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 0.614Sequence voltages and sequence currents for each type of fault;
Voltage (V) Current (mA)
1. Single line-to-earth fault
Positive Sequence 39.44 3.9
Negative Sequence -10.60 3.9
Zero Sequence -28.84 3.9
2. Double line-to-earth fault
Positive Sequence 20.60 9.0 Negative Sequence 20.60 -6.5 Zero Sequence 20.60 -2.5 3. Line-to-line fault Positive Sequence 24.52 7.0 Negative Sequence 24.50 -7.0
Theoretical Calculations
Single Line to Earth Fault
Assumptions; Fault impedance is zero
Load currents are negligible compared to fault current ∴ 𝑉𝑎 = 0 𝐼𝑏 , 𝐼𝑐 = 𝟎 𝑉𝑎 = 𝑉𝑎0 + 𝑉𝑎1 + 𝑉𝑎2 = 0 ... (1) 𝐼𝑎0 𝐼𝑎1 𝐼𝑎2 = 13 11 𝛼1 𝛼12 1 𝛼2 𝛼 𝐼𝑎 𝐼𝑏 = 0 𝐼𝑐 = 0 ⟹ 𝐼𝑎0 = 𝐼𝑎1 = 𝐼𝑎2 = 𝐼𝑎 3 𝑉𝑎0 𝑉𝑎1 𝑉𝑎2 = 0 𝐸𝑓 0 − 𝑍0 0 0 0 𝑍1 0 0 0 𝑍2 𝐼𝑎0 = 𝐼𝑎 3 𝐼𝑎1 =𝐼𝑎 3 𝐼𝑎2 =𝐼𝑎 3 𝐼𝑎 = 𝐼𝑓 = 3 𝐸𝑓 𝑍1 + 𝑍2 + 𝑍3 By observation data; 𝑍0 = 0.614 𝑝𝑢 𝑍1 = 0.240 𝑝𝑢 𝑍2 = 0.229 𝑝𝑢 𝐸𝑓 = 1 𝑝𝑢 ... (2) ... (3)
Fault Current;
Substituting values to equation (3),
𝐼𝑓,𝑝𝑢 = 3 × 1 0.614 + 0.240 + 0.229 𝑝𝑢 = 2.77 𝑝𝑢 𝐼𝑓 = 𝐼𝑓,𝑝𝑢 × 𝐼𝑏𝑎𝑠𝑒 𝑰𝒇 = 2.77 × 40 𝑀𝑉𝐴 132 𝑘𝑉 = 𝟖𝟑𝟗. 𝟑𝟏 𝑨 ∴ 𝐼𝑎0 = 𝐼𝑎1 = 𝐼𝑎2 = 839.31 3 𝐴 = 279.77 𝐴 Fault Voltages; 𝑍𝑏𝑎𝑠𝑒 = 132 𝑘𝑉 2 40 𝑀𝑉𝐴 = 435.6 Ω 𝑉𝑎0 = −𝑍0 𝐼𝑎0 = − 0.614 × 435.6 × 279.77 𝑉 = −74.83 𝑘𝑉 𝑉𝑎1 = 𝐸𝑓 − 𝑍1 𝐼𝑎1 = 1 × 132 × 103 − 0.24 × 435.6 × 279.77 𝑉 = 102.75 𝑘𝑉 𝑉𝑎2 = − 𝑍2 𝐼𝑎2 = −0.229 × 435.6 × 279.77 𝑉 = −27.91 𝑘𝑉 Phase Voltages; 𝑉𝑎 𝑉𝑏 𝑉𝑐 = 1 1 1 1 𝛼2 𝛼 1 𝛼 𝛼2 𝑉𝑎0 𝑉𝑎1 𝑉𝑎2 From equation (1), 𝑽𝒂 = 𝑉𝑎0 + 𝑉𝑎1 + 𝑉𝑎2 = 𝟎 𝑉𝑏 = 𝑉𝑎0 + 𝛼2 𝑉𝑎1 + 𝛼 𝑉𝑎2 𝑉𝑏 = −74.83 + 𝛼2 × 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 + 𝛼2 𝑉 𝑎2 𝑉𝑐 = −74.83 + 𝛼 × 102.75 + 𝛼2 × −27.91 𝑘𝑉 𝑉𝑐 = −74.83 + 102.75∠1200 − 27.91∠2400 𝑘𝑉 𝑽𝒄 = 𝟏𝟓𝟗. 𝟑𝟗∠𝟏𝟑𝟒. 𝟕𝟕𝟎 𝒌𝑽
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 0.24 + 0.229 × 0.6140.229 + 0.614 𝑝𝑢 = 2.46 𝑝𝑢 𝐼𝑎1 = 2.46 × 0.303 𝑘𝐴 = 0.745 𝑘𝐴 𝐼𝑎2 = − 𝐸𝑓 − 𝑍1 𝐼𝑎1 𝑍2 × 0.303 𝑘𝐴 = − 1 − 0.24 × 2.46 0.229 × 0.303 𝑘𝐴 = −0.542 𝑘𝐴 𝐼𝑎0 = − 𝐸𝑓 − 𝑍1 𝐼𝑎1 𝑍0 × 0.303 𝑘𝐴 = − 1 − 0.24 × 2.46 0.614 × 0.303 𝑘𝐴 = −0.202 𝑘𝐴 𝑉𝑎,𝑝𝑢 = 3 × 𝑉𝑎1 = 3 × 𝐸𝑓 − 𝑍1 𝐼𝑎1 = 3 × 1 − 0.24 × 2.46 𝑝𝑢 = 1.2288 𝑝𝑢 𝑉𝑎 = 1.2288 × 132 𝑘𝑉 = 𝟏𝟔𝟐. 𝟐 𝒌𝑽
Phase Currents; 𝐼𝑎 𝐼𝑏 𝐼𝑐 = 1 1 1 1 𝛼2 𝛼 1 𝛼 𝛼2 𝐼𝑎0 𝐼𝑎1 𝐼𝑎2 𝐼𝑏 = 𝐼𝑎0 + 𝛼2 𝐼 𝑎1 + 𝛼 𝐼𝑎2 𝐼𝑏 = −0.202 + 𝛼2× 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 + 𝛼2 𝐼 𝑎2 𝐼𝑐 = −0.202 + 𝛼 × 0.745 + 𝛼2× −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 = 1 0.24 + 0.229 𝑝𝑢 = 2.132 𝑝𝑢 𝐼𝑎1 = 2.132 × 0.303 𝑘𝐴 = 0.646 𝑘𝐴 𝐼𝑎2 = − 𝐼𝑎1 = −0.646 𝑘𝐴 𝐼𝑎0 = 0 𝐴 𝑉𝑎1 = 𝑉𝑎2 = 𝑍2 × 𝐼𝑎2 𝑉𝑎1 = 𝑉𝑎2 = 0.229 × 435.6 × 0.646 𝑘𝑉 = 64.44 𝑘𝑉
𝑉𝑎 𝑉𝑏 𝑉𝑐 = 1 1 1 1 𝛼2 𝛼 1 𝛼 𝛼2 𝑉𝑎0 𝑉𝑎1 𝑉𝑎2 𝑉𝑎 = 𝑉𝑎0+ 𝑉𝑎1+ 𝑉𝑎2 𝑉𝑎 = 0 + 64.44 + 64.44 𝑘𝑉 𝑽𝒂 = 𝟏𝟐𝟖. 𝟖𝟖 𝒌𝑽 𝑉𝑏 = 𝑉𝑎0+ 𝛼2 𝑉 𝑎1+ 𝛼 𝑉𝑎2 𝑉𝑏 = 0 + 64.44∠2400+ 64.44∠1200 𝑘𝑉 𝑽𝒃 = 𝑽𝒄 = 𝟔𝟒. 𝟒𝟒∠𝟏𝟖𝟎𝟎 𝒌𝑽 𝐼𝑎 𝐼𝑏 𝐼𝑐 = 11 𝛼12 1𝛼 1 𝛼 𝛼2 𝐼𝑎0 𝐼𝑎1 𝐼𝑎2 𝐼𝑏 = 𝐼𝑎0 + 𝛼2 𝐼 𝑎1 + 𝛼 𝐼𝑎2 𝐼𝑏 = 0 + 0.646∠2400− 0.646∠1200 𝑘𝐴 𝑰𝒃 = 𝟏. 𝟏𝟐∠−𝟗𝟎𝟎 𝒌𝑨 𝐼𝑐 = 𝐼𝑎0 + 𝛼 𝐼𝑎1 + 𝛼2 𝐼𝑎2 𝐼𝑐 = 0 + 0.646∠1200 − 0.646∠2400 𝑘𝐴 𝑰𝒄 = 𝟏. 𝟏𝟐∠𝟗𝟎𝟎 𝒌𝑨
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.
𝐴𝑐𝑡𝑢𝑎𝑙 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 = 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 × 132 𝑘𝑉 50 𝑉 132 𝑘𝑉 2 40 𝑀𝑉𝐴 4000 𝐴𝑐𝑡𝑢𝑎𝑙 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 = 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 × 24.2424 × 103 𝐴𝑐𝑡𝑢𝑎𝑙 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 = 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 × 132 𝑘𝑉 50 𝑉 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 𝐼𝑎 = 3.9 × 3 × 10−3× 24.2424 × 103 𝐴 = 283.64 𝐴 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 = 11 𝛼12 1𝛼 1 𝛼 𝛼2 −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 = 11 𝛼12 1𝛼 1 𝛼 𝛼2 −2.5 𝑚𝐴 9.0 𝑚𝐴 −6.5 𝑚𝐴 𝐼𝑎 = −2.5 + 9.0 − 6.5 × 24.2424 × 103 = 0 𝐼𝑏 = −2.5 + 9.0∠2400 − 6.5∠1200 × 24.2424 𝐴 𝐼𝑏 = 337.88∠−105.610 𝐴 𝐼𝑐 = −2.5 + 9.0∠1200− 6.5∠2400 × 24.2424 𝐴 𝐼𝑐 = 337.88∠105.610 𝐴 𝑉𝑎 𝑉𝑏 𝑉𝑐 = 11 𝛼12 𝛼1 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 = 0Line to Line Fault
𝐼𝑎 𝐼𝑏 𝐼𝑐 = 11 𝛼12 1𝛼 1 𝛼 𝛼2 0 𝑚𝐴 7 𝑚𝐴 −7 𝑚𝐴 𝐼𝑎 = 0 + 7 − 7 × 24.2424 𝐴 = 0 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 1 𝛼2 𝛼 1 𝛼 𝛼2 0 𝑉 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 Practical Value Theoretical Value
Fault Currents Ia 283.64 A 839.31 A Ib 0 0 Ic 0 0 Fault Voltages 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
Double Line to Earth Fault Practical Value Theoretical Value
Fault Currents 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 Fault Voltages Va 163.15 kV 162.2 kV Vb 0 0 Vc 0 0
Line to Line Fault Practical Value Theoretical Value
Fault Currents Ia 0 0 Ib 293.92∠-900 A 1.12∠-900 kA Ic 293.92∠900 A 1.12∠900 kA Fault Voltages 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
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.
