building science-2-report

BUILDING SCIENCE 2 [BLD 61303/ ARC 3413]

PROJECT 2 – INTEGRATION PROJECT

SENTUL COMMUNITY LIBRARY

REPORT & CALCULATION

NAME : LEE YAUE SHEN STUDENT ID : 0315381 TUTOR : MR SIVA

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TABLE OF CONTENT 1.0 LIGHTING ANALYSIS

1.1 STAFF REST AREA

1.1.1 DAYLIGHT 5-6 1.1.2 ARTIFICIAL LIGHT 7 1.1.3 PSALI 8

1.2 QUIET STUDY AREA

1.2.1 DAYLIGHT 10-11 1.2.2 ARTIFICIAL LIGHT 12 1.2.3 PSALI 13

2.0 ACOUSTIC ANALYSIS

2.1 STAFF REST AREA 2.1.1 SPL ( SOUND PRESSURE LEVEL ) 15-16 2.1.2 RT ( REVERBERATION TIME ) 17 2.1.3 SPL (SOUND PRESSURE LEVEL) 18-20


2.2.1 SPL ( SOUND PRESSURE LEVEL ) 21-22 2.2.2 RT ( REVERBERATION TIME ) 23 2.2.3 SPL (SOUND PRESSURE LEVEL) 24-26

3.0 REFERENCES 27

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LIGHTING

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1.0 LIGHTING ANALYSIS Daylight Factor (DF) according to MS 1525

1.1 STAFF REST AREA

Zone Daylight Factors (%) Distribution Very Bright > 6 Very large with thermal and glare

problems Bright 3 – 6 Good Average 1 – 3 Fair Dark 0 - 1 Poor

GROUND FLOOR PLAN

DAYLIGHT

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1.1.1 DAYLIGHT

Daylight Factor Calculation Floor Area (m2) = (4.1m x 3.2m) + (2.9m x 1.4m) = 13.1 + 4.1 = 17.2 m2 Area of facade exposed to sunlight (m2) = 2.2m x 3m = 6.6 m2

Area of skylight = 0 Exposed Facade & Skylight Area to floor area ratio/ Daylight Factor, DF = (6.6+0) / 17.2 = 0.39 = 3.9%

Natural Illumination Calculation Illuminance Example

120,000 lux Very Bright Sunlight

110,000 lux Bright Sunlight

20,000 lux Clear sky

1000 - 2000 lux Overcast day

400 lux Sunrise/ sunset on clear day

<200 lux Midday

DAYLIGHT FACTOR DIAGRAM

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40 lux Fully overcast

<1 lux Sunset, Storm cloud

Natural Illumination Calculation E external = 20000 lux DF = E internal / E external x 100 3.9 = E internal / 20000 x 100 E internal = 3.9 x 20000 / 100

= 780 lux The selected stuff rest area has a daylight factor of 3.9% and natural illumination of 780 lux. Based on the requirements of MS1525.The space is good daylight source. However, for the purpose of reducing the thermal problem, louvers facade system is proposed for the facade design of the library. The vertical louvers will provide a well shading and reduce the amount of sunlight penetrating into the space. Hence, this design should be augmented with internal shades for low perpendicular sun angles. The low-e laminated glass is using as the inner facade of the space to provide privacy and well sound properties for user.

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1.1.2 Artificial Light Utilization Factor Table

According to MS 1525, the recommendation illumination level for staff rest area is 200 lux.

Lumen Method Calculation

Dimension of room (L x W) (4.1m x 3.2m) + (2.9m x 1.4m)

Total floor Area (m2) 17.2 m2

Room cavity height (m) 3.5

Reflectance values Ceiling=0.7 Wall= 0.5 Floor= 0.3

Room index, K K =

4.5 𝑥 4.1

3.5(4.5 + 4.1)= 0.6

Utilization Factor, UF 0.27

Model : DN570B (low height recessed version) Input : 230 or 240v / 50-60Hz Lumen (lm) : 2600 Weight (kg) : 2.2 Power : 36w (3000k)

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1.1.3 Psali

This is not a huge space that can fit many user, therefore there is only one switch needed for this room. The opening is huge enough to allow daylight to penetrate in and brighten up the area.

Maintenance Factor, MF Illuminance Requirement

0.8 200

Number of Luminaires

N =𝐸 𝑥 𝐴

𝐹 𝑥 𝑈𝐹 𝑥 𝑀𝐹

N =200 𝑥 17.2

2600 𝑥 (0.8 𝑥 0.27)= 6.12 = ~7 𝑏𝑢𝑙𝑏𝑠

Spacing to height ration (SHR)

𝑆𝐻𝑅 = 1

𝐻𝑚 𝑥 √

𝐴

𝑁

𝑆𝐻𝑅 = 1

3.5 𝑥 √

17.2

7= 0.45

𝑆𝐻𝑅 = 𝑆

3.5= 0.45

S = 1.57

Fitting Layout Fitting required along 4.5m wall = 4.5/1.57 = 2.87 = 3 rows Fitting required along 4.1 m wall = 4.1/1.57 = 2.61 = 3 rows

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Daylight Factor (DF) according to MS 1525

Zone Daylight Factors (%) Distribution Very Bright > 6 Very large with thermal and glare

problems Bright 3 – 6 Good Average 1 – 3 Fair Dark 0 - 1 Poor

SECOND FLOOR PLAN

DAYLIGHT

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1.2.1 DAYLIGHT

Daylight Factor Calculation Floor Area (m2) = 5.7 m x 7.4 m = 42.18 m2 Area of facade exposed to sunlight (m2) = 2 x (5.7 m x 4 m) + ( 7.4 m x 4 m ) = 75.2 m2

Area of skylight = 0 Exposed Facade & Skylight Area to floor area ratio/ Daylight Factor, DF = (75.2+0) / 42.18 = 1.78 = 17.8%

Natural Illumination Calculation Illuminance Example

120,000 lux Very Bright Sunlight

110,000 lux Bright Sunlight

20,000 lux Clear sky

1000 - 2000 lux Overcast day

400 lux Sunrise/ sunset on clear day

DAYLIGHT FACTOR DIAGRAM

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<200 lux Midday

40 lux Fully overcast

<1 lux Sunset, Storm cloud

Natural Illumination Calculation E external = 20000 lux DF = E internal / E external x 100 17.8 = E internal / 20000 x 100 E internal = 17.8 x 20000 / 100

= 3560 lux The selected quiet study area has a daylight factor of 17.8% and natural illumination of 3560 lux. Based on the requirements of MS1525, the space is too bright and it might also cause thermal problems. For the purpose of reducing the thermal and glare problem, Louvers facade system and curtain are proposed for the library. The vertical louvers will provide a well shading and reduce the amount of sunlight penetrating into the space. Besides that, the curtain can also reduce the direct daylight penetrate in to the space and hence reduce the thermal and glare problem effectively.

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1.2.2 Artificial Light Utilization Factor Table

According to MS 1525, the recommendation illumination level for quiet study area is 300 - 500 lux.

Lumen Method Calculation

Dimension of room (L x W) 5.7 m x 7.4 m

Total floor Area (m2) 42.18 m2

Room cavity height (m) 4

Reflectance values Ceiling=0.7 Wall= 0.5 Floor= 0.3

Room index, K K =

5.7 𝑥 7.4

4(5.7 + 7.4)= 0.8

Utilization Factor, UF 0.33

Model : DN570B (low height recessed version) Input : 230 or 240v / 50-60Hz Lumen (lm) : 2600 Weight (kg) : 2.2 Power : 36w (3000k)

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1.2.3 Psali - Permanent supplementary Artificial Lighting for Interior

There is 2 switch in this space. As the back of the space is facing outside, the daylight during day time is effective enough. Therefore, switch 2 can be turn off during daytime for energy saving.

Maintenance Factor, MF Illuminance Requirement

0.8 300 - 500

Number of Luminaires

N =𝐸 𝑥 𝐴

𝐹 𝑥 𝑈𝐹 𝑥 𝑀𝐹

N =300 𝑥 42.18

2600 𝑥 (0.8 𝑥 0.33)= 18.4 = ~19 𝑏𝑢𝑙𝑏𝑠

Spacing to height ration (SHR)

𝑆𝐻𝑅 = 1

𝐻𝑚 𝑥 √

𝐴

𝑁

𝑆𝐻𝑅 = 1

4 𝑥 √

42.18

19= 0.37

𝑆𝐻𝑅 = 𝑆

4= 0.37

S = 1.48

Fitting Layout Fitting required along 5.7m wall = 5.7/1.48 = 3.85 = 4 rows Fitting required along 7.4 m wall = 7.4/1.48 = 5 = 5 rows

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ACOUSTIC

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2.0 ACOUSTIC ANALYSIS 2.1 STAFF REST AREA

2.1.1 SRI (SOUND REDUCTION INDEX )

Volume of Space

V = L x W x H = 4.1 x 4.5 x 3.5

= 64.58 m3

Assuming that the noise level of a staff rest area is 40 dB , Noise level of the external is 70 dB

The Sound Reduction index of the wall should be 70 – 40 dB , which is 30 dB

GROUND FLOOR PLAN

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Building

Component

Material

Surface Area

(m2)

SRI ( Decibels)

Transmission

Coefficient ( T )

Wall Glass 6.6 35 3.16 𝑥 10 −4

Wall Concrete 53.6 46 2.51 𝑥 10 −5

Door Glass 2 35 3.16 𝑥 10 −4

Concrete

SRI = 10 Log ( 1/T )

46 = 10 Log ( 1/ T)

3.98 𝑥 10 4 = 1 / T

T = 2.51 𝑥 10 −5

Glass

SRI = 10 Log ( 1/ T )

35 = 10 Log ( 1 / T)

3.16 𝑥 10 3 = 1 / T

T = 3.16 𝑥 10 −4

Average Transmission Coefficient

= [ 6.6 x (3.16 𝑥 10 −4) ] + [ 53.6 x ( 2.51 𝑥 10 −4) ] + [ 2 x (3.16 𝑥 10 −4 ) ] / ( 6.6 +

53.6 + 2 )

= 2.6 𝑥 10 −4

Overall SRI

= 10 Log ( 1 / 2.6 𝑥 10 −4 ) = 35.85 dB

Conclusion

The overall SRI of the wall is 35.85 dB , which fulfils the required transmission loss of 30 dB

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2.1.2 RT ( REVERBERATION TIME )

Volume of Space

V = L x W x H = 4.1 x 4.5 x 3.5

= 64.58 m3

Material Absorption Coefficient under 500 Hz , with 4 users

Building

Component

Material Area S ( m2) Absorption

Coefficient (a)

Sound Absorption

( SA)

Wall Concrete 53.6 0.02 1.07

Glass 6.6 0.06 0.4

Floor Carpet 18.45 0.30 5.54

Door Timber 2.0 0.13 0.26

Ceiling Concrete 18.45 0.02 0.37

Furniture Plywood Table 3.37 0.13 0.44

Plywood Seating 1.44 0.13 0.19

Users - 4 0.46 1.84

Total Absorption 10.11

Reverberation Time, RT = ( 0.16 V) / A

RT= (0.16 X 64.58) / 10.11

=1.00s


Building

Component


Coefficient (a)

Sound Absorption

( SA)


Glass 6.6 0.03 0.2

Floor Carpet 18.45 0.50 9.23





Users - 4 0.51 2.04


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RT= (0.16 X 64.58) / 13.59 = 0.76s

Conclusion

The reverberation time required for a staff rest area is 0.4 - 0.6s

However, based on calculation , the reverberation of the space under 500 Hz and 2000 Hz is

1.00s and 0.76s respectively. As they slightly differ from the requirement, acoustic panels and

exterior facade are introduced to fix the issue.

2.1.3 SPL (SOUND PRESSURE LEVEL) Sound Pressure Level Formula :

SPL = 𝟏𝟎 𝒍𝒐𝒈 𝟏𝟎𝑰

𝑰𝒐

Where SPL = sound pressure level (dB), I = sound power (intensity)(watt) and Io = reference power. Io is usually taken as 1 x 10-12 watts.

Peak Hour (Back Alley) Highest Reading : 70 dB (moderate sound) Lowest Reading : 55 dB (quite sound) Highest Reading

70 = 10 log 10 𝑰

𝑰𝒐

Antilog 7 = 𝐈

𝟏 𝒙 𝟏𝟎−𝟏𝟐

1 x 107 = 𝐈

𝟏 𝒙 𝟏𝟎−𝟏𝟐

I = 107 x (𝟏 𝒙 𝟏𝟎−𝟏𝟐) I = 1 x 10-5

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Lowest Reading

55 = 10 log 10 𝑰

𝑰𝒐

Antilog 5.5 = 𝐈

𝟏 𝒙 𝟏𝟎−𝟏𝟐

1 x 105.5 = 𝐈

𝟏 𝒙 𝟏𝟎−𝟏𝟐

I = 105.5 x (𝟏 𝒙 𝟏𝟎−𝟏𝟐) I = 1 x 10-6.5

Total Intensities, I : = (1 x 10-5) + (1 x 10-6.5) =1.03 x 10-5

Combined SPL :

SPL = 10 log 10 𝑰

𝑰𝒐

= 10 log 10 (𝟏.𝟎𝟑 𝒙 𝟏𝟎−𝟓

𝟏 𝒙 𝟏𝟎−𝟏𝟐 )

= 70dB

Non - Peak Hour (Back Alley) Highest Reading : 55 dB (quiet sound) Lowest Reading : 45 dB (very quiet sound) Highest Reading

55 = 10 log 10 𝑰

𝑰𝒐

Antilog 5.5 = 𝐈

𝟏 𝒙 𝟏𝟎−𝟏𝟐

1 x 105.5 = 𝐈

𝟏 𝒙 𝟏𝟎−𝟏𝟐

I = 105.5 x (𝟏 𝒙 𝟏𝟎−𝟏𝟐) I = 1 x 10-6.5

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Lowest Reading

45 = 10 log 10 𝑰

𝑰𝒐

Antilog 4.5 = 𝐈

𝟏 𝒙 𝟏𝟎−𝟏𝟐

1 x 104.5 = 𝐈

𝟏 𝒙 𝟏𝟎−𝟏𝟐

I = 104.5 x (𝟏 𝒙 𝟏𝟎−𝟏𝟐) I = 1 x 10-7.5

Total Intensities, I : = (1 x 10-6.5) + (1 x 10-7.5) =3.48 x 10-7

Combined SPL :


𝑰𝒐

= 10 log 10 (𝟑.𝟒𝟖 𝒙 𝟏𝟎−𝟕

𝟏 𝒙 𝟏𝟎−𝟏𝟐 )

= 55dB

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2.2.1 SRI (SOUND REDUCTION INDEX)

Volume of Space

V = L x W x H = 5.7 x 7.4 x 4

= 168.72 m3

Assuming that the noise level of a quiet study area is 45 dB , Noise level of the external is 75

dB The Sound Reduction index of the wall should be 75 - 45 dB , which is 30 dB.

SECOND FLOOR PLAN

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Building

Component

Material

Surface Area

(m2)

SRI ( Decibels)

Transmission

Coefficient ( T )

Wall Glass 75.2 35 3.16 𝑥 10 −4

Wall Concrete 29.6 46 2.51 𝑥 10 −5

Door Glass 2 35 3.16 𝑥 10 −4

Concrete

SRI = 10 Log ( 1/T )

46 = 10 Log ( 1/ T)

3.98 𝑥 10 4 = 1 / T

T = 2.51 𝑥 10 −5

Glass

SRI = 10 Log ( 1/ T )

35 = 10 Log ( 1 / T)

3.16 𝑥 10 3 = 1 / T

T = 3.16 𝑥 10 −4

Average Transmission Coefficient

= [ 75.2 x (3.16 𝑥 10 −4) ] + [ 29.6 x ( 2.51 𝑥 10 −4) ] + [ 2 x (3.16 𝑥 10 −4 ) ] / ( 75.2 +

29.6 + 2 )

= 2.56 𝑥 10 −4

Overall SRI

= 10 Log ( 1 / 2.56 𝑥 10 −4 ) = 35.92 dB

Conclusion

The overall SRI of the wall is 35.92 dB , which fulfils the required transmission loss of 30 dB

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2.2.2 RT ( REVERBERATION TIME )

Volume of Space

V = L x W x H = 5.7 x 7.4 x 4

= 168.72 m3


Building

Component


Coefficient (a)

Sound Absorption

( SA)


Glass 75.2 0.06 4.5

Floor Carpet 42.18 0.30 12.65

Door Glass 2.0 0.03 0.06




Users - 24 0.46 11.04



RT= (0.16 X 168.72) / 31.84

=0.85s


Building

Component


Coefficient (a)

Sound Absorption

( SA)


Glass 75.2 0.03 2.26

Floor Carpet 42.18 0.50 21.09





Users - 24 0.51 12.24


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RT= (0.16 X 168.72) / 38.88 = 0.69s

Conclusion

The reverberation time required for a quiet study area is 0.4 - 0.6s

However, based on calculation , the reverberation of the space under 500 Hz and 2000 Hz is

0.85s and 0.69s respectively. As they slightly differ from the requirement, acoustic panels and

exterior facade are introduced to fix the issue.

2.2.3 SPL (SOUND PRESSURE LEVEL) Sound Pressure Level Formula :

SPL = 𝟏𝟎 𝒍𝒐𝒈 𝟏𝟎𝑰

𝑰𝒐

Where SPL = sound pressure level (dB), I = sound power (intensity)(watt) and Io = reference power. Io is usually taken as 1 x 10-12 watts.

Peak Hour (Back Alley) Highest Reading : 70 dB (moderate sound) Lowest Reading : 55 dB (quite sound) Highest Reading

70 = 10 log 10 𝑰

𝑰𝒐

Antilog 7 = 𝐈

𝟏 𝒙 𝟏𝟎−𝟏𝟐

1 x 107 = 𝐈

𝟏 𝒙 𝟏𝟎−𝟏𝟐

I = 107 x (𝟏 𝒙 𝟏𝟎−𝟏𝟐) I = 1 x 10-5

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Lowest Reading

55 = 10 log 10 𝑰

𝑰𝒐

Antilog 5.5 = 𝐈

𝟏 𝒙 𝟏𝟎−𝟏𝟐

1 x 105.5 = 𝐈

𝟏 𝒙 𝟏𝟎−𝟏𝟐

I = 105.5 x (𝟏 𝒙 𝟏𝟎−𝟏𝟐) I = 1 x 10-6.5

Total Intensities, I : = (1 x 10-5) + (1 x 10-6.5) =1.03 x 10-5

Combined SPL :


𝑰𝒐

= 10 log 10 (𝟏.𝟎𝟑 𝒙 𝟏𝟎−𝟓

𝟏 𝒙 𝟏𝟎−𝟏𝟐 )

= 70dB

Non - Peak Hour (Back Alley) Highest Reading : 55 dB (quiet sound) Lowest Reading : 45 dB (very quiet sound) Highest Reading

55 = 10 log 10 𝑰

𝑰𝒐

Antilog 5.5 = 𝐈

𝟏 𝒙 𝟏𝟎−𝟏𝟐

1 x 105.5 = 𝐈

𝟏 𝒙 𝟏𝟎−𝟏𝟐

I = 105.5 x (𝟏 𝒙 𝟏𝟎−𝟏𝟐) I = 1 x 10-6.5

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Lowest Reading

45 = 10 log 10 𝑰

𝑰𝒐

Antilog 4.5 = 𝐈

𝟏 𝒙 𝟏𝟎−𝟏𝟐

1 x 104.5 = 𝐈

𝟏 𝒙 𝟏𝟎−𝟏𝟐

I = 104.5 x (𝟏 𝒙 𝟏𝟎−𝟏𝟐) I = 1 x 10-7.5

Total Intensities, I : = (1 x 10-6.5) + (1 x 10-7.5) =3.48 x 10-7

Combined SPL :


𝑰𝒐

= 10 log 10 (𝟑.𝟒𝟖 𝒙 𝟏𝟎−𝟕

𝟏 𝒙 𝟏𝟎−𝟏𝟐 )

= 55dB

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3.0 REFERENCES

1. (2016) (1st ed.). Retrieved from

http://www.utm.my/energymanagement/files/2014/07/MS- 1525-2007.pdf

2. Association of Australian Acoustical Consultants Guideline for Educational Facilities Acoustics. (2016) (1st ed.). Retrieved from http://file:///E:/Downloads/AAAC-Guideline-for-Educational- Facilities-Acoustics-2010%20(1).pdf