VRIJE UNIVERSITEIT BRUSSEL Acoustics - the Dept. of ...

72 CHAPTER 4. SOUND ABSORPTIONafter reflection I 1 = I 0 (1 − a). After n reflections the sound intensity willbe :I n = I 0 (1−a) n (4.19)In order to determine the number of reflections n, we define :n =ct = total path length in time tmean free path(4.20)One can prove that the mean free path in a space with volume V and wallsurface S is given by 4V (without proof). Therefore we can write :SI n= (1−a) cSt cSt4V = exp ln(1−a) (4.21)I 0 4Vbecause we know x = exp(blnq) with x given by x = q b .Bydefinitionofthereverberationtimeweknowthatatt = T is InI 0= 10 −6and so :10 −6 = exp cSt ln(1−a) (4.22)4VIf we take the natural logarithm of both members of this equation, we find :T = −6.3×4×VcSln(1−a)(4.23)For air, we can obtain :T =−V6Sln(1−a)(4.24)The different walls S i of the room shall have, in practice, different absorptioncoefficients a i . We define a mean value ā of the acoustic absorptioncoefficients as : ∑ā = ∑ ia iS i(4.25)According to the model of Eyring-Norris [14], Equation 4.24 will be :i S iT =−V6 ∑ i S iln(1−ā)(4.26)We note that this model is basically valid for both small as well as largevalues of ā, but it is assumed that the absorbing materials are spatially,fairly homogeneously distributed over the walls (if not, the mean value ā hasno physically sense).