Fault Analysis Fault is any failure that interferes with the normal flow of current to the loads. When the insulation of the system breaks down or a conducting object comes in touch with a live point, a short circuit or a fault occurs. Sources of Fault 1. Lightning 2. wires blowing together in the wind 3. animals or plants coming in contact with the wires 4. salt spray or pollution on insulators 5. Synchronous Generators 6. Synchronous Motors and Condensers 7. Induction Machines 8. Electrical Utility System Flashover is a condition which the voltage difference Flashover between the phase and the ground is large enough to ionize the air around the insulators, creating a current path from the phase to the structure of the tower, which is grounded. Fault Types 1. Balanced faults (symmetrical faults): A Fault involving all the three phases. This is less than 5 % in occurrence. 2. Unbalanced faults (unsymmetrical faults): A fault involving only one or two phases The most common type of fault on a three phase system is the single line – to – ground (SLG (60 – 75%)), followed by the double line – to – ground faults (DLG (15 – 25%)), line – to – line (LL (5 – 15%)) faults, and balanced three phase faults Fault studies studies form an important part of power system analysis and the problem consist of determining bus voltage and line current during faults. The three phase fault information is used to select and set phase relays. Fault studies are used for proper choice of circuit breakers and protective relaying. A power system network comprises of synchronous generators, transformers, transmission lines and loads. During fault, load current can be neglected because voltage dip very low so that current drawn by the loads can be neglected in comparison to fault currents. The magnitude of the fault current depends on the internal impedance of the synchronous generator and the impedance of the intervening circuit. Purpose of Fault Studies 1. To adjust and set the protective devices so we can detect any fault and isolate the faulty portion of the system. 2. To protect the human being and the equipment during the abnormal operating conditions. 3. To determine the maximum current to insure devices can survive the fault.
Symmetrical Faults are faults on the power system which give rise to symmetrical currents. They occur when all the three conductors of a three phase line are brought together simultaneously in to a short circuit condition. This type of f ault occurs infrequently but leads to the most severe fault current flow against which the system must be protected. Symmetrical Currents are equal fault currents in the line with 120° displacement. Fault currents are commonly known in these term. Limitations: 1. Short circuit currents are limited by the impedance of the system up to the point of the fault.
2.
The short circuit current from the generating station will have a limited value by the impedance of the generator and transformer and the point of the fault. Thus, the knowledge of the impedances of various equipment and circuits in the line of the system is very important for the determination of short circuit currents. 3. Per phase analysis can be considered, since the condition in the other phases are similar. The reactance of generators, transformers, reactors etc. is usually expressed in percentage reactance to permit rapid short circuit calculations. Percentage Reactances is the percentage of the total phase voltage drop in the circuit when full load current is flowing.
% = 100
where: IFL – full load current V – phase voltage X – reactance in ohms per phase It can also be expressed in terms of KVA and KV:
% = 10 10 100
where: X – reactance in ohms If X is the only reactance element in the circuit, then the short circuit current is given by:
Hence,
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
= = 100 %
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Fault Analysis Percentage reactance of an equipment depends upon its kVA rating. Generally, the various equipment used in the power system have different kVA ratings. Therefore, it is necessary to find the percentage reactances of all the elements on a common kVA rating. This common kVA rating is known as base kVA. The value of this base kVA is quite unimportant and may be: 1. equal to that of the largest plant 2. equal to the total plant capacity 3. any arbitrary value The conversion can be effected by using the following relation:
% @ = % @
Short Circuit KVA is the product of the normal system voltage and short circuit current at the point of fault and expressed in KVA. Let: V = nominal phase voltage in volts IFL = full load current at base KVA %X = % reactance of the system on base KVA up to the fault point but
= 3/1000 = =100/% 100⁄%
Another important point is that the circuit breakers rated MVA breaking capacity is based on three phase fault MVA. In fact high precision is not n ecessary when calculating the three phase fault level because circuit breakers are manufactured in standard sizes. Generally for three phase fault calculation, the following assumptions are made: 1. The emfs of all generators are 1˂0° pu. This assumption simplify the problem and it means that the voltage is at its nominal value and the system is operating at no load at the time if fa ult. Since all emfs are equal and in phase, all the generators can be replaced by a single generator. 2. Charging capacitances of the transmission line are ignored. 3. Shunt elements in the transformer model are neglected. 4. The system resistances are neglected and only the inductive reactance of different elements are taken into account. Three Phase Short Circuit Calculations Case 1: Simple Circuits and Loads ignored. The Calculations for the three phase fault are easy because the circuit is completely symmetrical and calculations can be done for only one phase. Steps in Solving Symmetrical Fault Calculations 1. Draw a single line diagram of the complete network indicating the rating, voltage and percentage reactance of each element of the network.
2.
Choose a numerically convenient value of base kVA and convert all percentage reactances to this base value. 3. Corresponding to the single line diagram of the network, draw the reactance diagram showing one phase of the system and the neutral. Indicate the % reactances on the base kVA in the reactance diagram. The transformer in the system should be represented by a reactance in series. 4. Find the total % reactance of the network using series, parallel and Delta – Wye transformations keeping the identity of the fault point intact. Find the total reactance of the system as seen from the fault point (Using Thevenin’s Theorem) . 5. Find the full load current corresponding to the selected base kVA and the normal system voltage at the fault point. Let it be I. 6. Solve for short circuit KVA and current. Case 2: Simple Circuits and Loads not ignored. Generally, the fault currents are much larger than the load currents. Therefore, the load current can be neglected during fault calculations. There are some cases where considering the load current is an essential factor in fault calculations. Superposition technique is proposed for such cases to compute the fault current. For such case, it is necessary to compute the terminal voltage at fault location before fault takes place. This terminal voltage is known as the pre – fault voltage. Connecting the load to the system causes the current to flow in the network. Voltage drop due to system impedance cause the voltage magnitude at different buses to be deviated from 1.0 pu. The pre – fault voltage and the Z TH are used in calculating the fault current. The fault current represents two compone nts: 1. The Load Current 2. The Short Circuit Current Consider a load (synchronous Motor) connected to a synchronous generator through a transmission line. The circuit model of the network could be represented as shown
Normal Operation Condition:
= =
During fault condition in the system, the generator as well as the load (synchronous motor) will supply the faulted
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
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Fault Analysis terminals with power from the energy stored in their windings.
Location: 1. In series with each generator 2. In series with each feeder 3. In busbars Types: 1. Generator Reactor – when the reactors are connected in series with each generator.
Fault Condition at Load Terminal:
= = = + + +
Case 3: Large Circuits For large circuits, the bus impedance matrix is used in solving for the three phase short circuit current and same steps are use like in case 1 and case 2. Reactor Control of Short Circuit Currents With the fast expanding power system, the fault level (i.e. the power available to flow into a fault) is also rising. The circuit breakers connected in the power system must be capable of dealing with maximum possible short circuit currents that can occur at their points of connection. Generally, the reactance of the system under fault conditions is low and fault currents may rise to a dangerously high value. If no steps are taken to limit the value of these short circuit currents, not only will the duty required of circuit breakers be excessively heavy, but also damage to lines and other equipment will almost certainly occur. In order to limit the short circuit currents to a value which the circuit breakers can handle, additional reactances known as reactors are connected in series with the system at suitable points. A reactor is a coil of number of turns designed to have a large inductance as compared to its ohmic resistance. The forces on the turns of these reactors under short circuit conditions are considerable and, therefore, the windings must be solidly braced. It may be added that due to very small resistance of reactors, there is very little change in the efficiency of the system. Advantages: 1. Reactors limit the flow of short circuit current and thus protect the equipment from overheating as well as from failure due to destructive mechanical forces. 2. Troubles are localised or isolated at the point where they originate without communicating their disturbing effects to other parts of the power system. This increases the chances of continuity of supply. 3. They permit the installation of circuit breakers of lower rating.
Disadvantages: 1. There is a constant voltage drop and power
2.
loss in the reactors even during normal operation. 2. If a bus bar or feeder fault occurs close to the bus bar, the voltage at the bus bar will be reduced to a low value, thereby causing the generators to fall out of step. 3. If a fault occurs on an y feeder, the continuity of supply to other is likely to be affected. Feeder Reactor – the reactors are connected in series with each feeder.
Advantages: 1. If a fault occurs on any feeder, the voltage drop in its reactors will not affect the busbar voltage, thus there is little tendency for the generator to lose synchronism. 2. The fault on the other feeder will not affect the other feeders and consequently the effects of faults are localized. Disadvantages: 1.
2.
There is a constant power loss and voltage drop in the reactors even during normal operation. If a short circuit occurs at the busbars, no protection is provided to the generators. However, this is of little importance because such faults are rare and modern generators
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
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Fault Analysis
3.
have considerable leakage reactance to enable them to withstand short circuit across their terminals. 3. If the number of generators is increased, the size of feeder reactors will have to be increased to keep the short circuit currents within the ratings of the feeder circuit breakers. Busbar Reactors – the reactors are located in the busbars a. Ring System In this system, busbar is divided into sections and these sections are connected through reactors as shown.
requiring changes in the existing reactors. Examples: 1. The Figure shows the single line diagram of a 3 phase system. The percentage reactance of each alternator is based on its own capacity. Find the short circuit current that will flow into a complete 3 phase short circuit at F.
2.
b.
Advantages: 1. Under normal operating conditions, each generator will supply its own section of the load and very little power will be fed by other generators. 2. Low power loss and voltage drop in the reactors 3. If a fault occurs on any feeder, only one generator mainly feeds the fault while the current fed from other generators is small due to the presence of the reactors. Tie Bar System
Advantages: 1. There are effectively two reactors in series between sections so that the reactors must have approximately half the reactance of those used in comparable ring system. 2. Additional generators may be connected to the system without
3.
4.
5.
A 3 phase, 20 MVA, 10 k V alternator has internal reactance of 5% and negligible resistance. Find the external reactance per phase to be connected in series with the alternator so that steady current on short circuit does not exceed 8 times the full load current. A 3 phase transmission line operating at 10 kV and having a resistance of 1Ω and reactance of 4 Ω is connecte d to the generating station busbars through 5 MVA step – up transformer having a reactance of 5%. The busbars are supplied by a 10 MVA alternator having 10% reactance. Calculate the short circuit kVA fed to symmetrical fault between phases if it occurs a. at the load end of transmission line b. at the high voltage terminals of the transformer The plant capacity of a 3 phase generating station consists of two 10,000 kVA generators of reactance 12% each and one 5000 kVA generator of reactance 18%. The generators are connected to the station busbars from which load is taken through three 5000 kVA step – up transformers each having a reactance of 5%. Determine the maximum fault MVA which the circuit breakers on low voltage side and high voltage side may have to deal with. The section busbars A and B are linked by a busbar reactor rated at 5000 kVA with 10% reactance. On busbar A, there are two generators each of 10,000 kVA with 10% reactance and on B two generators each of 8000 kVA with 12% reactance. Find the steady MVA fed into a dead short circuit between all phases on B with busbar reactor in the circuit.
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
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Fault Analysis
6.
A small generating station has two alternators of 3000 kVA and 4500 kVA and percentage reactances of 7% and 8% respectively. The circuit breakers have a rupturing capacity of 150 MVA. It is desired to extend the system by a supply from the grid via a transformer of 7500 kVA and 7·5% reactance. Find the reactance of the reactor connected in the busbar section to prevent the circuit breakers being overloaded, if a symmetrical short circuit occurs on an outgoing feeder connected to it. Assume the bus voltage = 3300 V. -
regarded as being composed of three separate sets of balanced vectors namely: 1. A balanced set of 3 phase vectors having the same phase sequence as the original unbalanced system of vectors (positive sequence). These vectors are known as “Positive Sequence System”. 2. A balanced set of 3 phase vectors having phase sequence which is opposite of the original unbalanced system of vectors (negative sequence). This balanced system of 3 phase vectors are known as “Negative Sequence System”. 3. A set of three (3) single phase vectors which are equal in magnitude and having zero displacement and which have exactly the same 3 phase position with respect to any given reference axis. This system of single phase vector is known as “Zero Sequence System”. hence in mathematical form:
== ++ ++ == == ++ ++ = + + = = + + = = = =
Zero Sequence Component
7.
The estimated short circuit MVA at the busbars of a generating station A is 1500 MVA and of another station B is 1200 MVA. The generated voltage at each station is 33 kV. If these stations are interconnected through a line having a reactance of 1Ω and negligible resistance, calculate the possible short circuit MVA at both stations.
Operator “A” is one which when multiplied to a vector, rotates through a 120° in the counter clockwise direction. Let:
1∠120° == =1∠360° 1∠240°= 1 == ==
Positive Sequence Component The great majority of faults on the power system are of unsymmetrical in nature; the most common type being a short circuit from one line to ground. When such a fault occurs, it gives rise to unsymmetrical currents i.e. the magnitude of fault currents in the three lines are different having unequal phase displacement. The calculation procedure known as method of symmetrical components is used to determine the currents and voltages on the occurrence of an unsymmetrical fault. Symmetrical Component Method Developed by Dr. C.L. Fortesque, an American scientist showed that any unbalanced system of three phase currents (or voltages) may be
Negative Sequence Component
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
== == 5
Fault Analysis Expressing Symmetrical Components of I B and I C in terms of IA symmetrical components:
= = + + ++ = + + + + = 3+ +1+ + +1+ 1+ + = 0 1+ + = 0 = / + + = + + == + + + + + 3 + 1+ + + =+1+ + = / + +
Adding them up:
But Therefore,
also
Also, Ia0 = Ib0 = Ic0 Multiply I a by 1, Multiply I b by A and Multiply I c by A2
Adding them up,
In the absence of path through the neutral of a 3 phase system, the neutral current is zero and the line currents contain no zero sequence components. A delta connected load provides no path to the neutral and the line currents flowing to delta connected load can contain no zero sequence components. 5. In a 3 phase unbalanced system, the magnitude of negative sequence components cannot exceed that of the positive sequence components. If the negative sequence components were the greater, the phase sequence of the resultant system would be reversed. 6. The current of a single phase load drawn from a 3 phase system comprises equal positive, negative and zero sequence components.
Sequence voltages and the voltage at the fault point for the various fault types
Also, Ib1 = A2Ia1 and I c1 = AIa1 Multiply I a by 1, Multiply I b by A2 and Multiply I c by A
= + + == + + + + Adding them up, + 3 + 1+ + + =+1+ + + = / + Also, I = AI and I = A I b2
a2
c2
2
a2
In matrix notation:
[] = [11 1 1 ][] 1
Some Facts about Sequence Currents 1. A balanced 3 phase system consists of positive sequence components only; the negative and zero sequence components being zero. 2. The presence of negative or zero sequence currents in a 3 phase system introduces unsymmetry and is indicative of an abnormal condition of the circuit in which these components are found. 3. The vector sum of the positive and negative sequence currents of an unbalanced 3 phase system is zero. The resultant solely consists of three zero sequence currents. 4.
+ +
In a 3 phase, 4 wire unbalanced system, the magnitude of zero sequence components is one third of the current in the neutral wire
= 13
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
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Fault Analysis Sequence currents and the fault current for the various fault types
4.
Find the phase voltages. The zero and positive sequence components of red phase are:
=0. =5 0. 866 2∠0° = 3∠0°
5.
6.
7.
If the phase voltage , find the negative sequence component of red phase and the phase voltages E Y and E B. The current from neutral to ground connection is 12 A. Calculate the zero phase sequence components in phases. A balanced star connected load takes 90 A from a balanced 3 phase, 4 wire supply. If the fuses in the Y and B phases are removed, find the symmetrical components of the line currents: a. before the fuses are removed b. after the fuses are removed A 3 φ, 4 wire system supplies loads which are unequally distributed in the three phases. An analysis of the current flowing in R, Y and B lines shows that in R line, positive phase sequence component is 200 0° A and the negative phase sequence component is 100 60° A. The total observed current flowing back to the supply in the neutral conductor is 300 300° A. Calculate the currents in the three lines. One conductor of a 3 phase line is open. The current flowing to the Δ – connected load through the line R is 10 A. With the current in line R as reference and assuming that line B is open, find the symmetrical components of the line currents.
∠
8.
Examples: 1. In a 3 phase, 4 wire system, the currents in R, Y and B lines under abnormal conditions of loading are as under:
∠ ∠
==100∠30° =50∠300° 30∠180°
2.
Calculate the positive, negative and zero sequence currents in the R – line and return current in the neutral wire. The currents in a 3 – phase unbalanced system are:
12+6 = ==1212 15+ 10
3.
The phase sequence in RYB. Calculate the zero, positive and negative sequence components of the currents. The sequence voltages in the red phase are:
= 100 = 200100 = 100
9.
Three resistors of 5Ω, 10Ω and 20Ω are connected in delta across the three phases of a balanced 100 volts supply. What are the sequence components in the resistors and in supply lines? 10. A delta connected load is supplied from a 3 phase supply. The fuse in the B line is removed and current in the other two lines is 20 A. Find the symmetrical components of line currents. 11. Three impedances of 5 — j10, 6 + j 5 and 3 + j15 ohms are connected in star to red, yellow and blue lines of a 3300 V, 3-phase, 3-wire supply.
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
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Fault Analysis The phase sequence is RYB. Calculate the line current I R. 12. A star connected load consists of three equal resistors of 1 Ω resistance. The load is assumed to be connected to an unsymmetrical 3 phase supply, the line voltages are 200 V, 346 V and 400 V. Find the magnitude of current in any phase by the method of symmetrical components. 13. The line currents in a 3 phase, 4 wire system are:
== 200+200 300+400 = 400200
Calculate the positive, negative and zero components of all the line currents. 14. The sequence components of phase a currents are:
496° == =18.3.0.24347+1. 31. 168.2°
Calculate the following: a. The line current I a b. The line current I b c. The line current I c 15. If the currents of a wye – connected transformer are:
==1012215°30° = 15 82°
What is the phase b positive sequence component? 16. Three equal impedances (8+j6) ohms are connected in wye across a 3 phase, 3 wire supply. The symmetrical components of the phase A voltage are
= 40+ 0 220+28.99 == 4028.
If there is no connection between the load neutral and the supply neutral. Calculate the line currents I a, Ib and Ic. Sequence Impedances In unsymmetrical fault calculations, each piece of equipment will have three values of impedance – one corresponding to each sequence current: i. Positive sequence impedance (Z 1) ii. Negative sequence impedance (Z 2) iii. Zero sequence impedance (Z 0) The following points may be noted: a. In a 3 – phase balanced system, each piece of equipment or circuit offers only one impedance – the one offered to positive or normal sequence current. This is expected because of the a bsence
of negative and zero sequence currents in the 3 – phase balanced system. b. In a 3 – phase unbalanced system, each piece of equipment or circuit will have three values of impedance: positive sequence impedance, negative sequence impedance and zero sequence impedance. c. The positive and negative sequence impedances of linear, symmetrical and static circuits (e.g. transmission lines, cables, transformers and static loads) are equal and are the same as those used in the analysis of balanced conditions. This is due to the fact that impedance of such circuits is independent of the phase order, provided the applied voltages are balanced. It may be noted that positive and negative sequence impedances of rotating machines (e.g. synchronous and induction motors) are normally different. d. The zero sequence impedance depends upon the path taken by the zero sequence current. As this path is generally different from the path taken by the positive and negative sequence currents, therefore, zero sequence impedance is usually different from positive or negative sequence impedance. Sequence Impedances of Power System Elements The concept of impedances of various elements of power system (e.g. generators, transformers, transmission lines etc.) to positive, negative and zero sequence currents is of considerable importance in determining the fault currents in a 3 – phase unbalanced system. a. Transmission Lines – are static devices and hence the phase sequence has no effect on the impedance because currents and voltages encounter the same geometry of the line. Therefore, positive and negative sequence impedances of transmission lines are equal.
=
Zero sequence currents are in phase and flow through the phases (a, b, c conductors) to return through the grounded neutral. The ground or any shielding wire are in the path of zero sequence and zero – sequence impedance Z TL0, which includes the effect of the return path through the ground, is different from Z 1 and Z2. Consider a 1 meter length of a three phase line shown in the figure. The ground surface is approximated to an equivalent conductor located at the average distance D n from each of the three phases. The phase conductors carry sequence currents with return paths through a grounded neutral.
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Fault Analysis and there is no reactance due to armature reaction. Hence, machine offers a very small reactance due to the leakage flux. Therefore, the rotor windings present leakage reactance only to the flow of zero sequence currents
+ + + = 0 = − = 1 1 = 3 1 1 = 2×10 ln + ln + ln + ln = = =3 = 2×10− ln ′ / = = 0.2ln ′ / = = 0.2ln ′ + 30.2 ln / = +
Since
then
Since
and
Since
Therefore, b.
Synchronous Machines is designed with symmetrical windings and it induces emfs of positive sequence only. The positive sequence generator impedance is the value found when positive sequence current flows due to an imposed positive – sequence set of voltages. Neglecting armature resistance, the positive sequence impedance of the machine is (Sub transient) (Transient) (Steady – state value) With the flow of negative sequence currents in the stator, the net flux in the air gap rotates at opposite direction to that of the rotor. Therefore, the net flux rotates twice the synchronous speed to the rotor. In this case, field winding has no influence because field voltage is associated with positive – sequence variables and only the damper winding produces an effect in the quadrantal axis. Therefore, the negative sequence impedance is close to the positive sequence sub transient impedance.
= ="′ =
=
c.
It may be worthwhile to mention here that any impedance Z e in the earth connection of a star connected system has the effect to introduce an impedance of 3Z e per phase. It is because the three equal zero – sequence currents, being in phase, do not sum to zero at the star point, but they flow back along the neutral earth connection. Transformers In power transformers, core losses and magnetizing current are on the order of 1 % of rated value and hence magnetizing branch is neglected. The transformer is modelled with equivalent series leakage impedance. Transformer is a static device and if the phase sequence is changed, leakage impedance will not change. Therefore, positive, negative and zero sequence impedances are the same and equal to leakage impedance of the transformer
= = =
The equivalent circuit for the zero – sequence impedance depends on the winding connection and also upon whether or not the neutrals are grounded. a. Wye (grounded) – Wye (grounded) connection: Both neutrals are grounded and there is a path for the zero sequence current to flow in the primary and secondary.
b.
Wye (grounded) – Wye connection: Primary neutral is grounded and secondary neutral is isolated. Zero – sequence current in the secondary is zero. Consequently, zero – sequence current in the primary is zero.
c.
Wye – Delta connection: In this case neutral is isolated. As neutral is isolated, zero sequence current cannot flow.
= "
In a synchronous machine, no zero sequence voltage is induced. Zero sequence impedance of the machine is due to the flow of zero – sequence current. The flow of zero sequence currents creates three mmfs which are in time phase but are distributed in space phase by 120°. Therefore, resultant air gap flux would be zero
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
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Fault Analysis Examples:
d.
Wye (grounded) – Delta connection: In this case neutral is grounded. Primary current can flow because there is zero sequence circulating current in the delta connected secondary and a ground return path for the wye connected primary. Also note that no zero sequence current can leave the delta terminals, therefore, there is an isolation between primary and secondary side.
e.
Delta – Delta connection: In this case zero sequence current circulate the delta connected windings, but no current can leave the delta terminals.
Sequence Network of Loaded Synchronous Machines
A – Positive, B – Negative and C – Zero Sequence Network Points to remember: 1. The three sequence networks are independent. 2. The neutral of the system is the reference for positive and negative sequence networks but ground is the reference for the zero sequence network. 3. There is no voltage source in the negative or zero – sequence networks. Only the positive sequence network has a voltage source. 4. The grounding impedance is reflected in the zero sequence network as 3Z n.
1.
A 50 MVA, 11 kV, synchronous generator has a sub transient reactance of 20%. The generator supplies two motors over a transmission line with transformers at both ends as shown below. The motors have rated inputs of 30 and 15 MVA, both 10 kV, with 25% sub transient reactance. The three – phase transformers are both rated 60 MVA, 10.8/121 kV, with leakage reactance of 10% each. Assume zero – sequence reactances for the generator and motors of 6% each. Current limiting reactors of 2.5 ohms each are connected in the neutral of the generator and motor No. 2. The zero sequence reactance of the transmission line is 300 ohms. The series reactance of the line is 100 Ω. Draw the positive, negative and zero sequence networks.
2.
Draw the positive sequence network of the power system.
3.
Draw the zero sequence network of the power system.
4.
Draw the Zero sequence network. Data is given below: G1: 100 MVA, 11 kV, Xg10 = 0.05 pu G2: 100 MVA, 11 kV, Xg20 = 0.05 pu T1: 100 MVA, 11/220 kV, XT1 = 0.06 pu T2: 100 MVA, 220/11 kV, XT2 = 0.07 pu Line 1: XL1 0 = 0.3 pu Line 2:XL2 0 = 0.3 pu
5.
Draw the zero – sequence network of the sample power system shown below. Data are given as: G: xg0 = 0.05 pu M: xm0 = 0.03 pu T1: xt1 = 0.12 pu T2: xt2 = 0.10 pu
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Fault Analysis Line 1: x l10 = 0.7 pu Line 2: x l20 = 0.7 pu
6.
Single Line to Ground Fault
Draw the positive, negative and zero sequence component of the power system. Line – to Ground Fault on Phase A
= = + + == 3 = =0 3 =3 = = but since thus,
Those faults on the power system which give rise to unsymmetrical fault currents (i.e. unequal fault currents in the lines with unequal phase displacement) are known as unsymmetrical faults. On the occurrence of an unsymmetrical fault, the currents in the three lines become unequal and so is the phase displacement among them. It may be noted that the term ‘unsymmetry’ applies only to the fault itself and the resulting line currents. However, the system impedances and the source voltages are always symmetrical through its main elements: generators, transmission lines, synchronous reactors etc. There are three ways in which unsymmetrical faults may occur in a power system: i. Single line – to – ground fault (L — G)
then
Sequence Network Connection for Single Line – to – Ground Fault By KVL:
( + 3) = 0 = + + + 3 = + + + = % + % + ×% + %
Therefore:
If %Z:
Line to Line Fault (2 – phase Fault) ii.
Line – to – line fault (L — L)
iii.
Double line – to – ground fault (L — L — G)
Line – to – Line Fault between Phase B and Phase C
= = = = = 0 but since
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
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Fault Analysis
= + + 3 + + 3 = + + 3 = + + 3+ + 3 + = + + +
By CDR:
Sequence Network Connection for Line – to – Line Fault By KVL:
If %Z:
Therefore:
Example:
Consider I b:
If %Z:
( + + + +)==00 = + + = + + = = √ 3 = √ = +√ + = %√ +% × +%
× × % = %% + % + % + %% + % 1.
2.
Double Line to Ground Fault
3.
4.
Double Line to Ground Fault between Phase B and Phase C
= =3 + = 0 but
5.
since
6. Sequence Network Connection for Double Line to Ground Fault
A 3 phase AC generator rated 125 MVA, 13.8 kV, has a positive, negative and zero reactances of 5%, 5% and 2% respectively. If this generator is grounded through a resistance of 0.2 Ω. Calculate the following: a. Fault current for a three phase fault b. Fault current for a two phase fault c. Fault current for a single line to ground fault d. Fault current for a double line to ground fault Three single phase transformer each rated 100 kVA, 24 kV primary and 277 V secondary with 4% impedance have their primary connected in delta and secondary in wye. What is the fault current delivered for a 2 phase fault? A 3 phase, 11 kV, 30 MVA turbo alternator has a positive equal to negative reactance of 0.15 pu and zero sequence reactance of 0.02 pu. Calculate the fault current for a line to line fault. At a certain point of a power system network of positive, negative and zero sequence impedances of 0.25 pu, 0.25 pu and 0.30 pu respectively. The base MVA is 100. The voltage level at that point is 34.5 kV. Determine zero sequence component for a one line to ground fault. A 50 MVA, 11 kV three – phase alternator was subjected to different types of faults. The fault currents are as under: 3 – Phase fault = 2000 A; Line – to – Line fault = 2600 A; Line – to – ground fault = 4200 A The generator neutral is solidly grounded. Find the values of the three sequence reactances of the alternator. Ignore resistances. A 20 MVA, 11 kV, 3 – phase, 50 Hz generator has its neutral earthed through a 5% reactor. It is in parallel with another identical generator having
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
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Fault Analysis isolated neutral. Each generator has a positive sequence reactance of 20%, negative sequence reactance of 10% and zero sequence reactance of 15%. If a line to ground short circuit occurs in the common bus bar, determine the fault current. 7. The per unit values of positive, negative and zero sequence reactances of a network at fault are 0·08, 0·07 and 0·05. Determine the fault current if the fault is double line – to – ground. 8. A 3 – phase, 3 – wire system has a normal voltage of 10·4 kV between the lines. It is supplied by a generator having positive, negative and zero sequence reactances of 0.6, 0.5 and 0.2 Ω per phase respectively. Calculate the fault current which flows when a line – to – line fault occurs at the generator terminals. 9. A 3 – phase, 11 kV, 25 MVA generator with X 0 = 0.05 pu, X1 = 0.2 pu and X 2 = 0.2 pu is grounded through a reactance of 0·3 Ω. Calculate the fault current for a single line to ground fault. 10. A 3 – phase, 11 kV, 10 MVA alternator has sequence reactances of X 0 = 0.05 pu, X 1 = 0.15 pu and X2 = 0.15 pu. If the generator is on no load, find the ratio of fault currents for L – G fault to that when all the 3 – phases are dead short – circuited.
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
13
