10.5Noise Reduction due to Acoustical Partitions

δ ¼ cSw

Direct Energy Density

The formula for direct energy density due to plane waves was derived in Sect. 10.1.3 and is shown below for reference:

where PRMS is the RMS pressure in the room radiated from the four walls. Because Pwall (RMS pressure on the wall) contributes to only part of PRMS (RMS

pressure on the room), PRMS can be related to Pwall with a ratio of SwallS as:

Therefore, the direct energy density δd can be formulated in terms of Pwall as:

Reverberant Energy Density

The reverberant energy density is a summation of all the reflected energy from the direct energy density:

δr ¼ δdhð1 αÞ þ ð1 αÞ2 þ . . . þ ð1 αÞni ¼ δd RS

where RS was derived in the previous section as:

S ¼ ð1 αÞ ¼ ð1 αÞ þ ð1 αÞ2 þ . . . þ ð1 αÞn

R α

Therefore:

δr ¼ δd RS

Total Energy Density

The total energy density δ is the sum of the direct and reverberant energy densities:

δ ¼ δd þ δr

¼δd þ δd RS

S

¼δd 1 þ R

Because the direct energy density δd can be formulated in terms of Pwall as:

therefore, the total energy density δ can be formulated in terms of Pwall as:

10.5.2 Sound Pressure Level due to a Partition Wall

Equating the energy density expression above to the energy density from the sound

pressure δ ¼ P2RMS (see Sect. 10.1.3) yields a relationship between the RMS pressure

ρoc2

in a room and the acoustic source, as shown below:

The ratio between the wall area Swall and the room surface S can be approximated as:

SwallS ¼ 14

Then:

Pr ¼ 20 10 6 [Pa], and take log base 10 of both sides of the above equation as:

where Lp, wall is the sound pressure level (SPL) near the wall without considering the reflection wave and LP is the SPL near the wall considering both the direct and reflected waves. Note that the first term 14 in the logarithmic function above is the result of the direct wave and the second term SwallR is the result of the reflection wave.

Sound Pressure Level Away from the Wall

The sound pressure level (SPL) located far from the wall is formulated based on the above formulas for SPL located near the wall. At locations far from the partition wall, considering only the reflection wave from the above equation, we obtain:

Pr ¼ 20 10 6 [Pa], and take log base 10 of both sides of the above equation to arrive at

LP ¼ Lp,wall þ 10 log Swall

R

10.5.3 Noise Reduction (NR)

The noise reduction of the enclosure is equal to the TL of the walls of the enclosure:

NR ¼ Lp1 Lp2

source

Example 10.2: Common Wall

Two rooms are separated by a common wall that has the dimensions 8 5 m2 with a TL of 30 dB. Room 1 contains a noise source that produces an SPL of 108 dB near the wall. The average absorption coefficient of room 2 is 0.4, and the total surface area of room 2 is 225 m2.

(a)What is the SPL near the wall in the second room?

(b)What is the SPL far from the wall in the second room?

Example 10.2: Solution

(a) The SPL near the wall in the second room is:

(a) The SPL far from the wall in the second room is:

Example 10.3

1 2 3

Example 10.4

Given:

Lp1 ¼ 100 dB

τ1 ¼ 0:01, τ2 ¼ 0:01

α1,2 ¼ 0:6, α2 ¼ 0:4

Sw ¼ 10 m2, S2 ¼ 200 m2

1 , 2

1 2

Determine the noise reduction of a double-leaf wall with sound absorption materials between the leaves.

Example 10.4: Solution

Lp1 ¼ 100 dB

τ1 ¼ 0:01

α1,2 ¼ 0:6

Sw ¼ 10 m2