FAULT LEVEL CALCULATIONS

(As per IS 13234 : 1992 &

IEC 60909 : 1995)

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?

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

FAULT LEVEL CALCULATIONS

TYPES OF FAULTS

SYMMETRICAL ASYMMETRICAL

LINE TO LINE

DOUBLE LINE TO EARTH

LINE TO EARTH

THREE PHASE FAULT

SOURCES OF SHORT CIRCUIT CURRENTS

ELECTRIC UTILITY SYSTEMS

D.G SETS

CONDENSERS

MOTORS

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

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

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 ?

CALCULATION OF SHORT CIRCUIT CURRENT

I S/C = 1 . 05 * LINE VOLTAGE

√ 3 * ( ZTR + Z CABLE )

)

Z TR = (in ohms)

% Z * 10 * KV2

KVA

CASE STUDY

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

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

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

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

STEP 4 : CALCULATION OF RL & XL

RL = 0.062 × 0.1 = 0.0062 OHMS

XL = 0.079 × 0.1 = 0.0079 OHMS

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

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

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

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

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)

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)

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.

THANK YOU