49143084-2-0-fault-Level
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Transcript of 49143084-2-0-fault-Level
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FAULT LEVEL CALCULATIONS
(As per IS 13234 : 1992 &
IEC 60909 : 1995)
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FAULT LEVEL AT ANY GIVEN POINT OF THE INSTALLATION IS THE MAXIMUM CURRENT THAT CAN FLOW IN CASE OF S/C AT THAT POINT
WHAT IS FAULT LEVEL?WHAT IS FAULT LEVEL?
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PURPOSE OF FAULT LEVEL CALCULATIONS
• FOR SELECTING S.C.P.Ds OF ADEQUATE S/C BREAKING CAPACITY
• FOR SELECTING BUSBARS, BUSBAR SUPPORTS, CABLES & SWITCHGEARS, DESIGNED TO WITHSTAND THERMAL & MECHANICAL STRESSES BECAUSE OF S/C
• TO DO CURRENT BASED DISCRIMINATION BETWEEN CBs
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FAULT LEVEL CALCULATIONS
TYPES OF FAULTS
SYMMETRICAL ASYMMETRICAL
LINE TO LINE
DOUBLE LINE TO EARTH
LINE TO EARTH
THREE PHASE FAULT
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SOURCES OF SHORT CIRCUIT CURRENTS
ELECTRIC UTILITY SYSTEMS
D.G SETS
CONDENSERS
MOTORS
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NATURE OF SHORT CIRCUIT CURRENT
THE SHORT CIRCUIT CURRENT WILL
CONSIST OF FOLLOWING COMPONENTS
THE AC COMPONENT WITH CONSTANT AMPLITUDE
THE DECAYING DC COMPONENT
SOURCE : UTILITY SYSTEMSOURCE : UTILITY SYSTEM
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NATURE OF SHORT CIRCUIT NATURE OF SHORT CIRCUIT CURRENTCURRENT
TOP ENVELOPE
DECAYING DC COMPONENT
CURRENT
BOTTOM ENVELOPE
2√2
I K =
2 √
2 I K
''
I P
A
2√2
I K ´
TIME
WAVEFORMWAVEFORM
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CALCULATION ASSUMPTIONS
TYPE OF SHORT CIRCUIT : THREE PHASE BOLTED SHORT CIRCUIT
IMPEDANCES OF BUSBAR/SWITCHGEAR/C.T. /JOINTS ARE NEGLECTED
TRANSFORMERS ARE CONNECTED TO INFINITE BUS ON H.T. SIDE
TRANSFORMER TAP IS IN THE MAX. POSITION
S/C CURRENT WAVEFORM IS A PURE SINE WAVE
DISCHARGE CURRENT OF CAPACITORS ARE NEGLECTED
WHAT ?WHAT ?
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CALCULATION OF SHORT CIRCUIT CURRENT
I S/C = 1 . 05 * LINE VOLTAGE
√ 3 * ( ZTR + Z CABLE )
)
Z TR = (in ohms)
% Z * 10 * KV2
KVA
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CASE STUDY
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STEP 1 : SINGLE LINE DIAGRAMSTEP 1 : SINGLE LINE DIAGRAM
°°)
°°)
°°)°
°) °
°)
°°)
F1
F2
PCCPCC
MCCMCC
M2M1 M3 M4
11
21
31
50HP 100HP 30HP 150HP
°°°°°°
G
°°)
° °)
12
13
STANDBY GENERATOR1250kVA
TRANSFORMER 1600kVA
350A 300A 300A
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STEP 2 : SYSTEM DATA
TRANSFORMERTRANSFORMER :: 11/0.433 KV11/0.433 KV 1600 KVA1600 KVA %R = 0.94%R = 0.94 %X = 5.46%X = 5.46 %Z = 5.54%Z = 5.54
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STEP 2 : SYSTEM DATA
CABLE :CABLE : R = 0.062 R = 0.062 ΩΩ /KM/KM X = 0.079 X = 0.079 ΩΩ /KM/KMLENGTH OF CABLE, 21 TO 31= 100mLENGTH OF CABLE, 21 TO 31= 100m
INDUCTION MOTORS :INDUCTION MOTORS : M1, IM1, IrMrM = 200A = 200A
M2, IM2, IrMrM = 135A = 135A
M3, IM3, IrM rM = 135A= 135A
M4, IM4, IrM rM = 200A = 200A
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STEP 3 : CALCULATION OF RT & XT
RT = 10 (%R)(SECONDARY KV)2
KVA
10 (0.94)(0.433)2
1600 = = 0.001102 OHMS
XT = 10 (%X)(SECONDARY KV)2
KVA10 (5.46)(0.433)2
1600 = = 0.006398 OHMS
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STEP 4 : CALCULATION OF RL & XL
RL = 0.062 × 0.1 = 0.0062 OHMS
XL = 0.079 × 0.1 = 0.0079 OHMS
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STEP 5 : CALCULATION OF ‘Z’ UP TO THE POINT OF FAULT
TOTAL ‘Z’ UP TO FAULT LOCATION F1 = √ (RT)2 + (XT)2 = √ (0.001102)2 + (0.006398)2 = 0.00649 Ω
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STEP 5 : CALCULATION OF ‘Z’ UP TO THE POINT OF FAULT
TOTAL ‘Z’ UP TO FAULT LOCATION F2
= √ (RT + RL)2 + (XT + XL)2
= √ (0.007302)2 + (0.01430)2
= 0.01606 Ω
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STEP 6 : CALCULATION OF RMS VALUE OF S/C CURRENT AT THE
POINT OF FAULT
IK″ AT FAULT LOCATION F1 =
c Un
√3 Z
= 1.05 × 415
√3 × 0.00649
= 38765 A OR 38.77 kA
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STEP 6 : CALCULATION OF RMS VALUE OF S/C CURRENT AT THE
POINT OF FAULT
IK″ AT FAULT LOCATION F2 =
c Un
√3 Z
= 1.05 × 415
√3 × 0.01606
= 15665 A OR 15.67 kA
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STEP 7 : CALCULATION OF MAKING CAPACITY AS PER
STANDARD IEC 60947-2
2.20.250< I
2.10.2520< I<=50
2.00.310< I<=20
1.70.56< I<=10
1.50.74.5<= I<=6
Multiplying factor
Power factorShort-circuit breaking
capacity (kA)
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STEP 8 : CALCULATION OF PEAK VALUE OF S/C CURRENT AT THE
POINT OF FAULT
iP AT FAULT LOCATION F1 = 2.1 × 38.77 = 81.41 kA (PEAK)
iP AT FAULT LOCATION F2 = 2 × 15.67 = 31.34 kA (PEAK)
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CALCULATION OF X AND R FOR GENERATOR
• The value of xd″ will be given in percentage terms.
• Calculate xd″ in ohms.• Calculate R in ohms as per data in standard:• Rg =0.15 xd″ for generators less than 1000V
• Rg =0.07 xd″ for generators up to 100MVA
• Rg =0.05 xd″ for generators 100MVA and above.• Apply correction factor and recalculate R and X.• Find Z and use in the formulae.
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THANK YOU