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NOISE REDUCTION IN SPACES BY THE ABSORPTION OF SOUND ENERGY

NOISE REDUCTION IN SPACES BY THE ABSORPTION OF SOUND ENERGY. ABSORPTION OF SOUND A WHEN PHYSICAL SOUND ENERGY STRIKES A SURFACE 1 The sound bounces to another direction if the surface is reflective. 2 The energy will cease if the material is absorptive.

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NOISE REDUCTION IN SPACES BY THE ABSORPTION OF SOUND ENERGY

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  1. NOISE REDUCTION IN SPACES BY THE ABSORPTION OF SOUND ENERGY

  2. ABSORPTION OF SOUND • A WHEN PHYSICAL SOUND ENERGY STRIKES A SURFACE • 1 The sound bounces to another direction if the surface is reflective. • 2 The energy will cease if the material is absorptive. • 3 A combination of 1 & 2, depending upon the reflective / absorptive nature of the surface. • 4 A portion of the energy will travel through the surface, depending upon the mass and elasticity of the material. • 5 The energy will scatter if the surface is refractive. • B REFLECTION OF SOUND ENERGY • 1 Energy that reflects from a smooth surface bounces away at the same angle with which it struck the surface, opposite the direction it came.

  3. 2 Angles of incidence and reflection: incidence reflection • 3 If sound energy reflects from one surface, moves to another and reflects, then moves to another and reflects, repeating until it decays, a garble of audibility is created within a space because of time lag, called REVERBERATION. • 4 Since sound travels at a constant speed through a given medium, like air, a time variance occurs between the original sound-to-recipient, and the collective reflected sounds-to-recipient. The recipient hears the sound numerous times at different periods in the form of an echo, or garble, and the sound may not be intelligible.

  4. 5 If a recipient hears the same sound several times because of reverberation, and the travel distance from direct path is different to reflected path, an echo will occur if the path difference is in excess of 68 feet. Or if the path difference is as much as 50 feet, the sounds may not be intelligible. • 6 Reverberation time is the number of seconds required for a sound to decay. • 7 In a situation involving a human voice, since intelligibility of words is constructed upon changes of pitch and loudness of syllables that form words, if the reverberation time within a room is more than 1 ½ seconds, the echoed syllables will mix, and the speech will not be intelligible, particularly if the speech is rapid or the overall pitch is low.

  5. SOUND RAY DIAGRAMMING is a graphic process of indicating the paths of sound from a point source by constructing straight lines that indicate the travel path until it contacts a surface and reflects

  6. Convex irregular shapes condense the effective ceiling area to help reflect sound energy. The specific shape is designed to cover a specific area.

  7. A uniform convex ceiling area will reinforce sound projection with a uniform pattern of reflected energy.

  8. REFLECTION – continued • 8 The reflection of sound is useful to reinforce sounds within a space, utilizing elements of the surrounding envelope when effective without creating echoes. • 9 The surfaces under consideration where reflection is useful should be made of reflective materials, and arranged in such way as to direct the reflected sound toward the audience. • 10 Reflected sounds that bounces from one surface to another, then to another, and another before reaching the audience become a nuisance in causing reverberation. • 11 Sound energy that goes directly to the audience, or is reflected off a surface to the audience will be absorbed, and will not bounce.

  9. C ABSORPTION • 1 The reverberation of sound is important to some types of facilities, such as spaces where an added ‘depth’ of sound or ring is desirable for dramatic affect. • 2 Where it is desirable to limit reverberation for the sake of audible clarity, such as the speaking voice, surfaces that do not contribute to reinforcement for the benefit of the audience should be absorptive or refractive. • 3 Sound energy that strikes an absorptive surface will die there and not bounce to cause reverberation. • 4 Spaces that have a variance in the number of people in an audience, present a potential problem with reverberation, since people, because of the softness of the body, clothing, etc. are absorptive.

  10. 5 If all the surfaces in a room are reflective and the room is filled with people, the reflective surfaces are limited to walls & ceiling, because audiences are absorptive, and audiences cover a floor. • 6 If in the same room, only half the people are present, an added floor area is available to cause reverberation. But in rooms with fixed seats, such as auditoriums and churches, it is important for the seatbacks and cushions to be of absorptive material so that a partially filled room will have the same absorptive area as a room filled with people. • 7 Floors beneath seating areas that are reflective do not contribute an appreciative amount of sound reflection within the room if the empty seats are absorptive, but will contribute to a slight reverberation and give depth to sound.

  11. D NOISE REDUCTION DUE TO ABSORPTION OF SOUND • 1 Noise in the form of unwanted sound caused by reverberation can be reduced by installing surface materials that stop the reflection of energy. • 2 All construction materials used on finish surfaces are laboratory tested for their ability to absorb sound energy. An absorption coefficient is a fractional number that represents the percentage of sound energy that will be absorbed by a material. • 3 Absorption coefficients are determined for a material at various sound frequencies, ranging from low sounds at 125 hertz to high sounds at 4000 hertz.

  12. 4 Consider a constant sound source such as a horn. Say the direct sound loudness of the horn is 30 decibels at a point 10 feet from the horn. Then place the horn within a room, say 30 feet square, that has hard, sound reflecting surfaces. Sound the horn and measure the loudness at the same 10’ distance. • 5 The level of sound will be higher than the initial 30 decibels because the measuring device records the direct sound PLUS the reflected sound from walls, floor, ceiling. • 6 Any room has a characteristic called TOTAL ROOM ABSORPTION which is a function of the summation of the absorption values of its surfaces. Surface absorption is measured in units called SABINS, and equals area x the absorption coefficient. TOTAL ROOM ABSORPTION equals the summation of the surface absorptions:

  13. SURFACE ABSORPTION Absorption = surface area x coefficient of absorption, or a = S x alpha, where a = sound absorption in units of Sabins S = area of surface in square feet alpha = coefficient of absorption • TOTAL ROOM ABSORPTION A = ∑ S x alpha, where A = total room absorption in Sabins ∑ S x alpha = the sum of all the surface absorptions that make up the enclosure of the room. ROOM NOISE REDUCTION, NR = 10 log [ a2 / a1 ] where a1 = absorption in Sabins BEFORE treatment is done a2 = absorption in Sabins AFTER treatment is done

  14. NOISE REDUCTION by the absorption of sound within the room is merely reducing the amount of sound that bounces from its source to adjacent floor, wall, ceilings, and any object within the room that would reflect sound. It is simply a process to make the room quieter. Recollect the times you have been to a restaurant, and the noise inside was practically unbearable. Some examples of this in Lubbock are Abuelos, Garcias, The Crab Shack, and perhaps numerous others I have not visited where the floors, walls, and ceilings are hard and sound reflective. A situation called the “cocktail hour affect” happens. One group of people visiting cannot hear their conversation, so they speak louder; another group does the same thing; and another group; and so on, until a din of noise occurs within the space. It is sound that reflects from the interior surfaces. In order to reduce the noise, treat the exposed surfaces so they will ABSORB sound rather than reflect. But realize that no more than about 11 to 12 decibel sound reduction can be realized, since no amount of absorption will reduce airborne sound.

  15. Consider a room 30’ x 30’ x 10’ in size. One wall is glass to the exterior, floor to ceiling. Assume the doors into the room have the same absorptive coefficient as the walls. The absorptive qualities of the room are thus: • The absorption coefficients, alpha, are as follows: (at 500 Hertz) unfinished concrete floor = .02 unfinished gyp. bd. walls = .05 unfinished gyp. bd. ceiling = .05 glass wall = .18 • Surface absorptions are as follows: floor = 30 x 30 x .02 = 18.00 sabins ceiling = 30 x 30 x .05 = 45.00 sabins walls = 3 (30 x 10) x .05 = 45.00 sabins glass = 30 x 10 x .18 = 54.00 sabins • TOTAL ROOM ABSORPTION = 162.00 sabins

  16. Consider that in the preceding room, reverberation would probably be a problem, since all the surfaces are reflective. However, since the room is relatively small, the distance involved for echoes may not be severe. • REVERBERATION TIME, the number of seconds it takes for a sound to decay, is a function of the volume of the room and the area of the room: • T = .05 x [ V / A ] where .05 is a constant V = volume of room in cu.ft. A = room absorption in sabins • Reverberation Time in the example room T = .05 x [ (30x30x10) / 162 ] = .05 x [ 9000 / 162 ] T = .05 x 55.55 = 2.78 seconds the room would have a definite ring.

  17. Consider the room as a classroom, and carpet is added to the floor – which may have the affect of the absorption rate of a classroom full of people . . . Absorption coefficient, alpha, of carpet = .14 Absorption = 30 x 30 x .14 = 126 If the walls, ceiling, & glass remain the same, then New ROOM ABSORPTION = 126 + 45 + 45 + 54 = 270 sabins • NOISE REDUCTION = 10 log [ 270 / 162 ] = 10 log 1.67 = 10 x .22 = 2.2 decibels • And REVERBERATION TIME = .05 x (9000 / 270) = .05 x 33.3 = 1.67 seconds. Noise is reduced slightly, & reverberation time still high

  18. NEXT, add acoustic tile to the ceiling, which in most cases where speech is important, is a no-no, as the ceiling is a handy place to reinforce sound for an audience - but this example is to demonstrate NOISE REDUCTION. • Absorption coefficient, alpha, for ceiling tile = .80 Ceiling surface absorption = 30 x 30 x .8 = 720 sabins • New ROOM ABSORPTION = 126 + 720 + 45 + 54 = 945 sabins. • NOISE REDUCTION = 10 log [ 945 / 162 ] = 10 log 5.83 = 10 x .77 = 7.7 decibels • And REVERBERATION TIME = .05 X (9000 / 945) = .05 x 9.52 = .47 seconds • Noise Reduction and Reverberation time is reduced significantly

  19. NEXT, add plywood paneling to the walls. The side walls are normally good for reinforcing sound to an audience. The wall opposite the speaker should be absorbent or refractive (broken surface to scatter reflected sound). Absorption coefficient, a, for paneling = .17 Surface absorption for walls = 30x10 x 3 x .17 = 153 sabins • New ROOM ABSORPTION = 153+126+720+54 = 1,053 sabins • NOISE REDUCTION = 10 log [ 1053 / 162 ] = 10 log 6.5 = 10 x .81 = 8.1 decibels ( 7.7) • REVERBERATION TIME = .05 X ( 9000 / 1053 ) = .05 X 8.55 = .43 second ( .47 ) • Wall paneling does not make a significant change.

  20. FINALLY, add drapes to the windows. Slatted blinds do not add significantly to absorption of sound, but refract energy to reduce reflection. Absorption coefficient, a, for drapes = .49 Surface absorption for drapes = 30x10 x .49 = 147 sabins • New ROOM ABSORPTION = 153+126+720+147 = 1146 sabins. • NOISE REDUCTION = 10 log [ 1146 / 162 ] = 10 log 7.07 = 10 x .85 = 8.5 decibels • REVERBERATION TIME = .05 x (9000 / 1146) = .05 x 7.85 = .39 second

  21. In both conditions from beginning left to right, it is significant that the progressive curves flatten quickly. Even if all the surfaces were 100 % sound absorptive, nothing can be done to reduce the direct sound through air. Noise reduction by absorption would probably never be above 11 to 12 decibels.

  22. 900 + 900 + 1200 = 3000 = surface area of room and would be the room absorption if all surfaces were 100 % absorptive – which is purely hypothetical. • Noise Reduction = 10 log [ 3000 / 162 ] = 10 log 18.52 = 10 x 1.267 = 12.67 decibels. • Reverberation Time = .05 x (9000/3000) = .05 x 3 = .15 sec • Perhaps the only surface that is 100 % absorptive is an open window. The sound energy continues without interruption until it decays. • The purpose of understanding sound projection and sound absorption is to realize the value of reinforcing sound where it is needed by utilizing reflection, and to diminish sound reverberation where it is not useful by using absorptive materials. Study and application will architecturally reveal where the applications can be made.

  23. PRACTICE PROBLEM ONE: A 16’ x 20’ x 9’ room has absorptive coefficients as follows: Ignore doors & windows. Walls .30 Floor .25 Ceiling .40 A. Find total room absorption B. Find reverberation time The finishes then change to the following coefficients: Walls .46 Floor .40 Ceiling .86 C. Find the noise reduction due to the new finishes D. Find the difference in reverberation time.

  24. PROBLEM SOLUTION • Wall area = 648 x .30 = 194.4 • Floor area = 320 x .25 = 80.0 • Ceiling area = 320 x .40 = 128.0 total = 402.4 sabins T = .05 x (2880 / 402.4) = .36 seconds NEW Wall area = 648 x .46 = 298.08 Floor area = 320 x .40 = 128.0 Ceiling area = 320 x .86 = 275.2 total = 701.28 Noise Reduction = 10log (701.28/402.4) = 10 log 1.7427 = 10 x .2412 = 2.41 db. T = .05 x 4.1068 = .205 sec Time difference = .36 - .205 = .1550 sec, = 43%

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