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Electrical Lighting &
Cable Selection
D.K.Pathirana
Part 1
Finding required number of luminaries
Types of Electrical Lighting
• Ambient lighting
• Task Lighting
• Accent lighting
Ambient lighting
Task lighting
Accent Lighting
Lumen(Luminous flux) [F]
• The lumen is defined such that…– At frequency of 540×1012 hz, (i.e., at wavelength
λ=555nm) 1 watt of radiant flux (power) will emit 683 lumens of light (luminous flux).
Illuminance (E)
• Defined as the luminous flux (lumens) falling on a surface per unit area normal to the surface.
• The unit of illuminance is the Lux
Luminaries
Required no. of Luminaries E x Lx W
N = _______________________ NL x x MF x UF x LLD
• E= Required Illuminance (Lux level)• L = Room length (m)• W = Room width (m)• NL = Number of lamps in each luminaries• N = Number of luminaries• = Lamp flux• M = Maintenance factor• UF = Utilization factor• LLD = Lamp Lumen Deterioration factor
Maintanance factor (MF)
• Can be defined as…
Average illuminance on the working plane
after a specified period of use of a lighting
installation to the average illuminance
obtained under the same conditions for a new
installation.
Lamps Lumen Deterioration (LLD)
• Manufacturer data….
• Examples.. Compact fluorescent 0.85 Mercury vapor 0.79 Metal halide 0.83 High pressure sodium 0.91
Reflectance factor
• the fraction of radiant energy that is reflected from a surface
• Examples- 70% , 0.8, ...• Here 10 times the reflection factor….. 1.Ceiling
2. Walls 3. Floor\Working plane.
Room Index
K = (LXW)/ [(H1-H0)X(L+W)]
Where,L = Room Length (m)
W = Room Width (m)
H1 = Height of luminaire’s light emitting surface
H0 = Height of the working plan
Utilization factor
Room Index(K)
Reflectance
873 753 731 551 531 311
0.60 0.39 0.31 0.27 0.29 0.26 0.24
0.80 0.45 0.37 0.32 0.35 0.32 0.29
1.00 0.49 0.42 0.36 0.38 0.36 0.33
1.25 0.53 0.46 0.40 0.42 0.39 0.37
1.50 0.56 0.49 0.42 0.44 0.41 0.39
2.00 0.59 0.53 0.46 0.47 0.45 0.43
2.50 0.61 0.56 0.48 0.49 0.47 0.45
3.00 0.63 0.57 0.49 0.50 0.49 0.47
4.00 0.65 0.60 0.51 0.52 0.51 0.49
5.00 0.66 0.62 0.53 0.53 0.52 0.50
Example for lighting calculation sheetLength Width
height of
height of the Room
Reflection
Utilization
Maintanance
Deterioration
Required
Installed
Lamp
Number of
Number of
Required Total Category of
luminaries
working plane index
factors factor factor factor
lux level flux
design lamps in
luminaires
luminaires load bulb
lumens
each luminaire
W
8.3857 5.318 3 0.5
1.3016967
7 3 10.52 0.85 0.8 400
50447.01
3350 2
7.5294038 8 288
LAMP, T8, 1200MM, 36W, 3500K
6.7564 4 3.75 0.5
0.7730811 7 3 1 0.42 0.85 0.8 400
37850.98
3350 2
5.6494001 6 216
15.2867 4.846 3 0.5
1.4718214
7 3 10.55 0.85 0.8 400
79229.25
3350 2
11.825261 12 432
4.0735 3.8 3.75 0.5
0.604923 7 3 1 0.36 0.85 0.8 400
25292.97
3350 2
3.7750707 4 144
3.3667 2.312 3.75 1
0.4984374
7 3 10.36 0.85 0.8 400
12718.64
3350 2
1.8983051 2 72
4.32 1.98 3 0.50.5430
8577 3 1
0.36 0.85 0.8 1505241.
176112
5 14.6588
235 5 90FT18W/2G11/830
4 3.22 3.75 0.50.5489
0267 3 1
0.36 0.85 0.8 1507892.
157261
0 13.0238
149 4 144FT36W/2G11/830
1386
Recommended Lighting LevelsActivity Illumination
(lux, lumen/m2)Public areas with dark surroundings 20 - 50Simple orientation for short visits 50 - 100
Working areas where visual tasks are only occasionally performed 100 - 150
Warehouses, Homes, Theaters, Archives 150
Easy Office Work, Classes 250
Normal Office Work, PC Work, Study Library, Groceries, Show Rooms, Laboratories 500
Supermarkets, Mechanical Workshops, Office Landscapes 750
Normal Drawing Work, Detailed Mechanical Workshops, Operation Theatres 1,000
Detailed Drawing Work, Very Detailed Mechanical Works 1500 - 2000
Performance of visual tasks of low contrast and very small size for prolonged periods of time 2000 - 5000
Load calculation
Part 2
• Design power socket outlets….. If not given• Then analyse….
Ground Floor QuantityLoad (Amp)
13A Socket Outlets 6 10*(1+(6-1)*0.2) 20.000
Florecent Fittings 2x36W 32 (36*2*32)/(0.7*230) 14.311Compact lamps (18W) 5 0.559Compact lamps (36W) 4 0.894Spot lights 4 0.745Shaver socket 1 4.000
Basic Lighting Load per Floor 40.509 Amp.
Per Phase load L-GR 13.503 Amp.
Total new Load for the new wing =49.29 Amp
Total KVA
=(49.29*1.732*400)/1000
=34.15 KVAAdd 10 % Extra Capacity = 37.6 KVA
AmpClosest Next is = 50 KVA 72.17
Total load for the new wing = Total load for ground floor+1st floor+2nd floor……
= 13.503+……+……
=49.29Amp
Bill Of Quantity (BOQ)• Example
Quantity Cost rate(Rs)
Total cost(Rs)
01 Phillips 20W CFL warm light
40 425 =40*425=17000
02 Pillips T8 Fluorescent lamp 18W
32 600 =32*600=19200
=17000+19200=36200
Part 3
Cable selection and Over load Protection
Basic types of MCB’s
• TYPE B
• TYPE C
• TYPE D
Sizing the cables• length of the wire • Check the following inequality
In
It ___________ and In Ib
Ca Ci Cg Cf
It = Maximum current capacity of cable for given temp.
Ib = Design current
In = Rated current of the protective device
Ca = Correction factor for ambient temperature.
CI = Correction factor for thermal insulation
Cg = Correction factor for grouping
Cf = Correction factor for semi enclosed fuse.
(for MCBs Cf = 1)
Contd…
• Refer IEE wiring regulation book• Lesser than 4% voltage drop….
Selecting a protective device
• Nature of the load• Rated current of the conductor• Environmental conditions
Contd..
• Ib rated current of protective divice Iz
Where,
Ib = Design service current
Iz = Maximum current carrying capacity of the conductor
• If 1.45 x Iz
Where,
Iz = Maximum current carrying capacity of the conductor
If = Thermal tripping current of MCCB
Single Line Diagram
Questions……
Thank you