FAULT LEVEL CALCULATIONS
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Transcript of FAULT LEVEL CALCULATIONS
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FAULT LEVEL CALCULATIONS
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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?
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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.
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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)
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SOURCES OF SHORT CIRCUIT CURRENTS
IN-HOUSE SYNCHRONOUS GENERATORS
SYNCHRONOUS MOTORS & SYNCHRONOUS CONDENSERS
ASYNCHRONOUS INDUCTION MOTORS
ELECTRIC UTILITY SYSTEM THROUGH THE TRANSFORMER
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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
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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
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NATURE OF SHORT CIRCUIT CURRENT
TOP ENVELOPE
DECAYING DC COMPONENT
CURRENT
BOTTOM ENVELOPE
TIME
WAVEFORM
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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
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CALCULATION ASSUMPTIONS
SIMPLIFIES CALCULATIONACCURACY IS NOT MUCH
AFFECTEDCALCULATED VALUES WILL BE
HIGHER THAN ACTUAL & HENCE SAFE
WHY ?
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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 ?
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CALCULATION METHODS
* DIRECT METHOD
* PER UNIT METHOD
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ADVANTAGES OF DIRECT METHOD# USES SYSTEM SINGLE LINE DIAGRAM
DIRECTLY# USES SYSTEM & EQUIPMENT DATA
DIRECTLY # USES BASIC ELECTRICAL EQUATIONS
DIRECTLY# EASIER TO COMPREHEND
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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)
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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.
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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).
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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).
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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
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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
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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
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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
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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,
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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
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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
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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
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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
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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
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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
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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
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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)
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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).
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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.
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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
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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
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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
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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)}].
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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
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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
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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].
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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.
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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
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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
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FAULT LEVEL CALCULATIONS- A CASE STUDY
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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
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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
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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)