Building Science 2 - Project 2

BUILDING SCIENCE 2 BLD61303 PROJECT 2:INTEGRATION

ACOUSTICS AND LIGHTING

NAME: JOLENE HOR STUDENT ID: 0313751 LECTURER: MR EDWIN YEAN

ALLEGR O I SENTUL COMMUNITY LIBRARY

LIGHTING

acoustics

TAYLORS LAKESIDE UNIVERSITY BUILDING SCIENCE 2 I PROJECT2 JOLENE HOR 0313751 MR EDWIN

SPACE 1 -PLAN -DAYLIGHT CONTOUR DIAGRAM -LIGHT SPECIFICATION -ARTIFICIAL LIGHTING DIAGRAM -PSALI SPACE 2 -PLAN -DAYLIGHT CONTOUR DIAGRAM -LIGHT SPECIFICATION -ARTIFICIAL LIGHTING DIAGRAM -PSALI

3 4 5 6 7

9 10 12 14

SPACE1 -PLAN -COMPONENTS OF SPACE -CALCULATION AT 125Hz, 500Hz AND 1000Hz

16 17 18

SPACE2 -PLAN -COMPONENTS OF SPACE -CALCULATION AT 125Hz, 500Hz AND 1000Hz

22 23 23

-PLAN - COMPONENT - INTERNAL PARTITION -EXTERNAL PARITION

30 30 31 32

1.1

DAYLIGHTING

Natural daylighting is a passive method of lighting up a space. It is the controlled admission of natural sunlight and diffuse skylight into a building to reduce electric lighting and saving energy. By providing a direct link to the dynamic and perpetually evolving patterns of outdoor illumination, daylighting helps create a visually stimulating and productive environment for building occupants, while reducing as much as one-third of total building energy costs.

DAYLIGHT FACTOR The daylight factor (DF) is commonly used to determine the ratio of internal light level to external light level and is defined as follows: DF =

Ei x 100% Eo

Where: DF : Daylight factor.

EO : simultaneous outdoor illuminance on a horizontal plane from an unobstructed hemisphere of overcast sky.

EI : illuminance due to daylight at a point on the indoors working plane.

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

Note: Figures are average daylight factors for windows without glazing Table 1.1 Distribution of Daylight Factor Source: MS1525, 2007

1.2

ARTIFICIAL LIGHTING

Artificial lighting by definition is any light that does not come from sunlight. Artificial lighting are technical instruments that produces light through the conversion of electrical energy into radiation and light. Artificial lighting have two types of light source which is the incandescent lamp whereby light is generated when the filament is radiated at high temperature and luminescent lamp when light is produced through excited electrons. We do not receive sunlight 24 hours and therefore it is important to have artificial lighting as a substitute.

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1.3

SPACE 1

E–BRARY The E-brary is a room where there all the computers as well and the printers are located. Controlled lighting in this space is important to ensure that there is not too much of sunlight that will cause glare but sufficient enough for users to do their work.

The artificial lighting needed to light up the library is 300 lux with the maximum lighting power of 18(W/m2) based on the MS 1525, 2007 standards.

PLAN

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The daylight factor that is aimed to be achieved for this space is 4%.

(Refer to appendix at the back)

DF: 4% EO : 14000 lux

EI=

EI x 100

EI=

4 x 14000

EO

100

EI : 560 lux

DAYLIGHTING CONTOUR DIAGRAM

The space does not get a lot of daylighting to reduce the glare that the sunlight causes in the morning. The room gets an average of 450 lux of daylighting. Where there is an opening, the room receives more light. Hence, this room depends more on artificial lighting to light the room up

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ARTIFICIAL LIGHTING The artificial lighting needed to light up the e-library is 300 lux with the maximum lighting power of 18(W/m2) based on the MS 1525, 2007 standards.

LIGHT SPECIFICATIONS

Name of Light Types of Light Type of Fixture Type of Light Bulb Used Light Bulb Brand Lighting Function Length of Tube Type of Luminaries Power, W Light Output, lm Lifetime of Lamp (hrs) Lumen Maintenance Factor

Tubular Florescent Light Tube Artificial Light Florescent Light Fixture High Efficient Florescent Light Philips Task Lighting 85.00 cm Cool Daylight 18 2100 25,000 0.8

Table 1.3.1 Distribution of Daylight Factor Source: MS1525, 2007

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LUMEN METHOD CALCULATIONS Total Floor Area (m2 ) Type of Lighting Fixtures Lumen of Lighting Fixtures/ F (lm)

5.0 m x 12.0 m 60 Florescent Light Fixture 2100

( refer table 1.2)

Work Level (m) Mounting Height/ H (m) Assumption of Reflectance Value Room Index / RI (k)

0.85 2.15 White Plaster Ceiling – 0.75 White Plaster Wall- 0.50 =

(5.0 x 12.0) 2.15 (5.0 + 12.0) = 1.6

Utilization Factor (UF) Standard Illuminance (lux) Number of Lamps required to reach the required illuminance

0.6 300 ExA N= F x UF x MF N=

300 x 60.0 2100 x 0.6 x 0.8 N=

18000 1008

N = 17.8 lamps đ??? = đ?&#x;?đ?&#x;– đ??Ľđ??šđ??Śđ??Šđ??Ź Table 1.3.2 Lumen Method Calculation

PROPOSED LAYOUT OF LIGHTIN G FIXTURES 21 lamps with the amount of 1760 lumens such as Panasonic Compact Fluorescent Spiral Bulb is needed to light the space up with artificial lighting.

Maximum spacing of each bulb = 1.5m x 2.15 m = 3.225 m 18 lamp = 3 rows of 6 lamps 1 fixture = 2 lamps = 3 rows of 3 fixtures Length of room = 5.0 m Width of room = 12.0 m Spacing between each bulb(W) =5.0 / 3 =1.67 m Spacing between each bulb(L) =12.0/3 =4.0 m

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PERMANENT SUPPLEMENTARY ARTIFICIAL LIGHTING INTERIORS (PSALI) Since one side of the room does not have openings where natural daylight can light that part of the room, the lights ( indicated in yellow) there is proposed to be independent from the rest of the switch. This is so that, during the day only those lights would be needed to switch on. According to the daylight contour, the middle area of the room and the corners of the room do not receive as much sunlight. Therefore, when it is less sunny, these lights (indicated in orange) can be switched on.

CEILING PLAN

ARTIFICIAL LIGHTING CONTOUR

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PSAL I SECTION IN THE DAY

PSAL I SECTION AT NIGHT

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1.4

SPACE 2

LIBRARY The daylighting of the library at the ground floor is selected to be studied for daylighting. This room requires ample daylight as it is double volume and a lot of light is needed to light the space up. If the daylight is sufficient to light the place up, less artificial light is needed and therefore less energy is needed.

PLAN

DAYLIGHTING 9|P A G E

The daylight factor that is aimed to be achieved for this space is 4%.

(Refer to appendix at the back)

DF: 5% EO : 14000 lux

EI=

EI x 100

EI=

5 x 14000

EO

100

EI : 700 lux

LIGHT CONTOUR DIAGRAM

R E A D I N G S A T 1500 M

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R E A D I N G S A T 2500

R E A D I N G S A T 3500 M

As there is a platform in this space, the working space height varies. The readings are measured at 1.5m, 2.5m and 3.5m. The result of the analysis shows that at all working plane, even at the darkest area, there is still sufficient natural daylighting.

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LIGHT SPECIFICATIONS

Name of Light Types of Light Type of Fixture Type of Light Bulb Used Light Bulb Brand Lighting Function Type of Luminaries Power, W Light Output, lm Lifetime of Lamp (hrs) Lumen Maintenance Factor

Round Pendant Light Artificial Light Vertical Lone Light Fixture Compact Fluorescent Spiral Bulb Philips Task Lighting Cool Daylight 12 1760 6000 0.8

Table 1.4.1 Light Specification

LUMEN METHOD CALCULATION Total Floor Area (m2 ) Type of Lighting Fixtures Lumen of Lighting Fixtures/ F (lm)

8.0 m x 12.0 m 96.00 Compact Fluorescent Spiral Bulb 1760

( refer table 1.2)

Work Level (m) Mounting Height/ H (m) Assumption of Reflectance Value Room Index / RI (k)

0.85 5.15 White Plaster Ceiling – 0.75 White Plaster Wall- 0.50 =

(8.0 x 12) 5.15 (8.0 + 12) = 0.932

Utilization Factor (UF) Standard Illuminance (lux) Number of Lamps required to reach the required illuminance

0.52 300 ExA N= F x UF x MF

12 | P A G E

N=

300 x 96 1760 x 0.52 x 0.8 N=

28,800 732.16

N = 39.3 lamps đ??? = đ?&#x;’đ?&#x;Ž đ??Ľđ??šđ??Śđ??Šđ??Ź

Table 1.4.2 Lumen Method Calculation

PROPOSED LAYOUT OF LIGHTIN G FIXTURES 40 lamps with the amount of 1760 lumens such as Panasonic Compact Fluorescent Spiral Bulb is needed to light the space up with artificial lighting.

Maximum spacing of each bulb = 1.5m x 5.15 m = 7.725 m 40 lamp = 4 rows of 10 lamps Length of room = 12.0 m Width of room = 8.0 m Spacing between each bulb(W) =12.0 /10 =1.2 m Spacing between each bulb(L) =8.0/4 =2 m

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PERMANENT SUPPLEMENTARY ARTIFICIAL LIGHTING INTERIORS (PSALI) Since the back of the room receives the least daylighting, the switches are independent of the lights on the front row that has higher exposure to natural daylight.

On a less sunny, or gloomy day, the back of the room might not receive as much sunlight therefore, the lights at the back of the room can be turn on without having to switch on the row of lights nearest to the opening which allows natural daylighting in.

The two rows of lights are separated to two switches. This is so that the light can be in better control should someone be only using part of the room at that time.

ARTIFICIAL LIGHTING CONTOUR

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PSAL I SECTION IN THE DAY

PSAL I SECTION AT NIGHT

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2.1

REVERBERATION TIME

The reverberant sound in an auditorium dies away with time as the sound energy is absorbed by multiple interactions with the surfaces of the room. Reverberation time is measured as the amount of time taken for sound to drop by 60 dB. In a more reflective room, it will take longer for the sound to die away and the room is said to be 'live'. In a very absorbent room, the sound will die away quickly and the room will be described as acoustically 'dead'. But the time for reverberation to completely die away will depend upon how loud the sound was to begin with, and will also depend upon the acuity of the hearing of the observer.

2.1.1

SELECTED SPACE 1

M U S I C W O R K S H OP The space selected to be studied for reverberation time is the Music Workshop, which is a space which can also act as an amphitheatre where musical performance or other performance would most probably be held. The ideal reverberation time for the music workshop is 1.9 seconds.

PLAN

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COMPONENTS OF ROOM

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A R E A (M2)

SURFACE

CEILING

6.2 x 11.775 =73.005

FLOOR

6.2 x 11.775 =73.005 Wall 1 (4.0 x 12.075 ) – 3(2.7 x 0.8) =48.3 – 6.48 = 41.82

ABSORPTION

125 (Hz) 0.03 Plaster, lime or gypsum on solid backing 0.15 Wood 0.01 Smooth unpainted concrete

TOTAL ABSOPTION 73.005 x 0.03 = 2.190

73.005 x 0.15 = 10.95 103.57 x 0.01 = 1.0357

Wall 2 (4.0 x 6.5)- (2.85 x 3.0) = 26 -8.55 = 17.45 WALL

Wall 3 (4.0 x 12.075 ) – (3.0 x 9.8) =48.3 – 29.4 = 18.9 Wall 4 4.0 x 6.35 = 25.4 Total wall =41.82 + 17.45 + 18.9 + 25.4 = 103.57

GLASS DOOR & WINDOW

0.3

37.95 x 0.3 = 11.385

0.35

6.48 x 0.35 =2.268

Wall 2 2.85 x 3.0 = 8.55 Wall 3 3.0 x 9.8 = 29.4

SOLID DOOR

Total 8.55 + 29.4 =37.95 Wall 1 3(2.7 x 0.8) 6.48

Acoustic door, steel frame, double seals, absorbent in airspace. Double sheet steel skin.

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HUMAN

25

0.21

25 x 0.21 = 5.25

Table 2.1.1 Reverberation Time for Music Workshop at 125 Hz

RT = 0.16 x Volume of room /Total absorption = 0.16 x (73.005 x 4) / 2.190 + 10.95 + 1.0357 + 11.385 + 2.268 + 5.25 = 0.16 x 292.02 / 33.0796 =46.7232 / 33.0796 = 1.4 seconds

A R E A (M2)

SURFACE

CEILING

6.2 x 11.775 =73.005

FLOOR

6.2 x 11.775 =73.005 Wall 1 (4.0 x 12.075 ) – 3(2.7 x 0.8) =48.3 – 6.48 = 41.82

ABSORPTION

500 (Hz) 0.02 Plaster, lime or gypsum on solid backing 0.1 Wood 0.02 Smooth unpainted concrete

TOTAL ABSOPTION 73.005 x 0.02 = 1.4601

73.005 x 0.1 = 7.3005 103.57 x 0.03 = 2.0714

Wall 2 (4.0 x 6.5)- (2.85 x 3.0) = 26 -8.55 = 17.45 WALL

Wall 3 (4.0 x 12.075 ) – (3.0 x 9.8) =48.3 – 29.4 = 18.9 Wall 4 4.0 x 6.35 = 25.4 Total wall =41.82 + 17.45 + 18.9 + 25.4 = 103.57

GLASS DOOR & WINDOW

0.1 Wall 2 2.85 x 3.0 = 8.55

37.95 x 0.1 = 3.795

Wall 3 3.0 x 9.8 19 | P A G E

= 29.4

SOLID DOOR HUMAN

Total 8.55 + 29.4 =37.95 Wall 1 3(2.7 x 0.8) 6.48 25

0.44

6.48 x 0.44 =2.8512

0.2

5 x 0.2 =5

Table 2.1.2 Reverberation Time for Music Workshop at 500 Hz

RT = 0.16 x Volume of room /Total absorption = 0.16 x (73.005 x 4) / 1.4601 + 7.3005 + 2.0714 + 3.795 + 2.8512 + 5 = 0.16 x 292.02 / 22.4782 =46.7232 / 22.4782 = 2.07 seconds

SURFACE

A R E A (M2)

CEILING

6.2 x 11.775 =73.005

FLOOR

6.2 x 11.775 =73.005 Wall 1 (4.0 x 12.075 ) – 3(2.7 x 0.8) =48.3 – 6.48 = 41.82

ABSORPTION

2000 (Hz) 0.04 Plaster, lime or gypsum on solid backing 0.1 Wood 0.02 Smooth unpainted concrete

TOTAL ABSOPTION 73.005 x 0.04 = 2.92 73.005 x 0.1 = 7.3005 103.57 x 0.0 =2.0714

Wall 2 (4.0 x 6.5)- (2.85 x 3.0) = 26 -8.55 = 17.45 WALL

Wall 3 (4.0 x 12.075 ) – (3.0 x 9.8) =48.3 – 29.4 = 18.9 Wall 4 4.0 x 6.35 = 25.4 Total wall =41.82 + 17.45 + 18.9 + 25.4 = 103.57

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GLASS DOOR & WINDOW

Wall 2 2.85 x 3.0 = 8.55

0.07

37.95 x 0.07 = 2.6565

0.54

6.48 x 0.54 =3.4992

0.21

5 x 0.21 = 5.25

Wall 3 3.0 x 9.8 = 29.4

SOLID DOOR HUMAN

Total 8.55 + 29.4 =37.95 Wall 1 3(2.7 x 0.8) 6.48 25

Table 2.1.3 Reverberation Time for Music Workshop at 2000 Hz

RT = 0.16 x Volume of room /Total absorption = 0.16 x (73.005 x 4) / 2.92 + 7.3005 + 2.0714 + 2.6565 + 3.4992 + 5.25 = 0.16 x 292.02 / 23.6978 =46.7232 / 23.6978 = 1.9716 seconds SUMMARY SURFACE AREA CEILING 73.005 FLOOR 73.005 WALL 103.57 GLASS WINDOW DOOR 37.95 SOLID DOOR 6.48 HUMAN 25 REVERBERATION TIME (S)

COEFFICIENT 2000Hz 125Hz 2000Hz 0.03 0.02 0.3 0.15 0.1 0.1 0.03 0.02 0.2 0.3 0.1 0.21

0.1 0.1 0.2

0.07 0.04 0.21

ABSORBTION 125Hz 2000Hz 125Hz 2.19015 1.4601 21.9015 10.95075 7.3005 7.3005 3.1071 2.0714 20.714 11.385 0.648 5.25

3.795 0.648 5

2.6565 0.2592 5.25

1.40

2.07

1.97

Table 2.1.4 Summary of Reverberation Time for Music Workshop

At different frequencies,the reverberation time differs. Low pitch sounds sch as the bass is most probably fade off first as the sound gets absorbed more. However, this would mean that vehicle sound from the street would be absorbed and therefore would not cause any unwanted noise in the music workshop. The reverberation time at 500 Hz and 2000Hz meets the requirement and therefore the space is suitable to be a music workshop that can be converted to an amphitheatrewhere music is played.

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2.2.2

SPACE 2

LIBRARY SHELFING The second space selected to be studied for reverberation time is the Library area, which is a where books are shelfed. This place is meant to be quiet so that it does not disrupt others who are concentrating. The ideal reverberation time is 0.8 seconds.

PLAN

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COMPONENTS OF ROOM

REVERBERATION

A T 125 H Z

A R E A (M2)

SURFACE

CEILING

TIME

142.57

ABSORPTIO N (K) at 125hz

0.3 Fibre Board 13mm, 25mm airspace

FLOOR

142.57 Wall 1 =3.0 x 17 = 51.00

WALL

Wall 2 (3.0 x 19.075) – (2.7 x 1.2) – (2.7 x 2.4) – (2.5 x 3.0) – (2.7 x 2.4) =57.225 – 3.24 – 6.48 – 7.5-6.48 =33.525

0.1 Carpet 0.01 Concrete Wall Unglazed, Painted

TOTAL ABSOPTION 142.57 x 0.3 =42.771 142.57 x 0.1 = 14.257 146.274x 0.01 = 1.4627

Wall 3 (3.0 x 8.225) – (2.1 x 1.7 ) =24.675 – 3.57 23 | P A G E

21.105 Wall 4 + 5 2 (3 x 0.835) =2 x 2.05 = 5.01 Wall 6 (3.0 x 8.075 ) – ( 2.7 x 0.9) – ( 2.7 x 2.9) =24.225 – 2.43 – 7.83 =13.965 Wall 7 (3 x 5.225) – (1.8 x 3.87) = 15.675 – 6.966 =8.709 Wall 8 (3 x 5.85) – (2.7 x 1.7) =17.55 x 4.59 =12.96 Total wall = 51 + 33.525 + 21.105 + 5.01 + 13.965 + 8.709 + 12.96 = 146.274 Wall 2 (2.7 x 1.2) + (2.7 x 2.4) + (2.5 x 3.0) + (2.7 x 2.4) =3.24 + 6.48 + 7.5 + 6.48 =23.7

0.3 Glass

64.191 x 0.3 = 19.257

Wall 3 2.1 x 1.7 =21.105 GLASS WINDOWS & DOORS

Wall 6 2.7 x 2.9 =7.83 Wall 7 1.8 x 3.87 = 6.966 Wall 8 2.7 x 1.7 = 4.59 Total glass windows and doors = 23.7 + 21.105 + 7.83 24 | P A G E

+ 6.966 + 4.59 = 64.191 WOODEN DOOR Human

Wall 6 2.7 x 0.9 =2.43 Number of Human : 50 People

0.1 Wooden

2.43 x 0.1 =0.243

0.21

50 x 0.21 = 10.5

Table 2.2.1 Reverberation Time for Library at 125Hz

RT = 0.16 x Volume of room /Total absorption = 0.16 x (142.57 x 3) / 42.771 + 14.257+ 1.4627+ 19.257+0.243+ 10.5 = 0.16 x 427.71 / 88.49104 =68.434 / 88.49104 = 0.77 seconds REVERBERATION

A T 500 H Z

A R E A (M2)

SURFACE

CEILING

TIME

142.57

ABSORPTION

(K) at 500hz 0.35 Fibre Board 13mm, 25mm airspace

FLOOR

142.57

Wall 1 =3.0 x 17 = 51.00

0.3 Carpet 0.02 Concrete Wall Unglazed, Painted

TOTAL ABSOPTION 142.57 x 0.35 =49.7 142.57 x 0.3 = 42.771 146.274x 0.02 = 2.925

Wall 2 (3.0 x 19.075) – (2.7 x 1.2) – (2.7 x 2.4) – (2.5 x 3.0) – (2.7 x 2.4) =57.225 – 3.24 – 6.48 – 7.5-6.48 =33.525 WALL Wall 3 (3.0 x 8.225) – (2.1 x 1.7 ) =24.675 – 3.57 21.105 Wall 4 + 5 2 (3 x 0.835) =2 x 2.05 = 5.01 Wall 6 25 | P A G E

(3.0 x 8.075 ) – ( 2.7 x 0.9) – ( 2.7 x 2.9) =24.225 – 2.43 – 7.83 =13.965 Wall 7 (3 x 5.225) – (1.8 x 3.87) = 15.675 – 6.966 =8.709 Wall 8 (3 x 5.85) – (2.7 x 1.7) =17.55 x 4.59 =12.96 Total wall = 51 + 33.525 + 21.105 + 5.01 + 13.965 + 8.709 + 12.96 = 146.274 Wall 2 (2.7 x 1.2) + (2.7 x 2.4) + (2.5 x 3.0) + (2.7 x 2.4) =3.24 + 6.48 + 7.5 + 6.48 =23.7

0.1 Glass

64.191 x 0.1 = 6.419

Wall 3 2.1 x 1.7 =21.105

GLASS WINDOWS & DOORS

Wall 6 2.7 x 2.9 =7.83 Wall 7 1.8 x 3.87 = 6.966 Wall 8 2.7 x 1.7 = 4.59 Total glass windows and doors = 23.7 + 21.105 + 7.83 + 6.966 + 4.59 = 64.191

WOODEN DOOR

Wall 6 2.7 x 0.9 =2.43

0.1 Wooden

2.43 x 0.1 =0.243

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Human

Number of Human : 50 People

0.2

50 x 0.46 = 23

Table 2.2.2 Reverberation Time for Library at 500Hz

RT = 0.16 x Volume of room /Total absorption = 0.16 x (142.57 x 3) / 49.7 + 35.64+ 2.925+ 0.6419+0.243+ 23 = 0.16 x 427.71 / 112.0285 =68.434 / 112.0285 = 0.61 seconds REVERBERATION

A T 2000 H Z

A R E A (M2)

SURFACE

CEILING

TIME

142.57

ABSORPTION

(K) at 2000hz 0.3 Fibre Board 13mm, 25mm airspace

FLOOR

142.57 Wall 1 =3.0 x 17 = 51.00

0.1 Carpet 0.2 Concrete Wall Unglazed, Painted

TOTAL ABSOPTION 142.57 x 0.3 =42.771 142.57 x 0.1 = 14.257 146.274x 0.02 = 2.925

Wall 2 (3.0 x 19.075) – (2.7 x 1.2) – (2.7 x 2.4) – (2.5 x 3.0) – (2.7 x 2.4) =57.225 – 3.24 – 6.48 – 7.5-6.48 =33.525 WALL

Wall 3 (3.0 x 8.225) – (2.1 x 1.7 ) =24.675 – 3.57 21.105 Wall 4 + 5 2 (3 x 0.835) =2 x 2.05 = 5.01 Wall 6 (3.0 x 8.075 ) – ( 2.7 x 0.9) – ( 2.7 x 2.9) =24.225 – 2.43 – 7.83 =13.965 27 | P A G E

Wall 7 (3 x 5.225) – (1.8 x 3.87) = 15.675 – 6.966 =8.709 Wall 8 (3 x 5.85) – (2.7 x 1.7) =17.55 x 4.59 =12.96 Total wall = 51 + 33.525 + 21.105 + 5.01 + 13.965 + 8.709 + 12.96 = 146.274 Wall 2 (2.7 x 1.2) + (2.7 x 2.4) + (2.5 x 3.0) + (2.7 x 2.4) =3.24 + 6.48 + 7.5 + 6.48 =23.7

0.07 Glass

64.191 x 0.07 = 4.4934

Wall 3 2.1 x 1.7 =21.105

GLASS WINDOWS & DOORS

Wall 6 2.7 x 2.9 =7.83 Wall 7 1.8 x 3.87 = 6.966 Wall 8 2.7 x 1.7 = 4.59 Total glass windows and doors = 23.7 + 21.105 + 7.83 + 6.966 + 4.59 = 64.191

WOODEN DOOR Human

Wall 6 2.7 x 0.9 =2.43 Number of Human : 50 People

0.04 Wooden 0.21

2.43 x 0.04 =0.0972 50 x 0.21 = 10.5

Table 2.2.3 Reverberation Time for Library at 2000Hz

28 | P A G E

RT = 0.16 x Volume of room /Total absorption = 0.16 x (142.57 x 3) / 42.771 + 14.257+ 2.925+ 4.4934 + 0.0972 + 25.5 = 0.16 x 427.71 / 75.044 =68.434 / 75.044 = 0.91 seconds

SUMMARY

SURFACE AREA CEILING 142.57 FLOOR 142.57 WALL 146.274 GLASS WINDOW DOOR 64.191 SOLID DOOR 2.43 HUMAN 50 REVERBERATION TIME (S)

COEFFICIENT 125Hz 500Hz 2000Hz 0.3 0.35 0.3 0.1 0.3 0.1 0.01 0.02 0.02 0.3 0.1 0.21

0.1 0.1 0.2

0.07 0.04 0.21

ABSORBTION 125Hz 500Hz 2000Hz 42.771 49.8995 42.771 14.257 42.771 14.257 1.46274 2.92548 2.92548 19.2573 0.243 10.5

6.4191 0.243 10

4.49337 0.0972 10.5

0.77

0.61

0.91

Table 2.2.4 Summary of Reverberation Time for Library

At all frequencies, the reverberation time meets the requirement of 0.8 seconds.

This means that the sound waves does not bounce on the surface too much and the space is maintained quiet. Should there be any noise from the external side of the library that enters inside, the sound can be easily absorbed. Besides that, noise from the music side of the library which have high frequencies such as violin sound would also be absorbed easily. Even though the reverberation time is low, itis not a dead space as sound does not get instantly absorbed.

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2.2

SOUND TRANSMISSION CLASS (STC)

The sound reduction index (SRI) or Transmission Loss (TL) of a partition measures the number of decibels lost when a sound of a given frequency is transmitted through the partition.

2.2.1

MUSIC PODS

The music pod is a room where music class are being carried. As it is located in a library, the partition that separates the music pod and the library needs to reduce as much noise as possible so that music made in the pod would not disturb other users in the library.

PLAN

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The sound made in the music room is about 90dB. The interior partition has to reduce the sound to 45dB which is the sound acceptable outside the music pods whereas the sound reduced into the plaza side does not really matter since there would be a lot of noise produced in the plaza that will not be an issue in the library and therefore causing disturbance.

Therefore internally the sound needs to be reduced by 90dB – 45dB = 45 dB.

COMPONENTS OF ROOM

INTERIOR PARTITION SRI STC of Wall- 8 inch (200mm) concrete block wall, painted with ½’’ drywall on independent still stud walls, each side, insulation in cavities = 72 STC of Door - Low Rage Acoustic Doors, with seals = 36 Area of Walls Wall 1 = 3.0 x 3.075 – (2.7 x 0.95) =6.66 m2 Wall 2 & 3 = 2( 3.0 x 2.075) =12.45 31 | P A G E

Total = 6.66 + 12.45 = 19.11 Area of door : 2.7 x 0.95 = 2.565 SRI For wall : 72 = 10 log ( 1/T ) anti log 7.2 = 1/T T = 6.3095 x 10−8 SRI For door : 36 = 10 log ( 1/T ) anti log 3.6 = 1/T T = 2.5118 x 10−4

TOTAL SRI = (19.11 x 6.3095 x 10−8 ) + (2.565 x 2.5118 x 10−4 )/ 19.11 + 2.565 =1.205 x 10−6 + 6.443 x 10−4 / 21.675 = 6.455 x 10−4 / 21.675 =2.978 x 10−5

SRI OVERALL = 10 log (1 / 2.978 x 10−5) =10 log 33579.58 = 45.26 dB

EXTERIOR PARTITION SRI

SRI of Window - Double Window, openable but weather-stripped. 150 mm airspace, = 35 SRI of Wall- 8 inch (203mm) Solid Concrete Wall = 58

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Area of Walls Walls 4 3.0 x 3.075 – (2.7 x 2.4 ) = 2.745 Wall 5 & 6 = 2( 3.0 x 2.075) =12.45 Total Wall area = 2.745 + 12.45 =15.195 Area of Window 2.7 x 2.4 =6.48 SRI For Walls: 58= 10 log ( 1/T ) anti log 5.8= 1/T 1/T = 630,957.00 T = 1.5849 x 10−6 For window: 35= 10 log ( 1/T ) anti log 3.5 = 1/T 1/T = 3162.27 T = 3.1623 x 10−4

TOTAL SRI = (15.195 x 1.5849 x 10−6 ) + (6.48 x 3.1623 x 10−4 )/ 15.195 + 6.48 =2.4083 x 10−5 + 2.0492 x 10−3 / 21.675 = 0.20733 x 10−3 / 21.675 =9.5653 x 10−5

SRI OVERALL = 10 log (1 / 9.5653 x 10−5) =10 log 10454. 434 = 40.19 dB

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After calculation, the internal partition is able to reduce the noise by 45dB, from 90dB to 45 dB. Therefore, the music pods do not cause disturbance to the library since the partitions reduces the noise when the doors are shut. The external partition reduces less noise from 90dB to 50dB as the external area which is the plaza has a higher NPC as there is noise from traffic at the street .

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Ander, G. (2003). Daylighting performance and design. Hoboken, N.J.: John Wiley & Sons.

Donkin, W. & Donkin, A. (1870). Acoustics. Oxford: Clarendon Press.

Energy Research Group, School of Architecture, University College Dublin, Richview, Clonskeagh,,. (2016). Daylighting in Buildings. Ireland: UCDOPET. Retrieved from http://cordis.europa.eu/pub/opet/docs/ucd_1.pdf Reverberation Time. (2016). Hyperphysics.phy-astr.gsu.edu. Retrieved 28 June 2016, from http://hyperphysics.phyastr.gsu.edu/hbase/acoustic/revtim.html Reinhart, C. & Stein, R. Daylighting handbook.

Templeton, C , (a995) Acoustic Design, Butterworth, London

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