Reverberation Time Problems (Solved)

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

L=40

'

;W =30

'

; H =20

'

; ´a=0.15

Required:

τ

Solution:

τ =

0.05 V

´

a S

s ft

−1

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

totalaverage absorption , ´a S=0.15 ×5200 ft

2

_{=780 ft}

2

_{OWU ∨sabin}

τ =

0.05 × 24000 ft

3

780 ft

2

s ft

−1

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:

τ =

0.05 × 24000 ft

3

(780+0.05 A)

=1.2

0.05 A=

1200

1.2

−780=1000−780=220

A=

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

´

a S

s m

−1

τ =

0.162× 360 m

3

0.3 × 2

[

(

15 ×8 )+(8× 3)+(3 ×15)

]

m

2

s m

−1

τ =

58.32

0.3 × 378

s

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

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 4.7ft2 _{OWU), find} reverberation time of the hall.

Given:

L=70

'

;W=40

'

; H =15

'

; ´a=0.1 ;n=100

Required:

τ

Solution:

∑

a

1

S

1

=0.1× 2

[

(

70 × 40)+(40 ×15 )+(15 ×70)

]

+4.7 ×100

∑

a

1

S

1

=0.1× 8900+4.7 × 100=890+470=1360 ft

2

OWU

τ =

0.05 V

∑

a

₁

S

₁

=

0.0 .5× 42000

1360

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

aS

s ft

−1

=

0.05× 64000 ft

3

∑

aS

s ft

−1

=1.5 s

∑

aS=

3200

1.5

ft

2

=2133 ft

2

(OWU )

With curtain of effective area A:

1.25 s=

0.05× 64000 ft

3

∑

a S +0.4 A

´

s ft

−1

1.25 s=

3200 ft

2

2133 ft

2

+0.4 A

s ft

−1

1.25× 2133 ft

2

_{+1.25 ×0.4 A=3200 ft}

2

2666 ft

2

_{+0.5 A=3200 ft}

2

A=

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

534 ft

2

=Lengt h ×10 ft

Curtainlengt h=53.4 ft

7. If a cubical hall of dimension a and average absorption coefficient

a

´

has reverberation time 2s, what will be the reverberation time if its dimension is doubled.

Required: τ w h en dimensionis doubled Solution:

τ =

0.05 V

aS

=

0.05 a

3

´

a(6 a

2

)

=2 s

Now ,dimension=2a

Hence ,V =(2 a)

3

=

8 a

3

∧S=6(2 a)

2

=4 ×6 a

2

So , τ=

0.05 V

´

a S

=

0.05(8 a

3

)

´

a (4 × 6 a

2

)

=2× 2 s

τ =4 s , i. e ., it isdoubled

8. If a cubical hall of dimension a and average absorption coefficient a´ has reverberation 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:

τ =

0.05 V

aS

=

0.05 a

3

´

a(6 a

2

)

=2 s

Now , volume=2V =2 a

3

=(2

1 3

_{× a)}

3

T h us , dimension (

¿

side)=2

1 3

_{× a , w h ic h gives S=6 (2}

1 3

_{× a)}

2

=2

2 3

_{(6 a}

2

)

So , τ=

0.05 V

´

a S

=

0.05(2 a

3

)

´

a 2

2 3

_{(6 a}

2

₎

=2

1 3

_{×2 s}

τ =2

1 3

_{× 2 s , i. e . , 2}

1 3

_×

_¿

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

L=50

'

;W=25

'

; H=15

'

; ´a=0.10

Required: τ Solution:

τ =

0.05 V

´

a S

s ft

−1

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

475 ft

2

s ft

−1

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.

Required: A Solution:

τ =

0.05 V

´

a S

s ft

−1

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

totalaverage absorption , ´a S=0.12× 6300 ft

2

=756 ft

2

OWU ∨sabin

τ

₁

=

0.05 ×27000 ft

3

756 ft

2

s ft

−1

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

τ

₂

=

0.05 ×27000 ft

3

(756+0.13 A)

=1.2

0.13 A=

1350

1.2

−756=1000−780=369

A=

369

0.13

A=2838.46 ft

2

11. A church has an internal volume of 90.05 ft3 _{(2550 m}3_{). 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 m}3

Total absorption, A = 186 metric sabins REQUIRED: Reverberation time, R = ? SOLUTION: 0.16V R A  0.16(2500) R 186  R 2.19sec 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: 0.16V R A  RA V 0.16  (2.26)(139.5) V 0.16  3

V=1970.44m

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

R(N-E) =

ln(1 A)





1.75

R(N-E) =

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

R(N-E) =

ln(1 A)





1.23

R(N-E) =

ln(1 0.65)





R(N E) 1.172s





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 = ?

0.16V R A  0.16(2034) R 111.6  R 2.92sec 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 α a (sabins) Bare ceiling Treated ceiling 1050x0.04= 1050x0.85= 42 892

Walls Floor 2850x0.30= 2100x0.10= 855 210 Total a = 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

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 sound-absorbing 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

Compute the total room absorption a using a = Σ S α . S α a (sabins) Ceiling Walls Floor 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

Material 125 Hz 250 Hz 500 Hz 1 kHz 2 kHz 4 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:

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(SABINE )

−ln ⁡(1− A)

RT

_{60( NORRIS−EYRING )}

=

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(h X l) SWALL = 2 X 15 X 35 SWALL = 2 X 15 X 60 SWALL = 1050 ft2 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 Walls Floor 2100 X 2850 X 2100 X 0.04 = 0.30 = 0.10= 84 855 210

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(h X l) SWALL = 2 X 5 X 75 SWALL = 2 X 5 X 100 SWALL = 750 m2 SWALL = 1000 m2 SFLOOR = l X w SFLOOR = 100 X 75 SFLOOR = 7500 ft2

S A A (sabins) Ceiling Walls Floor 7500 X 1750 X 7500 X 0.08 = 0.23 = 0.15= 600 402.5 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

S A SA (sabins)

Treated ceiling Walls Floor 1050 2850 2100 0.85 0.30 0.10 892 855 210 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

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:

RT

₆₀

=

0.161 V

−S ln ⁡(1− A)

V =l × w× h

V =5000 ×6000 × 4500

V =1.35 ×10

11

m

3 SCEILING = l X w SCEILING = 5000 X 6000 SCEILING = 30 X 106 m2 SWALL = 2(h X w) SWALL = 2(h X l) SWALL = 2 X 4500 X 6000 SWALL = 2 X 4500 X 5000 SWALL = 54 X 106 m2 SWALL = 45 X 106 m2 SFLOOR = l X w SFLOOR = 5000 X 6000 SFLOOR = 30 X 106 m2

RT

₆₀

=

0.161(1.35 ×10

11

)

−

(

159× 10

6

)

ln ⁡(1−.75)

RT

60

=34 . 9431 seconds

33. A church has an interval volume of 90.05 ft3 _{(2550 m}3_{). 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 ft}3₎

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 m3_{. When it contains 200 metric sabins,} what will be its reverberation time in seconds?

GIVEN:

Total absorption, α = 200 metric sabins REQUIRED: Reverberation time, RT60 SOLUTION: RT60

¿

0.161 V

α

=

0.161(4000)

200

RT60 = 3.22 seconds

35. 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 ft3

Total absorption,

α

= 1200 customary sabins REQUIRED: Reverberation time, RT60 SOLUTION: RT60

¿

0.049 V

α

=

0.049(5200)

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:

Absorption,

α

= 0.5 Reverberation time, RT = 2.5 s REQUIRED:

Norris-Eyering Reverberation time, RT60(N-E) SOLUTION: RT60(N-E)

¿

RT

−ln(1−α)

=

2.5

−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: RT60=

0.161 V

∑

(

S)(α)

=

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

¿

0.161(432)

(

144 x 0.15)+(144 x 0.7)+(150 x 0.1)

=

69.552

85.9

=

RT60 = 0.5062 seconds

38. _{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

¿

0.161 V

_α

V =

α(RT )

0.161

=

111.1(1.11)

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

RT(RTN −E)

_)=1+(e

1.23 1.1716

₎

α _{= 3.8572}

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

¿

0.161 V

_α

V =

α(RT )

0.161

=

238.1(1.35)

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 500ft3 _{and the coefficient of absorption is 229 .} GIVEN: Reverberation time, RT = 2.356seconds

Coefficient of absorption,

α

= 229 Area, A = 500ft3

REQUIRED: Volume, V SOLUTION:

RT

¿

0.049 V

_α

V

α(RT )

_0.049

=

229(2.35)

_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= 0.161 V

∑

(S)(α)

=

0.161 V

(S

_f

× α

_f

)+

₍

S

_c

× α

_c)

+(S

_w

× α

_w

)

α

_w

₌

S

(

¿¿

c × α

c

)

0.161V

RT

−

(

S

f

×α

f

)

−

¿

¿

¿

¿

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

α

w

₌

[

0.161(60)

2.34

−(

121 ×0.2)−(121 ×0.8)

]

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