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

260 11. Acoustics of Enclosed Spaces: Architectural AcousticsIt is assumed that reverberant sound comes from nearly all directions in a fairly evendistribution. The modes generated by standing waves must be rather insignificant;otherwise the assumption of uncorrelated sound is not valid and Equation (11.22)will not truly constitute the proper model for the actual sound field.The sound pressure level within the room can now be found from( ) [ ( I1L p = 10 log = 10 log WI ref 4πr + 4 2 R( 1= L W + 10 log4πr + 4 )2 R)]+ 120(11.23)for an ideal point source emanating equally in all directions. The reference soundintensity I ref is equal to 10 –12 W/m 2 , and the sound power level L W of the sourceis defined as( ) WL W = 10 log10 −12which is given in dB re 1 pW. For an ideal point source over an acousticallyreflective surface[ ( 1L P = 10 log W2πr + 4 )]( 1+ 120 = L 2 W + 10 logR2πr + 4 )(11.24)2 REquations (11.23) and (11.24) are based on the fact that the absorption coefficientsdo not vary radically from point to point in the room and the source is not closeto reflective surfaces. If the sound power and room absorption characteristics areassigned for each frequency band, the sound pressure level L P can be determinedfor each frequency band in dB/octave, dB/one-third octave, and so on. If the soundpower level is A-weighted, and if the room constant is based on frequencies in thesame range as the frequency content of the source, the sound power level will beexpressed in dB(A).Example Problem 3Predict the reading of a sound pressure level meter 12 m from a source havinga sound power level of 108 dB(A) re 1 pW in a room with a room constantR = 725 m 2 (at the source frequencies). The source is mounted directly on anacoustically hard floor.SolutionWe apply Equation (11.24) as follows:( 1L p = L W + 10 log2πr + 4 )2 R= 108 + 10 log[(2π × 12 2 m 2 ) −1 + 4/725 m 2 ]= 86.2 dB(A)