Noise Levels

[Daddy Cool]

Can anyone help or point me in the right direction?

i need to provide a noise level (in dba) for several condensing units of different noise levels.

i know it's not as easy as adding them all together, and that it's something to do with logarithams, but don't have any formula or simple calculator to do it for me.

Any assistance greatfully received.

[Brian_UK]

Daddy, have a look at this:-
http://www.nohsc.gov.au/OHSInformati.../NOISECONT.HTM
and see the paragraph on 'adding levels together', it may help...

[Argus]

Here?s another similar site that contains a basic rule of thumb addition table about halfway down.
http://www.ccohs.ca/oshanswers/phys_...ise_basic.html

But you need to know what you are adding, whether it is sound power or sound pressure.

If you are trying to predict sound from more than one source at a given distance, the surroundings and ambient noise will have an effect.

You are right ? the maths can be horrendous. There are professional acousticians available.
________
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[Daddy Cool]

Thanks Chaps that should be enough to give a reasonable idea.

[botrous]

Guys , just give 2 days to give u full formulaire for the noise calculation , i have just to put it together . if anyone interrested please say so ........

[Daddy Cool]

Botrous, we look forward to it. i did get a formula, but it was vast. i used to have one where i used to work, it was in ms excel and really easy to use. mainly used it for chillers when customers asked how they compared, data from one company was at 10 metres the other at 3 metres and so on.

[botrous]

I am working on it , as soon as it gets ready i'll host it and put a link to download it

[frank]

I found this in one of my acoustic books.

I tried to upload as a word document but the scan was too big. Send me an e-mail address and I'll attach a file

[frank]

OK so the attachment didn't work

So, the formula is as follows

Lav = 10 x Log10(10^n1+10^n2+10^n3 etc)/n

where
Lav = Average of the decibels
n = number of different sounds

Example:

you have 5 condensing units rated at say: 80db,82db,84db,86db and 88db.

Feeding these into the formula we get

10 x Log10 (10^8.0 + 10^8.2 + 10^8.4 + 10^8.6 + 10^8.8)/5

which gives 84.9db or 85db rounded.

Clear as mud to some no doubt

[frank]

Did anyone spot the quick "rule of thumb" way of working the formula?


Just add the indices then divide by the number of units and multiply by ten

or

add all the db levels up and divide by the number of units - same answer. easy isn't it

[botrous]

ok , this is an excel file i made , it's helpful in noise calculations , adding sources and comparing .
Please note , this is the first release , if any one have suggestions or comments just tell.
you can download it from :
www.kesserwan.net/Noise.xls

[frank]

Hi Botrous

The file says it's password protected and won't open

[botrous]

Hi Botrous

The file says it's password protected and won't open

OOPS maybe i've forgotten to unprotect it lololol

The password is : 10452km2

lololol

and guys i revised the file , in the "noise adding section" the dB unit isn't where it should be , just read and change it's places , a simple copy paste.

[frank]

Hi Botrous

I managed to open the file this time but the result showed a level of 7.98db when all the input boxes read 1.

This made all the results 7.98db too much when I put any data in. Also, is it supposed to convert any data to a lower figure? If I put in 86db it turns it into 0.86

I suppose I'm doing something wrong

[botrous]

dB should be written in the first colum as i mentioned before the dB unit should be sopied and pasted near the first colum eg. : Noise level of first type machine /85/dB.
quantity of first type machine/ 10 /
try this:
Noise level of first type machine 80 dB
quantity of first type machine 2

Noise level of 2nd type machine 90 dB
quantity of 2nd type machine 5

Noise level of 3rd type machine 30 dB
quantity of 3rd type machine 1

Noise level of 4th type machine 25 dB
quantity of 4th type machine 3

Noise level of 5th type machine 100 dB
quantity of 5tht type machine 4

And the result will be :
Resultant noise of all machines equals : 106.5513845 dB


As for the 7.98db when all inputs are one is because no input can be turned to zero because that the formula divides dy the inputs so dividing by zero will be an indeterminated number . the 7.98db will go when you input ur data , i've just tried it .
As for transforming for lower case , i didn't see that .
If you please send me the adress of cells where u inputed the data and where it appeared in lower case so i can fix the problem

[botrous]

Please note that for adding many noise levels from different sources the input noise level should be from the same distance for all the machines .
eg. machine one : 85 dB at 10 m
machine two can't be 90 dB at 20m
If you have such data and you want to calculate the resultant noise level just go to noise comparing section and transform the 90 dB at 20 m to what will be at 10 m then input it in the noise adding section.