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FAULT LEVEL CALCULATIONS
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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
NATURE OF SHORT CIRCUIT CURRENT
THE SHORT CIRCUIT CURRENT WILLCONSIST OF FOLLOWING COMPONENTS :
THE AC COMPONENT WITH DECAYING AMPLITUDE
THE DECAYING DC COMPONENT
SOURCE : SYNCHRONOUS GENERATORS & MOTORS / INDUCTION MOTORS
NATURE OF SHORT CIRCUIT CURRENT
TOP ENVELOPE
DECAYING DC COMPONENT
CURRENT
BOTTOM ENVELOPE
TIME
WAVEFORM
NATURE OF SHORT CIRCUIT CURRENT
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.
FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACH
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).
FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACH
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).
FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACH
Step 3: Convert %R into Ohmic values,to obtain RT.
10 x (%R) x (kV)2
RT (in Ω) = ------------------------kVA
FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACH
Step 4: Similarly, convert %X into Ohmic values, to obtain XT.
10 x (%X) x (kV)2
XT (in Ω) = ------------------------kVA
FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACH
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
FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACH
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
FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACH
Step 7: Calculate a correction factor ‘KG’.
Un cKG = ------ --------------------------
UrG 1 + [(x”d ) (Sin φrG)]
where,
FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACH
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
FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACH
Step 8: Now find out the Corrected Generator Resistance (RGK) & theCorrected Generator Reactance (XGK):
RGK = KG x RG
XGK = KG x XG
FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACH
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
FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACH
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
FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACH
Step 11: Now find out the initial symmetrical short circuit current, I”
k:
c Un
I”k = -------
√3 Zk
FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACH
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
FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACH
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)
FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACH
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.
FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACH
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
FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACH
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
FAULT LEVEL CALCULATIONS – DIRECTMETHOD – A STEP BY STEP APPROACH
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)