Noise Reduction (NR) is the percieved difference in sound intensity levels betwe

Question

Noise Reduction (NR) is the percieved difference in sound intensity levels between two enclosed spaces, due to the sound isolating properties of the separating barrier wall as well as the absorption present in the recieving room. If the transmitted noise is louder than the ambient background noise, it is percieved by the ear. If it is quieter, it is not percieved. NR is percieved in a room other than the source room. NR = TL + 10log(A/S)

Part A

Find the noise reduction (NR) in the receiving room.. The barrier wall has a Transmission Loss of 45 dB and is 16 ft wide and 12 ft tall and the total absorption in the receiving room is 725Sabins

Part B

Another type of noise reduction, independent of the barrier wall, is based on the acoustical improvement done to the room. An increase in the amount of acoustical absorption in a room will improve the noise reduction. The noise reduction is based on 10 times the log of the ratio of the final absorption divided by the initial absorption. NR* = 10*log(Af/Ai)

Find the noise reduction in a room if the acoustical absorption in the room increases from 450 Sabins to 1950 Sabins with new construction done for acoustical improvement.

Explanation / Answer

(Part A) We have given the formula NR = TL + 10log(A/S)...................(1)

here NR is noise reduction in dB , TL is transmission loss in dB which is given 45 dB

S is area of barrier wall=16*12ft2 =192ft2

A is total absorption of receiving room =725 sabins

putting all values in equation (1)

we get

NR=45+10 log (725/192)

    NR= 45 +10log(3.78)

NR=45+10*0.58 =50.8dB =51dB(say) answer

(Part B) the formula given is NR* = 10*log(Af/Ai).................(2)

here Af=1950 Sabins and Ai=450 Sabins

putting thevalues in equation (2) we get

NR=10 log(1950/450)

NR=10log(4.33)

NR= 10*0.636 =6.36dB answer

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