Bscience 2 Final Project Report

SCHOOL OF ARCHITECTURE, BUILDING & DESIGN Modern Architecture Studies in Southeast Asia (MASSA) Research Unit Bachelor of Science (Honours) (Architecture)

BUILDING SCIENCE 2 [ARD 3413] PROJECT 2: INTEGRATION

NAME: TAN WEI HOW STUDENT ID: 0310707 TUTOR: MR. SANJAY

Table Content 1.0

Introduction

2.0

Lighting Proposal 2.1

Natural Daylighting 2.1.1 Space 1: Young Adult Reading Zone 2.1.2 Space 2: Multifunction Hall 2.1.3 Space 3: Café 2

2.2

Artificial Lighting 2.1.1 Space 1: Multifunction Hall 2.1.2 Space 2: Kitchen 2.1.3 Space 3: Computer Lab

3.0

Acoustic Proposal 3.1

Sound Pressure Level: External Noise

3.2

Reverberation Time (RT) 3.2.1 Space 1: Multifunction Hall 3.2.2 Space 2: Lobby 3.2.3 Space 3: Staff Office

3.3

Sound Reduction Index (SRI) 3.3.1 Space 1: Discussion Room 3.3.2 Space 2: Café 2 3.3.3 Space 3: Multifunction Hall

1.0

Introduction This project is integration with our Architecture Design Studio 5 final project –

Community Library. The site of this project is located at an urban site, Brickfields, Kuala Lumpur. We are required to identify 3 spaces for artificial lighting and day lighting significance as well as integrating external and internal noises. Various calculations for both lighting and acoustic system for integration will be included into this report.

2.0

Lighting Proposal

2.1

Natural Lighting

2.1.1 Space 1: Young Adult Reading Zone

Formula: Daylight Factor (DF) = (Ei / Eo) x 100% Area of Space 1 (m²) 172 Area of Curtain Wall (m²) 102.8 Daylight Factor (%) (102.8 / 172) x 100% = 59.7% x 0.1 = 5.97%

Young Adult Reading Zone has a daylight factor of 5.97%. Based on the requirements of MS1525, this reading zone has a good natural daylighting as it is within 3-6%. Due to the façade design of the market, this reading zone is located at the front part of the first floor of library. Thus, the daylighting for the ground floor entrance of the library is slightly affected.

Zone

DF (%)

Distribution

Very Bright

>6

Very large with thermal and glare problem

Bright

3-6

Good

Average

1-3

Fair

Dark

0-1

Poor

Daylight factors and distribution (Department of Standards Malaysia, 2007)

Formula Given, Eo (unobstrusted sky of Malaysia)

EEi = (DF x Eo) / 100% 3200 EEi = (DF x Eo) / 100% = (5.97 x 32000) / 100% = 1910.4 lux

The general illuminance level of a reading area 300–400 lux. The final calculation of daylighting is 1910.4 lux, which is highly over the requirement range. This proves that the reading zone itself receives too much day light during the day time, and metal mesh panels will be added along the curtain wall to filter the day light.

Task Lighting for infrequently used area

Illuminance (Lux) 20 100 100 100 100 150 100 100 100 300 200

Lighting for working interiors

200 300 – 400 300 - 400 150 200 150 – 300 150 150 100 100 300 – 500 200 – 750 300

Localized lighting for exacting task

500 1000 2000

Example of Application Minimum service illuminance Interior walkway and car-park Hotel bedroom Lift interior Corridor, passageways, stairs Escalator, travelator Entrance and exit Staff changing room, locker and cleaner room Entrance hall, lobbies, waiting room Inquiry desk Gate house Infrequent reading and writing General offices, shops and stores, reading and writing Drawing office Restroom Restaurant, Canteen, Cafeteria Kitchen Lounge Bathroom Toilet Bedroom Classroom, Library Shop, Supermarket, Department store Museum and gallery Proof reading Exacting drawing Detailed and precise work

Recommended average illuminance levels

2.1.2 Space 2: Multifunction Hall

Formula: Daylight Factor (DF) = (Ei / Eo) x 100% Area of Space 1 (m²) 94.5 Area of Curtain Wall (m²) 37.45 Daylight Factor (%) (37.45 / 94.5) x 100% = 39.6% x 0.1 = 3.96%

Multifunction Hall has a daylight factor of 3.96%. Based on the requirements of MS1525, this hall has a good natural daylighting as it is within 3-6%. This hall is located at ground floor level.

Zone

DF (%)

Distribution

Very Bright

>6

Very large with thermal and glare problem

Bright

3-6

Good

Average

1-3

Fair

Dark

0-1

Poor

Daylight factors and distribution (Department of Standards Malaysia, 2007)

Formula Given, Eo (unobstrusted sky of Malaysia)

EEi = (DF x Eo) / 100% 3200 EEi = (DF x Eo) / 100% = (3.96 x 32000) / 100% = 1267.7 lux

The general illuminance level of a hall is 300 lux. The final calculation of daylighting is 1267.7 lux, which is highly over the requirement range. This proves that hall itself receives too much day light during the day time, and metal mesh panels will be added along the curtain wall to filter the day light.

2.1.3 Space 3: Café 2

Formula: Daylight Factor (DF) = (Ei / Eo) x 100% Area of Space 1 (m²) 34.5 Area of Curtain Wall (m²) 15.4 Daylight Factor (%) (15.4 / 34.5) x 100% = 44.6% x 0.1 = 4.46%

Café 2 has a daylight factor of 4.46%. Based on the requirements of MS1525, this area has a good natural daylighting as it is within 3-6%. This cafe is located at first floor level.

Zone

DF (%)

Distribution

Very Bright

>6

Very large with thermal and glare problem

Bright

3-6

Good

Average

1-3

Fair

Dark

0-1

Poor

Daylight factors and distribution (Department of Standards Malaysia, 2007)

Formula Given, Eo (unobstrusted sky of Malaysia)

EEi = (DF x Eo) / 100% 3200 EEi = (DF x Eo) / 100% = (4.46% x 32000) / 100% = 1427.2 lux

The general illuminance level of a cafe is 200 lux. The final calculation of daylighting is 1267.7 lux, which is highly over the requirement range. This proves that cafe itself receives too much day light during the day time, and metal mesh panels will be added along the curtain wall to filter the day light

2.2

Artificial Lighting

2.1.1 Space 1: Multifunction Hall

Multifunction hall is a space where required to lit up the entire space for event holding.

Philips Tornado High Lumen is chosen to be the light for the space.

Lamp Type Wattage Voltage Light Color Luminous Flux Color Temperature Color Rendering Index (CRI) Features

Philips Tornado High Lumen Fluorescent Lamp 45W 230V Cool White 2850lm 6500K 80 Long-life and energy saving

Lumen Method Calculation Dimension of room (m)

10.7m x 8.825m

Total floor area (m²)

94.5

Standard Illuminance Required (lux) according to MS1525 Type of lighting fixtures

300

Lumen of lighting fixtures (lm)

2850

Height of luminaire (m)

3.2

Work level (m)

0.8

Mounting Height (m)

2.4

Assumption of reflective value

Ceiling – 0.7

Room Index / RI (K)

10.7 x 8.825__ (10.7 + 8.825)2.4

Philips Tornado High Lumen

Wall – 0.5

Utilization Factor / UF

=1.93 0.62

Maintenance Factor / MF

0.80

Lumen Calculation

300 x 94.5___ 2850 x 0.62 x 0.8

Floor – 0.2

=20 bulbs Number of luminaires across

10.7 đ?‘Ľ 20 √ 8.825 =4.9 (take 5) So, each spacing is 10.7 / 5 = 2.14m

Number of luminaires along

8.825 đ?‘Ľ 20 √ 10.7 =4.06 (take 4) So, each spacing is 10.7 / 5 = 2.14m

2.1.2 Space 2: Kitchen

Kitchen is a space where requires sufficient artificial light for working. Philips 15W Mini Cool White Fluorescent Tube is chosen to be the light for the space.

Philips 15W Mini Cool White Fluorescent Tube Lamp Type Fluorescent Lamp Wattage 15W Voltage 230V Light Color Cool White Luminous Flux 960lm Color Temperature 2700K Color Rendering Index (CRI) 80 Features Low pressure mercury discarged

Dimension of room (m)

2m x 6.365m

Total floor area (m²)

12.73

Standard Illuminance Required (lux) according to MS1525 Type of lighting fixtures

300

Lumen of lighting fixtures (lm)

960

Height of luminaire (m)

3.2

Work level (m)

0.8

Mounting Height (m)

2.4

Assumption of reflective value

Ceiling – 0.7

Room Index / RI (K)

2 x 6.365__ (2 + 6.365)2.4

Philips 15W Mini Cool White Fluorescent Tube

Wall – 0.5

Utilization Factor / UF

=0.64 0.62

Maintenance Factor / MF

0.80

Lumen Calculation

300 x 12.73___ 960 x 0.62 x 0.8

Floor – 0.2

=8 lamps Number of luminaires across

2đ?‘Ľ8 √ 6.365 =1.58 (take 2) So, each spacing is 2 / 2 = 1m

Number of luminaires along

6.365 đ?‘Ľ 8 √ 2 =5.04 (take 5) So, each spacing is 6.365 / 5 = 1.27m

2.1.3 Space 3: Computer Lab

Computer Lab is a space where cannot receive too much of natural light because of the glare so it needs artificial light to lit up the entire space. Philips Corepro LED Lamp is chosen to be the light for the space.

Philips Corepro LED Lamp E27 6W (40W) Lamp Type Wattage Voltage Light Color Luminous Flux Color Temperature Color Rendering Index (CRI) Features

Fluorescent Lamp 15W 230V Cool White 1142lm 3000K 100 Emits without glare and harsh shadows

Dimension of room (m)

5.7m x 19.169m

Total floor area (m²)

109.2

Standard Illuminance Required (lux) according to MS1525 Type of lighting fixtures

300

Lumen of lighting fixtures (lm)

1142

Height of luminaire (m)

3.2

Work level (m)

0.8

Mounting Height (m)

2.4

Assumption of reflective value

Ceiling – 0.7

Room Index / RI (K)

5.7 x 19.169__ (5.7 + 19.169)2.4

Philips Corepro LED Lamp

Wall – 0.5

Utilization Factor / UF

=1.83 0.62

Maintenance Factor / MF

0.80

Lumen Calculation

300 x 109.2___ 1142 x 0.62 x 0.8

Floor – 0.2

=58 bulbs Number of luminaires across

5.7 đ?‘Ľ 58 √ 19.169 =4.15 (take 4) So, each spacing is 5.7 / 4 = 1.425m

Number of luminaires along

19.169 đ?‘Ľ 58 √ 5.7 =13.9 (take 14) So, each spacing is 19.169 / 14 = 1.37m

3.0

Acoustic Proposal

3.1

Sound Pressure Level: External Noise Sound Pressure Level Formula: SPL = 10log (l1/l0)

Noise Source:

Busy Traffic Noise Back lane Market Noise Normal Conversation

= 80dB = 90dB = 40dB

Busy Traffic Noise = 80dB 80 Antilog 8 8 x 108 l1

= 10log (l1/l0) = l1 / (1.0 x 10-12) = l1 / (1.0 x 10-12) = 1.0 x 10-4

Back lane Market Noise = 90dB 90 Antilog 9 9 x 109 l1

= 10log (l1/l0) = l1 / (1.0 x 10-12) = l1 / (1.0 x 10-12) = 1.0 x 10-3

Normal Conversation = 40dB 40 Antilog 4 4 x 104 l1

= 10log (l1/l0) = l1 / (1.0 x 10-12) = l1 / (1.0 x 10-12) = 1.0 x 10-8

Total Intensities

= (1.0 x 10-4) + (1.0 x 10-3) + (1.0 x 10-8) = 1.0 x 10-3

Combined SPL

= 10log (l1/l0) = 10log [(1.0 x 10-3) / (1.0 x 10-12)] = 90dB

Conclusion: The required sound pressure level for a library is 35dB but the sound pressure at that has extremely exceeded with 90dB. Action has to be taken to reduce the sound pressure level to the library.

3.2

Reverberation Time

3.2.1 Multifunction Hall

94.5 m2 378 m3 43

Total Floor Area (m²) Volume (m3) Occupancy

Components

Materials

Area (m²)

Wall Floor Ceiling Window Furniture Occupants Total

Brick Concrete Concrete Glass Fabric Chair

66.2 94.5 94.5 40 43 (pax) 43 (pax)

Absorption Coefficient (2000Hz) 0.05 0.02 0.02 0.07 0.82 0.51

Area of Absorption Coeeficient 3.31 1.89 1.89 2.8 35.26 21.93 67.08

RT = 0.16 V/A RT = 0.16 (378) / 67.08 = 0.90s

Conclusion: The reverberation time for the multifunction hall is 0.90s. The multifunction hall is used for public speaking so some acoustic panels are suggested to be installed to the hall.

3.2.2 Lobby

114 m2 399 m3 20

Total Floor Area (m²) Volume (m3) Occupancy

Components

Materials

Area (m²)

Wall Floor Ceiling Window Furniture

Brick Concrete Timber Glass Fabric Chair Sofa Timber Table Wood Panel

70.8 120 120 108 11 (pax) 4 (pax) 11 (pax) 12 30 (pax)

Occupants Total

Absorption Coefficient (2000Hz) 0.05 0.02 0.06 0.07 0.82 0.58 0.15 0.10 0.51

Area of Absorption Coeeficient 3.54 2.40 7.20 7.56 9.02 2.32 1.65 1.2 15.3 50.19

RT = 0.16 V/A RT = 0.16 (399) / 50.19 = 1.27s

Conclusion: The reverberation time for the lobby is 1.27s which has exceeded the requirement due to the size of the space and the users in the space.

3.2.3 Staff Office

15.8 m2 55.3 m3 3

Total Floor Area (m²) Volume (m3) Occupancy

Components

Materials

Area (m²)

Wall

Brick Glass Concrete Concrete Fabric Chair Timber Table

10.8 46.5 15.8 15.8 3 (pax) 3 (pax) 3 (pax)

Floor Ceiling Furniture Occupants Total

Absorption Coefficient (2000Hz) 0.05 0.07 0.02 0.02 0.82 0.58 0.51

RT = 0.16 V/A RT = 0.16 (55.3) / 10.17 = 0.87s

Conclusion: The reverberation time for the staff office is 0.87s.

Area of Absorption Coeeficient 0.54 3.26 0.32 0.32 2.46 1.74 1.53 10.17

3.3

Sound Reduction Index

3.3.1 Discussion Room

Components

Materials

Area (m²)

Wall

Glass Brick

32.5 7.5

Sound Reduction Index (SRI) 35 42

Transmission Coefficient of Materials a) Wall – Glass SRI (glass) = 10log (1 / Tglass) 35 = 10log (1 / Tglass) Antilog 35 = 1 / Tglass Tglass = 3.16 x 10-4 b) Wall – Brick SRI (brick) = 10log (1 / Tbrick) 42 = 10log (1 / Tbrick) Antilog 42 = 1 / Tbrick Tbrick = 6.31 x 10-5

Average Transmission Coefficient of Materials Tav =

Tav =

đ?‘†1đ?‘‡đ?‘?1+đ?‘†2đ?‘‡đ?‘?2+â‹Żđ?‘†đ?‘›đ?‘‡đ?‘?đ?‘› đ?‘‡đ?‘œđ?‘Ąđ?‘Žđ?‘™ đ?‘†đ?‘˘đ?‘&#x;đ?‘“đ?‘Žđ?‘?đ?‘’ đ??´đ?‘&#x;đ?‘’đ?‘Ž (3.16 đ?‘Ľ 10−4 )+(6.31 đ?‘Ľ 10−5 ) 32.5+7.5

= 9.48 x 10-6

Transmission Coefficient

Area x Tcn

Total sound reduction index, SRI SRI (overall) = 10 log (1 / Tav) = 10 log (1 / 9.48 x 10-6) = 52.23 dB

Noise level in Discussion Room = 90 dB – 52.23 dB = 37.77 dB

The overall transmission loss from outside to the discussion room is at 52.23dB. Assume that the sound pressure level in the outdoor is approximately 90dB, so the final sound that transmitted to the discussion room is approximately 37.77dB

3.3.2 CafĂŠ 2

Components

Materials

Area (m²)

Wall

Glass Brick

15 64.15

Sound Reduction Index (SRI) 35 42

Transmission Coefficient of Materials a) Wall – Glass SRI (glass) = 10log (1 / Tglass) 35 = 10log (1 / Tglass) Antilog 35 = 1 / Tglass Tglass = 3.16 x 10-4 b) Wall – Brick SRI (brick) = 10log (1 / Tbrick) 42 = 10log (1 / Tbrick) Antilog 42 = 1 / Tbrick Tbrick = 6.31 x 10-5

Average Transmission Coefficient of Materials Tav =

Tav =

đ?‘†1đ?‘‡đ?‘?1+đ?‘†2đ?‘‡đ?‘?2+â‹Żđ?‘†đ?‘›đ?‘‡đ?‘?đ?‘› đ?‘‡đ?‘œđ?‘Ąđ?‘Žđ?‘™ đ?‘†đ?‘˘đ?‘&#x;đ?‘“đ?‘Žđ?‘?đ?‘’ đ??´đ?‘&#x;đ?‘’đ?‘Ž (3.16 đ?‘Ľ 10−4 )+(6.31 đ?‘Ľ 10−5 ) 15+64.15

= 4.79 x 10-6

Transmission Coefficient

Area x Tcn

Total sound reduction index, SRI SRI (overall) = 10 log (1 / Tav) = 10 log (1 / 4.79 x 10-6) = 53.20 dB

Noise level in Discussion Room = 90 dB – 53.20 dB = 36.8 dB The overall transmission loss from outside to the café 2 is at 53.2dB. Assume that the sound pressure level in the outdoor is approximately 90dB, so the final sound that transmitted to the café is approximately 36.8dB

3.3.3 Multifunction Hall

Components

Materials

Area (m²)

Wall

Glass Brick

36.75 666.75

Transmission Coefficient of Materials a) Wall – Glass SRI (glass) = 10log (1 / Tglass) 35 = 10log (1 / Tglass) Antilog 35 = 1 / Tglass Tglass = 3.16 x 10-4 b) Wall – Brick SRI (brick) = 10log (1 / Tbrick) 42 = 10log (1 / Tbrick) Antilog 42 = 1 / Tbrick Tbrick = 6.31 x 10-5

Sound Reduction Index (SRI) 35 42

Transmission Coefficient

Area x Tcn

Average Transmission Coefficient of Materials Tav =

Tav =

đ?‘†1đ?‘‡đ?‘?1+đ?‘†2đ?‘‡đ?‘?2+â‹Żđ?‘†đ?‘›đ?‘‡đ?‘?đ?‘› đ?‘‡đ?‘œđ?‘Ąđ?‘Žđ?‘™ đ?‘†đ?‘˘đ?‘&#x;đ?‘“đ?‘Žđ?‘?đ?‘’ đ??´đ?‘&#x;đ?‘’đ?‘Ž (3.16 đ?‘Ľ 10−4 )+(6.31 đ?‘Ľ 10−5 ) 36.75+666.75

= 5.40 x 10-7

Total sound reduction index, SRI SRI (overall) = 10 log (1 / Tav) = 10 log (1 / 5.40 x 10-7) = 62.68 dB

Noise level in Discussion Room = 90 dB – 62.68 dB = 27.32 dB The overall transmission loss from outside to the multifunction hall is at 62.68dB. Assume that the sound pressure level in the outdoor is approximately 90dB, so the final sound that transmitted to the hall is approximately 27.32dB