s ANNEXURE-I for Doc. 13-3/14/SIE/16/0606

FAULT CURRENT CALCULATIONS FOR RELAY SETTINGS Different cases are analyzed depending on the fault location. The values for various impedances which will restrict the fault current are calculated first and after determining the parameters the following net work for 132kV and 66kV system is obtained.

Case- I (16.292KA)

X

X

X

66kV Ou

Source fault

132kV Bus at Gharafa (Fault Level 17.48kA), ZS= 0.878 + j 4.267

X

Line # 1 ZL1= 0.209+ j 0.261

Line # 2 ZL1= 0.209 + j 0.261

Case- II

132/66 kV 160MVA Transformer ZT= 0.34 + j23.57

CT Location (Typ.)

Case- III (4.69 kA on 132kV side) (9.38 kA on 66kV side)

X 66kV bus at Duhail (Fault level 9.38kA)

(0.114 kA on 132kV side) (0.228 kA on 66kV side)

X

X

66kV Outgoing fetgoing feeder

level at 132kV bus = 3996 MVA (17.48kA) (As informed by Clie

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132/66 kV 160MVA Transformer ZT= 0.34 + j23.57

eder

nt)

s ANNEXURE-I for Doc. 13-3/14/SIE/16/0606

FAULT CURRENT CALCULATIONS FOR RELAY SETTINGS Source X/R ratio = 4.86 (informed by the client) Source Impedance (p.u.) = Base MVA / Fault MVA Base MVA = 160MVA Positive sequence impedance (Zs1) = System Voltage √3 x 17480 = 132000 = 4.36 Ohms (√3x 17480) Min. Source resistance Rs1 = ( Zs1 )2

1 + (Xs1 / Rs1)2

= (4.36)2

1+ (4.86)2

Rs1 = 0.878 Ohms Min. Source reactance Xs1 = 4.86 x Rs = 4.86 x 0.878 = 4.267 Ohms 132kV power cable data (According to cable data) Positive sequence resistance (RL1) = 0.085 Ohms/Km Positive sequence reactance (XL1) = 0.106 Ohms/Km Line length = 2.466 km Positive sequence resistance (RL1) = 0.209 Ohms Positive sequence reactance (XL1) = 0.261 Ohms Transformer Data Voltage Ratio = 132/66 kV

Rating = 160 MVA Zpu = 0.12 (at 96MVA) Zpu = 0.22 (at 160MVA) Actual Impedance ZT = Zpu X Base kV2 X 1000 Base kVA

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s ANNEXURE-I for Doc. 13-3/14/SIE/16/0606

FAULT CURRENT CALCULATIONS FOR RELAY SETTINGS = 0.22 X 1322 X 1000 160 X 1000 = 23.96 Ohms XT = Reactance of the transformer RT = Resistance of the transformer XT / RT = 69.33 (According to Transformer data) RT = ( ZT )2

1 + (XT / RT)2

= (23.96)2

1+ (69.33)2

= 0.34 Ohm XT = 69.33 x RT

= 69.33 x 0.34 = 23.57 Ohms Detail calculations

Source

0.878

4.267

0.209 0.209

0.261 0.261

0.34 0.34

23.57 23.57

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s ANNEXURE-I for Doc. 13-3/14/SIE/16/0606

FAULT CURRENT CALCULATIONS FOR RELAY SETTINGS Equivalent

Source

0.878

4.267

0.889 0.209

0.261 47.401

Equivalent

0.878

4.267

0.169

0.259

Source

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s ANNEXURE-I for Doc. 13-3/14/SIE/16/0606

FAULT CURRENT CALCULATIONS FOR RELAY SETTINGS Three Phase fault condition Fault current will be limited by Source impedance (132kV Bus at Gharafa), impedance of both of 160MVA transformers and Line #2 Impedance. Reactance X = Source reactance Xs1 + (Line reactance XL2 + 2 X Trafo. Reactance XT)

in parallel with (Line reactance XL1) = 4.267 + (0.261 + 2X23.57)//0.261 = 4.526 Ohms Resistance R = Source resistance Rs1 + (Line resistance RL2 + 2 X Trafo. resistance RT)

in parallel with (Line resistance RL1) = 0.878 + (0.209 + 2X.345)//0.209 = 1.047 Ohms Impedance = 1.047 + j 4.526 Impedance Z1 = √X2 + R2

= √(4.526)2 + (1.047)2

= 4.645Ohms Fault current at the end of Feeder = 132000 = 16407.4 Amp. ( If ) (√3x 4.645)

Say 16.407 kA

Fault current in Case-1 Fault current through line 1 = If * (47.409) (0.334 + 47.409) = 16.407 * (47.409) 47.743 = 16.292 kA (132kV base) Fault current in Case-2 Fault current through line 2 = If * (0.334) (0.334 + 47.409) = 16.407 * (.334) 47.743

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s ANNEXURE-I for Doc. 13-3/14/SIE/16/0606

FAULT CURRENT CALCULATIONS FOR RELAY SETTINGS = 0.114 KA (132kV base) Case III: - Fault on the 66kV Bus of Duhail and Fault contribution from Source from Gharafa (Through fault current seen by 160 MVA transformers) Fault current will be limited by Source impedance (132kV Bus at Gharafa), Impedance of both the Lines and impedance of both the160MVA transformers (Impedance of both the lines and transformers will act in parallel).

Source

Rs

Xs

RL+RT RL+RT

XL+XT XL+XT

Three phase fault current calculation for 66kv Bus

eactance X= Source reactance Xs + Line reactance XL /2+Transformer Reactance XT /2

= 4.267+ 0.261/2+ 23.57/2

esistance R= Source resistance Rs +Line resistance RL/2+Transformer Resistance RT/2

= 0.878 + 0.209/2 + 0.345/2

pedance = 1.155 + j16.183

pedance Z1 = √X + R

= √ (16.183)2 + (1.155)2

R = 16.183 Ohms R = 1.155 Ohms Im

2 2Im

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s ANNEXURE-I for Doc. 13-3/14/SIE/16/0606

FAULT CURRENT CALCULATIONS FOR RELAY SETTINGS = 16.224 Ohms

ault current at the end of Feeder = 132000 F = 4697.5Amp.

Fault current seen by 66kV side CTs 0 x 4.69

(√3x 16.224) Say 4.69 kA

= 13200 = 9.38 kA

the current magnitude

e

66000 132kV and 66kV systems are solidly grounded system. Thereforeobtained above for phase fault current shall be used for earth fault relay settings also. Impedance for downstream of 66kV feeder is calculated based on 66kV Base voltagand 40MVA Base MVA and the following net work is obtained.

66kV Bus at Duhail (Fault level 9.38 kA), ZS= 0.289+ j 4.048

Line # 1 Impedance neglected due to small length

X X

Case IV o 66kV (9.38 kA n

side)

11kV Outgoing feeder 11kV O tgoing feeu

66/11 kV nsformer 40MVA Tra

ZT= 0.383 + j17.618

X X

X X 11kV bus at Duhail (Fault level 17.76 kA)

X

Case V on 66kV side)

(2.96 kA(17.76 kA on 11kV side)

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Line # 2 Impedance neglected due to small length

66/11 kV

CT ocation L(Typ.)

der

nsformer 40MVA TraZT= 0.383 + j17.618

s ANNEXURE-I for Doc. 13-3/14/SIE/16/0606

FAULT CURRENT CALCULATIONS FOR RELAY SETTINGS Source fault level at 66kV bus = 1072 MVA (9.38kA)

ource X/R ratio =14.01(Derived from the calculations performed above in Case-III)

ase MVA = 40MVA

ositive sequence impedance (Zs) = System Voltage

= 66000 = 4.062 Ohms

in. Source resistance Rs = 2

)2

= 1)2

R = 0.289 Ohms

in. Source reactance Xs = 14.01 x Rs89 = 4.048 Ohms

ransformer Data io = 66/11 kV

(at 32 MVA base)

0 MVA base = 0.13 x 40/32 = 0.1625p.u.

ctual Impedance ZT = Zpu X Base kV X 1000

= 0.1625 X 66 X 1000

= 17.69 Ohms

T = Reactance of the transformer

S B P √3 x 9380 (√3x 9380) M ( Zs1 ) 1 + (Xs1 / Rs1 (4.062)2

1+ (14.0 s M = 14.01 x 0.2 T Voltage Rat

Rating = 40 MVA Zpu = 0.13 p.u.

Impedance at 4

2A Base kVA

2 40 X 1000 XRT = Resistance of the transformer

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s ANNEXURE-I for Doc. 13-3/14/SIE/16/0606

FAULT CURRENT CALCULATIONS FOR RELAY SETTINGS XT / RT = 46

= 2

2

=

= 0.383 Ohm

T = 46 x RT

.383 = 17.618 Ohms

ase IV :- Fault on the 11kV Bus of Duhail and Fault contribution from Source

lt current will be limited by Source impedance (66kV Bus at Duhail) and parallel

RT ( ZT ) 1 + (XT / RT) (17.69)2

1+ (46)2

X

= 46 x 0

C(Through fault current seen by both of the 40 MVA transformers) Fauimpedance of both of the 40MVA transformers.

Rs

Xs

RT RT

X

Source

XT T

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s ANNEXURE-I for Doc. 13-3/14/SIE/16/0606

FAULT CURRENT CALCULATIONS FOR RELAY SETTINGS Three phase fault current calculation for 11kV Bus

eactance X= Source reactance Xs + Transformer Reactance XT /2

= 4.048+ 17.618/2

esistance R= Source resistance Rs +Transformer Resistance RT /2

= 0.289 + 0.383/2

pedance = 0.4805 + j12.857

pedance Z1 = √X2 + R2

= √(12.857)2 + (0. 4805)2

= 12.86 Ohms

ault current at the end of Feeder = 66000

R = 12.857 Ohms R = 0.4805 Ohms Im Im

F = 2963.1 Amp.

Say 2.96 kA (66KV base)

ault at 11KV bus = 66000 x 2.96

(√3x 12.86)

F = 17.76 kA

arth fault current magnitude for 11kV system is limited to 750A due to Earthing

Various impedances for 11kV system are calculated based on 11kV Base voltage and

11000 Etransformer. The same will be used for relay settings.

500 kVA Base kVA and the following network is obtained.

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s ANNEXURE-I for Doc. 13-3/14/SIE/16/0606

FAULT CURRENT CALCULATIONS FOR RELAY SETTINGS

Source fault leve

uen

Source X/R ratio Base MVA Positive seq

11kV Bus at Duhail (Fault level 17.76kA), ZS= 0.013+ j 0.348

l at 11kV bus

ce impedance

Line # 1 Impedance neglected

= 26.76

= 500kV

due to small length

11/0.433 kV 500KVA Transformer

3.22 ZT= 0.6565 + j2

Case VI (270A on 11kV side)

n 415V side) (7.15kA o

CT Location (Typ.)

= 338

tions performed above in Case-IV)

(Zs) = System Voltage √3 x 17760

= 0.357 Ohms (√3x 17760)

MVA (17.76 kA)

(Derived from the calcula

A

= 11000

415V Outgoing feeder

X

X

415V bus at Duhail

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s ANNEXURE-I for Doc. 13-3/14/SIE/16/0606

FAULT CURRENT CALCULATIONS FOR RELAY SETTINGS Min. Source resistance Rs = ( Zs1 )2

2

= (0.357)2

Rs

= 26.76 x 0.013 = 0.348 Ohms

ransformer Data Voltage Ratio = 11/0.415 kV

= 500kVA

e ZT V2 X 1000

Base kVA

0.5 X 1000

23.2

T = Reactance of the transformer T = Resistance of the transformer

T = ( ZT )2

2

= (23.23)

T

1 + (Xs1 / Rs1) 1+ (26.76)2

= 0.013 Ohms Min. Source reactance Xs = 26.76 x Rs T

Rating Zpu = 0.096 p.u.

Actual Impedanc = Zpu X Base k = 0.096 X 112 X 1000 = 3 Ohms XR XT / RT = 35.37 R 1 + (XT / RT)

2 1+ (35.37)2

= 0.6565 Ohm X = 35.37 x RT

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s ANNEXURE-I for Doc. 13-3/14/SIE/16/0606

FAULT CURRENT CALCULATIONS FOR RELAY SETTINGS

= 35.37 x 0.6565 = 23.22 Ohms

ase V: - Fault on the 415V Bus of Duhail and Fault contribution from Source hrough fault current seen by 500 kVA transformer)

us at Duhail) and impedance 00kVA transformer.

nt calculation for 415V bus

ctance XT

= 23.568 Ohms

Transformer Resistance RT

= 0.669 Ohms

68

1

)2 + (0.669)2

11000

C(T Fault current will be limited by Source impedance (11kV Bof 5 Three phase fault curre Reactance X= Source reactance Xs + Transformer Rea = 0.348+ 23.22 Resistance R= Source resistance Rs + = 0.013 + 0.6565 Impedance = 0.669 + j23.5 Impedance Z = √X2 + R2

= √(23.568

= 23.577 Ohms Fault current at the end of Feeder = = 269.45 Amp, say 270 Amp

(√3x 23.57)

ault current seen by 415kV side CTs = 11000 x 270

= 7.15kA 415

F

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