2.0 fault Level

8/7/2019 2.0 fault Level

1/23

FAULT LEVELCALCULATIONS

(As per IS 13234 : 1992 &

IEC 60909 : 1995)


2/23

FAULT LEVEL AT ANY

GIVEN POINT OF THE

INSTALLATION IS THEMAXIMUM CURRENT THAT

CAN FLOW IN CASE OF S/C

AT THAT POINT

WHAT IS FAULT LEVEL?WHAT IS FAULT LEVEL?


3/23

PURPOSE OF FAULT LEVEL

CALCULATIONS

FOR SELECTING S.C.P.Ds OFADEQUATE S/C BREAKING CAPACITY

FOR SELECTING BUSBARS, BUSBARSUPPORTS, CABLES &SWITCHGEARS, DESIGNED TOWITHSTAND THERMAL &MECHANICAL STRESSES BECAUSEOF S/C

TO DO CURRENT BASED

DISCRIMINATION BETWEEN CBs


4/23

FAULT LEVEL CALCULATIONS

TYPES OF FAULTS

SYMMETRICAL ASYMMETRICAL

LINE TO LINE

DOUBLE LINE TO

EARTH

LINE TO EARTH

THREE PHASE FAULT


5/23

SOURCES OF SHORT CIRCUIT

CURRENTS

ELECTRIC UTILITY SYSTEMS

D.G SETS

CONDENSERS

MOTORS


6/23

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


7/23

NATURE OF SHORT CIRCUITNATURE OF SHORT CIRCUIT

CURRENTCURRENT

TOP ENVELOPE

DECAYING DC COMPONENT

CURRENT

BOTTOM ENVELOPE

22I

K

=22

IK

''I

P

A2

2I K

TIME

WAVEFORMWAVEFORM


8/23

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 ?


9/23

CALCULATION OF SHORT CIRCUIT

CURRENT

IS/C

=1 . 05 * LINE VOLTAGE

3 * ( ZTR + Z CABLE )

Z TR =

(in ohms)

% Z * 10 * KV2

KVA


10/23

CASE STUDY


11/23

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


12/23

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


13/23

STEP 2 : SYSTEM DATA

CABLE :CABLE : R = 0.062R = 0.062 /KM/KMX = 0.079X = 0.079 /KM/KM

LENGTH 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, IrMrM = 135A= 135A

M4, IM4, IrMrM = 200A= 200A


14/23

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

KVA

10 (5.46)(0.433)2

1600= = 0.006398 OHMS


15/23

STEP 4 : CALCULATION OF RL &

XL

RL = 0.062 0.1 = 0.0062 OHMS

XL = 0.079 0.1 = 0.0079 OHMS


16/23

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


17/23

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


18/23

STEP 6 : CALCULATION OF RMS

VALUE OF S/C CURRENT AT THE

POINT OF FAULT

IK AT FAULT LOCATION F1 = c Un3 Z=

1.05 415

3 0.00649= 38765 A OR 38.77 kA


19/23

STEP 6 : CALCULATION OF RMS

VALUE OF S/C CURRENT AT THE

POINT OF FAULT

IK AT FAULT LOCATION F2 = c Un3 Z=

1.05 415

3 0.01606= 15665 A OR 15.67 kA


20/23

STEP 7 : CALCULATION OF

MAKING CAPACITY AS PER

STANDARD IEC 60947-2

2.20.250< I

2.10.2520< I


21/23


22/23

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.


23/23

THANK YOU

Documents

2.0 fault Level