Cable Voltage Drop for Different Size of Cables


ScreenHunter_01 Feb. 04 17.50

  • Calculate Voltage of Cable having Different Size
  • Calculate Starting Current
  • Calculate Running Current
  • Calculate Starting Voltage Drop
  • Calculate Running Voltage Drop

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Calculate Size of Solar Panel


Calculate Size of Solar Panel, No of Solar Panel and Size of Inverter for following Electrical Load

Electrical Load Detail:

  • 1 No’s of 100W Computer use for 8 Hours/Day
  • 2 No’s of 60W Fan use for 8 Hours/Day
  • 1 No’s of 100W CFL Light use for 8 Hours/Day

Solar System Detail:

  • Solar System Voltage (As per Battery Bank) = 48V DC
  • Loose Wiring Connection Factor = 20%
  • Daily Sunshine Hour in Summer = 6 Hours/Day
  • Daily Sunshine Hour in Winter = 4.5 Hours/Day
  • Daily Sunshine Hour in Monsoon = 4 Hours/Day

Inverter Detail:

  • Future Load Expansion Factor = 10%
  • Inverter Efficiency = 80%
  • Inverter Power Factor =0.8

Calculation:

Step-1: Calculate Electrical Usages per Day

  • Power Consumption for Computer = No x Watt x Use Hours/Day
  • Power Consumption for Computer = 1x100x8 =800 Watt Hr/Day
  • Power Consumption for Fan = No x Watt x Use Hours/Day
  • Power Consumption for Fan = 2x60x8 = 960 Watt Hr/Day
  • Power Consumption for CFL Light = No x Watt x Use Hours/Day
  • Power Consumption for CFL Light = 1x100x8 = 800 Watt Hr/Day
  • Total Electrical Load = 800+960+800 =2560 Watt Hr/Day

Step-2: Calculate Solar Panel Size

  • Average Sunshine Hours = Daily Sunshine Hour in Summer+ Winter+ Monsoon /3
  • Average Sunshine Hours = 6+4.5+4 / 3 =8 Hours
  • Total Electrical Load =2560 Watt Hr/Day
  • Required Size of Solar Panel = (Electrical Load / Avg. Sunshine) X Correction Factor
  • Required Size of Solar Panel =(2560 / 4.8) x 1.2 = 635.6 Watt
  • Required Size of Solar Panel = 635.6 Watt

Step-3: Calculate No of Solar Panel / Array of Solar Panel

If we Use 250 Watt, 24V Solar Panel in Series-Parallel Type Connection

  • In Series-Parallel Connection Both Capacity (watt) and Volt are increases
  • No of String of Solar Panel (Watt) = Size of Solar Panel / Capacity of Each Panel
  • No of String of Solar Panel ( Watt) = 635.6 / 250 = 2.5 No’s Say 3 No’s
  • No of Solar Panel in Each String= Solar System Volt / Each Solar Panel Volt
  • No of Solar Panel in Each String= 48/24 =2 No’s
  • Total No of Solar Panel = No of String of Solar Panel x No of Solar Panel in Each String
  • Total No of Solar Panel = 3×2 =6 No’s
  • Total No of Solar Panel =6 No’s

Step-4: Calculate Electrical Load:

  • Load for Computer = No x Watt
  • Load for Computer = 1×100 =100 Watt
  • Load for Fan = No x Watt
  • Load for Fan = 2×60 = 120 Watt
  • Load for CFL Light = No x Watt
  • Load for CFL Light = 1×100 = 100 Watt
  • Total Electrical Load = 100+120+100 =320 Watt

Step-5: Calculate Size of Inverter:

  • Total Electrical Load in Watt = 320 Watt
  • Total Electrical Load in VA= Watt /P.F
  • Total Electrical Load in VA =320/0.8 = 400VA
  • Size of Inverter =Total Load x Correction Factor / Efficiency
  • Size of Inverter = 320 x 1.2 / 80% =440 Watt
  • Size of Inverter =400 x 1.2 / 80% =600 VA
  • Size of Inverter = 440 Watt or 600 VA

Summary:

  • Required Size of Solar Panel = 635.6 Watt
  • Size of Each Solar Panel = 250 Watt. 12 V
  • No of String of Solar Panel = 3 No’s
  • No of Solar Panel in Each String = 2 No’s
  • Total No of Solar Panel =6 No’s
  • Total Size of Solar Panel = 750 Watt
  • Size of Inverter = 440 watt or 600 VA

Calculate Cable Trunking Size


Example: Calculate PVC Trunking size for following Power Cables running through Trunking .Consider 10% as Future expansion

  • 30 No’s of 1.5 Sq.mm Stranded Cables
  • 50 No’s of 2.5 Sq.mm Stranded Cables
  • 20 No’s of 4 Sq.mm Stranded Cables

Cable Factor for Trunking as per IEEE

Type of Cable Size of Cable Factor
Solid 1.5 7.1
Solid 2.5 10.2
Stranded 1.5 8.1
Stranded 2.5 11.4
Stranded 4 15.2
Stranded 6 22.9
Stranded 10 36.3
Stranded 16 50.3
Stranded 25 75.4
Stranded 35 95
Stranded 50 132.7
Stranded 70 176.7
Stranded 95 227
Stranded 120 284
Stranded 150 346

Trunking Factor As per IEEE

Trunking Size (mm) Factor
75 X 25 738
50 X 37.5 767
100 X 50 993
50 x 50 1037
75 X 37.5 1146
100 X 37.5 1542
75 X 50 1555
100 X 50 2091
75 X 75 2371
150 X 50 3161
100 X 75 3189
100 X 100 4252
150 X 75 4787
150 X 100 6414
150 X 150 9575

Calculations:

  • As per Table Cable Factor for 1.5 Sq.mm Stranded Cable=8.1
  • As per Table Cable Factor for 2.5 Sq.mm Stranded Cable=11.4
  • As per Table Cable Factor for 2.5 Sq.mm Stranded Cable=22.9
  • Total Cable Factor= No of Cables X Cable Factor
  • Total Cable Factor=(30X8.1)+(50X11.4)+(20X22.9)
  • Total Cable Factor=1271
  • Total Cable Factor After 10% Future Expansion=1271+127 =1398
  • As per Table Suitable Size for PVC Trunking for Cable Factor 1398 is 100 X 37.5

    Suitable PVC Trunking is 100 X 37.5 mm

Calculate Conduit Size for Wires / Cables


Example: Calculate Size of Conduit (Hume Pipe) for Following Size of Cables

  • 5 No’s of 3.5 Core 50 Sq.mm XLPE Cable. Diameter of cable is 28mm
  • 3 No’s of 3.5 Core 185 Sq.mm XLPE Cable. Diameter of cable is 54mm

% Fill Up Area of Conduit

No of Cables % Fill up Area of Conduit
Up to 2 No’s 53%
2 No’s 31%
More Than 2 No’s 40%

Calculations:

  • Area of Cable =3.14 X (Diameter/2)2
  • Area of 50 Cable =3.14X (28/2)2 =615.44 Sq.mm
  • Total Area of 50 Cable=No of Cable X Area of Each Cable
  • Total Area of 50 Cable=5X615.44 =3077.2
  • Area of 185 Cable =3.14X (54/2)2 = 2289 Sq.mm
  • Total Area of 185 Cable=No of Cable X Area of Each Cable
  • Total Area of 185 Cable=3X2289=6867.18
  • Total Area of Conductor= Total Area of 50 Cable+ Total Area of 185 Cable
  • Total Area of Conductor=3077.2+6867.18
  • Total Area of Conductor=9944 Sq.mm
  • Suppose We Select 150mm Diameter Hume Pipe
  • Total area of 150 mm Diameter Hume Pipe =3.14 X (Diameter/2)2 =3.14X(150/2)2.
  • Total area of 150 mm Diameter Hume Pipe=17662 Sq.mm.
  • % Fill up Area of Conduit due to Cables as per above Table is 40%
  • Actual Fill up Area of Hume Pipe =40% X Area of Conduit =40% X 17662
  • Actual Fill up Area of Hume Pipe =7065 Sq.mm
  • Required No of Conduit = Total Area of Cables / Actual Fill up Area of Conduit
  • Required No of Conduit = 9944 / 7065
  • Required No of Conduit= 2 No’s
  • Suppose We Select 3000mm Diameter Hume Pipe
  • Total area of 300 mm Diameter Hume Pipe =3.14 X (Diameter/2)2 =3.14X(300/2)2.
  • Total area of 150 mm Diameter Hume Pipe=70650 Sq.mm.
  • Actual Fill up Area of Hume Pipe =40% X Area of Conduit =40% X 70650
  • Actual Fill up Area of Hume Pipe =28260 Sq.mm
  • Required No of Conduit = Total Area of Cables / Actual Fill up Area of Conduit
  • Required No of Conduit = 9944 / 28260
  • Required No of Conduit= 1 No’s

Results:

  • Either We Select
  • 2 No’s of 150 mm Diameter Conduit /Hume Pipe or
  • 1 No’s 300 mm Diameter Conduit /Hume Pipe

Calculate Size of Diesel Generator


Calculate Size of Diesel Generator having following Electrical Load. Consider Future Expansion ratio is 10%. Average use of Equipment is 0.8 (1 is Full Time Use)

  • 4 No’s of 1Ph, 230V, 80Watt CFL Bulbs, Diversity Factor is 0.8; Starting & Running P.F is 0.8.
  • 2 No’s of 1Ph, 230V, 3000Watt Air Condition, Diversity Factor is 1, Starting & Running P.F is 0.8.
  • 2 No’s of 1Ph, 230V, 500Watt Halogen Lights Diversity Factor is 0.8 Starting & Running P.F is 0.8.
  • 1 No’s of 1Ph,230V, 10KW Motor with Y-D Starter, Diversity Factor is 0.8,Starting P.F is 0.7 & Running P.F is 0.8
  • 1 No’s of 3Ph,430V, 130KW Motor with Soft Starter , Diversity Factor is 0.8,Starting P.F is 0.7 & Running P.F is 0.8

Calculation:

Type of Load Equipment Starting Current
Linear Load General Equipment
100% of Full Load Current
Non-Linear Load UPS, Inverter, Computer, Ballast 160% of Full Load Current

 

Type of Starter Starting Current
DOL 6 X Full Load Current
Star-Delta 4 X Full Load Current
Auto Transformer 3 X Full Load Current
Soft Starter 2 X Full Load Current
VFD 1.5 X Full Load Current

 

Load Calculation-1:

  • Full Load KW of CFL Bulb=(No X Watt X Diversity Factor) /1000
  • Full Load KW of CFL Bulb=(4x80x0.8)/1000=0.3KW
  • Full Load KVA of CFL Bulb=KW / P.F
  • Full Load KVA of CFL Bulb=0.3 / 0.8=0.4KVA————(H)
  • Full Load current of CFL Bulb=(No X Watt X Diversity Factor) / (Volt x P.F)
  • Full Load current of CFL Bulb=(4x80x0.8) / (230×0.8) =2 Amp—-(M)
  • Type of Load=Linear
  • Starting KVA of CFL Bulb=1 X (KW / Starting P.F)
  • Starting KVA of CFL Bulb=0.3 / 0.8=0.4KVA———————-(1)
  • Starting Current=100% of Full Load Current.
  • Starting Current=1 X 2= 2 Amp.————–(A)

Load Calculation-2:

  • Total Full Load KW of A.C=(No X Watt X Diversity Factor) /1000
  • Total Full Load KW of A.C =(2x3000x0.8)/1000=4.8KW
  • Total Full Load KVA of A.C =KW / P.F
  • Total Full Load KVA of A.C =4.8 / 0.8=6KVA————(I)
  • Total Full Load current of A.C =(No X Watt X Diversity Factor) / (Volt x P.F)
  • Total Full Load current of A.C =(2x3000x0.8) / (230×0.8) =26 Amp——-(N)
  • Type of Load=Non Linear
  • Starting KVA of A.C=1.6 X (KW / Starting P.F)
  • Starting KVA of A.C =1.6 X (4.8 / 0.8)=9.6KVA———————-(2)
  • Starting Current=160% of Full Load Current.
  • Starting Current=1.6 X 26= 42 Amp. ————–(B)

Load Calculation-3:

  • Full Load KW of Halogen Bulb=(No X Watt X Diversity Factor) /1000
  • Full Load KW of Halogen Bulb=(2x500x0.8)/1000=0.8KW
  • Full Load KVA of Halogen Bulb=KW / P.F
  • Full Load KVA of Halogen Bulb=0.8 / 0.8=1KVA————(J)
  • Full Load current of Halogen Bulb=(No X Watt X Diversity Factor) / (Volt x P.F)
  • Full Load current of Halogen Bulb=(2x500x0.8) / (230×0.8) =4 Amp—-(O)
  • Type of Load=Non Linear
  • Starting KVA of Halogen Bulb =1.6 X (KW / Starting P.F) / Starting P.F
  • Starting KVA of Halogen Bulb =1.6 X (0.8 / 0.8)=1.6KVA———————-(3)
  • Starting Current=160% of Full Load Current.
  • Starting Current=1.6 X 4= 7 Amp .————–(C)

Load Calculation-4:

  • Full Load KW of Motor=(No X Watt X Diversity Factor) /1000
  • Full Load KW of Motor =(1x10000x0.8)/1000=8KW
  • Full Load KVA of Motor =KW / P.F
  • Full Load KVA of Motor =8 / 0.8=10KVA————(K)
  • Full Load current of Motor =(No X Watt X Diversity Factor) / (Volt x P.F)
  • Full Load current of Motor =(1x10000x0.8) / (230×0.8) =43 Amp—-(P)
  • Type of Starter=Star-Delta
  • Starting KVA of Motor =4 X (KW / Starting P.F)
  • Starting KVA of Motor=4X (8 / 0.7)=45.7KVA————————(4)
  • Starting Current=4 X Full Load Current
  • Starting Current=4 X 11.4 = 174 Amp.————–(D)

Load Calculation-5:

  • Full Load KW of Motor=(No X Watt X Diversity Factor) /1000
  • Full Load KW of Motor =(1x120000x0.8)/1000=96KW
  • Full Load KVA of Motor =KW / P.F
  • Full Load KVA of Motor =96 / 0.8=120KVA————(L)
  • Full Load current of Motor =(No X Watt X Diversity Factor) / (Volt x P.F)
  • Full Load current of Motor =(1x120000x0.8) / (1.732x430x0.8) =167 Amp—-(Q)
  • Type of Starter=Auto Transformer
  • Starting KVA of Motor =3 X (KW / Starting P.F)
  • Starting KVA of Motor=3 X (96 / 0.7)=411.4KVA—————(5)
  • Starting Current=3 X Full Load Current
  • Starting Current=3 X 167 = 501 Amp.————–(E)

Total Load Calculation:

  • Total Starting KVA = (1) + (2) +(3) + (4) + (5)
  • Total Starting KVA =0.4+9.6+1.6+45.7+411.4 =468.7 KVA
  • Total Starting Current = (A) + (B) +(C) + (D) + E
  • Total Starting Current =2+42+7+174+501= 725 Amp
  • Total Running KVA =(H)+(I)+(J)+(K)+(L)
  • Total Running KVA =4+6+1+10+120= 137KVA
  • Total Running Current=(M)+(N)+(O)+(P)+(Q)
  • Total Running Current=2+26+4+43+167= 242 Amp
  • Size of Diesel Generator= Starting KVA X Future Expansion X Average Use of Equipments
  • Size of Diesel Generator=468.7 X 1.1 X 0.8
  • Size of Diesel Generator= 412 KVA

Summary:

  • Total Starting KVA =468.7 KVA
  • Total Starting Current =725 Amp
  • Total Running KVA = 137KVA
  • Total Running Current= 242 Amp
  • Size of Diesel Generator= 412 KVA

Calculate Size of Inverter & Battery Bank


Calculate Size of Inverter for following Electrical Load .Calculate Size of Battery Bank and decide Connection of Battery.

Electrical Load detail:

  • 2 No of 60W,230V, 0.8 P.F Fan.
  • 1 No of 200W,230V, 0.8 P.F Computer.
  • 2 No of 30W,230V, 0.8 P.F Tube Light.

Inverter / Battery Detail:

  • Additional Further Load Expansion (Af)=20%
  • Efficiency of Inverter (Ie) = 80%
  • Required Battery Backup (Bb) = 2 Hours.
  • Battery Bank Voltage = 24V DC
  • Loose Connection/Wire Loss Factor (LF) = 20%
  • Battery Efficiency (n) = 90%
  • Battery Aging Factor (Ag) =20%
  • Depth of Discharge (DOD) =50%
  • Battery Operating Temp =46ºC
Temp. °C Factor
80 1.00
70 1.04
60 1.11
50 1.19
40 1.30
30 1.40
20 1.59

Calculation:

Step 1: Calculate Total Load:

  • Fan Load= No x Watt =2×60=120 Watt
  • Fan Load=(No x Watt)/P.F=(2×60)/0.8= 150VA
  • Computer Load= No x Watt =1×200=200 Watt
  • Computer Load=(No x Watt)/P.F =(1×200)/0.8= 250VA
  • Tube Light Load= No x Watt =2×30=60 Watt
  • Tube Light Load=(No x Watt)/P.F =(2×30)/0.8= 75VA
  • Total Electrical Load=120+200+60 =380 Watt
  • Total Electrical Load=150+250+75= 475VA

Step 2: Size of Inverter:

  • Size of Inverter=Total Load+(1+Af) / Ie VA
  • Size of Inverter= 475+(1+20%) / 80%
  • Size of Inverter= 712 VA

Step 3: Size of Battery:

  • Total Load of Battery Bank= (Total Load x Backup Capacity) / Battery Bank Volt
  • Total Load of Battery Bank=(380 x 2) / 24 Amp Hr
  • Total Load of Battery Bank= 32.66 Amp Hr
  • Temperature Correction Factor for 46ºC (Tp)=1
  • Size of Battery Bank=[ (Load) x (1+LF) x (1+Ag) x Tp] / [n x DOD] Amp/Hr
  • Size of Battery Bank= (32.66 x (1+20%) x (1+20%) x 1) / (90% x 50%)
  • Size of Battery Bank= 101.3 Amp/Hr

Step 4: Connection of Battery:

If We Select 120 Amp Hr , 12V DC Battery for Battery Bank:

Series Connection:

  • Series configurations will add the voltage of the two batteries but keep the amperage rating (Amp Hours) same.
  • Condition-I :
  • Selection of Battery for Voltage = Volt of Each Battery <= Volt of Battery Bank
  • Selection of Battery for Voltage =12< 24
  • Condition-I is O.K
  • No of Battery for Voltage = Volt of Battery Bank / Volt of Each Battery
  • No of Battery for Voltage =24/12 = 2 No’s
  • Condition-II :
  • Selection of Battery for Amp Hr = Amp Hr of Battery Bank <= Amp Hr of Each Battery
  • Selection of Battery for Amp Hr =3<=120
  • Condition-II is O.K
  • We can use Series Connection for Battery & No of Battery required 2 No’s

series_battery_configParallel Configuration

  • In Parallel connection, the current rating will increase but the voltage will be the same.
  • More the number of batteries more will be the amp/hour.  Two batteries will produce twice the amp/hour of a single battery.
  • Condition-I :
  • Selection of Battery for Amp Hr = Amp Hr of Battery Bank / Amp Hr of Each Battery <=1
  • Selection of Battery for Amp Hr =101/120 = 0.84=1 No’s
  • Condition-I is O.K
  • Condition-II :
  • Selection of Battery for Voltage = Volt of Battery Bank = Volt of Each Battery
  • Condition-II :Selection of Battery for Voltage for Amp Hr = 24<=12
  • Condition-II is Not Full Fill
  • We cannot use Parallel Connection for Battery as per our requirement But If We do Practically It is Possible and it will give more Hours of back

parallel_battery_configSeries-Parallel Connection:

  • Connecting the batteries up in series will increase both the voltage and the run time.
  • Condition-I :
  • Selection of Battery for Amp Hr = Amp Hr of Each Battery <= Amp Hr of Battery Bank
  • Selection of Battery for Amp Hr =120<=101
  • Condition-I is Not Full Fill
  • Condition-II :
  • Selection of Battery for Voltage = Volt of Each Battery <= Volt of Battery Bank
  • Selection of Battery for Voltage = 12<=24
  • Condition-II is OK
  • We cannot use Parallel Connection for Battery

If We Select 60 Amp Hr , 12V DC Battery for Battery Bank:

Series Connection:

  • Selection of Battery for Voltage = Volt of Each Battery <= Volt of Battery Bank
  • Selection of Battery for Voltage =12< 24
  • Condition-I is O.K
  • No of Battery for Voltage = Volt of Battery Bank / Volt of Each Battery
  • No of Battery for Voltage =24/12 = 2 No’s
  • Condition-II :
  • Selection of Battery for Amp Hr = Amp Hr of Battery Bank <= Amp Hr of Each Battery
  • Selection of Battery for Amp Hr =3<=60
  • Condition-II is Not Full Fill
  • We can use Series Connection for Battery

Parallel Configuration

  • Condition-I :
  • Selection of Battery for Amp Hr = Amp Hr of Battery Bank / Amp Hr of Each Battery <=1
  • Selection of Battery for Amp Hr =101/60 = 1.63=1 No’s
  • Condition-I is O.K
  • Condition-II :
  • Selection of Battery for Voltage = Volt of Battery Bank = Volt of Each Battery
  • Condition-II :Selection of Battery for Voltage for Amp Hr = 24=12
  • Condition-II is Not Full Fill
  • We cannot use Parallel Connection for Battery as per our requirement.

Series-Parallel Connection:

  • Condition-I :
  • Selection of Battery for Amp Hr = Amp Hr of Each Battery <= Amp Hr of Battery Bank
  • Selection of Battery for Amp Hr =120<=60
  • Condition-I is OK
  • No of Battery for Amp Hr = Amp Hr of Battery Bank / Amp Hr of Each Battery
  • No of Battery for Amp Hr = 120/60 =1.68 =2 No’s
  • Condition-II :
  • Selection of Battery for Voltage = Volt of Each Battery <= Volt of Battery Bank
  • Selection of Battery for Voltage = 12<=24
  • Condition-II is OK
  • No of Battery for Voltage = Volt of Battery Bank / Volt of Each Battery
  • No of Battery for Voltage = 24 / 12 =2 No’s
  • No of Battery Required = No of Battery Amp Hr x No of Battery for Voltage
  • No of Battery Required = 2 x 2= 4 No’s
  • We can use Series-Parallel Connection for Batteryseries-parallel_battery_config

Summary:

  • Total Electrical Load=380 Watt
  • Total Electrical Load=475VA
  • Size of Inverter= 712 VA
  • Size of Battery Bank= 101.3 Amp/Hr
  • For 120 Amp/Hr , 12V DC Battery : Series Connection & 2 No’s of Battery or
  • For 60 Amp/Hr , 12V DC Battery : Series-Parallel Connection & 4 No’s of Battery

Calculate Size of Main ELCB & Brach MCB of Distribution Box


Design Distribution Box of one House and Calculation of Size of Main ELCB and branch Circuit MCB as following Load Detail. Power Supply is 430V (P-P), 230 (P-N), 50Hz. Consider Demand Factor 0.6 for Non Continuous Load & 1 for Continuous Load for Each Equipment.

  • Branch Circuit-1: 4 No of 1Phase, 40W, Lamp of Non Continues Load + 2 No’s of 1Ph, 60W, Fan of Non Continues Load.
  • Branch Circuit-2: 2 No of 1Ph, 200W, Computer of Non Continues Load.
  • Branch Circuit-3: 1 No of 1Ph, 200W, Freeze of Continues Load.
  • Branch Circuit-4: 8 No of 1Ph, 40W, Lamp of Non Continues Load + 2 No’s of 1Ph  60W, Fan of Non Continues Load.
  • Branch Circuit-5: 4 No of 1Ph , 40W, Lamp of Non Continues Load + 1 No’s of 1Ph 60W, Fan of Non Continues Load.+ 1 No’s of 1Ph 150W, TV of Continues Load
  • Branch Circuit-6: 1 No of 1P , 1.7KW, Geyser of Non Continues Load.
  • Branch Circuit-7: 1 No of 1Ph, 3KW, A.C of Non Continues Load.
  • Branch Circuit-8: 1 No of 3Ph, 1HP, Motor-Pump of Non Continues Load.

Untitled

Fault Current
Voltage Fault Current
230V 6KA
430V 10KA
11KV 25KA
Class of MCB/ELCB/RCCB
Type of Load Class Sensitivity
Lighting B Class I∆n:30ma
Heater B Class I∆n:30ma
Drive C Class I∆n:100ma
A.C C Class I∆n:30ma
Motor C Class I∆n:100ma
Ballast C Class I∆n:30ma
Induction Load C Class I∆n:100ma
Transformer D Class I∆n:100ma
     
     
Size of MCB/ELCB
Current (Amp) Lighting Load MCB/ELCB (Amp) Heating/Cooling/Motor-Pump Load MCB/ELCB (Amp)
1.0 to 4.0 6 16
6.0 10 16
10.0 16 16
16.0 20 20
20.0 25 25
25.0 32 32
32.0 40 40
40.0 45 45
45.0 50 50
50.0 63 63
63.0 80 80
80.0 100 100
100.0 125 125
125.0 225 225
225.0 600 600
600.0 800 800
800.0 1600 1600
1600.0 2000 2000
2000.0 3000 3000
3000.0 3200 3200
3200.0 4000 4000
4000.0 5000 5000
5000.0 6000 6000
6000.0 6000 6000

Calculation:

Size of MCB for Branch Circuit-1:

  • Load Current of Lamp= (No X Watt X Demand Factor)/Volt =(4X40X0.6)/230=0.40Amp
  • Load Current of Fan= (No X Watt X Demand Factor)/Volt =(2X60X0.6)/230=0.31Amp
  • Branch Circuit-1 Current as per NEC = Non Continues Load+125% Continues Load
  • Branch Circuit-1 Current as per NEC =(0.4+0.31)+125%(0) =0.73Amp
  • Type of Load=Lighting Type
  • Class of MCB=B Class
  • Size of MCB=6 Amp
  • No of Pole of MCB=Single Pole

Size of MCB for Branch Circuit-2:

  • Load Current of Computer = (No X Watt X Demand Factor)/Volt =(2X200X0.6)/230=1.04Amp
  • Branch Circuit-2 Current as per NEC = Non Continues Load+125% Continues Load
  • Branch Circuit-2 Current as per NEC =(1.04)+125%(0) =1.04Amp
  • Type of Load=Lighting Type
  • Class of MCB=B Class
  • Size of MCB=6 Amp
  • Breaking Capacity: 6KA
  • No of Pole of MCB=Single Pole

Size of MCB for Branch Circuit-3:

  • Load Current of Freeze= (No X Watt X Demand Factor)/Volt =(1X200X0.6)/230=0.87Amp
  • Branch Circuit-3 Current as per NEC = Non Continues Load+125% Continues Load
  • Branch Circuit-3 Current as per NEC =(0.87)+125%(0) =0.87Amp
  • Type of Load=Lighting Type
  • Class of MCB=B Class
  • Size of MCB=6 Amp
  • Breaking Capacity: 6KA
  • No of Pole of MCB=Single Pole

Size of MCB for Branch Circuit-4:

  • Load Current of Lamp= (No X Watt X Demand Factor)/Volt =(8X40X0.6)/230=0.83Amp
  • Load Current of Fan= (No X Watt X Demand Factor)/Volt =(2X60X0.6)/230=0.31Amp
  • Branch Circuit-4 Current as per NEC = Non Continues Load+125% Continues Load
  • Branch Circuit-4 Current as per NEC =(0.83+0.31)+125%(0) =1.15Amp
  • Type of Load=Lighting Type
  • Class of MCB=B Class
  • Size of MCB=6 Amp
  • Breaking Capacity: 6KA
  • No of Pole of MCB=Single Pole

Size of MCB for Branch Circuit-5:

  • Load Current of Lamp= (No X Watt X Demand Factor)/Volt =(4X40X0.6)/230=0.42Amp
  • Load Current of Fan= (No X Watt X Demand Factor)/Volt =(1X60X0.6)/230=0.16Amp
  • Load Current of TV = (No X Watt X Demand Factor)/Volt =(1X150X1)/230=0.65Amp
  • Branch Circuit-5 Current as per NEC = Non Continues Load+125% Continues Load
  • Branch Circuit-5 Current as per NEC =(0.42+0.16)+125%(0.65) =0.57+0.82=1.39Amp
  • Type of Load=Lighting Type
  • Class of MCB=B Class
  • Size of MCB=6 Amp
  • Breaking Capacity: 6KA
  • No of Pole of MCB=Single Pole

Size of MCB for Branch Circuit-6:

  • Load Current of Geyser= (No X Watt X Demand Factor)/Volt =(1X1700X0.6)/230=4.43Amp
  • Branch Circuit-6 Current as per NEC = Non Continues Load+125% Continues Load
  • Branch Circuit-6 Current as per NEC =(4.43)+125%(0) =4.43Amp
  • Type of Load=Heating & Cooling Type
  • Class of MCB=C Class
  • Size of MCB=16 Amp
  • Breaking Capacity: 6KA
  • No of Pole of MCB=Single Pole

Size of MCB for Branch Circuit-7:

  • Load Current of A.C= (No X Watt X Demand Factor)/Volt =(1X3000X0.6)/230=7.83Amp
  • Branch Circuit-7 Current as per NEC = Non Continues Load+125% Continues Load
  • Branch Circuit-7 Current as per NEC =(7.83)+125%(0) =7.83Amp
  • Type of Load=Heating & Cooling Type
  • Class of MCB=C Class
  • Size of MCB=16 Amp
  • Breaking Capacity: 6KA
  • No of Pole of MCB=Single Pole

Size of MCB for Branch Circuit-8:

  • Load Current of Motor-Pump = (No X Watt X Demand Factor)/(1.732XVolt ) =(1X746X0.6)/(1.732X430)=0.60Amp
  • Branch Circuit-8 Current as per NEC =Non Continues Load+125% Continues Load
  • Branch Circuit-8 Current as per NEC =(0.60)+125%(0) =0.60Amp
  • Type of Load=Motor-Pump Type
  • Class of MCB=C Class
  • Size of MCB=16 Amp
  • Breaking Capacity: 10KA
  • No of Pole of MCB= Three Pole

Size of Main ELCB/RCCB:

  • Total Brach Circuit Load & MCB Detail is
Brach Circuit Total

Current (Amp)

Size of MCB (Amp) Class of MCB Breaking Capacity of MCB Pole of MCB
circuit-1 0.73 6 Amp B Class 6KA SP
circuit-2 1.04 6 Amp B Class 6KA SP
circuit-3 0.87 6 Amp B Class 6KA SP
circuit-4 1.15 6 Amp B Class 6KA SP
circuit-5 1.39 6 Amp B Class 6KA SP
circuit-6 4.43 16 Amp C Class 6KA SP
circuit-7 7.83 16 Amp C Class 6KA SP
circuit-8 0.63 16 Amp C Class 10KA TP
Total 18.04        
  • Total Load Current as per NEC= 18.04Amp——–(A)
  • Max Size of Branch circuit MCB=16Amp
  • Total Current of Panel as per Branch Circuit= 2X Max Size of Branch circuit MCB
  • Total Load Current of Panel as per Branch Circuit =2X16=32Amp——(B)
  • Total Load Current of Panel as per NEC = Maximum of (A) and (B)
  • Total Load Current of Panel as per NEC=32Amp
  • Min Size of ELCB/RCCB as per NEC=40Amp
  • Class of ELCB/RCCB= B or C Class
  • No of Pole of ELCB/RCCB=TP or FP
  • Sensitivity( I∆n)=30ma
  • Breaking Capacity=10KA

Size of Distribution Board:

  • No of Single Pole Branch Circuit MCB (SP)= 7 No’s
  • No of Three Pole Branch Circuit MCB (TP)= 1 No’s
  • Main ELCB (TP)=1 No’s
  • Total No of Way of D.B (SPN)= (SP)+3X(TP)=7+(3X2)=13Way SPN
  • Total No of Way of D.B (TPN)= (SP)/3+ (TP)=(7/3)+(2)=4+2=6Way SPN
  • Select Either 14Way SPN or 6 Way TPN

Load Balancing of Distribution Board:

  • To Balance Load We need to try distribute Single Phase load on each Phase.
  • Suppose We connect Branch Circuit load on following Phase
Brach Circuit Current (Amp) Type of Load Connection on
Branch circuit-1 0.73 Single Phase        Y Phase
Branch circuit-2 1.04 Single Phase B Phase
Branch circuit-3 0.87 Single Phase Y Phase
Branch circuit-4 1.15 Single Phase B Phase
Branch circuit-5 1.39 Single Phase Y Phase
Branch circuit-6 4.43 Single Phase B Phase
Branch circuit-7 7.83 Single Phase R Phase
Branch circuit-8 0.63 Three Phase RYB Phase
Summary of Load
R Phase Load 8.5 Amp
Y Phase Load 3.5 Amp
B Phase Load 7.23 Amp
Total Load 18.04 Amp

Summary of Distribution Box :

Untitled - Copy

  • Size of Distribution Box : 14Way SPN or 6 Way TPN
  • Size of Main ELCB: 40A,B or C Class,30ma,10KA
  • Size & No of Branch MCB: 5 No’s of 6A,SP, B Class,6KA
  • Size & No of Branch MCB: 2 No’s of 16A,SP, C Class,6KA
  • Size & No of Branch MCB: 1 No’s of 16A,TP, C Class,10KA
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