 mm = 0.60 m

    Required Information:

    (a) Acoustic Power Level = ?

    (b) Acoustic Particle Velocity = ?

    and Velocity level at a distance of 600 mm = ?

    Answer:

    Acoustic Power Level = Lw = 106.44 dB

    Acoustic Particle Velocity = v = 0.0106 m/s

    Velocity level = 60.25 dB

    Explanation:

    (a) Acoustic Power Level

    Acoustic Power = W = 4πr² I

    Acoustic Intensity = I = p ²/Z₀

    Where Z₀ is the characteristic impedance of air Z₀ = 409.8 rayl

    I = p ²/Z₀ = (2) ²/409.8 = 0.00976 W/m²

    W = 4πr² I = 4*π (0.60) ²*0.00976 = 0.0441 W

    Acoustic Power Level = Lw = 10log (W/Wref)

    Where Wref is Reference Acoustic Power Wref = 1x10⁻¹² W

    Lw = 10log (W/Wref) = 10log (0.0441/1x10⁻¹²) = 106.44 dB

    Lw = 106.44 dB

    (b) Acoustic Particle Velocity

    Acoustic Particle velocity = v = p ²/Zs

    Where Zs is specific acoustic impedance

    Zs = Z₀kr / (1 + k²r²) ⁰°⁵

    Where k = 2πf/c and c = 346.1 m/s is the speed of sound in air

    k = 2π*214/346.1 = 3.885 per m

    Zs = 409.8*3.885*0.60 / (1 + (3.885) ² (0.60) ²) ⁰°⁵

    Zs = 376.6 rayl

    v = p ²/Zs = 2²/376.6 = 0.0106 m/s

    v = 0.0106 m/s

    Velocity level = 10log (v/vref) where vref = 10x10⁻⁹ m/s

    Velocity level = 10log (0.0106/10x10⁻⁹) = 60.25 dB

    Velocity level = 60.25 dB
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