Report on detailed calculation in sync with the A1 board

1. Schoolof Architecture, Building & Design
Bachelor of Science (Hons) in Architecture
Building Science 2 [BLD 61303]
__________________________________________________________________
PROJECT 2: INTEGRATIONPROJECT
Name : Wong Kai Chiang
Student ID : 0323341
Lecturer : Mr. Edwin Chan

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Content
1.0 Introductiontosite…………………………………………………..………………………………………………..…………………3 - 4
2.0 Space 1: IT Section………………………………………………...………………………………………………..…………………5 - 11
3.0 Space 2: Staff’sOffice…………………………………………………………………………………………..………………..…12 - 16

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1.0 Introduction to Site
The site of thisassignmentislocatedatJalanTuankuAbdul Rahman(TAR),withcoordinatesof 3.153479,
101.696472, where it is an elongated site between two buildings. The one on the left is a 2 storey
traditional shophouse,where onthe rightisa modern13-storeybuilding.The difference inheightcanbe
referredtothe sunpath diagramsbelow.

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Figure:Sunpositionat10am
Figure:Sunpositionat2pm
Figure:Sunpositionat6pm

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2.0 Space 1: IT section
Figure:Secondfloorplan1:150 mts
Space 1

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Figure:Figure showssectionalperspective view of the selectedspace,the ITsection.
Daylightfactor= W/A x Tθ/ (1-R), where,
W (Areaof window)
W = 5m x 8.05m
= 40.25m2
A (Total area of internal surface)
A = (4.82 x 7.91 x 2) + (
5.6+7.91
2
x 3.23 x 2) + (8.05 x 5 x 2) + (4.82 x 5 x 2) + (7.91 x 5) + (5.6 x 2)
= 76.25 + 3.64 + 80.5 + 48.2 + 39.6 + 11.2
= 299.4m2
T (Transmission) –Double glazedwindow panel=0.7 approx.
Θ (Visible skyangle)=80°
R (Reflectance)=0.2 (woodmediumcolor)
Therefore,DF = W/A x Tθ/ (1-R)
= (40.25/ 299.4) x (0.7) (80)/ (1-0.2)
= 0.1344 x 70
= 9.408%

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The value of 9.4% isconsideredverybrightandmightcause uncomfortable glare andthermal problems.
After showing the daylight factor by using this calculation method, the internal illuminance is then
calculatedusinganotherformula,whichisstatedbelow,
DF =
Indoor illuminance,Ei
Outdoor illuminance,Eo
x 100%
Both indoorandoutdoorilluminance value isinlux,where the DFwill be the daylightfactorthatis
calculatedbefore this.Busubbinginthe values:
DF = 9.408%
Outdoorilluminance,Eo(External illuminationforovercastsky(lux)) =20,000 lux
Therefore,DF=
Indoor illuminance,Ei
Outdoor illuminance,Eo
x 100%
9.408% = Ei/20,000 x 100%
Ei = 200 x 9.408
= 1881.6 lux
Figure:Figure showsthe daylightingcontourof the space.

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The value of 1881.6 lux isobtained,whichisextremelyhigh,thismaybe due tothe bigareaof openingat
the surroundedside of the space,because itis a double skinfaçade system.Therefore,there are louvers
installedoutside thewindowpanels,whichformsanotherfaçadeandalsoashadingdevice tothe building
itself.Furthermore,the louverscanbe adjustedas well tomeetwiththe differentneedsof natural light
at differenttime.
Asdaylightwouldnot be alwaysthere,orthere mightbe achange of weather,artificiallightsare needed
as well. Through the lumen method calculation, the numbers of lights needed within the spacescan be
calculated,sothatthe artificial lightscanbe placesatthe right place anddistribute evenlyinaspace.The
lumenmethodcalculatedisasfollows:
Type of fixture Surface mountedLED downlight
Type of lightbulb LED
Material of fixture Aluminiumhousing
Polycarbonate reflector&opticmaterial
Productbrand PhilipsCorelineslimdownlight
Productcode DN135B LED20S/840 PSU II WH
Nominal life (hours) 50,000 hourswith2000 lux
Wattage range (W) available at13W &28W
CRI ≥80
Colortemperature (K) 4000K
Colordesignation Cool white

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First, some details of the lighting are researched and tabulated. Some details will be needed in the
calculationlater.Lumenmethodcalculationisshownbelow:
N = (E x A)/ F x UF x MF,where
N (Numberof lampsrequiredinthe space)
E (Requiredlux inaIT section) =300 lux
* StandardilluminationrecommendedbyMS1525
A (Areaat workingplane height)
A = (4.82 x 7.91) + (
5.6+7.91
2
x 3.23)
= 38.13 + 21.82
= 59.95 m2
≈ 60m2
F (Initial luminiousfluxfromeachlamp) =2000 lux
* Basedonthe productdetails
UF (Utilizationfactor),tofindthe value,roomindex(K) isneeded,formulaasfollows,
* L (Lengthof the space) W (Widthof the space) Hm (Mountedheightof fittingabove workingplane)
Roomindex = (L x W)/ Hm x (L+ W)
= (7.91 x 8.05)/ (4.5 -1) x (7.91 + 8.05)
= 63.68/ (3.5 x 15.96)
= 1.14
Reflectance forceiling(0.8),wall (0.8) andworkingplane (0.3)
With all the values obtained here, room index and also the reflectance of the surfaces, refer to the UF
table to obtainthe nearestvalue,inthiscase,UF= 0.63 approx.
MF (Maintenance factor) =assume 0.8

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So therefore,the numberof lightsneededinthe space is,
N = (E x A)/F x UF x MF
= (300 x 60)/ (2000 x 0.53 x 0.8)
= 18,000/ 848
= 21.22 ≈ 22 lampsneeded
It is shown in the calculation that 22 lamps are to be placed in the space to provide enough illuminance
for the user who’s using the space. To calculate the distance between each lamp,the following formula
can be used:
SHR (Space to heightratio) = 1/ Hm x √A/N
In which, Hm (Mounted height of fitting above working plane), A (Area of space), N (Number of light
neededinthe space).
Therefore,SHR= 1/3.5 x √63.68/ 22 SHR = s/ 3.5 = 0.486
= 0.2857 x 1.7 therefore s= 1.701 ≈ 1.7
= 0.486
The ratio will be 1.7.
Therefore,onthe 8045mm wall,
8.045/ 1.7 = 4.73 ≈ 5 rows
Where on the 8130mm wall,
8.13/ 1.7 = 4.78 ≈ 5 rows
For spacingalong8045mm wall,
8.045/ 5 = 1.609 ≈ 1.6m
For spacingalong8045mm wall,
8.13/ 5 = 1.63 ≈ 1.6m
Based on the result above, it is shown that there will be a total number of 25 lamps (5 x 5 rows), which
exceedsoutcalculatednumberof lightsbefore this.Therefore,thereshouldbe more thanenoughlamps
inthe space for userto comfortablyuse them.

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Figure: Plan above shows the lighting arrangement for IT section, each lamps are spaced with
1.6m distance.

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3.0 Space 2: Staff’s Office
Figure:Secondfloor
Space 2

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Figure:Figure showssectionalperspective of staff’soffice.
The staff’s office is totally enclosed, natural sunlight is hard to get through and lights up the space,
therefore, artificial lights are needed for the space. The numbers of lights needed within the spaces can
be calculatedthroughthe lumenmethodcalculationsothatthe artificial lightscanbe placesat the right
place and distribute evenlyinaspace.The lumenmethodcalculatedisasfollows:
Type of fixture Recessed LED lighting
Type of light bulb LED
Material of fixture Pre-painted steel housing
Product brand Philips Coreline panel office compliant
Productcode RC127V LED34S/840 PSU W60L60 OC
Nominal life (hours) 50,000 hours at 3400 lm
Wattage range (W) 41W
CRI ≥80
Color temperature (K) 4000K
Color designation Cool white

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The lighting chosen this time is the coreline panel office compliant, made just for the office
spaces and provide excellent lighting. Number of lights is calculated below:
N = (E x A)/ F x UF x MF,where
N (Numberof lampsrequiredinthe space)
E (Requiredlux inastaff’soffice) =300 lux
* StandardilluminationrecommendedbyMS1525
A (Areaat workingplane height)
A = 4 x 6.4
= 25.6 m2
F (Initial luminiousfluxfromeachlamp) =3400 lux
* Basedonthe productdetails
UF (Utilizationfactor),tofindthe value,roomindex(K) isneeded,formulaasfollows,
* L (Lengthof the space) W (Widthof the space) Hm (Mountedheightof fittingabove working plane)
Roomindex = (L x W)/ Hm x (L+ W)
= (6.4 x 4)/ (4.5 -1) x (6.4 + 4)
= 25.6/ (3.5 x 10.4)
= 0.703
Reflectance forceiling(0.8),wall (0.8) andworkingplane (0.3)
With all the values obtained here, room index and also the reflectance of the surfaces, refer to the UF
table to obtainthe nearestvalue,inthiscase,UF= 0.4 approx.
MF (Maintenance factor) =assume 0.8
So therefore,the numberof lightsneededinthe space is,
N = (E x A)/ F x UF x MF
= (300 x 25.6)/ (3400 x 0.4 x 0.8)
= 7680/ 1088
= 7.05 ≈ 8 lampsneeded

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It isshownin the calculationthat8 lampsare to be placedinthe space to provide enoughilluminance for
the userwho’susingthe space.To calculate the distance betweeneachlamp,the followingformulacan
be used:
SHR (Space to heightratio) = 1/ Hm x √A/N
In which, Hm (Mounted height of fitting above working plane), A (Area of space), N (Number of light
neededinthe space).
Therefore,SHR= 1/3.5 x √25.6/ 8 SHR = s/ 3.5 = 0.511
= 0.2857 x 1.789 therefore s= 1.789 ≈ 1.8
= 0.511
The ratio will be 1.8.
Therefore,onthe 6400mm wall,
6.4/ 1.8 = 3.56 ≈ 4 rows
Where on the 4000m wall,
4/ 1.8 = 2.22 ≈ 2 rows
For spacingalong6400mm wall,
6.4/ 4 = 1.6m
For spacingalong4000mm wall,
4/ 2 = 2m
Based on the result above, it is shown that there will be a total number of 8 lamps (4 x 2 rows), which
exceedsoutcalculatednumberof lightsbefore this.Therefore,thereshouldbe more thanenoughlamps
inthe space for userto comfortablyuse them.

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Figure: Figure shows the lighting arrangement for the staff’s office.