Short Circuit Current Calculation (Base KVA Method)


Example:

Calculate Fault current at each stage of following Electrical System SLD having details of.

  • Main Incoming HT Supply Voltage is 6.6 KV.
  • Fault Level at HT Incoming Power Supply is 360 MVA.
  • Transformer Rating is 2.5 MVA.
  • Transformer Impedance is 6%.

UntitledCalculation:

  • Let’s first consider Base KVA and KV for HT and LT Side.
  • Base KVA for HT side (H.T. Breaker and Transformer Primary) is 6 MVA
  • Base KV for HT side (H.T. Breaker and Transformer Primary) is 6.6 KV
  • Base KVA for LT side (Transformer Secondary and down Stream) is 2.5 MVA
  • Base KV for LT side (Transformer Secondary and down Stream) is 415V

Fault Level at HT Side (Up to Sub-station):

(1) Fault Level from HT incoming Line to HT Circuit Breaker

  • HT Cable used from HT incoming to HT Circuit Breaker is 5 Runs , 50 Meter ,6.6KV 3 Core 400 sq.mm Aluminum Cable , Resistance of Cable 0.1230 Ω/Km and Reactance of Cable is0.0990 Ω/Km.
  • Total Cable Resistance(R)= (Length of Cable X Resistance of Cable) / No of Cable.
  • Total Cable Resistance=(0.05X0.1230) / 5
  • Total Cable Resistance=0.001023 Ω
  • Total Cable Reactance(X)= (Length of Cable X Reactance of Cable) / No of Cable.
  • Total Cable Reactance=(0.05X0.0990) / 5
  • Total Cable Reactance =0.00099 Ω
  • Total Cable Impedance (Zc1)=√(RXR)+(XxX)
  • Total Cable Impedance (Zc1)=0.0014235 Ω——–(1)
  • U Reactance at H.T. Breaker Incoming Terminals (X Pu)= Fault Level / Base KVA
  • U Reactance at H.T. Breaker Incoming Terminals (X Pu)= 360 / 6
  • U. Reactance at H.T. Breaker Incoming Terminals(X Pu)= 0.01666 PU——(2)
  • Total Impedance up to HT Circuit Breaker (Z Pu-a)= (Zc1)+ (X Pu) =(1)+(2)
  • Total Impedance up to HT Circuit Breaker(Z Pu-a)=0.001435+0.01666
  • Total Impedance up to HT Circuit Breaker (Z Pu-a)=0.0181 Ω.——(3)
  • Fault MVA at HT Circuit Breaker= Base MVA / Z Pu-a.
  • Fault MVA at HT Circuit Breaker= 6 / 0.0181
  • Fault MVA at HT Circuit Breaker= 332 MVA
  • Fault Current = Fault MVA / Base KV
  • Fault Current = 332 / 6.6
  • Fault Current at HT Circuit Breaker = 50 KA

(2) Fault Level from HT Circuit Breaker to Primary Side of Transformer

  • HT Cable used from HT Circuit Breaker to Transformer is 3 Runs , 400 Meter ,6.6KV 3 Core 400 sq.mm Aluminium Cable , Resistance of Cable 0.1230 Ω/Km and Reactance of Cable is0.0990 Ω/Km.
  • Total Cable Resistance(R)= (Length of Cable X Resistance of Cable) / No of Cable.
  • Total Cable Resistance=(0.4X0.1230) / 3
  • Total Cable Resistance=0.01364 Ω
  • Total Cable Reactance(X)= (Length of Cable X Reactance of Cable) / No of Cable.
  • Total Cable Reactance=(0.4X0.0990) / 5
  • Total Cable Reactance =0.01320 Ω
  • Total Cable Impedance (Zc2)=√(RXR)+(XxX)
  • Total Cable Impedance (Zc2)=0.01898 Ω——–(4)
  • U Impedance at Primary side of Transformer (Z Pu)= (Zc2 X Base KVA) / (Base KV x Base KVx1000)
  • U Impedance at Primary side of Transformer (Z Pu)= (0.01898X6) /(6.6×6.6×1000)
  • U Impedance at Primary side of Transformer (Z Pu)= 0.0026145 PU——(5)
  • Total Impedance(Z Pu)=(4) + (5)
  • Total Impedance(Z Pu)=0.01898+0.0026145
  • Total Impedance(Z Pu)=0.00261——(6)
  • Total Impedance up to Primary side of Transformer (Z Pu-b)= (Z Pu)+(Z Pu-a) =(6)+(3)
  • Total Impedance up to Primary side of Transformer (Z Pu-b)= 0.00261+0.0181
  • Total Impedance up to Primary side of Transformer (Z Pu-b)=0.02070 Ω.—–(7)
  • Fault MVA at Primary side of Transformer = Base MVA / Z Pu-b.
  • Fault MVA at Primary side of Transformer = 6 / 0.02070
  • Fault MVA at Primary side of Transformer = 290 MVA
  • Fault Current = Fault MVA / Base KV
  • Fault Current = 290 / 6.6
  • Fault Current at Primary side of Transformer = 44 KA

(3) Fault Level from Primary Side of Transformer to Secondary side of Transformer:

  • Transformer Rating is 2.5 MVA and Transformer Impedance is 6%.
  • % Reactance at Base KVA = (Base KVA x % impedance at Rated KVA) / Rated KVA
  • % Reactance at Base KVA = (2.5X6)/2.5
  • % Reactance at Base KVA =6%
  • U. Reactance of the Transformer(Z Pu) =% Reactance /100
  • U. Reactance of the Transformer(Z Pu)= 6/100=0.06 Ω—–(8)
  • Total P.U. impedance up to Transformer Secondary Winding(Z Pu-c)=(Z Pu)+(Z Pu-b)=(7)+(8)
  • Total P.U. impedance up to Transformer Secondary Winding(Z Pu-c)=0.06+0.02070
  • Total P.U. impedance up to Transformer Secondary Winding(Z Pu-c)=0.0807 Ω—–(9)
  • Fault MVA at Transformer Secondary Winding = Base MVA / Z Pu-c
  • Fault MVA at Transformer Secondary Winding = 2.5/0.0807
  • Fault MVA at Transformer Secondary Winding =31 MVA
  • Fault Current = Fault MVA / Base KV
  • Fault Current = 31 / (1.732×0.415)
  • Fault Current at Transformer Secondary Winding = 43 KA

Fault Level at LT Side (Sub-station to Down stream):

(4) Fault Level from Transformer Secondary to Main LT Panel

  • LT Cable used from Transformer Secondary to Main LT Panel is 13 Runs , 12 Meter , 1KV, 3.5C x 400 Sq.mm Aluminium Cable , Resistance of Cable 0.1230 Ω/Km and Reactance of Cable is0.0618 Ω/Km.
  • Total Cable Resistance(R)= (Length of Cable X Resistance of Cable) / No of Cable.
  • Total Cable Resistance=(0.012X0.1230) / 13
  • Total Cable Resistance=0.00009 Ω
  • Total Cable Reactance(X)= (Length of Cable X Reactance of Cable) / No of Cable.
  • Total Cable Reactance=(0.012X0.0618) / 13
  • Total Cable Reactance =0.00006 Ω
  • Total Cable Impedance (Zc3)=√(RXR)+(XxX)
  • Total Cable Impedance (Zc3)=0.00011 Ω——–(10)
  • U Impedance at Main LT Panel (Z Pu)= (Zc3 X Base KVA) / (Base KV x Base KVx1000)
  • U Impedance at Main LT Panel (Z Pu)= (0.00011X2.5×1000)/(0.415×0.415X1000)
  • P P.U Impedance at Main LT Panel (Z Pu)= 001601 Ω ——(11)
  • Total Impedance up to Main LT Panel (Z Pu-d)= (Zc3)+ (Z Pu-c) =(11)+(9)
  • Total Impedance up to Main LT Panel (Z Pu-d)= 0.001601 +0.0807
  • Total Impedance up to Main LT Panel (Z Pu-d)=0.082306 Ω.——(12)
  • Fault MVA at Main LT Panel = Base MVA / Z Pu-a.
  • Fault MVA at Main LT Panel = 2.5 / 0.082306
  • Fault MVA at Main LT Panel = 30 MVA
  • Fault Current = Fault MVA / Base KV
  • Fault Current = 30 / (1.732X0.415)
  • Fault Current at Main Lt Panel = 42 KA

(5) Fault Level from Main LT Panel to Sub Panel:

  • LT Cable used from Main LT Panel to Sub Panel is 2 Runs , 160 Meter , 1KV, 3.5C x 400 Sq.mm Aluminium Cable , Resistance of Cable 0.1230 Ω/Km and Reactance of Cable is0.0618 Ω/Km.
  • Total Cable Resistance(R)= (Length of Cable X Resistance of Cable) / No of Cable.
  • Total Cable Resistance=(0.160X0.1230) / 2
  • Total Cable Resistance=0.008184 Ω
  • Total Cable Reactance(X)= (Length of Cable X Reactance of Cable) / No of Cable.
  • Total Cable Reactance=(0.160X0.0618) / 2
  • Total Cable Reactance =0.004944 Ω
  • Total Cable Impedance (Zc4)=√(RXR)+(XxX)
  • Total Cable Impedance (Zc4)=0.0095614 Ω——–(13)
  • U Impedance at Sub Panel (Z Pu)= (Zc4 X Base KVA) / (Base KV x Base KVx1000)
  • U Impedance at Sub Panel (Z Pu)= (0.0095614 X2.5×1000)/(0.415×0.415X1000)
  • P P.U Impedance at Sub Panel (Z Pu)= 13879 Ω ——(14)
  • Total Impedance up to Sub Panel (Z Pu-e)= (Zc4)+ (Z Pu-d) =(14)+(12)
  • Total Impedance up to Sub Panel (Z Pu-e)= 0.13879 +0.082306
  • Total Impedance up to Sub Panel (Z Pu-e)=0.2211 Ω.——(15)
  • Fault MVA at Sub Panel = Base MVA / Z Pu-a.
  • Fault MVA at Sub Panel = 2.5 / 0.2211
  • Fault MVA at Sub Panel = 11 MVA
  • Fault Current = Fault MVA / Base KV
  • Fault Current = 11 / (1.732X0.415)
  • Fault Current at Sub Panel = 16 KA

(6) Fault Level from Sub Panel to Motor Control Panel:

  • LT Cable used from Sub Panel to Motor Control Panel is 6 Runs , 150 Meter , 1KV, 3.5C x 400 Sq.mm Aluminium Cable , Resistance of Cable 0.1230 Ω/Km and Reactance of Cable is0.0739 Ω/Km.
  • Total Cable Resistance(R)= (Length of Cable X Resistance of Cable) / No of Cable.
  • Total Cable Resistance=(0.150X0.1230) / 6
  • Total Cable Resistance=0.003075 Ω
  • Total Cable Reactance(X)= (Length of Cable X Reactance of Cable) / No of Cable.
  • Total Cable Reactance=(0.150X0.0739) / 6
  • Total Cable Reactance =0.0018475 Ω
  • Total Cable Impedance (Zc5)=√(RXR)+(XxX)
  • Total Cable Impedance (Zc4)=0.003587 Ω——–(16)
  • U Impedance at Motor Control Panel (Z Pu)= (Zc5 X Base KVA) / (Base KV x Base KVx1000)
  • U Impedance at Motor Control Panel (Z Pu)= (0.003587 X2.5×1000)/(0.415×0.415X1000)
  • P P.U Impedance at Motor Control Panel (Z Pu)= 05207 Ω ——(17)
  • Total Impedance up to Motor Control Panel (Z Pu-f)= (Zc5)+ (Z Pu-e) =(17)+(15)
  • Total Impedance up to Motor Control Panel (Z Pu-e)= 0.13879 +0.2211
  • Total Impedance up to Motor Control Panel (Z Pu-e)=0.27317 Ω.——(15)
  • Fault MVA at Motor Control Panel = Base MVA / Z Pu-a.
  • Fault MVA at Motor Control Panel = 2.5 / 0.27317
  • Fault MVA at Motor Control Panel = 9 MVA
  • Fault Current = Fault MVA / Base KV
  • Fault Current = 9 / (1.732X0.415)
  • Fault Current at Motor Control Panel = 13 KA

 Summary of Calculation:

Sr.No Fault Location Fault MVA Fault Current (KA)

1

At HT Circuit Breaker

332

50

2

At Primary Side of Transformer

290

44

3

At Secondary Side of Transformer

31

43

4

At Main LT Panel

30

42

5

At Sub Main Panel

11

16

6

At Motor Control Panel

9

13

 

Motor Name Plate Terminology


General Terminology

(1) Service Factor:

  • The service factor is a multiplier that indicates the amount of overload a motor can be expected to handle. If a motor with a 1.15 service factor can be expected to safely handle intermittent loads amounting to 15% beyond its nameplate horsepower.
  • For example, many motors will have a service factor of 1.15, meaning that the motor can handle a 15% overload. The service factor amperage is the amount of current that the motor will draw under the service factor load condition.

(2) Slip:

  • Slip is used in two forms. One is the slip RPM which is the difference between the synchronous speed and the full load speed. When this slip RPM is expressed as a percentage of the synchronous speed, then it is called percent slip or just “slip”. Most standard motors run with a full load slip of 2% to 5%.

(3) Synchronous Speed:

  • This is the speed at which the magnetic field within the motor is rotating. It is also approximately the speed that the motor will run under no load conditions. For example, a 4 pole motor running on 60 cycles would have a magnetic field speed of 1800 RPM. The no load speed of that motor shaft would be very close to 1800, probably 1798 or 1799 RPM. The full load speed of the same motor might be 1745 RPM. The difference between the synchronous speed and the full load speed is called the slip RPM of the motor.

Untitled

Motor Torque:

(1) Pull Up Torque:

  • When the motor starts and begins to accelerate the torque in generally decrease until it reach a low point at a certain speed it called the pull-up torque.
  • The Pull-up Torque is the minimum torque developed by the electrical motor when it runs from zero to full-load speed (before it reaches the break-down torque point).
  • Pull-up torque is the minimum torque developed during the period of acceleration from locked-rotor to the speed at which breakdown torque occurs.
  • Some motor designs do not have a value of pull up torque because the lowest point may occur at the locked rotor point. In this case, pull up torque is the same as locked rotor torque.
  • For motors which do not have a definite breakdown torque (such as NEMA design D) pull-up torque is the minimum torque developed up to rated full-load speed. It is usually expressed as a percentage of full-load torque.

(2) Starting Torque (Locked Rotor Torque):

  • The amount of torque the motor produces when it is energized at full voltage and with the shaft locked in place is called starting torque.
  • The Locked Rotor Torque or Starting Torque is the torque the electrical motor develop when its starts at rest or zero speed.
  • It is the amount of torque available when power is applied to break the load away and start accelerating it up to speed.
  • A high Starting Torque is more important for application or machines hard to start – as positive displacement pumps, cranes etc. A lower Starting Torque can be accepted in applications as centrifugal fans or a pump where the start loads is low or close to zero.

 (3) Full Load Torque:

  • Full load torque is the rated continuous torque that the motor can support without overheating within its time rating.
  • In imperial units the Full-load Torque can be expressed as
  • T full-load torque (lb ft) = (Rated horsepower of Motor X 5252) / Rated rotational speed (rpm)       
  • In metric units the rated torque can be expressed as
  • Full-load torque (Nm) = (Rated KW of Motor X 9550) / Rated rotational speed (rpm)
  • Example :The torque of a 60 hp motor rotating at 1725 rpm can be expressed as
  • T full-load torque = 60 X 5,252 / 1725 (rpm)
  • T full-load torque = 182.7 lb ft

(4) Peak Torque:

  • Many types of loads such as reciprocating compressors have cycling torques where the amount of torque required varies depending on the position of the machine.
  • The actual maximum torque requirement at any point is called the peak torque requirement.
  • Peak torques is involved in things such as punch presses and other types of loads where an oscillating torque requirement occurs.

(5) Pull out Torque (Breakdown Torque):

  • Breakdown torque is the maximum torque the motor will develop with rated voltage applied at rated frequency without an abrupt drop in speed. Breakdown torque is usually expressed as a percentage of full-load torque
  • The load is then increased until the maximum point is reached.

Motor Current:

(1) Full Load Amps:

  • The amount of current the motor can be expected to draw under full load (torque) conditions is called Full Load Amps. It is also known as nameplate amps.

(2) Locked Rotor Amps:

  • Also known as starting inrush, this is the amount of current the motor can be expected to draw under starting conditions when full voltage is applied.
  • Lock Rotor Current (IL) Three Phase Motor: 1000x HP x (KVA/HP) / 1.732 x Volt
  • Lock Rotor Current (IL) Single Phase Motor: 1000x HP x (KVA/HP) / Volt

 

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 !!!!!!!!!!!!!!!

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