Fault Level Calculation_Methods

FAULT LEVEL CALCULATIONS

FAULT LEVEL AT ANY GIVENPOINT OF THE ELECTRIC POWERSUPPLY NETWORK IS THEMAXIMUM CURRENT THATWOULD FLOW IN CASE OF ASHORT CIRCUIT FAULT AT THATPOINT.

WHAT IS FAULT LEVEL?

PURPOSE OF FAULT LEVEL CALCULATIONS

FOR SELECTING SHORT CIRCUIT PROTECTIVEDEVICES OF ADEQUATE SHORT CIRCUITBREAKING CAPACITY.

FOR SELECTING CIRCUIT BREAKERS & SWITCHESOF ADEQUATE SHORT CIRCUIT MAKING CAPACITY.

FOR SELECTING BUSBARS, BUSBAR SUPPORTS,CABLE & SWITCHGEAR, DESIGNED TO WITHSTANDTHERMAL & MECHANICAL STRESSES BECAUSE OFSHORT CIRCUIT.

TO DO CURRENT BASED DISCRIMINATIONBETWEEN CIRCUIT BREAKERS.

TYPES OF SHORT CIRCUITS

L – E (SINGLE LINE TO EARTH) L – L (LINE TO LINE) L – L – E (LINE TO LINE TO EARTH) L – L – L (THREE PHASE)

SOURCES OF SHORT CIRCUIT CURRENTS

IN-HOUSE SYNCHRONOUS GENERATORS

SYNCHRONOUS MOTORS & SYNCHRONOUS CONDENSERS

ASYNCHRONOUS INDUCTION MOTORS

ELECTRIC UTILITY SYSTEM THROUGH THE TRANSFORMER

NATURE OF SHORT CIRCUIT CURRENT

THE SHORT CIRCUIT CURRENT WILLCONSIST OF FOLLOWING

COMPONENTS :

THE AC COMPONENT WITH CONSTANT AMPLITUDE

THE DECAYING DC COMPONENT

SOURCE : UTILITY SYSTEM


THE SHORT CIRCUIT CURRENT WILLCONSIST OF FOLLOWING COMPONENTS :

THE AC COMPONENT WITH DECAYING AMPLITUDE

THE DECAYING DC COMPONENT

SOURCE : SYNCHRONOUS GENERATORS & MOTORS / INDUCTION MOTORS


TOP ENVELOPE

DECAYING DC COMPONENT

CURRENT

BOTTOM ENVELOPE

TIME

WAVEFORM


IK = INITIAL SYMMETRICAL RMS S/C CURRENT

IK = STEADY STATE RMS S/C CURRENT

iP = PEAK S/C CURRENT

A = INITIAL VALUE OF DECAYING DC COMPONENT

Note: FOR S/C FAR FROM GENERATOR (e.g. L.V. SYSTEM GETTING POWER FROM UTILITY THROUGHTRANSFORMERS) : I K = IK

SYMBOLS USED

CALCULATION ASSUMPTIONS

SIMPLIFIES CALCULATIONACCURACY IS NOT MUCH

AFFECTEDCALCULATED VALUES WILL BE

HIGHER THAN ACTUAL & HENCE SAFE

WHY ?

CALCULATION ASSUMPTIONS

TYPE OF SHORT CIRCUIT : THREE PHASEBOLTED SHORT CIRCUIT

IMPEDANCES OF BUSBAR/SWITCHGEAR/C.T./JOINTS ARE NEGLECTED

FAULT CURRENT FROM THE TRANSFORMERWOULD BE LIMITED BY THE SOURCE FAULTLEVEL

TRANSFORMER TAP IS IN THE MAIN POSITION SHORT CIRCUIT CURRENT WAVEFORM IS A

PURE SINE WAVE DISCHARGE CURRENT OF CAPACITORS ARE

NEGLECTED

WHAT ?

CALCULATION METHODS

* DIRECT METHOD

* PER UNIT METHOD

ADVANTAGES OF DIRECT METHOD# USES SYSTEM SINGLE LINE DIAGRAM

DIRECTLY# USES SYSTEM & EQUIPMENT DATA

DIRECTLY # USES BASIC ELECTRICAL EQUATIONS

DIRECTLY# EASIER TO COMPREHEND

STANDARD

T H E F A U L T L E V E L C A L C U L A T I O NP R O C E D U R E F O L L O W E D I N T H I SP R E S E N T A T I O N I S A S G I V E N I NIS 13234 – 1992 (Indian Standard Guidefor Calculating Short Circuit Currents inAC Electrical Networks up to 220kV)

FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACH

Step 1: Prepare a single line diagram of the electrical power supply anddistribution network, clearly indicatingall the significant network elements,fault current contributors, short circuit protective devices, etc.


Step 2: Get the following data:

i) Transformer rated kVA, rated secondary voltage of the transformer (UrT), %R & %X values.

ii) Generator rated kVA, rated voltage (UrG), rated sub-transient reactance (%x”

d) & rated Power factor (Cos φrG).


Step 2: Get the following data:

iii) Cable Resistance (RC) & Cable Reactance (XC) per unit length and the actual length of the cable used.

iv) Motors’ rated voltages (UrM), rated currents (IrM) and locked rotor currents (ILR).


Step 3: Convert %R into Ohmic values,to obtain RT.

10 x (%R) x (kV)2

RT (in Ω) = ------------------------kVA


Step 4: Similarly, convert %X into Ohmic values, to obtain XT.

10 x (%X) x (kV)2

XT (in Ω) = ------------------------kVA


Step 5: Similarly, convert %x”d of the

generator into Ohmic values, to obtain XG.

10 x (%x”d ) x (kV)2

XG (in Ω) = ------------------------kVA


Step 6: Now, the resistance of the generator, ‘RG’ is normally given as a% of ‘XG’. For LV Generators, it is:

RG (in Ω) = 0.15 XG


Step 7: Calculate a correction factor ‘KG’.

Un cKG = ------ --------------------------

UrG 1 + [(x”d ) (Sin φrG)]

where,


KG = Generator Correction FactorUn = Nominal System Voltage, in VoltsURG = Generator Rated Voltage, in Voltsc = Voltage Correction Factor = 1.05 x”

d = Sub-transient Reactance of theGenerator, in p.u. form

Sin φrG = √(1 – Cos2φrG)CosφrG = Rated Power Factor of the Generator


Step 8: Now find out the Corrected Generator Resistance (RGK) & theCorrected Generator Reactance (XGK):

RGK = KG x RG

XGK = KG x XG


Step 9: Now find out the cable resistance & reactance for the actual length of cableused up to the point of fault:

RL = RC x LC

XL = XC x LC

FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACHwhere,

RL = Line or Lead Resistance, in ΩXL = Line or Lead Reactance, in ΩRC = Cable Resistance per km, in ΩXC = Cable Reactance per km, in ΩLC = Actual length of cable up to the

point of fault, in m


Step 10: Now add all ‘R’ & all ‘X’ values; find out ‘Zk’.

Zk = √[(Re)2 + (Xe)2]

Re = RT or RGK + RL

Xe = XT or XGK + XL


Step 11: Now find out the initial symmetrical short circuit current, I”

k:

c Un

I”k = -------

√3 Zk


where,

I”k = Initial Symmetrical short circuit

current, in AmperesUn = Nominal System Voltage, in voltsc = Voltage Correction factor = 1.05

for LV (for HV it is 1.10)Zk = Equivalent Impedance up to the

point of fault, in Ω


Step 12: Determine the ‘X’/’R’ ratio up to the point of fault.

Step 13: Calculate the Asymmetry Factor‘’ (pronounced as ‘KHI’).

= (1.02 + 0.98 e-3R/X)


Step 14: Now, calculate the peak current,

‘ip’ = √2 I”k

Step 15: Calculate the aggregate of the rated full load currents of all the motors ata particular location (ΣIrM).

FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACHStep 16: If (ΣIrM) at a particular location is less than 1% of the short circuit currentcontributed by other sources for a fault atthat particular location, then contribution tothe short circuit, from the motors in thatparticular group, need not be considered.Or else, the motors’ contribution to shortcircuit has to be calculated.


Step 17: Assuming that the motors’ contribution has to be considered, the R.M.S. Currentcontribution to the short circuit, from the groupof motors will be:

I”kM = c 6 ΣIrM, if the S/C is at the

motor terminals

I”kM = c 5 ΣIrM, if the S/C is away from

the motor terminals, involving a cable


Step 18: And, the peak current contribution from the motor group would be:

ipM = M √2 I”kM

where, M = 1.3 for an R/X ratio of 0.42, as in the

case of LV Induction motors


Step 19: Now, calculate the total fault level, both R.M.S. (I”

KT) and the Peak (ipT) at the fault location:

I”KT = I”

K + I”KM

and

ipT = ip + ipM

FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A FEW ‘IFS’ AND ‘BUTS’.

What if the transformer’s ‘%R’ & ‘%X’ values are not known?

Need not worry! However, one can get to know the transformer’s ‘%Z’ from the nameplate. Now, convert this ‘%Z’, into ZT in Ω,using the same formula: [ZT = {(10 x %Z x kV2)/(kVA)}].

FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A FEW ‘IFS’ AND ‘BUTS’.The resistance of the transformer in Ω (RT),can be got from one of the following ways:

i) From the manufacturer’s test certificate

ii) By actual measurements, using either a low resistance measuring meter or a Kelvin’s Double Bridge

FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A FEW ‘IFS’ AND ‘BUTS’.iii) Or from the formula:

RT = [(PkrT)/(3I2rT)], where,

PkrT = Transformer Full Load Copper Loss, in Watts

IrT = Rated full load secondary current of the transformer, in amperes

FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A FEW ‘IFS’ AND ‘BUTS’.Once you know ZT & RT, XT can be easily calculated by:

XT = √[(ZT)2 – (RT)2].

FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A FEW ‘IFS’ AND ‘BUTS’.What if I don’t have any of the above data regarding RT?

Transformer manufacturers give guidance values

of the no-load loss, the full load loss & %impedance for a wide range of transformers in their catalogues. They can be taken as a guidefor the calculations.

FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A FEW ‘IFS’ AND ‘BUTS’.For example, the following data is got from aleading transformer manufacturer

kVA NLL (W) FLL (W) %Z160 450 3000 4.75200 540 3200 4.75250 630 3800 4.75315 725 4400 5400 850 5500 5500 1040 6500 5

FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A FEW ‘IFS’ AND ‘BUTS’.

kVA NLL (W) FLL (W) %Z630 1200 8000 5800 1450 9500 5

1000 1800 11500 51250 1900 13500 6.251600 2300 17000 6.25

FAULT LEVEL CALCULATIONS- A CASE STUDY

STEP 1 : SINGLE LINE DIAGRAM

)

)

)

)

)

)

F1

F3

PCC

MCC BUSBAR

U/G CABLE

M2M1 M3 M4

11

21 22 23 24

31

50HP 100HP 30HP 150HP

) CB

SDF

STARTER

G

)

)

F2

12

13

STANDBY GENERATOR1250kVA

TRANSFORMER1600kVA

350 A 300 A 300 A

STEP 2 : SYSTEM DATA

TRANSFORMER : 11/0.433kV1600kVA%R = 0.94%X = 5.46%Z = 5.54

STANDBY GENERATOR : UrG = 415V1250kVA%x d = 20Cos rG = 0.8Un = 415V

STEP 2 : SYSTEM DATA

CABLE : R = 0.062 /kMX = 0.079 /kMLENGTH OF CABLE, 21 TO 31= 100M

INDUCTION MOTORS : M1, IrM = 70AM2, IrM = 135AM3, IrM = 40AM4, IrM = 200A

(UrM = 415V; ILR = 6 IrM)

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Fault Level Calculation_Methods