(R = Sαm/ (1- αm

مترمربع سابین 15.3= (R = A / (1- αm)=  R = (12.2 m2) / (1 - 0.2 

Lp = (90 dB) + 10 log (1 / (4 π (2 m)^2) + 4 / (15.3 m^2 Sabine) = 84.8 dB

A typical machine tool manufacturing area is 100 × 100 × 20 ft and has a room constant R of approximately 2000. From Fig. 9.4, the noise level due to the reverberant field at 5 ft from the machine is at most 2 dB above the free-field level. At 10 ft, the sound level of the reverberant field is about 4 dB above the free-field level, and at 20 ft the difference is nearly 10 dB

Consider now the effect of increasing the room constant to 5000, i.e. , adding absorption. Note that at 5 ft there is no difference between the free field and the reverberant field; at 10 ft the noise level is only 2 dB above the free field, and at 20 ft the level is only about 4 dB above the
free field. If, however, the room constant were increased to 10,000 (5 times the original absorption), noise level reductions of 6 to 8 dB could be obtained for distances beyond 10 ft

Some conclusions can now be drawn
Significant noise reduction can be obtained in manufacturing, assembly, or production areas by adding absorbing materials. However, little reduction is usually achieved within 10 ft of the source, which often includes a machine operator
 In many factory areas, the noise sources or machines are distributed over the whole floor area. As such, moving 10 ft from a machine merely puts one into the near field of another with little or no noise reduction achieved
 In summary, reverberation control in closely packed machine environments is usually only a secondary noise reduction measure. However, in areas remote to a noisy source, say 20 ft or more, the addition of absorptive treatment may reduce the reverberant noise buildup by 10 dB or more. Often these remote areas will contain a cluster of personnel such as inspectors, clerks, etc. , who will benefit significantly