5-1 sound transmission loss.ppt

1、CHAPTER FIVEACOUSTIC ENCLOSURES AND BARRIERS,5.1 SOUND TRANSMISSION LOSS,Airborne STL is defined as follows:,Where is:,1. Airborne STL,Absorption coefficient:,Transmission coefficient:,Incident: Ii Reflected: Ir Absorbed: Ia Transmitted: It,Sound Insertion loss (IL) The difference in sound pressure

2、measured in a single location with and without a noise-control device located between the source and receiver. The difference in the sound pressure measured in the occupied space with the wall versus without the wall is the IL of the wall,Sound transmission loss (TL) It equals to the ratio of the ai

3、rborne sound power incident on the partition to the sound power transmitted by the partition and radiated on the other side.,IL and TL,In a room,We have known:,i= absorption coefficient of i th surface,For i th surface:,For a whole room:,the room-averaged sound absorption coefficient , :,S is the to

4、tal surface area of the room.,Also, in a room,Also, in a room,The corresponding sound transmission loss is given by :,We assume :,Another form :,example,Wall: 20m2,Door: 2m2,Windows: 3m2,Calculate the STL.,example,Wall: 20m2,wall,door,windows,Door: 2m2,Windows: 3m2,STL=10log105=50 dB,STL=10log103=30

5、 dB,STL=10log102=20 dB,2. STL in a separated room,Sound transmission loss is calculated by:,L1,L2,The is obtained as follows:,The sound power impinging on the test specimen from the source side is,S= surface area of test specimen, m2,In the source room,In the receiver room,The rate of energy transmi

6、ssion into the room equals the rate of energy absorption, at steady-state conditions. Thus,Or,L1,Sound transmission loss is calculated:,L2,From the definition of transmission coefficient:,3. Sound transmission class STC,STC is a single-number rating used primarily to rate the speech privacy of a bar

7、rier or other structure. It is determined from a plot of STL data measured in 1/3-octave bands (or octave bands) from 125 to 4000 Hz. (NRC),STC following conditions:,1. The STL value at any frequency cannot be more than 8 dB below the STC contour. 2. The sum of all deficiencies (STL less than contou

8、r value) at all frequencies cannot exceed 32 dB. 3. The STC value is obtained from the STL value of the highest-valued contour at 500 Hz.,The sound transmission class number is used to rate the acoustic performance of walls, floors, ceilings, doors, folding partitions, panels and other architectural

9、 structures.,Can we always find the STC for sound transmission loss of a panel?,国家标准民用建筑隔声设计规范（GBJ118-88）中，对住宅分户墙的隔声性能规定是：一级住宅50dB，二级住宅45dB，三级住宅40dB .,参考发达国家相关标准，将我国住宅分户隔墙的空气声计权隔声量的实验室测量值均定为不应小于50dB，现场测量值不应小于45dB；套内分室隔墙的空气声计权隔声量，实验室测量值均定为不应小于35dB，现场测量值不应小于30dB。,表 各国住宅分户隔墙隔声最低要求对比表 单位： dB,4. Insulati

10、on value of combined structural elements,It is often necessary to calculate the net insulation value of two structural elements acting in combination. The net insulation of a wall with a window, a partition with a door or a roof with a rooflight, can be obtained by calculation if the insulation valu

11、es of each element are known.,example,Wall: 20m2,door: 2m2,windows: 3m2,Example:,Let us assume that a class rooms have 6mm fixed glazing of average insulation 26 dB set in a wall with an insulation value of 50 dB and that the ratio of window to wall is 2:1. calculate the Insulation value of combined

12、 windows and wall?,wall,Windows glazing,Ratio of areas: 2:1 Difference in insulation: 50 26 = 24 dB From graph, loss of insulation = 23 dB Therefore net insulation of wall and window = 50 23 =27 dB,Wall 50dB,Windows Glazing 26 dB,exercise,A class rooms have fixed glazing of average insulation 20 dB

13、set in a wall with an insulation value of 45 dB and that the ratio of window to wall is 1:4. calculate the Insulation value of combined windows and wall?,The hole effects of STL on the wall,For the wall and hole, we set:,Usually,Equal STL,In our design, We assume:,Which means if we want to improve t

14、he STL, we should improve the windows STL .,Ratio of areas: 2:1 Difference in insulation: 50 26 = 24 dB From graph, loss of insulation = 23 dB Therefore net insulation of wall and window = 50 23 =27 dB,Wall 50dB,Windows Glazing 26 dB,Attenuation versus Dissipation of Acoustic Energy,Note !,Absorption coefficient:,Transmission coefficient:,Incident: Ii Reflected: Ir Absorbed: Ia Transmitted: It,The guideline that a good absorber is a poor barrier is illustrated by the following example. Consider a sound wave impinging on