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Summer Semester, 2016
104450 Engineering Acoustics
MECH ENG 4115, 7027
Official Reading Time: 10 mins
Writing Time: 180 mins
Total Duration: 190 mins
Parts Questions Time Marks
A Answer all questions Recommended time 90 minutes 17 Marks
B Answer all questions Recommended time 90 minutes 19 Marks
Total: 36 Marks
Instructions for Candidates
It is recommended that you spend 10 minutes reading this paper.
Part A and Part B are Open Book.
Answer Part A on the Multiple Choice Answer Sheet using a BLACK ink pen only.
Answer Part B in the pink answer book.
Examination materials must not be removed from the examination room.
Begin each answer on a new page in the answer book.
Write your name and Student ID number on all loose diagrams/papers.
Permitted Materials
Part A and B – Open-book.
English language dictionaries are permitted.
A calculator without alphanumeric memory or remote communications capability is permitted.
Graphics calculator permitted.
DO NOT COMMENCE WRITING UNTIL INSTRUCTED TO DO SO
Course ID: 104450 Page 2 of 16
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Part A (Total: 17 Marks)
Answer all questions for Part A on the Multiple Choice Answer sheet.
Make sure your student ID number is completed on the multiple choice answer sheet.
Only use a pen with BLACK ink to mark the sheet. Do NOT use a pencil.
A sound level meter measures unweighted (linear) sound pressure levels. The SPL
measured in each octave band from 63 Hz to 8000 Hz as 100 dB re 20 micro-Pa
i.e. a flat spectrum. The A-weighted sound pressure level is approximately
A. 94 dB re 20 micro-Pa
B. 100 dB re 20 micro-Pa
C. 107 dB re 20 micro-Pa
D. 214 dB re 20 micro-Pa
E. None of the above.
[1 mark]
ANSWER: (C) 107 dB
The SPL is shown in the table
Freq
A-weight
correction
A
weighted
Level p^2
63 -26.2 73.8 23988329
125 -16.1 83.9 2.45E+08
250 -8.6 91.4 1.38E+09
500 -3.2 96.8 4.79E+09
1000 0 100 1E+10
2000 1.2 101.2 1.32E+10
4000 1 101 1.26E+10
8000 -1.1 98.9 7.76E+09
sum p^2 5E+10
SPL (A) 106.9871
Wind flowing over a power line generates “singing” tones that are caused by vortex
shedding and can be modelled as an aerodynamic dipole sound source. In order to double
the perceived loudness of the signing tone, the velocity of the wind would have to increase
by approximately:
A. Increase by 20%
B. Increase by 50%
C. Increase by 100%
D. Increase by 150%
E. None of the above.
[1 mark]
Answer: (B) increase by 50%
The sound power from a dipole source is proportional to the U^6.
In order to double the perceived loudness, the SPL would have to increase by
10 dB.
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Hence the sound power level would also have to increase by 10dB, which is
equivalent to an increase of 10^(10/10) =10x.
As sound power W is proportional to U^6, 10=U^6, then U=10^(1/6) = 1.467
A megaphone (also known as a speaking-trumpet, bullhorn, or loud hailer) is a hand-held,
cone-shaped acoustic horn used to amplify a person’s voice and direct it in a given
direction.
One of the acoustic principles by which the megaphone operates is:
A. improves the acoustic impedance matching between the inlet and exit of the
horn.
B. shortens the distance between the transmitter and receiver.
C. is an acoustic resonant device with low acoustic damping (i.e. no sound
absorptive material) so it amplifies sounds
D. simulates a high speed jet exhaust that has a sound power radiation proportional
to the power of eight of the jet velocity.
E. None of the above.
[1 mark]
Answer: (A) improves the impedance matching between the inlet and exit of the horn.
The gradual change in the open area helps to improve the coupling and also
directs sound in a particular direction.
https://en.wikipedia.org/wiki/Megaphone
A megaphone, speaking-trumpet, bullhorn, or loud hailer is a portable, usually hand-held,
cone-shaped acoustic horn used to amplify a person’s voice or other sounds and direct it
in a given direction. The sound is introduced into the narrow end of the megaphone, by
holding it up to the face and speaking into it, and the sound waves radiate out the wide
end. The megaphone increases the volume of sound by increasing the acoustic
impedance seen by the vocal cords, matching the impedance of the vocal cords to the air,
so that more sound power is radiated. It also serves to direct the sound waves in the
direction the horn is pointing. It somewhat distorts the sound of the voice because the
frequency response of the megaphone is greater at higher sound frequencies.
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Two adjoining offices have a partition that separates them. The transmission loss of the
partition is listed in the following table in one-third octave bands.
Freq
[Hz] TL [dB]
Freq
[Hz] TL [dB]
100 10 630 30
125 12 800 30
160 13 1000 30
200 14 1250 30
250 15 1600 30
315 16 2000 30
400 15 2500 30
500 30 3150 10
If the ambient noise level in the office is 50 dB(A) re 20 micro-Pa, what is the expected
speech privacy between the offices.
A. Intelligible
B. Ranging between intelligible and unintelligible
C. Audible but not obtrusive (unintelligible)
D. Inaudible
E. None of these.
[2 marks]
ANSWER: ( A ) Intelligible.
The average sound transmission loss between 100 Hz and 3150 Hz is the sum of levels
divided by the number of bands
(10 12 13 14 15 16 15 30 30 30
30 30 30 30 30 10 ) / 16
= 345 / 16 = 21.56 dB
Hence the average sound insulation plus 50 dB(A) =21.6 + 50 = 71. 6 dB
According to Table 4.10 in the textbook:
Sound as heard by occupant Average sound insulation a
plus ambient noise(dB(A))
Intelligible 70
Ranging between intelligible and unintelligible 75-80
Audible but not obtrusive (unintelligible) 80-90
Inaudible 90
Hence, speech from one office would be Intelligible by an occupant of the adjoining office.
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The part of the human ear that acts to convert motion of liquid into electrical signals is the
A. Ear drum
B. Stapes
C. Eustachian tube
D. Helicotrema
E. None of the above.
[1 mark]
Answer: None of the above.
The Stereocelia in the basilar membrane act to convert vibration in the fluid into electrical
signals.
An employee who works for 8 hours per day is exposed to the following A-weighted overall
sound pressure levels.
Duration
[ hours ]
SPL
[ dB(A) re 20 micro-Pa ]
1 80
2 87
3 95
1 85
1 70
TOTAL 8 hours.
What is the employee’s overall 8 hour A-weighted sound pressure level
A. 86.8 dB(A)
B. 91.3 dB(A)
C. 92.6 dB(A)
D. 100.4 dB(A)
E. None of these.
[1 mark]
ANSWER: B) 91.3 dB(A)
Use Eq (4.2) p139
,8 = 10 log10 [
1
8
× ( 110 1/10 + 210 2/10 + )]
,8 = 10 log10 [
1
8
× (
1 × 10
80
10 + 2 × 10
87
10 +
3 × 10
95
10 + 1 × 10
85
10 + 1 × 10
70
10)]
= 91.3 dB(A)
An employee is exposed to an LAeq,8hr sound pressure level of 90 dB(A) re 20 micro-Pa.
Using the Australian National Standard for Occupational Noise, the acceptable length of
time the employee can be exposed to this SPL is approximately:
A. 1.5 hours
B. 2.5 hours
C. 4.0 hours
D. 8 hours
E. None of the above.
[1 mark]
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ANSWER: (B) 2.5 hours
From Eq 4.42 in ENC,
= 8 × 2
(90 85)/ 3 = 2.51
The sound pressure levels in octave bands measured inside an office room are shown in
the following table.
Frequency
[Hz]
63 125 250 500 1000 2000 4000 8000
SPL
[dB re 20 micro-Pa]
45.0 35.0 30.0 30.0 35.0 35.0 40.0 35.0
The NCB rating of the room would be
A. 35 Hissy
B. 42
C. 45 Hissy
D. 48 Rumbly
E. None of the above.
[2 marks]
ANSWER: A) 35 Hissy
The rating is calculated using the arithmetic average of the values in the 4 octave
bands
500 1000 2000 4000
30.0 35.0 35.0 40.0
== 35.
To determine if the noise is “hissy”, the NCB curve which is the best fit of the
octave band sound levels between 125Hz and 500Hz is determined. If any of the
octave band sound levels between 1000Hz and 8000Hz inclusive exceed this
curve, then the noise is rated as “hissy”.
Use the plot on page 175 of the text book.
If one draws the levels and the line of best fit from 125Hz to 500Hz, the SPLs from
1000Hz to 8000Hz are much higher than this best fit lines, so it would be described
as Hissy.
Using ENC
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A large diesel engine is used to generate electricity. The exhaust pipe attached to the
diesel engine is 5m long, 100mm in diameter and points vertically upwards and radiates
noise. The end of the exhaust pipe has an acoustic condition that is best described by an
acoustic impedance equivalent to:
A. an anechoic termination.
B. a pressure release boundary condition.
C. a velocity release boundary condition.
D. a piston radiating into an infinite space.
E. None of the above.
[1 mark]
ANSWER: (D) a piston radiating into an infinite space.
Higher order acoustic modes can propagate in a circular duct when the frequency of
sound in the duct is above the “cut-on” frequency of the duct cross-section for each mode.
Select the statement that is true (correct) from the following:
A. The first circumferential mode of a circular duct occurs at a frequency that is
higher than the first diametral mode of the duct.
B. The first diametral mode of a circular duct occurs at a frequency that is higher
than the first circumferential mode.
C. The second diametral mode of a circular duct occurs at a frequency that is
higher than the first circumferential mode.
D. The second diametral mode of a circular duct occurs at a frequency that is
higher than the second circumferential mode.
E. None of the above.
[2 marks]
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ANSWER: A. The first circumferential mode of a circular duct occurs at a
frequency that is higher than the first diametral mode of the duct.
The resonance frequency of a pipe or cylindrical room is given by Eq (7.20)
=

2
√(

)
2
+ (

)
2
Hz
See page 297 of the text book for the table for the values of
mn 0 1 2 3 4
0 0 1.2197 2.2331 3.2383 4.2411
1 0.5861 1.6971 2.7172 3.7261 4.7312
2 0.9722 2.1346 3.1734 4.1923 5.2036
3 1.3373 2.5513 3.6115 4.6428 5.6623
4 1.6926 2.9547 4.0368 5.0815 6.1103
Where m = diametral pressure nodes
n=circumferential pressure nodes.
The axial modes nz are not important to this problem. The focus is to investigate
the term . To solve this problem one needs to sort the values of into
increasing order
m
diametral
n
circumferential
Comment
0 0 0 Plane wave mode
0.5861 1 0 1st radial mode
0.9722 2 0 2nd radial mode
1.2197 0 1 1st circumferential mode
1.3373 3 0
1.6926 4 0
1.6971 1 1
Going through the options:
A. The first circumferential mode ( = 1.2 ) of a circular duct occurs at a
frequency that is higher than the first diametral mode of the duct ( = 0.58
). HENCE STATEMENT A IS CORRECT
B. The first diametral mode ( = 0.58 )of a circular duct occurs at a
frequency that is higher than the first circumferential mode ( = 1.2 ).
HENCE THE STATEMENT IS INCORRECT
C. The second diametral mode ( = 0.97 )of a circular duct occurs at a
frequency that is higher than the first circumferential mode ( = 1.2 ).
HENCE THE STATEMENT IS INCORRECT
D. The second diametral mode ( = 0.97) of a circular duct occurs at a
frequency that is higher than the second circumferential mode ( =
1.69 2.23 ). HENCE THE STATEMENT IS INCORRECT.
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Note: Depending on how you interpret the response, 1,1 = 1.69 which
is the (1,1) mode is the second occurrence of a circumferential mode in
the table, or 2,0 = 2.23 is the (0,2) mode and the 0 radial and 2
circumferential mode. Either way, 0.97 < 1.69 or 0.97 < 2.23 hence the statement is incorrect. An impulse response method (balloon pop) was used to measure the reverberation time of a large room. The measured decay in sound pressure level versus time is shown in the graph below. The T60 reverberation time is closest to: A. 1.5 seconds B. 2.5 seconds C. 4.0 seconds D. 5.0 seconds E. Cannot be determined from the information supplied [2 marks] ANSWER: (B) 2.5 seconds Draw a line on the graph to determine the slope. -60 -50 -40 -30 -20 -10 0 0 0.5 1 1.5 2 Time [s] Sound Pressure Level of Impulse in Room Relative SPL [dB] Course ID: 104450 Page 10 of 16 PLEASE SEE NEXT PAGE X1 = 0.2 s y1 = -6 dB X2 = 1.2 s y2 = -30 dB Hence the slope is = 30 ( 6) (1.2 0.2) = 24 / Then the T60 value would be about 60 = 60 24 = 2.5 For interest, the accurately calculated T60 value over the full bandwidth using the RoomScope software was T60=2.35 s. A loud air compressor has a sound power rating of 110 dB re 1 pW in the 1000Hz octave band. An acoustic enclosure is placed over the compressor that has a cubic shape with 3m long edges, a transmission loss of 30 dB, and the internal acoustic conditions within the enclosure can be described as “average”. Assume that the speed of sound is 343 m/s, the density of air is 1.21 kg /m3 , and that the ground between the enclosure and the boundary is hard and reflects all sound. What is the expected sound pressure level at a distance of 30 m from the enclosure in the 1000 Hz octave band A. 42.5 dB re 20 micro-Pa B. 44.5 dB re 20 micro-Pa C. 47.5 dB re 20 micro-Pa [2 marks] -60 -50 -40 -30 -20 -10 0 0 0.5 1 1.5 2 Time [s] Sound Pressure Level of Impulse in Warehouse Relative SPL [dB] Course ID: 104450 Page 11 of 16 PLEASE SEE NEXT PAGE D. 68.5 dB re 20 micro-Pa E. None of the above. ANSWER: (C) 47.5 dB Immediately outside the enclosure the SPL is (ENC Eq. 8.85) 1 = 10 log10 + = 110 30 10 × log10(5 × (3 × 3)) + 5 = 68.46 At 30m distance from the enclosure the SPL is (ENC Eq. 8.87) 2 = 1 + 10 log10 + 10 log10 [ 4 2 ] = 68.46 + 10 log10(5 × 9) + 10 log10 [ 2 4 × 302 ] = 47.47 The incorrect (distractor) answers are: 1 = 10 log10 + 2 = 1 + 10 log10 + 10 log10 [ 4 2 ] 2 = + + 10 log10 [ 4 2 ] 2 = 110 30 + 5 + 10 log10 [ 1 4 × 302 ] = 44.47 2 = 68.46 Course ID: 104450 Page 12 of 16 Part B (Total: 19 Marks) Answer all questions in a new pink answer book. Clearly indicate “Part B” on the front of the answer book. Answer each question on a new page. Two uncorrelated spherically radiating monopole sources (in air) are 15m and 5m respectively from a receiver point. The sound pressure level at the receiver point is 80 dB re 20 micro-Pa. The sound power level of the source at 15m from the receiver is 110 dB re 10-12W. Assume that the density of air is = 1.21 / 3 and that the speed of sound is = 343 / . What is the sound power level of the source that is 5m from the receiver point [6 marks] ANSWER: The total sound level from two uncorrelated noise sources is determined by adding the square pressure contribution from each source. Hence 2 = 1 2 + 2 2 Where = 10 log [ 2 2 ] 2 = ( 2 ) × 10 10 For a monopole source the squared pressure is 2 = 4 2 Hence the total squared pressure from the two sources is 2 = 1 4 1 2 + 2 4 2 2 = 4 × [ 1 1 2 + 2 2 2 ] Rearranging this around for 1 1 = { [( 2 ) × 10 10] × 4 2 2 2 } × 1 2 2 = 10 log10 [ 2 ] 2 = × 10 2 10 1 = { [( 20 × 10 6 ) 2 × 1080 10] × 4 1.21 × 343 [10 12 × 10110 10] 152 } × 5 2 = 0.019167 1 = 10 log [ 0.019167 10 12 ] = 102.83 10 12 An alternative solution method is… The SPL at the receiver due to the source at 15m is ENC Eq 7.43, p305 1 = + 10 log10 [ 4 1 2 ] + 10 log10 [ 400] 1 = 110 + 10 log10 [ 4 1 152 ] + 10 log10 [ 1.21 × 343 400 ] = 75.646 dB We are told that the 2 sources are uncorrelated, hence incoherent, so the total SPL Lpt comprises Lp1 and Lp2 as ENC Eq1.98, p48 Course ID: 104450 Page 13 of 16 = 10 log10 [10 10 1 + 10 10 2 ] = 80 dB re 20 micro-Pa Hence this can be re-arranged to find Lp2 as 2 = 10 log10[1080/10 10 1/10] 2 = 10 log10[1080/10 1075.646/10] 2 = 78.014 dB re 20 micro-Pa So the sound power level of the second source which is 5m away is 1 = 78.014 = + 10 log10 [ 1 4 2 ] + 10 log10 [ 400] = 78.014 10 log10 [ 4 1 5 2 ] 10 log10 [ 1.21 × 343 400 ] = 102.826 dB re 10 12W A large room that is used for exhibiting art work has acoustically hard surfaces that results in a long reverberation time of 6.0 seconds, which makes the space acoustically uncomfortable. The architect has selected some decorative acoustic panels that can be used to reduce the reverberation time of the room, that have an absorption coefficient of 0.6. The volume of the room is 4000 m3 and the surface area of the room is 1890 m2 . What area of acoustic panels should be installed to reduce the reverberation time to 4.0 seconds Assume that the speed of sound c = 343 m/s, the acoustics of the room can be approximated using the Sabine model, and that when the absorptive panels are installed, they do not replace or remove the existing exposed surfaces. [6 marks] ANSWER: Dimensions of room: 25 x 25 x 6.4 m Volume of room V=4000 m3 Surface area of room S = 2x (25 x 25) + 2 x (25 x 6.4 + 25x 6.4) = 1890 m2 ENC Eq 7.52 60 = 55.25 × The original before modification is before before which is given by before before = 55.25 × 60,before And hence the average absorption coefficient is before = 55.25 × efore 60,before before = 55.25 × 4000 1890 × 343 × 6.0 before = 0.056818 After the modifications the new after after = 55.25 × 60,after Course ID: 104450 Page 14 of 16 after after = 55.25 × 4000 343 × 4.0 after after = 161.0787 Where the total surface area after modifications comprises the original surface area plus the surface area of the new absorptive panels after = before + panels. After installation of the panels the average Sabine absorption after is ENC Eq 7.79 = ∑ ∑ [∑ ] = ∑ [ after after ] = before before + panels panels Hence panels panels = [ after after ] before before panels = [ after after ] before before panels panels = [161.0787 ] [1890 × 0.056818] 0.6 panels = 89.49 m2 A large empty warehouse has a concrete floor with an absorption coefficient of = 0.03, and sheet metal walls and sheet metal roof that have an absorption coefficient of = 0.15. The warehouse has a height of 8.0 m, and the floor plan is drawn below showing the outline of the walls. The roof of the warehouse is flat and has the same area as the floor. Assume that the speed of sound of air is 343 m/s and the density of air is 1.21 kg / m3 . Using Sabine theory for room acoustics, estimate the T60 reverberation time. [4 marks] Course ID: 104450 Page 15 of 16 ANSWER: Reverberation Time of 3 Enterprise Court Dimensions of the warehouse Big area 33.6 m 14.4 m 23.2 483.84 m^2 Side rectangle 15.4 m 8.8 m 135.52 m^2 Total floor plan area 619.36 m^2 Height 8 m Volume 4954.88 m^3 Perimeter 113.600 m 113.6 CHECK Height 8 m Area of Walls 908.8 m^2 Area alpha Area*alpha Area of concrete floor 619.36 0.03 18.5808 Area of sheet metal wall 908.8 0.15 136.32 Area of sheet metal roof 619.36 0.15 92.904 Totals 2147.52 247.8048 0.115391 Average alpha T60 3.220784 seconds Norris Eyrring with air absorption 2.633262 seconds Measured over full bandwidth 2.35 seconds using Room Scope and balloon pop 23.2 m 15.4 m 8.8 m 33.6 m Course ID: 104450 Page 16 of 16 The measured reverberation time of the warehouse described in Question 15 was 2.35 seconds. The measurements were conducted properly and were independently verified. Provide 3 reasons why the calculated T60 reverberation time in Question 15 differs from the measured reverberation time. [3 marks] ANSWER: Q15 required the use of the Sabine theory to model the room acoustics, which makes some assumptions about the statistical nature of the room modes, that the acoustic absorption is evenly spread over the surfaces of the room, that the dimensions of the room are nearly cubic, which is not satisfied with the dimensions of the warehouse. There are several alternative formulations for determining the T60 value of unusual shaped rooms which are more appropriate such as Norris Erying and others. The dimensions of the warehouse are large, and air absorption is important, which the Sabine theory ignores. Alternative formulations incorporate air absorption. If one uses a Norris Eyring formulation with air absorption, the predicted T60 value is about 2.6 s, which is closer to the measured value of 2.35 seconds. The large warehouse is not air-tight and there are acoustic “leaks” around large doorways, which would alter the absorption coefficient used for the predictions. The absorption coefficients used for the predictions may not have been accurate. Although the values used for a concrete floor are likely to be accurate, the absorption coefficient for the corrugated steel walls and roofing, that comprise the majority of the Sabine area for the space, could be in error, leading to inaccurate predictions. END OF EXAMINATION