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

304 12. Walls, Enclosures, and BarriersWe note if room 2 is entirely absorbent, i.e., it supports no reverberation field or ifthe wall is an outside wall that it radiates outdoors, the value of R 2 approaches aninfinite value and Equation (12.37) revises toNR = TL + 6dB. (12.38)Example Problem 7A20ft× 8 ft wall with a transmission loss of 32 dB separates two rooms. Room 1contains a noise source that produces a reverberant field with a SPL = 110 dBnear the wall. Room 2 has a room constant R 2 of 1350 ft 2 . Determine the soundpressure level near the wall in room 2.SolutionWe apply Equation (12.32) to find the value of L p2 :( )1 20 × 8ft2L p2 = 110 − 32 + 10 log +4 1350 ft 2 dB= 73.7dB12.13 Sound Pressure Level at a Distance from the WallFor the most part we wish to predict the sound pressure level at some distancesfrom the wall. For appreciable distances the reverberant field will dominate overthe direct field. In Figure 12.14, it is desired to find L p3 at a point located somewhatfar from the wall, which results from the sound pressure level L p1 in room 1. Butit should be noted that the difference (L p1 − L p3 ) does not represent noise reduction,because region 3 in the chamber of room 2 is not directly contiguous to thewall.An expression for L p3 will be derived on the basis that it consists almost entirelyof the energy density in the reverberant field, i.e., the energy density is givenbybut since p 2 = ρ 0 c 2 δ, Equation (12.39) becomesδ r2 = 4W 2R 2 c = δ 3 (12.39)p 2 3 = 4W 2ρ 0 cR 2. (12.40)Here p3 2 represents the mean-square pressure in those regions of room 2 where thedirect field emerging directly from the wall as a plane wave has already dispersed