Bscience 2 Final Report

Building Science 2 (ARC 3413) Project 2 Integration with Architectural Design Studio 5

Sentul Community Library

Final Report & Calculation

Tutor: Mr. Edwin Student Name: Tan Kai Chong Student ID: 0314223

Table of Content Content

Page

1.0 Lighting 1.1 Daylighting (Printing Area) 1.2 Daylighting (Private Study Area) 1.3 Artificial Lighting (Printing Area) 1.4 Artificial Lighting (Private Study Area) 1.5 PSALI (Children Area) 1.6 PSALI (Meeting Room)

3~4 5~6 7~9 10 ~ 12 13 ~ 15 16 ~ 18

2.0 Acoustic 2.1 Sound Pressure Level 2.2 Reverberation Time (Auditorium) 2.3 Sound Transmission Loss (Auditorium) 2.4 Reverberation Time (Private Study Area) 2.5 Sound Transmission Loss (Private Study Area)

3.0 References

19 ~ 20 21 ~ 22 23 24 ~ 25 26

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1.1 Daylighting (Printing Area) According to MS 1525, Daylight Factor distribution as below: Daylight Factor, DF DF (%) >6 3~6 1~3 0~1

Distribution Very bright with thermal & glare problem Bright Average Dark

The selected Printing Area is located at the main entrance of the library as shown in floor plan. The faรงade design for the selected space is exposed to sunlight so there is minimal artificial lighting installed in this area.

Daylighting Printing Area

Ground Floor Plan 3|P a ge

Daylight Factor Calculation Floor Area (m²) Area of façade that exposed to sunlight Area of skylight (m²) Exposed Façade & Skylight Area to Floor Area Ratio / Daylight Factor, DF

32 35 0 (35+0)/32 = 1.093 = 109.3% x 0.1 = 10.93%

Natural Illumination Calculation Illumination 120,000 Lux 110,000 Lux 20,000 Lux 1000 – 2000 Lux < 200 Lux 400 Lux 40 Lux < 1 Lux

Example Brightest sunlight Bright sunlight Shade illuminated by entire clear blue sky Typical overcast day, midday Extreme of darkest storm clouds, midday Sunrise or sunset on clear day (ambient illumination) Fully overcast, sunset / sunrise Extreme of darkest storm clouds, sunset / rise

E external = 20,000 Lux DF = E internal / E external x 100 10 = E external / 20000 x 100 E external = 10 x 20000 / 100 = 2,000 Lux

Conclusion Printing Area where facing the street has a daylight factor of 10.93% and natural illumination of 2000 Lux. This will result thermal and glare problem. Therefore, low e-value glass can solve the glare problem and reduce heat gain to the area.

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1.2 Daylighting (Private Study Area) According to MS 1525, Daylight Factor distribution as below: Daylight Factor, DF DF (%) >6 3~6 1~3 0~1

Distribution Very bright with thermal & glare problem Bright Average Dark

This Private Study Area is located at second floor and facing directly to the main rood as shown in floor plan. The faรงade design for this study area is also exposed to sunlight as there are full curtain wall installed to ensure good lighting condition.

Daylighting Private Study Area

Second Floor Plan

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Daylight Factor Calculation Floor Area (m²) Area of façade that exposed to sunlight Area of skylight (m²) Exposed Façade & Skylight Area to Floor Area Ratio / Daylight Factor, DF

55 37 0 (55+0)/37 = 1.48 = 148% x 0.1 = 14.8%

Natural Illumination Calculation Illumination 120,000 Lux 110,000 Lux 20,000 Lux 1000 – 2000 Lux < 200 Lux 400 Lux 40 Lux < 1 Lux

Example Brightest sunlight Bright sunlight Shade illuminated by entire clear blue sky Typical overcast day, midday Extreme of darkest storm clouds, midday Sunrise or sunset on clear day (ambient illumination) Fully overcast, sunset / sunrise Extreme of darkest storm clouds, sunset / rise

E external = 20,000 Lux DF = E internal / E external x 100 10 = E external / 20000 x 100 E external = 10 x 20000 / 100 = 2,000 Lux

Conclusion The Study Area has daylight factor of 14.8% and natural illumination of 2000 Lux. This will cause thermal and glare problem too. So, low e-value glass will fix and solve the problem and reduce heat gain.

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1.3 Artificial Lighting (Printing Area) According to MS 1525, the minimum lighting level required for a Printing Area is 200 Lux. Printing Area

Ground Floor Plan Type of luminaire used as showed below: Type of fixture Type of light bulb

Disk Pendant Light

Material of fixture Product Brand & Code Nominal Voltage Wattage (W) CRI Color Temperature (K) Color Designation Lumens

Steel, Chrome Plated AM950LI20GVVZANUI 277.00 V 100 W 80 2800 K White 3380 lm

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Lumen Method Calculation

Location Dimension

Floor Area (A) Types of Lighting fixtures Lumen of lighting fixture (F) Height of work level Mounting Height (m) Reflection Factors

Room Index / RI (K)

Utilization Factor (UF) Maintenance Factor (MF) Standard Illuminance by MS1525 Number of light required

Spacing to Height Ratio (SHR)

Printing Area Length = 6.4m Width = 5.2m Height of the Ceiling = 3.7m 6.4 x 5.2 = 33.28 m² Disk Pendant Light 3380 lm 0.8 2.9m Ceiling : (0.7) Wall : (0.5) Floor : (0.2) 6.4 x 5.2_____ (6.4 + 5.2) x 2.9 = 0.98 0.5 0.8 200 N = E x A___ F x UF x MF N = 200 x 33.28___ 3380 x 0.5 x 0.8 N=5 1

đ??´

SHR = đ??ťđ?&#x2018;&#x161; đ?&#x2018;Ľ â&#x2C6;&#x161;đ?&#x2018; 1

= 2.9 đ?&#x2018;Ľ â&#x2C6;&#x161;

33.28 5

=1 SHR =

đ?&#x2018;&#x2020; 2.9

=1

S = 2.9 x 1 = 2.9 Fittings layout by approximately (m)

Fittings required along 6.4m wall 6.4 / 2.9 = 2.2 = 2 rows Therefore, approximately 2 x 3 = 6 Luminaires required Spacing along 5.2m wall 5.2 / 3 = 1.7m

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Fittings Layout

Conclusion According to the calculation. The total number of fitting that required to achieve standard lux for printing area is 6 units which arrange in 6 rows and 1.7m spacing.

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1.4 Artificial Lighting (Private Study Area) According to MS 1525, the minimum lighting level required for a Private Study Area is 200 Lux. Private Study Area

Second Floor Plan Type of luminaire used as showed below: Type of fixture Type of light bulb

Linear Pendant Light (2 Lamps)

Material of fixture Product Brand & Code Nominal Voltage Wattage (W) CRI Color Temperature (K) Color Designation Lumens

Steel Alcon 10106-4

277.00 V 100 W 80 4230 K White 5000 lm

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Lumen Method Calculation Location Dimension

Floor Area (A) Types of Lighting fixtures Lumen of lighting fixture (F) Height of work level Mounting Height (m) Reflection Factors

Room Index / RI (K)

Utilization Factor (UF) Maintenance Factor (MF) Standard Illuminance by MS1525 Number of light required

Spacing to Height Ratio (SHR)

Private Study Area Length = 10m Width = 6m Height of the Ceiling = 3m 10 x 6 = 60 m² Linear Pendant Light (2 Lamps) 5000 lm 0.8 2.2m Ceiling : (0.7) Wall : (0.5) Floor : (0.2) 10 x 6_____ (10 + 6) x 2.2 = 1.7 0.5 0.8 200 N = E x A___ F x UF x MF N= 200 x 60____ 5000 x 0.5 x 0.8 N=6 1

đ??´

1

60

SHR = đ??ťđ?&#x2018;&#x161; đ?&#x2018;Ľ â&#x2C6;&#x161;đ?&#x2018; = 2.2 đ?&#x2018;Ľ â&#x2C6;&#x161; 6 = 1.44 đ?&#x2018;&#x2020;

SHR = 2.2 = 1.44 S = 2.2 x 1.44 = 3.17 Fittings layout by approximately (m)

Fittings required along 10m wall 10 / 3.17 = 3.15 = 3 rows Therefore, approximately 3 x 2 = 6 Luminaires required Spacing along 6m wall 6 / 2 = 3m

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Fittings Layout

Conclusion According to the calculation. The total number of fitting that required to achieve standard lux for printing area is 6 units which arrange in 6 rows and 3m spacing.

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1.5 PSALI â&#x20AC;&#x201C; Permanent Supplementary Artificial Lighting of Interiors (Children Area) According to MS 1525, the minimum lighting level required for a Children Area is 300 Lux. Children Area

First Floor Plan Type of luminaire used as showed below: Type of fixture Type of light bulb

LED Downlight

Material of fixture Product Brand & Code Nominal Life (hours) Wattage Range (W) CRI Color Temperature (K) Color Designation Lumens

Aluminum ST422B 50, 000 40 80 3000K Warm White 5000 lm

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Lumen Method Calculation Location Dimension

Floor Area (A) Types of Lighting fixtures Lumen of lighting fixture (F) Height of work level Mounting Height (m) Reflection Factors

Room Index / RI (K)

Utilization Factor (UF) Maintenance Factor (MF) Standard Illuminance by MS1525 Number of light required

Spacing to Height Ratio (SHR)

Children Area Length = 10m Width = 9m Height of the Ceiling = 3m 10 x 9 = 90 m² LED Downlight 5000 lm 0.8 2.2m Ceiling : (0.7) Wall : (0.5) Floor : (0.2) 10 x 9_____ (10 + 9) x 2.2 = 2.15 0.5 0.8 300 N = E x A___ F x UF x MF N= 300 x 90____ 5000 x 0.5 x 0.8 N = 14 1

đ??´

1

90

SHR = đ??ťđ?&#x2018;&#x161; đ?&#x2018;Ľ â&#x2C6;&#x161;đ?&#x2018; = 2.2 đ?&#x2018;Ľ â&#x2C6;&#x161;14 = 1.15 đ?&#x2018;&#x2020;

SHR = 2.2 = 1.15 S = 2.2 x 1.15 = 2.53 Fittings layout by approximately (m)

Fittings required along 10m wall 10 / 2.53 = 3.95 = 4 rows Therefore, approximately 4 x 4 = 16 Luminaires required Spacing along 9m wall 9 / 4 = 2.25m

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Fittings Layout

Conclusion The lighting is controlled by 2 switches and divided into two part, the left and right. The left part is where children use for studying for getting better light source. For the right part is for the children playground. Switch on the left has to be open most of the time for indoor activity to fulfil the requirement of MS 1525.

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1.6 PSALI â&#x20AC;&#x201C; Permanent Supplementary Artificial Lighting of Interiors (Meeting Room) According to MS 1525, the minimum lighting level required for a Meeting Room is 200 Lux. Meeting Room

Second Floor Plan Type of luminaire used as showed below: Type of fixture Type of light bulb

LED Downlight

Material of fixture Product Brand & Code Nominal Life (hours) Wattage Range (W) CRI Color Temperature (K) Color Designation Lumens

Aluminum ST422B 50, 000 40 80 3000K Warm White 5000 lm

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Lumen Method Calculation Location Dimension

Floor Area (A) Types of Lighting fixtures Lumen of lighting fixture (F) Height of work level Mounting Height (m) Reflection Factors

Room Index / RI (K)

Utilization Factor (UF) Maintenance Factor (MF) Standard Illuminance by MS1525 Number of light required

Spacing to Height Ratio (SHR)

Meeting Room Length = 4m Width = 5m Height of the Ceiling = 3m 4 x 5 = 20 m² LED Downlight 5000 lm 0.8 2.2m Ceiling : (0.7) Wall : (0.5) Floor : (0.2) 4 x 5_____ (4 + 5) x 2.2 = 1.01 0.5 0.8 200 N = E x A___ F x UF x MF N= 200 x 20____ 5000 x 0.5 x 0.8 N=2 1

đ??´

1

20

SHR = đ??ťđ?&#x2018;&#x161; đ?&#x2018;Ľ â&#x2C6;&#x161;đ?&#x2018; = 2.2 đ?&#x2018;Ľ â&#x2C6;&#x161; 2 = 1.43 đ?&#x2018;&#x2020;

SHR = 2.2 = 1.43 S = 2.2 x 1.43 = 3.146 Fittings layout by approximately (m)

Fittings required along 10m wall 4 / 3.146 = 1.27 = 1 rows Therefore, approximately 1 x 3 = 3 Luminaires required Spacing along 5m wall 5 / 3 = 1.67m

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Fittings Layout

Conclusion The lighting is controlled by 1 switches to control the three main lights. This meeting room is facing street and having curtain wall installed which mean the light is switch off in the morning as the natural light provided sufficient light. It can be switch on when having bad weather condition such as cloudy or rainy which has not enough light for the room. Switch in the room has to be open most of the time for indoor activity to fulfil the requirement of MS 1525.

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2.0 Acoustic 2.1 External Noise Sound Pressure Level (Outdoor Reading Area) i) Peak Hour Highest Reading = 80db đ??ź

Use formula, L = 10 log10 (đ??ź ) 0

80 = 10 log10 (

đ??ź

1 đ?&#x2018;Ľ 10â&#x2C6;&#x2019;12

)

I = (108 8 ) (1x10â&#x2C6;&#x2019;12 ) = 1 x 10â&#x2C6;&#x2019;4 Lowest Reading = 70db đ??ź

Use formula, L = 10 log10 (đ??ź ) 0

đ??ź

70 = 10 log10 (1 đ?&#x2018;Ľ 10â&#x2C6;&#x2019;12 )

I = (107 0 ) (1x10â&#x2C6;&#x2019;12 ) = 1 x 10â&#x2C6;&#x2019;5

Total Intensities, I = (1x10â&#x2C6;&#x2019;4 ) + (1x10â&#x2C6;&#x2019;5 ) = 1.1x10â&#x2C6;&#x2019;4

Using formula Combined SPL = 10 log10 (đ?&#x2018;?2 /đ?&#x2018;?0 ²), where đ?&#x2018;?0=1x10â&#x2C6;&#x2019;12 Combined SPL = 10 log10 [(1/10â&#x2C6;&#x2019;4 )á (1x10â&#x2C6;&#x2019;5 )] = 80.43db

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ii) Non-peak hour Highest Reading = 65db đ??ź

Use formula, L = 10 log10 (đ??ź ) 0

đ??ź

65 = 10 log10 (1 đ?&#x2018;Ľ 10â&#x2C6;&#x2019;12 )

I = (106 5 ) (1x10â&#x2C6;&#x2019;12 ) = 3.16 x 10â&#x2C6;&#x2019;6 Lowest Reading = 66db đ??ź

Use formula, L = 10 log10 ( ) đ??ź0

đ??ź

57 = 10 log10 (1 đ?&#x2018;Ľ 10â&#x2C6;&#x2019;12 )

I = (105 7 ) (1x10â&#x2C6;&#x2019;12 ) = 5 x 10â&#x2C6;&#x2019;7

Total Intensities, I = (3.16x10â&#x2C6;&#x2019;6 ) + (5x10â&#x2C6;&#x2019;7 ) = 3.66x10â&#x2C6;&#x2019;6

Using formula Combined SPL = 10 log10 (đ?&#x2018;?2 /đ?&#x2018;?0 ²), where đ?&#x2018;?0=1x10â&#x2C6;&#x2019;12 Combined SPL = 10 log10 [(3.66/10â&#x2C6;&#x2019;6 )á (1x10â&#x2C6;&#x2019;12 )] = 65.63db

Conclusion The result show that at Outdoor Reading Area the average sound pressure level during Peak Hour and Non-peak Hour are 66.64db and 65.63db.

According to Acoustic Standard ANSI, standard acoustic level for a general library would be between 35db to 40db. So, some of the spaces used façade or other spaces as a butter must design away from the noise source. The acoustic panel must include to achieve optimum acoustic standard for library.

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2.2 Reverberation Time, RT (Auditorium)

Noise Auditorium

Space Volume = 10m x 6m x 3m = 180m³ Material absorption coefficient in 2000 Hz at non-peak hour with 10 persons in the space Building Component Wall Wall Ceiling Floor Door

Material

Glass Timber Wall Plaster Board Wood Flooring Stainless Steel with Glass Furniture Timber Marker board People (Peak) Total Sound Absorption by Material

Area (A/m²) 38.4 18 60 60 4.18

Absorption Coefficient (S) 0.10 0.05 0.40 0.06 0.10

Sound Absorption (Sa) 3.84 0.9 24 3.6 0.418

5.4

0.10

0.54

10

0.45

4.5 37.798

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Reverberation Time (RT) = (0.16 x V) / A V = Volume of Spaces A = Total Absorption = S1A1 + S2+A2 + S3+A3 + â&#x20AC;Ś S4+A4 Reverberation time (Non- Peak Hour) RT = (0.16 x V) / A = (0.16 x 311.22) / 37.798 =1.32 s Material absorption coefficient in 500 Hz at peak hour with 30 persons in the space Building Component Wall Wall Ceiling Floor Door

Material

Glass Timber Wall Plaster Board Wood Flooring Stainless Steel with Glass Furniture Timber Marker board People (Peak) Total Sound Absorption by Material

Area (A/m²) 40 104.4 54.6 54.6 4.18

Absorption Coefficient (S) 0.10 0.05 0.015 0.10 0.10

Sound Absorption (Sa) 4 5.22 0.819 5.46 0.418

5.4

0.06

0.324

30

0.44

13.2 29.441

Reverberation time (Peak Hour) RT = (0.16 x V) / A = (0.16 x 311.22) / 29.441 = 1.69 s

Conclusion: The optimum reverberation time for auditorium is between 1.4s to 2.0s. The material chose for the design is only 1.32 s during non-peak hour, but it successfully achieved 1.69s during the peak hour. The solution is to install acoustic panel or apply carpet.

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2.3 Sound Transmission Loss (Auditorium) Transmission Coefficient of Material Timber wall:

Glass Door:

1

1

SRI = 10log (đ?&#x2018;&#x2021; )

SRI = 10log (đ?&#x2018;&#x2021; )

SRI Timber wall = 50 dB

SRI Glass Door = 28 dB

1

1

50 dB = 10log( )

28 dB = 10log( )

đ?&#x2018;&#x2021;

đ?&#x2018;&#x2021;

T = 1 x 10-5

T = 1.58 x 10-3

Element

Material

Door Wall Total

Glass Timber

Tav = Tav =

đ?&#x2018;&#x2021;đ?&#x2018;&#x153;đ?&#x2018;Ąđ?&#x2018;&#x17D;đ?&#x2018;&#x2122; đ?&#x2018;&#x2021;đ??´ đ?&#x2018;&#x2021;đ?&#x2018;&#x153;đ?&#x2018;Ąđ?&#x2018;&#x17D;đ?&#x2018;&#x2122; đ?&#x2018;&#x2020;đ?&#x2018;˘đ?&#x2018;&#x;đ?&#x2018;&#x201C;đ?&#x2018;&#x17D;đ?&#x2018;?đ?&#x2018;&#x2019; đ??´đ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018;&#x17D; 7.64đ?&#x2018;Ľ10â&#x2C6;&#x2019;3 108.58

Tav = 7.03x10-5

Surface Area, A(m2) 4.18 104.4

SRI (dB) 28 50

Transmission Coefficient,T 1.58 x 10-3 1 x 10-5

Surface Area x Coefficient 6.6x10-3 1.04x10-3 7.64X10-3

1

SRIoverall = 10log (đ?&#x2018;&#x2021;) SRIoverall = 10log (

1 7.03 đ?&#x2018;Ľ 10â&#x2C6;&#x2019;5

)

SRIoverall = 41.53dB

Conclusion: The result of the calculation shows that auditorium has 41.53 dB of sound reduction during the transmission of external noise into the space. The sound level of external noise is 80 dB during peak hour and 65 dB during non-peak hour as shown in the previous calculation The transmissions of external noise into the auditorium are reduced by 41.53 dB which resulted in 38.47 dB during peak hour and 23.47 dB during non-peak hour.

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2.4 Reverberation Time, RT (Private Study Area)

Noise

Private Study Area

Space Volume = 10m x 6m x 3m = 180m³ Material absorption coefficient in 2000 Hz at non-peak hour with 5 persons in the space Building Component Wall Wall Ceiling Floor Door

Material

Glass Timber Wall Plaster Board Wood Flooring Stainless Steel with Glass Furniture Table People (Peak) Total Sound Absorption by Material

Area (A/m²) 40 48 60 60 4.18

Absorption Coefficient (S) 0.10 0.05 0.40 0.06 0.10

Sound Absorption (Sa) 0.4 2.4 24 3.6 0.418

5.04 5

0.10 0.45

0.504 2.25 33.572

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Reverberation Time (RT) = (0.16 x V) / A V = Volume of Spaces A = Total Absorption = S1A1 + S2+A2 + S3+A3 + â&#x20AC;Ś S4+A4 Reverberation time (Non- Peak Hour) RT = (0.16 x V) / A = (0.16 x 180) / 33.572 = 0.85s Material absorption coefficient in 500 Hz at peak hour with 15 persons in the space Building Component Wall Wall Ceiling Floor Door

Material

Glass Timber Wall Plaster Board Wood Flooring Stainless Steel with Glass Furniture Table People (Peak) Total Sound Absorption by Material

Area (A/m²) 40 48 60 60 4.18

Absorption Coefficient (S) 0.10 0.05 0.015 0.10 0.10

Sound Absorption (Sa) 0.4 2.4 0.9 6 0.418

5.04 15

0.06 0.44

0.302 6.6 17.02

Reverberation time (Peak Hour) RT = (0.16 x V) / A = (0.16 x 180) / 17.02 = 1.69s

Conclusion: The reverberation time for Private Study Area is 0.85 during the non-peak hour time and 1.69 during the peak hour time. It fall above the comfort reverberation time of 1.5s-2.5s in non-peak hour but successfully achieved in peak hour.

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2.5 Sound Transmission Loss (Private Study Area) Transmission Coefficient of Material Timber wall:

Glass Door:

1

1

SRI = 10log (đ?&#x2018;&#x2021; )

SRI = 10log (đ?&#x2018;&#x2021; )

SRI Timber wall = 50 dB

SRI Glass Door = 28 dB

1

1

50 dB = 10log( )

28 dB = 10log( )

đ?&#x2018;&#x2021;

đ?&#x2018;&#x2021;

T = 1 x 10-5

T = 1.58 x 10-3

Element

Material

Door Wall Total

Glass Timber

Tav = Tav =

đ?&#x2018;&#x2021;đ?&#x2018;&#x153;đ?&#x2018;Ąđ?&#x2018;&#x17D;đ?&#x2018;&#x2122; đ?&#x2018;&#x2021;đ??´ đ?&#x2018;&#x2021;đ?&#x2018;&#x153;đ?&#x2018;Ąđ?&#x2018;&#x17D;đ?&#x2018;&#x2122; đ?&#x2018;&#x2020;đ?&#x2018;˘đ?&#x2018;&#x;đ?&#x2018;&#x201C;đ?&#x2018;&#x17D;đ?&#x2018;?đ?&#x2018;&#x2019; đ??´đ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018;&#x17D; 7.08đ?&#x2018;Ľ10â&#x2C6;&#x2019;3 52.18

Tav = 1.35x10-4

Surface Area, A(m2) 4.18 48

SRI (dB) 28 50

Transmission Coefficient,T 1.58 x 10-3 1 x 10-5

Surface Area x Coefficient 6.6x10-3 4.8x10-4 7.08X10-3

1

SRIoverall = 10log (đ?&#x2018;&#x2021;) SRIoverall = 10log (

1 1.35 đ?&#x2018;Ľ 10â&#x2C6;&#x2019;4

)

SRIoverall = 38.7 dB

Conclusion: The result of the calculation shows that auditorium has 38.7 dB of sound reduction during the transmission of external noise into the space. The sound level of external noise is 80 dB during peak hour and 65 dB during non-peak hour as shown in the previous calculation The transmissions of external noise into the auditorium are reduced by 41.3 dB which resulted in 26.3 dB during peak hour and 23.47 dB during non-peak hour.

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3.0 References 1. How to choose the right downlight. (n.d.). Retrieved July 8, 2016, from http://www.beaconlighting.com.au/how-to-choose-the-right-led-downlight 2. Sound Absorption Coefficients. (n.d.). Retrieved July 9, 2016, from http://www.acousticalsurfaces.com/acoustic_IOI/101_13.htm

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