Thumb Rule-11


 

Size of Cable on Secondary Side of Transformer (11KV/433V)
Ref: KSEI Handbook
Rating of T/C (KVA) Primary current (Amp) Secondary Current (Amp) Min. Size of Neutral Earthing Conductor (mm2) Minimum Size of Cable (mm2)
63 3.3 84 25X3 50mm2
100 5.25 133.3 25X3 95mm2 or (2×50 mm2)
160 8.4 213.3 25X3 185mm2 or (2×95 mm2)
200 10.49 266.6 25X3 300mm2 or (2×120 mm2)
250 13.12 333 25X3 2×185 mm2
315 16.53 420 31X3 or 25X4 (2×300 mm2) or (3×185 mm2)
400 21.80 533 38X3 (3×300 mm2) or (2×400 mm2)
500 26.20 666.5 25X6 (3×400 mm2) or (4×240 mm2)
630 33 840 31X6 4×400 mm2
750 39.36 1000 50X4 Bus Bar Trucking (min. Isc 50KA)
1000 52.50 1333 210mm2 Bus Bar Trucking (min. Isc 50KA)
1250 65.50 1667 290mm2 Bus Bar Trucking (min. Isc 50KA)
1600 83.98 2133 380mm2 Bus Bar Trucking (min. Isc 50KA)
2000 105.00 2666 450mm2 Bus Bar Trucking (min. Isc 50KA)

 

HT Fuse on Primary Side of Transformer (11KV/433V)
Rating of T/C (KVA) Primary current (Amp) Secondary Current (Amp) HT Fuse
Min (Amp) Max(Amp)
63 3.3 84 10 16
100 5.25 133.3 16 25
160 8.4 213.3 16 40
200 10.49 266.6 25 40
250 13.12 333 32 40
315 16.53 420 40 63
400 21.80 533 40 63
500 26.20 666.5 50 100
630 33 840 63 100
750 39.36 1000 75 160
1000 52.50 1333 100 160
1250 65.50 1667 100 200
1600 83.98 2133 160 250
2000 105.00 2666 200 250

 

Capacitor Bank for Power Supply Voltage
System Voltage Minimum rating of capacitor bank
3.3 KV , 6.6KV 75 Kvar
11 KV 200 Kvar
22 KV 400 Kvar
33 KV 600 Kvar

 

Capacities of PVC conduits
Nominal conductor Size mm 16 mm 20 mm 25 mm 32 mm
Number of Cables (maximum)
1.0 6 5 19 30
1.5 5 4 15 24
2.5 3 3 11 17
4 2 2 8 13
6 2 - 6 10
10 - - 4 6
16 - - 3 4
25 - - 2 3
35 - - - 2

 

System Highest and Lower Voltage
Ref: NEC(India) :2011
System Voltage Highest Voltage Lowest Voltage
240 V 264 V 216 V
415 V 457 V 374 V
3.3 kV 3.6 kV 3.0 kV
6.6 kV 7.2 kV 6.0 kV
11 kV 12 kV 10 kV
22 kV 24 kV 20 kV
33 kV 36 kV 30 kV
66 kV 72.5 kV 60 kV
66 kV 72.5 kV 60 kV
132 kV 145 kV 120 kV
220 kV 245 kV 200 kV
400 kV 420 kV 380 kV

 

Number of Points for Dwelling Unit
Ref: NEC(India) :2011
No. Description Area for the Main Dwelling Unit (m2)
35 mm2 45 mm2 55 mm2 85 mm2 140 mm2
1 Light points 7 No 8 No 10 No 12 No 17 No
2 Ceiling fans Pont 2 No 3 No 4 No 5 No 7 No
3 Ceiling fans No’s 2 No 2 No 3 No 4 No 5 No
4 6A Socket outlets 2 No 3 No 4No 5 No 7 No
5 16A Socket outlets - 1 No 2 No 3 No 4No
6 Call-bell (buzzer) - - 1 No 1 No 1 No

 

Recommended Schedule of Socket-Outlets
Ref: NEC(India) :2011
Description Number of Socket
6A Socket 16A Socket
Bedroom 2 1
Living room 2 2
Kitchen 1 2
Dining room 2 1
Garage 1 1
For refrigerator - 1
For air-conditioner - 1 for each
Verandah 1 per 10mter2 1
Bathroom 1 1

 

Power requirements of the building
Ref: NEC(India) :2011
Part of ElectricalInstallation Part of the Total Power Requirement in % DiversityFactor
Ventilation, heating (air-conditioning) 45% 1.0
Power plant (drives) 52% 0.65
Lighting 30% 0.95
Lifts 20% 1.0
Kitchen 10% 0.6
Laundry 5% 0.6

 

Lift Car Speed
Ref: NEC(India) :2011
Occupancy No. of Floors Served Car Speed   m/s
Office building 4 to 5 0.5 to 0.75 m/sec
Office building 6 to 12 0.75 to 1.5 m/sec
Shops and departmental stores 13 to 20 More than 1.5 m/sec
Passenger lifts for low and medium lodging houses - 0.5 m/sec
Hotels 4 to 5 0.5 to 0.75 m/sec
Normal load carrying lifts - 2.0 to 2.5 m/sec
Hospital passenger Lift 4 to 5 0.5 to 0.75 m/sec
Hospital passenger Lift 13 to 20 More than 1.5 m/sec
Hospital bed lifts (Short travel lifts insmall hospitals) - 0.25 m/sec
Hospital bed lifts (Normal) - 0.5 m/sec
Hospital bed lifts (Long travel lifts inGeneral hospitals)   0.6 to 1.5 m/sec

 

Capacitor Ratings at Rated Voltage
Ref: NEC(India) :2011
Motor Rating(Kw) Capacitor Rating in kVAR for Motor Speed
3 000rev/min 1 500rev/min 1 000rev/min 750rev/min 600rev/min 500rev/min
2.25 1 1 1.5 2 2.5 2.5
3.7 2 2 2.5 3.5 4 4
5.7 2 3 3.5 4.5 5 5.5
7.5 3 4 4.5 5.5 6 6.5
11.2 4 5 6 7.5 8.5 9
15 5 6 7 9 11 12
18.7 6 7 9 10.5 13 14.5
22.5 7 8 10 12 15 17
37 11 12.5 16 18 23 25
57 16 17 21 23 29 32
75 21 23 26 28 35 40
102 31 33 36 38 45 55
150 40 42 45 47 60 67
187 46 50 53 55 68 76

 

:Maximum Current Demand for Motor:
Ref: NEC(India) :2011
Nature of supply Size of installation Maximum current demand
Single phaseor Three phase Up to and including 0.75 kW Six times the full load current
Above 0.75 kW and up to 7.5 kW Three times the full load current
Above 7.5 kW up to and up to11 kW Two times the full load current
Above 11 kW One and half times the full load current

 

Rated Basic Insulation Level (BIL)
Ref: NEC(India) :2011
Nominal System Voltage (kV) Rated BIL (kVp)
33 KV 170
22 KV 125
11 KV 75
6.6 KV 60
3.3 KV 40

 

Illumination Level
Ref: NEC(India) :2011
Location Illumination Level (Lux)
Residence
Entrance / Hallways 100
Living room 300
Dining Room 150
Bed Room (General) 300
Bed Room (Dressing , Bed Heads) 200
Kitchen 200
Kitchen sink 300
Bathroom 100
Sewing 700
Workshop 200
Staircase 100
Garage 70
Study Room 300
Office Building
Entrance hall / Reception 150
Conference Room / Executive Office 300
General Office Space 300
Business Machinery Operation 450
Drawing Office 450
Corridors 70
Stairs 100
Lift landing 150
Hospital Building
Reception & Waiting 150
General ward 100
Bed Side 150
Toilet 70
Stairs 100
Operation Theatre (General) 300
Operation Theatre (Operation Table) Special
Laboratories 300
Radiology 100
Causality 150
Dispensaries 300
Laundry 200
Dry Cleaning 200
Ironing 300
General Office 450
Kitchen 200
Assembly & Concert Halls
Foyers 100 to 150
Auditoria 100 to 150
Platform 450
Corridors 70
Stairs 100
Cinema Halls
Foyers 150
Auditoria 50
Corridors 70
Stairs 100
Theatres
Foyers 150
Auditoria 70
Corridors 70
Stairs 100
School / College Building
Assembly Halls  
General 150
Examination center 300
Platform 300
Classes  
Desktop 300
Blackboard 200 to 300
Libraries  
Shelves 70 to 150
Reading Room 150 to 300
Reading Table 300 to 700
Cataloguing 150 to 300
General  
Office 300
Staff Room 150
Corridors 70
Stairs 100

 

Lamp’s Lumen Data
Rating (Watt) Life (Hours) Initial Lumens
Incandescent Lamp
60 1000 870
100 750 1750
150 2000 1740
200 2000 2300
500 2000 6500
Fluorescent Lamp
18 7000 1120
20 7000 1020
36 7000 2800
40 7000 2700
2X40 7000 4000
Compact Fluorescent Lamp
5 10000 220
7 7000 380
11 7000 560
13 7000 680
15 7000 810
18 7000 1050
23 7000 1500
26 7000 1800
32 7000 2400
Mercury Vapour Lamp
100 18000 3700
175 24000 8600
250 24000 12100
400 24000 22500
1000 24000 57000
Metal Halide Lamp
50 15000 3400
70 15000 5600
100 15000 9000
150 10000 13500
175 10000 15000
250 10000 20500
400 20000 36000
1000 12000 110000
High Pressure Sodium Vapour Lamps
35 16000 2250
50 24000 4000
70 24000 5800
100 24000 9500
150 24000 16000
250 24000 27500
400 24000 47500
1000 24000 140000
Pulse Start Metal Halide Lamp
50 15000 3400
70 15000 5600
100 15000 9000
150 15000 15000
175 15000 17500
200 15000 21000
250 15000 26300
320 20000 34000
400 20000 44000
450 20000 50000

 

:Duty Type of Motor:
Ref: IS-325
Duty Type Symbol Duty Type Application
S1 Continuous Duty Pumps, Bowers, Compressors, Fans
S2 Short Time Duty Siren, Flood relief Gates
S3 Intermittent Periodic Duty Valve, Actuators, Wire drawing machine
S4 Intermittent Periodic Duty with starting Hoist, Cranes, Lifts, Escalators
S5 Intermittent Periodic Duty with starting / Breaking Hoist, Cranes (with electronics Breaks), Rolling mills
S6 Continuous Duty with Intermittent Periodic Duty Machine tools, Conveyors
S7 Continuous Duty with starting / Breaking Machine tools,
S8 Continuous Duty with periodic load changes Pole Channing Applications

 

Type of Distribution System
As per IEC 60364-3
Unearthed System Earthed System
IT TT / TN (TN-S,TN-C,TN-C-S)
First Letter (the neutral point in relation to earth):T= directly earthed neutral (from the French word Terre)I =unearthed or high impedance-earthed neutral (e.g. 2,000 Ω)
Second letter (Exposed conductive parts of the electrical installation in relation to earth):T =directly earthed exposed conductive partsN =exposed conductive parts directly connected to the neutral conductor

 

Panel Design & Calculate Size of Bus bar


Example: Calculate Size of Bus bar having Following Details

  • Bus bar Current Details:
  • Rated Voltage = 415V,50Hz ,
  • Desire Maximum Current Rating of Bus bar =630Amp.
  • Fault Current (Isc)= 50KA ,Fault Duration (t) =1sec.
  • Bus bar Temperature details:
  • Operating Temperature of Bus bar (θ)=85°C.
  • Final Temperature of Bus bar during Fault(θ1)=185°C.
  • Temperature rise of Bus Bar Bar during Fault (θt=θ1-θ)=100°C.
  • Ambient Temperature (θn) =50°C.
  • Maximum Bus Bar Temperature Rise=55°C.
  • Enclosure Details:
  • Installation of Panel= Indoors (well Ventilated)
  • Altitude of Panel Installation on Site= 2000 Meter
  • Panel Length= 1200 mm ,Panel width= 600 mm, Panel Height= 2400 mm
  • Bus bar Details:
  • Bus bar Material= Copper
  • Bus bar Strip Arrangements= Vertical
  • Current Density of Bus Bar Material=1.6
  • Temperature Co efficient of Material Resistance at 20°c(α20)= 0.00403
  • Material Constant(K)= 1.166
  • Bus bar Material Permissible Strength=1200 kg/cm2
  • Bus bar Insulating Material= Bare
  • Bus bar Position= Edge-mounted bars
  • Bus bar Installation Media= Non-ventilated ducting
  • Bus bar Artificial Ventilation Scheme= without artificial ventilation
  • Bus bar Size Details:
  • Bus bar Width(e)= 75 mm
  • Bus bar Thickness(s)= 10 mm
  • Number of Bus Bar per Phase(n)= 2 No
  • Bus bar Length per Phase(a)= 500 mm
  • Distance between Two Bus Strip per Phase(e)= 75 mm
  • Bus bar Phase Spacing (p)= 400 mm
  • Total No of Circuit= 3 No.
  • Bus bar Support Insulator Detail:
  • Distance between insulators on Same Phase(l)= 500 mm
  • Insulator Height (H)= 100 mm
  • Distance from the head of the insulator to the bus bar center of gravity (h)= 5 mm
  • Permissible Strength of Insulator (F’)=1000 Kg/cm2

 Untitled

Calculation:

(1) De rating Factors for Bus bar:

  • (1) Per Phase Bus Strip De rating Factor (K1):
  • Bus bar Width(e) is 75mm and Bus bar Length per Phase(a) is 500mm so e/a is 75/500=0.15
  • No of Bus bar per phase is 2 No’s.
  • From following table value of de rating factor is 1.83

Number of Bus Bar Strip per Phase (K1)

e/a No of Bus Bar per Phase
1 2 3
0.05 1 1.63 2.4
0.06 1 1.73 2.45
0.08 1 1.76 2.5
0.1 1 1.8 2.55
0.12 1 1.83 2.6
0.14 1 1.85 2.63
0.16 1 1.87 2.65
0.18 1 1.89 2.68

0.2

1 1.91 2.7

 

  • (2) Bus bar Insulating Material De rating Factor (K2)
  • Bus bar having No insulating material. It is Bare so following Table
  • De rating Factor is 1.
Bus Bar Insulating Material (K2): De rating Factor
Bare 1
PVC Sleeving 1.2
Painted 1.5

 

  • (3) Bus bar Position De rating Factor (K3)
  • Bus bar Position is Edge-mounted bars so following Table
  • De rating Factor is 1
Bus Bar Position(K3): De rating Factor
Edge-mounted bars 1
1 bar base-mounted 0.95
several base-mounted bars 0.75

 

  • (4) Bus bar Installation Media De rating Factor (K4)
  • Bus bar Installation Media is Non-ventilated ducting so following Table
  • De rating Factor is 0.8
Bus Bar Installation Media(K4): De rating Factor
Calm indoor atmosphere 1
Calm outdoor atmosphere 1.2
Non-ventilated ducting 0.8

 

  • (5) Bus bar Artificial Ventilation De rating Factor (K5)
  • Bus bar Installation Media is Non-ventilated ducting so following Table
  • De rating Factor is 0.9
Bus Bar Artificial Ventilation Scheme (K5): De rating Factor
without artificial ventilation 0.9
with artificial ventilation 1

 

  • (6) Enclosure & Ventilation De rating Factor (K6)
  • Bus bar Area per Phase = Bus width X Bus Thickness X Length of Bus X No of Bus bar per Phase
  • Bus bar Area per Phase = 75x10xX500X2= 750000mm
  • Total Bus bar Area for Enclosure= No of Circuit X( No of Phase + Neutral )X Bus bar Area per Phase
  • Here we used Size of Neutral Bus is equal to Size of Phase Bus
  • Total Bus bar Area for Enclosure=3X(3+1)X750000mm
  • Total Bus bar Area for Enclosure=9000000 Sq.mm
  • Total Enclosure Area= width X Height X Length
  • Total Enclosure Area=1200x600x2400=1728000000 Sq.mm
  • Total Bus bar Area for Enclosure / Total Enclosure Area =9000000/1728000000
  • Total Bus bar Area for Enclosure / Total Enclosure Area=0.53%
  • Bus bar Artificial Ventilation Scheme is without artificial ventilation so following Table
  • De rating Factor is 0.95
Volume of Enclosure & Ventilation De rating Factor (K6)
cross Section area of Bus bar/Total Bus Bar Area Indoors ( Panel is well Ventilated) Indoors ( Panel is Poorly Ventilated) Outdoor
0% 0.95 0.85 0.65
1% 0.95 0.85 0.65
5% 0.9 0.7 0.6
10% 0.85 0.65 0.5

 

  • (7) Proxy Effect De rating Factor (K7)
  • Bus bar Phase Spacing (p) is 400mm.
  • Bus bar Width (e) is 75mm and Space between each bus of Phase is 75mm so
  • Total Bus length of Phase with spacing (w) =75+75+75+75+75=225mm
  • Bus bar Phase Spacing (p) / Total Bus length of Phase with spacing (w) = 400 / 225 =2
  • From following Table De rating factor is 0.82
Proxy Effect (K7): De rating Factor
1 0.82
2 0.82
3 0.82
4 0.89
5 0.95
6 0.99
7 1

 

  • (8) Altitude of Bus Bar installation De rating Factor (K8)
  • Altitude of Panel Installation on Site is 2000 meter so following Table
  • De rating Factor is 0.88
Altitude of installation site (Meter) (K8) De rating Factor
2200 0.88
2400 0.87
2500 0.86
2700 0.85
2900 0.84
3000 0.83
3300 0.82
3500 0.81
4000 0.78
4500 0.76
5000 0.74
  • Total De rating Factor= K1XK2XK3Xk4Xk5Xk6Xk7Xk8
  • Total De rating Factor =1.83x1x1x0.8×0.9×0.95×0.82×0.88
  • Total De rating Factor =0.90

(2) Bus bar Size Calculation:

  • Desire Current Rating of Bus bar (I2) =630 Amp
  • Current Rating of Bus bar after De rating Factor (I1)= I2 X De rating Factor or I2 / De rating Factor
  • Current Rating of Bus bar after De rating Factor (I1)=630×0.9
  • Current Rating of Bus bar after De rating Factor (I1)=697Amp
  • Bus bar Cross Section Area as per Current= Current Rating of Bus bar / Current Density of Material
  • Bus bar Cross Section Area as per Current= 697 / 1.6
  • Bus bar Cross Section Area as per Current= 436 Sq.mm
  • Bus bar Cross Section Area as per Short Circuit= Isc X√ ((K/( θtx100)x(1+ α20xθ) xt
  • Bus bar Cross Section Area as per Short Circuit= 50000X√ ((1.166/( 100×100)x(1+ 0.00403×85) x1
  • Bus bar Cross Section Area as per Short Circuit=626 Sq.mm
  • Select Higher Size for Bus bar Cross section area between 436 Sq.mm and 626 Sq.mm
  • Final Calculated Bus Bar Cross Section Area =626 Sq.mm
  • Actual Selected Bus bar size is 75×10=750 Sq.mm
  • We have select 2 No’s of Bus bar per Phase hence.
  • Actual Bus bar cross section Area per Phase =750×2= 1500 Sq.mm
  • Actual Cross Section Area of Bus bar =1500 Sq.mm
  • Actual Bus bar Size is Less than calculated Bus bar size.

(3) Forces generated on Bus Bar due to Short Circuit Current

  • Peak electro-magnetic forces between phase conductors (F1) = 2X(l/d)X(2.5xIsc)2/100000000
  • Total width of Bus bar per Phase(w)=75+75+75=225mm =2.25cm
  • Bus bar Phase to Phase Distance (d)=400+225=625mm=6.25cm
  • Peak electro-magnetic forces between phase conductors (F1) =2x(50/63)x(2.5×50000)2/100000000
  • Peak electro-magnetic forces between phase conductors (F1)=250 Kg /cm2
  • Peak electro-magnetic forces between phase conductors (F1)=2.5 Kg /mm2
  • Actual Forces at the head of the Supports or Bus Bar (F)=F1X(H+h/H)
  • Actual Forces at the head of the Supports or Bus Bar (F)=2.5x(100+5/100)
  • Actual Forces at the head of the Supports or Bus Bar (F)= 3 Kg /mm2
  • Permissible Strength of Insulator (F’) is 10 Kg/mm2
  • Actual Forces at the head of the Supports or Bus Bar is less than Permissible Strength
  • Forces on Insulation is in within Limits

(4) Mechanical strength of the bus bars

  • Mechanical strength of the bus bars=(F1X i /12)x(1/ Modulus of inertia of a bus bar )
Value of Modulus of inertia of a bus bar or of a set of bus bars (i/v)
No of Bus Strip per Phase Vertical Bus Bar (cm3) Horizontal Bus Bar (cm3)
1 1.66 16.66
2 14.45 33.33
3 33 50

 

  • From above table Value of Modulus of inertia of a bus bar=14.45
  • Mechanical strength of the bus bars=(250×50/12)X(1/14.45)
  • Mechanical strength of the bus bars= 72 Kg/cm2
  • Mechanical strength of the bus bars= 0.72 Kg/mm2
  • Permissible Bus bar Strength is 12 Kg/mm2
  • Actual Mechanical Strength is less than Permissible Strength
  • Mechanical strength of Bus bar is in within Limit

(5) Temperature Rise Calculation

  • Specified Maximum Temperature Rise (T1) is 35°c
  • Calculated Maximum Temperature Rise (T2)=T/(log(I1/I2)1.64)
  • Calculated Maximum Temperature Rise (T2)=35/(Log(697/630)1.64)
  • Calculated Maximum Temperature Rise (T2)= 30°c
  • Calculated Bus bar Temperature rise is less than Specified Max Temperature rise
  • Temperature Rise is in within Limit

Results:

  • Size of Bus bar = 2No’s 75x10mm per Phase.
  • Total No of Feeder =3 No’s
  • Total No’s of Bus bar = 6 No’s 75x10mm for Phase and 1No’s 75x10mm for Neutral.
  • Forces at the head of the Supports or Bus Bar (F)= 3kg/mm2
  • Mechanical strength of the bus bars= 0.7 Kg/mm2
  • Maximum Temperature Rise=30°c

Apex of Technical Pirating


Now a days New version of technical pirating is discovered.

Not only New Technical Blogger, some Technical Service provider  Company also  start copy past of this blog and put these contains on their official Site !!!!!!!!!!!!!

Look at following totally copy past material of  this  Blog which published on Official site of  Lighting Protection  Solution provider Company . Company does not even care to change The Color and Style of the Font during copy past.

Original published Post: http://electricalnotes.wordpress.com/2011/03/30/lighting-arrester/

Copy Paste site of Company: http://www.radiantenergy.co.in

Copy Paste Material on Company’s Official Web Page:

http://www.radiantenergy.co.in/lightning-arrester-for-building-in-Dehradun-Uttarakhand-India/176/10

This post is published after lot of mail posted to above company to remove pirated contain but Still no any action taken from company !!!!!!!!!!!!!!!

Panel Design / Calculate Size of Bus bar


 

ScreenHunter_01 Jul. 21 20.06

  • Calculate Bus Bar Cross Section Area
  • Calculate Current Rating of Bus Bar
  • Calculate Peak elector-magnetic Force between Phase Conductors
  • Calculate Force on Insulator of Bus bar Support.
  • Calculate Mechanical Strength of Bus Bar.
  • Calculate Maximum Temperature Rise of Bus Bar.

Free Download

Calculate Voltage Drop and No’s of Street Light Pole


ScreenHunter_01 Jul. 14 20.14

  • Calculate Voltage drop in Street Light Pole
  • Calculate Distance between Each Street Light Pole
  • Calculate Wattage of each Street Light Luminaire
  • Calculate Required Street Light Watt as per Given Area.
  • Calculate Luminaire Payback Period

Free Download

Calculate Size of Neutral Earthing Transformer (NET)


  • Calculate Total Capacitive charging current (It).
  • Calculate Neutral Earthing Transformer  Secondary Current.
  • Calculate Required Grounding Resistor value at primary side (Rp).
  • Calculate Required Resistance value at secondary side (Rsec).
  • Calculate Neutral Earthing Transformer X/R Ratio
  • Calculate Fault current through Neutral (single line to ground fault) (If)
  • Calculate Rating of Neutral Earthing Transformer (P).

ScreenHunter_01 Jun. 06 22.20

Free Download

Calculate Street Light Pole’s Distance / Fixture Watt / Lighting Area


(1) Calculate Distance between each Street Light Pole:

Example: Calculate Distance between each streetlight pole having following Details,

  • Road Details: The width of road (w) is 11.5 Foot.
  • Pole Details: The height of Pole is 26.5 Foot.
  • Luminaire of each Pole: Wattage of Luminaries is 250 Watt, Lamp Out Put (LL) is 33200 Lumen, Required Lux Level (Eh) is 5 Lux, Coefficient of Utilization Factor (Cu) is 0.18, Lamp Lumen Depreciation Factor (LLD) is 0.8, Lamp Lumen Depreciation Factor (LLD) is 0.9.
  • Space Height Ratio should be less than 3.

Calculation:

  • Spacing between each Pole=(LL*CU*LLD*LDD) / Eh*W
  • Spacing between each Pole=(33200×0.18×0.8×0.9) / (5×11.5)
  • Spacing between each Pole= 75 Foot.
  • Space Height Ratio = Distance between Pole / Road width
  • Space Height Ratio = 3. Which is less than define value.
  • Spacing between each Pole is 75 Foot.

(2) Calculate Street Light Luminaire Watt:

Example: Calculate Streetlight Watt of each Luminaire of Street Light Pole having following Details,

  • Road Details: The width of road (w) is 7 Meter. Distance between each Pole (D) is 50 Meter.
  • Required Illumination Level for Street Light (L) is 6.46 Lux per Square Meter. Luminous efficacy is 24 Lumen/Watt.
  • Maintenance Factor (mf) 0.29, Coefficient of Utilization Factor (Cu) is 0.9.

Calculation:

  • Average Lumen of Lamp (Al) = 8663 Lumen.
  • Average Lumen of Lamp (Al) =(LxWxD) / (mfxcu)
  • Average Lumen of Lamp (Al)= (6.46x7x50) / (0.29×0.9)
  • Average Lumen of Lamp (Al)=8663 Lumen.
  • Watt of Each Street Light Luminar = Average Lumen of Lamp / Luminous efficacy
  • Watt of Each Street Light Laminar = 8663 / 24
  • Watt of Each Street Light Luminaire = 361 Watt

(3) Calculate Required Power for Street Light Area:

Example: Calculate Streetlight Watt of following Street Light Area,

  • Required Illumination Level for Street Light (L) is 6 Lux per Square Meter.
  • Luminous efficacy (En) is 20 Lumen per Watt.
  • Required Street Light Area to be illuminated (A) is 1 Square Meter.

Calculation:

  • Required Streetlight Watt = (Lux per Sq.Meter X Surface Area of Street Light) / Lumen per Watt.
  • Required Streetlight Watt = (6 X 1) / 20.
  • Required Streetlight Watt = 0.3 watt per Square Meter.

IP Rating for Electrical Enclosure


IP rating:

  •  IP letters stand for “International Protection” rating or “Ingress Protection” rating. IP ratings are defined in international standard (British BS EN 60529, IEC 60509).
  • It is used to explain levels of sealing effectiveness of electrical enclosures against foreign bodies (tools, dirt etc) and moisture.

 

Meaning of IP Rating:

  • The IP rating code is a two-digit (or optionally three-digit) designator to standardize the rating of protection level against intrusion of solids and liquids into mechanical and electrical enclosures. An enclosure can be a piece of equipment, an assembly unit, a cable or simply a connector.
  • The numbers of IP of each have a specific meaning.
  • First Number: The first Number indicates the degree of protection from moving parts, as well as the protection of enclosed equipment from foreign bodies.
  • Second Number: The second Number indicates the protection level that the enclosure enjoys from various forms of moisture (drips, sprays, submersion etc).
  • Third Number: The third digit in the designator is not part of the official IEC standard and is sometimes included (but more often omitted) to reference additional protections.

Abbreviation of IP Rating:

 

                                                                IP Rating Digits

IP Rating First Digit Second Digit Third Digit (Optional)
Solid Objects Protection Liquids Protection Mechanical impacts

0

No special protection No protection. No protection.
1 Protected against solid objects greater than 50mm in diameter ( such as large part of the body like hand) Protection against vertically falling drops of water e.g. condensation. Protects against impact of 0.225 joule (150 g weight falling from 15 cm height)
2 Protected against solid objects over 12 mm in diameter (person’s fingers) Protection against direct sprays of water up to 15° from the vertical. Protected against impact of 0.375 joule (250 g weight falling from 15 cm height)
3 Protected against solid objects not greater than 80mm in length and 12mm in diameter (tools and wires). Protected against direct sprays of water up to 60° from the vertical. Protected against impact of 0.500 joule (250 g weight falling from 20 cm height)
4 Protected against solid objects larger than 1 mm diameter (tools, wires, and small wires). Protection against water sprayed from all directions (limited ingress permitted). Protected against impact of 2.0 joule (500 g weight falling from 40 cm height)
5 Protected against dust limited ingress (no harmful deposit). Protected against low pressure jets of water from all directions (limited ingress). Protected against impact of 6.0 joule (1.5 kg weight falling from 40 cm height)
6 Totally dust tight. Protected against temporary flooding of water, e.g. for use on ship decks (limited ingress permitted). Protected against impact of 20.0 joule (5 kg weight falling from 40 cm height)
7 N/A Protected against the effect of immersion between 15 cm and 1 m. N/A
8 N/A Protects against long periods of immersion under pressure. N/A

 

Example:

  • IP65 Enclosure: IP rated as protection against dust (6) and protection from low water pressure (5). 
  • IP66 Enclosure – IP rated as protection against dust (6) and protected against heavy seas or powerful jets of water (6)

 

 

Calculate No of Street Light Poles


Typical Calculation of Road Lighting:

  •  Luminaries are properly selected and mounted on a location most feasible and effective with minimum cost. For a 230 volts system, a voltage drop of 5% is allowed although in extreme cases 15% voltage drop is sometimes tolerated.  3
  • Street illumination level in Lux (E)=(Al x (cu x mf)) / (w x d)
  • E = The illumination in Lux
  • w = Width of the roadway
  • d = Distance between luminaries
  • cu = Coefficient of utilization. Which is dependent on the type of fixture, mounting height, width of roadway and the length of mast arm of outreach?
  • Al = Average lumens, Al = (E x w x d) / Cu x mf
  • The typical value of Al is
  • 20500 lumens for 400 watts
  • 11500 lumens for 250 watts
  • 5400 lumens for 125 watts
  • The value of Al varies depending upon the type of lamp specified.
  • mf : It is the maintenance factor (Normally 0.8 to 0.9)

 (1) Calculate Lamp watt for street Light Pole:

  • Calculate Lamp Lumen for street Light Pole having Road width of 7 meter, distance between two Pole is 50 meter, Maintenance factor is 0.9, Coefficient of utilization factor is 0.29, light pedestrian traffic is medium and Vehicular traffic is very light and Road is concrete road.
    Solution:
    From Above table Recommended of illumination (E) in Lux is 6.46 per sq. meter.
    w = 7.00 meters , d = 50 meters , mf = 0.9, cu = 0.29
    To decide Lamp Watt It is necessary to calculate Average Lumens of Lamp (Al).
  • Average Lumen of Lamp (Al)=(E x w x d) / Cu x mf
  • Al=(6.46x7x50)/(0.29×0.9)= 8662.83 Average lumen
    Lamp lumen of a 250 watts lamp is 11,500 lm which is the nearest value to 8662.83 lumen. Therefore, a 250 watts lamp is acceptable.
    Let’s Computing for the actual illumination E for 250 Watt Lamp
  • Illumination (E)=(Al x (cu x mf)) / (w x d)
  • E= (11500×0.29×0.9) / (7×50) = 8.57 lumen per sq meter.
    Conclusion:
    Actual illumination (E) for 250 Watt is 8.57 lumen per sq meter which is higher than recommended illumination (E) 6.46.
  • Hence 250 watt gives adequately lighting.

 (2) Calculate Spacing between two Light Poles:

  • Calculate Space between Two Pole of Street Light having Fixture Watt is 250W , Lamp output of the Lamp (LL) is 33200 lumens , Required Lux Level (E) is 5 lux , Width of the road (W) = 11.48 feet (3.5 M),Height of the pole (H) = 26.24 feet (8 M) ,Coefficient of utilization (CU) = 0.18, Lamp Lumen Depreciation Factor (LLD) = 0.8 ,Luminaries dirt Depreciation Factor (LDD) = 0.9
    Solution:
  • Luminaries Spacing (S) = (LLxCUxLLDxLDD) / (ExW)
  • Luminaries Spacing (S) = (33200×0.18×0.9×0.8) / (5×11.48)
  • Luminaries Spacing (S) = 75 feet (23 Meters)

 (3) Calculation of the allowed illumination time:

  • The allowed illumination time in hours T = k.t.1000/E.
  • Where: k = extension factor
  • t = permissible time in hours at 1000 lux, unfiltered daylight
  • E = luminance (lx)

 

Extension Factor
Lamp Extension Factor
Incandescent lamps, 2.7 to 3.2
Halogen reflector lamps 2.5 to 3.5
Halogen capsules 2.5 to 3.5
High-pressure metal-halide 1.1 to 2.1
High-pressure sodium lamps 4
Fluorescent lamps 1.9 to 2.7

 

  • Example:
  • In sunlight (100000 lux) and extension factor 1: The permissible illumination time (T) =1 x 70 x 1000/100 000 = 0.7 hour.
  • In halogen light (200 lux) and extension factor 2.3: The permissible illumination time (T) = 2.3 x 70 x 1000/200 = 805 hours.
  • In UV-filtered halogen light (200 lux) and extension factor 3.5: The permissible illumination time (T) = 3.5 x 70 x 1000/200 = 1225 hours.

 (4) Calculate Uniformity Ratio:

  • Once luminaries spacing has been decided It is necessary to check the uniformity of light distribution and compare this value to the selected lighting
  • Uniformity Ratio ( UR) = Eav /Emin
  • Eav= average maintained horizontal luminance
  • Emin = maintained horizontal luminance at the point of minimum illumination on the pavement

 (5) Energy Saving Calculations:

  • At a simplistic level, the cost of running a light is directly related to the wattage of the globe plus any associated ballast or transformer. The higher the wattage, the higher the running cost and it is a straightforward calculation to work out the running cost of lamp over its lifetime:
  • Running cost = cost of electricity in $/kWh x wattage of lamp x lifetime in hours.

 Calculate Lux Level for Street Lighting

  • The Average Lux Level of Street Light is measured by 9 point method. Make two equal quadrants between two Street light poles. on the lane of light poles( one side pole to road).
  • We have 3 points P1,P2 and P3 under the light pole then P4 & P7 are points opposite pole 1 or Point P3 same is applicable for P6 and P9 for Pole 2.
  • The average lux = [(P1+P3+P7+P9)/16]+[(P2+P6+P8+P4)/8]+[P5/4]

 2

Calculate No of Lighting Fixtures / Lumen for Indoor Lighting


  • An office area is 20meter (Length) x 10meter (width) x 3 Meter (height). The ceiling to desk height is 2 meters. The area is to be illuminated to a general level of 250 lux using twin lamp 32 watt CFL luminaires with a SHR of 1.25. Each lamp has an initial output (Efficiency) of 85 lumen per watt. The lamps Maintenance Factor (MF) is 0.63 ,Utilization Factor is 0.69 and space height ratio (SHR) is 1.25

 Calculation:

 Calculate Total Wattage of Fixtures:

  • Total Wattage of Fixtures= No of Lamps X each Lamp’s Watt.
  • Total Wattage of Fixtures=2×32=64Watt.

 Calculate Lumen per Fixtures:

  •  Lumen per Fixtures = Lumen Efficiency(Lumen per Watt) x each Fixture’s Watt
  • Lumen per Fixtures= 85 x 64 = 5440Lumen

 Calculate No’s of Fixtures:

  •  Required No of Fixtures = Required Lux x Room Area / MFxUFx Lumen per Fixture
  • Required No of Fixtures =(250x20x10) / (0.63×0.69×5440)
  • Required No of Fixtures =21 No’s

 Calculate Minimum Spacing Between each Fixture:

  •  The ceiling to desk height is 2 meters and Space height Ratio is 1.25 so
  • Maximum spacing between Fixtures =2×1.25=2.25meter.

 Calculate No of Row Fixture’s Row Required along with width of Room:

  •  Number of Row required = width of Room / Max. Spacing= 10/2.25
  • Number of Row required=4.

 Calculate No of Fixture’s required in each Row:

  •  Number of Fixture Required in each Row = Total Fixtures / No of Row = 21/4
  • Number of Fixture Required in each Row = 5 No’s:

 Calculate Axial Spacing between each Fixture:

  •  Axial Spacing between Fixtures = Length of Room / Number of Fixture in each Row
  • Axial Spacing between Fixtures =20 / 5 = 4 Meter

 Calculate Transverse Spacing between each Fixture:

  •  Transverse Spacing between Fixtures = width of Room / Number of Fixture’s row
  • Transverse Spacing between Fixtures = 10 / 4 = 2.5 Meter.

   Untitled

 Conclusion:

  •  No of Row for Lighting Fixture’s= 4 No
  • No of Lighting Fixtures in each Row= 5 No
  • Axial Spacing between Fixtures= 4.0 Meter
  • Transverse Spacing between Fixtures= 2.5 Meter
  • Required No of Fixtures =21 No’s

 

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