Understanding symmetrical and asymmetrical faults using MVA and symmetrical component methods

Presented by Armstrong Okai Ababio
1
Armstrong Okai Ababio/ Electrical Engineer/
Electricity Company of Ghana9/9/2016

Goal of the Presentation-1
To Appreciate the use of Ohmic and MVA Methods
for Symmetrical Short Circuit Calculation.
A symmetrical or balanced fault affects each of the
three-phases equally (Three phase fault)
2

To appreciate the use of Symmetrical Components for
Asymmetrical Short Circuit Calculation.
An asymmetric or unbalanced fault does not affect each
of the three phases equally.
Common types of asymmetric faults:
line-to-line
line-to-ground
double line-to-ground
The analysis of this type of fault is often simplified by
using methods such as symmetrical components.
3

Introduction
The operation of a power system departs from normal after the
occurrence of a fault. The most common faults in any electrical
installation are short circuits, i.e. a breakdown of insullation
between conductive parts which normally are at different
potential.
Fault give rise to abnormal operating conditions such as excessive
current and voltages at certain points in the network.
The magnitude of fault current depends on the generated power in
the system, the distance to these sources, fault resistance and for
ground fault it depends also on the neutral point treatment of the
system.
4

Effect of Faults in a 3 phase Network
 A fault will cause voltage and current disturbances, including a complete
voltage collapse near the fault location.
 The voltage gradient in the ground and across the earthing resistance
create dangerous step and touch voltages near the fault.
 The dynamic forces of the fault current, the electric arc at the fault
location and the thermal effects of the current on the network elements in
the current path all cause damage to the plant equipment and personnel.
 The voltage and current disturbances interrupt the transmission of power
and thus also affect customers.
5

Effect of Faults-2
 A large amount of power is dissipated at the fault point, which means lost of revenue for
the utility.
 Failure to remove a fault will usually result in rapid expansion of damage to the system.
 Various protection equipment are used in guarding against such faults condition. An
idea of the magnitudes of such fault currents gives the Engineer the current
settings of the various protection devices and ratings of circuit breakers in the
network.
6

Effect of Faults-3
 In the light of the above discussed effects of faults, FAULT or SHORT CIRCUIT
ANALYSIS is highly needed.
 Performing short-circuit calculations requires an understanding of various
system components and their interaction.
 Components of Power System:
Generators, Transformers, Transmission lines, Protection Systems(control
and monitoring system), Switch gear, Load.
7

Types of Fault/ Short Circuits in a 3
phase network Three-phase fault with or without earth (5%)
 Phase-to- phase clear of ground (10 – 15%)
 Two-phase-to-earth fault (10 – 20%)
 Phase-to-earth fault (65 – 70%)
 Fault Incident:
85% of faults are overhead line.
50% of these are due to lightning strikes.

L1
L2
L3
L1
L2
L3
L1
L2
L3
L1
L2
L3
8

3 PHASE FAULTS-1
 A 3-ph fault affects the three-phase network equally (symmetrical fault). They rarely
occur. All three conductors are equally involved and carry the same rms short- circuit
current. There is the need to use only one conductor for the calculation.
 It is valid because system is maintained in a balanced state during fault.
 Voltages are equal and 120° apart.
 Currents are equal and 120° apart.
 Power system plant symmetrical:-
 Phase impedance equal
 Mutual impedance equal
 Shunt admittances equal
 Causes:
 System energization with maintenance earthing clamps still connected.
 1ø faults developing into 3ø faults. Etc.
9

3 PHASE FAULTS-2
 A 3-ph fault affects the three-phase network symmetrically. They rarely occur.
All three conductors are equally involved and carry the same rms short-
circuit current. There is the need to use only one phase (conductor ) for
the calculation.
The Per unit system or MVA method is used for the
analysis of Three phase fault.
10

3 PHASE FAULTS
11

3 PHASE FAULTS
12

Over View of Per Unit System
 Per- Unit Notations. It is used to simplify calculations on system with more than two
voltages. ( The √ are eliminated)
It can be seen by inspection of any power system diagram that:
a. several voltage levels exist in a system
b. it is common practice to refer to plant MVA in terms of per unit or percentage values
c. transmission line and cable constants are given in ohms/km
 Before any system calculations can take place, the system parameters must be referred
to 'base quantities' and represented as a unified system of impedances in either,
percentage, or per unit values.
 Per-unit analysis is based on "normalized" representations of the electrical quantities
(i.e., voltage, current, impedance, etc.). The per-unit equivalent of any electrical
quantity is dimensionless. It is defined as the ratio of the actual quantity in units (i.e.,
volts, amperes, ohms, etc.) to an appropriate base value of the electrical quantity.
 Per Unit (p.u.) value = Actual value / Base value
13

Referring Impedances
14
Electricity Company of Ghana
Can be derived from V1/V2 = I2/I1
Which translates into V1/V2=(V2/R2)/ (V1/R1)
9/9/2016

Formulae used in Per-Unit Calculations
 The base quantities are:
three-phase power in MVA (MVA rating of largest item or 100MVA) It is constant at all
voltages .
The line voltage in kV. Fixed at one part and it is transferred through transformers to
obtain base voltages of the other part of the system.
The following Equations are used in Per- Unit Calculation:
- Ip.u = Iactual/ Ibase
- Sp.u = Sactual / S base
-Vp.u = Vactual / Vbase
- Base Current (kA), Ib = MVAb (3ø)/ {1.732 x kVb (L-L)}
- Base Impedance (Ohm) Zb= (kV)2
(L-L)/ MVAb 3ø
- Zp.u
(new)
= Z p.u
(old)
x[MVAb
(N)
/ MVAb
(old)
] x[kVb
(old)
/ kVb
(new)
]2
15

Transformer Per Unit Impedance
16

Transformer Per Unit Impedance
17

Transformer Percentage Impedance
18

19

20

21

Transformer Base Voltage Selection
22

Example
23

Example
24

MVA METHOD FOR 3 PHASE SHORT CIRCUIT
CALCULATION
 The MVA Method is recognized and widely acceptable by industry in
calculating power system short circuits where the reactance of all circuit
components far exceeds resistance producing a consistently high X/R
ratio throughout the system.
 Combined MVA of components connected in series and parallel are
calculated using the following formulas:
 series: MVA1, 2 = MVA1 X MVA2 / (MVA1 + MVA2)
 parallel: MVA1, 2 = MVA1 + MVA2
 As can be seen from the formulas above, series MVA’s are being
calculated same as resistances in parallel. Parallel MVA combinations are
done same as resistances in series.
 MVA diagram undergoes same reduction process as impedance diagram,
only that MVA values are used instead of per unit impedances or
reactances.
25

CALCULATION
 The MVA method is a modification of the Ohmic method where the impedance of a circuit equals the sum of
the impedances of components constituting the circuit.
 In practice, the MVA method is used by separating the circuit into components and calculating each
component with its own infinite bus as shown in figures 1 and 2 below:
M
15 MVA
Xd = 0.2
20MVA
33 / 11kV
X = 10%
T2161 / 33 kV
X = 10%
161 kV T1
50MVA
Line 33kV
X = 4.0Ω
Sk
’’
= 700MVA
Figure 1-One Line Diagram
700 500 272.25
200
75
3ø F.
700 / 1 50 / 0.1 (33)2
/ 4
15 / 0.2
20 / 0.1
3ø F.
Figure 2- MVA Diagram
1 2 3
4
5
11kV
G
103.8 MVA
26

CALCULATION
 In our example:
MVA’s 1& 2 are in series
- MVA (1&2) = (700 x 500) / (700 + 500) = 291.67
MVA’s (1&2) & 3 are in series
- MVA = (291.67 x 272.25) / (291.67 + 272.25) = 140.81
MVA’s (1,2&3) // MVA 4
- MVA at the point = 140.81 + 75 = 215.81
MVA’s (1,2,3&4) in series with MVA 5
MVA at the Fault point = (251.81 x 200) /(251.81 + 200) = 103.802
Considering Voltage Factor of 1.1, MVA = 103.802 x 1.1 = 114.18
- 3Ø ISC =MVA /(1.732 x Un) =114.18 /(1.732 x 11) = 5.99 kA
Per Unit System is more accurate than the MVA method in the analysis of
three (3) phase faults.
27

D.C. Transient and Offsets
28

END OF PART 1
WAGE OPENER FOR PART 2
RIDDLE!!!
29

To appreciate the use of Symmetrical Components for
Asymmetrical Short Circuit Calculation.
An asymmetric or unbalanced fault does not affect each
of the three phases equally.
Common types of asymmetric faults:
line-to-line
line-to-ground
double line-to-ground
The analysis of this type of fault is often simplified by
using methods such as symmetrical components.
30

ASYMMETRICAL OR
UNBALANCED FAULTS
CALCULATIONS
Developed by Charles LeGeyt Fortescue (1876–1936) in
1918. Fortescue was an Electrical Engineer and he worked
for Westinghouse Corporation at East Pittsburgh,
Pennsylvania. He developed the method of Symmetrical
Components to resolve unbalance fault conditions.
31

UNBALANCED FAULTS-1
Unbalanced Faults may be classified into SHUNT
FAULTS and SERIES FAULTS
SHUNT FAULTS
Line to Ground
Line to Line
Line to Line to Ground
The above faults are described as single shunt faults
because they occur at one location and involve a
connection between one phase and another or to
earth. 32

UNBALANCED FAULTS-2
Unbalanced Faults may be classified into SHUNT
FAULTS and SERIES FAULTS
OPEN CIRCUIT/ SERIES FAULTS
Single phase open circuit
Double phase open circuit
33

CAUSES OF UNBALANCE FAULTS
CAUSES OF SHUNT FAULTS
Insulation Breakdown
Lightning Discharges
Mechanical Damage
CAUSES OF OPEN CIRCUIT FAULTS
Broken Conductor
Operation of Fuses
Mal-operation of single phase CB
During Unbalanced Faults, symmetry of system is
lost. Hence single phase representation is no longer
valid for the fault analysis.
34

Analysis of Unbalanced Faults
Unbalanced faults are analysed using :-
Symmetrical Components
Equivalent Sequence Networks of the Power System
Connection of Sequence Networks appropriate to the
Type of Fault
35

Sequence Components
The Symmetrical Components consist of three
subsystems for an unbalanced three-phase system.
They are;
Positive Sequence Subsystem-consisting of three
phasers of equal magnitude and 120° phase
displacement, and having the same phase sequence as
the original balanced system of phasers.
36

Sequence Components
The second subsystem is termed the 'negative
sequence' system, consisting of three phasers of
equal magnitude and 120° phase displacement, and
having a phase sequence which is the reverse of the
original balanced system of phasers.
The final subsystem is termed the 'zero sequence'
system, consisting of three phasers of equal
magnitude and zero phase displacement.
37

ILLustration of Symmetrical
Components
Orientation of the symmetrical components are as follows:
V A1
V B1V C1
V C2
V A2 VAo
V B2
VBo
120° 240°
V Co
Zero Sequence Phasors
Negative Sequence PhasorsPositive Sequence Phasors
38

Sequence Components-Three Phase
Fault
39

Sequence Components
PHASE TO PHASE FAULT
ONLY +VE AND –VE SEQUENCE COMPNENTS EXISTS
E
Z1 Z2
I a11 I a22
Zo
I a00 = 0= 0
Sequence Network Interconnection for phase to phase fault
40

Sequence Components-Single Phase-Earth Fault.
All sequence components are present
41

By definition, the magnitude and phase angle of
any phasor in an unbalanced three-phase system
is equal to the vector addition of the symmetrical
components from the respective sequence
subsystems.
Symmetrical Components
42

43

To provide a mathematical perspective on the
preceding discussion of phasor magnitudes and phase
displacements, it is necessary to define a phase
displacement operator 'a'.
a = 1<120° = -0.5 + j0.866
Successive applications of operator 'a' to a given
vector will result in rotation of that vector through
120°, 240° and 360°, degrees, as shown below.
44

a = 1<120° = 1ej2π/3
= -0.5 +j0.866
 a2
= 1<240° = 1ej4π/3
= -0.5 -j0.866
 a3
= 1<360° = 1e]2π
= 1
Therefore we have;
45

Converting from phase value toConverting from phase value to
sequence componentssequence components
46

Converting from phase value to
sequence components
47

Symmetrical Components-Similarly Equations for
sequence Currents can be derived
48
Electricity Company of Ghana
It can be deduced that
In= Ia + Ib + Ic = 3I0
9/9/2016

Phase Sequence Equivalent Circuits
E
a2
E
aE
I
a2
I
a I
P Q
P1 Q1
Z1 = E/I
Positive Sequence Impedance
9/9/2016
Electricity Company of Ghana 49

E
aE
a2
E
I
a2
I
a I
P Q
P2 Q2
Z2 = E/I
Negative Sequence Impedance
For static non-rotating plant :- Z2= Z1
9/9/2016

E
I
I
I
P Q
P0 Q0
Z0 = E/I
3I
Zero Sequence Impedance
9/9/2016

Sequence Networks
+ve, -ve and zero sequence networks are drawn for a
‘reference’ phase. This is usually taken as the ‘A’ phase.
Faults are selected to be ‘balanced’ relative to the
reference ‘A’ phase.
e.g. For Ø/E faults, we consider an A-E fault
For Ø/Ø faults, we consider a B-C fault
 Sequence network interconnection is the simplest for the
reference phase.
9/9/2016

1. Start with neutral point N1
-All generator and load neutrals are connected to N1
 2. Include all source voltages:- Phase-neutral voltage
 3. Impedance network:- Positive sequence impedance per
phase
 4. Diagram finishes at fault point F1
Positive Sequence Diagram
N1
E1
Z1
F1
9/9/2016

G
T Line F
R
N
E
N1
E1 Z G1 Z T1 Z L1 I1
F1
N1
V1
V1=Positive sequence PH-N voltage at fault point
I1=Positive sequence phase current flowing into F1
V1=E1 –I1 (ZG1+ ZT1 + ZL1)
System Single Line Diagram
9/9/2016

Negative Sequence Diagram
 Start with neutral point N2
-All generator and load neutrals are connected to N2
 No voltages included
-No negative sequence voltage is generated!
 Impedance network
-Negative sequence impedance per phase
 Diagram finishes at fault point F2
N2
Z2
F2
9/9/2016

G
T Line F
R
N
E
N2
Z G2 Z T2 Z L2 I2
F2
N2
V2
V2=Negative sequence PH-N voltage at fault point
I2=Negative sequence phase current flowing into F2
V2= –I2 (ZG2+ ZT2 + ZL2)
System Single Line Diagram
9/9/2016

Zero Sequence Diagram
For “In Phase” (Zero Phase Sequence) currents to flow in each phase
of the system, there must be a fourth connection (this is typically the
neutral or earth connection).
 IAo+ IBo+ICo=3IAo
IAo
IBo
ICo
N
9/9/2016

Zero Sequence Diagram
N
E
R
3IAo
Resistance Earthed System :-
Zero sequence voltage between N & E is given by
Vo= 3IAo * R
Zero sequence impedance of neutral to earth path
Zo= Vo/ IAo = 3R
9/9/2016

Zero Sequence Equivalent “D y” Transformer”
9/9/2016

Zero Sequence Equivalent “D y” Transformer
Thus, Equivalent single phase zero sequence diagram is as
shown:-
Side terminal Z To
I o
Y side
terminal
N o
(Eo)
9/9/2016

Zero Sequence Equivalent Circuits
9/9/2016

Equations Defining Shunt Fault
Conditions
It should be noted that for any type of fault there are
three equations that define the fault conditions. They
are as follows:
Single Phase-to-earth (A-E)
Ib= 0 (Ib= 0, Ic = 0, Because Phases B and C do not
contribute to fault current)
Ic = 0
Va = 0 (Va = 0, because voltage at the faulted phase
decreases to Zero)
62

Conditions
Phase-phase (B-C)
Ia= 0 (Ia= 0, Because Phase A does not
Ib =-Ic
Vb = Vc
63

Conditions
Phase-phase-to-earth (B-C-E)
Ib + Ic= In
Vb = Vc=0 (Because voltages at the faulted phases
decrease to Zero)
64

Conditions
Three Phase fault (A-B-C or A-B-C-E)
Ia + Ib + Ic= 0 (Because the system is balanced and
hence In= Ia + Ib + Ic=0)
Va = Vc
Vb = Vc
65

Single Phase to Earth Fault-1
We recall that for Single Phase to ground fault:
Ib=Ic = 0
Va = 0 and therefore we can write;
I1=1/3(Ia + aIb + a2
Ic)=1/3Ia
I2=1/3(Ia + a2
Ib + aIc)=1/3Ia Therefore
I0=I1=I2= 1/3Ia
I0=1/3(Ia + Ib + Ic)=1/3Ia 66

Also
Va = 0 and therefore we can write;
V1 +V2 +V0 =0 (Since Va =V1 +V2 +V0 =0)
But
V1 =V-I1Z1,V2=-I2Z2andV0=-I0Z0
Substituting we have;
V-I1Z1 -I2Z2-I0Z0=0 since I1=I2=I0, we have
V=I1(Z1 + Z2+ Z0)
67

The above analysis indicate that the equivalent circuit
for the fault is obtained by connecting the sequence
networks in series as shown below;
68

It also follows that;
I1= E/(Z1 + Z2+ Z0)
From the equation described earlier, I0=I1=I2= 1/3Iaandtherefore
Ia = 3I1 = 3E/(Z1 + Z2+ Z0)
69

In the more general case, with nonzero fault
resistance, the equality of I1; I2 and I0 is maintained, and
3I1 flows through the fault resistance. Therefore, it is
necessary that 3Zf exist in series with the zero
sequence subsystem to achieve the required effect.
The generalized equation is as follows:
Ia = 3I1 = 3E/(Z1 + Z2+ Z0 + 3Zf)
70

Phase to Phase Fault (B-C)-1
We recall that for a phase-phase fault,
Ia = 0, implying
I1=1/3(Ia + aIb + a2
Ic)=1/3(aIb + a2
Ic)
I2=1/3(Ia + a2
Ib + aIc)=1/3(a2
Ib + aIc)
Electricity Company of Ghana 719/9/2016

Also,
Ib= -Ic
I1=1/3(aIb + a2
Ic) = 1/3(alb-a2
lb) = l/3Ib(a-a2
)
I2=1/3(a2
Ib + aIc) = 1/3(a2
lb-alb) = l/3Ib(a2
-a)
The above equations illustrate that;
I 1 = - I 2

By inspection, the equations for this fault condition
are as follows:
I1=E/(Z1 + Z2)
I2=-E/(Z1 + Z2)
I0 = 0
It must be noted that no zero sequence current exist
since there is no connection to ground.Armstrong Okai Ababio/ Electrical Engineer/

From the equation below;
I1=1/3(aIb + a2
Ic) = 1/3(alb-a2
lb) = l/3Ib(a-a2
), Since Ib=-Ic
But
a = 1<120° = 1ej2π/3
= -0.5 +j0.866
a2
= 1<240° = 1ej4π/3
= -0.5 -j0.866
Thereforeitfollowsthat,a-a2
= j1.732= j√3 and I1= j√3/3Ib

Finally it implies that;
I1=1/3(aIb + a2
Ic) = 1/3(alb-a2
lb) = l/3Ib(a-a2
)
Since,
a-a2
= j1.732= j√3
I1 = j √3 /3Ib= jIb/ √3 by rationalization.
hence Ib =-j √3 I1

Similarly;
I2=1/3(a2
Ib + aIc) = 1/3(a2
lb-alb) = l/3Ib(a2
-a)
And since
a2
-a = -j1.732= -j√3, we have
I2 = -j √3 /3Ib= -jIb/ √3 by rationalization.
hence Ib =j √3 I2

By inspection, the equations for this fault condition
are as follows:
I1=E/(Z1 + Z2) but Ib =-j √3 I1
Therefore
Ib =-j √3 E/(Z1 + Z2)
Also,Ic =-Ib
Therefore Ic =j √3 E/(Z1 + Z2)

In the more general case, with nonzero fault
resistance, the generalized equation, based on the
fault impedance of Zf is expressed below:
f
cb
ZZZ
Ej
II
++
±
==
21
3
9/9/2016

Double Phase to Ground Fault (B-C-E)
We recall that for double Phase to Ground Fault
(B-C-E), the following equations can be written:
Ib + Ic= In
Vb = Vc=0 (Because voltages at the faulted phases
decrease to Zero)

From the general equation below, we have:
V-I1Z1 = -I2Z2 from(V1=V2=V0=Va/3) and Vb = Vc=0
Also, Since Ia= 0 (Ia= 0, Because Phase A does not
we have, I1 + I2 + I0= 0
And it follows that;
I1 = -( I2 + I0)

Vb = Vc=0 (Since voltages at the faulted phases
decrease to Zero)
Also implies, V1 = V2 = V0 = 1/3Va
Given that all three sequence voltages are equal in
phase and magnitude, and given that the sequence
currents sum vectorially to zero, it is obvious that the
three sequence networks are in parallelArmstrong Okai Ababio/ Electrical Engineer/

From the fact that
V1 = V2 = V0 = 1/3Va,wecanwrite
V2 = V0, and hence I2Z2 = I0Z0
The negative and zero sequence currents can be derived on the basis
of the current divider principle as shown below:
I2=-I1 Z0/(Z2+ Z0), and I0=-I1 Z2/(Z2+ Z0)

Equating V1 and V2, we have
V-I1Z1 = -I2Z2 orV = I1Z1 - I2Z2,whereV=E= phase voltage
Substituting I2, I2=-I1 Z0/(Z2+ Z0), into the above equation,
It gives,
 V = I1[Z1 + Z0Z2 /(Z0 + Z2)]
Therefore;
I1= E(Z0 + Z2)/(Z1Z0+Z1Z2 +Z0Z2)

Finally,
Substituting I2andI0into the above equation,
I2=-I1 Z0/(Z2+ Z0) , I0=-I1 Z2/(Z2+ Z0)
and
I1= E(Z0 + Z2)/(Z1Z0+Z1Z2 +Z0Z2)
021
2
IaIIaIb ++=
9/9/2016

We have;
020121
20
3
ZZZZZZ
aZZ
EIb
++
−
⋅=
9/9/2016

Three Phase Fault (A-B-C or A-B-C-E)
We recall that for a three phase fault;
Ia + Ib + Ic= 0 (Because the system is balanced and
hence In= Ia + Ib + Ic=0)
Va = Vc
Vb = Vc

It should be noted that, because this fault type is
completely balanced, there are no zero- or negative-
sequence currents;
 By inspection we can write;
I1= E/(Z1+ Zf) also I2 = I0 =0

It finally follows that for a three phase fault;
Ia= Ib = Ic = E/(Z1+ Zf) also I2 = I0 =0

CONCLUSION-1
It must be emphasized that in very large and complex
networks, system computer programs are used for
short circuit analysis.
In the Electricity Company of Ghana (ECG), ASPEN
software is used for short circuit analysis.
Electricity Company of Ghana9/9/2016 89

CONCLUSION-2
I would like to first of all thank God for His guidance and for the
wisdom and understanding He gave me to put this presentation
together.
Secondly, I appreciate the efforts of Ing. Godfred Mensah/ SM/System
Planning and Mr. Frank Osei Owusu of Protection Applications whom
I understudied when preparing for this presentation.
Finally, I very much appreciate your presence for this
presentation
THANK YOU ALL FOR COMING.!!!
Electricity Company of Ghana9/9/2016 90

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