B Science Report Project 2

School of Architecture, Building and Design

Modern Architecture Studies in Southeast Asia (MASSA) Research Unit

Bachelor of Science (Honours) (Architecture)

Building Science 2 [BLD 61303 / ARC 3413]

PROJECT 02

INTEGRATION PROJECT

Student : Phua Jing Sern 0314572

Tutor : Mr. Sivaraman Kuppusamy

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Table of Content Page

1.0 Lighting

1.1 Silent Area1.1.1 Daylight 21.1.2 Artificial Light 41.1.3 PSALI 5

1.2 Admin Office1.2.1 Daylight 61.2.2 Artificial Light 81.2.3 PSALI 9

2.0 Acoustics

2.1 Silent Area2.1.1 Sound Pressure Level (SPL) 102.1.2 Reverberation Time (RT) 122.1.3 Sound Reduction Index (SRI) 14

2.2 Admin Office2.2.1 Sound Pressure Level (SPL) 162.2.2 Reverberation Time (RT) 182.2.3 Sound Reduction Index (SRI) 20

3.0 References 21

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1.0 Lighting 1.1 Silent Area

1.1.1 Daylight

According to MS1525: Daylight Factor, DF

Zone Daylight Factors (%) Distribution

Very Bright > 6 Very large with thermal and glare problems

Bright 3 - 6 GoodAverage 1 - 3 Fair Dark 0 - 1 Poor

Daylight Factor Calculation

Floor Area (m2) 7m x 13m = 91Area of façade exposed to sunlight (m2)

50m

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Area of skylight 0Exposed façade & skylight area to floor area ratio / Daylight factor, DF

(50+0)/91= 0.55= 0.55 x 100%= 5.5%

Natural Illumination Calculation

Illuminance Example

120,000 lux Very Bright Sunlight110,000 lux Bright Sunlight20,000 lux Clear Sky1000 – 2000 lux Overcast day400 lux Sunrise / Sunset on clear day< 200 lux Midday40 lux Fully overcast<1 lux Sunset, storm cloud

E External = 20,000 lux

DF = E internal / E external x 100

5.5 = E internal / 20,000 x 100

E internal = 5.5 x 20,000 / 100

= 1100 lux

The silent area has a daylight factor of 5.5% and natural illumination of 1100 lux. Based on the requirements of MS 1525, this space has good daylight distribution as the value is in between 3-6%, but the illuminance value is higher than the required value of 300 lux. This can cause thermal and glare problems. Low-e coatings of the glass panels are proposed to minimize the penetration of ultraviolet and infrared rays into spaces.

1.1.2 Artificial light

Lumen Method Calculation

Model DN570B

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Input 230 or 240 v / 50-60HzLumen 4000Weight 2.2 kg

Power 36w (3000k)

Dimension of room ( L x W) 7m x 13mTotal floor Area (m2) 91Room cavity height (m) 4Reflectance values Ceiling = 0.7

Wall = 0.5Floor = 0.3

Room index, K K = (7 x 13) / 4(7+13) = 1

Utilization Factor, UF 0.38Maintenance Factor, MF 0.8Illuminance Requirement 300 - 500Number of Luminaries N =

300x 914000 x (0.8 x0.38)

= 22.45~ 23 bulbs

Spacing to height ration (SHR) SHR = 14 x √ 9123= 0.49

SHR = S4 = 0.49

S = 1.96Fitting Layout Fitting required along 7m wall

= 7/1.96 = 3.5= 4 rows

Fitting required along 13m wall= 13/1.96 = 6.6= 7 rows

Fitting layout and SPALI

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28 fluorescent lamps are used to illuminate the quiet reading area to achieve the minimum illuminance of 300 lux stated by MS 1525.With sufficient illuminance, users are able to utilize the space with comfort.

Referring to the daylight values obtained above, the silent area has a good daylight factor of 5.5%. The artificial lightings calculation also shows a result of the usage of 28 down lights to illuminate the silent area. PSALI is applied whereby the lightings are controlled with 2 switches; one that is closest to the window wall whereby it can be switched off during the day, while the rest can be operate with another switch when necessary as well.

1.2 Admin Office

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1.2.1 Daylight

According to MS1525: Daylight Factor, DF

Zone Daylight Factors (%) Distribution

Very Bright > 6 Very large with thermal and glare problems

Bright 3 - 6 GoodAverage 1 - 3 Fair Dark 0 - 1 Poor

Daylight Factor Calculation

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Floor Area (m2) 3.5m x 6m = 21Area of façade exposed to sunlight (m2) 4Area of skylight 0Exposed façade & skylight area to floor area ratio / Daylight factor, DF

(4+0)/21= 0.19= 0.19 x 100%= 1.9%

Natural Illumination Calculation

Illuminance Example120,000 lux Very Bright Sunlight110,000 lux Bright Sunlight20,000 lux Clear Sky1000 – 2000 lux Overcast day400 lux Sunrise / Sunset on clear day< 200 lux Midday40 lux Fully overcast<1 lux Sunset, storm cloud

E External = 20,000 lux

DF = E internal / E external x 100

1.9 = E internal / 20,000 x 100

E internal = 1.9 x 20,000 / 100

= 380 lux

The admin office has a daylight factor of 1.9% and natural illumination of 380 lux. Based on the requirements of MS 1525, this is considered as fair yet not adequately lit. Therefore, artificial lighting is required for the space. The illuminance value fits within the MS 1525 requirement of 300 – 400 lux.

1.2.2 Artificial light7

Lumen Method Calculation

Model LuxiLED G2Input 110 or 240 v / 50-60HzLumen 5000Weight 4.6

Power 32W (3000k)

Dimension of room ( L x W) 3.5m x 6mTotal floor Area (m2) 21Room cavity height (m) 4Reflectance values Ceiling = 0.7

Wall = 0.5Floor = 0.3

Room index, K K = (3.5 x 6) / 4(3.5+6) = 0.6

Utilization Factor, UF 0.27Maintenance Factor, MF 0.8Illuminance Requirement 300 - 500Number of Luminaries N =

300 x 215000 x (0.8 x 0.27)

= 5.83~ 6

Spacing to height ration (SHR) SHR = 14 x √ 216= 0.46

SHR = S4 = 0.46

S = 1.84Fitting Layout Fitting required along 3.5m wall

= 3.5/1.84 = 1.9= 2 rows

Fitting required along 6m wall= 6/1.84 = 3.2= 4 rows

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Fitting Layout and PSALI

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down lights are used to illuminate the admin office to achieve the minimum illuminance of 300 lux stated by MS 1525.With sufficient illuminance, office staffs are able to utilize the space with comfort.

Referring to the daylight values obtained above, the admin office has a fair yet not adequately lit daylight factor of 1.9%. Thus the artificial lightings calculation also shows a need of the usage of 8 down lights to illuminate the admin office. PSALI is applied whereby the lightings are controlled with 2 switches; one half controls the help desk and the corridor, while the other half controls the work desks of the staffs. To achieve a uniform lighting and to follow the requirements of MS 1525, both switches need to be turned on during operating hours.

2.0 Acoustic9

2.1 Silent Area

2.1.1 SPL – Sound Pressure Level

Sound Pressure Level Formula:

SPL = 10log10IIx SPL = sound pressure level (dB),

I = sound power (intensity)(Watts)

Ix = reference power (Ix is usually taken as 1x10-12 watts)

Peak Hour (Back Alley)

Highest Reading: 70 dB (Moderate sound)

Lowest Reading: 55 dB (Quiet sound)

Highest Reading:

70 = 10log10IIx

Antilog 7 = l

1x 10−12

1 X 10^7 = l

1x 10−12

L = 10^7 x (1x10^-12)

L = 1 x [10^7 + (-12))

L = 1 x 10^-5

Lowest Reading:

55 = 10log10IIx

Antilog 5.5 = l

1x 10−12

1 X 10^5.5 = l

1x 10−12

L = 10^5.5 x (1x10^-12)

L = 1 x [10^5.5+ (-12))

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Total Intensities:

Total intensities, I

= (1x10^-5) + (1 x 10^-6.5)

= 1.03x10^-5

Combined SPL:

SPL = 10log10IIx

= 10log101.03 x10−5

Ix 10−12

= 70 dB

L = 1 x 10^-6.5

Non -Peak Hour (Back Alley)

Highest Reading: 55 dB (Quiet sound)

Lowest Reading: 45 dB (Very Quiet sound)

Highest Reading:

55 = 10log10IIx

Antilog 5.5 = l

1x 10−12

1 X 10^5.5 = l

1x 10−12

L = 10^5.5 x (1x10^-12)

L = 1 x [10^5.5+ (-12))

L = 1 x 10^-6.5

Lowest Reading:

45 = 10log10IIx

Antilog 4.5 = l

1x 10−12

1 X 10^4.5 = l

1x 10−12

L = 10^4.5 x (1x10^-12)

L = 1 x [10^4.5+ (-12))

L = 1 x 10^-7.5

The noise criteria for a reading area is within the range of 35-40. According to the calculation above, the combined sound pressure level around the silent area during peak and non-peak hours are 70dB and 55 dB respectively, which exceeds the noise criteria for a

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Total Intensities:


= (1x10^-6.5) + (1 x 10^-7.5)

= 3.48x10^-7

Combined SPL:

SPL = 10log10IIx

= 10log103.48 x10−7

Ix 10−12

= 55 dB

reading area. This can be solved by proposing double layered channel glass which consists of an insulation layer, preventing the transmission of sound through the vibration of air.

2.1.2 Reverberation Time (Silent Area)

Volume of Space

V = L x W x H

V = 13m x 7m x 4m

V = 364m3

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Material Absorption Coefficient under 500 Hz, with 15 users

Building Component

Material Area S ( m2) Absorption Coefficient (a)

Sound Absorption( SA)

Wall Concrete 73.2 0.02 1.46Glass 47 0.06 2.82

Floor Carpet 91 0.30 27.3Door Timber 2 0.13 0.26Ceiling Concrete 91 0.02 1.82Furniture Desk 18 0.45 8.1

Padded Chair 6 0.77 4.62Fabric Barrel 96 0.30 28.8

Users - 15 0.46 6.9Total Absorption 81.88

RT= (0.16 X 364) / 81.88

= 0.71s

Material Absorption Coefficient under 2000 Hz, with 15 users

Building Component




Floor Carpet 91 0.50 45.5Door Timber 2 0.10 0.2Ceiling Concrete 91 0.02 1.82Furniture Desk 18 0.45 8.1

Padded Chair 6 0.77 4.62Fabric Barrel 96 0.50 48

Users - 15 0.51 7.62Total Absorption 118.73

RT= (0.16 X 364) / 118.73

= 0.49s

The reverberation time required for a library reading space is 0.4 – 0.6 seconds. Based on the calculation, the reverberation of the space under 500 Hz and 2000 Hz is 0.71s and 0.49s

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respectively. As they slightly differ from the requirement, acoustic panels are introduced to amend the issue.

2.1.3 Sound Reduction Index (Silent Area)

Volume of Space

V = L x W x H

V = 13m x 7m x 4m

V = 364m3

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Assuming that the noise level of a Silent Area is 40 dB, noise level of the external is 75dB. The sound reduction index of the wall should 75dB- 40dB = 35 dB.

Building Component Material Surface Area

(m2SRI

( Decibels)

Transmission Coefficient ( T

)Wall Concrete 73.2 46 2.51 x10−5

Wall Glass 47 35 3.16 x10−4

Door Plywood 2 31 7.94 x10−4

Concrete

SRI = 10 Log (1/T)

46 = 10 log (1/T)

3.98 x104 = 1 / T

T = 2.51 x10−5

Glass

SRI = 10 Log (1/T)

35 = 10 log (1/T)

3.16 x103 = 1 / T

T = 3.16 x10−4

Timber

SRI = 10 Log (1/T)

31 = 10 log (1/T)

103.1 = 10 log10 (1/T)

T = 7.94 x10−4

The overall transmission loss from the back lane to the silent area is 38.24 dB. Based on the previous assumption, according to the noise criteria environment perception, 35 dB is

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Average Transmission Coefficient

= [ 73.2 x (2.51 x10−5) ] + [ 47 x ( 3.16 x10−4) ]

+ [ 2 x (7.94 x10−4 ¿¿ / ( 73.2 + 47 + 2)

= 1.50 x10−4

Overall SRI

= 10 Log ( 1 / 1.50 x10−4 )

= 38.24 Db

External Sound Pressure Level

= 75dB – 38.24dB

= 36.76 dB

categorized as whispering sound. Thus the space has an ideal value for a silent reading area where users can focus on their reading with ease.

2.2 Admin Office

2.2.1 SPL – Sound Pressure Level

Sound Pressure Level Formula:

SPL = 10log10IIx SPL = sound pressure level (dB),

I = sound power (intensity)(Watts)

Ix = reference power (Ix is usually taken as 1x10-12 watts)

Peak Hour (Back Alley)

Highest Reading: 70 dB (Moderate sound)

Lowest Reading: 55 dB (Quiet sound)

Highest Reading:

70 = 10log10IIx

Antilog 7 = l

1x 10−12

1 X 10^7 = l

1x 10−12

L = 10^7 x (1x10^-12)

L = 1 x [10^7 + (-12))

L = 1 x 10^-5

Lowest Reading:

55 = 10log10IIx

Antilog 5.5 = l

1x 10−12

1 X 10^5.5 = l

1x 10−12

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Total Intensities:


= (1x10^-5) + (1 x 10^-6.5)

= 1.03x10^-5

Combined SPL:

SPL = 10log10IIx

= 10log101.03 x10−5

Ix 10−12

= 70 dB

L = 10^5.5 x (1x10^-12)

L = 1 x [10^5.5+ (-12))

L = 1 x 10^-6.5

Non -Peak Hour (Back Alley)

Highest Reading: 55 dB (Quiet sound)

Lowest Reading: 45 dB (Very Quiet sound)

Highest Reading:

55 = 10log10IIx

Antilog 5.5 = l

1x 10−12

1 X 10^5.5 = l

1x 10−12

L = 10^5.5 x (1x10^-12)

L = 1 x [10^5.5+ (-12))

L = 1 x 10^-6.5

Lowest Reading:

45 = 10log10IIx

Antilog 4.5 = l

1x 10−12

1 X 10^4.5 = l

1x 10−12

L = 10^4.5 x (1x10^-12)

L = 1 x [10^4.5+ (-12))

L = 1 x 10^-7.5

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Total Intensities:


= (1x10^-6.5) + (1 x 10^-7.5)

= 3.48x10^-7

Combined SPL:

SPL = 10log10IIx

= 10log10 3.48 x10−7

Ix 10−12

= 55 dB

The noise criteria for an admin office is within the range of 50 - 60. According to the calculation above, the combined sound pressure level around the silent area during peak and non-peak hours are 70dB and 55 dB respectively, which exceeds the noise criteria for a reading area. This can be solved by proposing double layered channel glass which consists of an insulation layer, preventing the transmission of sound through the vibration of air, or a double façade which acts as a sound barrier.

2.2.2 Reverberation Time (Admin Office)

Volume of Space

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V = L x W x H

V = 6m x 3.5m x 4m

V = 84m3

Material Absorption Coefficient under 500 Hz , with 5 users

Building Component




Floor Carpet 21 0.30 6.3Door Timber 2 0.13 0.26Ceiling Concrete 21 0.02 0.42Furniture Timber Desk 5 0.15 0.75

Padded Chair 3 0.77 2.31Users - 5 0.46 2.3Total Absorption 13.39

RT= (0.16 X 84) / 13.39

= 1s

Material Absorption Coefficient under 2000 Hz , with 5 users

Building Component




Floor Carpet 21 0.50 10.5Door Timber 2 0.10 0.2Ceiling Concrete 21 0.02 0.42Furniture Timber Desk 5 0.15 0.75

Padded Chair 3 0.77 2.31Users - 5 0.51 2.55Total Absorption 18.16

RT= (0.16 X 84) / 18.16

= 0.74s

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The reverberation time required for a library reading space is 0.6 – 0.8 seconds. Based on the calculation, the reverberation of the space under 500 Hz and 2000 Hz is 1s and 0.7s respectively. Acoustic panels are introduced to amend the issue.

2.2.3 Sound Reduction Index (Admin Office)

Volume of Space

V = L x W x H

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V = 6m x 3.5m x 4m

V = 84m3

Assuming that the noise level of an admin office is 40 dB, noise level of the external is 75dB. The sound reduction index of the wall should 75dB- 40dB = 35 dB.

Building Component Material Surface Area

(m2SRI

( Decibels)

Transmission Coefficient ( T

)Wall Concrete 65.6 46 2.51 x10−5

Wall Glass 4 35 3.16 x10−4

Door Plywood 2 31 7.94 x10−4

Concrete

SRI = 10 Log (1/T)

46 = 10 log (1/T)

3.98 x104 = 1 / T

T = 2.51 x10−5

Glass

SRI = 10 Log (1/T)

35 = 10 log (1/T)

3.16 x103 = 1 / T

T = 3.16 x10−4

Timber

SRI = 10 Log (1/T)

31 = 10 log (1/T)

103.1 = 10 log10 (1/T)

T = 7.94 x10−4

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Average Transmission Coefficient

= [65.6 x (2.51 x10−5) ] + [ 4 x ( 3.16 x10−4) ] +

[ 2 x (7.94 x10−4 ¿¿ / ( 65.6 + 4 + 2)

= 4.00 x 10−4

Overall SRI

= 10 Log ( 1 / 4.00 x 10−4)

= 33.97 Db

External Sound Pressure Level

= 75dB – 33.97dB

= 41.03 dB

The overall transmission loss from the back lane to the admin office is 33.97 dB. Based on the previous assumption, according to the noise criteria environment perception, the admin office exceeded the noise criteria, resulting in distracting noises interfering work environment. This can be solved by having double façade of wooden shutters which can filter and deflect noise coming from the back lane.

3.0 References

1. Cavanough, W.J. & Wikes, J.A. (1998). Architectural Acoustics: Principles and Practice. New York: Wiley and Sons.

2. Absorption Coefficient Building Material Finishes – Sengpiel Berlin (n.d.) Retrieved November 26 2015 fromhttp://www.sengpielaudio.com/calculator-RT60Coeff.htm

3. ACE Lamps Lighting (n.d.) Retreived December 6 2015 fromhttp://www.sengpielaudio.com/calculator-RT60Coeff.htm

4. Ir Izdihar, A. (n.d.) UBBL 2012 Amendments on EE Bylaw 38A and MS 1525:2014. Retrieved December 6 2016 fromhttp://www.sengpielaudio.com/calculator-RT60Coeff.htm

5. Ir Izdihar, A. (n.d.) UBBL 2012 Amendments on EE Bylaw 38A and MS 1525:2014. Retrieved December 6 2016 fromhttp://www.sengpielaudio.com/calculator-RT60Coeff.htm

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http://www.sengpielaudio.com/calculator-RT60Coeff.htm




B Science Report Project 2

Data & Analytics

Transcript of B Science Report Project 2