TCC Building Science 2 Project 2 Calculation Report
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Transcript of TCC Building Science 2 Project 2 Calculation Report
1. ACOUSTICSound Pressure Level, SPLSpace: Library Lobby
1. Peak Hour ( Jalan Tun Abdul Samad)
Highest Reading: 85 dB
Combined Sound Pressure level, SPL = 10 log 10( ll0 ( ref ) )
Where l = sound power (intensity) (Watts)
l = reference power (1 x 10−12Watts)
85 = 10 log10 (l / 1 x 10−12)
l = (108.5)( 1 x 10−12)
= 3.16 x 10−4
Lowest Reading: 75 dB
Combined Sound Pressure level, SPL = 10 log 10( ll0 ( ref ) )
Where l = sound power (intensity) (Watts)
l = reference power (1 x 10−12Watts)
75 = 10 log10 (l / 1 x 10−12)
l = (107.5)( 1 x 10−12)
= 3.16 x 10−5
Total intensities:
Total intensities, l = (3.16 x 10−4) + (3.16 x 10−5) = 3.48 x10−4
Combined SPL:
By using the formula, combined SPL = 10 log 10( ll0 (ref ) )
Combines SPL = 10 log 10( 3.48 x10−4
1 x 10−12 )
BUILDING SCIENCE 2 [ARC 3413]
Project 2 : Integration with Design Studio 5
Student’s Name: Toh Chee Cheng ID: 0311122
Tutor’s Name: Mr. Edwin
¿85.41 dB
2. Non - Peak Hour ( Jalan Tun Abdul Samad)
Highest Reading: 70 dB
Combined Sound Pressure level, SPL = 10 log 10( ll0 ( ref )
) Where l = sound power (intensity) (Watts)
l = reference power (1 x 10−12Watts)
70 = 10 log10 (l / 1 x 10−12)
l = (107)( 1 x 10−12)
= 1 x 10−5
Lowest Reading: 66 dB
Combined Sound Pressure level, SPL = 10 log 10( ll0 ( ref ) )
Where l = sound power (intensity) (Watts)
l = reference power (1 x 10−12Watts)
66 = 10 log10 (l / 1 x 10−12)
l = (106.6)( 1 x 10−12)
= 3.98 x 10−6
Total intensities:
Total intensities, l = (1 x 10−5+ (3.98 x 10−6 ¿= 1.4 x 10−5
Combined SPL:
By using the formula, combined SPL = 10 log 10( ll0 (ref )
) Combines SPL = 10 log 10( 1.4 x10−5
1 x 10−12 ) ¿ 71.46 dB
As a result, the average sound pressure level during Peak Hour and Non-Peak Hour are 85.41 dB and 71.46dB at Jalan Tun Abdul Samad. The sound pressure level of conversational speech is approximately 60 dB, so background noise levels greater than 45 dB will impair speech communication at exterior.
Sound Reduction of the Entry Façade at Ground Floor
Surface Area, Sn/m2 SRI/
dBT cn Sn
Wall Type:Curtain Wall
Sn=96.58 m2 34 2.51 ×10−3 0.242
Superstructure Type 2 : Glass Door
Sn=26.4 m2 25 3.16 ×10−3 0.083
TOTAL 122.98 0.325
SRI Glass Door = 25 dB
Sn=26.4 m2
Sound Reduction Index , SRI=10 log10( 1T av )
25=10 log10( 1T )
T=3.16 ×10−3
SRI Curtain Wall= 34 dB
Sn=96.58 m2
Sound Reduction Index , SRI=10 log10( 1T av )
34=10 log10( 1T )
T=2.511×10−3
Average transmission coefficient of materials
T av=( S1 ×T c1+S2 ×T c2+⋯+Sn ×T cn
Total Surface Area )¿( 0.325
122.98 )= 2.643 × 10−3
Sound Reduction Index
Sound Reduction Index , SRI=10 log10( 1T av )
¿10 log10( 12.643× 10−3 )
= 25.78 dB
85.41dB – 25.78 dB = 59.63dBAfter deducting the transmission loss, sound reduction of the entry facade is 25.78dB. Maximum sound pressure level of the street is 85.41dB, the sound that transmitted to lobby is 59.63dB. The sound pressure level of conversational speech is approximately 60 dB, The sound pressure level of conversational speech is approximately 60 dB, so background noise levels greater than 45 dB will impair speech communication.
Reverberation Time, RTSpace: Small Theatre
Standard reverberation time for theatre is 1.0 – 1.6s.
Space Volume of Small Theatre
Space Volume = 82m2 x 5m
= 410
Material Absorption Coefficient in 500 Hz at Peak Hours
Building ElementMaterials
Absorption Coefficient (500 Hz) Area (m2)
Wall Smooth Painted Concrete 0.06 200 12
Floor Carpet 0.14 82 11.48
Raw Concrete 0,01 82 0.82
Ceiling Plaster Board 0.02 82 1.64
Door Timber Door 0.06 7.02 0.42
Furniture Cushion Chair 0.33 172 56.76
Human - 0.42/person 84 35.28
TOTAL118.4
Reverberation Time, RT = 0.6 x SpaceVolumeTotal Absorption
=0.6 X 410118.4
=2.07 s
The reverberation time for small theatre in 500 Hz of absorption coefficient is 2.07s. According to the standard reverberation time, the standard comfort reverberation is between 1.0 – 1.6s. The reverberation time of the case study on 500 Hz is over the standard comfort reverberation.
Material Absorption Coefficient in 2000 Hz at Peak Hours
Building ElementMaterials
Absorption Coefficient (500 Hz) Area (m2)
Wall Smooth Painted Concrete 0.09 200 18
Floor Carpet 0.60 82 49.2
Raw Concrete 0,02 82 1.64
Ceiling Plaster Board 0.05 82 4.10
Door Timber Door 0.10 7.02 0.70
Furniture Cushion Chair 0.58 172 99.76
Human - 0.51/person 84 42.84
TOTAL198.24
Reverberation Time, RT = 0.6 x SpaceVolumeTotal Absorption
=0.6 X 410198.24
=1.24 s
The reverberation time for small theatre in 2000 Hz of absorption coefficient is 1.24s. According to the standard reverberation time, the standard comfort reverberation is between 1.0 – 1.6s. The reverberation time of the case study on 500 Hz is within the standard comfort reverberation.
Sound Reduction Index, SRI
Space: Kitchen and LoungeThe sound reduction index, SRI or the transmission loss, TL of a partition measure the number of decibels lost when a sound of a given frequency is transmitted through the partition.
Sound Reduction Index , SRI=10 log10( 1T av )
where T av=( S1× T c1+S2× T c 2+⋯+Sn× T cn
Total Surface Area )Sn=Surface area of material n
T cr=Transmission Coefficient of Material
SRI Concrete Painted Wall = 44 dB
Sn=20 m2
Sound Reduction Index , SRI=10 log10( 1T av )
44=10 log10( 1T )
T=3.98 ×10−5
SRI Timber Door = 22 dB
Sn=2.43 m2
Sound Reduction Index , SRI=10 log10( 1T av )
22=10 log10( 1T )
T=6.31 ×10−3
SRI Glass Door = 25 dB
Sn=2.43 m2
Sound Reduction Index , SRI=10 log10( 1T av )
25=10 log10( 1T )
T=3.16 ×10−3
Surface Area, Sn/m2 SRI/
dBT cn Sn
Wall Type:Concrete Painted
Sn=21.56 m2 44 3.98 ×10−5 8.58× 10−4
Superstructure Type 1 : Timber Door
Sn=2.43 m2 22 6.31 ×10−3 0.03
Superstructure Type 2 : Glass Door
Sn=2.43 m2 25 3.16 ×10−3 7.68 × 10−3
TOTAL 26.42 0.039
Average transmission coefficient of materials
T av=( S1 ×T c1+S2 ×T c2+⋯+Sn ×T cn
Total Surface Area )¿( 0.039
26.42 )= 1.476 × 10−3
Sound Reduction Index
Sound Reduction Index , SRI=10 log10( 1T av )
¿10 log10( 11.476 ×10−3 )
= 28.31 dB
70dB – 28.31 dB = 51.43dBAfter deducting the transmission loss, sound reduction of the highlighted wall is 28.31dB. Assume that the sound pressure level in kitchen is approximately 70dB, the sound that transmitted through walls into the lounge is 51.43dB.
2. Lighting Day LightingDaylight Factor distribution are as below, according to the MS 1525 standards.
Daylight Factor (%) Distribution>6 Very bright with thermal and glare problem
3-6 Bright
1-3 Average
0-1 Dark
The selected area is Young Adult Library Area located at third floor facing the street. This
area is exposed to sunlight. Therefore, it needs combination of day lighting and artificial
lighting to provide visual comfort, artificial lighting is not needed when daylight is sufficient,
so artificial light mainly use for night.
Daylight Factor Calculation
Floor Area (m²) 110Floor Area (m²) 210
Exposed Facade (m²) 96.8
Exposed Façade to Floor Area Ratio / Daylight Factor, DF
96.8210 = 0.46
=46% x 0.1 = 4.6%
ConclusionThe Young Adult Library Space at the Third Floor has daylight factor of 4.6% which is within the bright distribution level. Thus, solution such as double façade, using insulated glass has to be consider to resolve this problem and combination of daylight and artificial lighting are required.
Natural Illumination Calculation
DF = (Ei ÷ Eo) x 100%Given that Eo = 32000
(daylight level in Malaysia)
Ei= (DF x Eo) ÷ 100% =(4.6% x 32000) ÷ 100% =1472 lux
ConclusionThe natural illumination at a reference point for Young Adult Library Area is 1472 lux.
Artificial Lighting The minimum lighting level required for a theatre is 100 lux. Therefore the minimum
luminance for calculation has to be 100 lux.
Type of Fixture LED Recessed Lighting
Material of Fixture Aluminium
Wattage Range 10W
Colour Designation Warm Yellow
Lumens 700
Lumen Calculation
Location TheatreActivity Movie Showcase
Dimension of Room Length (L) = 16mWidth ( W) = 10mCeiling Height = 4.4m
Total Floor Area 82.8 m²
Standard Illuminance Required According to MS 1525
100 lux
Assumption of Reflectance Value Ceiling 0.7, Wall 0.5, Floor 0.2
Mounting Height (Hm) 5m – 0.6m = 4.4m
Room index, KL xW
h(L xW )
16 x 104.4 (16 x10) =
160704
=0.227Utilization Factor, UF 0.33
Maintenance Factor, MF N = LLMF X LSF X LMF X RSMF = 0.94 x 1 x 0.61 x 0.82 = 0.47
Lumen Calculation N= E x A
F xUF x MF
Conclusion
= 100 x 82.8
700 x 0.33 x0.47 =76.26 ≈76
Number of luminaires
across √ L x NW
√ L x NW
= √ 16 x 7610
= 11Number of luminaires
along √W x NL
√ W x NL
= √ 10 x7616
= 6. 89 ≈ 7
Conclusion