EE423
FAULT STUDY
Instructed By:
Name
: G. R. Raban
Index Number
:
Field
:
Group
:
Date of Performance
:
Date of Submission
:
Observations
Sequence Impedances; Impedance (pu) Positive Sequence
Z1
0.240
Negative Sequence
Z2
0.229
Zero Sequence
Z0
0.614
Sequence voltages and sequence sequence currents for each t ype of fault; Voltage (V)
Current (mA)
Positive Sequence
39.44
3.9
Negative Sequence
-10.60
3.9
Zero Sequence
-28.84
3.9
Positive Sequence
20.60
9.0
Negative Sequence
20.60
-6.5
Zero Sequence
20.60
-2.5
Positive Sequence
24.52
7.0
Negative Sequence
24.50
-7.0
1. Single line-to-earth fault
2. Double line-to-earth fault
3. Line-to-line fault
Theoretical Calculations
Single Line to Earth Fault
Assumptions; Fault impedance is zero
∴ ⟹ −
Load currents are negligible compared to fault current =
,
=
0
0
1
=
1
3
2
0
=
+
+
1
1 1 1
=
0
0
1
1
2
=
=
0 1 2
= = =
0.61 0.614 4 0.24 0.240 0 0.22 0.229 9
=
1
=
2
0
0
0
0 0
0
+
1
0 0
0
2
2
+
... (2)
3
3
1
... (1)
=0 =0
2
=
By observation data;
0
1
2
=
1
=
2
3
0
=
1
=
2
=
3 3 3
... (3)
Fault Current; Substituting values to equation (3),
∴ =
,
3 ×1
=
0.614 0.6 14 + 0.24 0.240 0 + 0.22 0.229 9
=
×
,
=
40
2.77 2.77 × =
0
=
132
=
1
2.77
.
=
2
839.31 3
=
279. 27 9.77 77
Fault Voltages;
Ω − − − − − − − − =
0
=
1
=
2
=
0
=
0
1
2
Phase Voltages;
1 × 132 × 103
=
2
=
= = = =
= = = =
435. 43 5.6 6 =
74.83
0.24 0.24 × 43 435. 5.6 6 × 27 279. 9.77 77
0.22 0.229 9 × 43 435.6 5.6 × 27 279.7 9.77 7
=
27.91
− − − ∠− ∠ ∠ − ∠ − − − ∠ ∠ − ∠ 1 1 1
=
From equation (1),
=
40
0.61 0.614 4 × 43 435.6 5.6 × 27 279.7 9.77 7
=
1
2
132 132
0
+
0
+
1
+
2
2
+
1
2
74.83 +
1
1
1
2
2
×
74.83 00 + 10 102. 2.75 75 2400
0
+
74.83 +
27.91
27.91 1200
.
1
+
2
2
× 10 102.7 2.75 5 +
74.83 74.8 3 + 102. 102.75 75 1200 .
2
=
× 102.7 2.75 +
.
0
2
.
2
×
27.91
27.91 2400
=
102. 10 2.75 75
Double Line to Earth Fault
Assumptions; Fault impedance is zero
∥ − − − − − − − − − − − − Load currents are negligible compared to fault current =
0
=
=
0
By observation data; 0
=
0.61 0.614 4
1
=
0.24 0.240 0
2
=
0.22 0.229 9
=
1
1,
1
=
2
=
=
0
,
=
1
=
+
2
0
2.46 2.46 × 0.30 0.303 3 1
1
=
1
1
× 0.30 0.303 3
1
=
× 0.30 0.303 3
3×
=
1.22 1.2288 88 × 13 132 2
1
=
1
=
3×
1
=
0.24 0.2 4 × 2.46 2.46 0.229
0.24 0.24 × 2.46 2.46
0.614
0
=
2 . 46
0.74 0.745 5
2
1
=
0.229 × 0.61 0.229 0.614 4 0.24 0.24 + 0.22 0.229 9 + 0.61 0.614 4
=
1
.
3× 1
× 0.30 0.303 3
× 0.30 0.303 3
0.24 0.24 × 2.46 2.46
=
=
=
0.542
0.202
1.22 1.2288 88
Phase Currents;
− − − ∠− ∠ ∠ − ∠ − − − ∠ ∠ ∠ − ∠ 1 1 1
1
× 0.74 0.745 5 +
×
=
= = = =
= = = =
0
+
2
+
1
2
0.202 +
0
2
2
0.542
0.542 1200
.
+
0.202 +
1
+
2
2
× 0.74 0.745 5 +
2
0.202 00 + 0.74 0.745 5 1200 .
0 1
2
0.202 00 + 0.74 0.745 5 2400 .
1
2
.
×
0.542
0.542 2400
Line to Line Fault
Assumptions; Fault impedance is zero
− − −
Load currents are negligible compared to fault current =
= =
By observation data; 0
=
0.61 0.614 4
1
=
0.24 0.240 0
2
=
0.22 0.229 9
=
1
=
1,
2
0.24 0.24 + 0.22 0.229 9
2.13 2.132 2 × 0.30 0.303 3
=
1
1
=
2
=
0
=
+
1
=
1
=
=
2.13 2.132 2
0.64 0.646 6
0.646
0
1
=
2
=
1
=
2
=
2
×
2
0.229 × 435.6 5.6 × 0.6 0.646
=
64.4 64 .44 4
∠ ∠ ∠ ∠ − ∠ ∠− ∠ − ∠ ∠ 1 1 1
=
= = =
= = =
0
+
1
+
= = =
= = =
1
2
0 1
2
2
2
0 + 64.44 + 64.44 .
0
+
2
1
+
2
0 + 64.44 2400 + 64 64.4 .44 4 1200 =
.
1 1 1
=
1
0
+
2
1
+
1
2
0 + 0.646 2400
0.646 1200
.
0
+
1
+
2
0 + 0.646 1200 .
1
2
2
0.646 2400
0 1
2
2
Practical Calculations
The following adjustments were made in the practical:
Resistances were multiplied by a factor of 4000
A 50 V DC supply was used instead of a 132 kV supply
The practical values must be adjusted accordingly.
=
132
×
=
50
2
132 1 32 40
4000
2 4.242 24.2 424 4 × 103
×
=
×
=
×
132 50
2640 2640
Single Line to Earth Fault
− ∠ ∠ ∠ ∠ − − − ∠ − ∠ ∠− ∠ − ∠ ∠ =
=
3.9 × 3 × 10
3
1 1 1
1
1
2
2
× 24.2 4.2424 × 103
=
3.9 3 .9 + 3.9 3.9 2400 + 3.9 1200
=
3.9 + 3.9 1200 + 3.9 2400
=
− − −
1 1 1
=
28.8 28 .84 4 + 39 39.4 .42 2
=
28.8 28 .84 4 + 39 39.4 .44 4 2400
=
= =
161. 16 1.65 65
3.9 3.9 3.9
10.6
1
=
283. 28 3.64 64
× 24.2 4.2424 × 103
× 24.2424 × 103
1
2
2
28.84 39.44 10.6
× 26 2640 40
=
0
10.6 1200
× 26 2640 40
10.6 2400
× 26 2640 40
134.950
28.8 28 .84 4 + 39 39.4 .44 4 1200
161. 16 1.65 65 134.950
= =
0 0
Double Line to Earth Fault
− − − ∠ − ∠ ∠− ∠ − ∠ ∠ ∠ ∠ ∠ ∠ =
= = = = =
− − −
2.5 + 9.0
6.5
2.5 + 9.0 2400
337. 33 7.88 88
1 1 1
1
1
2.5 9.0 6.5
2
2
× 24 24.2 .242 424 4 × 103
=
6.5 1200
× 24. 24.24 2424 24
6.5 2400
× 24. 24.24 2424 24
0
105.610
2.5 + 9.0 1200
337. 33 7.88 88 105.610
=
1 1 1
1
1
20.6 20.6 20.6
2
2
=
20.6 + 20.6 0.6 + 20.6 0.6 × 26 2640 40
=
2 0.6 + 20 20.6 20.6 .6 2400 + 20 20.6 .6 1200
× 26 2640 40
=
0
=
2 0.6 + 20 20.6 20.6 .6 1200 + 20.6 0.6 2400
× 26 2640 40
=
0
Line to Line Fault
163. 16 3.15 15
− − ∠ − ∠ ∠ − ∠ ∠ ∠ ∠ ∠ =
=
7
1 1 1
1
1
0 7 7
2
2
=
0+7
× 24. 24.24 2424 24
=
0 + 7 2400
7 1200
× 24. 24.24 2424 24
=
293. 29 3.92 92
=
0 + 7 1200
7 2400
× 24 24.24 .2424 24
=
293. 29 3.92 92 900
=
1 1 1
=
1
0
1
∠− ∠ 900
0 24.52 24.50
2
2
=
0 + 24.52 + 24.50 × 26 2640 40
=
129. 12 9.41 41
=
0 + 24.52 2400 + 24 24.5 .50 0 1200 × 26 2640 40
=
64.7 64 .71 1
=
0 + 24.52 1200 + 24 24.5 .50 0 2400
=
64.7 64 .71 1 179.960
× 26 2640 40
∠− ∠ 179.960
Results
Single Line to Earth Fault Fault Currents
Fault Voltages
Practical Value
Theoretical Value
Ia
283.64 A
839.31 A
Ib
0
0
Ic
0
0
Va
0
0
Fault Voltages
Fault Voltages
161.65 134.95 kV
0
159.39 134.77 kV
Practical Value
Theoretical Value
0
0
∠ ∠
∠ ∠
0
0
0
1.155 -105.23 kA
337.88 105.61 A
0
1.155 105.23 kA
Va
163.15 kV
162.2 kV
Vb
0
0
Vc
0
0
Practical Value
Theoretical Value
Ia
0
0
Ib
293.92 -90 A
Ic
293.92 90 A
1.12 90 kA
Va
129.41 kV
128.88 kV
Vb
64.71 -179.96 kV
Vc
64.71 179.96 kV
Ib
337.88 -105.61 A
Ic
Line to Line Fault Fault Currents
159.39 -134.77 kV
Vc
Ia
∠ ∠
0
161.65 -134.95 kV
Double Line to Earth Fault Fault Currents
∠ ∠
Vb
∠ ∠
∠ ∠
0
0
0
0
∠ ∠ ∠ ∠
0
1.12 -90 kA 0
0
64.44 -180 kV
0
0
64.44 180 kV
0
Discussion
The Importance of a Fault Study
A fault study emulates power system behavior during a fault by using a mock design of the power system. Fault calculations that result from a fault study are vital when deciding on protective gear. A fault study is important for the following functions; Designing power systems and their protective gear Enhancing existing power systems Simplified comprehension of system faults Deciding on backup protection Achieve efficient discrimination within the power system; fault levels at different locations
in the power system must be known.
Assumptions made in fault study and their validity
The following assumptions are made in this t his fault analysis;
All sources are balanced, and equal in magnitude and phase
Sources are represented by the Thevenin’s equivalent voltage prior to the fault at the faulty point
Large systems may be represented by infinite bus-bars
Transformers are on nominal tap position
Resistances are negligible compared to reactances
Transformer lines are assumed fully transposed and all there phase have same impedances
Load currents are negligible compared to the fault currents
Line charging currents can be completely neglected
According to the assumption all sources are balanced, and equal in magnitude and phase. Therefore prior to the fault, they consist only of positive sequence voltage components. This is in fact the equivalent Thevenin’s voltage at the point of fault before the occurrence of the fault. The stability of a large system is not affected by b y a single fault at a network. Therefore large systems can be represented by infinite bus-bars. The resistance of a transmission line is typically equal to one third of its reactance. Therefore resistance can be neglected.
Analogue methods of studying the fault flow in a system
There are two methods of analyzing a fault flow in a system;
Symmetrical components method This method can be used only for calculations of asymmetrical faults. The theory behind this
method is to analyze the unbalanced fault with symmetrical components of positive, negative and zero sequences.
Bus impedance method This method can be used to analyze both symmetrical and asymmetrical faults. The bus-bar
impedances of the system are used for calculations.
DC network analyzers used for analogue methods of fault studies
A DC network analyzer is a simulating device which can model all three sequence components separately. Impedances of the actual system are relatively small in magnitude; therefore by employing a multiplication scale factor, we have relatively large impedances which can be modeled easily. A DC power source is used as the voltage source and it represents the generators in the large system. The value of the DC source is selected by dividing the actual values by a suitable voltage scale factor. After connecting the three sequence circuits, this module can be used to take measurements for the calculation of any kind of faults. Actual fault values can be obtained by multiplying the relevant voltage or impedance by the above used scale factors. If it is required to find the Line-Ground fault level at a point in large system, all three sequence circuit must be connected in series. Similarly, if it is required to find out Line-Line fault, the positive and negative sequence circuits must be connected in parallel. This analyzer is also used to analyze symmetrical faults such as Line-Line-Line faults or Line-Line-Line-Ground faults. Fault values of large systems are very high and cannot be measured by conventional measuring instruments. The main advantage of using a DC analyzer is, it makes it possible to calculate faults of large magnitude by scaling them down into measurable quantities.
Importance of using sequence components co mponents
In the sequence components method, any kind of network either unbalanced or balanced, is reduced to three balanced symmetrical components. Analysis of a balanced system is fairly easy when compared with unbalanced systems. Fault values of a large system are very high, and are difficult to measure and dangerous to deal with. Large circuits can be represented by reducing them down to measurable values through symmetrical components. This is one of main advantage of a using the symmetrical components. Any kind of unbalanced system can be represented by symmetrical components of the positive, negative and zero sequences. sequences. Any element of the power system can be represented through a combination of impedances and voltage sources when symmetrical analyzing methods are employed. Calculations and the analysis of large scale systems become very easy due to reduction in complexity.
Relationship between the sequence impedance of generator, transformer and transmission lines
Generator The generator has different impedance values for positive sequence, negative sequence and zero sequence. sequence. This is because the impedance of rotating machines to the currents of the three sequences will generally be different for each sequence. The generator has a specific direction of rotation and the sequence considered may either have the same direction or the opposite direction. Thus the rotational EMF generated for the positive sequence and the negative sequence would also be different. Transmission Lines Transmission lines have same impedance values for both positive and negative sequences. sequences. The zero sequence impedance is different than positive or negative sequence values and zero sequence paths are involved with earth loop or earth return paths. Transformer The positive sequence, negative sequence & zero sequence impedances of transformer are equal regardless of transformer type as it does not have an inherent direction. However the zero sequence impedance of the unit may differ slightly from positive & negative sequence impedances as zero sequence paths across the windings of a transformer depend on the winding connection and even grounding impedance.
