WEBVTT
Kind: captions
Language: en
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Ah
welcome to the 33 lecture on the course
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NPTEL course on architectural acoustics, we
are in the se 7 th week and this is the
00:00:25.570 --> 00:00:31.920
third lecture of the 7 th week in the last
lecture on the airborne sound transmission.
00:00:31.920 --> 00:00:36.610
So, the lecture title is a airborne sound
transmission 3 .
00:00:36.610 --> 00:00:43.790
So, the learning objective of this lecture
or we will describe this lecture, what is
00:00:43.790 --> 00:00:48.510
the coincidence effect and what is the impact
of the coincidence effect in the transmission
00:00:48.510 --> 00:00:57.750
loss of a partition wall and we will try to
formulate in a methodology and the evaluate
00:00:57.750 --> 00:01:03.839
some value we have a partition wall that is
called the STC sound transmission loss value
00:01:03.839 --> 00:01:10.020
of a partition wall .
So, quickly go to the slide where we have
00:01:10.020 --> 00:01:15.160
already discuss this particular slide in the
very first lecture of our sound transmission
00:01:15.160 --> 00:01:24.530
airborne sound transmission. This transmission
loss TL is depend upon 2 physical parameter
00:01:24.530 --> 00:01:29.400
of course, there are various others also,
but this two a very important and very significant
00:01:29.400 --> 00:01:35.090
one is surface mass of the partition wall
and another one is the sound frequency. And
00:01:35.090 --> 00:01:44.219
as we understand that this per octave increase
in the octave for the per octave the sound
00:01:44.219 --> 00:01:51.710
frequency of if it is doubled from the lowered
one the transmission loss is increased by
00:01:51.710 --> 00:01:57.409
6 dB .
This is follow a kind of a stiff straight
00:01:57.409 --> 00:02:05.070
line graph kind of a think where we can plot
the frequency in the x axis and the sound
00:02:05.070 --> 00:02:10.560
transmission loss in the y axis and we can
see their transmission loss is something like
00:02:10.560 --> 00:02:17.959
this through this particular follow this particulars
straight line, but in actually in follow this
00:02:17.959 --> 00:02:24.090
green line where there is a sudden flat the
plateau and they again there is a increment
00:02:24.090 --> 00:02:28.000
.
So, if you produce this line the blue dotted
00:02:28.000 --> 00:02:35.260
line this is a TL the transmission loss as
per the mass law and this is green line indicate
00:02:35.260 --> 00:02:42.040
the there is a slight change in the TL as
per the frequency change and there is a kind
00:02:42.040 --> 00:02:48.130
of a plateau is available here and this is
also through that the sound transmissions
00:02:48.130 --> 00:02:56.870
loss is low in case of the low frequency.
So, the sound energy passes through in a low
00:02:56.870 --> 00:03:03.849
frequency less whereas, the high frequency
more sound energy passes through so, there
00:03:03.849 --> 00:03:10.640
is a high TL value are sound energy isolate
is isolation is very very high .
00:03:10.640 --> 00:03:17.629
So, next is the in the in this case again
if I see this try to find out why there
00:03:17.629 --> 00:03:23.959
is point depression, why this particular kind
of the change is occur and what are the significant
00:03:23.959 --> 00:03:30.989
role of this particular change in a partition
wall. We say there are we can see that third
00:03:30.989 --> 00:03:37.810
3 such the geometric parameter one is the
this depth which is called the coincidence
00:03:37.810 --> 00:03:45.470
dip which is almost about 5 to 15 dB and this
is a width of the plateau how much plateau
00:03:45.470 --> 00:03:53.439
will be long it shorter or the longer plateau.
It depends upon the damping of the trasmi
00:03:53.439 --> 00:04:00.469
the this partition wall how much damping
of the sound is going to create and this plateau
00:04:00.469 --> 00:04:09.239
height or this coincidence dip is 2 plateau
height in this particular coincidence depend
00:04:09.239 --> 00:04:14.890
upon the stiffness of the partition wall how
much it is stiff, how much it is rigid .
00:04:14.890 --> 00:04:26.600
So, this 3 phenomena gives us a kind of a
the kind of a understanding that the
00:04:26.600 --> 00:04:33.240
even if the frequency if a frequency increases
the transmission loss is going to increase
00:04:33.240 --> 00:04:42.130
, but it is not same for all the range of
the frequency some where it is kind of a
00:04:42.130 --> 00:04:48.530
a jerk and there is again it is follow a particular
straight line and this particular phenomena
00:04:48.530 --> 00:04:53.270
is called coincidence effect.
The coincidence effect is that suppose there
00:04:53.270 --> 00:05:00.190
is a partition wall and by virtue of the partition
wall the mass and the stiffness and the
00:05:00.190 --> 00:05:09.430
damping factor an all those the physical parameters
it has a shear wave capacity or the some vibration
00:05:09.430 --> 00:05:18.169
mode and this vibration mode can be describe
with it is natural frequency and if the incident
00:05:18.169 --> 00:05:25.120
frequency of the sound which is actually the
incident sound wave is matches with this shear
00:05:25.120 --> 00:05:32.990
wave or the face of this particular vibration
then this frequency the incident frequency
00:05:32.990 --> 00:05:41.080
and it is own natural frequency matches together
then there is a resonating effect then the
00:05:41.080 --> 00:05:47.610
sound will going to pass through the a maximum
amount. So, that is why there is a sudden
00:05:47.610 --> 00:05:52.180
dip in this particular the par the TL value
of the partition wall .
00:05:52.180 --> 00:06:04.090
So, the what happened is that. So, in any
particular object has it is own natural frequency
00:06:04.090 --> 00:06:12.020
and if that matches with the incidence sound
wave it is passes a huge amount of sound because
00:06:12.020 --> 00:06:19.990
of the resonance and this vibration of this
particular partition wall will coincident
00:06:19.990 --> 00:06:26.270
when it is coincident with the incident wave
this called as a coincident effect coincident
00:06:26.270 --> 00:06:34.050
effect an due to that there is a gradual change
or there is a sudden dip in this frequency
00:06:34.050 --> 00:06:39.139
and TL carve.
So, a normal frequency TL curve will look
00:06:39.139 --> 00:06:49.350
like this, the initial in short frequency
range there is a lots of dips and undulations.
00:06:49.350 --> 00:06:55.400
This can be described as there are some kind
of a stiffness control and there some kind
00:06:55.400 --> 00:07:02.199
of a localized resonated resonance is can
be happened. So, the that is whether is a
00:07:02.199 --> 00:07:07.860
some undulations .
Stiffness control means suppose a wall having
00:07:07.860 --> 00:07:16.629
the same surface mass, but it is stiffen by
some regular interval by some structurally
00:07:16.629 --> 00:07:24.720
strengthen material then this particular the
partition wall can be act as a very stiffen
00:07:24.720 --> 00:07:32.160
varied rigid and it can have a kind of a control
in the particular frequency particular
00:07:32.160 --> 00:07:38.040
transmission loss. But if it is not stiffen
the it is not in regular only in the ages
00:07:38.040 --> 00:07:43.730
or may be only one in between only in the
ages then it is not that much of stiffen and
00:07:43.730 --> 00:07:50.830
there is a decrease in the the stiffness and
the TL values also, but after certain amount
00:07:50.830 --> 00:07:57.729
of the frequency after certain initial frequency
band if you leave this kind of a disturbances
00:07:57.729 --> 00:08:03.879
then it is follow a straight line.
A follow straight line and this is your mass
00:08:03.879 --> 00:08:12.139
law, as per the mass law 6 dB increase as
per the octave or the 6 dB increase as per
00:08:12.139 --> 00:08:18.801
the doubling of the mass also. It will continue
in some portion and then there is a dip there
00:08:18.801 --> 00:08:25.129
is a coincidence effect , but is coincident
effect will initiate in certain point and
00:08:25.129 --> 00:08:31.360
it will follow a dip follow a plateau and
that again it will follow the curve which
00:08:31.360 --> 00:08:38.440
is 6 dB per octave or the 6 dB is the doubling
of the mass curve. So, this line and this
00:08:38.440 --> 00:08:44.260
line is parallel, but there is a offset between
these two line there is offset between this
00:08:44.260 --> 00:08:48.670
two line .
Now, this frequency where this particular
00:08:48.670 --> 00:08:55.400
coincidence effect is initiated it is called
the critical frequency or that is the frequency
00:08:55.400 --> 00:09:03.070
of the natural frequency or the face frequency
of the particular wall which matches with
00:09:03.070 --> 00:09:09.570
the incident frequency. So, if we have a different
partition wall and different material this
00:09:09.570 --> 00:09:17.440
critical frequency point may be somewhere
here may be 500 or may be in 4000 . So, this
00:09:17.440 --> 00:09:22.510
may change this may change depending upon
the critical frequency point the changing
00:09:22.510 --> 00:09:27.590
because of the nature and the material of
the partition wall .
00:09:27.590 --> 00:09:36.270
So, we are landed of in a problem the problem
is the number one is the I need a single value
00:09:36.270 --> 00:09:44.500
of a transmission loss to design a particular
noise beerier or a partition wall single
00:09:44.500 --> 00:09:51.480
value, why single number average value, because
that is help me to decide to give it is some
00:09:51.480 --> 00:09:57.820
kind of decision, but which partition wall
I will use for my purpose for my noise reduction
00:09:57.820 --> 00:10:06.520
, but a single value average value of the
transmission loss is always going to be misleading,
00:10:06.520 --> 00:10:13.160
why it is misleading, because it is depend
upon various frequency and not only it is
00:10:13.160 --> 00:10:19.090
depend upon various frequency there is a critical
frequency also where there is a coincidence
00:10:19.090 --> 00:10:28.200
dip and initially there is some kind of a
disturbance because of the stiffness and the
00:10:28.200 --> 00:10:32.220
damping.
So, if I calculate the average value for all
00:10:32.220 --> 00:10:39.910
the TL over the range of the frequency it
will definitely going to misleading, but what
00:10:39.910 --> 00:10:48.000
is the way I can find out. So, still it is
we are depending upon a single numbers transmission
00:10:48.000 --> 00:10:56.940
loss , it is some methodology as to establish
that can the decrease this particular misleading
00:10:56.940 --> 00:11:04.040
phenomenal or it will just dilute some way
may not it will be accurate in nature.
00:11:04.040 --> 00:11:10.680
But still is single number transmission loss
can give me a kind of understanding, it
00:11:10.680 --> 00:11:20.210
is some method that yes this particular partition
wall if I use average in a term in a average
00:11:20.210 --> 00:11:27.520
term it may be a 10 percent error 10 to 15
percent error maximum, that give me this much
00:11:27.520 --> 00:11:30.880
amount of the sound reduction or the transmission
loss .
00:11:30.880 --> 00:11:39.380
So, for that a stranded system has been develop
in united states that is called the sound
00:11:39.380 --> 00:11:48.010
transmission class STC via a STC contours
and this particular STC contours was develop
00:11:48.010 --> 00:11:56.700
with a proper straight forward a worry
the step by step order methodology by the
00:11:56.700 --> 00:12:04.350
ASTM, the American society for testing an
material . So, will discuss that how this
00:12:04.350 --> 00:12:10.920
particular the steps the way is STC
value of a partition wall can be formulated
00:12:10.920 --> 00:12:20.060
can be found out an any kind of the product
the partition wall product manufacturer and
00:12:20.060 --> 00:12:26.600
lot of the organization the those who are
actually providing those kind of a solutions
00:12:26.600 --> 00:12:32.600
for the partition wall solutions or may
be the noise acoustical solution for the room,
00:12:32.600 --> 00:12:38.600
they actually go for this STC number or the
STC rating of the partition wall while
00:12:38.600 --> 00:12:45.190
designing any kind of the encloser.
So, there are 6 or 5 or 6 steps, the first
00:12:45.190 --> 00:12:52.020
one we have to test a particular partition
wall in a laboratory for various frequencies
00:12:52.020 --> 00:12:56.350
also.
An the frequencies as 16 frequencies it starts
00:12:56.350 --> 00:13:05.910
from 125 hertz to 4000 hertz. So, this one
this 16 frequencies are all one - third octave
00:13:05.910 --> 00:13:16.810
band of frequencies in between 125 to 4000
hertz. Then we have to plot a graph in between
00:13:16.810 --> 00:13:24.260
the this frequency and the TL , then we have
overlay the STC contour because this graph
00:13:24.260 --> 00:13:30.090
which is will plot in the state number 3 will
be a very very haphazard kind of a graph . So,
00:13:30.090 --> 00:13:36.840
I have to actually slice of some of the
points and I have to go with a smoothening
00:13:36.840 --> 00:13:41.810
this graph with kind of a contour, with kind
of a the different segment of a straight line
00:13:41.810 --> 00:13:48.200
and I find out the STC contour.
And the fifth and final the step is that
00:13:48.200 --> 00:13:54.300
from that STC contour I can get the single
value that single value I am wanting for the
00:13:54.300 --> 00:14:02.790
STC value of a wall . So, all the 4 steps
or the this 5 steps has some typical methods
00:14:02.790 --> 00:14:08.590
or some kind of a rules and we will describe
that rules one after another .
00:14:08.590 --> 00:14:14.110
So, what is the first step is that you have
to test the partition wall in a laboratory.
00:14:14.110 --> 00:14:19.130
So, we use that particular formula what we
are use in the second lecture, the previous
00:14:19.130 --> 00:14:31.830
lecture, TL is equal to L S minus L R minus
the 10 log alpha into S. So, this
00:14:31.830 --> 00:14:39.980
is the particular equation and if I have a
microphone at the particular receiving
00:14:39.980 --> 00:14:45.570
room and a loudspeaker in the source room.
So, I can find out with a d b meter, but what
00:14:45.570 --> 00:14:50.760
is the L S, that is a source room intensity
of the sound and what is a level of the sound
00:14:50.760 --> 00:14:56.940
in the receiving room, I can find out this
I no those values and A a S and alpha A s
00:14:56.940 --> 00:15:03.260
every think I can find out the TL values .
An if I know the TL values I now have a table
00:15:03.260 --> 00:15:11.780
where all the 16 frequencies are listed down
and the I will I will find out all the
00:15:11.780 --> 00:15:17.560
corresponding actual TL value of all the 16
frequencies and all the frequencies 16 is
00:15:17.560 --> 00:15:27.200
form 125 to 4000 hertz actually cover in the
one- third octave band and out of that this
00:15:27.200 --> 00:15:33.610
blue colour was at the octave band, the 125
250 in between that there are 2 one - third
00:15:33.610 --> 00:15:39.590
and 250 to 5000 there are 2 octaves us one
- third octave like that .
00:15:39.590 --> 00:15:45.150
So, this 16 frequencies are those TL actual
value has to be plotted.
00:15:45.150 --> 00:15:54.800
So, after plot in the frequency verses the
TL plot for those 18 sorry 16 octave frequency,
00:15:54.800 --> 00:16:02.230
we will get the graph of this nature very
haphazard kind of a graph, but even though
00:16:02.230 --> 00:16:08.560
this is a very haphazard there is a tendency
of moving up I mean which is a in the positive
00:16:08.560 --> 00:16:15.210
kind of a graph. So, as in way in the frequencies
increases you have TL value is increases,
00:16:15.210 --> 00:16:20.740
but it may not follow a straight line, probably
at this particular point there is a follow
00:16:20.740 --> 00:16:26.270
is straight line there is a kind of a disturbance
over here there is a deep and disturbance
00:16:26.270 --> 00:16:31.640
and there is another disturbance because of
maybe there is coincidence effect. So, there
00:16:31.640 --> 00:16:37.700
may be one or two coincidence effect in also.
So, there are by virtue of the actual plot
00:16:37.700 --> 00:16:46.170
in grade those graph, but as I told in earlier
first taking the averaging of all those values
00:16:46.170 --> 00:16:52.750
will not going to help me, because there are
some the phenomena the coincidence phenomenon
00:16:52.750 --> 00:16:56.790
and all.
So, we have to go to the step number 4.
00:16:56.790 --> 00:17:03.600
Where I will going to overlay a STC contour
on that I will just going to cut off some
00:17:03.600 --> 00:17:08.520
of the points and try to fix some straight
line in a some the method with their some
00:17:08.520 --> 00:17:15.110
method. What are the method . So, the first
in this 4 the state number 4 the calculate
00:17:15.110 --> 00:17:21.079
the average TL value for the last 6 one - third
octave that is 1250 to 4000.
00:17:21.079 --> 00:17:29.970
So, what I did is that you have to find out
the what are the average of this 1 2 3 4 5
00:17:29.970 --> 00:17:39.750
6 this last 6 TL value and find out the what
is the average of that and you plot the you
00:17:39.750 --> 00:17:45.290
plot the line. So, draw a straight line parallel
to x axis of that particular and find out
00:17:45.290 --> 00:17:50.360
the this A B line is actually the average
of those. So, some points should be above
00:17:50.360 --> 00:17:59.000
A B, some points should be below A B .
Now, let us go to the point number 4 c, plot
00:17:59.000 --> 00:18:07.090
the point c over 4000 hertz frequency and
5 dB below A B line so; that means, .
00:18:07.090 --> 00:18:16.940
Now, you plot a point C over a 4000 frequency
line and this will be below B how much below
00:18:16.940 --> 00:18:25.789
now 5 dB below B . So, you have you see this
the from 1250 this blue line to this 400 this
00:18:25.789 --> 00:18:35.140
blue line there are 1 2 3 4 5 gaps . So, the
5 gaps means per octave pa per frequency one
00:18:35.140 --> 00:18:44.230
- third octave there is a decrease of 1 dB
. So, I got the point C the I will join the
00:18:44.230 --> 00:18:50.870
line B, C also the line segment B C is inclined
and having 1 dB reduction per octave one - third
00:18:50.870 --> 00:18:55.510
octave.
Now, I have to go to the point D, point D
00:18:55.510 --> 00:19:04.910
will be at this 125 the last line and the
drop will be now 15 dB now this is again incline
00:19:04.910 --> 00:19:14.860
line to C to D and again from 400 to 125 there
are 1 2 3 4 5 gaps. So, per one third octave
00:19:14.860 --> 00:19:23.340
there is a 3 d b decrease per one third
octave. So, if u go to that you see plot the
00:19:23.340 --> 00:19:31.820
D over 125 15 dB below and join this point
C D the live segment C D is also inclined
00:19:31.820 --> 00:19:39.010
having 3 dB reduction per one - third octave
. So, you are A B C D is the STC contour is
00:19:39.010 --> 00:19:48.190
completed . So, what I did is that, all those
haphazard point is now strictly in a discipline
00:19:48.190 --> 00:19:53.550
manner a formulated a contour green contour
line which is STC contour.
00:19:53.550 --> 00:19:59.251
But this is not final you have to see some
criteria, it is see a get a single number
00:19:59.251 --> 00:20:04.890
of this STC we have to find out some of the
criteria, what you have to do in the state
00:20:04.890 --> 00:20:11.600
number 5 a is that, identify the frequency
having the magnitude of the deficiency . What
00:20:11.600 --> 00:20:16.900
is the deficiency? The deficiency is that
the difference between the STC contour and
00:20:16.900 --> 00:20:24.090
the TL value.
So, I got this contour ABCD and this line
00:20:24.090 --> 00:20:31.270
is somewhere it is plotting above and somewhere
may be it is dipping below you are actual
00:20:31.270 --> 00:20:38.700
TL points . So, if it is plotting above so,
there is a deficiency . So, deficiency is
00:20:38.700 --> 00:20:46.010
defined as STC minus actual TL. So, these
are the deficiencies point, these are the
00:20:46.010 --> 00:20:54.350
deficiency point, but this is not the deficiency
this is where the TL is above in your STC
00:20:54.350 --> 00:20:57.840
contour .
So, you have to find out this deficiency in
00:20:57.840 --> 00:21:03.950
the seve 5 b state number 5 b, sum of the
deficiency is has to be calculated, what is
00:21:03.950 --> 00:21:11.620
the total amount of deficiency for all the
16 and we have to find out the total amount
00:21:11.620 --> 00:21:18.350
of deficiency is and you have to check this
2 criteria. The first criteria is that, the
00:21:18.350 --> 00:21:26.350
sum of the deficiencies is less than or equal
to 32 dB and second criteria is that the maximum
00:21:26.350 --> 00:21:32.980
deficiencies in any one band does not exceed
8 dB this two criteria as to be full filled.
00:21:32.980 --> 00:21:39.660
So, if these two criteria has to be full filled
like so, the first some of the deficiency.
00:21:39.660 --> 00:21:46.400
So, this sum the amount of this deficiencies
plus this plus this. So, this all sum of the
00:21:46.400 --> 00:21:53.430
deficiencies should be less than or equal
to 32 and any one I mean you have to check
00:21:53.430 --> 00:21:58.150
everyone. So, the maximum deficiency it will
be this one or may be this one, should not
00:21:58.150 --> 00:22:07.211
be less than 6 dB less than 8 dB.
And at the final stage in the 5 5 c we
00:22:07.211 --> 00:22:13.990
have to there are 2 situation, if all criteria
both the criteria is satisfied a both the
00:22:13.990 --> 00:22:23.491
criteria satisfied we go to next trail of
the STC contour we will increase the trail
00:22:23.491 --> 00:22:30.080
STC contour by 1 dB and we will recalculated
the recheck the criteria are the step 5 a
00:22:30.080 --> 00:22:37.510
and 5 b and come again and sometime is criteria
does not match with the criteria the actual
00:22:37.510 --> 00:22:44.130
this deficiency or may not match.
So, then we will decrease the STC by 1 dB
00:22:44.130 --> 00:22:54.210
and again we recalculate or re do the step
number 5 a and 5 b an it is a again a trail
00:22:54.210 --> 00:22:59.520
and in our process and finally, we have to
stop at a particular point where both the
00:22:59.520 --> 00:23:08.110
criteria is full filled matches and we will
find out the STC value for the 500.
00:23:08.110 --> 00:23:13.630
So, suppose this criteria is matches. So,
I have remove those points and this is the
00:23:13.630 --> 00:23:22.450
final STC contour A B C D and then correspondingly
500 hertz frequency the value corresponding
00:23:22.450 --> 00:23:27.710
TL value will be my single value STC value
of the wall.
00:23:27.710 --> 00:23:34.750
So, let us have a small example.
In this example what I did is that I have
00:23:34.750 --> 00:23:43.440
written down the actual TL value of a partition
wall from 125 hertz to 4000 hertz all the
00:23:43.440 --> 00:23:50.020
16 one - third octave band . So, by virtue
of that if you plot will get a plot like this
00:23:50.020 --> 00:23:54.550
and you will see in this plot where some disturbance
there is a straight line portion and there
00:23:54.550 --> 00:24:00.680
is again a straightening got this may be a
coincidence dip is a coincidence dip and I
00:24:00.680 --> 00:24:06.840
have written down all the values for all the
16 frequencies for and in the x axis I
00:24:06.840 --> 00:24:14.070
just mention the number of the frequency number
the band number.
00:24:14.070 --> 00:24:20.550
Now, you if you remember the first what
I have to did is that, this last 6 this 40
00:24:20.550 --> 00:24:30.050
36 33 33 45 and 47 just find out the average
the average is 39 . So, average of the last
00:24:30.050 --> 00:24:39.150
6 frequencies 39. So, the next is that next
point is that this 400 frequency here, the
00:24:39.150 --> 00:24:47.130
400 frequency TL decrease the 5. So, this
35 minus 5 and 1 dB decrease per octave . So,
00:24:47.130 --> 00:24:58.790
38 37 36 like that and this is 34 and then
at 125 there is 15 dB decrease from the earlier
00:24:58.790 --> 00:25:06.640
and that is 19 why, 19 because 34 minus 15
is 19. So, there is a 6 dB decrease per one
00:25:06.640 --> 00:25:13.570
- third octave band so, 34 31 28 25 22 and
19 .
00:25:13.570 --> 00:25:23.500
So, I can plot my STC contour overlay over
my actual TL plot .
00:25:23.500 --> 00:25:30.710
And now next step is that step of finding
are the deficiencies see here , this is above
00:25:30.710 --> 00:25:36.980
this is also above this is again above there
are again this 2 are above the STC. So, the
00:25:36.980 --> 00:25:41.760
above I do not want to take above, I do not
have any interest I have only interest in
00:25:41.760 --> 00:25:49.000
this dips, this is a deficiency from the STC
line to the this the actual TL line, this
00:25:49.000 --> 00:25:53.500
is the point this is the coordinate where
I have to find out .
00:25:53.500 --> 00:26:02.720
So, I found out for all the this is the frequency
one third octave band actual TL value by first
00:26:02.720 --> 00:26:11.190
proposed STC first trail values and this 19
and the this difference between 17 minus 19
00:26:11.190 --> 00:26:18.600
suppose this is the first 17 and this is 19.
So, deficiency is minus 2, minus 6, minus
00:26:18.600 --> 00:26:25.490
7, minus 3, but here it is not deficiency
is some increment of suppose it is the 315.
00:26:25.490 --> 00:26:32.490
So, this 33 and 31 33 and 31 so, those are
positive and not at all interested with this
00:26:32.490 --> 00:26:37.640
positive values this positive values only
I am interested in those in negative values.
00:26:37.640 --> 00:26:46.630
So, what I did is that, I got the full range
of the deficiencies and all those read or
00:26:46.630 --> 00:26:58.050
the ahh the this coloured cell I got the summation
of that . So, the in the first trail my all
00:26:58.050 --> 00:27:06.220
the summation or minus x. So, total deficiency
is minus 32 dB and the maximum deficiency
00:27:06.220 --> 00:27:12.850
among this is minus 7. So, what are the criteria,
if you remember the criteria the first criteria
00:27:12.850 --> 00:27:20.320
is that sum of the deficiencies is less than
or equal to 32 is 34 minus 34.
00:27:20.320 --> 00:27:25.870
So, I have to read do the things, but the
second criteria is fulfilled it is not exceeded
00:27:25.870 --> 00:27:34.370
minus 8 it is just minus 7, but the first
criteria is failed over here it is minus 34
00:27:34.370 --> 00:27:41.030
I should have I mean there should be it is
should be restricted to minus 32 dB also.
00:27:41.030 --> 00:27:47.530
So, what I have to do, the option is we have
to go to the next option next trail second
00:27:47.530 --> 00:27:53.480
trail of STC contour. So, this STC line, the
grill line I have developed that has to be
00:27:53.480 --> 00:28:01.100
lower down by 1 dB and if I lower down by
1 dB the deficiency is will be less and we
00:28:01.100 --> 00:28:05.929
have to recalculate.
So, I have the second trail with me it was
00:28:05.929 --> 00:28:13.230
a first trail and the first trail I had deficiency
is are calculated minus 34 minus 7 and all
00:28:13.230 --> 00:28:20.710
. So, my second trail what I did, I have just
decrease the value the STC contour by 1 dB
00:28:20.710 --> 00:28:31.940
c, this is 39 over here, this is 38 last 36
38 in this 34 that is to 4 sorry this 35
00:28:31.940 --> 00:28:38.400
I go on decreasing by 1 dB and from 400 onwards
decreasing by 3.
00:28:38.400 --> 00:28:46.400
There is similarly what I did over here and
let us see in this case these are the deficiency
00:28:46.400 --> 00:28:54.760
for the second case and if you add up this
is minus 26 and the highest is minus 6 both
00:28:54.760 --> 00:29:03.550
satisfied the criteria. Now both satisfied
criteria and I can stop here because I do
00:29:03.550 --> 00:29:09.450
not want to go to the further one less because
definitely that will be again classify
00:29:09.450 --> 00:29:16.310
satisfy the criteria. So, my second trail
of the STC contour satisfy the criteria. So,
00:29:16.310 --> 00:29:24.120
on that lying second trail the corresponding
the TL value corresponding to 500 hertz is
00:29:24.120 --> 00:29:28.929
my is single digit STC value and that is 34
.
00:29:28.929 --> 00:29:37.830
So, 500 line hertz line striking to my this
34 this STC contour that is give me
00:29:37.830 --> 00:29:48.261
the STC value this 34 dB that is how I found
out this particular transmission sound transmission
00:29:48.261 --> 00:29:53.240
class of a wall.
The next one is if there is a whole in the
00:29:53.240 --> 00:30:00.210
wall. So, this whole will actually try to
leak the noise. So, it is depend upon how
00:30:00.210 --> 00:30:07.840
much amount of noise will be leaked through
the hole, suppose it is a very wonderful partition
00:30:07.840 --> 00:30:15.240
wall made by double leaf and all those and
there is a hole that miss failure total the
00:30:15.240 --> 00:30:23.112
amount of effort an all to the reduce the
noise from one source room to the receiving
00:30:23.112 --> 00:30:31.820
room. So, it is depend upon the percentage
of whole area S H by S 1 into 100 what is
00:30:31.820 --> 00:30:36.570
this S H, the area of the hole. What is S
1, the total area.
00:30:36.570 --> 00:30:44.050
The total area is S 1 and the S H is the amount
of hole is ratio into 100 and what is TL 1,
00:30:44.050 --> 00:30:51.480
TL 1 is the transmission loss of the partition
wall this partition wall and what is TL C,
00:30:51.480 --> 00:30:59.000
is a combined transmission loss of hole when
it is a hole. So, the TL wall and the partition
00:30:59.000 --> 00:31:06.130
transmission loss of the partition wall
is placed in the x axis an a combine partition
00:31:06.130 --> 00:31:12.870
the transmission loss of the partition
wall with hole is plotted in the y axis. It
00:31:12.870 --> 00:31:19.540
is again the sum set of graph has been
standardized by virtue of some experiment.
00:31:19.540 --> 00:31:27.370
Suppose there is a graph so, this is for the
whole area is 0.1 percent. So, if it is 0.1
00:31:27.370 --> 00:31:34.860
percent so, if can go with this graph suppose
25 percent of the 20 TL of the partition wall
00:31:34.860 --> 00:31:41.870
is 25 if there is a 0.1 percent of the hole
it is little less than 25 may be around may
00:31:41.870 --> 00:31:48.960
be around 22 20 24 or something like that
if it is 10. So, this is almost 10 there is
00:31:48.960 --> 00:31:54.990
no significant change.
So, as the whole area is increase and the
00:31:54.990 --> 00:32:00.810
percentage whole area also increases this
particular graph is now lower down. So, if
00:32:00.810 --> 00:32:06.420
it is 1 percent you have to followed take
this graph 2 percent this think. So, 50 percent
00:32:06.420 --> 00:32:12.610
of you see there is a 50 percent whole. So,
then there is a huge reduction with 30 is
00:32:12.610 --> 00:32:19.630
equivalent almost 3. So, if there is a 5 percent
of the hole is there a 30 actual gives you
00:32:19.630 --> 00:32:26.241
almost about 12 and also .
So, let us have a numerical problem we
00:32:26.241 --> 00:32:33.950
have a door of 2 meter by one meter and in
this door the that is door is have a 30 dB
00:32:33.950 --> 00:32:40.360
TL and there is a size hole size there is
a hole in that particular door which is having
00:32:40.360 --> 00:32:47.250
20 centimeter by 12 centimeter. So, the total
area of the door is 2 meter square is one
00:32:47.250 --> 00:32:54.450
and total area of hole is 200 centimeter square
and transmission loss of the partition wall
00:32:54.450 --> 00:33:04.910
without the hole is 30 dB, there is a small
the typological area this should be TL
00:33:04.910 --> 00:33:14.710
1 is 30 dB and this percentage of the whole
area is this so, this is almost 1 percent.
00:33:14.710 --> 00:33:22.740
So, in that I have to see this 1 percentage
curve percentage of whole area is 1 percent
00:33:22.740 --> 00:33:29.540
curve and as this is 30 dB this transmission
loss of the actual the wall is 30 dB. So,
00:33:29.540 --> 00:33:39.100
this TL 1 you have to actually this is TL
1 TL values 30 dB. So, 30 dB is this and corresponding
00:33:39.100 --> 00:33:47.179
to that percentage of the hole area is 1 percent
hole area gives you the combine transmission
00:33:47.179 --> 00:33:55.160
loss as 20 dB this is 20 dB . So, actually
this the this particular hole of 20 centimeter
00:33:55.160 --> 00:34:03.250
by 10 centimeter over a door of 2 meter by
1 meter destroy almost 10 dB sound
00:34:03.250 --> 00:34:09.460
there is a leakage of 10 dB sound . So, that
is all that this particular portion.
00:34:09.460 --> 00:34:15.260
So, I have noted down some of the the partition
wall significant character the we have to
00:34:15.260 --> 00:34:20.580
increase the surface mass of the partition
wall to established reasonable maximum to
00:34:20.580 --> 00:34:27.020
get a good amount of TL, these are the some
of the partition wall design criteria or the
00:34:27.020 --> 00:34:31.240
guide lines double leaf partition wall is
better than the single leave because there
00:34:31.240 --> 00:34:36.550
is aircushion in between. So, doubling of
the mass is also and happening there is a
00:34:36.550 --> 00:34:43.511
internal aircushions that also improve the
quality of the partition wall are the TL value
00:34:43.511 --> 00:34:48.460
of the partition wall. So, this is very important
if we go with a double leave kind of think
00:34:48.460 --> 00:34:52.710
the air gap between 2 leaves may be filled
with some kind of a acoustical blanket.
00:34:52.710 --> 00:34:58.530
So, it also in hence the total amount of the
transmission loss the proper stiffeners
00:34:58.530 --> 00:35:04.250
as you know the stiffener at attachment
will actually reduce of those kind of initial
00:35:04.250 --> 00:35:09.369
disturbance. So, the stiffener arrangement
is required and this fixing of the partition
00:35:09.369 --> 00:35:16.030
wall with the wall or the ceiling has to be
proper because of the minimizing the coincidence
00:35:16.030 --> 00:35:22.910
effect in the zone of the actual middle fix
frequency and noise leaks that has to be checked
00:35:22.910 --> 00:35:27.670
because as you know if there is any kind of
hole or a any kind of a noise leak there is
00:35:27.670 --> 00:35:30.590
a decrease in the potential amount of the
TL value .
00:35:30.590 --> 00:35:37.030
So, let us go to the last page I have again
to homework for you, one is you please write
00:35:37.030 --> 00:35:43.851
down or try to understand the are what
is the significance of the STC sound transmission
00:35:43.851 --> 00:35:49.900
class value and the second one is that if
the average TL value of the last 6 control
00:35:49.900 --> 00:35:58.150
the last 6 the the frequency one - third
frequency of a partition wall is 35 dB then
00:35:58.150 --> 00:36:05.980
what could be it is STC value if all the criteria
is full filled. So, that is all for today.
00:36:05.980 --> 00:36:13.370
So, the air borne sound transmission is
within with this 3 consecutive lecture number
00:36:13.370 --> 00:36:21.680
31 32 and 33 is now the to an end.
The next lecture on what 34 lecture and
00:36:21.680 --> 00:36:30.340
the 35th lecture we will discuss about
the structure borne sound in in this week
00:36:30.340 --> 00:36:32.599
so.
Thank you very much.