Fault Current Calculation 2f2r25

1 BASIC FOR CALCULATION 1 In the calculation only three phase symmetrical fault has been considered. 2 A simplified procedure for calculating the symmetrical short circuit current called the E/X method is used, which disregards all resistance,all static load and all prefault current. 3 The PU system is adopted for fault calculations as it simplifies the calculations 4 Since X/R ratio of the effective impedance at the bus is unknown,It si considered that the calculated current should not more than 80% of the allowed valve for the breaker at the existing bus voltage. 5 Motor Contribution has been considered as 2 times the rated current of HT motors and negligible in case of LT motors. 6 For claculation,the impedances of cables has been neglected. 7 The distribution transformers in different load centres of the plant shall operate in isolation and not in parallel. 8 The utility fault level has been considered as 600 MVA.

2 DESIGN DATA : a Utility fault level at plant grid

=

600

MVA

b Base MVA chosen

=

25

MVA

c) Distribution Transformer 1 Rated capacity

=

3.15

2 Voltage at HT side

=

11

KV

3 Voltage at LT side(full load)

=

0.380

KV

4 Percentage impedance

=

8

%

d) Rating of HT motors‐ 1 HT motor rating

MVA

=

4150

Kw

2 Motor Power factor

=

0.85

red

3 Motor efficiency

=

0.95

%

4 Maximum starting torque considered for the motor at 11kv

=

200

% FLT

5 Maximum Running torque considered for the motor at 11kv Cable Size XLPE,Armoured

=

100

% FLT

240

Sqmm

Cable no of run

5

No.

5 HT motor fed from cable 5Rx3x240Sqmm XLPE CU (Z)

=

0.23

ohm/km

6 HT motor fed from cable length

=

150

ohm/km

e) Rating of LT motors‐ 1 LT motor rating

=

350

Kw

2 Motor Power factor

=

0.85

red

3 Motor efficiency

=

0.90

%

4 Maximum starting torque considered for the motor at 11kv

=

200

% FLT

4 Maximum Running torque considered for the motor at 11kv

=

100

% FLT

1.5

Sqmm

Cable Size XLPE,Armoured Cable no of run

1

5 LT motor fed from cable 3Rx4x300Sqmm XLPE CU (Z)

=

0.020

6 LT motor fed from cable Length

=

70

No. ohm/km mtr

A

FAULT CURRENT CALCULATION FOR HT (UP TO NEAREST MOTOR TERMINAL FROM SOURCE)

1

FAULT CURRENT AT 33KV SWITCHGEAR AT APTRANSCO END

1.1

Calculation for source impedance (Zo) System Voltage (KV)

=

33

Sorce Fault MVA

=

500

Source Impedance (Z0) in Ohm

=

ZO 1.2

Fault Current at 33KV Switchgear at APTRANSCO End (IF0)

IFO

KV MVA 2

System KV Short Circuit MVA

ohm

=

Square (33) 500

ohm

=

2.178

ohm

=

1.1 X KV √ 3 X ZO

kA

=

1.1 x 33 √ 3 X 2.178

kA

=

9.62

kA

2

FAULT CURRENT AT 33KV SWITCHGEAR AT PROPOSED SUBSTATION

2.1

Main Incoming power supply through 33kV, 1R X 3C X 300Sq.mm. Aluminium armoured Cable from APTRANSCO Substation to Plant Substation Resistance of cable ( R ) (Resistance / Km) = 0.130 ohm/Km Length of Cable (L)

=

1500

No.of Run (N)

=

1

=

R x L (N x 1000)

ohm

=

0.195

ohm

=

0.100

ohm/Km

=

X x L (N x 1000)

ohm

=

0.150

ohm

=

√(Ra2 + Xa2)

ohm

=

0.2460

ohm

=

Za + Zo

=

2.425

Actual Resistance of Cable (Ra) Ra 2.2

Reactance of cable (X) (Reactance/Km) Actual Reactance of Cable (Xa)

Xa 2.3

Impedance of Cable (Za)

2.4

Total Impedance at 33KV SWGR at Substation‐1(Z1)

Za

Z1 2.5

Fault Current at 33KV SWGR at Proposed Substation (IF1)

IF1

3

=

=

1.1 X KV

√ 3 X Z1

Meter No.

ohm kA

( 1.1 X 33) (√ 3 X 2.425)

kA

8.643

kA

Selected Fault KA at 33KV Bus 25 kA

FAULT CURRENT AT PRIMARY OF 8MVA, 33/6.9KV TRANSFORMER AT PROPOSED SUBSTATION 33KV, 1R X 3C X 300Sqmm (E) Aluminium armoured Cable is connected between 33KV HT & 8 MVA Transformer

3.1

Resistance of cable (Resistance / Km)

=

0.130

Length of Cable (L)

=

25

Meter

No.of Run (N)

=

1

No.

=

R x L (N x 1000)

ohm

=

0.00325

ohm

=

0.100

=

X x L (N x 1000)

ohm

=

0.0025

ohm

=

√(Rb2 + Xb2)

ohm

Actual Resistance of Cable (Rb) Rb 3.2

Reactance of cable (X) (Reactance/Km) Actual Reactance of Cable (Xb)

Xb 3.3

Impedance of Cable (Zb)

ohm/km

ohm/Km

Zb 3.4

Total Impedance at 33KV SWGR at Substation‐1(Z1) Z2

3.5

Fault Current at 33KV SWGR at Proposed Substation (IF1)

IF2

4

=

0.0041

ohm

=

Zb + Z1

ohm

=

2.430

ohm

=

1.1 X KV

kA

√ 3 X Z2

=

( 1.1 X 0.15) (√ 3 X 2.43)

kA

=

8.625

kA

FAULT CURRENT AT SECONDARY OF 8MVA TRANSFORMER AT SUBSTATION Power Transformer 8MVA, 33/6.6 KV , 8.35% Impedance (as per IS‐2026)

4.1

4.2

Primary Voltage of Transformer

Vp

=

33

KV

Secondary Voltage of Transformer

Vs

=

6.6

KV

Transformer Capacity

=

8000

KVA

% Impedance

=

8.35

%

Transformer Impedance (Zc)

=

(% Imp/100) x KVs ohm (KVA / 1000)

Zc

=

(8.35/100) x 6.6) (8000/1000)

Zc

=

Total Impedance on secondary side (Z3 )

Z3 4.3

Fault Current at Secondary side of 8MVA Transformer(IF3)

IF3

5

2

0.455

2

ohm

ohm

=

{((Vs)²/ (Vp)²)x Z2 }+ Zc

ohm

=

((6.6) / (33) X 2.43) + 0.455

ohm

=

0.553

=

1.1 X KVs √ 3 X Z3

ohm KA

=

( 1.1 X X x L) (√ 3 X 0.553)

kA

=

7.580

KA

FAULT CURRENT AT 6.6 KV HT AT PROPOSED SUBSTATION Incoming power supply through 6.6kV, 3R X 3C X 300Sq.mm. Aluminium armoured Cable from 6.6KV Terminal of Power Transformer

5.1

Resistance of cable (Resistance / Km)

=

0.129

Length of Cable (L)

=

25

Meter

No.of Run (N)

=

3

No.

=

R x L (N x 1000)

Actual Resistance of Cable (Rd)

Rd 5.2

Reactance of cable (X) (Reactance/Km) Actual Reactance of Cable (Xd)

Xd 5.3

Impedance of Cable (Zd) Zd

5.4

Total Impedance at 6.6KV HT (Z4) Z4

5.5

Fault Current at Proposed Substation 6.6KV HT (IF4)

IF4

ohm/km

ohm

=

0.001075

=

0.0830

=

X x L (N x 1000)

ohm

=

0.000691667

ohm

2

ohm ohm/Km

2

=

√(Rd + Xd )

ohm

=

0.0013

ohm

=

Zd + Z3

=

0.555

=

1.1 X KV

√ 3 X Z4

ohm kA

=

( 1.1 X ) (√ 3 X 0.555)

kA

=

7.553

kA

Selected Fault KA at 6.6KV Bus 26.3 kA

6

FAULT CURRENT AT MOTOR TERMINAL (IBC‐3 ‐ nearest) 6.6KV, 1R X 3C X 150Sqmm Aluminium armoured Cable is connected between 6.6kV HT & HT Motor terminal

6.1

Resistance of cable (Resistance / Km)

=

0.264

Length of Cable (L)

=

180

No.of Run (N) Actual Resistance of Cable (Re)

Re 6.2

Reactance of cable (X) (Reactance/Km) Actual Reactance of Cable (Xe)

Xe 6.3

Impedance of Cable (Ze) Ze

6.4

Total Impedance at HT Motor terminal (Z5) Z5

6.5

Fault Current at IBC‐3 Motor Terminal (IF5)

=

1

=

R x L (N x 1000)

No. ohm

=

0.04752

ohm

=

0.0890

ohm/Km

=

X x L (N x 1000)

ohm

=

0.01602

ohm

=

√(Re2 + Xe2)

ohm

=

0.0501

ohm

=

Ze + Z4

=

0.606

=

IF5

ohm/km Meter

1.1 X KV

√ 3 X Z5

ohm kA

=

( 1.1 X X x L) (√ 3 X 0.606)

kA

=

6.917

kA

3 PU IMPEDANCES OF VARIOUS EQUIPMENT/SYSTEMS: 1 Base MVA chosen 2 PU impedance of Utility

=

25

MVA (Base MVA) (Utility MVA)

=

=

25 600

=

25 0.042

= 0.041667 Pu 3 PU Impedance of equipment

= =

i) Distribution transformer(3.15 MVA)

(Base MVA) x (Percentage Impedance) ( Equipment MVA) 25 x ( 8/ 100 ) 3.15

= 0.634921 Pu

4 FAULT LEVEL AT 11 KV BUS Effective impedance of utility

=

0.041667 Pu

Total impedance at 11 kV bus

=

0.041667 Pu

Fault level corresponding to above value

(Base MVA) Total Pu Impedance

= =

600

MVA (Total kW of HT motors x 2) (1000 x pf x efficiency)

Contribution of HT motors

=

Contribution of HT motors assuming a p.f. 0.9 and efficiency 95%

=

Fault Level corresponding to impedance at 11kV bus + Contribution of HT motor

= 600 + 10.28

=

Hence Fault current at 11kv Bus

( 4150 x 2 ) (1000 x 0.85 x 0.95) 10.28

=

610

=

610 (√ 3 x 11)

MVA

MVA

= 32.03233 kA The 11kv Switchgears with fault with stand capacity of up to 40kA is feasible a) Conclusion: Fault Level at 11 kv Bus is 32.0KA

5 FAULT LEVEL AT 0.380 KV BUS 1 Base MVA chosen 2 The pu impedance up to 11kV bus to 11kV bus

Effiective impedance of distribution transformer Fault level at 0.380 Kv bus

= =

25

Base MVA Hight fault level at 11kV bus

=

25 610

=

0.041

pu

=

0.635

pu

=

Base MVA (The pu imedance up to 11kV bus + inpedance of distribution Transformer) 25 (0.041 + 0.635)

=

Hence fault current at 0.380Kv bus

MVA

=

37

=

37 (√ 3 x 0.38)

MVA

=

56.19

MVA

=

37

MVA

b) Conclusion: Hence , the fault level at 0.380kV bus rounded of to the nearest integer Value

6 CALCULATION FOR VOLTAGE DROP

The per unit impedance of the system components are calculated as follows : i) Per Unit impedance of the system Components: System Component Sr.No. i) pu impedance of utility

P.U.Impedance 25 = 600

=

0.23 x (150/ 1000 ) x 25 3 x (11^ 2 )

0.0417

Pu

=

0.0024

Pu

ii) HT main motor fed from cable Impedance 5Rx3x240Sqmm XLPE CU for 150mm length

=

iii) Starting of HT motor impedance

4150kW

=

(25 x 0.85 x 0.95) (4150 / 1000 ) x (200/ 100 )

=

2.4323

pu

iii) Running of HT motor impedance

4150kW

=

(25 x 0.85 x 0.95) (350 / 1000 ) x (100/ 100 )

=

57.679

pu

iv) Distribution transformer Impedance 3.15MVA

=

(25 x ( 8/ 100) 3.15

=

0.635

pu

v) LT main motor fed from cable Impedance 3Rx4x300Sqmm XLPE CU for 70mm length

=

0.02 x (70/ 1000 ) x 25 2 x (0.38^ 2 )

=

0.121

Pu

vi) Starting of LT motor impedance

350kW

=

(25 x 0.85 x 0.9) (350 / 1000 ) x (200/ 100 )

=

27.32

Pu

vi) Running of LT motor impedance

350kW

=

(25 x 0.85 x 0.9) (350 / 1000 ) x (100/ 100 )

=

54.64

Pu

2.4764

Pu

Total Pu Impedance

7

= (0.0417 + 0.0024 + 2.4323)

=

100

Voltage requirement for stable starting of HT motor 11kV ,4150kW System Component

% Voltage Drop

i) Utility System

ii) 5‐3x240mm2 XLPE Cable

=

0.0417 2.476

x 100%

=

1.68

%

=

0.0024 2.4764

x 100%

=

0.097

%

=

1.6839

%

=

1.68

%

=

1.78

%

=

98.22

%

=

193.15

%

iii) Voltage Drop upto

11KV Bus

iv) Voltage drop upto

4150kW motor terminals= (1.6839 + 0.097)

v) Voltage avaliable to the motor

= 100 ‐ 1.7809

vi) Hence maximum starting torque normally avaliable with 10.81 Kv

=

(200 x 10.81 ^ 2 ) (11^2 )

Conclusion Higher starting torque,however,can be achived by adjustment of electrolyte strength of LRS. For starting of 4150 kw motor,the starting surrent have to be resticted to less than assumption the voltage drop is 1.78 %

8

200

% Even with above

Voltage requirement for stable Running of HT motor 150kV ,0.9kW System Component

% Voltage Drop

i) Utility System

ii) 5‐3x240mm2 XLPE Cable

=

0.0417 2.476

x 100%

=

1.68

%

=

0.0024 2.4764

x 100%

=

0.097

%

=

1.6839

%

=

1.68

%

=

1.78

%

=

98.22

%

iii) Voltage Drop upto

11KV Bus

iv) Voltage drop upto

4150kW motor terminals= (1.6839 + 0.097)

v) Voltage avaliable to the motor

= 100 ‐ 1.7809

vi) Hence maximum starting torque normally avaliable with 10.81 Kv

=

(100 x 10.81 ^ 2 ) (11^2 )

=

96.58

%

Conclusion Higher starting torque,however,can be achived by adjustment of electrolyte strength of LRS. For Running of 0.9 kw motor,the starting surrent have to be resticted to less than assumption the voltage drop is 1.78 %

8

0

% Even with above

Voltage requirement for stable starting of LT motor 0.38kV ,350kW squirrel cage induction motor with DOL starter. Total Pu Impedance

= (0.0417 + 0.635 + 0.1212+27.3215)

System Component

=

28.119

Pu

% Voltage Drop

i) Utility System

ii) Distribution transformer 3.15MVA

iii) 3Rx4x300Sqmm XLPE CU Cable

=

0.0417 28.119

x 100%

=

0.15

%

=

0.6350 28.119

x 100%

=

2.26

%

=

0.1212 28.119

x 100%

=

0.440

%

=

0.15

%

=

0.15

%

=

2.85

%

=

97.15

%

=

189

%

iv) Voltage Drop upto

0.38 KV Bus

v) Voltage drop upto

350kW motor terminals = (0.15 + 0.44 +2.26)

vi) Voltage avaliable to the motor

= 100 ‐ 2.85

vii) Hence maximum starting torque normally avaliable with 369 V

=

(200 x 369.17 ^ 2 ) ((0.38 x 100 )^ 2)

368.6

viii) Higher staring torque, however can be achieved by VVVF drive 380

8

Voltage requirement for stable Running of LT motor kV ,3.15kW squirrel cage induction motor with DOL starter. Total Pu Impedance

= (0.0417 + 0.635 + 0.1212+54.6429)

System Component

54.81

Pu

% Voltage Drop

i) Utility System

ii) Distribution transformer

=

MVA

iii) 3Rx4x300Sqmm XLPE CU Cable

=

0.0417 54.806

x 100%

=

0.08

%

=

0.6350 54.806

x 100%

=

1.16

%

=

0.1212 54.806

x 100%

=

0.230

%

=

0.08

%

=

0.08

%

=

1.47

%

=

98.53

%

=

94

%

iv) Voltage Drop upto

KV Bus

v) Voltage drop upto

3.15kW motor terminals = (0.08 + 0.23 +1.16)

vi) Voltage avaliable to the motor

= 100 ‐ 1.47

vii) Hence maximum starting torque normally avaliable with 374.41 V

=

(100 x 374.414 ^ 2 ) ((0.38 x 100 )^ 2)

viii) Higher staring torque, however can be achieved by VVVF drive

CALCULATION FOR SELECTION OF CABLE SIZES FOR VARIOUS RATING OF LT MOTORS PU impedance of Utility

=

0.0417

=

0.0417 pu

Distribution transformer Impedance 3.15MVA

=

0.6350

=

0.6350 pu

=

0.6767 pu

Therefor ,pu impedance upto

0.38 kV Bus = 0.0417+0.635

Base MVA

=

25

MVA

Motor Rating

=

350

kw

Motor Voltage rating

=

380

v

Motor full load current

=

Motor full load current

Power Factor

25 MVA

(350 x 1000) (√ 3 x (380 x 1000) x 0.85 x 0.95) 695.13

Motor Efficiency

=

A

0.90 Cos θ

=

0.85

rad

Sin θ

=

0.53

rad

Al =

1.5

Sqmm

Cable From MCC to Motor Cable Size Cable Impedance z%

Z % =

0.020

Cable Length

(L) =

70

ohm/mtr mtr

No.of Run of the Cable

(N) =

1

No.

AC resistance of the cable at 90°C

=

Reactance of the cable at 50Hz

= 0.000107 Ω /mtr

0.02317 Ω /mtr

The pu impedance of 150m (typical) cable of different sizes is calculated as follows : = %Z of Cable x Base MVA x Length of cable in kms 2 (kV) =

( % Z Ohm/km x 25 x 70/ 1000 ) (380x 1000^2)

= ( Z% Ohm/km x 12.12) Pu Impedance of the Cable Voltage drop of the selected cable

Voltage drop of the selected cable

=

242.40

pu

=

695.13 1

( x 0.02 x 70)

=

16.43

V

= =

16.43 x 100 (380 / √3 ) 2.50

%

The starting KVA of motors is calculated as: For squirrel cage induction motors Starting KVA

=

Kw x 6 Efficiency x Power factor

=

(k W x 6) (0.9 x 0.85)

=

kw X

7.84

= 2745.098 KVA For Slip ring induction motors Starting KVA

=

Kw x 2.5 Efficiency x Power factor

= (k W x 2.5) (0.9 x 0.85) =

kw X

3.27

= 1143.791 KVA For squirrel cage induction motors Running KVA

=

Kw x 1 Efficiency x Power factor

=

(k W x 1) (0.9 x 0.85)

=

kw X

1.31

= 457.5163 KVA

=

12.12