THE SCIENCE AND APPLICATIONS OF ACOUSTICS - H. H. Arnold ...

12.15 Small Enclosures 309walls. According to Crèmer (1961), the latter two effects may be taken into accountin terms of sound frequency f by the expressionα f = 1.8 × 10 −4√ f .The average excess air absorption coefficient ᾱ ex due to air absorption in a roomor enclosure isα ex = k 4V Swhere k is an experimentally determined constant, V is the enclosed volume, and Sthe room or enclosure area. Combining the last two expressions, we get a minimumᾱ min expressed byᾱ min = α ex + α f = 4kVS + 1.8 × 10−4√ f (12.49)which is also valid for reverberation rooms. Equation (12.49) helps to establishthe upper limit of the reverberation time that can be achieved in an echo chamber.The two limiting cases for Equation (12.48) areIL = 0 dB for ᾱ = ¯τ, IL = 10 log(1/τ) = TL dB for ᾱ = 1.The first case of IL = 0 obviously represents the worst case, and the second casewhere IL = TL represents the best case for the insertion loss of the hood. Thisconnotes that ᾱ should be near unity as possible and ¯τ be made much less thanunity for the most effective noise attenuation by a hood.The above analysis is premised on the presence of high frequencies with a diffusefield inside the enclosure. For a more complete treatment that includes the effectsof low frequencies, the reader is referred to Ver (1973).Example Problem 9Determine the insertion loss of a hood with an average absorption coefficient of0.30 and a transmission coefficient of 0.002.SolutionApplying Equation (12.48)(ᾱ )IL = 10 log¯τ12.15 Small Enclosures( ) 0.30= 10 log = 21.8 ≈ 22 dB0.002A small enclosure is one that fits closely around the noise source. If the noisesource’s geometry contains flat planes that are parallel to the walls of the enclosure,standing wave resonances can occur at frequencies that are integer multiples of