# EPCIV Module 2

ELECTRICAL PROTECTION IV
Dr. Oriedi Akumu
Building 20-G01M
[email protected]
Consultation Hours:
Monday: 7:00-10:00
Wednesday:7:00-12:00
Thursday:7:00-10:00
ELECTRICAL PROTECTION IV
LEARNING OUTCOMES
Student should be able to: explain the concepts and
methods for calculating unsymmetrical short circuit
currents
ASSESSMENT CRITERIA
 Unsymmetrical faults are defined
 Use of sequence networks in fault current calculation is
explained
 Calculation of line-to-ground faults, line-to-line faults, and
line-to-line-to-ground faults is explained
 Effect of delta-star transformer on fault currents is determined
ELECTRICAL PROTECTION IV
Text Book: Power System Analysis & Design
Authors: J. D. Glover, M. S. Sarma and T. J. Overbye
Assessment Type:
Semester Tests
Class Tests
Assignments
Design Project using Power World
ELECTRICAL PROTECTION IV
Symmetrical Components
 A mathematical model for representing an unbalanced system of ‘n’
phasors as ‘n’ systems of balanced phasors.
N/B: This is only a model for analysis since symmetrical components do not
exist in real life
Unsymmetrical fault calculation is important for:
 Determination of protective relay setting
 Transient stability studies for interconnected systems
ELECTRICAL PROTECTION IV
Symmetrical Components
 The key idea of symmetrical component analysis is to decompose
the system into three sequence networks. The networks are then
coupled only at the point of the unbalance (i.e., the fault)
 The three sequence networks are known as the
 positive sequence (this is the one we’ve been using)
 negative sequence
 zero sequence
ELECTRICAL PROTECTION IV
Symmetrical Components
Three phase voltages (or currents) can be resolved into three sets of
sequence components:
Phase “a”
 Zero sequence components
Phase “b”
 Positive sequence components
Same phase sequence as original
phasor
Phase “c”
 Negative sequence components
Opposite phase sequence to original
phasor
ELECTRICAL PROTECTION IV
Symmetrical Components
Defined by:
or
Where
Also
or
ELECTRICAL PROTECTION IV
Symmetrical Components
Similarly for currents:
and
In 3-phase Y connected systems:
 In a balanced Y-connected system, and in any three-phase system
with no neutral path, such as a -connected system or a three-wire
Y-connected system with an ungrounded neutral, line currents have
no zero-sequence component.
ELECTRICAL PROTECTION IV
Symmetrical Components
Example
Solution
ELECTRICAL PROTECTION IV
Symmetrical Components
Example
Solution
ELECTRICAL PROTECTION IV
Sequence Impedances
 Each piece of equipment has 3 values of impedance: positive,
negative and zero sequence impedances (Z1, Z2 and Z0,
respectively)
 Z1 and Z2 of linear, symmetrical and static circuits (transformers,
static loads, transmission lines …) are equal, and are the same as
those used in analysis of balanced conditions
 The positive and negative sequence impedances of rotating
machines are normally different
 The zero sequence impedance depends on the path taken by the
zero sequence current
ELECTRICAL PROTECTION IV
 Synchronous generators:
Z2 < Z1
Z0 = variable value
= Z1 if its value is not given
Any impedance Ze in the earth connection introduces an impedance
of 3Ze per phase
 Transformers:
Z1 = Z2 = impedance of transformer
Z0 = Z1, if there is circuit for earth current
= infinity, if there is no through circuit for earth current
 Transmission lines:
Z1 = Z2 = impedance of the line
Z0 = variable value
= 3Z1, if its value is not given
ELECTRICAL PROTECTION IV
Symmetrical components of generator
Zero
sequence
network
Positive
sequence
network
=   = 3 0
Negative
sequence
network
Voltage drop in neutral due only
to zero sequence current
ELECTRICAL PROTECTION IV
Zero sequence
circuits for three
phase transformers
ELECTRICAL PROTECTION IV
Unsymmetrical Fault Analysis
The following assumptions are made to compute subtransient fault current for short circuits in power systems:
 Transformers are represented by their leakage reactances. Winding
resistances, shunt admittances, and delta/star phase shifts are
neglected.
 Transmission lines are represented by their equivalent series
reactances. Series resistance and shunt admittances are neglected.
 Synchronous machines are represented by constant voltage sources
behind subtransient reactances. Armature resistance, saliency and
saturation are neglected
 Non-rotating impedance loads are neglected
 Induction motors are either neglected (especially for small motors
rated less than 40kW), or represented in the same way as
synchronous machines.
ELECTRICAL PROTECTION IV
Unsymmetrical Fault Analysis
In addition to previous assumptions, the following assumptions should be
made with regards to unsymmetrical fault analysis:
 The power system operates under balanced steady-state
conditions before the fault occurs. Thus the zero-, positive-,
and negative sequence networks are uncoupled before the
fault occurs. During unsymmetrical faults they are
interconnected only at the fault location depending on the
type of fault.
 Prefault load current is neglected. Because of this, the
positive sequence internal voltages of all machines are equal
to the prefault voltage VF. Therefore, the prefault voltage at
each bus in the positive-sequence network equals VF.
ELECTRICAL PROTECTION IV
Unsymmetrical Fault Analysis
Development of three sequence networks:
ELECTRICAL PROTECTION IV
Positive Sequence Network
ELECTRICAL PROTECTION IV
Negative Sequence Network
ELECTRICAL PROTECTION IV
Zero Sequence Network
How?
ELECTRICAL PROTECTION IV
 To do further analysis we first need to calculate the Thevenin
equivalents as seen from the fault location. In this example the
fault is at the terminal of the right machine (bus 2) so the
Thevenin equivalents are:
Zth  j 0.2 in parallel with j0.455
Zth  j 0.21 in parallel with j0.475
ELECTRICAL PROTECTION IV
Single-Line-Ground Faults

= 0
0
Constraint 1
0
+
−
1 1 1
= 1
3
1 2
1
2

0
+
−
0 ⟹  =  =  = 3
0
ELECTRICAL PROTECTION IV
Single-Line-Ground Faults
f
Va
Vaf
 f
Vb
 f
Vc

f
Z f Ia

1 1


2
  1 

1 


= 3 +
0

V
1 f
 
  V f 
2 
  V f 
 
Constraint 2
This means Vaf  V f0  V f  V f
The only way these two constraints can be satisified
is by coupling the sequence networks in series
ELECTRICAL PROTECTION IV
Single-Line-Ground Faults
With the sequence networks in series
we can solve for the fault currents
(assume Zf=0 for bolted fault)
0
=
+
=
−
=

0 +  + +  − + 3
1.050
I I I 
  j1.96427 p.u.
j (0.25  0.13893  0.14562  3Z f )
0
f

f

f
I af  3  I 0f   j 5.8928 p.u.
ELECTRICAL PROTECTION IV
Line-Line Faults
I af  0, I bf   I cf , I 0f  0
Vbg  Vcg if Z F  0
−  =
ELECTRICAL PROTECTION IV
Line-Line Faults
Using the current relationships we get
 I 0f 
1  0
1 1
 
1

2 f
 I f   1     I b
3
 
1  2     I f

 I f 

 b
0
If
 0

If
1 f
2
 Ib   
3

Hence I f   I f


If






1 f 2
 Ib   
3
Constraint 1
ELECTRICAL PROTECTION IV

Line-Line Faults
Using the voltage relationships we get
f 
V f0 

V
1  ag
1 1
 
 


1
V f   1   2  Vbgf  
3
 
 
2


  Vcgf 
V f 
1 
 
Hence




1 f

Vag     2 Vbgf 

3
1 f

V f  Vag   2   Vbgf 
3
V f
Constraint 2

V f  V f
ELECTRICAL PROTECTION IV
Line-Line Faults
To satisfy I f   I f
& V f  V f
the positive and negative sequence networks must
be connected in parallel
Doesn’t play role
as I0=0
Why?
ELECTRICAL PROTECTION IV
Line-Line Faults
+
=
−− =
Solving the network for the currents we get
1.050

If 
 3.691  90
j 0.1389  j 0.1456
 I af 
1 
1 1
0
  0 
 f


2
   6.39 
I

1


3.691


90

 b


 

 f
1   2   3.69190   6.39 
I


 c 
ELECTRICAL PROTECTION IV

+ + −
Line-Line Faults
Solving the network for the voltages we get

Vf
Why “-”?
V f
 1.050  j 0.1389  3.691  90  0.5370
  j 0.1452  3.69190  0.5370
Vaf
 f
Vb
 f
Vc

1   0   1.074 
1 1



2
  0.537    0.537 
  1 

 


1   2  0.537   0.537 



ELECTRICAL PROTECTION IV
Double Line-Ground Faults
I af  0
Vbgf  Vcgf  Z f ( I bf  I cf )
ELECTRICAL PROTECTION IV
Double Line-Ground Faults
From the current relationships we get
 I af
 f
 Ib
 f
 I c
0

I

1 f
1 1
 



2



1


I


 f

1   2    

  I f 

Since I af  0

Constraint 1
I 0f  I f  I f  0
Note, because of the path to ground the zero
sequence current is no longer zero.
ELECTRICAL PROTECTION IV
Double Line-Ground Faults
From the voltage relationships we get
f 
V f0 

V
1  ag
1 1
 
 


1
V f   1   2  Vbgf  
3
 
1  2    f 

V f 

 Vbg 
Since
f
Vbg
f
 Vcg


Vf

 Vf
Constraint 2
Then Vbgf  V f0  ( 2   )V f
But since 1     2  0 
 2    1
Vbgf  V f0  V f
ELECTRICAL PROTECTION IV
Double Line-Ground Faults
Vbgf  V f0  V f

 Zf I I
f
b
f
c

Also since
I bf  I 0f   2 I f  I f
I cf  I 0f  I f   2 I f
Adding these together (with     1)
2


Vbgf  Z f 2 I 0f  I f  I f with I 0f   I f  I f
V f0  V f  3Z f I 0f
Constraint 3
ELECTRICAL PROTECTION IV
Double Line-Ground Faults
Taking all constraints into consideration:
+

= +
+  −  0 + 3
ELECTRICAL PROTECTION IV
Double Line-Ground Faults

V
1
.
05

0
F
I f  

  j 4.547 p.u.

0
Z  Z // Z
j 0.13893  j 0.092


V f  1.05  4.547  90  j 0.1389  0.4184
I f  0.4184 / j 0.1456  j 2.874
Why “-”?
I 0f   I f  I f  j 4.547  j 2.874  j1.673
Converting to phase: I bf  1.04  j 6.82
I cf  1.04  j 6.82
ELECTRICAL PROTECTION IV
Unsymmetrical Fault Summary
 SLG: Sequence networks are connected in series, and also in
parallel to three times the fault impedance
VF
I I I 
j ( Z   Z   Z 0  3Z f )
0
f

f
I  3 I
f
a

f
0
f
ELECTRICAL PROTECTION IV
 LL: Positive and negative sequence networks are connected in parallel;
zero sequence network is not included since there is no path to ground
3VF
I  I 


j(Z  Z )
f
b
f
c
ELECTRICAL PROTECTION IV
 DLG: Positive, negative and zero sequence networks are connected in
parallel, with the zero sequence network including three times the fault
impedance
VF
I  

0
Z  Z // Z  3Z F

f



ELECTRICAL PROTECTION IV