Fault Study

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 currents for each type 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 ∴ 𝑉𝑎

=

𝐼𝑏 , 𝐼𝑐

𝑉𝑎

=

𝐼𝑎0 𝐼𝑎1 𝐼𝑎2 ⟹

0 =

𝟎

𝑉𝑎0 + 𝑉𝑎1 + 𝑉𝑎2 1 1 1

1

=

3

1 𝛼 𝛼2

=

𝑍0 0 𝐸𝑓 − 0 0 0

=

... (1)

𝐼𝑎 𝐼𝑏 = 0 𝐼𝑐 = 0

1 𝛼2 𝛼

𝐼𝑎0 = 𝐼𝑎1 = 𝐼𝑎2 =

𝑉𝑎0 𝑉𝑎1 𝑉𝑎2

0

0 𝑍1 0

𝐼𝑎 3

0 0 𝑍2

... (2)

𝐼𝑎1 𝐼𝑎2

𝐼𝑎

=

𝐼𝑓

=

By observation data; 𝑍0 = 0.614 𝑝𝑢 𝑍1 = 0.240 𝑝𝑢 𝑍2 = 0.229 𝑝𝑢 𝐸𝑓 = 1 𝑝𝑢

3 𝐸𝑓 𝑍1 + 𝑍2 + 𝑍3

𝐼𝑎 3 𝐼𝑎 = 3 𝐼𝑎 = 3

𝐼𝑎0 =

... (3)

Fault Current; Substituting values to equation (3), 𝐼𝑓,𝑝𝑢

=

3 ×1 𝑝𝑢 0.614 + 0.240 + 0.229

𝐼𝑓

=

𝐼𝑓,𝑝𝑢 × 𝐼𝑏𝑎𝑠𝑒

𝑰𝒇

=

2.77 ×

∴ 𝐼𝑎0

=

=

40 𝑀𝑉𝐴 132 𝑘𝑉

=

𝟖𝟑𝟗. 𝟑𝟏 𝑨

𝐼𝑎1

𝐼𝑎2

=

=

2.77 𝑝𝑢

839.31 𝐴 3

=

279.77 𝐴

Fault Voltages; 𝑍𝑏𝑎𝑠𝑒 𝑉𝑎0

=

−𝑍0 𝐼𝑎0

=

𝑉𝑎1

=

𝐸𝑓 − 𝑍1 𝐼𝑎1

𝑉𝑎2

=

− 𝑍2 𝐼𝑎2

=

132 𝑘𝑉 2 40 𝑀𝑉𝐴

=

=

435.6 Ω

− 0.614 × 435.6 × 279.77 𝑉

=

−74.83 𝑘𝑉

= 1 × 132 × 103 − 0.24 × 435.6 × 279.77 𝑉 = 102.75 𝑘𝑉 −0.229 × 435.6 × 279.77 𝑉

=

−27.91 𝑘𝑉

Phase Voltages; 𝑉𝑎 𝑉𝑏 𝑉𝑐

1 1 1 𝛼2 1 𝛼

=

1 𝛼 𝛼2

𝑉𝑎0 𝑉𝑎1 𝑉𝑎2

From equation (1), 𝑽𝒂

=

𝑉𝑎0 + 𝑉𝑎1 + 𝑉𝑎2

=

𝑉𝑏

=

𝑉𝑎0 + 𝛼 2 𝑉𝑎1 + 𝛼 𝑉𝑎2

𝑉𝑏

=

−74.83 + 𝛼 2 × 102.75 + 𝛼 × −27.91 𝑘𝑉

𝑉𝑏

=

−74.83∠00 + 102.75∠2400 − 27.91∠1200 𝑘𝑉

𝑽𝒃

=

𝟏𝟓𝟗. 𝟑𝟗∠ − 𝟏𝟑𝟒. 𝟕𝟕𝟎 𝒌𝑽

𝑉𝑐

=

𝑉𝑎0 + 𝛼 𝑉𝑎1 + 𝛼 2 𝑉𝑎2

𝑉𝑐

=

−74.83 + 𝛼 × 102.75 + 𝛼 2 × −27.91 𝑘𝑉

𝑉𝑐

=

−74.83 + 102.75∠1200 − 27.91∠2400 𝑘𝑉

𝑽𝒄

=

𝟏𝟓𝟗. 𝟑𝟗∠𝟏𝟑𝟒. 𝟕𝟕𝟎 𝒌𝑽

𝟎

Double Line to Earth Fault

Assumptions; Fault impedance is zero Load currents are negligible compared to fault current 𝐼𝑎

=

0

𝑉𝑏

=

𝑉𝑐

=

0

By observation data; 𝑍0 = 0.614 𝑝𝑢 𝑍1 = 0.240 𝑝𝑢 𝑍2 = 0.229 𝑝𝑢 𝐸𝑓 = 1 𝑝𝑢

𝐼𝑎1,𝑝𝑢

𝐸𝑓 𝑍1 + 𝑍2 ∥ 𝑍0

=

=

1 𝑝𝑢 0.229 × 0.614 0.24 + 0.229 + 0.614

=

0.745 𝑘𝐴

𝐼𝑎1

=

2.46 × 0.303 𝑘𝐴

𝐼𝑎2

=

− 𝐸𝑓 − 𝑍1 𝐼𝑎1 × 0.303 𝑘𝐴 = 𝑍2

𝐼𝑎0

=

− 𝐸𝑓 − 𝑍1 𝐼𝑎1 × 0.303 𝑘𝐴 = 𝑍0

𝑉𝑎,𝑝𝑢

=

3 × 𝑉𝑎1 =

3 × 𝐸𝑓 − 𝑍1 𝐼𝑎1

𝑉𝑎

=

1.2288 × 132 𝑘𝑉

=

=

2.46 𝑝𝑢

− 1 − 0.24 × 2.46 × 0.303 𝑘𝐴 = −0.542 𝑘𝐴 0.229 − 1 − 0.24 × 2.46 × 0.303 𝑘𝐴 = −0.202 𝑘𝐴 0.614

=

𝟏𝟔𝟐. 𝟐 𝒌𝑽

3 × 1 − 0.24 × 2.46 𝑝𝑢 =

1.2288 𝑝𝑢

Phase Currents; 𝐼𝑎 𝐼𝑏 𝐼𝑐

=

1 1 1

1 𝛼2 𝛼

1 𝛼 𝛼2

𝐼𝑎0 𝐼𝑎1 𝐼𝑎2

𝐼𝑏

=

𝐼𝑎0 + 𝛼 2 𝐼𝑎1 + 𝛼 𝐼𝑎2

𝐼𝑏

=

−0.202 + 𝛼 2 × 0.745 + 𝛼 × −0.542

𝐼𝑏

=

−0.202∠00 + 0.745∠2400 − 0.542∠1200 𝑘𝐴

𝑰𝒃

=

𝟏. 𝟏𝟓𝟓∠ − 𝟏𝟎𝟓. 𝟐𝟑𝟎 𝒌𝑨

𝐼𝑐

=

𝐼𝑎0 + 𝛼 𝐼𝑎1 + 𝛼 2 𝐼𝑎2

𝐼𝑐

=

−0.202 + 𝛼 × 0.745 + 𝛼 2 × −0.542

𝐼𝑐

=

−0.202∠00 + 0.745∠1200 − 0.542∠2400 𝑘𝐴

𝑰𝒄

=

𝟏. 𝟏𝟓𝟓∠𝟏𝟎𝟓. 𝟐𝟑𝟎 𝒌𝑨

Line to Line Fault

Assumptions; Fault impedance is zero Load currents are negligible compared to fault current 𝑰𝒂

=

𝟎

𝑉𝑏

=

𝑉𝑐

𝐼𝑏

=

− 𝐼𝑐

By observation data; 𝑍0 = 0.614 𝑝𝑢 𝑍1 = 0.240 𝑝𝑢 𝑍2 = 0.229 𝑝𝑢 𝐸𝑓 = 1 𝑝𝑢

𝐼𝑎1,𝑝𝑢

=

𝐸𝑓 𝑍1 + 𝑍2

𝐼𝑎1

=

2.132 × 0.303 𝑘𝐴

𝐼𝑎2

=

− 𝐼𝑎1

𝐼𝑎0

=

0𝐴

=

=

1 𝑝𝑢 0.24 + 0.229 =

=

2.132 𝑝𝑢

0.646 𝑘𝐴

−0.646 𝑘𝐴

𝑉𝑎1

=

𝑉𝑎2

=

𝑍2 × 𝐼𝑎2

𝑉𝑎1

=

𝑉𝑎2

=

0.229 × 435.6 × 0.646 𝑘𝑉

=

64.44 𝑘𝑉

𝑉𝑎 𝑉𝑏 𝑉𝑐

=

1 1 1 𝛼2 1 𝛼

1 𝛼 𝛼2

𝑉𝑎

=

𝑉𝑎0 + 𝑉𝑎1 + 𝑉𝑎2

𝑉𝑎

=

0 + 64.44 + 64.44 𝑘𝑉

𝑽𝒂

=

𝟏𝟐𝟖. 𝟖𝟖 𝒌𝑽

𝑉𝑏

=

𝑉𝑎0 + 𝛼 2 𝑉𝑎1 + 𝛼 𝑉𝑎2

𝑉𝑏

=

0 + 64.44∠2400 + 64.44∠1200 𝑘𝑉

𝑽𝒃

=

𝑽𝒄

=

𝑉𝑎0 𝑉𝑎1 𝑉𝑎2

𝟔𝟒. 𝟒𝟒∠𝟏𝟖𝟎𝟎 𝒌𝑽

𝐼𝑎 𝐼𝑏 𝐼𝑐

=

1 1 1

1 𝛼2 𝛼

1 𝛼 𝛼2

𝐼𝑏

=

𝐼𝑎0 + 𝛼 2 𝐼𝑎1 + 𝛼 𝐼𝑎2

𝐼𝑏

=

0 + 0.646∠2400 − 0.646∠1200 𝑘𝐴

𝑰𝒃

=

𝟏. 𝟏𝟐∠−𝟗𝟎𝟎 𝒌𝑨

𝐼𝑐

=

𝐼𝑎0 + 𝛼 𝐼𝑎1 + 𝛼 2 𝐼𝑎2

𝐼𝑐

=

0 + 0.646∠1200 − 0.646∠2400 𝑘𝐴

𝑰𝒄

=

𝟏. 𝟏𝟐∠𝟗𝟎𝟎 𝒌𝑨

𝐼𝑎0 𝐼𝑎1 𝐼𝑎2

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 𝑘𝑉 50 𝑉 132 𝑘𝑉 2 4000 40 𝑀𝑉𝐴

𝐴𝑐𝑡𝑢𝑎𝑙 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 = 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 ×

𝐴𝑐𝑡𝑢𝑎𝑙 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 = 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 × 24.2424 × 103

132 𝑘𝑉 50 𝑉

𝐴𝑐𝑡𝑢𝑎𝑙 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 = 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 ×

𝐴𝑐𝑡𝑢𝑎𝑙 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 = 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 × 2640

Single Line to Earth Fault 𝐼𝑎 𝐼𝑏 𝐼𝑐

=

1 1 1

1 𝛼2 𝛼

1 𝛼 𝛼2

3.9 𝑚𝐴 3.9 𝑚𝐴 3.9 𝑚𝐴

3.9 × 3 × 10−3 × 24.2424 × 103 𝐴

𝐼𝑎

=

𝐼𝑏

=

3.9 + 3.9∠2400 + 3.9∠1200 𝑚𝐴 × 24.2424 × 103

=

0

𝐼𝑐

=

3.9 + 3.9∠1200 + 3.9∠2400 𝑚𝐴 × 24.2424 × 103

=

0

𝑉𝑎 𝑉𝑏 𝑉𝑐

=

1 1 1 𝛼2 1 𝛼

1 𝛼 𝛼2

=

283.64 𝐴

−28.84 𝑉 39.44 𝑉 −10.6 𝑉

𝑉𝑎

=

−28.84 + 39.42 − 10.6 × 2640 𝑉

=

0

𝑉𝑏

=

−28.84 + 39.44∠2400 − 10.6∠1200 × 2640

𝑉𝑏

=

161.65∠−134.950 𝑘𝑉

𝑉𝑐

=

−28.84 + 39.44∠1200 − 10.6∠2400 × 2640

𝑉𝑐

=

161.65∠134.950 𝑘𝑉

Double Line to Earth Fault 𝐼𝑎 𝐼𝑏 𝐼𝑐

=

1 1 1 𝛼2 1 𝛼

1 𝛼 𝛼2

−2.5 𝑚𝐴 9.0 𝑚𝐴 −6.5 𝑚𝐴

𝐼𝑎

=

−2.5 + 9.0 − 6.5 × 24.2424 × 103

𝐼𝑏

=

−2.5 + 9.0∠2400 − 6.5∠1200 × 24.2424 𝐴

𝐼𝑏

=

𝐼𝑐

=

𝐼𝑐

=

=

0

337.88∠−105.610 𝐴 −2.5 + 9.0∠1200 − 6.5∠2400 × 24.2424 𝐴 337.88∠105.610 𝐴 𝑉𝑎 𝑉𝑏 𝑉𝑐

=

1 1 1 𝛼2 1 𝛼

1 𝛼 𝛼2

20.6 𝑉 20.6 𝑉 20.6 𝑉

𝑉𝑎

=

20.6 + 20.6 + 20.6 × 2640 𝑉

=

163.15 𝑘𝑉

𝑉𝑏

=

20.6 + 20.6∠2400 + 20.6∠1200 × 2640 𝑉

=

0

𝑉𝑐

=

20.6 + 20.6∠1200 + 20.6∠2400 × 2640 𝑉

=

0

Line to Line Fault 𝐼𝑎 𝐼𝑏 𝐼𝑐

=

1 1 1 𝛼2 1 𝛼

1 𝛼 𝛼2

𝐼𝑎

=

0 + 7 − 7 × 24.2424 𝐴

𝐼𝑏

=

0 + 7∠2400 − 7∠1200 × 24.2424 𝐴

=

293.92∠−900 𝐴

𝐼𝑐

=

0 + 7∠1200 − 7∠2400 × 24.2424 𝐴

=

293.92∠900 𝐴

𝑉𝑎 𝑉𝑏 𝑉𝑐

=

1 1 1 𝛼2 1 𝛼

=

0 𝑚𝐴 7 𝑚𝐴 −7 𝑚𝐴 0

1 𝛼 𝛼2

0𝑉 24.52 𝑉 24.50 𝑉

𝑉𝑎

=

0 + 24.52 + 24.50 × 2640 𝑉

=

129.41 𝑘𝑉

𝑉𝑏

=

0 + 24.52∠2400 + 24.50∠1200 × 2640 𝑉

=

64.71∠−179.960 𝑘𝑉

𝑉𝑐

=

0 + 24.52∠1200 + 24.50∠2400 × 2640 𝑉

=

64.71∠179.960 𝑘𝑉

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

Vb

161.65∠-134.950 kV

159.39∠-134.770 kV

Vc

161.65∠134.950 kV

159.39∠134.770 kV

Practical Value

Theoretical Value

Ia

0

0

Ib

337.88∠-105.610 A

1.155∠-105.230 kA

Ic

337.88∠105.610 A

1.155∠105.230 kA

Va

163.15 kV

162.2 kV

Vb

0

0

Vc

0

0

Practical Value

Theoretical Value

Ia

0

0

Ib

293.92∠-900 A

1.12∠-900 kA

Ic

293.92∠900 A

1.12∠900 kA

Va

129.41 kV

128.88 kV

Vb

64.71∠-179.960 kV

64.44∠-1800 kV

Vc

64.71∠179.960 kV

64.44∠1800 kV

Double Line to Earth Fault Fault Currents

Fault Voltages

Line to Line Fault Fault Currents

Fault Voltages

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 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 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 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. 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. 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. 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.