Key Features of MATLAB
INTRODUCTION Any abnormal conditions which causes flow of huge current in the conductors or cable through inappropriate paths in the circuit can be defined as a fault. In normal operating conditions all the circuit elements of an electrical system carry currents whose magnitude depends upon the value of the generator voltage and the effective impedances of all the power transmission and distribution system elements including the impedances of the loads usually relatively larger than other impedances. Modern electric systems may be of great complexity and spread over large geographical area. An electric power system consists of generators, transformers, transmission lines and consumer equipment. The system must be protected against flow of heavy short-circuit currents, which can cause permanent damage to major equipments, by disconnecting the faulty section of system by means of circuit breaker and protective relaying. Such conditions are caused in the system accidentally through insulation failure of equipment or flashover of lines initiated by a lightning stroke or through accidental faulty operation. The safe disconnection can only be guaranteed if the current does not exceed the capability of the circuit breaker. Therefore, the short circuit currents in the network must be computed and compared with the ratings of the circuit breakers at regular intervals as part of the normal operation planning. The short circuit currents in an AC system are determined mainly by the reactance of the alternators, transformers and lines upto the point of the fault in the case of phase to phase faults. When the fault is between phase and earth, the resistance of the earth path play an important role in limiting the currents. Balanced three phase faults may be analyzed using an equivalent single phase circuit. With asymmetrical three phase faults, the use of symmetrical components help to reduce the complexity of the calculations as transmission lines and components are by and large symmetrical, although the fault may be asymmetrical. Fault analysis is usually carried out in per-unit quantities as they give solutions which are somewhat consistent over different voltage and power ratings, and operate on values of the order of unity. In case of circuit breakers, their rupturing capacities are based on the symmetrical short circuit current which is most easy to calculate among all types of circuit currents. But for the determination of relay settings, it is absolutely necessary to know fault current due to unsymmetrical condition too for which knowledge of symmetrical components is required. Depending on the location, the type, the duration, and the system grounding, short circuits may lead to ? electromagnetic interference with conductors in the vicinity (disturbance of communication lines),? stability problems,? mechanical and thermal stress (i.e. damage of equipment, personal danger)? danger for personnel In high voltage networks, short circuits are the most frequent type of faults. Short circuits may be solid or may involve an arc impedance. Figure 1 illustrates different types of short circuits. . FIGURE 1- Examples for different types of short circuits A power network comprises synchronous generators, transformers, lines,and loads. Though the operating conditions at the time of fault are important, the loads can usually be neglected during short circuits, as voltages dip very low so that currents drawn by loads can be neglected in comparison with short circuit currents. The synchronous generator during short circuit has a characteristic time varying behavior. In the event of a short circuit, the flux per pole undergoes dynamic change with associated transients in damper and field windings. The reactance of the circuit model of the machine changes in the first few cycles from a low subtransient reactance to a higher transient value, finally settling at a still higher synchronous (steady state) value. Depending upon the arc interruption time of the circuit breakers, an appropriate reactance value is used for the circuit model of synchronous generators for the short circuit analysis. In a modern large interconnected power system, heavy currents flowing during a short circuit must be interrupted much before the steady state conditions are established. Furthermore, from the considerations of mechanical forces that act on the circuit breaker components, the maximum current that a breaker has to carry momentarily must also be determined. Therefore, for selecting a circuit breaker, the initial current that flows on occurrence of a short circuit and also the current in the transient that flows at the time of circuit interruption must be determined. There are two different approaches to calculate the short circuits in a power system:? Calculation of transient currents? Calculation of stationary currentsTRANSIENTS ON A TRANSMISSION LINE Let us consider the short circuit transient on a transmission line. Certain simplifying assumptions are made at this stage:1. The line is fed from a constant voltage source.2. Short circuit takes place when the line is unloaded.3. Line capacitance is negligible and the line can be represented by alumped RL series circuit. FIGURE 2 –Transmission line model With the above assumptions the line can be represented by the circuit model shown above. The short circuit is assumed to take place at t = 0. The parameter α controls the instant on the voltage wave when short circuit occurs. It is known from circuit theory that the current after short circuit is composed of two parts, i.e. where is represents the steady state alternating current and it represents the transient direct current With A plot of i = is + it is shown in figure 3. In power system terminology, the sinusoidal steady state current is called the symmetrical short circuit current and the unidirectional transient component is called the DC off-set current, which causes the total short circuit current to be unsymmetrical till the transient decays. FIGURE 3 - Waveform of a short circuit current on a transmission line It follows easily from figure 3 that the maximum momentary short circuit current imm corresponds to the first peak. If the decay of transient current in this short time is neglected, then: Since transmission line resistance is small, θis nearly 90?. This has the maximum possible value for α = 0, i.e. short circuit occurring when the voltage wave is going through zero. Thus imm may be a high as twice the maximum of the symmetrical short circuit current: For the selection of circuit breakers, momentary short circuit current is takencorresponding to its maximum possible value. Modern circuit breakers are designed to interrupt the current in the first few cycles (five cycles or less). With reference to Figure, it means that when the current is interrupted, the DC off-set it has not yet died out and contributes thus to the current to be interrupted. Rather than computing the value of the DC off-set at the time of interruption (this would be highly complex in a network of even moderately large size),the symmetrical short circuit current alone is calculated. This current is then increased by an empirical multiplying factor to account for the DC off-set current. SHORT CIRCUIT OF A SYNCHRONOUS MACHINE Under steady state short circuit conditions, the armature reaction of a synchronous generator produces a demagnetizing flux. In terms of a circuit this effect is modelled as a reactance Xa in series with the induced emf. This reactance when combined with the leakage reactance Xl of the machine is called synchronous reactance Xd. The index d denotes the direct axis. Since the armature reactance is small, it can be neglected. The steady state short circuit model of a synchronous machine is shown in figure shown below. FIGURE 4- Steady state short circuit model of a synchronous machine Consider now the sudden short circuit of a synchronous generator that has initially been operating under open circuit conditions. The machine undergoes a transient in all the three phases finally ending up in the steady state condition described above. The circuit breaker must interrupt the current long before the steady condition is reached. Immediately upon short circuit, the DC off-set currents appear in all three phases, each with a different magnitude since the point on the voltage wave at which short circuit occurs is different for each phase. These DC off-set currents are accounted for separately on an empirical basis. Therefore, for short circuit studies, we need to concentrate our attention on the symmetrical short circuit current only. In the event of a short circuit, the symmetrical short circuit current is limited initially only by the leakage reactance of the machine. Since the air gap flux cannot change instantaneously, to counter the demagnetization of the armature short circuit current, currents appear in the field winding as well as in the damper winding in a direction to help the main flux. These currents decay in accordance with the winding time constants. The time constant of the damper winding which has low X/R-ratio is much less than the one of the field winding, which has high leakage inductance with low resistance. Thus, during the initial part of the short circuit, the damper and field windings have transformer currents induced in them. In the circuit model their reactances—Xf of field winding and Xdw of damper winding—appear in parallel with Xa as shown in figure below. FIGURE 5 - Approximate circuit model during subtransient period of short circuit FIGURE 6 - Approximate circuit model during transient period of short circuit As the damper winding currents are first to die out, Xdw effectively becomes open circuited and at a later stage Xf becomes open circuited. The machine reactance thus changes from the parallel combination of Xa, Xf , and Xdw during the initial period of the short circuit to Xa and Xf in parallel (Figure ) during the middle period. The machine reactance finally becomes Xa in steady state (Figure 7.8). The reactance presented bythe machine in the initial period of the short circuit, i.e.is called the subtransient reactance of the machine; while the reactance effectiveafter the damper winding currents have died out, i.e.is called the transient reactance. Of course, the reactance under steady conditions is the synchronous reactance. Obviously X′′d < X′d < Xd. The machine thus offers a time-varying reactance which changes from X′′d to X′d and finally to Xd.FIGURE 7 -Symmetrical short circuit armature current in synchronous machine. SYMMETRICAL THREE PHASE FAULT ANALYSIS In normal operating conditions, a three-phase power system can be treated as a single-phase system when the loads, voltages, and currents are balanced. If we postulate plane-wave propagation along the conductors (it is, however, known from the Maxwell equations that in the presence of losses this is not strictly true), a network representation with lumped elements can be made when the physical dimensions of the power system, or a part of it, are small as compared with the wavelength of the voltageand current signals. When this is the case, one can successfully use a single line lumped-element representation of the three-phase power system for calculation. A fault brings the system to an abnormal condition. Short-circuit faults are especially of concern because they result in a switching action, which often results in transient overvoltages. In the case of a symmetrical three-phase fault in a symmetrical system, one can still use a single-phase representation for the short-circuit and transient analysis. A three phase fault is a condition where either (a) all three phases of the system are short circuited to each other, or (b) all three phase of the system are earthed.FIGURE 8 – (a) Balanced three phase fault (b) Balanced three phase to earth fault This is in general a balanced condition, and we need to only know the positive-sequence network to analyze faults. Further, the single line diagram can be used, as all three phases carry equal currents displaced by 120?. Typically, only 5% of the initial faults in a power system, are three phase faults with or without earth. Of the unbalanced faults, 80 % are line-earth and 15% are double line faults with or without earth and which can often deteriorate to 3 phase fault. Broken conductor faults account for the rest.Fault Level Calculations In a power system, the maximum the fault current (or fault MVA) that can flow into a zero impedance fault is necessary to be known for switch gear solution. This can either be the balanced three phase value or the value at an asymmetrical condition. The Fault Level defines the value for the symmetrical condition. The fault level is usually expressed in MVA (or corresponding per-unit value), with the maximum fault current value being converted using the nominal voltage rating. MVAbase =√ 3 . Nominal Voltage(kV) . Ibase (kA) MVAfault =√ 3 . Nominal Voltage(kV) . Isc (kA)whereMVAfault – Fault Level at a given point in MVAIbase – Rated or base line currentIsc – Short circuit line current flowing in to a faultThe per unit value of the fault Level may thus be written asThe per unit voltage for nominal value is unity, so that The Short circuit capacity (SCC) of a busbar is the fault level of the busbar. The strength of a busbar (or the ability to maintain its voltage) is directly proportional to its SCC. An infinitely strong bus (or Infinite bus bar) has an infinite SCC, with a zero equivalent impedance and will maintain its voltage under all conditions. Magnitude of short circuit current is time dependant due to synchronous generators. It is initially at its largest value and decreasing to steady value. These higher fault levels tax Circuit Breakers adversely so that current limiting reactors are sometimes used. The Short circuit MVA is a better indicator of the stress on CBs than the short circuit current as CB has to withstand recovery voltage across breaker following arc interruption. The currents flowing during a fault is determined by the internal emfs of machines in the network, by the impedances of the machines, and by the impedances between the machines and the fault. The following figure shows a part of a power system, where the rest of the system at two points of coupling have been represented by their Thevenin’s equivalent circuit (or by a voltage source of 1 pu together its fault level which corresponds to the per unit value of the effective Thevenin’s impedance).FIGURE 9 – CIRCUIT FOR FAULT CALCULATIONWith CB1 and CB2 open, short circuit capacities are SCC at bus 1 = 8 p.u. gives Zg1 = 1/8 = 0.125 pu SCC at bus 2 = 5 p.u. gives Zg2 = 1/5 = 0.20 puEach of the lines are given to have a per unit impedance of 0.3 pu. Z1 = Z2 = 0.3 p.u.Suppose with CB1 and CB2 closed,the SCCs (or Fault Levels) of the busbars in the system is to be determined.FIGURE 10 – Determination of short circuit capacitiesThe circuit can be reduced and analysed as shown in the figure 11.FIGURE 11 – Determination of short circuit capacity at bus 3 Thus, the equivalent input impedance is given by Zin=0.23 pu at bus 3, so that the short circuit capacity at busbar 3 is given as | SCC3 |= 1/0.23 = 4.35 p.uThe network may also be reduced keeping the identity of Bus 1 as in the following figure.FIGURE 12 – Determination of short circuit capacity at bus 1 Thus, the equivalent input impedance is given by Zin=0.108 pu at bus 1, so that the short circuit capacity at busbar 1 is given as | SCC1 |= 1/0.108 = 9.25 p.u This is a 16% increase on the short circuit capacity of bus 1 with the circuit breakers open. The network may also be reduced keeping the identity of Bus 2. This would yield a value of Zin as 0.157 pu, giving the short circuit capacity at busbar 2 as| SCC2 |= 1/0.157 = 6.37 p.u This is a 28% increase on the short circuit capacity of bus 2 with the circuit breakers open. SYMMETRICAL COMPONENT ANALYSIS (FOR UNSYMMETRICAL SYSTEMS) For the majority of the fault situations, the power system has become unsymmetrical. Symmetrical components and, especially, the sequence networks are an elegant way to analyse faults in unsymmetrical three-phase power systems because in many cases the unbalanced portion of the physical system can be isolated for a study, the rest of the system being considered to be in balance. This is, for instance, the case for an unbalanced load or fault. In such cases, we attempt to find the symmetrical components of the voltages and the currents at the point of unbalance and connect the sequence networks, which are, in fact, copies of the balanced system at the point of unbalance (the fault point). The method of symmetrical components is a very powerful approach and has simplified the procedure for solving problems on the unbalanced polyphase systems. The method of symmetrical components was suggested by C.L. Fortesque in the year 1918. This method can be applied to any number of phases but three phase system is of main interest. According to Fortesque theorem, any unbalanced three phase system of currents, voltages or other sinusoidal quantities can be resolved into there balanced systems of phasors which are called symmetrical components of the original unbalanced system. Such three phase unbalanced systems constitute three sequence networks which are solved separately on a singe phase basis. Once the problem is solved in terms of the symmetrical components, it can be transferred back to the actual circuit condition by superposition or phasor additions of these quantities (currents or voltages) easily. Symmetrical components of three phase systems The symmetrical components differ in the phase sequence ,that is, the order in which the phase quantities go through a maximum. There may be a positive phase sequence, negative phase sequence and a zero phase sequence. Thus the balanced set of components can be given as positive sequence component, negative sequence component and zero sequence component. These are shown below in the figure:\ FIGURE 13- Symmetrical components of unbalanced three phase system The positive sequence system is that system in which the phase or line currents or voltages attain a maximum in the same cyclic order as those in a normal supply e.g. assuming conventional counter clockwise rotation, then the positive phase sequence phasors are as shown above in the figure. A balanced system corresponding to normal conditions contains a phase sequence only. It is also the condition for 3 phase fault.The positive sequence components are marked by subscript 1.The three phasors of positive sequence system are of equal magnitude, spaced 120 degrees apart. The negative sequence system is that system in which phasors still rotate anti-clockwise but attain maximum value in the reverse order as shown in the figure. This sequence only arises in the case of occurrence of an unsymmetrical fault. Such faults also contain the positive sequence system. The negative components are marked by subscript 2.The three phasors of positive sequence system are of equal magnitude, spaced 120 degrees apart . The zero phase sequence system a single phasor system combining three equal phasors in phase as illustrated in the figure given above and represents the residual current or voltage present under fault conditions on a 3 phase system with a fourth wire or earth return present. Clearly the zero phase sequence embraces the ground , therefore in addition to the three line wires and represents a fault condition to ground or to a fourth wire if present. Its presence arise only where fault to earth currents can return to the system via the star point of that system or via an artificial neutral point provided to earth a delta system. In an earth fault, positive and negative phase sequences are also present. The zero phase sequence components are marked by the subscript 0. The phase components are the addition of the symmetrical components and can be written as follows. The unknown unbalanced system has three unknown magnitudes and three unknown angles with respect to the reference direction. Similarly, the combination of the 3 sequence components will also have three unknown magnitudes and three unknown angles with respect to the reference direction. Thus the original unbalanced system effectively has 3 complex unknown quantities a, b and c (magnitude and phase angle of each is independent), and that each of the balanced components have only one independent complex unknown each, as the others can be written by symmetry. Thus the three sets of symmetrical components also have effectively 3 complex unknown quantities. These are usually selected as the components of the first phase a (i.e. a0, a1 and a2) . One of the other phases could have been selected as well, but all 3 components should be selected for the same phase. Thus it should be possible to convert from either sequence components to phase components or vice versa. Definition of the operator α When the balanced components are considered, it is seen that that the most frequently occurring angle is 120?. In complex number theory, j is defined as the complex operator which is equal to √-1 and a magnitude of unity, and more importantly, when operated on any complex number rotates it anti-clockwise by an angle of 90?. i.e. j = √-1 = 1 ∠90?In like manner, we define a new complex operator α which has a magnitude of unity. α when operated on any complex number rotates it anti-clockwise by an angle of 120? and square of α rotates it by 240o . i.e. α = 1 ∠120? = - 0.500 + j 0.866 and α 2 = 1∠2400 = -0.500 – j 0.866The three phasors (1+j0), α 2 and α (taken in this order) form a balanced, symmetrical , set of phasors of positive sequence rotation since the phasors are of equal length displaced by equal angles of 1200 from each other, and cross the reference line in the order 1, α 2 and α (following the usual convention of counter-clockwise rotation for the phasor diagram). The phasors 1, α and α 2 (taken in order) form the balanced, symmetrical, set of phasors of negative phase-sequence, since the phasors do not cross the reference line in the order named, keeping the same convention of counter-clockwise rotation, but third name following the first etc. Some Properties of α Phasor Addition Since α is complex, it cannot be equal to 1, so that α - 1 cannot be zero. ∴ α2 + α + 1 = 0This also has the physical meaning that the three sides of an equilateral triangles must close as in figure 2.11.Also α?1 = α2 and α?2 = α Analysis of decomposition of phasors The sequence components of the unbalanced quantity are again examined, with each of the components written in terms of phase a components, and the operator α, as in figure shown below. Figure 14 – Expressing components in terms of phase All the sequence components can be expressed in terms of the quantities for phase a using the properties of rotation of 0?, 120? or 240? .Thus This can be written in matrix form as follows: This gives the basic symmetrical component matrix equation, which shows the relationship between the phase component vector Ph and the symmetrical component vector Sy using the symmetrical component matrix [Λ]. Both the phase component vector Ph and the symmetrical component vector Sy can be either voltages or currents, but in a particular equation, they must of course all be of the same type. Since the matrix is a [3×3] matrix, it is possible to invert it and express Sy in terms of Ph.The symmetrical component matrix [Λ] can be inverted as follows.and the discriminent Δ = 3(α – α2) = 3α (1-α)Substituting, the matrix equation simplifies to giveSince α?1 = α2, α?2 = α and 1 + α + α2 = 0, the matrix equation further simplifies toIt is seen that α is the complex conjugate of α2, and α2 is the complex conjugate of α.Thus the above matrix [Δ]-1 is one-third of the complex conjugate of [Δ]. Thus,This can now be written in the expanded form asSEQUENCE IMPEDANCES AND SEQUENCE NETWORKS The impedance of the network offered to the flow of positive sequence currents is called positive sequence impedance. Similarly if only negative sequence currents flow, the impedance of the network offered to these currents is called negative sequence impedance, also, the impedance offered to the flow of zero sequence currents is called zero sequence impedance. If Za, Zb and Zc are the impedance of the load between phases a, b and c to neutral n, then the sequence impedances are given as: Positive sequence impedance- Z1= 13 (Za + α Zb + Zc)Negative sequence impedance – Z2 = 13 (Za + Zb + α Zc)Zero sequence impedance – Z3= 13 (Za + Zb + Zc) For a three phase symmetrical static circuit without internal voltages like transformers and transmission lines, the impedances offered to the currents of any sequences are the same in the three phases; also the current of a particular sequence will cause voltage drop of the same sequence or a voltage of a particular sequence will give rise to current of the same sequence only which means that there there is no mutual coupling between the sequence networks. Since in case of a static device, the sequence has no significance, the positive and negative sequence impedances are equal. But the zero sequence im pedance which includes impedance of the return path through the ground is usually different from the positive and negative sequence impedances. The impedances offered by rotating machine to positive sequence components of currents is usually different from those offered to the negative sequence components of currents. The single phase equivalent circuit composed of the impedance to current of any one sequence only is called the sequence network for that particular sequence. Hence, corresponding to positive-, negative- and zero- sequence currents, we have positive-, negative- and zero- sequence networks. Thus, for every power system, three sequence networks can be formed and these sequence networks are the interconnected in different ways to represent different unbalanced fault conditions. The sequence networks and voltages during the fault are then calculated from which actual fault currents and voltages can be determined. Negative sequence network differs from the positive sequence network in the following respects:-Normally, there are no negative sequence emf sources.Negative sequence impedances of rotating machines are generally differ from their positive sequence impedances. The zero sequence network likewise will be free of internal voltages, the flow of current being caused by the voltage at fault point. Zero sequence reactance of the transmission line is higher than for positive sequence. The impedances of transformers or generators will depend upon the type of connections ( delta or star (grounded or isolated) )(a)Synchronous machines (motor or generator) An unloaded generator may, in general, be represented by the star-connected equivalent with possibly a neutral to earth reactance as shown in figure. The induced emf’s in the three phases are E, E∠1200 and E∠2400. When an unsymmetrical fault occurs on the machine terminals, unbalanced currents Ia, Ib and Ic flow in the lines. Unbalanced line currents can be resolved into their symmetrical components.FIGURE 15 – Sequence networks of generator Since a synchronous machine is designed with symmetrical windings, it has induced emf’s of positive sequence only. The positive sequence network for a synchronous machine can be represented by the source emf on no load and positive sequence impedance Zg1 in series with it. The neutral impedance Zn does not appear in the circuit because phasor sum of Ia1, Ib1 and Ic1 is zero and no positive sequence current can flow through Zn. Since it is a balanced network, so it can be drawn on single phase basis. The reference bus for positive sequence network is at neutral potential. Synchronous machine does not generate any negative sequence voltage. The negative sequence network can be represented by a negative sequence impedance Zg2. In this case also, no neutral impedance appears as there is no negative sequence current through Zn. Since it is a balanced network, so it can be drawn on single phase basis. The reference bus for negative sequence network is also at neutral potential. No zero sequence voltage is induced in the machine. As the current flowing in the reactor impedance Zn is the sum of zero sequence currents in all the three phases , hence voltage drop caused by it will be 3Ia0 Zn. So, net zero sequence impedance of the machine will be:Z0=Zg0 + Zn. The reference bus is at ground potential. FIGURE 16- Positive, negative and zero sequence networks of generator (b)Transmission lines and cables Transmission lines are assumed to be symmetrical in all three phases and therefore positive and negative sequence impedances are independent of phase sequence and are equal. However, this assumption would not be valid for long un-transposed lines (say beyond 500 km) as the mutual coupling between the phases would be unequal, and symmetrical components then cannot be used. The transmission line (or cable) may be represented by a single reactance in the single-line diagram. Typically, the ratio of the zero sequence impedance to the positive sequence impedance would be of the order of 2 for a single circuit transmission line with earth wire, about 3.5 for a single circuit with no earth wire or for a double circuit line.For a single core cable, the ratio of the zero sequence impedance to the positive sequence impedance would be around 1 to 1.25. When only zero sequence currents flow in a transmission line, the currents in each phase are identical in both magnitude and phase. Such currents return partly through ground and the rest through overhead ground wires. The magnetic field due to the flow of zero sequence currents is very different from that set up by the flow of positive or zero sequence currents.(c)Single windings Each of the simple types of windings for the zero sequence path is considered. These diagrams are shown, along with the zero sequence single line diagram in figure.FIGURE 17- Zero sequence network of various types of windingsThe unearthed star connection does not provide a path for the zero sequence current to pass across, and hence in the single line diagram, there is no connection to the reference. With an earthed star connection, the winding permits a zero sequence current to flow, and hence is shown with a direct connection to the reference. The earthed star with impedance, is similar except that 3 times the neutral impedance appears in the zero sequence path. The delta connection on the other hand does not permit any zero sequence current in the line conductors but permits a circulating current. This effect is shown by a closed path to the reference.(d)Transformers The positive sequence impedance of a transformer is equal to its leakage reactance ( the resistance of winding is usually small in comparision to the leakage reactance ). Transformer, being a static device, the positive and negative sequence impedances are equal because the impedance is independent of phase order , provided the applied voltages are balanced. Thus, for a transformer Z1 = Z2 = Zleakage The situation with 3 phase transformer is more complex with regard to zero sequence impedance because of the possibility of variety of connections. The zero sequence currents can flow through the winding connected in star only if the star point is grounded. Moreover, the zero sequence currents cannot flow in the windings if the star point is isolated. No zero sequence currents can flow in the lines connected to a delta star winding as no return path is available for them. The zero sequence currents can , however flow through the delta connected winding themselves if any zero sequence voltages are induced in delta.Two-winding transformers Two winding (primary and secondary), three phase transformers may be categorised into (i) star-star, (ii) earthed star – star, (iii) earthed star – earthed star, (iv) delta – star, (v) delta – earthed star, (vi) delta – delta. There are also zig-zag windings in transformers which has not been dealt with in the following sections.FIGURE 18- Single line diagram of a two winding transformerThe figure 19 shows the zero-sequence diagrams of the transformers.FIGURE 19 – Zero sequence networks of a two winding transformer Considering the transformer as a whole, it can be seen that the single-line diagramsindicate the correct flow of the zero-sequence current from primary to secondary.Three-winding transformersFIGURE 20- Single line diagram of a three winding transformer Three phase, three winding have an additional tertiary winding, and may be represented by a single line diagram corresponding to the ampere-turn balance, or power balance.which in per unit quantities would yield the common equationThis may be represented by three reactances connected in T, giving the general single line diagram for fault studies for the 3 winding transformer, as shown in figure. FIGURE 21 - Three winding transformer The positive sequence and negative sequence diagrams have a direct connection to the T connection of reactances from P, S and T.The zero sequence network is built up from the single winding arrangements described and yields the single line diagrams given in the following section, and other combinations. FIGURE 22 – Single line diagrams for sequences of three winding transformer Generally available from measurements for a 3 winding transformer are the impedances across a pairs of windings. (ie. ZPS, ZPT, and ZST ), with the third winding is open circuited. Thus we could relate the values to the effective primary, secondary and tertiary impedances (ZP, ZS and ZT ) as follows, with reference to figure. The values of ZP, ZS and ZT can then be determined as As in the case of the 2 winding transformer, 3Zn is included wherever earthing of a neutral point is done through an impedance Zn. ANALYSIS OF FAULTS1.Three phase fault Three-phase faults, when all three lines touch each other or fall to ground, occur in only a small percentage of the cases but are very severe faults for the system and its components. The three-phase-to-ground fault is, in fact, a symmetrical fault because the power system remains in balance after the fault occurs. It is the most severe fault type and other faults, if not cleared promptly, can easily develop into it. Assumptions Commonly Made in Three Phase Fault Studies 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. The generated voltages in the transmission system are assumed balanced prior to the fault, so that they consist only of the positive sequence component Ef (pre-fault voltage). This is in fact the Thevenin’s equivalent at the point of the fault prior to the occurrence of the fault.Va0 = 0 – Z0 Ia0Va1 = Ef – Z1 Ia1Va2 – 0 – Z2 Ia2This may be written in matrix form as follows- Va0Va1Va2=0Ef0- Z0000Z1000Z2Ia0Ia1Ia2These may be expressed in network form as shown in the figureFIGURE 23 – Sequence networks representing three phase fault 2.Single Line to Ground faults (L – G faults) Line-to-ground faults are faults in which an overhead transmission line touches the ground because of wind, ice loading, or a falling tree limb. A majority of transmission-line faults are single line-to-ground faults.The single line to ground fault can occur in any of the three phases. However, it is sufficient to analyse only one of the cases. Looking at the symmetry of the symmetrical component matrix, it is seen that the simplest to analyse would be the phase a. Consider an L-G fault with zero fault impedance as shown in figure. FIGURE 24 – L-G fault on phase aSince the fault impedance is zero, on occurrence of the the fault. Va = 0 , Ib = 0 , Ic = 0since load currents are neglected.These can be converted to equivalent conditions in symmetrical components as follows. Va = Va0 + Va1 + Va2 =0 and Ia0Ia1Ia2=13 1111αα2 1α2 αIaIb=0Ic=0giving Ia0 = Ia1 = Ia2 = Ia3 Mathematical analysis using the network equation in symmetrical components would yield the desired result for the fault current If = Ia. Va0Va1Va2=0Ef0- Z0000Z1000Z2Ia0Ia1Ia2 where Ia0 = Ia1 = Ia2 = Ia/3Thus, we getT Simplification, with If = Ia, gives If = 3Ef Z1+ Z2+ Z0 Also, the equationsVa0 + Va1 + Va2 = 0, and Ia0 = Ia1 = Ia2 indicate that the three networks (zero, positive and negative) must be connected in series (same current, voltages add up) and short-circuited, giving the circuit as shown in figure. FIGURE 25 – Sequence networks representing L-G fault on phase a with Zf =0 In this case, Ia corresponds to the fault current If, which in turn corresponds to 3 times any one of the components (Ia0 = Ia1 = Ia2 = Ia/3). Thus the network would also yield the same fault current as in the mathematical analysis. In this example, the connection of sequence components is more convenient to apply than the mathematical analysis. Thus for a single line to ground fault (L-G fault) with no fault impedance, the sequence networks must be connected in series and short circuited. Now an L-G fault is considered with fault impedance Zf as shown in figure 25.FIGURE 26 – L-G fault on phase a with Zf present Under the the fault condition, Va = Ia Zf, Ib = 0, Ic = 0 These can be converted to equivalent conditions in symmetrical components asfollows Ia0Ia1Ia2=13 1111αα2 1α2 αIaIb=0Ic=0 giving Ia0 = Ia1 = Ia2 = Ia3 Also, Va = Va1 + Va2 + Va0 = Ia Zf Mathematical analysis using the network equation in symmetrical components would yield the desired result for the fault current If as If = 3Ef Z1+ Z2+ Z0+3ZfSimilarly, the basic equations,Ia0 = Ia1 = Ia2 = Ia3Va1 + Va2 + Va0 = Ia Zf = 3 Ia0 Zf would yield a circuit connection of the 3 sequence networks in series with an effective impedance of 3Zf.FIGURE 27 – Sequence network representing L-G fault on phase a with Zf 3.Line to Line faults (L – L faults) Line-to-line faults are usually the result of galloping lines because of highwinds or because of a line breaking and falling on a line below. Line-to-Line faults may occur in a power system, with or without the earth, and with or without fault impedance. Solution of the L-L fault gives a simpler solution when phases b and c are considered as the symmetrical component matrix is similar for phases b and c. The complexity of the calculations reduce on account of this selection. Under the fault condition, FIGURE 28 – L-L fault on phases b-c with no Zf Mathematical analysis may be done by substituting these conditions to the relevant symmetrical component matrix equation. However, the network solution after converting the boundary conditions is more convenient and is therefore considered here.Ia = 0 and Ib = – Ic when substituted into the matrix equation giveIa0Ia1Ia2=13 1111αα2 1α2 αIa=0IbIc=-Ibwhich on simplification gives Ia0 = 0, and Ia1 = – Ia2 or Ia1 + Ia2 = 0Similarly,Vb = Vc on substitution givesVa0Va1Va2= 13 1111αα2 1α2 αVaVbVc=Vbwhich on simplification yields Va1 = Va2. The boundary conditions Ia0 = 0, Ia1 + Ia2 = 0, and Va1 = Va2 indicate a sequence network where the positive and negative sequence networks are in parallel and the zero sequence is open circuited, as shown in following figureFIGURE 29 – Sequence network representing L-L fault on phases b and c with no Zf Mathematical analysis using the network equation in symmetrical components would yield the desired result for the fault current If asIf= -j3EfZ1 +Z24.Double line to ground faults (L-L-G faults)L-L-G fault with earth and no ZfFIGURE 30 – L-L-G fault on phases b-c with no Zf Under the fault condition ,Ia = 0, Vb = Vc = 0 Ia0 + Ia1 + Ia2 = Ia = 0It can be shown that Va0=Va1 = Va2 These conditions taken together, can be seen to correspond to all three sequence networks connected in parallel as shown in the figure . FIGURE 31 – Sequence network representing L-L-G fault on phases b and c (no Zf) From above figure, it can be shown that Ia1= Ef Z1+Z2Z0/(z1+Z0)(B)L-L-G fault with earth and Zf If Zf appears in the earth path, it could be included as 3Zf, giving (Z0 + 3Zf) in the zero sequence path.(C)L-L-G fault with Zf and no earthIf Zf appears in the fault path, between phases b and c, it could be included as ? Zf in each of b and c. Inclusion of ? Zf in phase a having zero current would not affect it, so that in effect, ? Zf can be added to each of the three phases and hence to each of the 3 sequence networks as (Z1+? Zf), (Z2+? Zf) and (Z0+? Zf). The normal circuit analysis yields the positive and negative sequence networks in parallel with a connecting impedance of Zf, which is effectively the same.5. Broken conductor faults In broken conductor (or open conductor) faults, the load currents cannot be neglected, as these are the only currents that are flowing in the network. The load currents prior to the fault are assumed to be balanced. Single conductor open on phase “a”FIGURE 32 –Open conductor fault on phase “a” In the case of open conductor faults, the voltages are measured across the break, such as a-a′. For the single conductor broken on phase “a” condition, shown in figure 32. the boundary conditions are Ia = 0, Vb = Vc = 0 This condition is mathematically identical to the condition in the L-L-G fault in the earlier section, except that the voltages are measured in a different manner. The connection of sequence networks will also be the same except that the points considered for connection are different.Two conductors open on phases “b” and “c”FIGURE 33 –Open conductor fault on phases b and c For the two conductors broken on phases “b” and “c” condition, the boundary conditions are Va = 0 , Ib = Ic =0 This condition is mathematically identical to the condition in the L-G faultin the earlier section. The connection of sequence networks will also be the same except that the points considered for connection are different.MATLAB THEORYWhat Is MATLAB? The name MATLAB stands for MATrix LABoratory. MATLAB was written originally to provide easy access to matrix software developed by the LINPACK (linear system package) and EISPACK (Eigen system package) projects. MATLAB is a high-performance language for technical computing. It integrates computation, visualization, and programming environment where problems and solutions are expressed in familiar mathematical notation. Furthermore, MATLAB is amodern programming language environment: it has sophisticated data structures, contains built-in editing and debugging tools, and supports object-oriented programming. These factors make MATLAB an excellent tool for teaching and research. MATLAB has many advantages compared to conventional computer languages (e.g., C, FORTRAN) for solving technical problems. Matlab program and script files always have filenames ending with ".m"; the programming language is exceptionally straightforward since almost every data object is assumed to be an array. Graphical output is available to supplement numerical results. MATLAB is an interactive system whose basic data element is an array that does not require dimensioning. This allows us to solve many technical computing problems, especially those with matrix and vector formulations, in a fraction of the time it would take to write a program in a scalar non-interactive language such as C or Fortran. The software package has been commercially available since 1984 and is now considered as a standard tool at most universities and industries worldwide. MATLAB supports the entire data analysis process, from acquiring data from external devices and databases, through preprocessing, visualization, and numerical analysis, to producing presentation-quality output. It has powerful built-in routines that enable a very wide variety of computations. It also has easy to use graphics commands that make the visualization of results immediately available. Specific applications are collected in packages referred to as toolbox. There are toolboxes for signal processing, symbolic computation, control theory, simulation, optimization, and several other fields of applied science and engineering.Key Features of MATLAB -High-level language for technical computingDevelopment environment for managing code, files, and dataInteractive tools for iterative exploration, design, and problem solvingMathematical functions for linear algebra, statistics, Fourier analysis, filtering, optimization, and numerical integration2-D and 3-D graphics functions for visualizing dataTools for building custom graphical user interfacesFunctions for integrating MATLAB based algorithms with external applications and languages, such as C, C++, Fortran, Java, COM, and Microsoft ExcelPARTS OF The MATLAB System:The MATLAB system consists of five main parts:Desktop Tools and Development Environment This is the set of tools and facilities that help you use MATLAB functions and files. Many of these tools are graphical user interfaces. It includes the MATLAB desktop and Command Window, a command history, an editor and debugger, and browsers for viewing help, the workspace, files, and the search path.The MATLAB Mathematical Function Library This is a vast collection of computational algorithms ranging from elementary functions, like sum, sine, cosine, and complex arithmetic, to more sophisticated functions like matrix inverse, matrix eigenvalues, Bessel functions, and fast Fourier transforms.The MATLAB Language This is a high-level matrix/array language with control flow statements, functions, data structures, input/output, and object-oriented programming features. It allows both "programming in the small" to rapidly create quick and dirty throw-away programs, and "programming in the large" to create large and complex application programs.Graphics MATLAB has extensive facilities for displaying vectors and matrices as graphs, as well as annotating and printing these graphs. It includes high-level functions for two-dimensional and three-dimensional data visualization, image processing, animation, and presentation graphics. It also includes low-level functions that allow you to fully customize the appearance of graphics as well as to build complete graphical user interfaces on your MATLAB applications.The MATLAB External Interfaces/API This is a library that allows you to write C and Fortran programs that interact with MATLAB. It includes facilities for calling routines from MATLAB (dynamic linking), calling MATLAB as a computational engine, and for reading and writing MAT-files.WHAT IS SIM POWER SYSTEMS Sim Power Systems and Sim Mechanics of the Physical Modeling product family work together with Simulink to model electrical, mechanical, and control systems.The Role of Simulation in Design Electrical power systems are combinations of electrical circuits and electromechanical devices like motors and generators. Engineers working in this discipline are constantly improving the performance of the systems. Requirements for drastically increased efficiency have forced power system designers to use power electronic devices and sophisticated control system concepts that tax traditional analysis tools and techniques. Further complicating the analyst's role is the fact that the system is often so nonlinear that the only way to understand it is through simulation. Land-based power generation from hydroelectric, steam, or other devices is not the only use of power systems. A common attribute of these systems is their use of power electronics and control systems to achieve their performance objectives. SimPowerSystems is a modern design tool that allows scientists and engineers to rapidly and easily build models that simulate power systems. SimPowerSystems uses the Simulink environment, allowing you to build a model using simple click and drag procedures. Not only can you draw the circuit topology rapidly, but your analysis of the circuit can include its interactions with mechanical, thermal, control, and other disciplines. This is possible because all the electrical parts of the simulation interact with the extensive Simulink modeling library. SIMULATION MODELThe circuit shown below is designed for the simulation of various types of faults in the transmission lines. .SIMULATION GRAPHSTHREE PHASE TO GROUND FAULTDOUBLE LINE TO GROUND FAULTSINGLE LINE TO GROUND FAULTDESCRIPTION OF VARIOUS BLOCKSSimplified Synchronous MachineModels the dynamics of a simplified three-phase synchronous machineDescriptionThe Simplified Synchronous Machine block models both the electrical and mechanical characteristics of a simple synchronous machine. The electrical system for each phase consists of a voltage source in series with an RL impedance, which implements the internal impedance of the machine. The value of R can be zero but the value of L must be positive.Three-Phase Series RLC LoadImplementing a three-phase series RLC load with selectable connectionDescriptionThe Three-Phase Series RLC Load block implements a three-phase balanced load as a series combination of RLC elements. At the specified frequency, the load exhibits a constant impedance. The active and reactive powers absorbed by the load are proportional to the square of the applied voltage.Three-Phase Transformer (Two Windings) Implementing a three-phase transformer with configurable winding connections DescriptionThe Three-Phase Transformer (Two Windings) block implements a three-phase transformer using three single-phase transformers. You can simulate the saturable core or not simply by setting the appropriate check box in the parameter menu of the block.Three-Phase Breaker Implementing a three-phase circuit breaker opening at the current zero crossing DescriptionThree-Phase Breaker block implements a three-phase circuit breaker where the opening and closing times can be controlled either from an external Simulink signal (external control mode), or from an internal control timer (internal control mode). The Three-Phase Breaker block uses three Breaker blocks connected between the inputs and the outputs of the block. This block can be used in series with the three-phase element that one wants to switch. If the Three-Phase Breaker block is set in external control mode, a control input appears in the block icon. The control signal connected to this input must be either 0 or 1, 0 to open the breakers, 1 to close them. If the Three-Phase Breaker block is set in internal control mode, the switching times are specified in the dialog box of the block. The three individual breakers are controlled with the same signal.Distributed Parameter Line Implementing an N-phase distributed parameter transmission line model with lumped losses DescriptionThe Distributed Parameter Line block implements an N-phase distributed parameter line model with lumped losses. The model is based on the Bergeron's traveling wave method used by the Electromagnetic Transient Program (EMTP) . In this model, the lossless distributed LC line is characterized by two values (for a single-phase line): the surge impedance Zc = (L/C) and the phase velocity v= 1/√(LC). The model uses the fact that the quantity e+Zi (where e is line voltage and i is line current) entering one end of the line must arrive unchanged at the other end after a transport delay of τ= d/v, where d is the line length.Three-Phase V-I Measurement Measures three-phase currents and voltages in a circuit DescriptionThe Three-Phase V-I Measurement block is used to measure three-phase voltages and currents in a circuit. When connected in series with three-phase elements, it returns the three phase-to-ground or phase-to-phase voltages and the three line currents. The block can output the voltages and currents in per unit (p.u.) values or in volts and amperes.Three-Phase Sequence Analyzer Measures the positive-, negative-, and zero-sequence components of a three-phase signalDescriptionThe Three-Phase Sequence Analyzer block outputs the magnitude and phase of the positive- (denoted by the index 1), negative- (index 2), and zero-sequence (index 0) components of a set of three balanced or unbalanced signals. The signals can contain harmonics or not.ScopeDisplays signals generated during a simulation DescriptionThe Scope block displays its input with respect to simulation time. The Scope block can have multiple axes (one per port); all axes have a common time range with independent y-axes. The Scope allows you to adjust the amount of time and the range of input values displayed. The Scope window can be moved and resized and the Scope's parameter values can be modified during the simulation. When the simulation is started, Simulink does not open Scope windows, although it writes data to connected Scopes. As a result, if a Scope is opened after a simulation, the Scope's input signal or signals will be displayed. If the signal is continuous, the Scope produces a point-to-point plot. If the signal is discrete, the Scope produces a stair-step plot. The Scope provides toolbar buttons that enables to zoom in on displayed data, display all the data input to the Scope, preserve axis settings from one simulation to the next, limit data displayed, and save data to the workspace.Three-Phase Fault Implementing a programmable phase-to-phase and phase-to-ground fault breaker system DescriptionThe Three-Phase Fault block implements a three-phase circuit breaker where the opening and closing times can be controlled either from an external Simulink signal (external control mode), or from an internal control timer (internal control mode). The Three-Phase Fault block uses three Breaker blocks that can be individually switched on and off to program phase-to-phase faults, phase-to-ground faults, or a combination of phase-to-phase and ground faults.CONCLUSION After the MATLAB simulation for faults, it was observed that the voltage and current waveforms were transient in nature in the initial period after the occurrence of faults. During the initial part of short circuit, the short circuit current was limited by subtransient reactance of synchronous machine and impedance of transmission line between the machine and point of fault. After that, it was limited by transient reactance of synchronous machine and impedance of line. Finally, the short circuit current settled down to steady state short circuit value limited by synchronous reactance of the machine and line impedance. The negative and zero sequence components were present initially only and they disappeared after the circuit breaker cleared the fault. BIBLIOGRAPHYELECTRICAL POWER SYSTEMS BY C.L WADHWA.ELEMENTS OF POWER SYSTEM ANALYSIS BY W.D. STEVENSON.SYMMETRICAL COMPONENTS BY C.F.WAGNER & R.D.EVANS.THE TRANSMISSION & DISTRIBUTION OF ELECTRICAL ENERGY BY H.COTTON.MODELLING AND ANALYSIS OF ELECTRIC POWER SYSTEMS BY GORAN ANDERSSON JOURNALS ................
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