B Science Report Project 2

1. 0
School of Architecture, Building and Design
Modern Architecture Studies in Southeast Asia (MASSA) Research Unit
Bachelor of Science (Honours) (Architecture)
Building Science 2 [BLD 61303 / ARC 3413]
PROJECT 02
INTEGRATION PROJECT
Student : Phua Jing Sern 0314572
Tutor : Mr. Sivaraman Kuppusamy

2. 1
Table of Content Page
1.0 Lighting
1.1 Silent Area
1.1.1 Daylight 2
1.1.2 Artificial Light 4
1.1.3 PSALI 5
1.2 Admin Office
1.2.1 Daylight 6
1.2.2 Artificial Light 8
1.2.3 PSALI 9
2.0 Acoustics
2.1 Silent Area
2.1.1 Sound Pressure Level (SPL) 10
2.1.2 Reverberation Time (RT) 12
2.1.3 Sound Reduction Index (SRI) 14
2.2 Admin Office
2.2.1 Sound Pressure Level (SPL) 16
2.2.2 Reverberation Time (RT) 18
2.2.3 Sound Reduction Index (SRI) 20
3.0 References 21

3. 2
1.0Lighting
1.1 Silent Area
1.1.1 Daylight
According to MS1525: Daylight Factor, DF
Zone Daylight Factors (%) Distribution
Very Bright > 6 Very large with thermal and glare
problems
Bright 3 - 6 Good
Average 1 - 3 Fair
Dark 0 - 1 Poor

4. 3
Daylight Factor Calculation
Floor Area (m2) 7m x 13m = 91
Area of façadeexposed to sunlight
(m2)
50m
Area of skylight 0
Exposed façade & skylightarea to
floor area ratio / Daylightfactor, DF
(50+0)/91
= 0.55
= 0.55 x100%
= 5.5%
Natural Illumination Calculation
Illuminance Example
120,000 lux Very Bright Sunlight
110,000 lux Bright Sunlight
20,000 lux Clear Sky
1000 – 2000 lux Overcastday
400 lux Sunrise/ Sunset on clear day
< 200 lux Midday
40 lux Fully overcast
<1 lux Sunset, stormcloud
E External = 20,000 lux
DF = E internal / E external x 100
5.5 = E internal / 20,000 x100
E internal = 5.5 x 20,000 /100
= 1100 lux
The silent area has a daylight factor of 5.5% and natural illumination of 1100 lux. Based on
the requirements of MS 1525, this space has good daylight distribution as the value is in
between 3-6%, but the illuminance value is higher than the required value of 300 lux. This
can cause thermal and glare problems. Low-e coatings of the glass panels are proposed to
minimize the penetration of ultraviolet and infrared rays into spaces.

5. 4
1.1.2 Artificial light
Lumen Method Calculation
Dimension of room( L x W) 7m x 13m
Total floor Area (m2) 91
Room cavity height (m) 4
Reflectance values Ceiling = 0.7
Wall = 0.5
Floor = 0.3
Room index, K K = (7 x 13) / 4(7+13)
= 1
Utilization Factor, UF 0.38
Maintenance Factor, MF 0.8
IlluminanceRequirement 300 - 500
Number of Luminaries N =
300 𝑥 91
4000 𝑥 (0.8𝑥0.38)
= 22.45
~ 23 bulbs
Spacing to height ration (SHR)
SHR =
1
4
x √
91
23
= 0.49
SHR =
𝑆
4
= 0.49
S = 1.96
Fitting Layout Fitting required along 7m wall
= 7/1.96 =3.5
= 4 rows
Fitting required along 13mwall
= 13/1.96 =6.6
= 7 rows
Model DN570B
Input 230 or 240 v / 50-60Hz
Lumen 4000
Weight 2.2 kg
Power 36w (3000k)

6. 5
Fitting layout and SPALI
28 fluorescent lamps are used to illuminate the quiet reading area to achieve the minimum
illuminance of 300 lux stated by MS 1525.With sufficient illuminance, users are able to
utilize the space with comfort.
Referring to the daylight values obtained above, the silent area has a good daylight factor of
5.5%. The artificial lightings calculation also shows a result of the usage of 28 down lights to
illuminate the silent area. PSALI is applied whereby the lightings are controlled with 2
switches; one that is closest to the window wall whereby it can be switched off during the
day, while the rest can be operate with another switch when necessary as well.

7. 6
1.2 Admin Office
1.2.1 Daylight
According to MS1525: Daylight Factor, DF
Zone Daylight Factors (%) Distribution
Very Bright > 6 Very large with thermal and glare
problems
Bright 3 - 6 Good
Average 1 - 3 Fair
Dark 0 - 1 Poor

8. 7
Daylight Factor Calculation
Floor Area (m2) 3.5m x 6m = 21
Area of façade exposed to sunlight (m2) 4
Area of skylight 0
Exposed façade & skylight area to floor area
ratio / Daylight factor, DF
(4+0)/21
= 0.19
= 0.19 x 100%
= 1.9%
Natural Illumination Calculation
Illuminance Example
120,000 lux Very Bright Sunlight
110,000 lux Bright Sunlight
20,000 lux Clear Sky
1000 – 2000 lux Overcast day
400 lux Sunrise / Sunset on clear day
< 200 lux Midday
40 lux Fully overcast
<1 lux Sunset, storm cloud
E External = 20,000 lux
DF = E internal / E external x 100
1.9 = E internal / 20,000 x100
E internal = 1.9 x 20,000 /100
= 380 lux
The admin office has a daylight factor of 1.9% and natural illumination of 380 lux. Based on
the requirements of MS 1525, this is considered as fair yet not adequately lit. Therefore,
artificial lighting is required for the space. The illuminance value fits within the MS 1525
requirement of 300 – 400 lux.

9. 8
1.2.2 Artificial light
Lumen Method Calculation
Dimension of room( L x W) 3.5mx 6m
Total floor Area (m2) 21
Room cavity height (m) 4
Reflectance values Ceiling = 0.7
Wall = 0.5
Floor = 0.3
Room index, K K = (3.5 x 6) / 4(3.5+6)
= 0.6
Utilization Factor, UF 0.27
Maintenance Factor, MF 0.8
IlluminanceRequirement 300 - 500
Number of Luminaries N =
300 𝑥 21
5000 𝑥 (0.8𝑥0.27)
= 5.83
~ 6
Spacing to height ration (SHR)
SHR =
1
4
x √
21
6
= 0.46
SHR =
𝑆
4
= 0.46
S = 1.84
Fitting Layout Fitting required along 3.5mwall
= 3.5/1.84 =1.9
= 2 rows
Fitting required along 6m wall
= 6/1.84 =3.2
= 4 rows
Model LuxiLED G2
Input 110 or 240 v / 50-60Hz
Lumen 5000
Weight 4.6
Power 32W (3000k)

10. 9
Fitting Layout and PSALI
8 down lights are used to illuminate the admin office to achieve the minimum illuminance of
300 lux stated by MS 1525.With sufficientilluminance,office staffs areableto utilizethe space
with comfort.
Referring to the daylight values obtained above, the admin office has a fair yet not adequately
lit daylight factor of 1.9%. Thus the artificial lightings calculation also shows a need of the
usage of 8 down lights to illuminate the admin office. PSALI is applied whereby the lightings
are controlled with 2 switches; one half controls the help desk and the corridor, while the
other half controls the work desks of the staffs. To achieve a uniform lighting and to follow
the requirements of MS 1525, both switches need to be turned on during operating hours.

11. 10
2.0 Acoustic
2.1 Silent Area
2.1.1 SPL – Sound Pressure Level
Sound Pressure Level Formula:
SPL = 10log10
𝐼
𝐼𝑥
SPL = sound pressure level (dB),
I = sound power (intensity)(Watts)
Ix = reference power (Ix is usually taken as 1x10-12 watts)
Peak Hour (Back Alley)
Highest Reading: 70 dB (Moderate sound)
Lowest Reading: 55 dB (Quiet sound)
Highest Reading:
70 = 10log10
𝐼
𝐼𝑥
Antilog 7 =
𝑙
1𝑥10 ^−12
1 X 10^7 =
𝑙
1𝑥10^−12
L = 10^7 x (1x10^-12)
L = 1 x [10^7 + (-12))
L = 1 x 10^-5
Lowest Reading:
55 = 10log10
𝐼
𝐼𝑥
Antilog 5.5 =
𝑙
1𝑥10 ^−12
1 X 10^5.5 =
𝑙
1𝑥10 ^−12
L = 10^5.5 x (1x10^-12)
L = 1 x [10^5.5+ (-12))
L = 1 x 10^-6.5
Total Intensities:
Total intensities, I
= (1x10^-5) + (1 x 10^-6.5)
= 1.03x10^-5
Combined SPL:
SPL = 10log10
𝐼
𝐼𝑥
= 10log10
1.03𝑥10^−5
𝐼𝑥10 ^−12
= 70 dB

12. 11
Non -Peak Hour (Back Alley)
Highest Reading: 55 dB (Quiet sound)
Lowest Reading: 45 dB (Very Quiet sound)
Highest Reading:
55 = 10log10
𝐼
𝐼𝑥
Antilog 5.5 =
𝑙
1𝑥10^−12
1 X 10^5.5 =
𝑙
1𝑥10^−12
L = 10^5.5 x(1x10^-12)
L = 1 x [10^5.5+(-12))
L = 1 x 10^-6.5
LowestReading:
45 = 10log10
𝐼
𝐼𝑥
Antilog 4.5 =
𝑙
1𝑥10^−12
1 X 10^4.5 =
𝑙
1𝑥10^−12
L = 10^4.5 x(1x10^-12)
L = 1 x [10^4.5+(-12))
L = 1 x 10^-7.5
The noise criteria for a reading area is within the range of 35-40. According to the calculation
above, the combined sound pressure level around the silent area during peak and non-peak
hours are 70dB and 55 dB respectively, which exceeds the noise criteria for a reading area.
This can be solved by proposing double layered channel glass which consists of an insulation
layer, preventing the transmission of sound through the vibration of air.
Total Intensities:
Total intensities, I
= (1x10^-6.5) +(1 x 10^-7.5)
= 3.48x10^-7
Combined SPL:
SPL = 10log10
𝐼
𝐼𝑥
= 10log10
3.48𝑥10^−7
𝐼𝑥10^−12
= 55 dB

13. 12
2.1.2 Reverberation Time (Silent Area)
Volume of Space
V = L x W x H
V = 13m x 7m x 4m
V = 364m3

14. 13
Material Absorption Coefficient under 500 Hz, with 15 users
Building
Component
Material Area S ( m2) Absorption
Coefficient (a)
Sound Absorption
( SA)
Wall Concrete 73.2 0.02 1.46
Glass 47 0.06 2.82
Floor Carpet 91 0.30 27.3
Door Timber 2 0.13 0.26
Ceiling Concrete 91 0.02 1.82
Furniture Desk 18 0.45 8.1
Padded Chair 6 0.77 4.62
Fabric Barrel 96 0.30 28.8
Users - 15 0.46 6.9
Total Absorption 81.88
RT= (0.16 X 364) /81.88
= 0.71s
Material Absorption Coefficient under 2000 Hz, with 15 users
Building
Component
Material Area S ( m2) Absorption
Coefficient (a)
Sound Absorption
( SA)
Wall Concrete 73.2 0.02 1.46
Glass 47 0.03 1.41
Floor Carpet 91 0.50 45.5
Door Timber 2 0.10 0.2
Ceiling Concrete 91 0.02 1.82
Furniture Desk 18 0.45 8.1
Padded Chair 6 0.77 4.62
Fabric Barrel 96 0.50 48
Users - 15 0.51 7.62
Total Absorption 118.73
RT= (0.16 X 364) /118.73
= 0.49s
The reverberation time required for a library reading space is 0.4 – 0.6 seconds. Based on the
calculation, the reverberation of the space under 500 Hz and 2000 Hz is 0.71s and 0.49s
respectively. As they slightly differ from the requirement, acoustic panels are introduced to
amend the issue.

15. 14
2.1.3 Sound Reduction Index (Silent Area)
Volume of Space
V = L x W x H
V = 13m x 7m x 4m
V = 364m3
Assuming that the noise level of a Silent Area is 40 dB, noise level of the external is 75dB. The
sound reduction index of the wall should 75dB- 40dB = 35 dB.

16. 15
Building
Component Material Surface Area
(m2
SRI
( Decibels)
Transmission
Coefficient
( T )
Wall Concrete 73.2 46 2.51 𝑥 10 −5
Wall Glass 47 35 3.16 𝑥 10 −4
Door Plywood 2 31 7.94 𝑥 10 −4
Concrete
SRI = 10 Log (1/T)
46 = 10 log (1/T)
3.98 𝑥 10 4
= 1 / T
T = 2.51 𝑥 10 −5
Glass
SRI = 10 Log (1/T)
35 = 10 log (1/T)
3.16 𝑥 10 3
= 1 / T
T = 3.16 𝑥 10 −4
Timber
SRI = 10 Log (1/T)
31 = 10 log (1/T)
10 3.1
= 10 log10 (1/T)
T = 7.94 𝑥 10 −4
The overall transmission loss from the back lane to the silent area is 38.24 dB. Based on the
previous assumption, according to the noise criteria environment perception, 35 dB is
categorized as whispering sound. Thus the space has an ideal value for a silent reading area
where users can focus on their reading with ease.
Average Transmission Coefficient
= [ 73.2 x ( 2.51 𝑥 10 −5
) ] + [ 47 x
( 3.16 𝑥 10 −4
) ] + [ 2 x (7.94 𝑥 10 −4
) ] /
( 73.2 + 47 + 2)
= 1.50 𝑥 10 −4
Overall SRI
= 10 Log ( 1 / 1.50 𝑥 10 −4
)
= 38.24 Db
External Sound PressureLevel
= 75dB– 38.24dB
= 36.76 dB

17. 16
2.2 Admin Office
2.2.1 SPL – Sound Pressure Level
Sound Pressure Level Formula:
SPL = 10log10
𝐼
𝐼𝑥
SPL = sound pressure level (dB),
I = sound power (intensity)(Watts)
Ix = reference power (Ix is usually taken as 1x10-12 watts)
Peak Hour (Back Alley)
Highest Reading: 70 dB (Moderate sound)
Lowest Reading: 55 dB (Quiet sound)
Highest Reading:
70 = 10log10
𝐼
𝐼𝑥
Antilog 7 =
𝑙
1𝑥10 ^−12
1 X 10^7 =
𝑙
1𝑥10^−12
L = 10^7 x (1x10^-12)
L = 1 x [10^7 + (-12))
L = 1 x 10^-5
Lowest Reading:
55 = 10log10
𝐼
𝐼𝑥
Antilog 5.5 =
𝑙
1𝑥10 ^−12
1 X 10^5.5 =
𝑙
1𝑥10 ^−12
L = 10^5.5 x (1x10^-12)
L = 1 x [10^5.5+ (-12))
L = 1 x 10^-6.5
Total Intensities:
Total intensities, I
= (1x10^-5) + (1 x 10^-6.5)
= 1.03x10^-5
Combined SPL:
SPL = 10log10
𝐼
𝐼𝑥
= 10log10
1.03𝑥10^−5
𝐼𝑥10 ^−12
= 70 dB

18. 17
Non -Peak Hour (Back Alley)
Highest Reading: 55 dB (Quiet sound)
Lowest Reading: 45 dB (Very Quiet sound)
Highest Reading:
55 = 10log10
𝐼
𝐼𝑥
Antilog 5.5 =
𝑙
1𝑥10^−12
1 X 10^5.5 =
𝑙
1𝑥10^−12
L = 10^5.5 x(1x10^-12)
L = 1 x [10^5.5+(-12))
L = 1 x 10^-6.5
LowestReading:
45 = 10log10
𝐼
𝐼𝑥
Antilog 4.5 =
𝑙
1𝑥10^−12
1 X 10^4.5 =
𝑙
1𝑥10^−12
L = 10^4.5 x(1x10^-12)
L = 1 x [10^4.5+(-12))
L = 1 x 10^-7.5
The noise criteria for an admin office is within the range of 50 - 60. According to the
calculation above, the combined sound pressure level around the silent area during peak and
non-peak hours are 70dB and 55 dB respectively, which exceeds the noise criteria for a
reading area. This can be solved by proposing double layered channel glass which consists of
an insulation layer, preventing the transmission of sound through the vibration of air, or a
double façade which acts as a sound barrier.
Total Intensities:
Total intensities, I
= (1x10^-6.5) +(1 x 10^-7.5)
= 3.48x10^-7
Combined SPL:
SPL = 10log10
𝐼
𝐼𝑥
= 10log10
3.48𝑥10^−7
𝐼𝑥10^−12
= 55 dB

19. 18
2.2.2 Reverberation Time (Admin Office)
Volume of Space
V = L x W x H
V = 6m x 3.5m x 4m
V = 84m3

20. 19
Material Absorption Coefficient under 500 Hz , with 5 users
Building
Component
Material Area S ( m2) Absorption
Coefficient (a)
Sound Absorption
( SA)
Wall Concrete 65.6 0.02 1.31
Glass 4 0.06 0.24
Floor Carpet 21 0.30 6.3
Door Timber 2 0.13 0.26
Ceiling Concrete 21 0.02 0.42
Furniture Timber Desk 5 0.15 0.75
Padded Chair 3 0.77 2.31
Users - 5 0.46 2.3
Total Absorption 13.39
RT= (0.16 X 84) / 13.39
= 1s
Material Absorption Coefficient under 2000 Hz , with 5 users
Building
Component
Material Area S ( m2) Absorption
Coefficient (a)
Sound Absorption
( SA)
Wall Concrete 65.6 0.02 1.31
Glass 4 0.03 0.12
Floor Carpet 21 0.50 10.5
Door Timber 2 0.10 0.2
Ceiling Concrete 21 0.02 0.42
Furniture Timber Desk 5 0.15 0.75
Padded Chair 3 0.77 2.31
Users - 5 0.51 2.55
Total Absorption 18.16
RT= (0.16 X 84) / 18.16
= 0.74s
The reverberation time required for a library reading space is 0.6 – 0.8 seconds. Based on the
calculation, the reverberation of the space under 500 Hz and 2000 Hz is 1s and 0.7s
respectively. Acoustic panels are introduced to amend the issue.

21. 20
2.2.3 Sound Reduction Index (Admin Office)
Volume of Space
V = L x W x H
V = 6m x 3.5m x 4m
V = 84m3
Assuming that the noise level of an admin office is 40 dB, noise level of the external is 75dB.
The sound reduction index of the wall should 75dB- 40dB = 35 dB.

22. 21
Building
Component Material Surface Area
(m2
SRI
( Decibels)
Transmission
Coefficient
( T )
Wall Concrete 65.6 46 2.51 𝑥 10 −5
Wall Glass 4 35 3.16 𝑥 10 −4
Door Plywood 2 31 7.94 𝑥 10 −4
Concrete
SRI = 10 Log (1/T)
46 = 10 log (1/T)
3.98 𝑥 10 4
= 1 / T
T = 2.51 𝑥 10 −5
Glass
SRI = 10 Log (1/T)
35 = 10 log (1/T)
3.16 𝑥 10 3
= 1 / T
T = 3.16 𝑥 10 −4
Timber
SRI = 10 Log (1/T)
31 = 10 log (1/T)
10 3.1
= 10 log10 (1/T)
T = 7.94 𝑥 10 −4
The overall transmission loss from the back lane to the admin office is 33.97 dB. Based on the
previous assumption, according to the noise criteria environment perception, the admin
officeexceeded the noise criteria, resulting in distracting noises interfering work environment.
This can be solved by having double façade of wooden shutters which can filter and deflect
noise coming from the back lane.
AverageTransmission Coefficient
= [65.6 x ( 2.51 𝑥 10 −5
) ] + [ 4 x
( 3.16 𝑥 10 −4
) ] + [ 2 x (7.94 𝑥 10 −4
) ] /
( 65.6 + 4 + 2)
= 4.00 𝑥 10 −4
OverallSRI
= 10 Log ( 1 / 4.00 𝑥 10 −4
)
= 33.97 Db
External Sound PressureLevel
= 75dB– 33.97dB
= 41.03 dB

23. 22
3.0 References
1. Cavanough, W.J. & Wikes, J.A. (1998). ArchitecturalAcoustics: Principles
and Practice. New York: Wiley and Sons.
2. Absorption Coefficient Building Material Finishes – Sengpiel Berlin (n.d.)
Retrieved November 26 2015 from
http://www.sengpielaudio.com/calculator-RT60Coeff.htm
3. ACE Lamps Lighting (n.d.) Retreived December 6 2015 from
http://www.sengpielaudio.com/calculator-RT60Coeff.htm
4. Ir Izdihar, A. (n.d.) UBBL 2012 Amendments on EE Bylaw 38A and MS
1525:2014. Retrieved December 6 2016 from
http://www.sengpielaudio.com/calculator-RT60Coeff.htm
5. Ir Izdihar, A. (n.d.) UBBL 2012 Amendments on EE Bylaw 38A and MS
1525:2014. Retrieved December 6 2016 from
http://www.sengpielaudio.com/calculator-RT60Coeff.htm