Wednesday, March 12, 2014

Electrical Thumb Rules

Current rating of Transformer / Generator (I)  

 KVA  X  1.4

 

Impedence of  Transformer  (Z) 

5 %

 

Short circuit rating of Transformer   

Current Rating / % Impedence  =  I / Z

 

No Load Current of Transformer

< 2 % of rated current of Transformer

 

Impedence of Generator

15 %

 

Short circuit rating of Generator

Current Rating / % Impedence  =  I / Z

 

Full load current ( FLC ) of 3 Phase Motor  

HP  X  1.5

  

No load current of 3 Phase Motor

30%  of FLC 

 

Full load current of Single Phase Motor 

HP  X  6

 

KW  rating of Motor 

HP  X   0.75

 

OLR  selection

 DOL  STARTER :-   FLC  X  1

STAR & DELTA  STARTER  :-   FLC  X  0.6

 

Fuse Rating for  Motor  protection

 DOL    :-  FLC  X  2

 STAR & DELTA   :-   FLC  X  1.5

 SLIP RING MOTOR  :-    FLC  X   1.25

Electrical Engineering Thumb Rules

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.

 

Cable Capacity:

For Copper  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 Aluminum wire 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)

 

 

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).

 

Tuesday, March 11, 2014

Choice of a circuit-breaker

Choice of a circuit-breaker

 

The choice of a CB is made in terms of:

 

·         Electrical characteristics of the installation for which the CB is intended

·         Its eventual environment: ambient temperature, in a kiosk or switchboard enclosure, climatic conditions, etc.

·         Short-circuit current breaking and making requirements

·         Operational specifications: discriminative tripping, requirements (or not) for remote control and indication and related auxiliary contacts, auxiliary tripping coils, connection

·         Installation regulations; in particular: protection of persons

 

 

 

Low Voltage Circuit Breakers

Based on IEC 60947-2 (LV switchgear and controlgear - Part 2: Circuit breakers):

 

In (Rated current)

 

The rated continuous / uninterrupted current that the circuit breaker can carry.

 

Icm (Rated short-circuit making current)

 

The short-circuit current that the circuit breaker can withstand as it is closing where the act of closing initiates the fault.

 

Icu (Rated ultimate short-circuit current)

 

The maximum symmetrical short-circuit current the circuit breaker can interrupt.

 

Ics (Rated service short-circuit current)

 

The breaking capacity such that the circuit breaker is tested to carry its current continuously. The test sequence verifies that the breaker can be returned to service. Ics is the maximum current the breaker can interrupt multiple times and be returned to service without being damaged and is expressed as a % of Icu.

 

Icw (Rated short time withstand current)

 

This is the steady state symmetrical fault current the breaker has to be able to carry for a duration of 0.05s to 3s without exceeding its thermal integrity.

Sunday, March 9, 2014

Power factor calculation

The power factor correction capacitor should be connected in parallel to each phase load.

 

Single phase circuit calculation

 

Power factor calculation:

PF = |cos φ| = 1000 × P(kW) / (V(V) × I(A))

 

Apparent power calculation:

|S(kVA)| = V(V) × I(A) / 1000

 

Reactive power calculation:

Q(kVAR) = √(|S(kVA)|2 - P(kW)2)

 

Power factor correction capacitor's capacitance calculation:

C(F) = 1000 × Q(kVAR) / (2πf(Hz)×V(V)2)

 

Three phase circuit calculation

 

For three phase with balanced loads:

 

Calculation with line to line voltage

 

Power factor calculation:

PF = |cos φ| = 1000 × P(kW) / (√3 × VL-L(V) × I(A))

 

Apparent power calculation:

|S(kVA)| = √3 × VL-L(V) × I(A) / 1000

 

Reactive power calculation:

Q(kVAR) = √(|S(kVA)|2 - P(kW)2)

 

Power factor correction capacitor's capacitance calculation:

C(F) = 1000 × Q(kVAR) / (2πf(Hz)×VL-L(V)2)

 

Calculation with line to neutral voltage

 

Power factor calculation:

PF = |cos φ| = 1000 × P(kW) / (3 × VL-N(V) × I(A))

 

Apparent power calculation:

|S(kVA)| = 3 × VL-N(V) × I(A) / 1000

 

Reactive power calculation:

Q(kVAR) = √(|S(kVA)|2 - P(kW)2)

 

Power factor correction capacitor's capacitance calculation:

C(F) = 1000 × Q(kVAR) / (3×2πf(Hz)×VL-N(V)2)

 

Wednesday, March 5, 2014

Electronics & Electrical Engineering

BY
C. JAMES ERICKSON Retired Principal Consultant, E. I. du Pont de Nemours & Co., Inc.
CHARLES D. POTTS Retired Project Engineer, E. I. du Pont de Nemours & Co., Inc.
BYRON M. JONES Consulting Engineer, Assistant Professor of Electrical Engineering,
University of Wisconsin—Platteville.

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