Building Science 2 (ARC3413) Project 2: Integration with Design Studio 4 Nature Appreciation Centre (NAC_BELUM)
Final Report & Calculation
Tutor: Mr. Edwin Yang Ge Shen 0315960
Table of Content
Content
Page
1.0 Lighting 1.1 Daylight 1.2 Artificial Lighting 1.3 PSALI
1-2 3-6 7-9
2.0 Acoustic 2.1 Sound Pressure Level (Exterior Noise)
10 - 11
2.2 Reverberation Time. RT
12 - 13
2.3 Sound Reduction Index, SRI
14 - 15
References
16
1.1 Daylighting (Administration Room) 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 area (Indoor Reading Space) is located at Lower Ground floor and different height. The indoor reading space is the square area shown in floor plan. Faรงade designed for this space is totally completely exposed to sunlight with several shading device located on the facade. Therefore, there is minimal artificial lighting will be using in this area.
Daylighting
Daylighting
Daylight Factor Calculation Floor Area (m2) Area of faรงade that exposed to sunlight Area of skylight (m2) Exposed Faรงade & Skylight Area to Floor Area Ratio/ Daylight Factor, DF
100 100 0 (100+0)/100 =1 = 100% x 0.1 = 10%
Natural Illumination Calculation Illuminance
Example
120,000 lux
Brightest sunlight
110,000 lux
Bright sunlight
20,000 lux
Shade illuminated by entire clear blue sky
1000-2000 lux
Typical overcast day, midday
<200 lux
Extreme of darkest storm clouds, midday
400 lux
Sunrise or sunset on clear day (ambient illumination)
40 lux
Fully overcast, sunset/sunrise
<1 lux
Extreme of darkest storm clouds, sunset/ rise
E external = 20000 lux DF= E internal/E external x 100 10 = E external /20000 x 100
E external = 10 x 20000 / 100 = 2000 lux
Conclusion The Indoor Reading Space has a daylight factor of 10% and natural illumination of 2000 lux. This will result in thermal and glare problem. Hence, a glazing roof or wall of double-glazed low e-value glass will be proposed to solve the glare problem and to reduce the heat gain in this area.
1.2 Artificial Lighting (Audio-visual Room)
According to MS1525, the minimum lighting level required for a Public Living Area is 200 lux. 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 Colour Temperature, K Colour Designation Lumens Type of fixture
Aluminium DN571B
25,000 36 80 3000 Warm White 4000 LED Spotlight
Type of light bulb
Material of fixture Product Brand & Code Nominal life (hours) Wattage Range (W) CRI Colour Temperature, K Colour Designation Lumens
Aluminium RIO LED L4T706
25,000 30 80 2700 Warm White 1414.6
Lumen Method Calculation Location
Audio-Visual Room
Dimension
Floor Area (A)
Length = 16m Width = 9m Height of the ceiling = 4m 16m x 9m = 144m2
Types of lighting fixture
LED Downlight
Lumen of lighting fixture F(Lux)
4000 Lm
Height of work level
0.8
Height of Luminaire (m)
4m
Mounting height (Hm)
3.2m
Reflection Factors
Ceiling: (0.7) Wall: (0.5) Floor: (0.2) 9 x 16 (9+16) x 3.2 = 1.8
Room Index / RI (K)
Utilisation Factor (UF)
0.5
Maintenance Factor (MF)
0.8
Standard Illuminance by MS1525
200
Number of light required
N=
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N = (4000)đ?&#x2018;Ľ
0.5 đ?&#x2018;Ľ 0.8
N = 18 Spacing to Height Ratio(SHR)
1
đ??´
SHR = đ??ťđ?&#x2018;&#x161; x â&#x2C6;&#x161;đ?&#x2018; 1
144
= 3.2 x â&#x2C6;&#x161; 18 = 0.88 đ?&#x2018;&#x2020;
SHR = 3.2 = 0.88 S = 3.2 x 0.88 = 2.816 Fittings layout by approximately (m)
Fittings required along 16m wall, 16/ 2.816 = 5.681 â&#x2030;&#x2C6; 6 rows Therefore, approximately 6 x 3 = 18 luminaires required Spacing along 9m wall 9/3rows = 3.0m
Fittings Layout
Conclusion 18 LED downlights are used to illuminate public living area to achieve minimum of 200 lux that stated in MS1525. With the sufficient level of illumination, the users can have clear visualization inside the space comfortably.
1.3 PSALI â&#x20AC;&#x201C; Permanent Supplementary Artificial Lighting of Interiors (Public Reading Space)
According to MS1525, the minimum lighting level required for a reading area is 300 lux. 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 Colour Temperature, K Colour Designation Lumens
Aluminium ST422B
50,000 40 80 3000 Warm White 5000
Lumen Method Calculation Location
Public Reading Space
Dimension
Floor Area (A)
Length = 11m Width = 10m Height of the ceiling = 4m 11m x 10m = 110m2
Types of lighting fixture
LED Downlight
Lumen of lighting fixture F(Lux)
5000 Lm
Height of work level
0.8
Height of Luminaire (m)
4m
Mounting height (Hm)
3.2m
Reflection Factors
Ceiling: (0.7) Wall: (0.5) Floor: (0.2) 12 x 12 (12+12) x 5.2 = 1.15
Room Index / RI (K)
Utilisation Factor (UF)
0.5
Maintenance Factor (MF)
0.8
Standard Illuminance by MS1525
300
Number of light required
N= N=
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N = 17 Spacing to Height Ratio(SHR)
1
đ??´
SHR = đ??ťđ?&#x2018;&#x161; x â&#x2C6;&#x161;đ?&#x2018; =
1 3.2
110
xâ&#x2C6;&#x161;
17
= 0.79
đ?&#x2018;&#x2020;
SHR = 3.2 = 0.79 S = 3.2 x 0.79 = 2.53
Fittings layout by approximately (m)
Fittings required along 11m wall, 11/ 2.53 = 4.3 â&#x2030;&#x2C6; 4 rows Therefore, approximately 4 x 5 = 20 luminaires required Spacing along 10m wall 10/5 =2m
Fittings Layout
Conclusion Basically, this big public reading area is controlled with 2 switches. One for the upper part, one for the bottom part. Switch 2 can be turned off due to sufficient daylight from outdoor. Hence, the energy can be saved. However, switch 1 has to be open most of the time for indoor activity to fulfil the requirement of MS 1525.
2.1 External Noise Sound Pressure Level (Administration Room) i) Peak Hour (Jalan Lebuhraya Timur-Barat) Highest reading: 75dB Use the formula, L = 10log10 (I/Io), 70 = 10log10 (I/ 1x10-12) I
= (107) (1x10-12) = 3 x 10-5
Lowest reading: 60dB Use the formula, L = 10log10 (I/Io), 60 = 10log10 (I/ 1x10-12) I
= (106) (1x10-12) = 1 x 10-6
Total Intensities, I = (3 x 10-5) + (1 x 10-6) = 3.1 x 10-5 Using the formula Combined SPL = 10log10 (p2/po2), where po = 1x10-12 Combined SPL = 10log10 [(3.1 x 10-5) รท (1 x 10-12)] = 74.91dB
ii) Non-peak Hour (Administration Room) Highest reading: 53dB Use the formula, L = 10log10 (I/Io), 53 = 10log10 (I/ 1x10-12) I
= (105.3) (1x10-12) = 2 x 10-7
Lowest reading: 45dB Use the formula, L = 10log10 (I/Io), 45 = 10log10 (I/ 1x10-12) I
= (104.5) (1x10-12) = 3.2 x 10-8
Total Intensities, I = (2 x 10-7) + (3.2 x 10-8) = 2.3 x 10-7
Using the formula Combined SPL = 10log10 (p2/po2), where po = 1x10-12 Combined SPL = 10log10 [(2.3 x 10-7) รท (1x10-12)] = 53.62dB As a result, at Administration Room the average sound pressure level during Peak Hour and Non-peak Hour are 74.91dB and 53.62dB.
2.2 Reverberation Time, RT (Audio-Visual Room)
Space volume, V= 16m x 9m x 5m = 720 m3 Material absorption coefficient in 500Hz at Peak hour with 40 persons in the space Building Component Wall
Material
Area, S/m2
Sound Absorption, Sa 2.4
140
0.06
8.4
Door
Precast Concrete Wall Concrete Strain Glass
47.5
0.10
4.8
Ceiling
Plaster Finish
140
0.015
2.1
People [Peak]
-
40
0.46
18.4
Floor
120
Absorption Coefficient, a 0.02
Total Absorption, A
Reverberation Time, RT
36.1
= (0.16 x V) /A = (0.16 x 720) / 36.1 = 3.19s
The reverberation time for the Audio-Visual room at 500Hz during peak hours is 3.19s. This doesnâ&#x20AC;&#x2122;t fall above the comfort reverberation of in between 1.5 to 2.5s. However higher reverberation time is good for an audio-visual space. This indicates how there is inadequate acoustic absorption within the space during peak hours.
Material Absorption Coefficient at 2000Hz, Non-Peak Hour with 20 persons in the space. Building Component Wall
Material
Area, S/m2
Sound Absorption, Sa 6.0
140
0.02
2.8
Door
Precast Concrete Wall Concrete Strain Glass
47.5
0.10
4.8
Ceiling
Plaster Finish
140
0.04
5.6
People [Peak]
-
20
0.51
10.2
Floor
120
Absorption Coefficient, a 0.05
Total Absorption, A
Reverberation Time, RT
29.4
= (0.16 x V) /A = (0.16 x 720) / 29.4 = 3.91s
The reverberation time for the Audio-Visual Room at 2000Hz during peak hour is 3.91s. Again, it doesnâ&#x20AC;&#x2122;t fall above the comfort reverberation of between1.5-2.5s which indicates how there is inadequate acoustic absorption within the space during peak hours.
2.3 Sound Reduction Index, SRI Calculation (Nature Gallery)
Transmission coefficient of materials Building Element Wall Wall 2
Material Glass Panel Precast Concrete Wall
Surface Area, (m2) 40
SRI (dB) 26
Transmission Coefficient, T 2.51 x 10-3
89.2
50
1 x 10-5
a) Wall- Glass Panel 1
SRI Glass = 10log10 T glass 1
26 = 10log10 T glass 1
102.6 = T glass T glass = 2.51 x 10-3
b) Wall – Precast Concrete Wall 1
SRI Concrete = 10log10 T Concrete 1
50 = 10log10 T Concrete 105.0 = T
1 Concrete
T concrete = 1 x 10-5
Average Transmission Coefficient of Materials Tav = (40 x 2.51 x 10-3) + (89.2 x 1 x 10-5) / (40+89.2) = [0.1004 + (8.92 x 10-4)] / 172.5 = 5.872 x 10-4 1
SRI overall = 10log10 T av 1
= 10log10 5.872 x 10−4 = 32.31 dB 60dB – 32.31dB = 27.69dB The overall transmission loss from Nature Gallery to the Public Living area is at 32.31dB. Assume that the sound pressure level in the quiet study area is approximately 60 dB, the sound that is transmitted through partitions into the public living area is approximately 27.69dB. According to the noise criteria environment perception, 27.69 dB is the category between soft whisper noises. It is considering an ideal value for public living area where people read, rest and chatting.
References Kotzen, B., & English, C. (1999). Environmental noise barriers a guide to their acoustic and visual design. London: E & FN Spon. Lumens, watts and lux: A user's guide to basic lighting design. (2012). Nelson, N.Z.: Karen Wardell. Lux Europae: Outdoor light installations by 35 European artists across the city of Edinburgh. (1993). Edinburgh: Lux Europae Trust. Pritchard, D. (1969). Lighting. New York: American Elsevier Pub.