SCHOOL OF ARCHITECTURE, BUILDING & DESIGNBachelor of Science (Honours) (Architecture)

Building Science 2 [BLD 61303/ ARC 3413]PROJECT 2 - Integration Project

NAME: GARNETTE DAYANG ROBERT STUDENT ID: 0315491 TUTOR: MR.EDWIN

TABLE OF CONTENT

1.0 Introduction

2.0 Lighting Proposal

2.1 Natural Daylighting (Daylighting Factor) 2.1.1 Reading Area 1 2.1.2 Reading Area 2

2.2 Artificial Lighting (Lumen Method)

2.2.1 Multipurpose Hall 2.2.2 Computer Room

3.0 Acoustic Proposal

3.1 External Noise Calculations

3.2 Management Office 3.2.1 Reverberation Time 3.2.2 Transmission Loss

4.0 References

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1.0 Introduction

Integrating with the Architectural Design Studio 5, this project requires us to calculate the optimum lighting and acoustic proposal for the spaces in the building. The building is a Community Library which is located in an urban site, Sentul, Kuala Lumpur. Within the building design, we are required to identify spaces to integrate our external lighting and day-lighting. As for the acoustics, we are required to identify spaces to integrate our external noises and internal noises. For each category, at least one space is being chosen.

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Task Illuminance (Lux)

Example of Application

Lighting for infrequently used area

20

100

100

100

100

150

100

100

100

300

200

Minimum service illuminance

Interior walkway and car-park

Hotel bedroom

Lift Interior

Corridor, passageway, stairs

Escalator. Travellator

Entrance and exit

Staff changing room, locker and cleaner room

Entrance hall, lobbies, waiting room

Inquiry desk

Gate house

Lighting for working interiors 200

300 – 400

300 – 400

150

200

150 – 300

150

150

100

100

300 – 500

200 – 750

300

Infrequent reading and writing

General offices, shops, and stores, reading and writing

Drawing office

Restroom

Restaurant, Canteen, Cafeteria

Kitchen

Lounge

Bathroom

Toilet

Bedroom

Classroom, Library

Shop, Supermarket, Department store

Museum and gallery

Table 1 Recommended Average Illuminance Levels

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2.0 Lighting Proposal

2.1 Natural Lighting 2.1.1 Reading Area 1

Formula : Daylight Factor (DF) = (Ei / Eo) x 100%

Area of Space (m2) 170

Area of Curtain Wall (m2) (18.6 x 4)+(6.1 x 4)+(5.9 x 4)+(17.8 x 4) =172.6

Daylight Factor (%) (172.6 / 170) x 100%

= 101.5 % x 0.1

= 10.15 %

The reading area receives 10.15% amount of daylight factor. This reading area receives more light than it is required (based on the requirement of MS1525), which is between 3% to 6%. This is most likely to be caused by the the majority of curtain walls wrapping this area. Curtain or blinds can be used in order to reduce the amount of daylight coming through. In my studio design, I have incorporated the vertical louvers on the façade of my building so the amount of daylight penetrating can be controlled through the rotating of the louvers. Another possible way to reduce the thermal and glare issues is by replacing some parts of the curtain wall with solid wall and thus reducing the curtain wall area.

Zone DF (%) Distribution

Very Bright >6 Very large with therrmal and glare problem

Bright 3 – 6 Good

Average 1 – 3 Fair

Dark 0 - 1 Poor

Table 2 Daylight factors and distribution (Department of Standards Malaysia, 2007)

DAYLIGHT FACTOR CALCULATION

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Formula EEi = (DF x Eo) /100%

Given, Eo (unobstructed sky of Malaysia) 20000

EEi = (DF x Eo) / 100%

= (10.15 x 20000) / 100%

= 2030 lux

The general illuminance level of an exact proof reading place is 500 lux. The final measure-ment of daylighting for this reading area is set at 2030 lux, which is considerably higher than the required range. This result shows that that this particular reading area receives a little too much daylight during the day. So there are a number of solutions that can be taken in order to rectify the issues that have been mentioned.

NATURAL ILLUMINATION CALCULATION

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2.1.2 Reading Area 2

Formula : Daylight Factor (DF) = (Ei / Eo) x 100%

Area of Space (m2) 198.6

Area of Curtain Wall (m2)

Daylight Factor (%)

This reading area receives 5.0% amount of daylight factor. It complies with the current requirement of MS 1525 (between 3% to 6%), which can be considered to be a good distribution. This is mainly due to the positioning of the windows where most are facing either to the north or to the south. This naturally reduces the amount of direct sunlight that penetrates through into the space area.

Zone DF (%) Distribution

Very Bright >6 Very large with therrmal and glare problem

Bright 3 – 6 Good

Average 1 – 3 Fair

Dark 0 - 1 Poor

Table 3 Daylight factors and distribution (Department of Standards Malaysia, 2007)

(1.7 x 4) + (12.1 x 4) + (1.9 x 4) + ( 6.8 x 4) + (2.4 x 4) = 99.6

(99.6 / 198.6) x 100%= 50.15% x 0.1= 5.0 %

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Formula EEi = (DF x Eo) /100%

Given, Eo (unobstructed sky of Malaysia) 20000

EEi = (DF x Eo) / 100%

= ( 5.0 x 20000) / 100%

= 1000 lux

The general illuminance level of an exact proof reading place is 500 lux. As shown above, the final measurement of daylighting for this reading area is 1000 lux, which is twice as much as the general illuminance level of the required range. This result shows that this particular reading area receives a little too much daylight during the day. Despite that, this is still considered as a good result because they are still within the acceptable range of the general illuminance level.

NATURAL ILLUMINATION CALCULATION

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2.2 Artificial Lighting 2.2.1 Multipurpose Hall

The Multipurpose Hall is a space that requires multi-functional lighting. Both natural and artificial lighting should be sufficient provided for various possible uses. It is located on the highest floor with some parts enclosed with solid walls and most of the other parts enclosed by curtain walls. Artificial lights are still an important element to be considered in order to have an efficient lighting setting. The type of light to lit up this space would be the White Fluorescent Tube with Reflector.

White Fluorescent Tube with Reflector

Type of Light Bulb T5 High Output

Material of Fixture Aluminium

Wattage 39W

Light Color Cool White

Nominal Life (Hours) 36000

Color Temperature 3000 K

Color Rendering Index (CRI) 85

Lumens 3500

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CALCULATIONS

Location

Dimensions (m)

Total Floor Area,

Standard Illuminance Required (lux) according to MS1525, E

Multipurpose Hall

Length (L) = 15.8Width (W) = 12.9Height of the ceiling = 5.1 m

15.8 x 12.9 = 203.82

500

Lumen of lighting fixtures, lm 3500

Height of luminaire (m)

Work level (m)

Mounting Height, Hm (m)

3.8

0.8

3

Assumption of reflective

Room Index, RI (K)

K=(L x W)/(L+W)hm

Ceiling: 0.3 Wall: 0.5 =(15.8 x 12.9 )/(15.8+12.9)3

=2.36

Utilization Factor, UF 0.65

Maintenance Factor, MF 0.8

Lumen CalculationN= (E x A)/(F x UF x MF)

=(500 x 203.82)/(3500 x 0.62 x 0.8)

=59 bulbs

Number of luminaires across

√((L x N)/W)

=√((15.8 x 59)/12.9)

=8.5 (take 9)

Therefore, each spacing would be 15.8÷9=1.75m

Number of luminaires along

√((W x N)/L)

=√((12.9 x 59)/15.8)

=6.9 (take 7)

Therefore, each spacing would be 12.9÷7=1.84m

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2.2.2 Computer Room

This computer room is a combination of a few different spaces, which are the discussion area and the study area. As a result, the aid from artificial light is needed to provide sufficient amount of illuminance. Artificial light is important as it needs to deliver a suitable reading environ-ment for the users and can be used to complement the lighting of the area when natural daylight is not sufficient. Philips Corepro LED Lamp is used to light up this space.

Philips 15W Cool White Fluorescent Tube

Type of Light Bulb T5 High Output

Voltage 230V

Wattage 15W

Light Color Cool White

Luminous Flux 960 lm

Color Temperature 27000 K

Color Rendering Index (CRI) 80

Light Output 960

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CALCULATIONS

Location

Dimensions (m)

Total Floor Area,

Standard Illuminance Required (lux) according to MS1525, E

Computer Room

Length (L) = 11.5Width (W) = 10.2Height of the ceiling = 5.1 m

11.5 x 10.2 = 117.3

500

Lumen of lighting fixtures, lm 2600

Height of luminaire (m)

Work level (m)

Mounting Height, Hm (m)

3.8

0.8

3

Assumption of reflective

Room Index, RI (K)

K=(L x W)/(L+W)hm

Ceiling: 0.3 Wall: 0.5 =(11.5 x 10.2)/(11.5+ 10.2 )3

=1.8

Utilization Factor, UF 0.63

Maintenance Factor, MF 0.8

Lumen CalculationN= (E x A)/(F x UF x MF)

=(500 x 117.3)/(2600x 0.62 x 0.8)

=46 bulbs

Number of luminaires across

√((L x N)/W)

=√((11.5 x 46)/10.2)

=7.2 (take 7)

Therefore, each spacing would be 11.5÷7=1.64m

Number of luminaires along

√((W x N)/L)

=√((10.2 x 46)/11.5)

=6.4 (take 6)

Therefore, each spacing would be 10.2÷6=1.7m

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3.0 ACOUSTIC Proposal

3.1 External Noise Calculation Assuming the external noise sources,

Traffic Noise : 70dBActivity Noise from 5 foot path : 40dB Back Lane : 30dB

Intensity for Traffic Noise

Intensity for Activity Noise

Intensity for Back Lane Noise

70 = 10 log (Itraffic/ Io)

Antilog 7.0 = Itraffic/ 1 x 10-12

Itraffic = 1 x 10-5

40 = 10 log (Itraffic/ Io)

Antilog 4.0 = Itraffic/ 1 x 10-12

Itraffic = 1 x 10-8

30 = 10 log (Itraffic/ Io)

Antilog 3.0 = Itraffic/ 1 x 10-12

Itraffic = 1 x 10-9

Total Intensities, ITotal = 1 x 10-5 + 1 x 10-8 + 1 x 10-9 = 1.1 x 10-5 Combined SPL, = 10 log (ITotal/ Io) = 10 log (1.1 x 10 -5 / 1 x 10-12)= 70dB The external noise with a combined SPL of 70dB generated from traffic noise, street activity noises and noises from the back lane travels into the community library compound without passing through walls that will cause transmission loss. Therefore, the combined SPL of the external noise is approximately the total sound pressure level in the library.

In conclusion, the required sound pressure level for a library is set at 35dB, but the sound pressure level calculated has extremely exceeded that level at 70dB. Appropriate and necessary actions has to be taken in order to reduce the sound pressure level into the community library.

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3.2 Management Office 3.2.1 Reverberation Time Calculation

Reverberation time is deliberated to define the space quality in relation to human comfort. Hence, the space chosen here is the management office because it is located next to an area with much noise which is the loading bay and the back alley. Thus, an optimum acoustic surrounding must be provided to reduce the noise level in this area.

Management Office

Room Height : 3.5 mPeak Hour Capacity : 10 peopleVolume of Office, (V): 67.9m x 3.5m = 237.6 m3

Components Materials Area (m2) Absorption Coefficient (2000Hz)

Area of Absorption

Coefficient, (A)

Wall Concrete 63.72 0.05 3.2

Floor Concrete 67.9 0.02 1.4

Ceiling Plaster 67.9 0.02 1.4

Window and Door Glass 59.5 0.07 4.2

Chair Fabric 0.98 (10pax) 0.28 (per seat) 3

Occupants 10 pax 0.46 4.6

Total 17.8

RT = 0.16 V/A = 0.16 (237.6 / 17.8) = 2.0 s

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Conclusion

The reverberation time for the management office during peak hours is calculated to 0.75 s, which has exceeded the optimum reverberation time for the management office. Thus, to reduce the reverberation time, drapery curtains can be added against the glass wall or carpets also placed on the floors. Another alternative is to place and plant some trees and other various plants on the exterior of the building or install a few panel absorbers to effectively to lower the frequencies.

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3.2.2 Transmission Loss Calculation

The management office is located on the ground floor surrounding the building from the back, to provide some kind of protection to the community library. As a result, it is exposed to the noises coming from the back lane and therefore requires a good partition to filter and absorb the sound waves before it enters into the space.

TL = 10 Log (1 /Tav )

Tav = [( T1 × A1 )+(T2 × A2)+(T3 × A3)+...(Tn × An)] /Total Surface Area Tcn = the transmission coefficient of material

Sn = the surface area of the material n

Assuming the external noise is 70dB, the minimum interior noise required for an office is 40dB. Therefore, the transmission loss should be 30dB.

Components Materials Area (m2)

Curtain Wall Glass 59.5

Wall Concrete 63.7

Sound Reduction Index (SRI)

26

42

Transmission Coefficient of Materials

Curtain Wall

SRI (glass) = 10 log (1/ Tglass) 26 = 10 log (1/ Tglass) Antilog 2.6 = 1/ Tglass

Tglass = 2.5 x 10-3

Wall

SRI (concrete) = 10 log (1/ Tconcrete) 42 = 10 log (1/ Tconcrete) Antilog 4.2 = 1/ Tconcrete

Tconcrete = 6.31 x 10-5

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Average Transmission Coefficient of Materials

Tav = (S1Tc1 + S2Tc2......SnTcn)/(Total Surface Area)

Tav = ((2.5 x 10-3) + (6.31 x 10-5)/(59.2+63.7)

= 2.08 x 10-5

Total Sound Reduction Index, SRI

SRI = 10 log (1/Tav)

= 10 log (1/2.08 x 10-5)

= 47 dB

As proven in the calculation, the external noise from the back lane, which was initially at 70dB is reduced by 47dB during the transmission, which resulted in a sound level of 23dB when it reaches the office interior. This noise level (23 dB) is lower than the range of recommended level for an office. Hence, the acoustical comfort can be achieved by having walls as external sound barriers.

Noise Level In the Office

70dB - 47dB = 23dB

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LAYOUT OF LIGHTING FIXTURES

Conclusion

The reverberation time for the multipurpose hall during peak hours is 2.0 s, which has exceeded the optimum reverberation time for a multipurpose hall. Thus, to reduce the reverberation time, drapery curtains can be added against the glass wall or carpets placed on the floors. The other alternative is to install a few acoustic panels since the hall would probably be used as a space involving some form of public speaking using mcirophones and speakers as medium to amplify sound.

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