Electrical Thumb Rules

Title: Illuminating Insights: Electrical Basics and General Thumb Rules

Introduction:
Understanding the fundamentals of electricity is crucial for anyone working with electrical systems, whether it’s in a professional capacity or simply for household maintenance. In this guide, we’ll explore some essential electrical basics and general thumb rules to help you navigate the world of electricity safely and effectively.

1. Voltage, Current, and Resistance:

• Voltage (V): Voltage is the force that pushes electrical charges through a conductor. It’s measured in volts (V) and represents the potential energy difference between two points in an electrical circuit.
• Current (I): Current is the flow of electrical charges through a conductor. It’s measured in amperes (A) and represents the rate of flow of electric charge.
• Resistance (R): Resistance is the opposition to the flow of electrical current in a circuit. It’s measured in ohms (Ω) and is determined by the material, length, and cross-sectional area of the conductor.

1. Ohm’s Law:

• Ohm’s Law states that the current (I) flowing through a conductor between two points is directly proportional to the voltage (V) across the two points and inversely proportional to the resistance (R) between them. Mathematically, it’s expressed as: V = I * R or I = V / R or R = V / I.

1. Power:

• Power (P) in an electrical circuit is the rate at which work is done or energy is transferred. It’s measured in watts (W) and is calculated as the product of voltage (V) and current (I): P = V * I.

1. Basic Safety Rules:

• Always turn off the power before working on any electrical circuit or device.
• Use insulated tools and wear appropriate personal protective equipment (PPE) when working with electricity.
• Never overload electrical circuits or outlets.
• Keep electrical cords away from heat sources, water, and sharp objects.
• Regularly inspect electrical cords, outlets, and switches for signs of wear or damage.

1. General Thumb Rules:

• The National Electrical Code (NEC) recommends that electrical outlets should be spaced no more than 12 feet apart in residential buildings.
• For safety, it’s generally advised to limit the load on a circuit to 80% of its maximum capacity.
• The standard voltage for residential buildings in the United States is 120 volts AC, while commercial buildings often use 240 volts AC.
• The resistance of a wire increases with its length and decreases with its cross-sectional area.

•   Cable Capacity:

•   For Cu Wire Current Capacity (Up to 30 Sq.mm) = 6X Size of Wire in Sq.mm

•   Ex. For 2.5 Sq.mm=6×2.5=15 Amp, For 1 Sq.mm=6×1=6 Amp, For 1.5 Sq.mm=6×1.5=9

Amp

•   For Cable Current Capacity = 4X Size of Cable in Sq.mm ,Ex. For 2.5

Sq.mm=4×2.5=9 Amp.

•   Nomenclature for cable Rating = Uo/U

•   where Uo=Phase-Ground Voltage, U=Phase-Phase Voltage, Um=Highest Permissible

Voltage

•   Current Capacity of Equipments:

•   1 Phase Motor draws Current=7Amp per HP.

•   3 Phase Motor draws Current=1.25Amp per HP.

•   Full Load Current of 3 Phase Motor=HPx1.5

•   Full Load Current of 1 Phase Motor=HPx6

•   No Load Current of 3 Phase Motor =30% of FLC

•   KW Rating of Motor=HPx0.75

•   Full Load Current of equipment =1.39xKVA (for 3 Phase 415Volt)

•   Full Load Current of equipment =1.74xKw (for 3 Phase 415Volt)

•   Earthing Resistance:

•   Earthing Resistance for Single Pit=5Ω ,Earthing Grid=0.5Ω

•   As per NEC 1985 Earthing Resistance should be <5Ω.

•   Voltage between Neutral and Earth <=2 Volts

•   Resistance between Neutral and Earth <=1Ω

•   Creepage Distance=18 to 22mm/KV (Moderate Polluted Air) or

•   Creepage Distance=25 to 33mm/KV (Highly Polluted Air)

•   Minimum Bending Radius:

•   Minimum Bending Radius for LT Power Cable=12xDia of Cable.

•   Minimum Bending Radius for HT Power Cable=20xDia of Cable.

•   Minimum Bending Radius for Control Cable=10xDia of Cable.

•   Insulation Resistance:

•   Insulation Resistance Value for Rotating Machine= (KV+1) MΩ.

•    Insulation Resistance Value for Motor (IS 732) = ((20xVoltage (L-L)) / (1000+ (2xKW)).

•   Insulation Resistance Value for Equipment (<1KV) = Minimum 1 MΩ.

•   Insulation Resistance Value for Equipment (>1KV) = KV 1 MΩ per 1KV.

•   Insulation Resistance Value for Panel = 2 x KV rating of the panel.

•   Min Insulation Resistance Value (Domestic) = 50 MΩ / No of Points. (All Electrical

Points with Electrical fitting & Plugs). Should be less than 0.5 MΩ

•   Min Insulation Resistance Value (Commercial) = 100 MΩ / No of Points. (All

Electrical Points without fitting & Plugs).Should be less than 0.5 MΩ.

•   Test Voltage (A.C) for Meggering = (2X Name Plate Voltage) +1000

•   Test Voltage (D.C) for Meggering = (2X Name Plate Voltage).

•   Submersible Pump Take 0.4 KWH of extra Energy at 1 meter drop of Water.

•   Lighting Arrestor:

•   Arrestor have Two Rating=

•   (1) MCOV=Max. Continuous Line to Ground Operating Voltage.

•   (2) Duty Cycle Voltage. (Duty Cycle Voltage>MCOV).

•   Transformer:

•   Current Rating of Transformer=KVAx1.4

•   Short Circuit Current of T.C /Generator= Current Rating / % Impedance

•   No Load Current of Transformer=<2% of Transformer Rated current

•   Capacitor Current (Ic)=KVAR / 1.732xVolt (Phase-Phase)

•    Typically the local utility provides transformers rated up to 500kVA For maximum connected load of 99kW,

•   Typically the local utility provides transformers rated up to 1250kVA For maximum

connected load of 150kW.

•   The diversity they would apply to apartments is around 60%

•   Maximum HT (11kV) connected load will be around 4.5MVA per circuit.

•   4No. earth pits per transformer (2No. for body and 2No. for neutral earthing),

•    Clearances, approx.1000mm around TC allow for transformer movement for replacement.

•   Diesel Generator:

•   Diesel Generator Set Produces=3.87 Units (KWH) in 1 Litter of Diesel.

•   Requirement Area of Diesel Generator = for 25KW to 48KW=56 Sq.meter,

100KW=65 Sq.meter.

•   DG less than or equal to 1000kVA must be in a canopy.

•    DG greater 1000kVA can either be in a canopy or skid mounted in an acoustically treated room

•   DG noise levels to be less than 75dBA @ 1meter.

•    DG fuel storage tanks should be a maximum of 990 Litter per unit Storage tanks above this level will trigger more stringent explosion protection provision.

•   Current Transformer:

•   Nomenclature of CT:

•   Ratio: input / output current ratio

•   Burden (VA): total burden including pilot wires. (2.5, 5, 10, 15 and 30VA.)

•   Class: Accuracy required for operation (Metering: 0.2, 0.5, 1 or 3, Protection: 5, 10, 15,

20, 30).

•   Accuracy Limit Factor:

•   Nomenclature of CT: Ratio, VA Burden, Accuracy Class, Accuracy Limit

Factor.Example: 1600/5, 15VA 5P10  (Ratio: 1600/5, Burden: 15VA, Accuracy Class:

5P, ALF: 10)

•    As per IEEE Metering CT: 0.3B0.1 rated Metering CT is accurate to 0.3 percent if the connected secondary burden if impedance does not exceed 0.1 ohms.

•   As per IEEE Relaying (Protection) CT: 2.5C100 Relaying CT is accurate within 2.5

percent if the secondary burden is less than 1.0 ohm (100 volts/100A).

Electrical Thumb Rules (Part-2)

August 6, 2013  14 Comments

Useful Equations:

•   For Sinusoidal Current : Form Factor = RMS Value/Average Value=1.11

•   For Sinusoidal Current : Peak Factor = Max Value/RMS Value =1.414

•   Average Value of Sinusoidal Current(Iav)=0.637xIm (Im= Max.Value)

•   RMS Value of Sinusoidal Current(Irms)=0.707xIm (Im= Max.Value)

•   A.C Current=D.C Current/0.636.

•   Phase Difference between Phase= 360/ No of Phase (1

Phase=230/1=360°,2Phase=360/2=180°)

•   Short Circuit Level of Cable in KA (Isc)=(0.094xCable Dia in Sq.mm)/√ Short Circuit

Time (Sec)

•   Max.Cross Section Area of Earthing Strip(mm2) =√(Fault Current x Fault Current x

Operating Time of Disconnected Device ) / K

•   K=Material Factor, K for Cu=159, K for Alu=105, K for steel=58 , K for GI=80

•   Most Economical Voltage at given Distance=5.5x√ ((km/1.6)+(kw/100))

•   Cable Voltage Drop(%)=(1.732xcurrentx(RcosǾ+jsinǾ)x1.732xLength

(km)x100)/(Volt(L-L)x Cable Run.

•    Spacing of Conductor in Transmission Line (mm) = 500 + 18x (P-P Volt) + (2x (Span in Length)/50).

•   Protection radius of Lighting Arrestor = √hx (2D-h) + (2D+L). Where h= height of

L.A, D-distance of equipment (20, 40, 60 Meter), L=Vxt (V=1m/ms, t=Discharge Time).

•   Size of Lighting Arrestor= 1.5x Phase to Earth Voltage or 1.5x (System Voltage/1.732).

•   Maximum Voltage of the System= 1.1xRated Voltage (Ex. 66KV=1.1×66=72.6KV)

•   Load Factor=Average Power/Peak Power

•   If Load Factor is 1 or 100% = This is best situation for System and Consumer both.

•   If Load Factor is Low (0 or 25%) =you are paying maximum amount of KWH

consumption. Load Factor may be increased by switching or use of your Electrical

Application.

•   Demand Factor= Maximum Demand / Total Connected Load (Demand Factor <1)

•   Demand factor should be applied for Group Load

•   Diversity Factor= Sum of Maximum Power Demand / Maximum Demand (Demand

Factor >1)

•   Diversity factor should be consider for individual Load

•   Plant Factor(Plant Capacity)= Average Load / Capacity of Plant

•   Fusing Factor=Minimum Fusing Current / Current Rating (Fusing Factor>1).

•   Voltage Variation(1 to 1.5%)= ((Average Voltage-Min Voltage)x100)/Average Voltage

•   Ex: 462V, 463V, 455V, Voltage Variation= ((460-455) x100)/455=1.1%.

•   Current Variation(10%)= ((Average Current-Min Current)x100)/Average Current

•   Ex:30A,35A,30A, Current Variation=((35-31.7)x100)/31.7=10.4%

•   Fault Level at TC Secondary=TC (VA) x100 / Transformer Secondary (V) x

Impedance (%)

•   Motor Full Load Current= Kw /1.732xKVxP.FxEfficiency

Electrical Thumb Rules-(Part-3)

August 8, 2013  15 Comments

Size of Capacitor for P.F Correction:

Earthing Resistance value:

Earthing Resistance Value

Voltage Limit (As per CPWD & Kerala Elect.Board):

Voltage Variation

Insulation Class:

Standard Voltage Limit:

Standard Electrical Connection (As per GERC):

Standard Meter Room Size (As per GERC):

Approximate Load as per Sq.ft Area (As per DHBVN):

Contracted Load in case of High-rise Building:

Electrical Thumb Rules-(Part-4)

August 18, 2013  7 Comments

Sub Station Capacity & Short Circuit Current Capacity:

Sub Station Capacity & Short Circuit Current Capacity:

Minimum Ground Clearance and Fault Clearing Time:

Bus bar Ampere Rating:

Bus bar Spacing:

Bus bar Support between Two

Insulator

250mm.

Sound Level of Diesel Generator (ANSI 89.2&NEMA

51.20):

IR Value of Transformer:

Standard Size of MCB/MCCB/ELCB/RCCB/SFU/Fuse:

Electrical Thumb Rules-(Part-5)

September 1, 2013  4 Comments

Standard Size of Transformer (IEEE/ANSI 57.120):

Standard Size of Motor (HP):

Electrical Motor (HP)

1,1.5,2,3,5,7.5,10,15,20,30,40,50,60,75,100,125,150,200,250,300,400,450,500,600,700,

800,900,1000,1250,1250,1500,1750,2000,2250,3000,3500,4000

Approximate RPM of Motor

Standard Size of Motor (HP):

Electrical Motor (HP)

1,1.5,2,3,5,7.5,10,15,20,30,40,50,60,75,100,125,150,200,250,300,400,450,500,600,700,

800,900,1000,1250,1250,1500,1750,2000,2250,3000,3500,4000

Motor Line Voltage:

Motor Starting Current:

Motor Starter:

Impedance of Transformer (As per IS 2026):

Standard Size of Transformer:

Electrical Thumb Rules-(Part-6)

September 5, 2013  9 Comments

Transformer Earthing Wire / Strip Size:

Motor Earthing Wire / Strip Size:

Panel Earthing Wire / Strip Size:

Electrical Equipment Earthing:

Earthing Wire (As per BS 7671)

Area for Transformer Room: (As per NBC-2005):

•    The Capacitor Bank should be automatic Switched type for Sub Station of 5MVA and

Higher.

•   Transformer up to 25KVA can be mounted direct on Pole.

•   Transformer from 25KVA to 250KVA can be mounted either on “H” Frame of Plinth.

•   Transformer above 250KVA can be mounted Plinth only.

•   Transformer above 100MVA shall be protected by Drop out Fuse or Circuit Breaker.

Span of Transmission Line (Central Electricity Authority):

Three Phase Motor Code (NEMA)

Service Factor of Motor:

Type of Contactor:

Contactor Coil:

Electrical Thumb Rules-(Part-7)

September 16, 2013  10 Comments

Overhead Conductor /Cable Size:

Type of Tower:

Tower Swing Angle Clearance (Metal Part to Live Part):

Cable Coding (IS 1554) 🙁 A2XFY / FRLS / FRPVC / FRLA

/ PILC)

Corona Ring Size:

Load as per Sq.Ft:

Size of Ventilation Shaft:

Electrical Thumb Rules-(Part-8)

October 1, 2013  6 Comments

Accuracy Class of Metering CT:

Accuracy Class Letter of CT:

Accuracy Class of Protection CT:

Calculate IDMT over Current Relay

Setting (50/51)

October 11, 2013  18 Comments

•   Calculate setting of  IDMT over Current Relay for following Feeder and CT Detail

•   Feeder Detail: Feeder Load Current 384 Amp, Feeder Fault current Min11KA and Max

22KA.

•   CT Detail:  CT installed on feeder is 600/1 Amp. Relay Error 7.5%, CT Error 10.0%, CT

over shoot 0.05 Sec, CT interrupting Time is 0.17 Sec and Safety is 0.33 Sec.

•   IDMT Relay Detail:

•   IDMT Relay Low Current setting: Over Load Current setting is 125%, Plug setting of

Relay is 0.8 Amp and Time Delay (TMS) is 0.125 Sec, Relay Curve is selected as

Normal Inverse Type.

•   IDMT Relay High Current setting :Plug setting of Relay is 2.5 Amp and Time Delay

(TMS) is 0.100 Sec, Relay Curve is selected as Normal Inverse Type

Calculation of Over Current Relay Setting:

(1)  Low over Current Setting: (I>)

•   Over Load Current (In) = Feeder Load Current X Relay setting = 384 X 125% =480

Amp

•   Required Over Load Relay Plug Setting= Over Load Current (In) / CT Primary

Current

•   Required Over Load Relay Plug Setting = 480 / 600 = 0.8

•   Pick up Setting of Over Current Relay (PMS) (I>)= CT Secondary Current X Relay

Plug Setting

•   Pick up Setting of Over Current Relay (PMS) (I>)= 1 X 0.8 = 0.8 Amp

•   Plug Setting Multiplier (PSM) = Min. Feeder Fault Current / (PMS X (CT Pri.

Current / CT Sec. Current))

•   Plug Setting Multiplier (PSM) = 11000 / (0.8 X (600 / 1)) = 22.92

•   Operation Time of Relay as per it’s Curve

•   Operating Time of Relay for Very Inverse Curve (t) =13.5 / ((PSM)-1).

•   Operating Time of Relay for Extreme Inverse Curve (t) =80/ ((PSM)2 -1).

•   Operating Time of Relay for Long Time Inverse Curve (t) =120 / ((PSM) -1).

•   Operating Time of Relay for Normal Inverse Curve (t) =0.14 / ((PSM) 0.02 -1).

•   Operating Time of Relay for Normal Inverse Curve (t)=0.14 / ( (22.92)0.02-1) = 2.17

Amp

•   Here Time Delay of Relay (TMS) is 0.125 Sec so

•   Actual operating Time of Relay (t>) = Operating Time of Relay X TMS =2.17 X

0.125 =0.271 Sec

•    Grading Time of Relay = [((2XRelay Error)+CT Error)XTMS]+ Over shoot+ CB Interrupting Time+ Safety

•    Total Grading Time of Relay=[((2X7.5)+10)X0.125]+0.05+0.17+0.33 = 0.58 Sec

•    Operating Time of Previous upstream Relay = Actual operating Time of Relay+ Total Grading Time Operating Time of Previous up Stream Relay = 0.271 + 0.58 = 0.85

Sec

(2)  High over Current Setting: (I>>)

•    Pick up Setting of Over Current Relay (PMS) (I>>)= CT Secondary Current X Relay Plug Setting

•   Pick up Setting of Over Current Relay (PMS) (I>)= 1 X 2.5 = 2.5 Amp

•   Plug Setting Multiplier (PSM) = Min. Feeder Fault Current / (PMS X (CT Pri.

Current / CT Sec. Current))

•   Plug Setting Multiplier (PSM) = 11000 / (2.5 X (600 / 1)) = 7.33

•   Operation Time of Relay as per it’s Curve

•   Operating Time of Relay for Very Inverse Curve (t) =13.5 / ((PSM)-1).

•   Operating Time of Relay for Extreme Inverse Curve (t) =80/ ((PSM)2 -1).

•   Operating Time of Relay for Long Time Inverse Curve (t) =120 / ((PSM) -1).

•   Operating Time of Relay for Normal Inverse Curve (t) =0.14 / ((PSM) 0.02 -1).

•   Operating Time of Relay for Normal Inverse Curve (t)=0.14 / ( (7.33)0.02-1) = 3.44

Amp

•   Here Time Delay of Relay (TMS) is 0.100 Sec so

•   Actual operating Time of Relay (t>) = Operating Time of Relay X TMS =3.44 X

0.100 =0.34 Sec

•    Grading Time of Relay = [((2XRelay Error)+CT Error)XTMS]+ Over shoot+ CB Interrupting Time+ Safety

•   Total Grading Time of Relay=[((2X7.5)+10)X0.100]+0.05+0.17+0.33 = 0.58 Sec

•    Operating Time of Previous upstream Relay = Actual operating Time of Relay+ Total Grading Time.

•    Operating Time of Previous up Stream Relay = 0.34 + 0.58 = 0.85 Sec

Conclusion of Calculation:

•   Pickup Setting of over current Relay (PMS) (I>) should be satisfied following Two

Condition.

•    (1) Pickup Setting of over current Relay (PMS)(I>) >= Over Load Current (In) / CT Primary Current

•   (2) TMS <= Minimum Fault Current / CT Primary Current

•   For Condition (1) 0.8 > =(480/600) = 0.8 >= 0.8, Which found  OK

•   For Condition (2) 0.125 <=  11000/600 = 0.125 <= 18.33,  Which found  OK

•   Here Condition (1) and (2) are satisfied so

•   Pickup Setting of Over Current Relay = OK

•   Low Over Current Relay Setting: (I>) = 0.8A X In Amp

•   Actual operating Time of Relay (t>) = 0.271 Sec

•   High  Over Current Relay Setting: (I>>) = 2.5A X In Amp

•   Actual operating Time of Relay (t>>) = 0.34 Sec

Selection of 3P-TPN-4P MCB & Distribution Board

November 1, 2013  28 Comments

Type of breakers based on number of pole:

•   Based on the number of poles, the breakers are classified as

1.   SP – Single Pole

2.   SPN – Single Pole and Neutral

3.   DP – Double pole

4.   TP – Triple Pole

5.   TPN – Triple Pole and Neutral

6.   4P – Four Pole

1.     SP ( Single Pole ) MCB:

•   In Single Pole MCCB, switching & protection is affected in only one phase.

•   Application: Single Phase Supply to break the Phase only.

2.     DP ( Double Pole ) MCB:

•   In Two Pole MCCB, switching & protection is affected in phases and the neutral.

•   Application: Single Phase Supply to break the Phase and Neutral.

3.     TP ( Triple Pole) MCB:

•    In Three Pole MCB, switching & protection is affected in only three phases and the neutral is not part of the MCB.

•    3pole MCCB signifies for the connection of three wires for three phase system (R-Y-B Phase).

•   Application: Three Phase Supply only (Without Neutral).

4.     TPN (3P+N) MCB:

•    In TPN MCB, Neutral is part of the MCB as a separate pole but without any protective given in the neutral pole (i.e.) neutral is only switched but has no protective element incorporated.

•    TPN for Y (or star) the connection between ground and neutral is in many countries not allowed. Therefore the N is also switches.

•   Application: Three Phase Supply with Neutral

5.     4 Pole MCB:

•    4pole MCCB for 4 wires connections, the one additional 4th pole for neutral wire connection so that between neutral and any of the other three will supply.

•   In 4-Pole MCCBs the neutral pole is also having protective release as in the phase poles.

•   Application: Three Phase Supply with Neutral

Difference between TPN and 4P (or SPN and DP):

•    TPN means a 4 Pole device with 4th Pole as Neutral. In TPN opening & closing will open & close the Neutral.

•    For TPN, protection applies to the current flows through only 3 poles (Three Phase) only; there is no protection for the current flow through the neutral pole. Neutral is just an isolating pole.

•    TP MCB is used in 3phase 4wire system. It is denoted as TP+N which will mean a three pole device with external neutral link which can be isolated if required.

•    For the 4 pole breakers, protection applies to current flow through all poles. However when breaker trips or manually opened, all poles are disconnected.

•   Same type of difference also applies for SPN and DP.

Where to Use TP, TPN and 4P in Distribution panel:

•   For any Distribution board, the protection system (MCB) must be used in the incomer.

For a three phase distribution panel either TP or TPN or 4P can be used as the incoming protection.

•   TP MCB: It is most commonly used type in all ordinary three phase supply.

•   TPN MCB: It is generally used where there are dual sources of incomer to the panel

(utility source and emergency generator source).

•    4P MCB: It is used where is the possibility of high neutral current (due to unbalance loads and /or 3rd and multiple of 3rd harmonics current etc) and Neutral / Earth Protection is provided on Neutral.

Where to use 4 Pole or TPN MCB instead of 3 Pole (TP) MCB.

•   Multiple Incoming Power System:

•    When we have a transformer or a stand-by generator feeding to a bus, it is mandatory that at least either of the Incomers or the bus coupler must be TPN or 4-Pole Breaker please

refers IS 3043.

•   In multi incomer power feeding systems, we cannot mix up the neutrals of incoming

powers to other Power Source so we can use TPN or 4P breakers or MCB instead of TP MCB to isolate the Neutral of other power sources from the Neutral of incomer power in use.

•   We can use 4 Pole ACB instead of TP for safety reasons .If there is power failure and DG

sets are in running condition to feed the loads, if there is some unbalance in loads(which

is practically unavoidable in L.V. distribution system ), depending of quantum of

unbalance, there will be flow of current through Neutral. During this time, if Power Supply Utility Technicians are working, and if they touch the neutral conductors(which is earthed at their point ) they will likely to get electric shock depending on the potential

rise in common neutral due flow of current through Neutral conductor as stated above. Even fatal accident may occur due the above reason. As such, it is a mandatory practice to isolate the two Neutrals.

•    We can use 4-pole breakers or TPN Breakers when the system has two alternative sources and, in the event of power failure from the mains, change-over to the standby

generator is done. In such a case, it is a good practice to isolate the neutral also.

•    4 pole circuit breakers have advantages in the case when one of the poles of the device will get damage, and it also provides isolation from neutral voltage.

•   Normally, Neutral is not allowed to break in any conditions, (except special applications)

for human & equipment safety. So for single incomer power fed systems, 3P breaker is

used, where only phases are isolated during breaking operations.

•   Where We have dual Power like in DG & other electricity supply sources ,it is required

to isolate neutral, where neutral needs to be isolated  in internal network TPN MCB or 4P MCB can be used.

Where to use 4 Pole MCB instead of TPN MCB

•   Any Protection Relay used on Neutral (Ground Fault Protection of Double ended

System):

•    The use of four poles or three poles CB will depend on system protection and system configuration.

•    Normally in 3phase with neutral we just use 3pole CB and Neutral is connected on common Neutral Link but if application of 3pole will affect the operation of protective

relay then we must use 4pole CB.

•   System evaluation has to be required to decide whether three-pole circuit breakers plus

neutral link can be used or four-pole breakers are required.

•    If unrestricted ground fault protection is fitted to the transformer neutral, then the bus section circuit breaker should have 4-poles and preferably incomer circuit breakers should also have 4-poles because un cleared ground fault located at the load side of a

feeder have two return paths. As shown in fig a ground fault on a feeder at the bus section

“A” will have a current return path in both the incomers, thus tripping both Bus. The

sensitivity of the unrestricted ground fault relay is reduced due to the split current paths.

•   For System Stability :

•    In an unbalanced 3phase system or a system with non-linear loads, the neutral gives the safety to the unbalanced loads in the system and therefore It must not be neglected. In

perfectly balanced conditions the neutral functions as a safety conductor in the unforeseen short-circuit and fault conditions. Therefore by using 4-pole MCB will enhance the system stability.

•   4 Poles will be decided after knowing the Earthing Systems (TT, TN-S, TN-C, IT).

(1) IT (with distributed neutral) System:

•   The Neutral should be switched on & off with phases.

•   Required MCB: TPN or 4P MCB.

(2) IT (without distributed neutral) System:

•   There is no neutral.

•   Required MCB: TP MCB.

(3) TN-S System:

•    Required MCB: TP MCB because even when neutral is cut off system remains connected with Ground.

(4) TN-C System:

•    Required MCB: TPN or 4P only, because we cannot afford to cut neutral doing so will result in system loosing contact with Ground.

(5) TN-C-S System:

•   Neutral and Ground cable are separate

•   Required MCB: TP MCB Because Neutral and Ground cable are separate.

(6) TT System:

•   Ground is provided locally

•   Required MCB: TP MCB because ground is provided locally.

•   Conclusion: Its compulsory to use TPN in TN-C system rest everywhere you can use

MCB.

Nomenclature of Distribution Board:

•   Distribution Box can be decided by “way” means how many how many single phase

(single pole) distribution. Circuit and Neutral are used.

1)     SPN Distribution Board (Incoming+ Outgoing)

•   4way (Row) SPN = 4 X 1SP= 4Nos (Module) of single pole MCB as outgoing feeders.

•   6way (Row) SPN = 6 X 1SP= 6Nos (Module) of single pole MCB as outgoing feeders.

•   8way (Row) SPN = 8 X 1SP= 8Nos (Module) of single pole MCB as outgoing feeders.

•    10way (Row) SPN = 10 X 1SP= 10Nos (Module) of single pole MCB as outgoing feeders.

•   12way (Row) SPN = 12 X 1SP= 12Nos (Module) of single pole MCB as outgoing

feeders.

•    Normally single phase distribution is mainly used for small single phase loads at house wiring or industrial lighting wiring.

2)     TPN Distribution Board (Incoming, Outgoing)

•    4way (Row) TPN = 4 X TP= 4nos of 3pole MCB as outgoing feeders =12 No of single pole MCB.

•    6way (Row) TPN = 6 X TP= 6nos of 3pole MCB as outgoing feeders =18 No of single pole MCB.

•    8way (Row) TPN = 8 X TP= 8nos of 3pole MCB as outgoing feeders =24 No of single pole MCB.

•   10way (Row) TPN = 10 X TP= 10nos of 3pole MCB as outgoing feeders =30 No of

single pole MCB.

•    12way (Row) TPN =12 X TP= 12nos of 3pole MCB as outgoing feeders =36 No of single pole MCB

33)Transformer Losses-Regulation-

Efficiency(TC Name Plate)

December 31, 2013  9 Comments

Calculate Transformer Losses- Regulation- Efficiency (From TC Name Plate Data)

•   Calculate No Load Losses at various Loading of Transformer.

•   Calculate Full Load Losses at various Loading of Transformer.

•   Calculate Percentage Impedance

•   Calculate Transformer regulation at various Power Factor.

•   Calculate Transformer Efficiency at Unity P.F at various Loading condition.

•     Calculate Transformer Efficiency at various P.F at various Loading condition

Calculate Size and Short Circuit Capacity of

D.G Synchronous Panel

January 17, 2014  7 Comments

•   Calculate Size of D.G Synchronous Panel.

•   Calculate Total Fault  Current of D.G Synchronous Panel.

•   Total Equivalent Impedance of  D.G Synchronous Panel.

•   Calculate Short Circuit Current of D.G Synchronous Panel.

Electrical Thumbs Rules (Part-9)

March 15, 2014  7 Comments

Load in Multi-storied Building (Noida Power Company

Limited)

Distribution Losses (Gujarat Electricity Board)

Electrical Thumbs Rules (Part-10)

March 18, 2014  9 Comments

Economical Voltage for Power Transmission:

•    Economic generation voltage is generally limited to following values (CBIP Manual).

•    Generally terminal voltage of large generators is 11 kV in India. Step up voltage depends upon Length of transmission line for interconnection with the power system and Power to be transmitted.

•   High voltage increases cost of insulation and support structures for increased clearance

for air insulation but decreases size and hence Cost of conductors and line losses.

•    Many empirical relations have been evolved to approximately determine economic voltages for power evacuation. An important component in transmission lines is labor costs which are country specific.

•   An empirical relation is given below.

•   Voltage in kV (line to line) = 5.5x√0.62L + kVA/150

•   where kVA is total power to be transmitted;

•   L is length of transmission line in km.

•   American practice for economic line to line voltage kV (based on empirical formulation)

is

•   Voltage in kV line to line = 5.5x√0.62L + 3P/100

•    For the purpose of standardization in India transmission lines may be classified for operating at 66 kV and above. 33 kV is sub transmission, 11 kV and below may be classified as distribution.

•    Higher voltage system is used for transmitting higher amounts of power and longer lengths and its protection is important for power system security and requires complex

relay systems.

Factor affected on Voltage Level of system:

•    Power carrying capability of transmission lines increases roughly as the square of the voltage. Accordingly disconnection of higher voltage class equipment from bus bars get increasingly less desirable with increase in voltage levels.

•    High structures are not desirable in earthquake prone areas. Therefore in order to obtain lower structures and facilitate maintenance it is important to design such sub-stations

preferably with not more than two levels of bus bars.

Size of Cable according to Short circuit (for 11kV,3.3kV

only)

•   Short circuit verification is performed by using following formula:

•   Cross Section area of Cable (mm2)S = I x√t / K

•   Where:

•   t = fault duration (S)

•   I = effective short circuit current (kA)

•   K = 0.094 for aluminum conductor insulated with XLPE

•   Example: Fault duration(t)= 0.25sec,Fault Current (I) = 26.24 kA

•   Cross Section area of Cable = 26.24 x √ (0.25) / 0.094= 139.6 sq. mm

•   The selected cross sectional area is 185 sq. mm.

Ground Clearance:

•   Ground Clearance in Meter = 5.812 + 0.305 X K

•   Where K= (Volt-33) / 33

Voltage Rise in Transformers due to Capacitor Bank:

•    The voltage drop and rise on the power line and drop in the transformers. Every transformer will also experience a voltage rise from generating source to the capacitors. This rise is independent of load or power factor and may be determined as follows:

•   % Voltage Rise in Transformer=(Kvar / Kva)x Z

•   Kvar =Applied Kvar

•   Kva = Kva of the transformer

•   z = Transformer Reactance in %

•   Example: 300 Kvar bank given to 1200 KVA transformer with 5.75% reactance.

•   % Voltage Rise in Transformer=(300/1200)x 5.75 =1.43%

Calculate Size of Capacitor Bank / Annual

Saving & Payback Period

April 1, 2014  5 Comments

•   Calculate Size of Capacitor Bank Annual Saving in Bills and Payback Period for

Capacitor Bank.

•   Electrical Load of (1) 2 No’s of 18.5KW,415V motor ,90% efficiency,0.82 Power Factor

,(2) 2 No’s of 7.5KW,415V motor ,90% efficiency,0.82 Power Factor,(3) 10KW ,415V

Lighting Load. The Targeted Power Factor for System is 0.98.

•   Electrical Load is connected 24 Hours, Electricity Charge is 100Rs/KVA and 10Rs/KW.

•   Calculate size of Discharge Resistor for discharging of capacitor Bank. Discharge rate of

Capacitor is 50v in less than 1 minute.

•    Also Calculate reduction in KVAR rating of Capacitor if Capacitor Bank is operated at frequency of 40Hz instead of 50Hz and If Operating Voltage 400V instead of 415V.

•   Capacitor is connected in star Connection, Capacitor voltage 415V, Capacitor Cost is

60Rs/Kvar. Annual Deprecation Cost of Capacitor is 12%.

 Calculation:

•   For Connection (1):

•   Total Load KW for Connection(1) =Kw / Efficiency=(18.5×2) / 90=41.1KW

•   Total Load KVA (old) for Connection(1)= KW /Old Power Factor= 41.1 /0.82=50.1

KVA

•   Total Load KVA (new) for Connection(1)= KW /New Power Factor= 41.1 /0.98=

41.9KVA

•   Total Load KVAR= KWX([(√1-(old p.f)2) / old p.f]- [(√1-(New p.f)2) / New p.f])

•   Total Load KVAR1=41.1x([(√1-(0.82)2) / 0.82]- [(√1-(0.98)2) / 0.98])

•   Total Load KVAR1=20.35 KVAR

•   OR

•   tanǾ1=Arcos(0.82)=0.69

•   tanǾ2=Arcos(0.98)=0.20

•   Total Load KVAR1= KWX (tanǾ1- tanǾ2) =41.1(0.69-0.20)=20.35KVAR

•   For Connection (2):

•   Total Load KW for Connection(2) =Kw / Efficiency=(7.5×2) / 90=16.66KW

•   Total Load KVA (old) for Connection(1)= KW /Old Power Factor= 16.66 /0.83=20.08

KVA

•   Total Load KVA (new) for Connection(1)= KW /New Power Factor= 16.66 /0.98=

17.01KVA

•   Total Load KVAR2= KWX([(√1-(old p.f)2) / old p.f]- [(√1-(New p.f)2) / New p.f])

•   Total Load KVAR2=20.35x([(√1-(0.83)2) / 0.83]- [(√1-(0.98)2) / 0.98])

•   Total Load KVAR2=7.82 KVAR

•   For Connection (3):

•   Total Load KW for Connection(2) =Kw =10KW

•   Total Load KVA (old) for Connection(1)= KW /Old Power Factor= 10/0.85=11.76 KVA

•   Total Load KVA (new) for Connection(1)= KW /New Power Factor= 10 /0.98=

10.20KVA

•   Total Load KVAR3= KWX([(√1-(old p.f)2) / old p.f]- [(√1-(New p.f)2) / New p.f])

•   Total Load KVAR3=20.35x([(√1-(0.85)2) / 0.85]- [(√1-(0.98)2) / 0.98])

•   Total Load KVAR1=4.17 KVAR

•   Total KVAR=KVAR1+ KVAR2+KVAR3

•   Total KVAR=20.35+7.82+4.17

•   Total KVAR=32 Kvar

 Size of Capacitor Bank:

•    Site of Capacitor Bank=32 Kvar.

•   Leading KVAR supplied by each Phase= Kvar/No of Phase

•   Leading KVAR supplied by each Phase =32/3=10.8Kvar/Phase

•   Capacitor Charging Current (Ic)= (Kvar/Phase x1000)/Volt

•   Capacitor Charging Current (Ic)= (10.8×1000)/(415/√3)

•   Capacitor Charging Current (Ic)=44.9Amp

•   Capacitance of Capacitor = Capacitor Charging Current (Ic)/ Xc

•   Xc=2×3.14xfxv=2×3.14x50x(415/√3)=75362

•   Capacitance of Capacitor=44.9/75362= 5.96µF

•   Required 3 No’s of 10.8 Kvar Capacitors and

•   Total Size of Capacitor Bank is 32Kvar

 Protection of Capacitor Bank

Size of HRC Fuse for Capacitor Bank Protection:

•    Size of the fuse =165% to 200% of Capacitor Charging current.

•   Size of the fuse=2×44.9Amp

•   Size of the fuse=90Amp

Size of Circuit Breaker for Capacitor Protection:

•    Size of the Circuit Breaker =135% to 150% of Capacitor Charging current.

•   Size of the Circuit Breaker=1.5×44.9Amp

•   Size of the Circuit Breaker=67Amp

•   Thermal relay setting between 1.3 and 1.5of Capacitor Charging current.

•   Thermal relay setting of C.B=1.5×44.9 Amp

•   Thermal relay setting of C.B=67 Amp

•   Magnetic relay setting between 5 and 10 of Capacitor Charging current.

•   Magnetic relay setting of C.B=10×44.9Amp

•   Magnetic relay setting of C.B=449Amp

Sizing of cables for capacitor Connection:

•    Capacitors can withstand a permanent over current of 30% +tolerance of 10% on capacitor Current.

•   Cables size for Capacitor Connection= 1.3 x1.1 x nominal capacitor Current

•   Cables size for Capacitor Connection = 1.43 x nominal capacitor Current

•   Cables size for Capacitor Connection=1.43×44.9Amp

•   Cables size for Capacitor Connection=64 Amp

Maximum size of discharge Resistor for Capacitor:

•    Capacitors will be discharge by discharging resistors.

•    After the capacitor is disconnected from the source of supply, discharge resistors are required for discharging each unit within 3 min to 75 V or less from initial nominal peak voltage (according IEC-standard 60831).

•    Discharge resistors have to be connected directly to the capacitors. There shall be no switch, fuse cut-out or any other isolating device between the capacitor unit and the discharge resistors.

•   Max. Discharge resistance Value (Star Connection) = Ct / Cn x Log (Un x√2/ Dv).

•   Max. Discharge resistance Value (Delta Connection)= Ct / 1/3xCn x Log (Un x√2/

Dv)

•   Where Ct =Capacitor Discharge Time (sec)

•   Cn=Capacitance  Farad.

•   Un = Line Voltage

•   Dv=Capacitor Discharge voltage.

•   Maximum Discharge resistance =60 / ((5.96/1000000)x log ( 415x√2 /50)

•   Maximum Discharge resistance=4087 KΩ

Effect of Decreasing Voltage & Frequency on Rating of Capacitor:

•    The kvar of capacitor will not be same if voltage applied to the capacitor and frequency changes

•   Reduced in Kvar size of Capacitor when operating 50 Hz unit at 40 Hz

•   Actual KVAR = Rated KVAR x(Operating Frequency / Rated Frequency)

•   Actual KVAR = Rated KVAR x(40/50)

•   Actual KVAR = 80% of Rated KVAR

•   Hence 32 Kvar Capacitor works as 80%x32Kvar= 26.6Kvar

•   Reduced in Kvar size of Capacitor when operating 415V unit at 400V

•   Actual KVAR = Rated KVAR x(Operating voltage / Rated voltage)^2

•   Actual KVAR = Rated KVAR x(400/415)^2

•   Actual KVAR=93% of Rated KVAR

•   Hence 32 Kvar Capacitor works as 93%x32Kvar= 23.0Kvar

Annual Saving and Pay Back Period

Before Power Factor Correction:

•    Total electrical load KVA (old)= KVA1+KVA2+KVA3

•   Total electrical load= 50.1+20.08+11.76

•   Total electrical load=82 KVA

•   Total electrical Load KW=kW1+KW2+KW3

•   Total electrical Load KW=37+15+10

•   Total electrical Load KW =62kw

•   Load Current=KVA/V=80×1000/(415/1.732)

•   Load Current=114.1 Amp

•   KVA Demand Charge=KVA X Charge

•   KVA Demand Charge=82x60Rs

•   KVA Demand Charge=8198 Rs

•   Annual Unit Consumption=KWx Daily usesx365

•   Annual Unit Consumption=62x24x365 =543120 Kwh

•   Annual charges =543120×10=5431200 Rs

•   Total Annual Cost= 8198+5431200

•   Total Annual Cost before Power Factor Correction= 5439398 Rs

After Power Factor Correction:

•    Total electrical load KVA (new)= KVA1+KVA2+KVA3

•   Total electrical load= 41.95+17.01+10.20

•   Total electrical load=69 KVA

•   Total electrical Load KW=kW1+KW2+KW3

•   Total electrical Load KW=37+15+10

•   Total electrical Load KW =62kw

•   Load Current=KVA/V=69×1000/(415/1.732)

•   Load Current=96.2 Amp

•   KVA Demand Charge=KVA X Charge

•   KVA Demand Charge=69x60Rs =6916 Rs————-(1)

•   Annual Unit Consumption=KWx Daily usesx365

•   Annual Unit Consumption=62x24x365 =543120 Kwh

•   Annual charges =543120×10=5431200 Rs—————–(2)

•   Capital Cost of capacitor= Kvar x Capacitor cost/Kvar = 82 x 60= 4919 Rs—(3)

•   Annual Interest and Deprecation Cost =4919 x 12%=590 Rs—–(4)

•   Total Annual Cost= 6916+5431200+4919+590

•   Total Annual Cost After Power Factor Correction =5438706 Rs

 Pay Back Period:

•    Total Annual Cost before Power Factor Correction= 5439398 Rs

•   Total Annual Cost After Power Factor Correction =5438706 Rs

•   Annual Saving= 5439398-5438706 Rs

•   Annual Saving= 692 Rs

•   Payback Period= Capital Cost of Capacitor / Annual Saving

•   Payback Period= 4912 / 692

•   Payback Period = 7.1 Years

Calculate No of Lighting Fixtures / Lumen for Indoor Lighting

April 8, 2014  6 Comments

•    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.

 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