Building Science Project 2

BUILDING SCIENCE 2 ARC 3413

PROJECT 2 INTEGRATION WITH DESIGN STUDIO 5

CYNTHIA LEE SIAW WUN 0306112 MR RIZAL

TABLE OF CONTENT 1.0 LIGHTING 1.1 NATURAL LIGHTING CLASSROOM THEATRE 1.2 ARTIFICIAL LIGHTING CLASSROOM THEATRE 2.0 ACOUSTIC 2.1 EXTERNAL NOISE 2.2 REVERBERATION TIME CLASSROOM THEATRE 3.0 APPENDIX AND REFERENCE

1.0 LIGHTING ANALYSIS 1.1 NATURAL DAYLIGHTING CLASSROOM

According to MS1525, daylight factor distribution as follow: Zone

DF (%)

Distribution

Very Bright

>6

Very large with thermal and glare problems.

Bright

3-6

Good

Average

1-3

Fair

dark

0-1

Poor

The classroom is located at the first floor of the building, facing the main street. One side of the classroom is facing natural daylighting so that the cost of artificial lighting is reduced.

Daylight factor, DF Floor area, m2

33.3 m2

Open faรงade area, m2

5.3m x 3.5m = 18.55m2

Open faรงade area to floor area ratio,

18.55 33.3

x 100%

Daylight factor (DF)

Natural illuminance calculation Formula Given Eo, unobstructed sky in Chow Kit

= 55.7% x 0.1 = 5.57%

DF = (Ei x Eo) x 100% 2720 lux DF = (Ei x Eo) x 100% E i = (DF x Eo) รท 100% = (5.57% x 2720) รท 100% = 151.50 Lux

CONCLUSION The daylight factor of this area is 5.57% which is consider being a good natural daylighting. The standard illuminance level for a classroom is 300-500lux. The final calculation for the classroom is 151.50 lux.

THEATRE

According to MS1525, daylight factor distribution as follow: Zone

DF (%)

Distribution

Very Bright

>6

Very large with thermal and glare problems.

Bright

3-6

Good

Average

1-3

Fair

dark

0-1

Poor

The theatre is located on the second floor where it is not exposed to any natural daylighting at all. Thus there is a lot of artificial lighting used in the space.

Daylight factor, DF Floor area, m2

44.6 m2

Open faรงade area, m2

0m x 5m = 0 m2

Open faรงade area to floor area ratio, Daylight factor (DF)

44.6

0

x 100%

=0 % x 0.1 = 0%

Natural illuminance calculation Formula

DF = (Ei x Eo) x 100%

Given Eo, unobstructed sky in Chow Kit

2720 DF = (Ei x Eo) x 100% E i = (DF x Eo) รท 100% = (0% x 2720) รท 100% = 0 Lux

CONCLUSION There is no daylight factor in this space as there are no open faรงade that is exposed to natural daylighting.

1.2 ARTIFICIAL LIGHTING CLASSROOM According to MS1525, the lighting level required for a classroom is 300-500 lux. Therefore, the minimum luminance for calculation below has to be 300lux. Types of luminaire are as shown below:

Fluorescent tube light bulb

Lamp type

Fluorescent tube light bulb

Light colour

Cool white

Wattage

24W

Initial Lumen

2000

CRI

85

Lumen calculation Location

Classroom

Activity

Educational activities

Dimension of room

L= 5.3m, W= 5.3m (approximate)

Total floor area

33.3 m2

Standard illuminance required according to MS1525 Assumption of reflectance value

300-500 lux

Mounting height (Hm)

3.5m – 0.6m = 2.9m

70, 50, 20

5.3 đ?‘Ľ 5.3 2.9 (5.3 + 5.3)

Room index, K ��� � (� + �)

= 0.91

Utilization factor, UF

0.51

Maintenance factor, MF

N= LLMF X LSF X LMF X RSMF = 0.94 X 1 X 0.61 X 0.79 = 0.45

Lumen Calculation

N = đ??š đ?‘Ľ đ?‘ˆđ??š đ?‘Ľ đ?‘€đ??š

đ??¸đ?‘Ľđ??´

300 đ?‘Ľ 33.3

= 2000 đ?‘Ľ 0.51 đ?‘Ľ 0.45 =21.76 ≈ 22 đ?‘łđ?’™đ?‘ľ

Number of luminaires across√

đ?‘ž

���

Number of luminaires along√

đ?‘ł

5.3 đ?‘Ľ 22

√

5.3 5.3 đ?‘Ľ 22

√

5.3

= 4.7 (take 5) = 4.7 (take 5)

CONCLUSION The fluorescent lights are used to achieve the required lux level in the classroom. Total of 22 fluorescent tube lights is needed to achieve the sufficient luminance for the comfortability of the users in the cooking classroom, that is 5 lights both sides of the space.

THEATRE According to MS1525, the lighting level required for a theatre is 100 lux. Therefore, the minimum luminance for calculation below has to be 100 lux. Types of luminaire are as shown below:

LED RECESSED LIGHTING

Lamp type

LED Recessed Lighting

Light colour

Warm Yellow

Wattage

10W

Initial Lumen

700

Lumen calculation Location

Theatre

Activity

Movie showcase

Dimension of room

L = 9.6m, W= 5.7m (approximate)

Total floor area

44.6 m2

Standard illuminance rquired according to MS1525 Assumption of reflectance value

100 lux

Mounting height (Hm)

5m – 0.6m = 4.4m

70, 50, 20

9.6 đ?‘Ľ 5.7 4.4 (9.6 + 5.7)

Room index, K ��� � (� + �)

= 0.81

Utilization factor, UF

0.33

Maintenance factor, MF

N= LLMF X LSF X LMF X RSMF = 0.94 X 1 X 0.74 X 0.92 = 0.64

Lumen Calculation

N = đ??š đ?‘Ľ đ?‘ˆđ??š đ?‘Ľ đ?‘€đ??š

đ??¸đ?‘Ľđ??´

100 đ?‘Ľ 44.6

= 700 đ?‘Ľ 0.33 đ?‘Ľ 0.64 =30.16 ≈ 30 đ?‘łđ?’™đ?‘ľ

Number of luminaires across√

đ?‘ž

���

Number of luminaires along√

đ?‘ł

9.6 đ?‘Ľ 30

√

5.7 5.7 đ?‘Ľ30

√

9.6

= 7.1 ( take 7)

= 4.2 ( take 4)

CONCLUSION The LED recessed lights are used to achieve the required lux level in the classroom. Total of 30 halogen spot lights are needed to achieve the sufficient luminance for the comfortability of the customers in the cafĂŠ that is 7 lights on the long side and 4 sides on the shorter side of the space.

2.0 ACOUSTIC ANALYSIS 2.1 EXTERNAL NOISE CALCULATION External Noise sources : Traffic Noise : 80 dB Intensity: 80 = 10 log (Itraffic/Io) Log (Itraffic/1x10-12) = 8.0 Log-1Log (Itraffic/1x10-12) = Log-18 (Itraffic/1x10-12)=108 Itraffic = 1 x 10-4

Conversation Noise : 40 dB Intensity: 40 = 10 log (Iconv/Io) Log (Iconv/1x10-12) = 4.0 Log-1Log (Iconv/1x10-12) = Log-14 (Iconv/1x10-12)=104 Iconv = 1 x 10-8

Total Intensities = (1 x 10-4) + (1 x 10-8) = 1 x 10-4 Combined SPL = 10og (Itotal/Io) = 10 log (1 x 10-4/1x10-12) = 80 dB The external noise with a combined SPL of 80dB is generated from the vehicles and conversation. The SPL has exceeded the requirement for ground floor in the market. To reduce the noise produced by the street towards the building, a concrete wall with insulation is placed on the faรงade of the building to act as a buffer zone.

2.2 REVERBERATION TIME CALCULATION CLASSROOM Room height = 3.5m Standard Reverberation Time for Classroom: 1s Peak Hour Capacity : 25 people Volume of Classroom: 33.3 m2 x 3.5m = 116.55 m3 Material Plaster Concrete Aluminium Glass Timber People

Function Ceiling Wall Floor Glass Frame Wall Furniture

Area [A] (m2) 33.30 48.65 33.30 2.10 29.75 50.0 25

Absorption Coefficient in 2000Hz [S] 0.04 0.1 0.1 0.01 0.07 0.1 0.46

Sound Absorption [SA] 1.332 4.865 3.3 0.021 1.873 5.0 11.5

Total Sound Absorption by materials 27.90 Table x.x.x : Material Absorption Coefficient at 2000 Hz.

RT

= (0.16 x V) / A

Where V = Volume of Space A = Total Absorption = S1A1 + S2A2 + S3A3 + ‌ + SnAn Reverberation Time [Peak Hour] RT = (0.16 x V) / A = 0.16 x 116.55 / 27.90 = 0.67 s The reverberation time for the classroom during peak hour is 0.67s. According to Acoustic Standard ANSI (2008), the reverberation time of the classroom during non-peak hour exceeds the maximum reverberation time for classroom that is 0.6s. To reduce the reverberation time, drapery curtains are added on the glass walls.

Material

Function

Area [A] m2

Plaster Concrete

Ceiling Wall Floor Glass Frame Wall Furniture Curtain

33.30 48.65 33.30 2.10 29.75 50.0 29.75 25

Aluminium Glass Timber Fabric People

Absorption Coefficient in 2000Hz [S] 0.04 0.1 0.1 0.01 0.07 0.1 0.7 0.46

Sound Absorption [SA] 1.332 4.865 3.3 0.021 1.873 5.0 20.825 11.5

Total Sound Absorption by materials 48.716 Table 3.3.1.2: Material Absorption Coefficient at 2000 Hz [updated]

Reverberation Time [Peak Hour] RT = (0.16 x V) / A = 0.16 x 116.55 / 48.716 = 0.38 s

CONCLUSION After adding in the drapery curtains, the reverberation time for the peak hour is successfully reduced, and it has achieved the standard required reverberation time.

THEATRE Room height = 5m Standard Reverberation Time for Theatre: 1.0 – 1.5s Peak Hour Capacity : 30 people Volume of Theatre: 44.6 m2 x 5m = 223 m3 Material Plaster Concrete Fabric

Function Ceiling Wall Floor Furniture

People

Area [A] (m2) 44.60 100.45 44.60 10.8 30

Absorption Coefficient in 2000Hz [S] 0.04 0.1 0.7 0.7 0.46

Sound Absorption [SA] 1.784 10.045 31.22 7.56 13.8

Total Sound Absorption by materials 64.409 Table x.x.x : Material Absorption Coefficient at 2000 Hz.

RT

= (0.16 x V) / A

Where V = Volume of Space A = Total Absorption = S1A1 + S2A2 + S3A3 + ‌ + SnAn Reverberation Time [Peak Hour] RT = (0.16 x V) / A = 0.16 x 223 / 64.409 = 0.55 s The reverberation time for the theatre during peak hour is 0.4s. According to Acoustic Standard ANSI (2008), the reverberation time of the theatre during non-peak hour exceeds the maximum reverberation time for classroom that is 1.0-1.5s. To increase the reverberation time, the materials of the floor and furniture must be changed to concrete and timber respectively.

Material

Function

Area [A] m2

Plaster Concrete

Ceiling Wall Floor Furniture

44.60 100.45 44.60 10.8 30

Timber People

Absorption Coefficient in 2000Hz [S] 0.04 0.1 0.1 0.1 0.46

Sound Absorption [SA] 1.784 10.045 4.46 1.08 13.8

Total Sound Absorption by materials 31.169 Table 3.3.1.2: Material Absorption Coefficient at 2000 Hz [updated]

Reverberation Time [Peak Hour] RT = (0.16 x V) / A = 0.16 x 223 / 31.169 = 1.1 s

CONCLUSION After changing the materials of the floor and furniture, the reverberation time for the peak hour is successfully increased, and it has achieved the standard required reverberation time.

SOUND TRANSMISSION LOSS CALCULATION CLASSROOM

Wall type 1 : Concrete Wall with insulation

Wall type 2 : Glass Panel

SRI = 10 log (1/T)

SRI=10 log (1/T)

SRIconcrete= 46.5

SRIglass=26

46.5 = 10 log (1/Tconc)

26 = 10 log (1/Tglass)

Log-14.65=(1/Tconc)

Log-12.6 =(1/Tglass)

Tconc = 2.24 x 10-5

Tglass =2.51x10-3

Surface Material Surface Area, m2 [S] Concrete Wall 48.65 Glass Panel 29.75 Total 78.4 Table 3.4.2.1: STC calculation table

Transmission Coefficient [Tcn] 2.24x10-5 2.51x10-3

Surface Area x Transmission Coefficient [ST] 1.09 x 10-3 7.47 x 10-2 7.58 x 10-2

Tav = (0.00758/78.4) = 9.67x10-4 SRIoverall = 10 log (1/Tav) = 10 log(1/9.67x10-4) =30.15 dB

CONCLUSION As shown in calculations above, 30.15dB of noise level can be reduce during transmission from the street to the interior of the classroom. Therefore, the external noise (80dB) during peak hour, is reduced by 30.15dB during transmission, resulting in a sound level of 49.85dB when it reaches the interior of the classroom. 49.85dB is within the range of recommend level for classroom. Hence, acoustical comfort can be achieved.

THEATRE

Wall type 1 : Concrete Wall with insulation SRI = 10 log (1/T) SRIconcrete= 46.5 46.5 = 10 log (1/Tconc) Log-14.65=(1/Tconc) Tconc = 2.24 x 10-5

Surface Material Surface Area, m2 [S] Concrete Wall 100.45 Table 3.4.2.1: STC calculation table

Transmission Coefficient [Tcn] 2.24x10-5

Surface Area x Transmission Coefficient [ST] 2.26 x 10-3

Tav = (0.00226/100.45) = 2.24x10-5 SRIoverall = 10 log (1/Tav) = 10 log(1/2.24x10-5) = 46.49 dB

CONCLUSION As shown in calculations above, 46.49 dB of noise level can be reduce during transmission from the street to the interior of the classroom. Therefore, the external noise (80dB) during peak hour, is reduced by 46.49 dB during transmission, resulting in a sound level of 33.5 dB when it reaches the interior of the classroom. 33.5 dB is within the range of recommend level for classroom. Hence, acoustical comfort can be achieved.

3.0

APPENDIX AND REFERENCE

ARCHITECTS' DATA. (2012). CHICESTER: JOHN WILEY AND SONS. ASHRAE. (1995). ASHRAE HANDBOOK 1984 SYSTEMS. ATLANTA, GA: AMERICAN SOCIETY HEATING, REFRIGERATING &.

MALAYSIA. (2007). CODE OF PRACTICE ON ENERGY EFFICIENCY AND USE OF RENEWABLE ENERGY FOR NONRESIDENTIAL BUILDINGS (FIRST REVISION). PUTRAJAYA: DEPARTMENT OF STANDARD MALAYSIA. SOUND ABSORPTION COEFFICIENTS OF ARCHITECTURAL ACOUSTICAL MATERIALS. (1957). NEW YORK.