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
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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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