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

302 12. Walls, Enclosures, and BarriersThe partition absorbs a certain amount of power W α from the reverberant fieldin room 1, namely( )Sw α wW α = W r (12.26)S 1 α 1whereW α = power absorbed by the dividing wallW r = power in the reverberant fieldS w = area of the wallα w = absorption coefficient of the wallS 1 = total surface area of room 1α 1 = average absorption coefficient.Let us assume that all the power incident upon the wall will be absorbed, i.e.,α w = 1. The portion of the power W 1 of the source in room 1 that becomes thepower in the reverberant field is given byW r = (1 − ᾱ)W 1Substituting the above into Equation (12.26) givesW α = (1 − ᾱ)W 1S wS 1. (12.27)The power W α absorbed by the wall can be expressed in terms of the room constantR 1 , so Equation (12.27) becomes1 − ᾱ 1W α = W 1 S w = W 1S wSᾱ 1 R 1which can now be combined with Equation (12.2) to yieldW 2 = W 1S w τ(12.28)R 1where W 2 represents the power transmitted into room 2 and τ the transmissioncoefficient of the wall.We can consider the direct field in the region near the partition in room 2 to bea uniform plane wave progressing outward from the radiating wall. The energy inthe direct field equals the product of the power transmitted into the room multipliedby the time required for the plane wave to transverse the room. This time is givenby t = L/c, where L denotes the length of room 2. The energy density δ d2 ofthe direct field is given by the directed field energy divided by the room volumeV = S w L:δ d2 = W 2 LV c . (12.29)From Equation (11.21), the reverberant energy is given byδ r2 = 4W 2cR 2. (12.30)