1. Find reverberation time for a hall of dimensions 40’ x 30’ x 20’ having average absorption coefficient of 0.15.
L=40' ; W =30' ; H =20' ; a´ =0.15
Given:
τ
Required: Solution:
τ=
0.05 V −1 s ft a´ S
Volume of t h e hall ,V =40 × 30× 20=24000 ft
3
Surface area of hall , S=2 [ ( 40 × 30 ) + ( 30 ×20 ) + ( 20 × 40 ) ] ¿ 2 ( 1200+ 600+ 800 )=5200 ft 2 total average absorption , a´ S=0.15 ×5200 ft 2=780 ft 2 OWU ∨sabin
τ=
0.05 × 24000 ft 3 −1 s ft 2 780 ft
τ =1.54 s
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:
a´ =0.20 ; τ=1.2 s
Required:
A
Solution:
tot al average absorption , a´ S=0.15 × ( 5200− A )+ 0.20 A=780+0.05 A
τ=
0.05 × 24000 ft 3 =1.2 (780+ 0.05 A)
0.05 A=
A=
1200 −780=1000−780=220 1.2
220 0.05
A=4400 ft
2
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:
L=15 m; W =8 m ; H =3 m; ´a=0.3
Required:
τ
Solution:
τ=
0.162 V −1 sm a´ S 3
0.162× 360 m τ= s m−1 2 0.3 × 2 [ ( 15 ×8 )+ ( 8× 3 ) + ( 3 ×15 ) ] m
τ=
58.32 s 0.3 × 378
τ =0.51 s 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:
L=15 m; W =8 m ; H =3 m; ´a=0.3
Required:
τ
Solution:
τ=
0.162 V −Sln(1− a´ )
τ=
0.162 ×360 −378 ln ( 1−0.3)
τ=
58.32 378 × 0.357
τ =0.43 s
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 4.7ft 2 OWU), find reverberation time of the hall. Given:
'
'
'
L=70 ; W =40 ; H =15 ; a´ =0.1 ; n=100 τ
Required: Solution:
∑ a 1 S1=0.1× 2 [ ( 70 × 40 ) +( 40 ×15 )+(15 ×70)] +4.7 ×100 ∑ a 1 S1=0.1× 8900+4.7 × 100=890+ 470=1360 ft 2 OWU τ=
0.05 V 0.0 .5× 42000 = 1360 ∑ a1 S1
τ ≅1.54 s
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:
'
'
'
L=80 ; W =40 ; H =20 ; ´a=0.4 ; ∆ τ =0.25 s ; W × H =breadt h=10'
Required:
Curtainlengt h
Solution: Without curtain
τ=
0.05 V 0.05× 64000 ft 3 −1 s ft−1= s ft =1.5 s aS ∑ aS
∑ aS=
3200 2 2 ft =2133 ft (OWU ) 1.5
With curtain of effective area A:
1.25 s=
0.05× 64000 ft 3 −1 s ft ∑ a´ S +0.4 A
1.25 s=
3200 ft s ft−1 2 2133 ft +0.4 A
2
1.25× 2133 ft 2 +1.25 ×0.4 A=3200 ft 2 2
2666 ft +0.5 A=3200 ft
A=
2
534 0.5
A=1068 ft 2 Area includes both sides of the curtain, thus actual curtain area will be half of this area, i.e., 534 ft2 Now
Curtainarea=Lengt h × Breadt h 2
534 ft =Lengt h ×10 ft Curtainlengt h=53.4 ft
7. If a cubical hall of dimension a and average absorption coefficient
a´
time 2s, what will be the reverberation time if its dimension is doubled. Given:
τ =2 s ; dimension a , absorption coefficient , a´
has reverberation
Required:
τ w h en dimensionis doubled
Solution:
τ=
0.05 V 0.05 a3 = =2 s aS a´ (6 a2)
Now ,dimension=2a
Hence ,V =(2 a)3=8 a3∧S=6(2 a)2=4 ×6 a 2
So , τ=
3 0.05 V 0.05(8 a ) = =2× 2 s a´ S a´ (4 × 6 a2 )
τ =4 s , i. e ., it isdoubled
8. If a cubical hall of dimension a and average absorption coefficient
a´
time 2s, what will be the reverberation time if its volume is doubled. Given:
τ =2 s ; dimension a , absorption coefficient , a´
Required:
τ w h en volume is doub led
Solution: 3
τ=
0.05 V 0.05 a = =2 s 2 aS a´ (6 a ) 1 3
3
Now , volume=2V =2 a =(2 × a)3 1
1
2
T h us , dimension ( ¿ side )=2 3 × a , w h ic h gives S=6 (2 3 × a)2=2 3 (6 a2 )
has reverberation
3
So , τ=
1
0.05 V 0.05(2 a ) 3 = 2 =2 ×2 s a´ S 2 a´ 2 3 (6 a )
1
1
τ =2 3 × 2 s , i. e . , 2 3 ׿ 9. Find the reverberation time for a hall of dimensions 50’ x 25’ x 15’ having average absorption coefficient of 0.10. ' ' ' L=50 ; W =25 ; H=15 ; a´ =0.10
Given:
Required:
τ
Solution:
τ=
0.05 V s ft−1 a´ S
Volume of t h e hall ,V =50 ×25 ×15=18750 ft 3 Surface area of hall , S=2 [ ( 50 ×25 ) + ( 25 ×15 )+ ( 15× 50 ) ] ¿ 2 ( 1250+375+ 750 )=4750 ft 2 ¿ tal average absorption , a´ S=0.10 × 4750 ft 2=475 ft 2 OWU ∨sabin
τ=
0.05 ×18750 ft 3 −1 s ft 475 ft 2
τ =1.97 s 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:
L=60' ; W =30 ' ; H=15' ; a´ 1=0.12; a´ 2=0.25 ; τ 2=1.2 s
Required:
A
Solution:
τ=
0.05 V s ft−1 a´ S
For reverberation time
Volume of t h e hall ,V =60 ×30 ×15=27000 ft 3 Surface area of hall , S=2 [ ( 60 ×30 )+ ( 30 ×15 )+ ( 15× 60 ) ] ¿ 2 ( 1800+ 450+900 )=6300 ft 2 2
2
total average absorption , a´ S=0.12× 6300 ft =756 ft OWU ∨sabin
τ 1=
0.05 ×27000 ft 3 −1 s ft 756 ft 2
τ 1=1.79 s For area for reduced reverberation time
total average absorption , a´ S=0.12× ( 6300− A )+ 0.25 A=756+0.13 A 0.05 ×27000 ft 3 τ 2= =1.2 (756+ 0.13 A)
0.13 A=
A=
369 0.13
1350 −756=1000−780=369 1.2
A=2838.46 ft 2 11. A church has an internal volume of 90.05 ft 3 (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: R
0.16V A
R
0.16(2500) 186
R 2.19 sec onds
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: R
0.16V A
V
RA 0.16
V
(2.26)(139.5) 0.16
V=1970.44m 3
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:
R(N-E) =
R ln(1 A)
R(N-E) =
1.75 ln(1 0.8)
R(N E) 1.087s 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:
R(N-E) =
R ln(1 A)
R(N-E) =
1.23 ln(1 0.65)
R(N E) 1.172s 15. A room has an internal volume of 2034 m 3. 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:
R
0.16V A
R
0.16(2034) 111.6
R 2.92 sec onds
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 α = 0.9: T60 = (-0.161V)/(S ln(1-α) = (-0.161 x 42m-3)/(75 m2 x ln (1-0.9) ) = 0.042s (42 x10-3s) For a = 0.2 we get: T60 = (-0.161V)/(S ln(1-α) = (-0.161 x 42m-3)/(75 m2 x ln (1-0.2) ) = 0.43 s which 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= 2 so the old reverberation times are increased by a factor of 2: T60 doubled = T60 X 2 which gives a reverberation time of: T60 doubled = T60 X 2 = 0.042 x 2 = 0.084 s when α = 0.9 and: T60 doubled = T60 X 2 = 0.43 x 2 = 0.86 s 17. Compute the reverberation time T using T = 0.05 (V/a). Solution: T = 0.05 (V÷a) = 0.05 X 31,500÷1149 = 1575÷1149 = 1.37s at 500 Hz
Find the reverberation time T if 50 percent of the ceiling surface (along the perimeter of the room) is treated with acoustical panels at α 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 = Σ S α. S Bare ceiling Treated ceiling
1050x0.04= 1050x0.85=
α
a (sabins) 42 892
Walls Floor Total a =
2850x0.30= 2100x0.10=
855 210 1999 sabins
Compute new reverberation time T. T = 0.05 (V÷a) = 0.05x31,500÷1999 = 1575÷1999 = 0.79s at 500Hz The 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 α 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 areas S. S = 5 x 10 x 10=500 ft^2 of concrete S = 10 X 10 = 100 ft^2 of spray-on material Compute the total room absorption a with spray-on material on the ceiling. a1 = Σ S α = (500 X 0.02) + .(100 X 0.70) = 10 + 700 sabin’ Compute 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 sabin Compute the noise reduction NR. NR= 10log (a2÷a1) = 10log (246/80) = 10log (3.075 X 10^0) = 10(0.4878) = 5dB This 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 provides NR = 10 log (80/12)= 10 log 6.67 = 10(0.8241) = 8 dB
which 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 α 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 = Σ S α = (400X0.85) + (100 X0.70) + (100X0.50) = 340 + 70 + 50 = 460 sabins 2. 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 dB Surfaces Treated ( addition to ceiling)
Room NR (at 500 Hz)
Two walls Four walls and floor
5 dB 8 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 α of 0.30 for walls, 0.04 for ceiling, and 0.10 for floor. All α are at 500 Hz.
Find the reverberation time T at 500 Hz in this space with no occupants and no soundabsorbing treatment. Compute the room volume V. V=60X35X15=31,500ft Compute the surface areas S. Ceiling S =60X35=2100ft Walls S = 2 X 35 X 15 = 1050 ft S = 2X60X 15= 1800ft Floor S = 60 X 35 = 2100 ft
Compute the total room absorption a using a = Σ S α .
Ceiling Walls Floor
S
α
a (sabins)
2100 X 2850 X 2100 X
0.04 = 0.30 = 0.10=
84 855 210
Total a = 1149 sabins Note: Include air absorption in total for large rooms at frequencies greater than 1000 Hz. Compute the reverberation time T using T = 0.05 (V/a) T = 0.05 (V÷a) = 0.05 X 31,500÷1149 = 1575÷1149 = 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 = (70×45×20) = 63000 ft3 Next, identify the room's surfaces that will contribute to the total absorption: Atot = Scac + Sflafl + Srsars + Slsals + Sftaft + Sbkabk Now calculate each surface area in square feet: Sc = Sfl = (70x45) = 3150 Srs = Sls = (70x20) = 1400 Sft = Sbk = (45x20) = 900 So our expression for total absorption becomes: Atot = (3150)ac + (3150)afl + (1400)ars + (1400)als + (900)aft + (900)abk or, 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:
Material
125 250 500 1 2 4 Hz Hz Hz kHz kHz kHz
plaster on lath
0.14 0.10 0.06 0.05 0.04 0.03
heavy drapes
0.14 0.35 0.55 0.72 0.70 0.65
carpet on pad
0.08 0.24 0.57 0.69 0.71 0.73
susp acoust tile
0.76 0.93 0.83 0.99 0.99 0.99
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 ft2 and the reverberation time, using the constant 0.049 with our foot units, will be: TR, 125 Hz= (0.049)(63000)/(3290) = 0.94 sec Now 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 ft2 Giving 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)2 For 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)2 Using 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
RT 60=
0.049 V A
RT 60=
0.049(90.05) 2000 x 10−3
RT 60=2 . 2062 sec 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 seconds Total absorption (A) = 279 metric sabins Find: Volume (V) Solution: Using the formula for Metric system
RT 60=
0.161V A
V=
RT 60 × A 0.161
V=
2.23 ×279 0.161
V =3864 . 4099m
3
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) = 0.69 Sabine Reverberation time (RT60(SABINE)) = 1.23 sec Find: Norris-Eyring reverberation time (RT60(NORRIS-EYRING)) Solution: Using the relationship between Sabine and Norris-Eyring Equation
RT 60( NORRIS−EYRING )= RT 60( NORRIS−EYRING )=
RT 60(SABINE ) −ln ( 1− A) 1.23 sec −ln ( 1−0.69)
RT 60( NORRIS−EYRING )=1 .0502 sec
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 ft Width (w) = 35 ft Height (h) = 15 ft Sound absorption coefficients (A): Wall = 0.30 Ceiling = 0.04 Floor = 0.10 500 Hz Find: Reverberation time (RT60) Solution: Compute the room volume (V) V= l X w X h V = 60 X 35 X 15 V =31,500 ft3 Compute the surface areas (S) SCEILING = l X w SCEILING = 60 X 35 SCEILING = 2100 ft2 SWALL = 2(h X w) SWALL = 2 X 15 X 35 SWALL = 1050 ft2
SWALL = 2(h X l) SWALL = 2 X 15 X 60 SWALL = 1800 ft2
SFLOOR = l X w SFLOOR = 60 X 35 SFLOOR = 2100 ft2 Compute the total room absorption a using A= Σ S A
S
A
A (sabins)
Ceiling
2100 X
0.04 =
84
Walls
2850 X
0.30 =
855
Floor
2100 X
0.10=
210
Total A = 1149 sabins Using the formula for English system
RT 60=
0.049 V A
RT 60=
0.049(31500) 1149
RT 60=1 .3433 sec at 500 Hz 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 m Width (w) = 75 m Height (h) = 5 m Sound absorption coefficients (A): Wall = 0.23 Ceiling = 0.08 Floor = 0.15 Find: Reverberation time (RT60) Solution: Compute the room volume (V) V= l X w X h V = 100 X 75 X 5 V =37,500 m3 Compute the surface areas (S) SCEILING = l X w SCEILING = 100 X 75 SCEILING = 7500 m2 SWALL = 2(h X w) SWALL = 2 X 5 X 75 SWALL = 750 m2
SWALL = 2(h X l) SWALL = 2 X 5 X 100 SWALL = 1000 m2
SFLOOR = l X w SFLOOR = 100 X 75 SFLOOR = 7500 ft2 Compute the total room absorption a using A= Σ S A
S
A
A (sabins)
Ceiling
7500 X
0.08 =
600
Walls
1750 X
0.23 =
402.5
Floor
7500 X
0.15=
1125
Total A = 2127.5 sabins Using the formula for Metric system
RT 60=
0.161V A
RT 60=
0.161(37500) 2127.5
RT 60=2 . 8378 sec 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 ft Width (w) = 35 ft Height (h) = 15 ft Sound absorption coefficients (A): Wall = 0.30 Ceiling = 0.04 Floor = 0.10 500 Hz Find: Reverberation time (RT60) Solution: Compute the total room absorption a using A = Σ S A
Bare ceiling
S
A
SA (sabins)
1050
0.04
42
Treated ceiling
1050
0.85
892
Walls
2850
0.30
855
Floor
2100
0.10
210
Total A =
1999 sabins
Using the formula for English system
RT 60=
0.049 V A
RT 60=
0.049(31500) 1999
RT 60=0 . 7721 sec at 500 Hz 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) = 0.69 Sabine Reverberation time (RT60(SABINE)) = 2.23 sec Find: Norris-Eyring reverberation time (RT60(NORRIS-EYRING)) Solution: Using the relationship between Sabine and Norris-Eyring Equation
RT 60( NORRIS−EYRING )=
RT 60(SABINE ) −ln ( 1− A)
RT 60(SABINE )=RT 60(NORRIS− EYRING) ×−ln ( 1− A) RT 60(SABINE )=2.23 sec ×−ln ( 1−0.69)
RT 60(SABINE )=2 . 6117 sec 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 seconds Volume (V) = 2300 m3 Find: Total absorption (A)
Solution: Using the formula for Metric system
RT 60=
0.161V A
A=
0.161 V RT 60
A=
0.161 ×2300 1.8
A=205. 7222 metric sabins 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 seconds Volume (V) = 5000 m3 Find: Total absorption (A) Solution: Using the formula for Metric system
RT 60=
0.161V A
A=
0.161 V RT 60
A=
0.161 ×5000 1.3
A=619.2308metric sabins ×
10.7527 sabin 1 metric sabin
A=6658. 4030 sabin 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 m Width (w) = 6000 m Height (h) = 4500 m Total sound absorption coefficient (A) = 0.98 Find: Reverberation time (RT60) Solution: Using the formula for Metric system
RT 60=
0.161 V −S ln ( 1− A)
V =l × w× h
V =5000 ×6000 × 4500 V =1.35 ×1011 m3 SCEILING = l X w SCEILING = 5000 X 6000 SCEILING = 30 X 106 m2 SWALL = 2(h X w) SWALL = 2 X 4500 X 6000 SWALL = 54 X 106 m2
SWALL = 2(h X l) SWALL = 2 X 4500 X 5000 SWALL = 45 X 106 m2
SFLOOR = l X w SFLOOR = 5000 X 6000 SFLOOR = 30 X 106 m2 11
0.161(1.35 ×10 ) RT 60= 6 −( 159× 10 ) ln (1−.75) RT 60=34 . 9431 seconds 33. A church has an interval volume of 90.05 ft 3 (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, RT60 SOLUTION: RT60=
0.161 V A
=
0.161(2550) 186
RT60=2.19 seconds 34. An enclosed room has an internal volume of 4000 m 3. When it contains 200 metric sabins, what will be its reverberation time in seconds? GIVEN: Volume, V = 4000 m3
Total absorption,
α
= 200 metric sabins
REQUIRED: Reverberation time, RT60 SOLUTION: RT60
¿
0.161 V 0.161(4000) = α 200
RT60 = 3.22 seconds 35. A church has an internal volume of 5200 ft 3. When it contains 1200 customary sabins of absorption, what will be its reverberation time in seconds? GIVEN: Volume, V = 5200 ft3 Total absorption,
α
= 1200 customary sabins
REQUIRED: Reverberation time, RT60 SOLUTION: RT60
¿
0.049 V 0.049(5200) = α 1200
RT60 = 0.2123333 seconds 36. 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:
α
Absorption,
= 0.5
Reverberation time, RT = 2.5 s REQUIRED: Norris-Eyering Reverberation time, RT60(N-E) SOLUTION: RT60(N-E)
¿
RT 2.5 = −ln( 1−α ) −ln(1−0.5)
RT60(N-E) = 3.6067 seconds
37. 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 mounted GIVEN: Frequency = 500Hz Width of the wall, W = 9m. Length of the wall, L = 16m. Height of the wall, H = 3m. Surface area of floor, Sf = 144 m2 Surface area of ceiling, Sc = 144 m2
Absorption coefficient of wall , α w =0.1 Absorption coefficient of floor , α f =0.15 Absorption coefficient of ceiling, α c =0.7
REQUIRED: Reverberation time, RT60 SOLUTION:
0.161 V ∑ (S)(α )
RT60=
=
0.161 V (S f × α f )+ ( S c × α c ) +(S w × α w )
V = L x W x H = 16m x 9m x 3m = 432 m3 Sw = 2(W x H) + 2(L x H) = 2(9 x 3m) + 2(16m x 3m ) = 150m2
¿
RT60 38.
0.161( 432) 69.552 = ( 144 x 0.15 ) + ( 144 x 0.7 ) +(150 x 0.1) 85.9
=
RT60 = 0.5062 seconds 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,
α
= 111.1 metric sabins
Reverberation time, RT = 1.11 seconds REQUIRED: Volume, V SOLUTION: RT
V=
¿
0.161 V α
α ( RT ) 111.1(1.11) = 0.161 0.161
V = 765.9689441 m3 39. 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 seconds Reverberation time, RT = 1.23 s REQUIRED: Absorption,
α
SOLUTION:
α
¿ 1+(e
α
= 3.8572
RT RT ( N −E )
)=1+(e
1.23 1.1716
)
40. 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,
α
= 238.1 metric sabins
Reverberation time, RT = 1.35 seconds REQUIRED: Volume, V SOLUTION: RT
¿
V=
0.161 V α
α ( RT ) 238.1(1.35) = 0.161 0.161
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 500ft 3 and the coefficient of absorption is 229 . GIVEN: Reverberation time, RT = 2.356seconds Coefficient of absorption, Area, A = 500ft3 REQUIRED: Volume, V SOLUTION:
α
= 229
RT
V
¿
0.049 V α
α ( RT ) 229(2.35) = 0.049 0.049
V = 10982.65 ft3 42. 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 = 500Hz Width of the wall, W = 5m. Length of the wall, L = 4m. Height of the wall, H = 3m. Surface area of floor, Sf = 121 m2 Surface area of ceiling, Sc = 121 m2
Absorption coefficient of floor , α f =0.2 Absorption coefficient of ceiling, α c =0.8
Reverberation time, RT = 2.34 seconds REQUIRED:
Absorption coefficient of wall , α w
SOLUTION: RT=
αw
0.161 V ∑ (S)(α )
=
=
0.161 V (S f × α f )+ ( S c × α c ) +(S w × α w )
S (¿ ¿ c × α c ) 0.161V −( S f ×α f ) −¿ RT ¿ ¿ ¿
V = L x W x H = 5m x 4m x 3m = 60 m3 Sw = 2(W x H) + 2(L x H) = 2(5 x 3m) + 2(4m x 3m ) = 54m2
αw
=
[
0.161( 60) −( 121 ×0.2 ) −(121 ×0.8) 2.34 138
α w = -2.1643
]
PAMANTASAN NG LUNGSOD NG MAYNILA College of Engineering and Technology Department of Electronics Engineering Intramuros, Manila
Broadcasting and Acoustics: Compilation of Sample Problems in Reverberation Time
Prepared by: BSECE IV SY 2014-2015 Prepared to: Engr. Carlos C. Sison PECE, DEM
