WEBVTT
Kind: captions
Language: en
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.
So, we were continuing with lecture 11 and
00:00:23.321 --> 00:00:32.700
lecture 12 where we elaborated on Acoustical
Absorbers, but what I did not touch upon is
00:00:32.700 --> 00:00:36.270
where you are placing is also very important.
00:00:36.270 --> 00:00:44.489
So, we have applications of these absorbers
in different areas; it can be a big hall,
00:00:44.489 --> 00:00:49.760
it can be a small classroom, it can be a small
very small recording room, it can be a very
00:00:49.760 --> 00:00:57.269
big auditorium, it may be medium sized lecture
hall we have given you some examples.
00:00:57.269 --> 00:01:04.269
But which is more effective putting it on
the ceiling, putting it on the wall we have
00:01:04.269 --> 00:01:12.060
seen the phenomena of reflection how reflection
helps, to what extent it helps we have seen
00:01:12.060 --> 00:01:20.030
that further is the point the sound travels
and comes back takes creates delay and these
00:01:20.030 --> 00:01:22.320
long delays are to be trapped in.
00:01:22.320 --> 00:01:31.930
So, we have a back wall, we have a ceiling
far, we have side walls ends where it is more
00:01:31.930 --> 00:01:37.040
effective where to put absorbers.
00:01:37.040 --> 00:01:43.299
Obviously, we will put absorbers so that it
does not disturb or disturb the quality of
00:01:43.299 --> 00:01:44.299
sound.
00:01:44.299 --> 00:01:52.649
So, we will have to put absorbers in the room
after having an understanding whether ceilings
00:01:52.649 --> 00:01:59.770
are better, whether walls are better, whether
50 percent ceiling 50 percent wall is better,
00:01:59.770 --> 00:02:04.840
what is the volume of the room whether volume
has an effect of it effect on it because all
00:02:04.840 --> 00:02:10.670
this things are all inbuilt in the equations
which were taught by Professor Bhattacharya.
00:02:10.670 --> 00:02:21.950
So, will see today and you will also calculate
with what I present; so, that you can also
00:02:21.950 --> 00:02:23.790
make your own understanding.
00:02:23.790 --> 00:02:31.769
So, deriving the relationship of noise reduction
and application of absorbers in ceiling versus
00:02:31.769 --> 00:02:38.760
wall for different volume of spaces is today's
concern.
00:02:38.760 --> 00:02:45.370
So, we will try to generate the curves to
plot sound reduction; how it reduces the sound
00:02:45.370 --> 00:02:54.849
in a enclosure ok . So, let us proceed.
00:02:54.849 --> 00:03:01.299
We consider we are going with a very simple
thing, we are considering a 10 feet by 10
00:03:01.299 --> 00:03:08.420
feet by 10 feet room; when all the surfaces
are plastered with coefficient of absorption
00:03:08.420 --> 00:03:12.180
around 0.02.
00:03:12.180 --> 00:03:20.430
And hence the total absorption if we consider
even the floor is 600 that is 10 into 10 each
00:03:20.430 --> 00:03:24.830
of the faces, there are 6 faces of this particular
cube.
00:03:24.830 --> 00:03:33.370
And we get in 2.02 that much of sabine that
is 12 sabines is getting absorbed.
00:03:33.370 --> 00:03:40.200
All other sound produced in this particular
space that is 10 cross 10 cross 10; apart
00:03:40.200 --> 00:03:50.080
from 12 sabine is reflected back, this is
our test condition that is our reference.
00:03:50.080 --> 00:04:00.790
And then that all surfaces remain plastered
that is that 10 cross 10 cross 10 room.
00:04:00.790 --> 00:04:11.270
So, the walls are and the floor all are free
and only the ceiling is having absorber and
00:04:11.270 --> 00:04:23.380
with 0.7 as its coefficient that is the NRC
value that is the alpha is 0.7 for this figure
00:04:23.380 --> 00:04:25.360
2, it is written as alpha 2.
00:04:25.360 --> 00:04:33.360
So, that total absorption A 2 is 500 into
that is one of the faces that in the ceiling
00:04:33.360 --> 00:04:42.680
is left out and all other 5 faces are reflecting
are absorbing 500 into that 0.02 units.
00:04:42.680 --> 00:04:52.630
And the ceiling is observing quite a lot that
is 100 into that is 10 cross 10 that is the
00:04:52.630 --> 00:04:59.230
ceiling area into 0.7; the unit is sabine.
00:04:59.230 --> 00:05:08.230
So, it is absorbing seventy sabine the ceiling
where as the all other 5 walls are absorbing
00:05:08.230 --> 00:05:11.569
80 sabine 10 sabine.
00:05:11.569 --> 00:05:18.530
So, it is total into 80 sabine and what is
the comparison between these 2 rooms, what
00:05:18.530 --> 00:05:26.690
will be the noise reduction once you are producing
a sound here and producing a sound here what
00:05:26.690 --> 00:05:30.690
will be the reduction of the sound intensity
between the 2 rooms?
00:05:30.690 --> 00:05:38.199
It is given by the formula 10 log total absorption
of A 2 divided by the total absorption of
00:05:38.199 --> 00:05:42.789
a one these has been discussed at link in
earlier classes.
00:05:42.789 --> 00:05:55.830
So, it is 10 log A 2 is 80 divided by 12 you
can calculate it out it will be 0; 6.66 something
00:05:55.830 --> 00:06:05.759
and log of that will be around 0.8 8 and in
10 times of it will be 8 decimal.
00:06:05.759 --> 00:06:13.789
So, the noise reduction whatever sound is
produced here and whatever sound is produced
00:06:13.789 --> 00:06:24.010
in this figure 2, with this is as our reference
the first figure as our reference I 0 that
00:06:24.010 --> 00:06:33.820
is intensity at 0 that is figure 1 and figure
2 we get a reduction of sound by 8 decibels
00:06:33.820 --> 00:06:43.750
in case of figure 2 .
Now, we will put in another case when all
00:06:43.750 --> 00:06:57.470
surfaces are treated with acoustical tiles
of coefficient of absorptions say 0.7 and
00:06:57.470 --> 00:07:01.520
floor is only having a absorption of 0.02.
00:07:01.520 --> 00:07:09.860
So, after leaving out the floor we have covered
all the walls along with the ceiling with
00:07:09.860 --> 00:07:11.849
absorbers.
00:07:11.849 --> 00:07:18.759
And we see here A 3 that is the total unit
of absorption which is 500 that is the area
00:07:18.759 --> 00:07:26.629
into 0.7 and the floor is left free that is
point that is 100 that is 10 cross 10 into
00:07:26.629 --> 00:07:32.650
0.02 that comes out to be 352 sabines.
00:07:32.650 --> 00:07:39.020
And again we compare it with our reference
here the noise reduction between figure 3
00:07:39.020 --> 00:07:48.479
and figure 1 is 10 log A 3 by A 1 that is
that will come out to be 15 decimals.
00:07:48.479 --> 00:07:59.830
So, if you consider the previous case figure
2; by covering the ceiling only you had a
00:07:59.830 --> 00:08:13.050
reduction of 8 decibels, when you are covering
all sides; all walls leaving only the floor
00:08:13.050 --> 00:08:23.129
you are getting a reduction of 15 dB; so,
where so 4 of the walls together could
00:08:23.129 --> 00:08:31.590
make a difference of 7 dB .
So, the inference is when the ceiling only
00:08:31.590 --> 00:08:37.170
covered by absorber there is 80 dB reduction,
when the ceiling and one wall say the back
00:08:37.170 --> 00:08:44.160
wall is covered by absorber if you again workout
another step, another case it is only one
00:08:44.160 --> 00:08:51.100
wall that is ceiling and the back wall is
covered by absorber; you will see 11 dB reduction
00:08:51.100 --> 00:08:52.960
with the test case.
00:08:52.960 --> 00:08:58.510
When ceiling and all walls are absorbers are
covered by a absorbers leaving the floor only,
00:08:58.510 --> 00:09:07.610
you will see it is a 15 dB reduction .
So, is it that simple?
00:09:07.610 --> 00:09:15.030
That only ceiling is effective only ceiling
and one wall is also is little more effective
00:09:15.030 --> 00:09:23.060
and covering all will be further effective
and we will keep on trying iterating.
00:09:23.060 --> 00:09:32.890
We also did not consider the volume; here
it is 10 cross 10 cross 10 and if we think
00:09:32.890 --> 00:09:43.570
that it increases or it decreases, the volume
also changes and the surface areas also changes.
00:09:43.570 --> 00:09:49.950
So, we cannot confine ourselves within a 10
cross 10 cross 10; we have to think beyond
00:09:49.950 --> 00:09:50.950
it.
00:09:50.950 --> 00:09:58.500
So, this was the starting and the if we assume
that the ceiling height in all our cases is
00:09:58.500 --> 00:10:08.260
10 and absorbers which are applied all are
having 0.7 as their coefficient, we will try
00:10:08.260 --> 00:10:20.200
to generate the craft how ceiling and the
wall absorption behaves in different volumes
00:10:20.200 --> 00:10:24.080
of room.
00:10:24.080 --> 00:10:31.650
And what is the ratio between the wall area
and the ceiling area considering that also
00:10:31.650 --> 00:10:38.560
we can make our understanding so, that we
can make our choices.
00:10:38.560 --> 00:10:49.640
So, you can actually write down this table
for your purpose and we will be filling it
00:10:49.640 --> 00:10:57.320
up with your help; maybe you are virtually
connected to us, but you just check whether
00:10:57.320 --> 00:11:03.270
you are doing the right whether your calculation
is right maybe I can also put in some wrong
00:11:03.270 --> 00:11:07.090
data you can inform, you that the system is
that friendly.
00:11:07.090 --> 00:11:16.940
So, we have on the first column the room size
in feet length, breadth and height we are
00:11:16.940 --> 00:11:17.940
reporting.
00:11:17.940 --> 00:11:24.180
The wall area that is the summation of the
all the walls we are not considering the floor
00:11:24.180 --> 00:11:28.590
that is the base in our case in this calculation.
00:11:28.590 --> 00:11:38.960
So, we have 4 walls we have one ceiling we
have absorption reported with no absorber
00:11:38.960 --> 00:11:47.760
that was that is our reference that is alpha
is 0.02; that is very minimal which is our
00:11:47.760 --> 00:11:50.540
start or our reference level.
00:11:50.540 --> 00:11:59.670
And wall absorbers when walls having absorbers
when alpha is 0.7; ceiling with absorbers
00:11:59.670 --> 00:12:07.090
when alpha is 0.7 same., we have the wall
to ceiling ratio.
00:12:07.090 --> 00:12:17.860
Then the noise reduction in decimals with
only wall absorption, next column is on reduction
00:12:17.860 --> 00:12:26.990
in decibels with ceiling absorbers and then
total noise reduction that is considering
00:12:26.990 --> 00:12:28.630
ceiling as well as the noise.
00:12:28.630 --> 00:12:36.620
So, if you cover all the ceiling all the all
the wall then what is the where do we stand
00:12:36.620 --> 00:12:38.290
for different volumes.
00:12:38.290 --> 00:12:48.130
So, we start with the first we have 10 cross
10 cross 10; as we had already shown the wall
00:12:48.130 --> 00:12:50.440
area without the floor area is 400.
00:12:50.440 --> 00:12:58.870
So, 10 into 10 into 4 walls height of the
entire calculation height is constant as 10
00:12:58.870 --> 00:13:14.000
feet . So, the absorption with no absorbers
it is 50 into 0.02 that is 10 10 sabine with
00:13:14.000 --> 00:13:23.770
walls having absorbers of 0.7 and ceiling
with no absorber that is 0.02, you will come
00:13:23.770 --> 00:13:36.190
across 400 into 0.7 plus 0.02
00:13:36.190 --> 00:13:37.570
into 100.
00:13:37.570 --> 00:13:49.900
So, hopefully you all have got this 282 units;
similarly with ceiling only that is 100 into
00:13:49.900 --> 00:14:01.880
0.7 plus 400 into 0.02, you will get 70 plus
8 that is 78 sabine.
00:14:01.880 --> 00:14:08.480
In this case, the wall to ceiling ratio is
4 that is 400 divided by 100 that is 4; we
00:14:08.480 --> 00:14:13.930
have the first second and third column the
wall area and the ceiling area; so, the ratio
00:14:13.930 --> 00:14:16.080
is 4.
00:14:16.080 --> 00:14:27.620
And noise reduction in dB with wall absorption
only is 14 dB, reduction in dB with ceiling
00:14:27.620 --> 00:14:36.610
absorption is 9 dB, it varies from the previous
one a little bit because of we have omitted
00:14:36.610 --> 00:14:42.350
the floor here .
So, the what is the total noise reduction
00:14:42.350 --> 00:14:52.050
considered in this 500 square feet of absorber
of absorber with 0.7?
00:14:52.050 --> 00:15:06.560
You will see it will come around 16 dB, it
will be 10 log 36 . So, hopefully you have
00:15:06.560 --> 00:15:08.941
understood what we are trying to do.
00:15:08.941 --> 00:15:16.570
So, we have here we see our volume is also
reported in the first column.
00:15:16.570 --> 00:15:23.980
So, we have for the first case the volume
is 1000, in the next case it is 2250, where
00:15:23.980 --> 00:15:29.320
we get the wall area as 600, ceiling area
as 225.
00:15:29.320 --> 00:15:38.940
And if you can calculate it out; you will
get the absorption with no absorber as 16.5
00:15:38.940 --> 00:15:50.440
which is the reference, then with wall absorbers
only 424.5, next is 169.5.
00:15:50.440 --> 00:16:04.960
And the wall to ceiling ratio is 600 divided
by 225 that is 2.66 and it gives 14 dB of
00:16:04.960 --> 00:16:14.240
noise reduction and 10 dB of noise reduction,
when it is ceiling.
00:16:14.240 --> 00:16:24.130
And if you total it out and find out for both
the ceiling and the floor; that is summing
00:16:24.130 --> 00:16:44.541
these 2, summing these 2 into point 0.7, you
will get and then dividing it with 16.5 you
00:16:44.541 --> 00:16:54.500
will again get 16 dB.
00:16:54.500 --> 00:17:04.790
If you go for 20 into 20 into 10; the volume
is 4000 cubic feet; so, we get wall area and
00:17:04.790 --> 00:17:08.760
ceiling area as 800 and 400.
00:17:08.760 --> 00:17:20.850
So, the ceiling has gone bigger
and the absorption with no absorber that is
00:17:20.850 --> 00:17:24.610
24 sabine.
00:17:24.610 --> 00:17:32.630
And with wall as absorber you see 296, with
ceiling as absorber you will get 568 you must
00:17:32.630 --> 00:17:41.910
also account for the wall absorption of 0.02.
00:17:41.910 --> 00:17:52.230
And then report this value and see the wall
to ceiling ratio is just 2 and the noise reduction
00:17:52.230 --> 00:18:01.890
coefficient; noise reduction sorry noise reduction
in decibel is 11 dB.
00:18:01.890 --> 00:18:13.440
Whereas, it has changed with the ceiling absorption
it has gone high to 14 dB, but the total noise
00:18:13.440 --> 00:18:23.890
reduction with ceiling and floor is still
16; I am reporting the value 16 it may be
00:18:23.890 --> 00:18:32.920
15.5, it maybe 15.3 in that case I am reporting
15; otherwise I am reporting if it is more
00:18:32.920 --> 00:18:38.190
than 15.5, 15.7; I am reporting it at as 16.
00:18:38.190 --> 00:18:46.050
So, you may find it exact values can be written,
but you know in decibels we do vomit the beyond
00:18:46.050 --> 00:18:50.130
the first decimal place.
00:18:50.130 --> 00:19:03.110
So, if we have 25 into 25 into 10; we see
1000 is the wall area that is 25 into 10;
00:19:03.110 --> 00:19:29.660
250 into 4 walls
00:19:29.660 --> 00:19:42.720
that makes 1000 and you have 25 cross 25 as
your ceiling that makes 625; that is the ceiling
00:19:42.720 --> 00:19:48.420
is 25 cross 25 whereas, your box is height
is 10.
00:19:48.420 --> 00:20:04.270
So, in that case it is like this what comes
next is; for this particular area with 0.02
00:20:04.270 --> 00:20:19.670
as absorption coefficient, you get 32.5 that
is that is oh oh that is 1000 plus 625 that
00:20:19.670 --> 00:20:29.190
is 16 16; 25 into 0.02 that make the 32.5,
then with only absorber for the walls that
00:20:29.190 --> 00:20:41.110
is 1000 into 0.7 plus 625 into 0.02, you get
712.5 sabines.
00:20:41.110 --> 00:21:03.370
For the ceiling, you get 625 into 0.7 625
into 0.7 plus 1000 into 0.02 that gives you
00:21:03.370 --> 00:21:16.270
the next value 457.5; wall to ceiling ratio
is 1000 divided by 625 which comes to 1.6.
00:21:16.270 --> 00:21:24.300
And if you go for the noise reduction in decibels
with wall absorption only it is 13 dB, with
00:21:24.300 --> 00:21:35.050
ceiling absorption it is 11 dB and the total
noise reduction will again be around 16 dB.
00:21:35.050 --> 00:22:00.430
So, you can carry out the same calculation
with some more figures, for 40 into 40 into
00:22:00.430 --> 00:22:09.880
10; you have the ratio as 1, where you see
the wall absorption and the ceiling absorption
00:22:09.880 --> 00:22:15.810
are matching and the noise reduction is both
12 dB and 12 dB.
00:22:15.810 --> 00:22:23.750
But and the ceiling and the floor together
its reporting together 15 little over 15;
00:22:23.750 --> 00:22:34.950
it is 15 dB, for 42 by 42 you see the ratio
of the wall to ceiling goes below 1.
00:22:34.950 --> 00:22:42.480
And also you see it is almost similar giving
similar results, you can work out for 45 into
00:22:42.480 --> 00:23:03.380
45 you will get little changes and for 60
into 60, where the ratio is 1.5 that is again
00:23:03.380 --> 00:23:12.870
increasing you see it is 13 dB and 11 dB reduction
for wall and ceiling and you see the overall
00:23:12.870 --> 00:23:16.770
reduction is by 15 dB.
00:23:16.770 --> 00:23:25.320
For all these cases you will see on this on
the last column; it is always 10 log 36 hopefully
00:23:25.320 --> 00:23:29.070
you have got similar kind of result.
00:23:29.070 --> 00:23:38.140
So, after getting this if we want to plot
this what we will see?
00:23:38.140 --> 00:23:40.240
Let us try to find out.
00:23:40.240 --> 00:23:51.180
So, this is your graph where we are reporting
1600 square feet floor area that is 40 cross
00:23:51.180 --> 00:23:52.180
40.
00:23:52.180 --> 00:24:04.400
And 400 here that is 20 cross 20 where the
ratio of wall to ceiling is 2 and beyond 1600
00:24:04.400 --> 00:24:11.550
square feet, we are talking telling these
are classifying them as large rooms, below
00:24:11.550 --> 00:24:20.450
1600 square feet of floor area we are classifying
them as small rooms or small room zones.
00:24:20.450 --> 00:24:27.780
And you have at the age end in the first test
case that is 10 by 10 room where the ceiling
00:24:27.780 --> 00:24:31.240
is to the wall is to ceiling ratio was 4.
00:24:31.240 --> 00:24:38.690
So, on the x axis you see apart from the square
feet area of the ceiling you get the ratio
00:24:38.690 --> 00:24:44.940
of the wall is ceiling that is 0 starting
from 1 2 3 and 4.
00:24:44.940 --> 00:24:51.110
So, we have all the cases in our hand that
is we have 1 we have 2 and I think we have
00:24:51.110 --> 00:24:58.090
4 and let us try to plot it plot whatever
value we had got in the previous slide; we
00:24:58.090 --> 00:25:08.100
will see that if we plot the last column that
is wall and ceiling treatment; you will see
00:25:08.100 --> 00:25:15.980
it is coming like from wall to ceiling wall
and ceiling it is gradually growing and getting
00:25:15.980 --> 00:25:29.330
static at 16; starting from 15 little less
than 15 it is going, you can go further down.
00:25:29.330 --> 00:25:36.520
And if we look into the wall absorption, if
you look into the ceiling absorption only
00:25:36.520 --> 00:25:44.570
you will see gradually with smaller areas
the curve is decaying down
00:25:44.570 --> 00:25:57.221
So, the ceiling is not effective when we are
going for smaller and smaller size rooms and
00:25:57.221 --> 00:26:08.670
when we are looking into walls only, we are
seeing that walls are more effective when
00:26:08.670 --> 00:26:13.690
we are looking into looking for smaller size
rooms.
00:26:13.690 --> 00:26:21.190
And for bigger size rooms the ceiling is more
effective; however, if we can make a mix and
00:26:21.190 --> 00:26:29.620
match this is for the entire room getting
treated, but in actual we do not treat the
00:26:29.620 --> 00:26:30.650
entire room.
00:26:30.650 --> 00:26:37.750
So, we can actually develop such curves considering
how much part of the ceiling or how much part
00:26:37.750 --> 00:26:43.740
of the wall surface we are actually treating
and what will be the total reduction.
00:26:43.740 --> 00:26:51.180
So, if we can plot for particularly for big
spaces, auditoriums and all we can see how
00:26:51.180 --> 00:26:54.511
effective is the application.
00:26:54.511 --> 00:27:02.180
So, what we see here is that at one the wall
area and the ceiling area is same so; obviously,
00:27:02.180 --> 00:27:07.480
it is giving it is merging at 12 dB at around
12 dB.
00:27:07.480 --> 00:27:15.300
And we will see that around 400 that is 2
when the ratio is 2 between wall and ceiling,
00:27:15.300 --> 00:27:22.170
we see that ceiling is effect feeling is less
effective, but as wall is more effective.
00:27:22.170 --> 00:27:27.170
And hence the inference is treating ceiling
is more effective for big room whereas, treating
00:27:27.170 --> 00:27:34.750
the walls is with absorbers is a better solution
for rooms of smaller volume and we can leave
00:27:34.750 --> 00:27:36.000
the ceiling in cases.
00:27:36.000 --> 00:27:45.120
But for each and every case by if we can develop
this particular table we will be in a position
00:27:45.120 --> 00:27:52.090
of increasing or decreasing the area or increasing
or decreasing the absorption coefficients
00:27:52.090 --> 00:27:59.559
such that we can compromise with the we do
not make any compromise with the design criteria
00:27:59.559 --> 00:28:05.950
and we can actually find out what is the total
absorption happening within that room and
00:28:05.950 --> 00:28:07.929
leading to the noise reduction.
00:28:07.929 --> 00:28:12.440
So, our objective is to fulfill the reverberation
time.
00:28:12.440 --> 00:28:23.400
And if that is our objective then we can actually
control all these factors through excel sheets,
00:28:23.400 --> 00:28:28.290
simple running excel sheets and we can solve
our purpose.
00:28:28.290 --> 00:28:35.610
So, you task would be to generate similar
curves changing the height from 10 feet to
00:28:35.610 --> 00:28:47.920
20 feet; may be one case changing the absorption
coefficient 2.8 you can go for even 50 percent
00:28:47.920 --> 00:28:54.420
of the wall being covered you can see the
behavior because you know first part of the
00:28:54.420 --> 00:28:59.490
wall is helping to reinforce sound through
reflection.
00:28:59.490 --> 00:29:06.940
So, you can work out another with 50 percent
coverage and these will actually enrich you
00:29:06.940 --> 00:29:15.110
with or develop you help you develop the understanding
how the walls and the ceiling should be treated
00:29:15.110 --> 00:29:20.890
to get a good acoustical quality; when you
know what is your target.
00:29:20.890 --> 00:29:27.350
Whether it is for speech, whether it is for
music what is the performance going to happen
00:29:27.350 --> 00:29:29.070
in that particular space.
00:29:29.070 --> 00:29:38.540
So, depending on so many parameters you have
to create this table . So, hope you understood
00:29:38.540 --> 00:29:46.830
the purpose of this lecture and we will again
continue with some more of the absorbers which
00:29:46.830 --> 00:29:57.100
we can actually make use without taking help
of the structural support that is of the boundary
00:29:57.100 --> 00:29:58.100
condition.
00:29:58.100 --> 00:30:00.810
So, we will discuss on those in the next lecture.