PREPARED BY: DR. ZULKIFFLI BIN ABDUL HAMID
1
Part 1 Per unit system
2
The per-unit system In the power systems analysis field of electrical engineering, a per-unit system is the expression of system quantities as fractions of a defined base unit quantity. Calculations are simplified because quantities expressed as per-unit do not change when they are referred from one side of a transformer to the other. In per-unit system, all parameters (voltage, current, impedance and power) are unit less or precisely, they are expressed in per-unit (p.u.) basis. Generally, the per-unit of any parameter is determined as follows:
The quantity in per-unit can be voltage, current, impedance or power. This is given as follows:
3
The per-unit system ď‚— In the per-unit system, all circuit theories are valid. ď‚— For instance, the complex power and voltage expressed in per-unit can still be found using the following equation:
ď‚— Observe that there is NO 3 or surd(3) in the above equations; this is one of the advantages of using per-unit system over conventional methods.
4
The per-unit system – base parameters Whereas, the base impedance is found as:
And, the base current is found as:
IB =
SB 3VB
5
The per-unit system – change of base Sometimes, the equipment such as generator, transformer and motor have their own per-unit impedance (Zpu) or reactance (Xpu) which are based on their rated MVA and rated voltage. When performing analysis involving per-unit, all parameters must be based on the new base MVA and new base voltage. This means that the old per-unit impedance or reactance have to be converted in accordance with the new base using the following equation: 2
Z
new pu
SB Vrated old S =Z @ Zpu S Srated VB old pu
new B old B
V V
old B new B
2
Where, SB and VB are the new base MVA and base voltage, while Srated and Vrated are the old base MVA (or rated MVA) and old base voltage (or rated voltage) of the equipment.
6
The per-unit system – load impedance The load in power system can be a solid impedance or motor. If it is a motor, then the per-unit reactance based on the new base MVA and voltage can be found using the “change-of-base” equation. However, if an impedance represents the load, then the per-unit impedance of that load is found as: VL2_ pu Z L _ pu = * S L _ pu Where, VL_pu is the per-unit voltage at the load bus and SL_pu is the per-unit load complex power
7
The per-unit system Advantages of per-unit system are:
It gives us a clear idea of relative magnitudes of various quantities, such as
voltage, current, power and impedance.
The per-unit impedance of equipment (generator or transformer) fall in a
narrow range regardless of the equipment rating. Whereas their impedance in ohms vary greatly with the rating.
The per-unit values of impedance, voltage and current of a transformer are
the same regardless of whether they are referred to primary or secondary side.
The per-unit systems are ideal for computerized analysis and simulation of
complex power system.
The circuit laws are valid in per-unit systems, and the power and voltage
equations are simplified since the factors of 3 or surd(3) are eliminated.
8
The per-unit system – analysis  Analysis of per-unit system can be summarized in the following steps: 1. 2.
3.
4.
5.
Divide the system into different voltage zones created by each transformer. Determine the new base voltage (VB) for each zone. The given new base MVA (SB) are the same for all zones in the power system. Different sides of transformer (primary or secondary) form two different zones. Convert all reactance or impedance into their per-unit values based on the new base MVA (SB) and new base voltage (VB). Use appropriate per-unit equation for each component (such as generator, transmission line, transformer, motor or load). Convert the given single-line diagram of power system into the single-phase equivalent circuit. At this stage, replace all components (such as generator, transmission line, transformer, motor or load) with their equivalent perunit reactance or impedance. Analyze the circuit using any circuit theories (Ohm’s Law, KVL, KCL, nodal analysis, CDR, VDR and etc.). Begin the analysis according to the load requirement; i.e. the voltage and current must be according to the load.
9
The per-unit system – analysis  Assumptions made in symmetrical fault analysis 1. All sources are balanced and equal in magnitude & phase. 2. Sources represented by the Thevenin’s voltage prior to fault 3. 4. 5. 6. 7. 8.
at the fault point. Large systems may be represented by an infinite bus-bars. Transformers are on nominal tap position. Resistances are negligible compared to reactances. Transmission lines are assumed fully transposed and all 3 phases have same Z. Loads currents are negligible compared to fault currents. Line charging currents can be completely neglected.
10
Per-unit equivalent circuit Load
Load
Transformer
Transmission line
M
_ Motor
_ Generator
M
+
+
11
Example 1 A power system consists of a generator, two transformers, a transmission line and a motor is shown below. By taking the MVA and rated voltage of generator G1 as the base MVA and base voltage: i) Convert all impedance according to the new base MVA and base voltage ii) Draw the per-unit equivalent circuit of the following diagram.
G1
T1
L1
T2
M
12 + j75 Ί 100 MVA 12 kV 0.1 + j0.9 pu
50 MVA 11/120 kV 0.01 + j0.05 pu
M1
50 MVA 130/11 kV 0.01 + j0.05 pu
50 MVA 11 kV 0.1 + j0.9 pu
12
Solution (division of zones) Zone 1 G1
Zone 3
Zone 2 T1
L1
T2
M1
M
12 + j75 Ω 100 MVA 12 kV 0.1 + j0.9 pu
50 MVA 11/120 kV 0.01 + j0.05 pu
VB1 = 12 kV
50 MVA 130/11 kV 0.01 + j0.05 pu
VB2 11/120 kV
50 MVA 11 kV 0.1 + j0.9 pu
VB3 130/11 kV
13
Solution (base voltages) VB1 = 12 kV VB 2
T1 VHV = VB1 T 1 V LV
= (12 kV) 120 = 130.91 kV 11
VB 3
T2 VLV = VB 2 T 2 V HV
= (130.91 kV) 11 = 11.077 kV 130
14
Solution (change-of-base) S G : Z G = Z G _ old B SG
2
2
VG 100 12 ( 0 . 1 0 . 9 ) j = + = 0.1 + j 0.9 pu V 100 12 B1 2
2
T1 : Z T 1
S V 100 11 = Z T 1 _ old B T 1 = (0.01 + j 0.05) = 0.017 + j 0.084 pu 50 S V 12 T 1 B1
T2 : Z T 2
S = Z T 2 _ old B ST 2
VT 2 VB 2
M : ZM
SB = Z M _ old SM
VM 100 11 = (0.1 + j 0.9) = 0.197 + j1.775 pu V 50 11 . 077 B 3
2
2
100 130 = (0.01 + j 0.05) = 0.0197 + j 0.0986 pu 50 130.91 2
2
15
Solution (line impedance) Z B2
VB22 (130.91 kV) 2 = = = 171.374 â„Ś 100 MVA SB
Line 1 : Z L1 =
actual impedance 12 + j 75 = = 0.07 + j 0.438 pu base impedance 171.374
16
Solution (per-unit equivalent circuit) 0.017 + j0.084 pu
0.07 + j0.438 pu
0.0197 + j0.0986 pu
T1
L1
T2
j0.9 pu
G1 EG
0.1 pu
+ _
j1.775 pu
M1
0.197 pu
+
M
_
EM
17
Example 2
18
Example 4
19
Solution (division of zones) Zone 1
22/220 kV
VB1 = 22 kV
Zone 2
220/11 kV
VB2
Zone 4
VB4
VB3 22/110 kV
Zone 3
110/11 kV
20
Solution (change-of-base) 2
2
G : XG
S = X G _ old B SG
T1 : X T 1
S V 100 22 = X T 1 _ old B T 1 = 0.10 = 0.20 pu 50 22 ST 1 VB1
T2 : X T 2
S = X T 2 _ old B ST 2
T3 : X T 3
S V 100 22 = X T 3 _ old B T 3 = 0.064 = 0.16 pu 40 S V 22 T 3 B1
T4 : X T 4
S = X T 4 _ old B ST 4
VT 4 100 110 = 0.08 = 0.20 pu V 40 110 B 3
M : XM
S = X M _ old B SM
VM VB 4
VG 100 22 0 . 18 = = 0.20 pu V 90 22 B1 2
VT 2 VB 2
2
2
2
100 220 = 0.06 = 0.15 pu 40 220 2
2
2
2
2
2
100 10.45 = 0.185 = 0.25 pu 66.5 11
21
Solution (line impedance) Z B2
VB22 (220 kV) 2 = = = 484 â„Ś 100 MVA SB
Z B3
VB23 (110 kV) 2 = = = 121 â„Ś S B 100 MVA
Line 1 : X L1 = Line 2 : X L 2
actual impedance 48.4 = = 0.10 pu base impedance 484
actual impedance 65.43 = = = 0.54 pu base impedance 121 22
Solution (load impedance) V4 (in pu ) =
S Load
V4 10.45 = = 0.95 pu VB 4 11
S Load (in pu ) = ∠ + cos −1 PFLoad SB =
57 ∠ + cos −1 0.6 100
= 0.57 pu ∠ + 53.13°
Z Load =
V42
S *Load
(0.95) 2 = 0.57∠ − 53.13 = 0.95 + j1.267 pu 23
Solution (per-unit equivalent circuit)
24
Part 2 Symmetrical Fault
25
Fault Faults can be defined as any failure that interferes with the normal flow
of current to the loads They can occur in power systems when equipment insulation fails due to: System over voltages due to lightning or switching surges Insulation contamination Mechanical and natural stress In most faults, a current path forms between: Two or more phases One or more phases and the neutral (ground) This current path has low impedance (represented by Zf) resulting in excessive current flows.
26
Fault In an electric power system, a fault is any abnormal flow of electric
current to the loads due to the failure or breakdown in power system. For example, a short circuit is a fault in which current bypasses the normal load. In three-phase systems, a fault may involve one or more phases and ground, or may occur only between phases. If a fault occurs between a phase and ground (such as between a conductor of transmission line and tower structure), it is called as flashover. When a fault occurs, the high current flow in the line are detected by protective devices (such as circuit breakers). Circuit breakers on affected transmission lines will be automatically opened isolating the transmission line until the fault can be located and cleared by repair crew. 27
Categories of fault Fault can be categorized in two ways; in terms of: How long the fault happens
Transient fault Permanent fault
How the fault happens
Symmetrical fault Three phase fault Unsymmetrical fault Single line-to-ground (SLG) fault Line-to-line (LL) fault Double line-to-ground (DLG) fault
Main focus of Chapter 3
28
Categories of fault Symmetrical faults (three-phase faults / balanced fault) The magnitudes of the AC currents in each phase are the same All three phases of a transmission lines are shorted together The most severe fault, but happens infrequently.
Unsymmetrical faults (unbalanced fault) The magnitudes of the AC currents in each phase differ Can be subdivided into 3: I. II. III.
Single line-to-ground fault (SLG) - One line touch the ground Line-to-line fault (LL) - Two lines of a transmission line touch Double line-to-ground fault (DLG) - Two lines touch and also touch the ground 29
Symmetrical and unsymmetrical fault Symmetrical fault (3-phase fault):
The most severe fault
Unsymmetrical fault:
SLG
LL
DLG
30
Causes of fault The causes of fault are: Momentary tree contact Bird or other animal contact Lightning strike Conductor clashing Weather condition (heavy rain, heavy winds, snow and ice) Equipment failure Ionization of surrounding air (due to smoke particles)
31
Effects of fault Over current flow:
When fault occurs it creates a very low impedance path for the current flow. This results in a very high current being drawn from the supply, causing tripping of relays, damaging insulation and components of the equipments. Danger to operating personnel:
Fault occurrence can also cause shocks to individuals. Severity of the shock depends on the current and voltage at fault location and even may lead to death. Loss of equipment:
Heavy current due to short circuit faults result in the components being burnt completely which leads to improper working of equipment or device. Sometimes heavy fire causes complete burnout of the equipments. Electrical fires:
Short circuit causes flashovers and sparks due to the ionization of air between two conducting paths which further leads to fire as we often observe in news such as building and shopping complex fires.
32
33
Importance of fault analysis To determine the size of protective devices (such as circuit breakers)
needed for installation.
Proper relay setting and coordination of protective devices. To determine the capability of insulating materials (such as high
voltage bushing) in sustaining breakdown.
As a preparation for breakdown in power system.
34
Fault analysis Generally, there are two types of fault analysis: Symmetrical fault or balanced fault analysis Unsymmetrical fault or unbalanced fault analysis
Under symmetrical fault analysis, there are:
Analysis with no-load assumption where pre-fault voltage = 1 pu ii. Analysis by considering load – pre-fault voltage must be calculated iii. Analysis by using bus impedance matrix – for large power system i.
For the purpose of this chapter, only the analysis with no-load
assumption is covered. Because symmetrical fault is balanced, it is solved on a per-phase basis, using Thevenin’s method. 35
Fault analysis ď‚— Thevenin Theorem:
States that the change in the network voltage caused by the added branch (the fault impedance) is equivalent to those caused by an added voltage (V TH ) with all other sources short circuited. ď‚— Important terms:
Short-circuit current: Ifault = 1 / Zeq Zeq = ZTh + Zf
Short-circuit MVA: MVAfault = MVAbase / Zeq
36
Fault analysis The following assumptions are usually made in fault analysis in three phase transmission lines. All sources are balanced and equal in magnitude & phase. Sources represented by the Thevenin’s voltage prior to fault at the fault
point. Large systems may be represented by an infinite bus-bars. Transformers are on nominal tap position. Resistances are negligible compared to reactances. Transmission lines are assumed fully transposed and all 3 phases have same Z. Loads currents are negligible compared to fault currents. Line charging currents can be completely neglected
37
Symmetrical Fault analysis The steps for symmetrical fault analysis are as follows: 1. Convert all reactances according to the new base MVA and voltages. If the rated MVA and voltage of each component equal to base MVA and voltage (i.e. SB = Srated and VB = Vrated), use directly the given reactances without using change-of-base equation. 2. Draw the per-unit equivalent circuit between the faulted bus and common bus. Label all the reactances. 3. Determine the per-unit Thevenin equivalent impedance ZTh looking from the faulted bus. Turn off all voltage sources (short circuit the sources). 4. Calculate the total per-unit fault current using the equation: Ifault = 1 / (Zfault + ZTh) where Zfault is the fault impedance connected on the faulted bus. For bolted or solid fault, Zfault = 0 pu. 5. Determine the actual Ifault in Ampere (times with Ibase) and MVAfault. 6. Determine the other fault currents (such as the generator current or transmission line current under fault condition) if necessary. 38
Symmetrical Fault analysis Important terms:
39
Example 3 ď‚— Based on the per-unit equivalent circuit as in Example 4, determine the
magnitude of fault current in per-unit and Ampere if a bolted threephase fault happens at bus 3 as follows. G1
Bus 1
100 MVA 12 kV 0.1 + j0.9 pu
T1
Bus 2
L1
Bus 3
T2
Bus 4
12 + j75 Ί
50 MVA 11/120 kV 0.01 + j0.05 pu
M1
M 50 MVA 130/11 kV 0.01 + j0.05 pu
50 MVA 11 kV 0.1 + j0.9 pu
Fault
40
Solution – equivalent circuit under fault condition Z G1 Common bus
1 pu
0.1 + j0.9 pu
Z T1
Z L1 0.017 + j0.084 pu 0.07 + j0.438 pu
0 Z T2 ZM 0.197 + j1.775 pu 0.0197 + j0.0986 pu
Bus 3
Ifault
Fault
41
0.187 + j1.422 pu
Common bus
1 pu
0 0.2167 + j1.8736 pu
Bus 3
Ifault
Fault Common bus
1 pu
0
Bus 3
0.1008 + j0.8085 pu
ZTh
Ifault
Fault
42
Fault current and short circuit MVA Base current at fault location:
SB
I B2 =
3VB 2
=
100 MVA 3 (130.91 kV)
= 441.028 A
Short circuit MVA is:
MVA sc = S B × I fault (in pu ) = 100 MVA × 1.2274 pu = 122.74 MVA
So, the three-phase fault current is:
I fault =
1 1 = Z Th 0.1008 + j 0.8085
= 1.2274 pu ∠ − 82.89° = 541.318 A ∠ − 82.89°
43
Example 4 Based on the per-unit equivalent reactances calculated in Example 2, determine the magnitude of fault current in per-unit and Ampere if a solid three-phase fault happens at bus 5 as follows.
Fault
44
Solution – equivalent circuit under fault condition
Common bus
1 pu
XG j0.2 pu XM j0.25 pu
XT2 j0.15 pu
Bus 1
XT3 j0.16 pu
Bus 5
XL1 XT1 j0.10 pu j0.20 pu
0
Bus 4 Z Load 0.95 + j1.267 pu
XT4 j0.20 pu
XL2 j0.54 pu Fault
45
Common bus
1 pu
XG j0.2 pu XM j0.25 pu
Bus 1
XT2 j0.15 pu
XT3 j0.16 pu
Bus 5
XL1 XT1 j0.10 pu j0.20 pu
0
Bus 4 Z Load 0.95 + j1.267 pu
Common bus
XT4 j0.20 pu
XL2 j0.54 pu Fault
Bus 4 1 pu
Bus 1
j0.2 pu
j0.16 pu
Bus 5
j0.45 pu
0
0.01853 + j0.2204 pu
j0.74 pu Fault
46
Common bus
Bus 4 1 pu
Bus 1
j0.2 pu
j0.16 pu
Bus 5
j0.45 pu
0 j0.74 pu
0.01853 + j0.2204 pu
Fault
Common bus
j0.2 pu
Bus 1
j0.16 pu
Bus 5
j0.45 pu 1 pu
0
Bus 4
0.01853 + j0.2204 pu
j0.74 pu Fault
47
j0.2 pu
Common bus
Bus 1
j0.16 pu
Bus 5
j0.45 pu 1 pu
0
Bus 4
0.01853 + j0.2204 pu
j0.74 pu Fault
Bus 1
j0.16 pu
Bus 5
Bus 5
Bus 1 j0.0533 pu
j0.45 pu
j0.74 pu
j0.0877 pu j0.2467 pu Bus 4
48
49
Common bus
j0.2 pu
Bus 1
j0.16 pu
Bus 5
j0.45 pu 1 pu
0
Bus 4 j0.74 pu
0.01853 + j0.2204 pu
Fault
Common bus
1 pu
0
j0.2 pu
Bus 5
Bus 1 j0.0533 pu
Bus 4
j0.0877 pu j0.2467 pu
0.01853 + j0.2204 pu Fault
50
Common bus
1 pu
0
j0.2 pu
Bus 5
Bus 1 j0.0533 pu
Bus 4
j0.0877 pu j0.2467 pu
0.01853 + j0.2204 pu Fault
Common bus
j0.2533 pu
Bus 5
j0.0877 pu 1 pu
0
0.01853 + j0.4671 pu Fault
51
Common bus
j0.2533 pu
Bus 5
j0.0877 pu 1 pu
0
0.01853 + j0.4671 pu Fault
Common bus
Bus 5
j0.1643 pu 1 pu
j0.0877 pu
Ifault
0 j0.252 pu
ZTh
Fault
52
Fault current and short circuit MVA Base current at fault location:
I B3 =
SB 3VB 3
=
100 MVA 3 (110 kV)
= 524.86 A
Short circuit MVA is:
MVA sc = S B × I fault (in pu ) = 100 MVA × 3.968 pu = 396.8 MVA
So, the three-phase fault current is:
I fault =
1 1 = Z Th j 0.252
= 3.968 pu ∠ − 90° = 2082.64 A ∠ − 90°
53
Example 5
54
Part 3 Unsymmetrical fault
55
Unsymmetrical Faults Most faults that occur on power systems are unsymmetrical faults. The most common is single line-to-ground fault Since the magnitudes of the voltages & currents in each phase differ, a technique known as the method of symmetrical components is used to analyse unsymmetrical faults.
Positive sequence Unsymmetrical fault
Symmetrical components
Negative sequence Zero sequence
56
Symmetrical Components
Using the method, any unsymmetrical set of voltages or current could be broken down into three symmetrical sets of balanced three-phase components. Positive-sequence components Consist of three phasors equal in magnitude, displaced from each other by 120
degrees, and have the same phase sequence as the original power system. It has abc sequence Negative-sequence components
Consist of three phasors equal in magnitude, displaced from each other by 120
degrees and have the opposite phase sequence as the original power system It has acb sequence Zero-sequence components Consist of three phasors equal in magnitude and phase No phase sequence i.e. all three phases peak at the same time
57
Symmetrical Components - voltage
Representation of symmetrical components
58
Symmetrical Components - current
Positive sequence
negative sequence
zero sequence
59
Relationship of voltage and current with symmetrical components The unsymmetrical three-phase voltages (VA , VB and VC) and currents (IA, IB and IC) can be determined by adding together their symmetrical components
60
Relationship with symmetrical components – [derivation] The unsymmetrical voltage in each phase (A, B and C) is the sum of the three components, given by,
(1)
Let:
61
Relationship with symmetrical components – [derivation]  From the phasor diagrams of positive, negative and zero sequence, the voltages in phase B and C are expressed in terms of that of phase A as follows:
62
Relationship with symmetrical components – [derivation] Substitute (2), (3) and (4) into (1), yields:
In matrix form, the equations become:
rearrange
63
Relationship with symmetrical components – [derivation] If we make the definition:
So, equation (5) can be written as:
64
Relationship with symmetrical components – [derivation]
Therefore, the symmetrical components of the unbalanced three-phase voltage can be expressed as:
Where:
65
Relationship with symmetrical components – [derivation] Thus, equation (6) becomes:
Finally, the symmetrical components of voltage in phase A can then be represented as:
66
Relationship with symmetrical components – [derivation] Similarly, currents in each phase can be represented as:
Then, substituting the a constant, we get:
67
Relationship with symmetrical components – [derivation]  The symmetrical components of current in phase A can then be represented as:
68
Relationship with symmetrical components – [derivation] Note that:
Therefore: if a component has no neutral current, there can be no zero sequence
currents in the component. thus, Δ-connected & ungrounded Y-connected components cannot have zero-sequence components of currents because they have no neutral connection.
69
Summary: unsymmetrical voltage and current with their symmetrical components Unsymmetrical voltage and current
Symmetrical components
70
Sequence Impedance and Sequence Networks In general, the impedance of a circuit differs for positive, negative & zero sequences currents. When only positive-sequence current is flowing, the impedance is known as the positive sequence impedance of the circuit. Similarly for negative and zero sequences. To analyze an unsymmetrical fault, three different per-phase equivalent circuits must be constructed.
71
Sequence Impedance and Sequence Networks Each circuit represents each type of symmetrical components: Positive-sequence network – a per-phase equivalent circuit containing only
the positive sequence impedances and sources
Negative-sequence network – a per-phase equivalent circuit containing only
the negative sequence impedances
Zero-sequence network – a per-phase equivalent circuit containing only the
zero-sequence impedances
72
Sequence Networks for generator ď‚— The equivalent circuit of a three-phase, Y-connected synchronous generator with grounded neutral is shown below:
73
Sequence Networks for generator – positive sequence network Phase A
No grounded neutral
74
Sequence Networks for generator – negative sequence network Phase A
No grounded neutral
75
Sequence Networks for generator – zero sequence network Phase A
Grounded neutral is considered
76
Sequence Networks for generator ď‚— So, the symmetrical components of synchronous generators considering phase A are:
Positive
Negative
Zero
77
Sequence Networks for transformer The equivalent circuit of a transformer for positive & negative sequences networks is identical. It consists of just the series impedances of the transformer, regardless of primary and secondary connections (Y or Δ) For zero-sequence network, however, there are 5 possible cases, depending on their connections at primary and secondary sides. Table below shows the connection diagram and zero-phase sequence network for various types of transformer’s connection.
78
Sequence Networks for transformer – zero sequence networks
79
Sequence Networks for transformer – zero sequence networks
80
Sequence Networks for transformer – zero sequence networks
81
Types of unsymmetrical fault As mentioned before, three types of unsymmetrical fault are: Single line-to-ground (SLG) fault Line-to-line fault Double line-to-ground (DLG) fault
SLG
LL
DLG
82
Single line-to-ground (SLG) fault ď‚— A synchronous generator with SLG fault on phase A is shown below:
SLG fault
83
Single line-to-ground (SLG) fault ď‚— Since IB = 0 and IC = 0, the symmetrical components of current in phase A are:
ď‚— It can be observed that:
84
Single line-to-ground (SLG) fault ď‚— Recall that the symmetrical components of voltage in phase A of generator is given by:
ď‚— Substitute IA1 into VA1, VA2 and VA0, the voltage equations become:
85
Single line-to-ground (SLG) fault The unsymmetrical voltage in phase A can also be expressed as:
In SLG fault, VA = 0
So:
Hence, the positive sequence current is:
86
Single line-to-ground (SLG) fault Z0
In general, SLG fault is modeled as:
IA0 +
VA0
I A 0 = I A1 = I A 2
_
Z1
EA I A1 = Z 0 + Z 1 + Z 2 + 3Z f The fault current is:
EA
IA1 +
+
3Z F
VA1
_
_
Z2
IA2 +
I A = I A 0 + I A1 + I A 2 = 3I A1
VA2 _
87
Line-to-line (LL) fault ď‚— A synchronous generator with LL fault on phase B and C is shown below:
LL fault
88
Line-to-line (LL) fault ď‚— Since VB = VC, the symmetrical components of voltage in phase A are as follows:
ď‚— From the above derivation, we see that:
VA1 = VA2
89
Line-to-line (LL) fault ď‚— Also, since IA = 0 and IB = - IC, the symmetrical components of current in phase A become:
ď‚— From the above derivation:
90
Line-to-line (LL) fault ď‚— Recall that the symmetrical components of voltage in phase A of generator is given by:
ď‚— Substitute the currents (IA0, IA1 and IA2) and voltages (VA0, VA1 and VA2) into the above equations:
91
Line-to-line (LL) fault ď‚— Equating the equations:
ď‚— Therefore, the positive sequence current is:
92
Line-to-line (LL) fault In general, LL fault is modeled as:
EA I A1 = Z1 + Z 2 + Z f
Z1
EA
+ _
Z2
IA1
IA2
+
+
VA1
VA2
_
_
The fault current is:
I B = − I C = − j 3I A1
ZF
93
Double line-to-ground (DLG) fault ď‚— A synchronous generator with DLG fault on phase B and C is shown below:
DLG fault
94
Double line-to-ground (DLG) fault
95
Double line-to-ground (DLG) fault ď‚— Recall that the symmetrical components of voltage in phase A of generator is given by:
ď‚— SinceVA0 = VA1 =VA2 , equating the above equations:
and (1)
(2)
96
Double line-to-ground (DLG) fault In DLG fault:
By the fact that IN = 3IA0 , the following derivation is obtained:
IN
IB
IC
(3)
97
Double line-to-ground (DLG) fault ď‚— Substitute (1) and (2) into (3):
ď‚— Therefore, the positive sequence current is:
98
Double line-to-ground (DLG) fault In general, DLG fault is modeled as:
Z1
EA
+ _
Z2
IA1
Z0
IA2
IA0
+
+
+
VA1
VA2
VA0
_
_
_
3ZF
EA I A1 = Z 2 ( Z 0 + 3Z f ) Z1 + Z 2 + Z 0 + 3Z f
The fault current is:
I f = 3I A 0
Z1 I A1 − E A = 3 Z + 3Z f 0
99
Unsymmetrical fault analysis Unsymmetrical fault analysis is about to convert the unsymmetrical faulted power system to the equivalent symmetrical one. This is done by breaking down the unsymmetrical system into its three symmetrical components, namely: Positive sequence component Negative sequence component Zero sequence component
As a result, the analysis can be performed in analogous to that of symmetrical fault analysis.
Positive sequence Unsymmetrical fault
Symmetrical components
Negative sequence Zero sequence
100
Unsymmetrical fault analysis Procedures for unsymmetrical fault analysis: 1.
2.
3.
4.
5.
Convert all reactances according to the new base MVA and voltages. If the rated MVA and voltage of each component equal to base MVA and voltage (i.e. SB = Srated and VB = Vrated), use directly the given reactances without using change-of-base equation. Draw the equivalent circuit for positive sequence between the faulted bus and common bus. Ignore the transformer connection. Simplify the circuit and determine its total reactance. This will be the total reactance of positive sequence, labeled as Z1. Repeat step 2 for negative sequence network to determine the total reactance Z2. No voltage source in this network. If the individual components have the same values of their positive and negative reactance, then Z2 = Z1. Repeat step 2 for zero sequence network to determine the total reactance Zo. No voltage source in this network. The transformer connection (Y or Δ) must be considered in this step. Find the voltage and current based on the type of fault (SLG, LL or DLG) using the calculated reactances Z0, Z1 and Z2.
101