Building Science II (Studio V Integration)

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

Building Science 2 (ARC3413)

Project 2: Integration Project with Design Studio 5 [Community Library, in Sentul West]

Final Report & Calculation

Name: Leong Huiyi ID: 0319280 Tutor: Mr. Siva

Table of Content 1.0 Lighting 1.1 Day lighting 1.1.1 Reading pod 1.1.2 Informal reading area 1.2 Artificial Lighting 1.2.1 Café 1.2.2 Admin / Lobby 1.3 PSALI 1.3.1 Café 1.3.2 Gallery

2.0 Acoustic 2.1 Sound Pressure Level (Exterior Noise) 2.1.1 Large print collection 2.1.2 Reading pod 2.2 Reverberation Time, RT 2.2.1 Workshop 2.2.2 Discussion room 2.3 Sound Reduction Index, SRI 2.3.1 Indoor studio

2.3.2 Admin / lobby Appendix I Appendix II References

1 . 0 L i g h t i n g

1.1 Day lighting According to MS1525:2007, the daylight factors are categorized in table as below: Zone Very bright

DF (%) > 6

Distribution Very large with thermal and glare problems

Bright

3-6

Good

Average

1-3

Fair

Dark

0-1

Poor

Fig 1.1: Daylight factors and distribution

1.1.1 Reading Pod In this community library, the selected area for analyzing day lighting is reading pod. Reading pod is located on the second floor facing towards the back lane where the façade design is totally exposed to sunlight. Thus, there will be concerns for external glare and minimal artificial lighting required in this space.

Fig 1.2: CafĂŠ Plan

Daylight Factor (DF) Calculation Floor Area, m2

12.2

Area of faรงade exposed to sunlight, m2

12.2

Open Area to Floor Area Ratio (Daylight

[ (area of faรงade exposed / (floor area) ] x 100%

Factor, DF)

=

!".! !".!

x100%

= 100 x 0.1 = 10%

Nature Illumination Table

Illuminance (lux)

Example

120,000

Brightest sunlight

110,000

Bright sunlight

20,000

Shade illuminated by entire clear blue sky

1000-2000

Typical overcast day, midday

400

Sunrise or sunset on clear day

100

Very dark overcast day

40

Fully overcast, sunset/sunrise

< 1

Extreme of darkest storm clouds, sunset/sunrise Fig 1.3: Examples of different effects under each illuminance

Using the formula,

DF =

Internal illuminance, Ei External illuminance, Eo

x 100%

10 = Ei / 20,000 x 100 Ei = 10 x 20000 / 100 = 2000 lux Conclusion The selected reading pod has a daylight factor of 10% and a natural illumination of 2000lux. Based on the requirements of MS1525, the reading pod should be around 500 lux. Hence it is too bright and faces thermal and glares problems.

In order to solve the issues, the double faรงade system is proposed as the faรงade design by incorporating the inner skin as brick facades, arranged in a way to let appropriate amount of sunlight in. The outer skin will be perforated aluminum panels to reduce large amount of sunlight penetrating into the space. Besides that, double-glazed low-e glass will be used for the windows in this space to reduce heat gain at the same time.

1.1.2 Informal Reading Area

Fig1.4: informal reading plan

The second selected area for analyzing day lighting is informal reading area. It is located on the second floor at the inner side of the building where a light well is situated in the middle to let more sunlight in. It is an open-space design hence the data calculated is approximate. The only concern in this area is whether the light well produced enough or too little sunlight in the area.

Daylight Factor (DF) Calculation Floor Area, m2

86.14

Area of faรงade exposed to sunlight, m2

19.2

Open Area to Floor Area Ratio (Daylight

[ (area of faรงade exposed / (floor area) ] x 100%

Factor, DF)

=

!".! !".!"

x100%

= 22.3 x 0.1 = 2.2%

Nature Illumination Table

Illuminance (lux)

Example

120,000

Brightest sunlight

110,000

Bright sunlight

20,000

Shade illuminated by entire clear blue sky

1000-2000

Typical overcast day, midday

400

Sunrise or sunset on clear day

100

Very dark overcast day

40

Fully overcast, sunset/sunrise

< 1

Extreme of darkest storm clouds, sunset/sunrise Fig 1.5: Examples of different effects under each illuminance

Using the formula,

DF =

Internal illuminance, Ei External illuminance, Eo

x 100%

2.2 = Ei / 20,000 x 100 Ei = 2.2 x 20000 / 100 = 440 lux

Conclusion The informal reading area has a daylight factor of 2.2% and a natural illumination of 440lux. Based on the standard requirements [1], the informal reading area should be 500 lux which in this case it has a fairly distribution of daylight.

Due to the analyzed area is open space hence the result is calculated approximately. However, to enhance the experience within this space, artificial lightings can be implemented to create an adequate lux for the users while reading.

1.2 Artificial lighting 1.2.1 Cafe The selected space to study the artificial lighting is cafe. It is located on the ground floor facing the back lane. Although natural light can be acquired, in order to create a suitable ambience, artificial lighting is vital to light up the area evenly.

Fig 1.6: cafĂŠ plan with light fittings

According to MS1525:2007, the minimum lighting level requirement for cafe is 200 lux.

Table of fixture used Type of fixture

Linear Fluorescent light bulb

Picture of lighting

Product brand

Philips

Material of fixture

Aluminum

Nominal life (hours)

36000

Luminous flux (lm)

3800

Colour Rendering Index (CRI)

85

Colour Temperature (Kelvin)

3000 Warm White

Wattage (W)

40

Lumen Method Calculation Selected space

Cafe

Room dimension, m

Length (L) = 6 Width (W) = 10 Height of the ceiling (h) = 4

Floor area (A), m2

6 x 10 = 60

Type of lighting fixture

Fluorescent light bulb

Lumen of lighting fixture (F), lm

3800

Height of work level, m

0.8

Mounting height (Hm), m

4.0 â&#x20AC;&#x201C; 0.8 = 3.2

Reflectance value

Ceiling = 0.3 Wall = 0.35 Floor = 0.1

Room Index (RI), K L x W RI = (L + W) x Hm

6 x 10

(6 + 10) x 3.2 = 1.17

Utilization factor (UF) [2]

0.32

Maintenance factor (MF)

0.8

Recommended illuminance level (E) according to MS1525:2007

200

Numbers of lights required (N) E x A

200 x 60 3800 x 0.32 x 0.8

N = F x UF x MF

= 12.34 = 12 light bulbs

Spacing to height ratio (SHR)

SHR =

SHR =

1 Hm

x √

A

N

1 SHR =

3.2

= 0.7 SHR =

Spacing between lightings

!

!.!

Mounting height

x √

60

12

S = 0.7 x 3.2 = 2.24

Approximate light fittings layout, m

Fittings required along 6m wall: 6 ÷ 2.24 = 2.68 = 3 rows

No. of lamps required in each row total number of lights required (N) =

numbers of rows = 12 ÷ 3 = 4 Spacing along 10m wall: 10 ÷ 4 = 2.5 = 3m Hence, approximately 3 x 4 = 12 light fittings

Conclusion There are total of 12 florescent light fittings to illuminate the cafĂŠ in order to achieve optimum reading of 300 lux required by MS1525. Under sufficient ambience, users are able to have a better experience when dining inside the cafĂŠ.

1.2.2 Admin / Lobby The selected space to study the artificial lighting is admin/lobby. It is located on the ground floor facing the front walkway and center courtyard. In order to create a suitable ambience, artificial lighting is vital to light up the area evenly even though natural light can be acquired.

Fig 1.7: Admin/lobby plan with light fittings

According to MS1525:2007, the minimum lighting level requirement for admin/lobby is 100 lux.

Table of fixture used Type of fixture

LED Downlight

Picture of lighting

Product brand

DN571B

Material of fixture

Aluminum

Nominal life (hours)

25000

Luminous flux (lm)

4000

Colour Rendering Index (CRI)

80

Colour Temperature (Kelvin)

3000 Warm white

Wattage (W)

36

Lumen Method Calculation Selected space

Admin/Lobby

Approximate room dimension, m

Length (L) = 9 Width (W) = 4.5 Height of the ceiling (h) = 4

Floor area (A), m2

9 x 4.5 = 40.5

Type of lighting fixture

LED Downlight

Lumen of lighting fixture (F), lm

4000

Height of work level, m

0.8

Mounting height (Hm), m

4.0 â&#x20AC;&#x201C; 0.8 = 3.2

Reflectance value

Ceiling = 0.3 Wall = 0.35

Floor = 0.14 Room Index (RI), K L x W

9 x 4.5

RI = (L + W) x Hm

(9 + 4.5) x 3.2 = 0.94

Utilization factor (UF) [2]

0.29

Maintenance factor (MF)

0.8

Recommended illuminance level (E) according to MS1525:2007 Numbers of lights required (N)

100

E x A

100 x 40.5

N = F x UF x MF

4000 x 0.29 x 0.8 = 4.36 = 4 light bulbs

Spacing to height ratio (SHR)

SHR =

SHR =

1 Hm

x √

A

N

Spacing between lightings Mounting height

1

SHR =

3.2

= 0.99

SHR =

!

!.!

x √

40.5

4

S = 0.99 x 3.2 = 3.19

Approximate light fittings layout, m

Fittings required along 4.5m wall: 4.5 ÷ 3.19 = 1.4 = 1 row No. of lamps required in each row total number of lights required (N) numbers of rows

= 6 รท 1 = 6

Spacing along 9m wall: 9 รท 6 = 1.5m

Hence, approximately 1 x 6 = 6 light fittings

Conclusion There are total of 6 LED down light fittings to illuminate the admin/lobby in order to achieve optimum reading of 100 lux required by MS1525. Under sufficient ambience, users are able to have a clearer vision when entering lobby.

1.3 PSALI Permanent Supplementary Artificial Lighting of the Interior (PSALI) is based on three principles: a. Utilization of daylight as far as practicable b. Use of electric lighting to supplement the daylight in the interior parts of the room c. Installation of the electric lighting in such a way that the daylight character of the room is retained The most important factors when looking at providing a PSALI system are to ensure that there is good lighting levels and even light distribution to the inner corner of the room.

Fig 1.8: PSALI scheme

The required quantity of artificial light input is to achieve an illuminance at the rear of the room comparable to, but slightly less than, the daylight illuminance nears the window.

For example: Daylight Factor, %

Artificial light, lux

0.5

250

1.0

500

1.5

750

2.0

1000

1.3.1 Quiet Study Area The selected space to study the PSALI is quiet study area, which located on the second floor. It has front and side faรงade exposing to daylight. At the same time, the quiet reading area required artificial lighting in the middle to achieve sufficient and uniform lighting.

Fig 1.9: Quiet study area light fittings

According to MS1525:2007, the minimum lighting level requirement for quiet study area is 400 lux.

Daylight Factor (DF) Calculation Floor Area, m2

76.8

Area of faรงade exposed to sunlight, m2

46.4

Open Area to Floor Area Ratio (Daylight

[ (area of faรงade exposed / (floor area) ] x 100%

Factor, DF)

=

!".! !".!

x100%

= 60.4 x 0.1 = 6%

Nature Illumination Table

Illuminance (lux)

Example

120,000

Brightest sunlight

110,000

Bright sunlight

20,000

Shade illuminated by entire clear blue sky

1000-2000

Typical overcast day, midday

400

Sunrise or sunset on clear day

100

Very dark overcast day

40

Fully overcast, sunset/sunrise

< 1

Extreme of darkest storm clouds, sunset/sunrise Fig 1.10: Examples of different effects under each illuminance

Using the formula,

DF =

Internal illuminance, Ei External illuminance, Eo

x 100%

6 = Ei / 20,000 x 100 Ei = 6 x 20000 / 100 = 1200 lux

Conclusion The quiet study area has a daylight factor of 6% and a natural illumination of 1200lux. Based on the standard requirements, the quiet study area has a very good distribution of daylight.

To make sure when there is no daylight at night, artificial lightings can be implemented to create an adequate lux for the users while reading.

Table of fixture used Type of fixture

Surface mounted fluorescent batten

Picture of lighting

Product brand

SLT5128

Material of fixture

Aluminum

Nominal life (hours)

10000

Luminous flux (lm)

3000

Colour Rendering Index (CRI)

85

Colour Temperature (Kelvin)

3200 Cool white

Wattage (W)

28

Lumen Method Calculation Selected space

Quiet Study Area

Approximate room dimension, m

Length (L) = 12 Width (W) = 7.5 Height of the ceiling (h) = 3.2

Floor area (A), m2

12 x 7.5 = 90

Type of lighting fixture

Surface mounted fluorescent batten

Lumen of lighting fixture (F), lm

2600

Height of work level, m

0.8

Mounting height (Hm), m

3.2 â&#x20AC;&#x201C; 0.8 = 2.4

Reflectance value

Ceiling = 0.3 Wall = 0.35 Floor = 0.1

Room Index (RI), K L x W RI = (L + W) x Hm

12 x 7.5

(12 + 7.5) x 2.4 = 1.92

Utilization factor (UF) [2]

0.40

Maintenance factor (MF)

0.8

Recommended illuminance level (E) according to MS1525:2007 Numbers of lights required (N)

400

E x A N = F x UF x MF

400 x 90

3000 x 0.4 x 0.8 = 37.5 = 38 light bulbs

Spacing to height ratio (SHR)

SHR =

SHR =

1 Hm

x √

A

N

1 SHR =

2.4

= 0.64 SHR =

Spacing between lightings

!

!.!

Mounting height

x √

90

38

S = 0.64 x 2.4 = 1.54

Approximate light fittings layout, m

Fittings required along 12m wall: 12 ÷ 1.54 = 7.8 = 8 rows

No. of lamps required in each row total number of lights required (N) =

numbers of rows = 38 ÷ 8 = 4.75 = 5

Spacing along 7.5m wall: 7.5 ÷ 5 = 1.5m

Hence, approximately 8 x 5 = 40 light fittings

Conclusion There are total of 40 surface mounted fluorescent battens to illuminate the quiet study area in order to achieve optimum reading of 400 lux required by MS1525. The area is supplemented by artificial lighting which will automatically switch off if there is enough daylight in the day.

1.3.2 Gallery The selected space to study the PSALI is gallery, which located on the second floor (mezzanine level). It has front and side faรงade exposing to daylight. At the same time, the gallery required artificial lighting in the middle to achieve sufficient and uniform lighting.

Fig 1.10: Gallery plan with light fittings

According to MS1525:2007, the minimum lighting level requirement for gallery is 300 lux.

Daylight Factor (DF) Calculation Floor Area, m2

107.25

Area of faรงade exposed to sunlight, m2

52.8

Open Area to Floor Area Ratio (Daylight

[ (area of faรงade exposed / (floor area) ] x 100%

Factor, DF)

=

!".! !"#.!"

x100%

= 49.2 x 0.1 = 4.9%

Nature Illumination Table

Illuminance (lux)

Example

120,000

Brightest sunlight

110,000

Bright sunlight

20,000

Shade illuminated by entire clear blue sky

1000-2000

Typical overcast day, midday

400

Sunrise or sunset on clear day

100

Very dark overcast day

40

Fully overcast, sunset/sunrise

< 1

Extreme of darkest storm clouds, sunset/sunrise Fig 1.11: Examples of different effects under each illuminance

Using the formula,

DF =

Internal illuminance, Ei External illuminance, Eo

x 100%

4.9 = Ei / 20,000 x 100 Ei = 4.9 x 20000 / 100 = 980 lux

Conclusion The gallery has a daylight factor of 4.9% and a natural illumination of 980lux.

Based on the standard requirements, the gallery has a very good distribution of daylight.

To make sure when there is no daylight at night, artificial lightings can be implemented to create an adequate lux for the users while reading.

Table of fixture used Type of fixture

Surface mounted fluorescent batten

Picture of lighting

Product brand

SLT5128

Material of fixture

Aluminum

Nominal life (hours)

10000

Luminous flux (lm)

3000

Colour Rendering Index (CRI)

85

Colour Temperature (Kelvin)

3200 Cool white

Wattage (W)

28

Lumen Method Calculation Selected space

Gallery

Approximate room dimension, m

Length (L) = 19.5 Width (W) = 5.5 Height of the ceiling (h) = 3.2

Floor area (A), m2

19.5 x 5.5 = 107.25

Type of lighting fixture

Surface mounted fluorescent batten

Lumen of lighting fixture (F), lm

2600

Height of work level, m

0.8

Mounting height (Hm), m

3.2 â&#x20AC;&#x201C; 0.8 = 2.4

Reflectance value

Ceiling = 0.3 Wall = 0.35 Floor = 0.14

Room Index (RI), K L x W RI = (L + W) x Hm

19.5 x 5.5

(19.5 + 5.5) x 2.4 = 1.79

Utilization factor (UF) [2]

0.40

Maintenance factor (MF)

0.8

Recommended illuminance level (E) according to MS1525:2007 Numbers of lights required (N)

300

E x A N = F x UF x MF

300 x 107.25 3000 x 0.4 x 0.8

= 33.5 = 34 light bulbs

Spacing to height ratio (SHR)

SHR =

SHR =

1 Hm

x √

A

N

1 SHR =

2.4

= 0.74 SHR =

Spacing between lightings

!

!.!

Mounting height

x √

107.25

34

S = 0.74 x 2.4 = 1.78

Approximate light fittings layout, m

Fittings required along 19.5m wall: 19.5 ÷ 1.78 = 10.9 = 11 rows

No. of lamps required in each row total number of lights required (N) =

numbers of rows = 34 ÷ 11 = 3.1 = 3 Spacing along 5.5m wall: 5.5 ÷ 3 = 1.8m Hence, approximately 11 x 3 = 33 light fittings Conclusion There are total of 33 surface mounted fluorescent battens to illuminate the gallery in order to achieve optimum reading of 300 lux required by MS1525. The area is supplemented by artificial lighting which will automatically switch off if there is enough daylight in the day.

2 . 0 A c o u s t i c

2.1 Sound Pressure Level, SPL (External Noise) Using the formula,

Swl =

10 log10

Sound power (intensity) (Watts), Ii Reference power (1 x 10-12), Io

Spl = 10 log10

Root mean squared pressure (n/m2), p2 Reference pressure (1 x 10-12 N/m2), po2

2.1.1 Large Print Collection

Fig 1.12: Large print collection plan

Located in front of busy and noisy main road of Jalan Sultan Azlan Shah, Sentul West, the community library faces challenges of external noise source from the vehicles and pedestrians walking by.

Sound Pressure Level Calculation

Highest reading, dB Lowest reading, dB Intensity for highest reading (Ii1), W

Swl = 10 log10

Peak hour 85 63 8.5 = log10

!! !.! ! !"!!"

Log-1 8.5 =

!! !.! ! !"!!"

Non-peak hour 58 52 !!

5.8 = log10

Log-1 5.8 =

!! !.! ! !"!!"

!.! ! !"!!"

Swl = 10 log10

!! !.! ! !"!!"

= 3.16 x 10-4 W

= 6.31 x 10-7 W

!!

!.! ! !"!!"

Log-1 6.3 =

!! !.! ! !"!!"

Ii = log-1 6.3 x (1.0 x 10-12) = 1.99 x 10-6 W

Total intensity (I), W I = (Ii1) + (Ii2) Sound pressure level (SPL)

!! !.! ! !"!!"

Log-1 5.2 =

!! !.! ! !"!!"

Ii = log-1 5.2 x (1.0 x 10-12) = 1.58 x 10-7 W I = (6.31 x 10-7 ) + (1.58 x 10-7 )

= 3.18 x 10-4 W

= 7.89 x 10-7 W

SPL = 10 log10 (

3.18 đ?&#x2018;Ľ 10â&#x2C6;&#x2019;4 1.0 đ?&#x2018;Ľ 10

â&#x2C6;&#x2019;12 )

SPL = 10 log10 (

7.89 đ?&#x2018;Ľ 10â&#x2C6;&#x2019;7 1.0 đ?&#x2018;Ľ 10â&#x2C6;&#x2019;12

= 59 dB

5.2 = log10

I = (3.16 x 10-4 ) + (1.99 x 10-6 )

= 85 dB

Ii = log-1 5.8 x (1.0 x 10-12)

6.3 = log10

!! !.! ! !"!!"

Ii = log-1 8.5 x (1.0 x 10-12)

Intensity for lowest reading (Ii2), W

)

Conclusion The average sound pressure level in large print collection during peak and non-peak hour are 85dB and 59dB. The standard sound pressure level for large print collection, which acts as a reading area for senior citizens, is 40dB. However, the combined SPL reading of both peak and nonpeak hour could not achieve the low sound pressure level.

Thus, the design solution to solve the issue is proposed by creating a buffer zone in front of the area. Besides that, the faรงade of the community library has double layer skin faรงade using inner skin of bricks and outer skin of perforated aluminum panels to block out external noise from entering large print collection.

2.1.2 Reading Pod

Fig 1.13: Reading pod plan

Located at the back lane where vehicles and pedestrian will occasionally pass by, the community library faces challenges of mild external noise source.

Sound Pressure Level Calculation

Highest reading, dB Lowest reading, dB Intensity for highest reading (Ii1), W

Swl = 10 log10

Peak hour 62 55 62 = log10

!!

!.! ! !"!!"

6.0 = log10

!.! ! !"!!"

Log-1 6.2 =

!!

Non-peak hour 60 52

!! !.! ! !"!!"

!! !.! ! !"!!"

Log-1 6.0 =

!! !.! ! !"!!"

Ii = log-1 6.2 x (1.0 x 10-12)

Ii = log-1 6.0 x (1.0 x 10-12)

= 1.58 x 10-6 W

= 1.0 x 10-6 W

Intensity for lowest reading (Ii2), W Swl = 10 log10

!! !.! ! !"!!"

55 = log10

!! !.! ! !"!!"

Log-1 5.5 =

5.2 = log10

!! !.! ! !"!!"

Ii = log-1 5.5 x (1.0 x 10-12) = 3.16x 10-7 W

Total intensity (I), W I = (Ii1) + (Ii2) Sound pressure level (SPL)

!! !.! ! !"!!"

Log-1 5.2 =

!! !.! ! !"!!"

Ii = log-1 5.2 x (1.0 x 10-12) = 1.58 x 10-7 W

I = (1.58 x 10-6) + (3.16x 10-7)

I = (1.0 x 10-6) + (1.58 x 10-7 )

= 1.89 x 10-6 W

= 1.16 x 10-6 W

SPL = 10 log10 (

1.89 đ?&#x2018;Ľ 10â&#x2C6;&#x2019;6 1.0 đ?&#x2018;Ľ 10

= 62.8 dB

â&#x2C6;&#x2019;12 )

SPL = 10 log10 (

1.16 đ?&#x2018;Ľ 10â&#x2C6;&#x2019;6 1.0 đ?&#x2018;Ľ 10â&#x2C6;&#x2019;12

)

= 60.6 dB

Conclusion The average sound pressure level in large print collection during peak and non-peak hour are 62.8dB and 60.6dB. The standard sound pressure level for reading pod is 40dB. However, the combined SPL reading of both peak and non-peak hour could not achieve the low sound pressure level.

Thus, the design solution to solve the issue is proposed by locating the reading pod on a higher level, such as second floor, to get as far away from external noise as possible.

Besides that, the faรงade of the community library has double layer skin faรงade using inner skin of bricks and outer skin of perforated aluminum panels to block out external noise from entering large print collection.

2.2 Reverberation time, RT The prolongation of sound wave in a space even though the source of noise is cut off is referred as the reverberation time.

The total room absorption (A) is the sum of the absorption of a surface determined by multiplying its surface area (S) by its absorption coefficient (a), with the inclusion of audience absorption plus other room contents.

Using the formula,

A = S1a1 + S2a2 + S3a3 + S4a4 + … Snan

Where A S1......Sn a2……an

Total absorption

Area of each surface from 1 to n

Absorption coefficient of each surface 1 to n

(!.!" ! !)

RT =

Where V

A

!

Total space volume (m3)

Total absorption

2.2.1 Workshop

Fig 1.14: workshop plan

Volume (m3) = 6.2m x 5.5m x 4m = 136.4 m3

Material Absorption Coefficient at 500Hz with 18 people contained within the space.

Building Component

Material

Area (S), m2

Absorption Coefficient (a)

Sound absorption (Sa)

Wall

Concrete brick

74.4

0.02

1.49

Concrete slab

34.1

0.05

1.71

Timber floor

34.1

0.04

1.36

wall Floor

finish Bi-fold door

Aluminum

21.6

0.99

21.38

Plywood panel

1.8

0.17

0.31

Ceiling

Plaster finish

34.1

0.015

0.51

Furniture

Steel chair with

4.5

0.26

1.17

Timber table

3.12

0.04

0.12

-

18

0.46

8.28

frame

timber finish

Occupants

Total absorption, A

36.33

(!.!" ! !)

Reverberation Time, RT =

!

(!.!" ! !"#.!)

=

= 0.6s

!".!!

Material Absorption Coefficient at 2000Hz with 18 people contained within the space.

Building Component

Material

Area (S), m2

Absorption Coefficient (a)

Sound absorption (Sa)

Wall

Concrete brick

74.4

0.05

3.72

Concrete slab

34.1

0.01

0.34

Timber floor

34.1

0.04

1.36

21.6

0.02

0.43

1.8

0.24

0.43

wall Floor

finish Bi-fold door

Aluminum frame Plywood panel

Ceiling

Plaster finish

34.1

0.04

1.36

Furniture

Steel chair with

4.5

0.46

2.07

Timber table

3.12

0.02

0.06

-

18

0.51

9.18

timber finish

Occupants

Total absorption, A

18.95

(!.!" ! !)

Reverberation Time, RT =

!

(!.!" ! !"#.!)

=

= 1.15s

!".!"

Conclusion The standard reverberation time for workshop in 500Hz and 2000 Hz of absorption coefficient is 0.8-1.0s. In this case, during 500 Hz the RT is 0.6s and during 2000 Hz has a longer RT of 1.15s. Both the data falls around the standard requirement. Even though during 2000 Hz the RT is slightly higher, it is because the analyzed area is a workshop.

Hence, noises of users talking loudly or having activities are allowed to have longer time for it to transfer throughout the space. The proposed solution to reduce the RT of workshop under 2000 Hz is to apply more absorptive materials or replace hard surfaces with soft surfaces on the furniture, ceiling, floor or wall.

2.2.2 Discussion room

Fig 1.15: Discussion room plan

Volume (m3) = 2.8m x 9m x 3.2m = 80.64 m3

Material Absorption Coefficient at 500Hz with 12 people contained within the space.

Building Component

Material

Area (S), m2

Absorption Coefficient (a)

Sound absorption (Sa)

Wall

Concrete brick

17.92

0.02

0.36

Concrete slab

25.2

0.05

1.26

Timber floor

25.2

0.04

1.01

0.3

0.99

0.3

Glass panel

0.9

0.03

0.03

Steel frame

1.5

0.25

0.38

Large glass

57.6

0.06

3.46

wall Floor

finish Door

Aluminum frame

Window

panel Ceiling

Plaster finish

25.2

0.015

0.38

Furniture

Steel chair with

3

0.26

0.78

Timber table

2.08

0.04

0.08

-

12

0.46

5.52

timber finish

Occupants

Total absorption, A

(!.!" ! !)

Reverberation Time, RT =

!

13.56

(!.!" ! !".!")

=

= 0.95s

!".!"

Material Absorption Coefficient at 2000Hz with 12 people contained within the space.

Building Component

Material

Area (S), m2

Absorption Coefficient (a)

Sound absorption (Sa)

Wall

Concrete brick

17.92

0.05

0.9

Concrete slab

25.2

0.01

0.25

Timber floor

25.2

0.06

1.51

0.3

0.02

0.01

Glass panel

0.9

0.02

0.02

Steel frame

1.5

0.38

0.19

Large glass

57.6

0.02

1.15

wall Floor

finish Door

Aluminum frame

Window

panel Ceiling

Plaster finish

25.2

0.04

1.01

Furniture

Steel chair with

3

0.46

1.38

Timber table

2.08

0.02

0.04

-

12

0.51

6.12

timber finish

Occupants

Total absorption, A

(!.!" ! !)

Reverberation Time, RT =

!

12.58

(!.!" ! !".!")

=

= 1.03s

!".!"

Conclusion The standard reverberation time for discussion room in 500Hz and 2000 Hz of absorption coefficient is 0.8-1.0s. In this case, during 500 Hz the RT is 0.95s and during 2000 Hz has a longer RT of 1.03s. Both the data falls around the standard requirement.

Even though during 500 Hz the RT is slightly higher, it is because the analyzed area is a discussion room. Hence, noises of users discussing and talking are allowed to have longer time for it to transfer throughout the space. The proposed solution to reduce the RT of discussion room under 2000 Hz is to apply acoustic tiles on ceiling or add another layer of carpet in the area.

2.3 Sound Reduction Index, SRI Sound reduction Index (SRI), which also referred as Transmission Loss (TL) of a partition measures the number of decibels lost when a sound of a given frequency is transmitted through the partition.

Using the formula,

SRI = 10 log10

Where Tcn

Sn Tav

! !!"

Transmission Coefficient of Material

Surface area of material n !! ! !!" ! !! ! !!" !....!! ! !!" ) !"#$% !"#$%&' !"#!

(

2.3.1 Indoor studio

Fig 1.1:6: Indoor studio plan

Transmission coefficient of materials (Tcn) calculation

1. Brick wall SRI = 10 log10 42 = 10 log10 4.2 = log10

!!" !

!!"

! !!"

15848.93 =

!

! !!"

Tav = 6.31 x 10-5

2. Glass wall SRI = 10 log10 26 = 10 log10 2.6 = log10 398.11 =

!!"

!

!!"

!!" !

!!"

!

!

Tav = 2.51 x 10-3

3. Tempered glass window SRI = 10 log10 35 = 10 log10 3.5 = log10 3162.28 =

!

!!" !

!!"

!!"

!

!!"

!

Tav = 3.16 x 10-4

4. Aluminum window frame SRI = 10 log10 42 = 10 log10 4.2 = log10

!!" !

!!"

! !!"

15848.93 =

!

! !!"

Tav = 6.31 x 10-5

5. Glass door SRI = 10 log10 26 = 10 log10 2.6 = log10 398.11 =

! !!"

! !!"

! !!"

! !!"

Tav = 2.51 x 10-3

Building

Material

Surface area

SRI (dB)

(Sn), m2

component

Transmission

Sn x Tcn

coefficient of material (Tcn)

Wall

Brick

54.4

42

6.31 x 10-5

3.43 x 10-3

Window

Tempered

16

35

3.16 x 10-4

5.06 x 10-3

10.98

42

6.31 x 10-5

6.93 x 10-4

3.1

26

2.51 x 10-3

7.78 x 10-3

glass Aluminum frame Door

Glass 2

Total Surface Area (TSn) = 84.48

Average transmission coefficient of materials calculation

!.!" ! !"!! ! !.!" ! !"!! ! !.!" ! !"!! !(!.!" ! !"!! )

Tav =

= 2.01 x 10-4 SRI = 10 log10

!".!"

! (!.!" ! !"!! )

= 36.97 dB

Conclusion The standard requirement from Appendix, the requirement for indoor studio is approximately 40dB to contain the stereo music sound. Hence in this case, the indoor studio has fulfilled the requirement as the SRI is 36.97dB.

Assuming the sound pressure level in indoor studio is 62dB, after extracting calculated SRI in the area which is 36.97dB, the transmission loss through the partitions into the passage and outdoor garden area is 25.03dB (62dB â&#x20AC;&#x201C; 36.97dB). From the Appendix, 25dB of sound falls under the category of whisper and quiet learning area. In this case, it is an ideal value since it will not affect the users talking or chatting in neighboring spaces of the indoor studio.

2.3.2 Admin / lobby

Fig 1.17: admin plan area

Transmission coefficient of materials (Tcn) calculation

1. Brick wall SRI = 10 log10 42 = 10 log10 4.2 = log10

!!" !

!!"

! !!"

15848.93 =

!

! !!"

Tav = 6.31 x 10-5

2. Glass wall SRI = 10 log10 26 = 10 log10 2.6 = log10 398.11 =

!!"

!

!!"

!!" !

!!"

!

!

Tav = 2.51 x 10-3

3. Tempered glass window SRI = 10 log10 35 = 10 log10 3.5 = log10 3162.28 =

!

!!" !

!!"

!!"

!

!!"

!

Tav = 3.16 x 10-4

4. Aluminum window frame SRI = 10 log10 42 = 10 log10 4.2 = log10

!!" !

!!"

! !!"

15848.93 =

!

! !!"

Tav = 6.31 x 10-5

5. Glass door SRI = 10 log10 26 = 10 log10 2.6 = log10 398.11 =

! !!"

! !!"

! !!"

! !!"

Tav = 2.51 x 10-3

Building

Material

Surface area

SRI (dB)

(Sn), m2

component

Transmission

Sn x Tcn

coefficient of material (Tcn)

Wall

Brick

47.2

42

6.31 x 10-5

2.98 x 10-3

Window

Tempered

79.2

35

3.16 x 10-4

2.5 x 10-2

14.4

42

6.31 x 10-5

9.09 x 10-4

5

26

2.51 x 10-3

1.26 x 10-2

glass Aluminum frame Door

Glass 2

Total Surface Area (TSn) = 145.8

Average transmission coefficient of materials calculation

!.!" ! !"!! ! !.! ! !"!! ! !.!" ! !"!! !(!.!" ! !"!! )

Tav =

= 2.86 x 10-4 SRI = 10 log10

!"#.!

! (!.!" ! !"!! )

= 35.45 dB

Conclusion The standard requirement from Appendix, the requirement for admin/lobby is approximately 44-64dB to contain the noises from visitors coming in. Hence in this case, the SRI of admin/lobby is 35.45 dB, which is lower than requirement. Thus, it is an idea value since lobby consist much loud noises which does not penetrate into the center courtyard and affects any on-going activities within the area.

Assuming the sound pressure level in admin/lobby is 66dB, after extracting calculated SRI in the area which is 35.45 dB, the transmission loss through the partitions into the passage and outdoor garden area is 30.55dB (66dB â&#x20AC;&#x201C; 35.45dB). From the Appendix, 31dB of sound falls under the category of quiet learning area and soft stereo in residence. In this case, it is will not let much noises penetrate to the center courtyard or neighboring planting area.

Appendix I

Table 1.1: Recommended light level in different workspaces

Component Ceiling

Material Concrete

Colour Grey

Area, m2 60

Reflectance Value 0.3

Wall

Brick wall with plaster finish Glass

Grey

44

0.25

Transparent

84

0.1

Brownish red

60

0.1

Floor

Carpet

Table 1.2: Reflectance value in cafĂŠ and quiet study area

Component Ceiling

Material Concrete

Colour Grey

Area, m2 60

Reflectance Value 0.3

Wall

Brick wall with plaster finish Glass

Grey

44

0.25

Transparent

84

0.1

Timber

Brown

60

0.14

Floor

Table 1.3: Reflectance value in admin/lobby and gallery

Table 1.4: Indicates how much of the luminous flux produced by the lamp in the fixture enters the work plane under a variety of conditions

Table 1.5: Room illumination Level

Appendix II

Table 1.6: Sound absorption coefficient table under 500 Hz and 2000 Hz

Table 1.7: Reverberation Criteria.

(Source: Engineering Acoustics & Noise Control)

Table 1.8: Standard requirement of dB in different rooms and areas

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Table chart sound pressure levels SPL level test normal voice sound levels pressure sound intensity ratio decibel comparison chart conversion of sound pressure to sound intensity noise sound units decibel level comparison of common sounds calculation compression rarefaction loudness decibel dB scale ratio factor unit examples - sengpielaudio Sengpiel Berlin. (n.d.). Retrieved July 04, 2016, from http://www.sengpielaudio.com/TableOfSoundPressureLevels.htm

Chartered Institution of Building Services Engineers (CIBSE). 1993. Lighting Guide.

London: CIBSE. Sound Pressure - recommended Maximum Level in Rooms. (n.d.). Retrieved July 04, 2016, from http://www.engineeringtoolbox.com/sound-pressure-d_66.html

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