1. Find reverberation time for a hall of dimensions 40 x 30 x 20 having average absorption coefficient of 0.15.

Given: Required: Solution:

2. In problem 1, how much area we should treat with an absorbing material of absorption coefficient, 0.20, to reduce its reverberation time to 1.2 s?

Given: Required: Solution:

3. If a University lecture hall (15m x 8m x 3m) is heavily damped with absorption coefficient 0.3. Calculate its reverberation time if approximation is applied.

Given: Required: Solution:

4. If a University lecture hall (15m x 8m x 3m) is heavily damped with absorption coefficient 0.3. Calculate its reverberation time if approximation is NOT applied

Given: Required: Solution:

5. If a Concert hall of 70 x 40 x 15 has plastered surface of absorption oefficient, 0.1, and a capacity of an audience of 100 adults (each having an absorption of 4.7ft2 OWU), find reverberation time of the hall.

Given: Required: Solution:

6. A hall (80 x 40 x 20) has reverberation time of 1.5 second. If the hall is partitioned into two halves by hanging a curtain (absorption coefficient = 0.4), and consequently, the reverberation time reduces by 0.25 s, find the curtain length if the breadth is 10.

Given: Required: Solution:

Without curtain

With curtain of effective area A:

Area includes both sides of the curtain, thus actual curtain area will be half of this area, i.e., 534 ft2

Now

7. If a cubical hall of dimension a and average absorption coefficient has reverberation time 2s, what will be the reverberation time if its dimension is doubled.

Given: Required: Solution:

8. If a cubical hall of dimension a and average absorption coefficient has reverberation time 2s, what will be the reverberation time if its volume is doubled.

Given: Required: Solution:

9. Find the reverberation time for a hall of dimensions 50 x 25 x 15 having average absorption coefficient of 0.10.

Given: Required: Solution:

10. For a hall (60 x 30 x 15), find the reverberation time if its absorption coefficient is 0.12. Also find the area to be treated with a material of absorption coefficient 0.25 to reduce its reverberation time 1.2s.

Given: Required: Solution:

For reverberation time

For area for reduced reverberation time

11. A church has an internal volume of 90.05 ft3 (2550 m3). When it contains 2000 customary sabins of absorption (186 metric sabins), what will be its reverberation time in seconds?

GIVEN: Volume, V = 90.05 ft3 = 2550 m3 Total absorption, A = 186 metric sabins

REQUIRED: Reverberation time, R = ?

SOLUTION:

12. A ministry contains 1500 customary sabins of absorption (139.5 metric sabins). Its reverberation time in seconds is 2.26 . What is its internal volume in meters. ?

GIVEN: Total absorption, A = 139.5 metric sabins Reverberation time, R = 2.26 seconds

REQUIRED: Volume, V = ?SOLUTION:

13. Calculate the Norris-Eyring reverberation time of uniformly diffuse sound in a live room if the average absorption is 0.8 and the reverberation time is 1.75 s.

GIVEN: Absorption, A = 0.8 Reverberation time, R = 1.75 s

REQUIRED: Norris-Eyring Reverberation time, R(N-E) = ?

SOLUTION:

14. Calculate the Norris-Eyring reverberation time of uniformly diffuse sound in a live room if the average absorption is 0.65 and the reverberation time is 1.23 s.

GIVEN: Absorption, A = 0.65 Reverberation time, R = 1.23 s

REQUIRED: Norris-Eyring Reverberation time, R(N-E) = ?

SOLUTION:

15. A room has an internal volume of 2034 m3. When it contains 1200 customary sabins of absorption (111.6 metric sabins), what will be its reverberation time in seconds?

GIVEN: Volume, V = 2034 m3 Total absorption, A = 111.6 metric sabins

REQUIRED: Reverberation time, R = ?

SOLUTION:

16. What is the reverberation time of a room whose surface area is 75 m2, whose volume is 42 m3, and whose average absorption coefficient is 0.9, 0.2? What would be the effect of doubling all the dimensions of the room while keeping the average absorption coefficients the same?Solution: for = 0.9:T60= (-0.161V)/(S ln(1-) = (-0.161 x 42m-3)/(75 m2x ln (1-0.9) ) = 0.042s (42 x10-3s)For a = 0.2 we get:T60= (-0.161V)/(S ln(1-) = (-0.161 x 42m-3)/(75 m2x ln (1-0.2) ) = 0.43 swhich would correspond well with the typical T6o of a living room, which is in fact what it is.If the room dimensions are doubled then the ratio of volume with respect to the surface area also doubles so the new reverberation times are given by:Vdoubled/S doubled= Linear Dimension doubled= 2so the old reverberation times are increased by a factor of 2:T60doubled = T60X 2which gives a reverberation time of:T60doubled = T60X 2 = 0.042 x 2 = 0.084 swhen = 0.9 and:T60doubled = T60X 2 = 0.43 x 2 = 0.86 s

17. Compute the reverberation time T using T = 0.05 (V/a).Solution:T = 0.05 (Va) = 0.05 X 31,5001149 = 15751149 = 1.37s at 500 Hz

Find the reverberation time T if 50percent ofthe ceiling surface (along the perimeter of the room) is treated withacousticalpanels at of 0.85. The central area remains sound-reflecting to help distribute sound energy from lectern end toward rear of the room. Compute the total room absorption a using a = S .Sa (sabins)

Bare ceilingTreated ceilingWallsFloor1050x0.04=1050x0.85=2850x0.30=2100x0.10=42892855210

Total a =1999 sabins

Compute new reverberation time T.T = 0.05 (Va) = 0.05x31,5001999 = 15751999 = 0.79s at 500HzThe reverberation time is reduced to below 1 s with 50 percent ceiling treatment for unoccupied conditions. This represents a reduction of (1.37 - 079)/1.37 x 100 = 42 percent, which is a clearly noticeable change. Absorption provided by teachers and students will further reduce reverberation depending on the number of occupants, their distribution throughout the room, and the clothing worn.

18. A small room 10 ft by 10 ft by 10 ft has all walls and floor finished in ex posed concrete. The ceiling is completely covered with sound-absorbing spray- on material. Sound absorption coefficients are 0.02 for concrete and 0.70 for spray-on material, both at 500 Hz.

Find the noise reduction NR in this room if sound-absorbing panels are added to two adjacent walls. The sound absorption coefficient a is 0.85 for panels at 500 Hz.Solution:Compute the surface areasS.S= 5 x 10 x 10=500 ft^2 of concreteS= 10 X 10 = 100 ft^2 of spray-on materialCompute the total room absorption a with spray-on material on the ceiling.a1= S = (500 X 0.02) + .(100 X 0.70) = 10 + 700 sabinCompute the total room absorption a2 with sound-absorbing panels covering two walls and spray-on material on ceiling.a2= (300X 0.02) + (200X 0.85) + (100 X 0.70) = 6 + 170 + 70 = 246 sabinCompute the noise reduction NR.NR= 10log (a2a1) = 10log (246/80) = 10log (3.075 X 10^0)= 10(0.4878) = 5dBThis would be a noticeable improvement. With no treatment, the total absorption in the room would only be 600 X 0.02 = 12 sabins. Therefore, treating the ceiling alone providesNR = 10 log (80/12)= 10 log 6.67 = 10(0.8241) = 8 dBwhich is a significant reduction. However, initial conditions of all hard surfaces in unfurnished rooms rarely occur.

19. Find the noise reduction NR if all four wall surfaces are treated with fabric- covered panels and the floor is carpeted. The sound absorption coefficient of the carpet is 0.50 at 500 Hz.Solution:Compute the total room absorption a3 with sound-absorbing panels on all walls, spray-on material on ceiling, and carpet on floor.a3= S = (400X0.85) + (100 X0.70) + (100X0.50)= 340 + 70 + 50 = 460 sabins2. Compute the noise reduction NR for these improvements compared to room conditions of spray-on ceiling treatment alone.NR = 10log (a3/a1) = 10log(460/80) = 10 log (5.75 x 10^0)= 10(0.7597) = 8 dBSurfaces Treated ( addition to ceiling)Room NR (at 500 Hz)

Two wallsFour walls and floor5 dB8 dB

The results from both parts of the problem are summarized below.Note: The NRs given in the above table would not be as great at low frequencies be cause sound absorption coefficients usually are smaller at low frequencies than at mid- or high frequencies.

20. A classroom 60 ft long by 35 ft wide by 15 ft high has sound absorption coefficients of 0.30 for walls, 0.04 for ceiling, and 0.10 for floor. All are at 500 Hz.

Find the reverberation time T at 500 Hz in this space with no occupants and no sound-absorbing treatment.Compute the room volume V.V=60X35X15=31,500ftComputethe surface areas S.CeilingS=60X35=2100ftWallsS= 2 X 35 X 15 = 1050 ftS= 2X60X 15= 1800ftFloorS= 60 X 35 = 2100 ftCompute the total room absorption a using a = S .Sa (sabins)

CeilingWallsFloor2100 X2850 X2100 X0.04 =0.30 =0.10=84855210

Total a = 1149 sabinsNote: Include air absorption in total for large rooms at frequencies greater than1000 Hz.Compute the reverberation time T using T = 0.05 (V/a)T = 0.05 (Va) = 0.05 X 31,5001149 = 15751149 = 1.37s at 500 Hz

21. Here are the dimensions of a room to use in an example calculation:

[c=ceiling, fl=floor, rs=rightside, ls=leftside, ft=front, bk=back]Begin by calculating the volume of the room in cubic feet:V = (704520) = 63000 ft3Next, identify the room's surfaces that will contribute to the total absorption:Atot= Scac+ Sflafl+ Srsars+ Slsals+ Sftaft+ SbkabkNow calculate each surface area in square feet:Sc= Sfl= (70x45) = 3150Srs= Sls= (70x20) = 1400Sft= Sbk= (45x20) = 900So our expression for total absorption becomes:Atot= (3150)ac+ (3150)afl+ (1400)ars+ (1400)als+ (900)aft+ (900)abkor, more simply:Atot= (3150)( ac+ afl) + (1400)( ars+ als) + (900)( aft+ abk)Now we need absorption coefficients for each surface. Lets choose plaster on lath for the front and side walls, a suspended "acoustic" tile ceiling, heavy drapery covering the back wall, and a carpet on a pad on the floor. Here are the coefficients for those materials for different frequencies. The frequency labels are the center frequencies of octave bands:Material125 Hz250 Hz500 Hz1 kHz2 kHz4 kHz

plaster on lath0.140.100.060.050.040.03

heavy drapes0.140.350.550.720.700.65

carpet on pad0.080.240.570.690.710.73

susp acoust tile0.760.930.830.990.990.99

First, we'll plug in the coefficients for low pitched sounds -- the 125 Hz octave band:Atot= (3150)( .76 + .08) + (1400)( .14 + .14) + (900)( .14 + .14)= (2646) + (392) + (252) = 3290 ft2and the reverberation time, using the constant 0.049 with our foot units, will be:TR, 125 Hz= (0.049)(63000)/(3290) = 0.94 secNow we'll plug in the coefficients for the same materials but for high pitched sounds -- the 2 kHz octave band:Atot= (3150)( .99 + .71) + (1400)( .04 + .04) + (900)( .04 + .70)= (5355) + (112) + (666) = 6133 ft2Giving us a very different reverberation time at these frequencies:TR, 2 kHz= (0.049)(63000)/(6133) = 0.50 sec

22. Refer to the given in problem number 4 and use the Fitzroy Equation.TR= cV((Sx2/Ax) + (Sy2/Ay) + (Sz2/Az))/(Sx+ Sy+ Sz)2For our example, this becomes:= cV[((Sft+ Sbk)2/(Aft+ Abk)) + ((Sls+ Srs)2/(Als+ Ars)) + ((Sc+ Sfl)2/(Ac+ Afl))]/(Sft+ Sbk+ Sls+ Srs+ Sc+ Sfl)2Using the same surface materials as with the Sabine equation above, for the 125 Hz octave band we get:= (.049)(63000)((900+900)2/(126+126)) + (1400+1400)2/(196+196) + (3150+3150)2/(2394+252))/(900+900+1400+1400+3150+3150)2= (.049)(63000)((12857 + 20000 + 15000)/(118810000)TR= 1.24 sec

23. A gymnasium has an internal volume of 90.05 ft3. When it contains 2000 customary sabins of absorption what will be its reverberation time in seconds? Given: Volume (V) = 90.05 ft3 Total absorption (A) = 2000 sabins Find:Reverberation time (RT60)Solution:Using the formula for English system

24. A concert hall contains 3000 customary sabins of absorption (279 metric sabins). Its reverberation time in seconds is 2.23. What is its internal volume in meters?Given: Reverberation time (RT60) = 2.23 secondsTotal absorption (A) = 279 metric sabins Find:Volume (V)Solution:Using the formula for Metric system

25. Calculate the Norris-Eyring reverberation time of uniformly diffuse sound in a live room if the average absorption is 0.8 and the Sabine reverberation time is 1.23 sec.Given: Average absorption (A) = 0.69Sabine Reverberation time (RT60(SABINE)) = 1.23 secFind:Norris-Eyring reverberation time (RT60(NORRIS-EYRING))Solution:Using the relationship between Sabine and Norris-Eyring Equation

26. A classroom 60 ft long by 35 ft wide by 15 ft high has sound absorption coefficients (A) of 0.30 for walls, 0.04 for ceiling, and 0.10 for floor. All A are at 500 Hz. Find the reverberation time (RT60) at 500 Hz in this space with no occupants and no sound-absorbing treatment. Given: Length (l) = 60 ftWidth (w) = 35 ftHeight (h) = 15 ftSound absorption coefficients (A):Wall = 0.30Ceiling = 0.04Floor = 0.10500 HzFind:Reverberation time (RT60)Solution:Compute the room volume (V)V= l X w X h V = 60 X 35 X 15V =31,500 ft3Compute the surface areas (S)SCEILING = l X w SCEILING = 60 X 35SCEILING = 2100 ft2

SWALL = 2(h X w) SWALL = 2(h X l)SWALL = 2 X 15 X 35 SWALL = 2 X 15 X 60 SWALL = 1050 ft2SWALL = 1800 ft2

SFLOOR = l X w SFLOOR = 60 X 35SFLOOR = 2100 ft2Compute the total room absorption a using A= S A S AA (sabins)

CeilingWallsFloor2100 X2850 X2100 X0.04 =0.30 =0.10=84855210

Total A = 1149 sabinsUsing the formula for English system

27. A studio 100 m long by 75 m wide by 5 m high has sound absorption coefficients (A) of 0.23 for walls, 0.08 for ceiling, and 0.15 for floor. Find the reverberation time (RT60) in this space with no occupants and no sound-absorbing treatment. Given: Length (l) = 100 mWidth (w) = 75 mHeight (h) = 5 mSound absorption coefficients (A):Wall = 0.23Ceiling = 0.08Floor = 0.15Find:Reverberation time (RT60)Solution:Compute the room volume (V)V= l X w X h V = 100 X 75 X 5V =37,500 m3Compute the surface areas (S)SCEILING = l X w SCEILING = 100 X 75SCEILING = 7500 m2

SWALL = 2(h X w) SWALL = 2(h X l)SWALL = 2 X 5 X 75 SWALL = 2 X 5 X 100 SWALL = 750 m2SWALL = 1000 m2

SFLOOR = l X w SFLOOR = 100 X 75SFLOOR = 7500 ft2Compute the total room absorption a using A= S A S AA (sabins)

CeilingWallsFloor7500 X1750 X7500 X0.08 =0.23 =0.15=600402.51125

Total A = 2127.5 sabinsUsing the formula for Metric system

28. Same given with problem number 4. Find the reverberation time (RT60) if 50 percent of the ceiling surface (along the perimeter of the room) is treated with acoustical panels at A of 0.85. The central area remains sound-reflecting to help distribute sound energy from lectern end toward rear of the room.Given: Length (l) = 60 ftWidth (w) = 35 ftHeight (h) = 15 ftSound absorption coefficients (A):Wall = 0.30Ceiling = 0.04Floor = 0.10500 HzFind:Reverberation time (RT60)Solution:Compute the total room absorption a using A = S A S ASA (sabins)

Bare ceilingTreated ceilingWallsFloor10501050285021000.040.850.300.1042892855210

Total A = 1999 sabins

Using the formula for English system

29. Calculate the Sabine reverberation time of uniformly diffuse sound in a studio-type room if the average absorption is 0.69 and the Norris-Eyring reverberation time is 2.23 sec.Given: Average absorption (A) = 0.69Sabine Reverberation time (RT60(SABINE)) = 2.23 secFind:Norris-Eyring reverberation time (RT60(NORRIS-EYRING))Solution:Using the relationship between Sabine and Norris-Eyring Equation

30. A cinema has an internal volume of 2300 m3. Its reverberation time in seconds is 1.8. What is its total absorption in metric?Given: Reverberation time (RT60) = 1.8 secondsVolume (V) = 2300 m3Find:Total absorption (A) Solution:Using the formula for Metric system

31. A theatre hall has an internal volume of 5000 m3. Its reverberation time in seconds is 1.3. What is its internal volume in English system?Given: Reverberation time (RT60) = 1.3 secondsVolume (V) = 5000 m3Find:Total absorption (A) Solution:Using the formula for Metric system

32. A theatre hall 5000 m long, 6000 m wide and 4500 m high has a total sound absorption coefficient of 0.98. Find its reverberation time (RT60) using Norris-Eyring Equation.Given: Length (l) = 5000 mWidth (w) = 6000 mHeight (h) = 4500 mTotal sound absorption coefficient (A) = 0.98Find:Reverberation time (RT60) Solution:Using the formula for Metric system

SCEILING = l X w SCEILING = 5000 X 6000SCEILING = 30 X 106 m2

SWALL = 2(h X w) SWALL = 2(h X l)SWALL = 2 X 4500 X 6000SWALL = 2 X 4500 X 5000SWALL = 54 X 106 m2SWALL = 45 X 106 m2

SFLOOR = l X w SFLOOR = 5000 X 6000SFLOOR = 30 X 106 m2

33. A church has an interval volume of 90.05 ft3 (2550 m3). When it contains 2000 customay sabins of absorption (186 metric sabins), what will be its reverberation time in seconds.GIVEN:V=2550 m3 (90.05 ft3)A=186 metric sabins (2000 customay sabins of absorption)REQUIRED:Reverberation time, RT60SOLUTION:RT60= = RT60=2.19 seconds34. An enclosed room has an internal volume of 4000 m3. When it contains 200 metric sabins, what will be its reverberation time in seconds? GIVEN:Volume, V = 4000 m3Total absorption, = 200 metric sabinsREQUIRED: Reverberation time, RT60 SOLUTION: RT60 RT60 = 3.22 seconds35. A church has an internal volume of 5200 ft3. When it contains 1200 customary sabins of absorption, what will be its reverberation time in seconds? GIVEN:Volume, V = 5200 ft3Total absorption, = 1200 customary sabinsREQUIRED: Reverberation time, RT60SOLUTION: RT60 RT60 = 0.2123333 seconds36. Calculate the Norris-Eyering reverberation time of uniformly diffuse sound in a live room if the average absorption is 0.5 and the reverberation time is 2.5 s. GIVEN: Absorption, = 0.5Reverberation time, RT = 2.5 sREQUIRED: Norris-Eyering Reverberation time, RT60(N-E)SOLUTION: RT60(N-E) RT60(N-E) = 3.6067 seconds37. Calculate the reverberation time at 500Hz, for a classroom that is 9m wide by 16m long by 3m high. The floor is with heavy carpet on concrete, the walls are made of plywood sheet, 1/4" on studs, and the ceiling is acoustic tile rigidly mountedGIVEN:Frequency = 500HzWidth of the wall, W = 9m.Length of the wall, L = 16m.Height of the wall, H = 3m.Surface area of floor, Sf = 144 m2Surface area of ceiling, Sc = 144 m2REQUIRED:Reverberation time, RT60SOLUTION:RT60= = V = L x W x H = 16m x 9m x 3m = 432 m3Sw = 2(W x H) + 2(L x H) = 2(9 x 3m) + 2(16m x 3m ) = 150m2RT60 = RT60 = 0.5062 seconds38. An office contains 111.1 metric sabins of absorption. Its reverberation time in seconds is 1.11. What is its internal volume in meters? GIVEN: Total absorption, = 111.1 metric sabinsReverberation time, RT = 1.11 secondsREQUIRED:Volume, VSOLUTION: RT V = V = 765.9689441 m339. Calculate the absorption of uniformly diffuse sound in a live room if the the Norris-Eyering reverberation time is 1.1716 seconds and the reverberation time is 1.23 s. GIVEN: RT60(N-E) = 1.1716 secondsReverberation time, RT = 1.23 sREQUIRED: Absorption, SOLUTION: = 3.857240. An office contains 1567 customary sabins of absorption (238.1 metric sabins). Its reverberation time in seconds is 1.35 . What is its internal volume in meters. ?GIVEN:Total absorption, = 238.1 metric sabinsReverberation time, RT = 1.35 secondsREQUIRED:Volume, VSOLUTION: RT V = V = 1996. 49 m3

41. In a music studio, there is said to be a reverberation time of 2.356seconds, compute for the volume of the room in in ft when the area is 500ft3 and the coefficient of absorption is 229 .GIVEN:Reverberation time, RT = 2.356secondsCoefficient of absorption, = 229Area, A = 500ft3REQUIRED:Volume, VSOLUTION: RT V V = 10982.65 ft342. Find the absorption coefficient of the wall at 500Hz, for a classroom that is 9m wide by 14m long by 3m high. The floor is with heavy carpet on concrete, and the ceiling is acoustic tile rigidly mounted with a 1.23 seconds reverberation time.GIVEN:Frequency = 500HzWidth of the wall, W = 5m.Length of the wall, L = 4m.Height of the wall, H = 3m.Surface area of floor, Sf = 121 m2Surface area of ceiling, Sc = 121 m2Reverberation time, RT = 2.34 secondsREQUIRED:SOLUTION:RT= = = V = L x W x H = 5m x 4m x 3m = 60 m3Sw = 2(W x H) + 2(L x H) = 2(5 x 3m) + 2(4m x 3m ) = 54m2 = = -2.1643

PAMANTASAN NG LUNGSOD NG MAYNILACollege of Engineering and TechnologyDepartment of Electronics EngineeringIntramuros, Manila

Broadcasting and Acoustics:Compilation of Sample Problems in Reverberation Time

Prepared by:BSECE IVSY 2014-2015

Prepared to:Engr. Carlos C. SisonPECE, DEM