LOL

1. BUILDING SCIENCE 2 [BLD 61303/ ARC 3413]
PROJECT 2 – INTEGRATION PROJECT
SENTUL COMMUNITY LIBRARY
REPORT & CALCULATION
NAME : LEE YAUE SHEN
STUDENT ID : 0315381
TUTOR : MR SIVA

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TABLE OF CONTENT
1.0 LIGHTING ANALYSIS
1.1 STAFF REST AREA
1.1.1 DAYLIGHT 5-6
1.1.2 ARTIFICIAL LIGHT 7
1.1.3 PSALI 8
1.2 QUIET STUDY AREA
1.2.1 DAYLIGHT 10-11
1.2.2 ARTIFICIAL LIGHT 12
1.2.3 PSALI 13
2.0 ACOUSTIC ANALYSIS
2.1 STAFF REST AREA
2.1.1 SPL ( SOUND PRESSURE LEVEL ) 15-16
2.1.2 RT ( REVERBERATION TIME ) 17
2.1.3 SPL (SOUND PRESSURE LEVEL) 18-20
2.2 QUIET STUDY AREA
2.2.1 SPL ( SOUND PRESSURE LEVEL ) 21-22
2.2.2 RT ( REVERBERATION TIME ) 23
2.2.3 SPL (SOUND PRESSURE LEVEL) 24-26
3.0 REFERENCES 27

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LIGHTING

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1.0 LIGHTING ANALYSIS
Daylight Factor (DF) according to MS 1525
1.1 STAFF REST AREA
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
GROUND FLOOR
PLAN
DAYLIGHT

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1.1.1 DAYLIGHT
Daylight Factor Calculation
Floor Area (m2)
= (4.1m x 3.2m) + (2.9m x 1.4m)
= 13.1 + 4.1
= 17.2 m2
Area of facade exposed to sunlight (m2)
= 2.2m x 3m
= 6.6 m2
Area of skylight
= 0
Exposed Facade & Skylight Area to floor
area ratio/ Daylight Factor, DF
= (6.6+0) / 17.2
= 0.39
= 3.9%
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
DAYLIGHT FACTOR DIAGRAM

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40 lux Fully overcast
<1 lux Sunset, Storm cloud
Natural Illumination Calculation
E external = 20000 lux
DF = E internal / E external x 100
3.9 = E internal / 20000 x 100
E internal = 3.9 x 20000 / 100
= 780 lux
The selected stuff rest area has a daylight factor of 3.9% and natural illumination
of 780 lux. Based on the requirements of MS1525.The space is good daylight
source. However, for the purpose of reducing the thermal problem, louvers
facade system is proposed for the facade design of the library. The vertical
louvers will provide a well shading and reduce the amount of sunlight
penetrating into the space. Hence, this design should be augmented with internal
shades for low perpendicular sun angles. The low-e laminated glass is using as
the inner facade of the space to provide privacy and well sound properties for
user.

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1.1.2 Artificial Light
Utilization Factor Table
According to MS 1525, the recommendation illumination level for staff rest
area is 200 lux.
Lumen Method Calculation
Dimension of room (L x W) (4.1m x 3.2m) + (2.9m x 1.4m)
Total floor Area (m2) 17.2 m2
Room cavity height (m) 3.5
Reflectance values Ceiling=0.7
Wall= 0.5
Floor= 0.3
Room index, K
K =
4.5 𝑥 4.1
3.5(4.5 + 4.1)
= 0.6
Utilization Factor, UF 0.27
Model : DN570B (low height recessed version)
Input : 230 or 240v / 50-60Hz
Lumen (lm) : 2600
Weight (kg) : 2.2
Power : 36w (3000k)

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1.1.3 Psali
This is not a huge space that can fit many user, therefore there is only one switch
needed for this room. The opening is huge enough to allow daylight to penetrate
in and brighten up the area.
Maintenance Factor, MF
Illuminance Requirement
0.8
200
Number of Luminaires
N =
𝐸 𝑥 𝐴
𝐹 𝑥 𝑈𝐹 𝑥 𝑀𝐹
N =
200 𝑥 17.2
2600 𝑥 (0.8 𝑥 0.27)
= 6.12 = ~7 𝑏𝑢𝑙𝑏𝑠
Spacing to height ration (SHR)
𝑆𝐻𝑅 =
1
𝐻𝑚
𝑥 √
𝐴
𝑁
𝑆𝐻𝑅 =
1
3.5
𝑥 √
17.2
7
= 0.45
𝑆𝐻𝑅 =
𝑆
3.5
= 0.45
S = 1.57
Fitting Layout Fitting required along 4.5m wall
= 4.5/1.57 = 2.87
= 3 rows
Fitting required along 4.1 m wall
= 4.1/1.57 = 2.61
= 3 rows

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1.2 QUIET STUDY AREA
Daylight Factor (DF) according to MS 1525
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
SECOND FLOOR
PLAN
DAYLIGHT

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1.2.1 DAYLIGHT
Daylight Factor Calculation
Floor Area (m2)
= 5.7 m x 7.4 m
= 42.18 m2
Area of facade exposed to sunlight (m2)
= 2 x (5.7 m x 4 m) + ( 7.4 m x 4 m )
= 75.2 m2
Area of skylight
= 0
Exposed Facade & Skylight Area to floor
area ratio/ Daylight Factor, DF
= (75.2+0) / 42.18
= 1.78
= 17.8%
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
DAYLIGHT FACTOR DIAGRAM

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<200 lux Midday
40 lux Fully overcast
<1 lux Sunset, Storm cloud
Natural Illumination Calculation
E external = 20000 lux
DF = E internal / E external x 100
17.8 = E internal / 20000 x 100
E internal = 17.8 x 20000 / 100
= 3560 lux
The selected quiet study area has a daylight factor of 17.8% and natural
illumination of 3560 lux. Based on the requirements of MS1525, the space is too
bright and it might also cause thermal problems. For the purpose of reducing the
thermal and glare problem, Louvers facade system and curtain are proposed for
the library. The vertical louvers will provide a well shading and reduce the
amount of sunlight penetrating into the space. Besides that, the curtain can also
reduce the direct daylight penetrate in to the space and hence reduce the
thermal and glare problem effectively.

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1.2.2 Artificial Light
Utilization Factor Table
According to MS 1525, the recommendation illumination level for quiet
study area is 300 - 500 lux.
Lumen Method Calculation
Dimension of room (L x W) 5.7 m x 7.4 m
Total floor Area (m2) 42.18 m2
Room cavity height (m) 4
Reflectance values Ceiling=0.7
Wall= 0.5
Floor= 0.3
Room index, K
K =
5.7 𝑥 7.4
4(5.7 + 7.4)
= 0.8
Utilization Factor, UF 0.33
Model : DN570B (low height recessed version)
Input : 230 or 240v / 50-60Hz
Lumen (lm) : 2600
Weight (kg) : 2.2
Power : 36w (3000k)

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1.2.3 Psali - Permanent supplementary Artificial Lighting for Interior
There is 2 switch in this space. As the back of the space is facing outside, the
daylight during day time is effective enough. Therefore, switch 2 can be turn off
during daytime for energy saving.
Maintenance Factor, MF
Illuminance Requirement
0.8
300 - 500
Number of Luminaires
N =
𝐸 𝑥 𝐴
𝐹 𝑥 𝑈𝐹 𝑥 𝑀𝐹
N =
300 𝑥 42.18
2600 𝑥 (0.8 𝑥 0.33)
= 18.4 = ~19 𝑏𝑢𝑙𝑏𝑠
Spacing to height ration (SHR)
𝑆𝐻𝑅 =
1
𝐻𝑚
𝑥 √
𝐴
𝑁
𝑆𝐻𝑅 =
1
4
𝑥 √
42.18
19
= 0.37
𝑆𝐻𝑅 =
𝑆
4
= 0.37
S = 1.48
Fitting Layout Fitting required along 5.7m wall
= 5.7/1.48 = 3.85
= 4 rows
Fitting required along 7.4 m wall
= 7.4/1.48 = 5
= 5 rows

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ACOUSTIC

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2.0 ACOUSTIC ANALYSIS
2.1 STAFF REST AREA
2.1.1 SRI (SOUND REDUCTION INDEX )
Volume of Space
V = L x W x H = 4.1 x 4.5 x 3.5
= 64.58 m3
Assuming that the noise level of a staff rest area is 40 dB , Noise level of the external is 70 dB
The Sound Reduction index of the wall should be 70 – 40 dB , which is 30 dB
GROUND FLOOR
PLAN

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Building
Component Material Surface Area
(m2)
SRI ( Decibels)
Transmission
Coefficient ( T )
Wall Glass 6.6 35 3.16 𝑥 10 −4
Wall Concrete 53.6 46 2.51 𝑥 10 −5
Door Glass 2 35 3.16 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
Average Transmission Coefficient
= [ 6.6 x (3.16 𝑥 10 −4
) ] + [ 53.6 x ( 2.51 𝑥 10 −4
) ] + [ 2 x (3.16 𝑥 10 −4
) ] / ( 6.6 +
53.6 + 2 )
= 2.6 𝑥 10 −4
Overall SRI
= 10 Log ( 1 / 2.6 𝑥 10 −4
) = 35.85 dB
Conclusion
The overall SRI of the wall is 35.85 dB , which fulfils the required transmission loss of 30 dB

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2.1.2 RT ( REVERBERATION TIME )
Volume of Space
V = L x W x H = 4.1 x 4.5 x 3.5
= 64.58 m3
Material Absorption Coefficient under 500 Hz , with 4 users
Building
Component
Material Area S ( m2) Absorption
Coefficient (a)
Sound Absorption
( SA)
Wall Concrete 53.6 0.02 1.07
Glass 6.6 0.06 0.4
Floor Carpet 18.45 0.30 5.54
Door Timber 2.0 0.13 0.26
Ceiling Concrete 18.45 0.02 0.37
Furniture Plywood Table 3.37 0.13 0.44
Plywood Seating 1.44 0.13 0.19
Users - 4 0.46 1.84
Total Absorption 10.11
Reverberation Time, RT = ( 0.16 V) / A
RT= (0.16 X 64.58) / 10.11
=1.00s
Material Absorption Coefficient under 2000 Hz , with 4 users
Building
Component
Material Area S ( m2) Absorption
Coefficient (a)
Sound Absorption
( SA)
Wall Concrete 53.6 0.02 1.07
Glass 6.6 0.03 0.2
Floor Carpet 18.45 0.50 9.23
Door Timber 2.0 0.1 0.2
Ceiling Concrete 18.45 0.02 0.37
Furniture Plywood Table 3.37 0.10 0.34
Plywood Seating 1.44 0.10 0.14
Users - 4 0.51 2.04
Total Absorption 13.59

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Reverberation Time, RT = ( 0.16 V) / A
RT= (0.16 X 64.58) / 13.59 = 0.76s
Conclusion
The reverberation time required for a staff rest area is 0.4 - 0.6s
However, based on calculation , the reverberation of the space under 500 Hz and 2000 Hz is
1.00s and 0.76s respectively. As they slightly differ from the requirement, acoustic panels and
exterior facade are introduced to fix the issue.
2.1.3 SPL (SOUND PRESSURE LEVEL)
Sound Pressure Level Formula :
SPL = 𝟏𝟎 𝒍𝒐𝒈 𝟏𝟎
𝑰
𝑰𝒐
Where SPL = sound pressure level (dB), I = sound power (intensity)(watt) and Io
= reference power. Io is usually taken as 1 x 10-12 watts.
Peak Hour (Back Alley)
Highest Reading : 70 dB (moderate sound)
Lowest Reading : 55 dB (quite sound)
Highest Reading
70 = 10 log 10
𝑰
𝑰𝒐
Antilog 7 =
𝐈
𝟏 𝒙 𝟏𝟎−𝟏𝟐
1 x 107 =
𝐈
𝟏 𝒙 𝟏𝟎−𝟏𝟐
I = 107 x (𝟏 𝒙 𝟏𝟎−𝟏𝟐
)
I = 1 x 10-5

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Lowest Reading
55 = 10 log 10
𝑰
𝑰𝒐
Antilog 5.5 =
𝐈
𝟏 𝒙 𝟏𝟎−𝟏𝟐
1 x 105.5 =
𝐈
𝟏 𝒙 𝟏𝟎−𝟏𝟐
I = 105.5 x (𝟏 𝒙 𝟏𝟎−𝟏𝟐
)
I = 1 x 10-6.5
Total Intensities, I :
= (1 x 10-5) + (1 x 10-6.5)
=1.03 x 10-5
Combined SPL :
SPL = 10 log 10
𝑰
𝑰𝒐
= 10 log 10 (
𝟏.𝟎𝟑 𝒙 𝟏𝟎−𝟓
𝟏 𝒙 𝟏𝟎−𝟏𝟐
)
= 70dB
Non - Peak Hour (Back Alley)
Highest Reading : 55 dB (quiet sound)
Lowest Reading : 45 dB (very quiet sound)
Highest Reading
55 = 10 log 10
𝑰
𝑰𝒐
Antilog 5.5 =
𝐈
𝟏 𝒙 𝟏𝟎−𝟏𝟐
1 x 105.5 =
𝐈
𝟏 𝒙 𝟏𝟎−𝟏𝟐
I = 105.5 x (𝟏 𝒙 𝟏𝟎−𝟏𝟐
)
I = 1 x 10-6.5

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Lowest Reading
45 = 10 log 10
𝑰
𝑰𝒐
Antilog 4.5 =
𝐈
𝟏 𝒙 𝟏𝟎−𝟏𝟐
1 x 104.5 =
𝐈
𝟏 𝒙 𝟏𝟎−𝟏𝟐
I = 104.5 x (𝟏 𝒙 𝟏𝟎−𝟏𝟐
)
I = 1 x 10-7.5
Total Intensities, I :
= (1 x 10-6.5) + (1 x 10-7.5)
=3.48 x 10-7
Combined SPL :
SPL = 10 log 10
𝑰
𝑰𝒐
= 10 log 10 (
𝟑.𝟒𝟖 𝒙 𝟏𝟎−𝟕
𝟏 𝒙 𝟏𝟎−𝟏𝟐
)
= 55dB

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2.2 QUIET STUDY AREA
2.2.1 SRI (SOUND REDUCTION INDEX)
Volume of Space
V = L x W x H = 5.7 x 7.4 x 4
= 168.72 m3
Assuming that the noise level of a quiet study area is 45 dB , Noise level of the external is 75
dB The Sound Reduction index of the wall should be 75 - 45 dB , which is 30 dB.
SECOND FLOOR
PLAN

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Building
Component Material Surface Area
(m2)
SRI ( Decibels)
Transmission
Coefficient ( T )
Wall Glass 75.2 35 3.16 𝑥 10 −4
Wall Concrete 29.6 46 2.51 𝑥 10 −5
Door Glass 2 35 3.16 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
Average Transmission Coefficient
= [ 75.2 x (3.16 𝑥 10 −4
) ] + [ 29.6 x ( 2.51 𝑥 10 −4
) ] + [ 2 x (3.16 𝑥 10 −4
) ] / ( 75.2 +
29.6 + 2 )
= 2.56 𝑥 10 −4
Overall SRI
= 10 Log ( 1 / 2.56 𝑥 10 −4
) = 35.92 dB
Conclusion
The overall SRI of the wall is 35.92 dB , which fulfils the required transmission loss of 30 dB

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2.2.2 RT ( REVERBERATION TIME )
Volume of Space
V = L x W x H = 5.7 x 7.4 x 4
= 168.72 m3
Material Absorption Coefficient under 500 Hz , with 24 users
Building
Component
Material Area S ( m2) Absorption
Coefficient (a)
Sound Absorption
( SA)
Wall Concrete 29.6 0.02 0.6
Glass 75.2 0.06 4.5
Floor Carpet 42.18 0.30 12.65
Door Glass 2.0 0.03 0.06
Ceiling Concrete 42.18 0.02 0.84
Furniture Plywood Table 12.24 0.13 1.59
Plywood Seating 4.32 0.13 0.56
Users - 24 0.46 11.04
Total Absorption 31.84
Reverberation Time, RT = ( 0.16 V) / A
RT= (0.16 X 168.72) / 31.84
=0.85s
Material Absorption Coefficient under 2000 Hz , with 4 users
Building
Component
Material Area S ( m2) Absorption
Coefficient (a)
Sound Absorption
( SA)
Wall Concrete 29.6 0.02 0.6
Glass 75.2 0.03 2.26
Floor Carpet 42.18 0.50 21.09
Door Timber 2.0 0.1 0.2
Ceiling Concrete 42.18 0.02 0.84
Furniture Plywood Table 12.24 0.10 1.22
Plywood Seating 4.32 0.10 0.43
Users - 24 0.51 12.24
Total Absorption 38.88

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Reverberation Time, RT = ( 0.16 V) / A
RT= (0.16 X 168.72) / 38.88 = 0.69s
Conclusion
The reverberation time required for a quiet study area is 0.4 - 0.6s
However, based on calculation , the reverberation of the space under 500 Hz and 2000 Hz is
0.85s and 0.69s respectively. As they slightly differ from the requirement, acoustic panels and
exterior facade are introduced to fix the issue.
2.2.3 SPL (SOUND PRESSURE LEVEL)
Sound Pressure Level Formula :
SPL = 𝟏𝟎 𝒍𝒐𝒈 𝟏𝟎
𝑰
𝑰𝒐
Where SPL = sound pressure level (dB), I = sound power (intensity)(watt) and Io
= reference power. Io is usually taken as 1 x 10-12 watts.
Peak Hour (Back Alley)
Highest Reading : 70 dB (moderate sound)
Lowest Reading : 55 dB (quite sound)
Highest Reading
70 = 10 log 10
𝑰
𝑰𝒐
Antilog 7 =
𝐈
𝟏 𝒙 𝟏𝟎−𝟏𝟐
1 x 107 =
𝐈
𝟏 𝒙 𝟏𝟎−𝟏𝟐
I = 107 x (𝟏 𝒙 𝟏𝟎−𝟏𝟐
)
I = 1 x 10-5

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Lowest Reading
55 = 10 log 10
𝑰
𝑰𝒐
Antilog 5.5 =
𝐈
𝟏 𝒙 𝟏𝟎−𝟏𝟐
1 x 105.5 =
𝐈
𝟏 𝒙 𝟏𝟎−𝟏𝟐
I = 105.5 x (𝟏 𝒙 𝟏𝟎−𝟏𝟐
)
I = 1 x 10-6.5
Total Intensities, I :
= (1 x 10-5) + (1 x 10-6.5)
=1.03 x 10-5
Combined SPL :
SPL = 10 log 10
𝑰
𝑰𝒐
= 10 log 10 (
𝟏.𝟎𝟑 𝒙 𝟏𝟎−𝟓
𝟏 𝒙 𝟏𝟎−𝟏𝟐
)
= 70dB
Non - Peak Hour (Back Alley)
Highest Reading : 55 dB (quiet sound)
Lowest Reading : 45 dB (very quiet sound)
Highest Reading
55 = 10 log 10
𝑰
𝑰𝒐
Antilog 5.5 =
𝐈
𝟏 𝒙 𝟏𝟎−𝟏𝟐
1 x 105.5 =
𝐈
𝟏 𝒙 𝟏𝟎−𝟏𝟐
I = 105.5 x (𝟏 𝒙 𝟏𝟎−𝟏𝟐
)
I = 1 x 10-6.5

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Lowest Reading
45 = 10 log 10
𝑰
𝑰𝒐
Antilog 4.5 =
𝐈
𝟏 𝒙 𝟏𝟎−𝟏𝟐
1 x 104.5 =
𝐈
𝟏 𝒙 𝟏𝟎−𝟏𝟐
I = 104.5 x (𝟏 𝒙 𝟏𝟎−𝟏𝟐
)
I = 1 x 10-7.5
Total Intensities, I :
= (1 x 10-6.5) + (1 x 10-7.5)
=3.48 x 10-7
Combined SPL :
SPL = 10 log 10
𝑰
𝑰𝒐
= 10 log 10 (
𝟑.𝟒𝟖 𝒙 𝟏𝟎−𝟕
𝟏 𝒙 𝟏𝟎−𝟏𝟐
)
= 55dB

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3.0 REFERENCES
1. (2016) (1st ed.). Retrieved from
http://www.utm.my/energymanagement/files/2014/07/MS-
1525-2007.pdf
2. Association of Australian Acoustical Consultants Guideline for
Educational Facilities Acoustics.
(2016) (1st ed.). Retrieved from http://file:///E:/Downloads/AAAC-
Guideline-for-Educational-
Facilities-Acoustics-2010%20(1).pdf