# Abatement and Distance from Source

## The disruption of sound pressures waves which reduces the noise is called attenuation - Sound Pressure Level Calculator

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### Spherical Distance

The sound pressure in a spherical distance from a source can be expressed as:

p^{2}= ρ c N / (4 π r^{2})(1)

where

p= sound pressure (Pa, N/m^{2})

ρ= density of air (kg/m^{3})

c= speed of sound (m/s)

N= sound power (W)

π= 3.14

r= distance from source (m)

### Half Spherical Distance

The sound pressure in a half spherical distance from a source can be expressed as:

p^{2}= ρ c N / (4 π r^{2}/ 2) = 2 ρ c N / (4 π r^{2})(2)

A more generic expression can be formulated to:

p^{2}=Dρ c N / (4 π r^{2})(3)

where

D= directivity coefficient (1 spherical, 2 half sperical)

The directivity coefficient depends on several parameters - the position and direction of the source, the room or the surrounding area, etc.

The Sound Pressure Level - *L _{p}*

*-*can be expressed logarithmic as:

L_{p}= 20 log( p / p_{ref}) =20 log ((Dρ c N / (4 π r^{2}))^{1/2}/ p_{ref})

= 20 log( 1 / r (Dρ c N / (4 π))^{1/2}/ p_{ref})(4)

where

L_{p}= sound pressure level (dB)

p_{ref}= 2 10^{-5}- reference sound pressure (Pa)

Note! that for every doubling of the distance from the noise source, the sound pressure levels - *L*_{p,} will be reduced by *6 decibels*.

### Sound Pressure Level Calculator

### Example - Sound Pressure from Wood Planer

If the sound power generated from a wood planer can be estimated to *0.01 W - *the sound pressure in distance *10 m* can be calculated as

*L _{p} = *

*20 log ((D*

*ρ c N / (4 π r*

^{2}))^{1/2}/ p_{ref})* = 20 log(2 (1 kg/m ^{3}) (331.2 m/s) (0.01 W) / (4 π (10 m)^{2}))^{1/2} / (2 10^{-5} Pa))*

* = 71 dB *

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