With reference to IEM Penang Training dated 20 March 2012, Noise Prevention in Factories

1. Noise Prediction/Prevention, An Industrial
Calculation Approach
Prepared by
Gan Chun Chet
MSc (UK), BSc (UK), PEng (MAL)
20 Nov 2020
Email : chun_gan@hotmail.com
Whatsapp (019-2938364) (Malaysia)

2. Sound Level (Noise) Equations
• Sound Power Level SWL
SWL = 10 log(
𝑊
𝑊𝑜
) [dB] ; 𝑊𝑜 = 10−12 𝑊 (watts)
• Sound Intensity Level IL
IL = 10 log(
𝐼
𝐼 𝑜
) [dB] ; 𝐼 𝑜 = 10−12 𝑊/ 𝑚2
• Sound Pressure Level SPL
SWL = 20 log(
𝑝
𝑝 𝑜
) [dB] ; 𝑃𝑜 = 20 μPa

3. Noise Equations – Point Source
Inverse Square Law
• Spherical Wave
I =
𝑊
4𝜋𝑟2 { I α
1
𝑟2 }
• Hemisphere
I =
𝑊
2𝜋𝑟2 { I α
1
𝑟2 }
(Note : Sound Intensity Level IL
reduces by 6.02 dB by doubling
the distance, Example 1 is used to
derived this decibel figure )
W
r
r
W

4. Noise Equations – Line Source
Inverse Distance Law
• Distance
I =
𝑊
𝜋𝑟
{ I α
1
𝑟
}
(Note : Sound Intensity Level IL
reduces by 3.01 dB by doubling the
distance, similarly, Example 1 is used
to derived this decibel figure)
W
r
I

5. Example 1
Calculate noise at source if SPL at 1 metre from source is 70dB
• Intensity at 1m :-
IL = 10 log(
𝐼1
𝐼 𝑜
) = 70
𝐼1
10−12 = 107
𝐼1= 10−5 𝑊/ 𝑚2
I =
𝑊
4𝜋𝑟2 ; spherical wave

6. Example 1 (con’t)
• Point Source, spherical wave pattern
W = 4 π 𝑟2 𝐼1
= 4 π 12 10−5
Therefore, SWL = 10 log(
𝑊1
𝑊𝑜
)
= 10 log(
4 π 12 10−5
10−12 )
= 81 dB

7. Example 1 (con’t)
• Intensity at 10m :-
By Inverse Square Law
I α
1
𝑟2
𝐼10 𝑟10
2
= 𝐼1 𝑟1
2
𝐼10
𝐼1
= (
𝑟1
𝑟10
)2
𝐼10
𝐼1
= (
1
10
)2
=
1
100

8. Example 1 (con’t)
IL = 10 log (
𝐼1
𝐼 𝑜
)
= 10 log (
𝐼10
𝐼1
) ; 𝐼10relative to 𝐼1
= 10 log (
1
100
)
= 10 𝑙𝑜𝑔10 10−2
= -20 dB
So, IL = 𝐼𝐿1 – 𝐼𝐿10 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑡𝑜 1
= 70 – 20 = 50 dB
81 70 50
1m 10msource

9. Example 2
Calculate noise level for given multiple (three) Sound Intensity Level IL [dB]
• 𝐼𝐿1= 77 dB
• 𝐼𝐿2= 80 dB
• 𝐼𝐿3= 88 dB
𝐼𝐿1 = 10 log(
𝐼1
𝐼 𝑜
) [dB]
= 10 log(
𝐼1
𝐼 𝑜
) = 77
𝐼1
𝐼 𝑜
= 107.7
Similarly,
𝐼2
𝐼 𝑜
= 108.0
𝐼1
𝐼 𝑜
+
𝐼2
𝐼 𝑜
= 107.7
+ 108.0
𝐼1+ 𝐼2
𝐼0
= 107.7+ 108.0

10. Example 2 (con’t)
10 log (
𝐼1+ 𝐼2
𝐼0
) = 10 log ( 107.7+ 108.0)
𝐼𝐿1+2= 10 log (
𝐼1+𝐼2
𝐼0
)
= 10 𝑙𝑜𝑔10 ( 107.7+108.0 )
= 81.76 dB ≈ 82 dB

11. Example 2 (con’t)
• For 𝐼𝐿1+2 = 81.76 dB
𝐼𝐿3= 88 dB
𝐼𝐿1+2 = 10 log(
𝐼1+2
𝐼 𝑜
) [dB]
= 10 log(
𝐼1+2
𝐼 𝑜
) = 81.76
𝐼1+2
𝐼0
= 108.176
Similarly,
𝐼3
𝐼 𝑜
= 108.8
𝐼1+2
𝐼 𝑜
+
𝐼3
𝐼 𝑜
= 108.176
+ 108.8
𝐼1+2+ 𝐼3
𝐼0
= 108.176
+ 108.8

12. Example 2 (con’t)
10 log (
𝐼1+2+ 𝐼3
𝐼0
) = 10 log ( 108.176+ 108.8)
𝐼𝐿1+2+3= 10 log (
𝐼1+2+𝐼3
𝐼0
)
= 10 𝑙𝑜𝑔10 ( 108.176+108.8 )
= 88.92 dB ≈ 89 dB

13. Example 3
Calculate noise level for given multiple Sound Intensity Level IL [dB] at various
frequency
• 𝐼𝐿1= 60 dB @ 8000Hz
• 𝐼𝐿2= 60 dB @ 4000Hz
• 𝐼𝐿3= 61 dB @ 2000Hz
{work/move from highest frequency to lower}
𝐼1
𝐼0
= 106.0
,
𝐼2
𝐼 𝑜
= 106.0
𝐼1 + 𝐼2 = (106.0 + 106.0) 𝐼0

14. Example 3 (con’t)
10 log (
𝐼1+ 𝐼2
𝐼0
) = 10 log (106.0 + 106.0)
𝐼𝐿1+2 = 10 𝑙𝑜𝑔 (106.0 + 106.0)
= 63 dB
𝐼1+2
𝐼0
= 106.3
,
𝐼3
𝐼 𝑜
= 106.1
𝐼1+2 + 𝐼3 = (106.3
+ 106.1
) 𝐼0

15. Example 3 (con’t)
10 log (
𝐼1+2+ 𝐼3
𝐼0
) = 10 log (106.3 + 106.1)
𝐼𝐿1+2 = 10 𝑙𝑜𝑔 (106.3 + 106.1)
= 65.1 dB

16. The Total Energy of Noise Source for unsteady
noise over a time period
• Equivalent Steady Sound Level 𝐿 𝑒𝑞
𝐿 𝑒𝑞 = 10 log (
1
𝑇 0
𝑇
10
𝐼𝐿
10 dt )
Cause IL = 10 log (
𝐼
𝐼0
) , then (
𝐼
𝐼0
) = 10
𝐼𝐿
10
in linear scale
in decibel scale

17. • Equivalent Noise Exposure Level for 8 hours
𝐿 𝑒𝑞,8 = 10 log ( 𝑖=1
𝑛 Δ𝑡
8
10
𝑆𝑃𝐿
10 )
Where 𝐿 𝑒𝑞,8 is equivalent sound exposure level in 8 hours
∑ is the sum of the exposure for i=1 to i=n
i is discrete value of each exposure, for a specific duration
Δt is the exposure for that particular duration
SPL is sound pressure level in decibel sound meter reading
dB(A), weighted “A” scale

18. Example 4
1 Discrete exposure period, 10 minutes of Leq 120 dB(A), what is the Equivalent
Steady Sound Level for 40 hours duration Leq,40
SPL = 120 dB(A) for 10 minutes
𝐿 𝑒𝑞,40 = 10 log ( 𝑖=1
𝑛 Δ𝑡
40
10
𝑆𝑃𝐿
10 )
= 10 log (
10/60
40
∗ 10
120
10 )
= 96.1 dB(A)

19. • Partial Noise Exposure Index
𝐸𝑖 =
Δ𝑡𝑖
𝑇
100.1(𝐿 𝑖−70)
then 𝐿 𝑒𝑞,𝑇 = 70 + 10 log ( ∑ 𝐸𝑖) [dB(A)] – [1]
𝐸𝑖 =
Δ𝑡 𝑖
𝑇
∗
10
𝐿 𝑖
10
10
70
10
Substituting into equation [1]
𝐿 𝑒𝑞,𝑇 = 70 + 10 log (∑
Δ𝑡 𝑖
𝑇
* 10
𝐿𝑖
10 ) - 10 log ( 10
70
10 )
= 10 log (∑
Δ𝑡 𝑖
𝑇
* 10
𝐿𝑖
10 )

20. Example 5
Calculate the equivalent noise exposure level for the exposure duration
i SPL dBA (𝐿𝑖) Duration (Δ𝑡𝑖) ∑ 𝐸𝑖 (120)
1 100 1 25
2 95 9 71.2
3 85 30 23.7
𝐸1 =
1
40
100.1(100−70)
𝐸2 =
9
40
100.1(95−70)
𝐸3 =
30
40
100.1(85−70)
𝐿 𝑒𝑞,𝑇 = 70 + 10 log ( ∑ 𝐸𝑖)
𝐿 𝑒𝑞,𝑇 = 70 + 10 log ( 120 )
= 90.8 dB(A) approximately 91 dB(A)

21. Summary / Conclusion
• Figures used in Example 1, 2, 3, 4 and 5 are use for illustration only,
for the equations as derived, to the Equivalent Steady Sound Level
Leq.
• Noise Prediction/Prevention in this situation, is to reduce Sound
Pressure Level SPL dB(A) at noise source.
• With this, noise calculation shall be with reference to Factories and
Machinery Act 1967 (ACT 139) (Noise Exposure) Regulations 1989.