calculations - integration with studio 5
TRANSCRIPT
Project 2 : Lighting and Acoustics Integration with Design Studio 5
Tutor : Mr. Sanjey Raman
Nadia Othman 0303423
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1.0 Lighting
1.1 Daylighting _____________________________________________________________________ 3
I) Market
II) Reading Area
III) Spa
1.2 Artificial lighting _________________________________________________________________4
I) Market
II) Reading Area
III) Spa
2.0 Acoustic
2.1 Reverberation time ______________________________________________________________10
I) Market
II) Reading Area
III) Spa
2.2 External noise sound pressure level _________________________________________________13
2.3 Sound transmission class _________________________________________________________14
I) Market
II) Reading Area
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1.1 Daylighting
DF =
EiEo X 100%
I) Market
DF = 800/32000 X 100%
= 2.5%
This space has a fair daylight factor, the usage of artificial light would be sufficient to the space.
II) Reading
DF = 700/32000 X 100%
= 2.19%
This space has a fair daylight factor, the usage of artificial light would be sufficient to the space.
III) Spa
DF = 600/32000 X 100%
= 1.88%
This space has a fair daylight factor, the usage of artificial light would be sufficient to the space.
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1.2 Artificial Lighting
I) Market
Referring to MS 1525, the lighting level required for a market is 500lux.
Location Ground floorRoom Name MarketRoom Dimension L=3.0 m , W=3.0mTotal Floor Area (m²) 9.0Standard Illuminance required (lux) 500Assumption of Reflectance valueMounting Height (hm) 4.5 – 1.0 = 3.5mRoom Index (K) (3 x 3) / (3 + 3) x (4.5-1.0)
= 0.43Utilization Factor (UF) 0.43Maintenance Factor (MF) 0.31
N =
E×AF×UF×MF
= (500 x 9) / (3500x 0.43 x 0.31)
= 9.6 ≈ 10
Type of Lighting Quantity Luminous Flux (lm)
LUMILUX XT T8Wattage : 58WColor temperature : 3000K
10 3500
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II) Reading Area
Referring to MS 1525, the lighting level required for a reading area is 300.
Location First FloorRoom Name Reading AreaRoom Dimension L=6.5m , W=7.5mTotal Floor Area (m²) 48.75Standard Illuminance required (lux) 300Assumption of Reflectance valueMounting Height (hm) 3.5 – 0.8 = 2.7mRoom Index (K) (6.5 x 7.5) / (6.5 + 7.5) x 2.7
= 1.29Utilization Factor (UF) 0.46Maintenance Factor (MF) 0.35
N =
E×AF×UF×MF
= (300 x 48.75) / (3500x 0.46 x 0.35)
= 25.9 ≈ 26
Type of Lighting Quantity Luminous Flux (lm)
Luminous flux : 11000 lm
Color temp: 3000 K
Lifespan: 300 h
26 3500
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III) Spa
Referring to MS 1525, the lighting level required for a spa/bedroom is 150
Location Second FloorRoom Name SpaRoom Dimension L=5.6m , W=3mTotal Floor Area (m²) 16.8Standard Illuminance required (lux) 150Assumption of Reflectance valueMounting Height (hm) 3.5- 1.0 = 2.5 mRoom Index (K) (5.6 x 3)/ (5.6 + 3) x (3.5-1.0)
= 0.8Utilization Factor (UF) 0.40Maintenance Factor (MF) 0.35
N =
E×AF×UF×MF
= (150 x 16.8) / (1100 x 0.4 x 0.35)
= 16.3 ≈ 18
Type of Lighting Quantity Luminous Flux (lm)
OSRAM DULUX INTELLIGENT DIM CFLLuminous flux: 1100 lm
Color temp: 2500 K
Lifespan: 20000 h
18 1100
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Acoustics
Market
Component Material Function Area(m²) Absorption Coefficient (2000Hz)
Sound Absorption (SA)
CEILING Glass Ceiling 255 0.05 12.75WALL Brick
GlassGlass
WallWindowDoor
127.276.838.7
0.040.070.03
5.0885.3761.161
FLOOR Concrete Floor 255 0.06 15.3FURNITURE Wood Chairs
Tables 1616
0.280.06
4.482.76
Total Absorption
46.92
RT =
0 .16×VA
RT = 0.16 x 2805 / 46.92
RT = 9.6s
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Reading Room
Component Material Function Area(m²) Absorption Coefficient (2000Hz)
Sound Absorption (SA)
CEILING Concrete Ceiling 48.75 0.06 2.93WALL Concrete
GlassWallWindow
76.916.2
0.060.07
4.61.1
FLOOR Carpet Floor 48.75 0.60 29.25FURNITURE
PEOPLE
Fabric Chair ChairsTables
10210
0.960.060.46
9.60.124.6
Total Absorption
52.2
RT =
0 .16×VA
RT = 0.16x 170.63 / 52.2
RT = 0.52s
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Spa
Component Material Function Area(m²) Absorption Coefficient (2000Hz)
Sound Absorption (SA)
CEILING Concrete Ceiling 16.8 0.06 1.008WALL Concrete
Glasswood
WallWindowDoor
7.533
0.060.070.03
0.450.210.09
FLOOR Carpet Floor 16.8 0.60 10.08FURNITURE
PEOPLE
Fabric Wood
ChairsTables
212
0.960.060.46
1.920.060.92
Total Absorption
14.74
RT =
0 .16×VA
RT = 0.16 x 58.8 / 14.74
RT = 0.64s
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External noise
Sound pressure level
Traffic noise: 83dB
Activity noise: 60dB
Traffic
83 = 10 log (I₁/I₀)
log¯ⁱ 8.3 = I₁/I₀
log¯ⁱ 8.3 = I₁/(1.0 x 10¯ⁱ²)
I₁ = 1.995 x 10¯⁴
Activity
65 = 10 log (I₁/I₀)
log¯ⁱ 6.5 = I₁/I₀
log¯ⁱ 6.5 = I₁/(1.0 x 10¯ⁱ²)
I₁ = 3.16 x 10¯⁶
Combine intensity
(1.995 x 10¯⁴) + (3.16 x 10¯⁶)
= 2.03 x 10¯⁴
Combined SPL
10 log (I₁/I₀)
= 10 log (2.03 x 10¯⁴/1 x 10¯ⁱ²)
= 10 log 2.03 x 10⁸
=83dB
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Sound Transmission Class
Reading Room
TL for brick wall TL for glass window
42 = 10log
1T 37 = 10log
1T
log¯ⁱ4.2 =
1T log¯ⁱ3.7 =
1T
T = 1/ (1.585 x 10⁴) T = 1/ (5.01 x 10³)
= 6.309 x 10¯⁵ T = 1.995 x 10¯⁴
Average Transmission Coefficient of Materials
Toverall = (7.164 x 10¯4) + (1.407 x 10¯3)/25.9
= 8.19x 10¯⁵
Total SRI of Wall
SRI = 10log 1/8.19 x 10¯⁵
= 10log (12210)
= 40.87
83 – 40.87
= 42.13dB
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Market
TL for brick wall TL for glass window
42 = 10log
1T 37 = 10log
1T
log¯ⁱ4.2 =
1T log¯ⁱ3.7 =
1T
T = 1/ (1.585 x 10⁴) T = 1/ (5.01 x 10³)
= 6.309 x 10¯⁵ T = 1.995 x 10¯⁴
Average Transmission Coefficient of Materials
Toverall = (2.47 x 10¯2) /187
= 1.32 x 10¯4
Total SRI of Wall
SRI = 10log 1/1.32 x 10¯4
= 10log (7575)
= 38.79
83 – 38.79
= 44.21dB
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References
1. LED light colour, CRI and experiments. (n.d.). LED light colour, CRI and experiments. Retrieved December 4, 2014, from http://swhs.home.xs4all.nl/fiets/tests/verlichting/experimenten/index_en.html#cri
2. Sanjeh R. (2014). Lumen Method, Lecture Notes. Taylor’s University.
3. Sivaraman K. (2014). Acoustic Calculations, Lecture Notes. Taylor’s University.
4. Binggeli, C. (2003). Building System for Interior Designers. New Jersey: John Wiley & Sons.
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