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

12.14 Enclosures 307and similarly for the external side of the wallδ 2 = E 2= W 2LV 2 Vcand because the transmission coefficient τ = W 2 /W 1δ 2 = τ W 1L= τδ 1 .VcUsing the relationship p1 2 = ρ 0c 2 δ 1 and p2 2 = ρ 0c 2 δ 2 , and the fact L p =10 log (p/p ref ) 2 we readily obtain(L p1 = 10 logρ 0 c 2 δ 1p 2 refNoise reduction NR is given by)andL p2 = 10 logNR = L p1 − L p2 = 10 log(1/τ)(ρ 0 c 2 δ 2p 2 refwhich we note is also the definition for transmission loss TL. Hence,NR = TLwhich indicates that the noise reduction from just within the hood (located ina virtual free field) to a region very near the external hood wall is equal to thetransmission loss of the wall alone. It should be realized here the value of L p1 issupposed to be the value obtained with the hood in place, not the original lowervalue of L p1 measured in the vicinity of the noise source before the hood is placedover it. L p2 will be correspondingly higher, as the result of the hood insertion. Asa result, it is more useful to know the insertion loss IL of the hood rather than thenoise reduction. The insertion loss IL is given by).IL = L p0 − L p2 (12.42)where L p0 denotes the sound pressure level at a selected location without the hood;and L p2 , the sound pressure level at the same point with the hood enshrouding thesound source.In deriving an approximate expression for insertion loss, we shall assume thatthe room is considerably larger than the hood. The sound pressure level L p0 at anylocation in the room may be found from Equation (11.21)(Q0L p0 = L w0 + 10 log4πr + 4 ). (12.43)2 RIn a similar fashion we can express SPL at the same location with the hood in placeby(Q2L p2 = L w2 + 10 log4πr + 4 ). (12.44)2 RWe now need to determine the total power W 2 emitted by the noise source andits hood acting as a single unit. Since the combination of noise source (usually