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
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
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
5 LT motor fed from cable
3Rx4x300Sqmm XLPE CU (Z)
0.020
6 LT motor fed from cable Length
70
No.
ohm/km
mtr
FAULT CURRENT CALCULATION FOR HT (UP TO NEAREST MOTOR TERMINAL FROM SOURCE)
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
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)
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 Substation1(Z1)
Za
Z1
2.5
Fault Current at 33KV SWGR at Proposed Substation (IF1)
IF1
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 Panel & 8 MVA Transformer
3.1
Resistance of cable (Resistance / Km)
0.130
Length of Cable (L)
25
Meter
No.of Run (N)
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 Substation1(Z1)
Z2
3.5
Fault Current at 33KV SWGR at Proposed Substation (IF1)
IF2
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 IS2026)
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
0.455
ohm
ohm
{((Vs)/ (Vp))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 PANEL 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)
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 Panel (Z4)
Z4
5.5
Fault Current at Proposed Substation 6.6KV HT Panel (IF4)
IF4
ohm/km
ohm
0.001075
0.0830
X x L
(N x 1000)
ohm
0.000691667
ohm
ohm
ohm/Km
(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
FAULT CURRENT AT MOTOR TERMINAL (IBC3 nearest)
6.6KV, 1R X 3C X 150Sqmm Aluminium armoured Cable is connected between 6.6kV HT Panel & 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 IBC3 Motor Terminal (IF5)
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
= (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) 53x240mm2 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
%
200
% Even with above
Voltage requirement for stable Running of HT motor 150kV ,0.9kW
System Component
% Voltage Drop
i) Utility System
ii) 53x240mm2 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
%
% 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
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
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
0.90
Cos
0.85
rad
Sin
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) =
No.
AC resistance of the cable at 90C
Reactance of the cable at 50Hz
= 0.000107 /mtr
0.02317 /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
=
=
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
