How to calculate the sound pressure from the sound power
As will be seen, the power and sound pressure are two very different concepts that are often confused very easily. Sometimes, there are manufacturers that contribute to this confusion by commercial interests, since the value of sound pressure is always equal to or less than the value of sound power. In anticipation of what we'll see in this article, should we have very clear that to compare two units acoustically, always have to ask the values of sound power, as the values of sound pressure are not directly comparable .
This article is the second part of the series "Working with Decibels". In the first part we talked about the concepts of Decibel, A-weighting and how to add sound spectra and get the dB(A), if you have not read the first article, or if you simply want to refresh the concepts, you can access the article at this link .
We'll start by seeing the differences between power and sound pressure and later explain how to calculate the sound pressure from the sound power.
Sound power, acoustic power or acoustic power level (LW)
The acoustic power is the amount of energy per unit time (power) emitted from an acoustic source in the form of sound waves.
The acoustic power is independent of the environment, in ther words, it is an intrinsic value of the source itself and not depend on the location and the distance we are from the sound source, therefore the value to be used to compare units from different manufacturers because it is independent of the place and conditions in which they have carried out acoustic tests.
Sound pressure, acoustic pressure or acoustic pressure level (LP)
The sound pressure is a pressure disturbance in the atmosphere, the intensity of which depends on the sound source itself, the environment in which the source is located and the distance between the sound source and receiver.
Our ears and acoustic measurement instruments are able to perceive the sound pressure (LP), but not the sound power (LW).
Relation between sound power (LW) and sound pressure (LP)
As seen, measuring instruments capture the sound pressure level (LP), for the sound power level (LW) you have to "add" to the measured values, the acoustic attenuation of the room in which you are making the measurement, and the effect of the distance between the sound source and the measuring device. Therefore, the sound pressure value is always equal to or less than the sound power value .
Directivity of the sound source
Depending on the location of the sound source with respect to the premises, there are the following four cases for the directivity of the source ():
. : When the sound source is at the absolute center of the room, ie, equidistant from the floor and ceiling and equidistant between the walls. In this case it said sound radiation is spherical.
. : When the sound source is equidistant from the walls and is supported on the floor or ceiling. Thus, sound radiation is semi-spherical.
. : When the sound source is located in the center of a wall and resting on the floor or ceiling. That is, there are two sides supported on the premises. In this case, the sound radiation will be about 1/4 of a sphere.
. : When the sound source is in a corner of the room, that is, having three sides supported on the spot. In this situation, the sound radiation will be about 1/8.
Getting sound pressire (LP) from the sound power(LW)(LW)
The relationship between sound pressure and sound power can be obtained using the formula Bies and Hansen (1996):
. is the Sound Pressure dB
. is the Sound Power dB
. Sound source directivity can be 1, 2, 4 or 8
. is the distance to the sound source (m).
. is the constant of the site
. is the sum of the surfaces forming the site (m2)
. is the mean value of the acoustic absorption coefficient of the site
. is the number of equal sound sources in the site
Suppose you have the following information:
- We have an AHU installed in the center of the floor of the room ( Q = 2) whose radiated sound power is 70 dB.
- A 10 meters long room, 8 wide and 4 high.
- The floor of the room has an absorption coefficient of 0.7, the ceiling 0.2 and the walls 0.05.
You want to know the sound pressure in these conditions a a distance of 1.5 m from the AHU.
We begin by calculating the area of the room and the average sound absorption coefficient:
a) Ceiling area = Superficie de suelo = 10 x 8 = 80 m2.
b) Walls area = 2 x (10 x 4 + 8 x 4) = 144 m2
c) Room total area .
d) = (80 x 0.7) + (80 x 0.2) + (72 x 0.05) / (304) = 75,6 / 304 = 0.25
Now we calculate the room constant:
With these data, we can calculate the acoustic attenuation of the room:
So the Sound Pressure is:
70 - 9.6 = 60.4 dB