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