Chapter Two- power system analysis Symmetrical Faults.pdf
- 1. Dr. Javed Ahmed Laghari PhD, University of Malaya, Malaysia Associate Professor javed@quest.edu.pk Department of Electrical Engineering QUEST, Nawabshah, Sindh, Pakistan 1
- 2. There may be accidental faults such as falling of a tree along a line, vehicles colliding with supporting structures, airplane crashing with the line, birds shorting line. 2 Def: A fault on the power system which gives rise to symmetrical fault currents (i.e. equal fault currents in the lines with 120o displacement) is called symmetrical fault. Explanation: A fault in a circuit is any failure which interferes with the normal flow of current. A short circuit is a fault in which current bypasses the normal load. An open-circuit fault occurs if a circuit is interrupted by some failure. A fault may occur on a power system due to a number of reasons. Some of the common causes have their origins in natural disturbances like lightning, high speed winds, earthquakes. Generators, transformers, and other protective switchgear may fail due to insulation breakdown.
- 3. The relays should immediately detect the existence of the fault and initiate circuit breaker operation to disconnect the faulty section. 3 Sometime sabotage also results in creating faults. Contamination of insulators may also result in a fault. Sometimes, small animals like rats, lizards enter switch gear to create faults. Thus, symmetrical faults can also be defined as: A fault in which all three phase are shorted to one another and to ground is known as three phase short circuit fault or symmetrical faults. A three phase short circuit occurs rarely but it is most severe type of fault involving largest currents. For this reason the balanced short circuit calculations are performed to determine these large currents to be used to determine the rating of the circuit breakers. Faults can cause system to become unstable. Hence, it is necessary that upon the occurrence of fault , the faulty section should be disconnected as rapidly as possible in order that the normal operation of the rest of the system is not affected. If this is not done, the equipment may be damaged and the power supply is disrupted.
- 4. When a short circuit occurs at any point in a system, the short-circuit current is limited by the impedance of the system up to the point of fault. 4 The purpose of fault analysis is to determine the values of voltages and currents at different points of the system during the fault. Such an analysis enables us to select appropriate protective schemes, relays, and circuit breakers in order to save the system from the abnormal condition within minimum time. Note: The assumptions for short circuit calculations are same as for per unit system reactance diagram. Hence, 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.
- 5. Experiences has shown that between 70% and 80% of transmission line faults are single line to ground faults, which arise from the flashover of only one line to the tower and ground. Roughly 5% of all faults involve all three phases. These are the so called symmetrical three phase faults. 5 Typical relative frequencies of occurrence of different kinds of faults in a power system (in order of decreasing severity) are: Type of Fault Percentage Occurrence Three phase (3 ɸ) Faults 5% Double line to ground (LLG) faults 10% Double Line or Line to Line (LL) faults 15% Single Line to ground faults 70% Experience in the operation of transmission lines have shown that ultra-high speed reclosing breakers successfully reclose after most faults. Of those cases where reclosure is not successful, many are caused by permanent faults where reclosure would be impossible regardless of the interval between opening and reclosing.
- 6. The short circuit capacity of a bus network is defined as: “The product of the magnitudes of the prefault voltage and the fault current”. 6 Where, V0 = the prefault voltage in Volts, IF = the short circuit current in amperes. VA I V SCC F o phase VA Z V Z V V I V SCC T T T T T F T / / / 2 1 …Equation (A) The short circuit capacity is also known as the fault level. Mathematically For a solid fault, the fault impedance ZF = 0, and the fault current is given by: IF = VT / ZT Where VT =Thevenin voltage per phase in volts and ZT =Thevenin impedance in ohms. In our case VT = V0, therefore,
- 7. 7 If VT is chosen as base voltage, VT = Vb. Therefore, Equation (B) can be written as: T B T u p T V S Z Z 2 . u p T B T T Z S Z V . 2 …………………Equation (C) Combining Equations (A) and (C) phase VA Z S SCC u p T b / / ) . ( 1 MVA Z S SCC u p T b ) . ( 3 3 / ) ( Where, Sb =Base volt-amperes in VA, Vb = Base voltage in volts, ZT p.u = Thevenin impedance in per unit. B B T u p T V S Z Z 2 . ……………………Equation (B) We know that,
- 8. 8 Consider the Figure which shows the equivalent circuit of three phase generator on no load, running at its synchronous speed and carrying a constant field current. It can be noticed that the equivalent circuit resembles to series RL circuit. The synchronous generator offers time varying reactance which changes from X’’d to X’d and finally to Xd. The currents which flow in different parts of a power system immediately after the occurrence of a fault differ from those flowing a few cycles later just before circuit breakers are called upon to open the line on both sides of the fault. All of these current differ widely from the currents which would flow under steady state conditions. Now, suppose, suddenly, the three phases of the synchronous generator are short circuited. The short circuit in each phase consists of a steady state AC component and a transient DC offset.
- 9. The current flowing immediately after a fault occurs in a power network is determined by the impedances of the network components and the synchronous machines. 9 The synchronous generator during short circuit has a characteristic time varying behaviour. 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 sub-transient reactance to a higher transient value, finally setting at a still higher synchronous (steady state) value. Depending upon the arc interruption time of circuit breakers, a suitable reactance value is used for the circuit model of synchronous generators for short circuit analysis. The initial symmetrical rms fault current can be determined by representing each machine by its sub-transient reactance in series with its sub-transient internal voltage.
- 10. 10 Figure shows the short circuit for one phase of a synchronous generator with DC offset component neglected. The dashed envelope is called the symmetrical short circuit armature current. The wave may be divided into three distinct time periods. Figure: Symmetrical short circuit current for one phase of synchronous generator
- 11. 11 The corresponding reactance of the winding is called the direct-axis sub-transient reactance X’’d. This reactance is essentially due to the presence of damper winding. The period lasts for only about 2 cycles. During this period the current decays very rapidly. The transient period lasts for about 20 to 30 cycles. During this period the current decreases somewhat slowly. The current reaches its steady state value. The rms value of initial current (that is, the current at the instant of short circuit) is called sub transient current I’’.
- 12. 12 The corresponding reactance of the winding is called the direct-axis transient reactance Xd. The rms value of current after two cycles to 30 cycles is called transient current I’. The corresponding reactance of the winding is called the direct-axis transient reactance X’d. This reactance is essentially due to the presence of field winding. The rms value of current after 20 cycles is called steady state current (I).
- 13. For oil circuit breakers above 5 kV, the sub-transient current multiplied by 1.6 is considered to be the rms value of the current whose disruptive forces the breaker must withstand during the first half cycle after the fault occurs. This current is called Momentary current. Selection of Circuit Breakers: 13 The electric utility company furnishes data to a customer who must determine the fault current in order to specify circuit breakers properly for an industrial plant or industrial power distribution system connected to the utility system at a certain point. Up to this point, we have devoted most of our attention to the sub-transient current called the initial symmetrical current, which does not include the DC component. Inclusion of the DC component results in a rms value of current immediately after the fault, which is higher than the sub-transient current. This interrupting current is ofcourse lower than the momentary current and depends on the speed of the breaker such as 8, 5, 3 or 2 cycles, which is a measure of the time from the occurrence of the fault to the extinction of the arc.
- 14. 14 Two of the rated circuit breaker ratings which require the computation of short circuit current are: Rated momentary current and Rated symmetrical interrupting current. Symmetrical short circuit current is obtained by using the sub-transient reactances for synchronous machines. Momentary current (rms) is then calculated by multiplying the symmetrical momentary current by a factor of 1.6 to account for the presence of DC-off set current. Selection of Circuit Breakers: The DC off set value to be added to obtain the current to be interrupted is accounted for by multiplying the symmetrical short circuit current by a factor as tabulated below: Circuit Breaker Speed Multiplying Factor 8 cycles or slower (0.16 seconds) 1.0 5 cycles (0.1 seconds) 1.1 3 cycles (0.06 seconds) 1.2 2 cycles (0.04 seconds) 1.4
- 15. The rated interrupting time of a circuit breaker is the period between the instant of energizing the trip circuit and the arc extinction on an opening operation. 15 Obviously the rated MVA interrupting capacity of a circuit breaker is to be more than (or equal to ) the Short circuit MVA required to be interrupted. Preceding this period is the tripping delay time which is usually assumed to be ½ cycle for relays to pick up. Selection of Circuit Breakers:
- 16. 16 A reactor is a coil of number of turns designed to have a large inductance as compared to its ohmic resistance. 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. Reactor Control of Short Circuit Currents: 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. 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.
- 17. 17 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. They permit the installation of circuit breakers of lower rating. Reactor Control of Short Circuit Currents: Short circuit current limiting reactors may be connected in series: With each generator With each feeder and With Bus-bars. Depending upon the location of reactors, following are the main types of Reactors Generator Reactors Feeder Reactors Bus-bar Reactors Each installation has its own particular demands which must be carefully considered before a choice of reactor location can be made. Note: The information about Bus-bar can be found in Power system Analysis by V. K. Mehta, Chapter 16, Topic 16.4 Bus bar arrangements Page Number 391.
- 18. When the reactors are connected in series with each generator, they are known as generator reactors. 18 Modern generators are designed to have sufficiently large reactance to protect them in even dead short circuits at their terminals. Thus, these reactors may only be used in old generators having low values of reactance. Generator Reactors:
- 19. Due to these disadvantages and also since modern power station generators have sufficiently large leakage reactance to protect them against short-circuit, it is not a common practice to use separate reactors for the generators. 19 There is a constant voltage drop and power loss in the reactors even during normal operation. 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 with the result that generators may lose synchronism and supply may be interrupted. If a fault occurs on any feeder, the continuity of supply to other is likely to be affected. Disadvantages of Generator Reactors:
- 20. Since most of the short-circuits occur on feeders, a large number of reactors are used for such circuits. 20 When the reactors are connected in series with each feeder, they are known as feeder reactors. Feeder Reactors:
- 21. 21 Two principal advantages are claimed for feeder reactors: Firstly, if a fault occurs on any feeder, the voltage drop in its reactor will not affect the bus bars voltage so that there is a little tendency for the generator to lose synchronism. Secondly, the fault on a feeder will not affect other feeders. There is a constant power loss and voltage drop in the reactors even during normal operation. 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. Advantages of Feeder Reactors:
- 22. When the reactors are connected in series with each Bus-bar, they are known as Bus- bar reactors. Ring System: 22 In this system, bus-bar is divided into sections and these sections are connected through reactors. There are two methods for this purpose, namely; The limitations of generator and feeder reactors can be overcome by locating the reactors in the bus-bars. Ring system and Tie-Bar system. This is the most common method of connection of reactors. Bus-bar Reactors:
- 23. Under normal operation, each generator supplies feeder connected to its own section and there will be no current through the reactors. 23 Therefore, only that section of bus-bar is affected to which the feeder is connected, the other sections being able to continue in normal operation Thus, there is no voltage drop or power loss or very little power loss in the reactor during normal operation. Generally, one feeder is fed from one generator only. The principal advantage of the system is that if a fault occurs on any feeder, only that bus bar section is affected to which it is connected. The other sections continue to operate normally. One generator (to which the particular feeder is connected) mainly feeds the fault current while the current fed from other generators is small due to the presence of reactors. Ring System:
- 24. In this system, the generators are connected to the common bus-bar (tie-bar) through the reactors but the feeders are fed from the generator side of the reactors. 24 The tie-bar system is better and more flexible than the ring system. Tie-Bar System:
- 25. The Tie-bar system has the disadvantage that it requires an additional bus-bar i.e. the tie-bar. 25 In the ring system, the short circuit current due to a fault on any bus bar section, is fed from the generators connected to other sections through one reactor, whereas in the tie- bar system the current flows through two reactors in series. Therefore this system requires only half the reactance compared to the ring system. Tie-Bar System:
- 26. Whatever value of base kVA we may choose, the actual value of short-circuit current will be the same. 26 The common kVA rating assumed for calculation of per unit or short circuit current is known as base kVA. The value of this base kVA is quite unimportant and may be : Equal to that of the largest plant Equal to the total plant capacity Any arbitrary value The fact is that the value of base kVA does not affect the actual short circuit current. It can be proved with the help of an example as shown in the next slides Selection of Base kVA and Its Importance:
- 27. Consider a 3-phase transmission line operating at 66 kV and connected through a 1000 kVA transformer with 5% reactance to a generating station bus-bar as shown in Figure. The generator is of 2500 kVA with 10% reactance. Suppose a short-circuit fault between three phases occurs at the high voltage terminals of transformer. Prove that whatever value of base kVA we may choose, the value of short-circuit current will be the same. 27 This example is taken from Book Principles of Power System V K Mehta, Chapter 17, Topic 17.4 Percentage Reactance and Base kVA Example:
- 28. ISC Calculation at 2500 kVA as Base kVA: Suppose we have chosen 2500 kVA and 5000 kVA, two common base kVA. Now first, short circuit current is calculated with 2500 kVA as below: On this base value, the reactances of the various elements in the system are calculated as below: New Per unit Reactance of Generator G: u p j j u p G X new . 1 . 0 10 2500 10 2500 10 11 10 11 1 . 0 ) . ( 3 3 2 3 3 New Per unit Reactance of Transformer T: u p j j j u p T X new . 125 . 0 5 . 2 05 . 0 10 1000 10 2500 10 11 10 11 05 . 0 ) . ( 3 3 2 3 3 Total Per Unit Reactance Up to Point of Fault: u p j j j u p Total X . 225 . 0 125 . 0 1 . 0 ) . ( J.A.Laghari way of solution is different than V.K.Mehta Solution 28 Solution:
- 29. Short Circuit Current ISC Per unit : u p X V u p SC I u p u p . 44 . 4 225 . 0 / 0 . 1 / ) . ( . . Base Short Circuit Current ISC : Base V kVA Base Base SC I 3 ) ( Actual Short Circuit Current ISC : A Base SC I u p SC I Actual SC I 1 . 97 87 . 21 44 . 4 ) ( ) . ( ) ( Equation …1 29 3 3 10 66 3 10 2500 66 3 2500 A 87 . 21 3153 . 114 2500 Solution:
- 30. New Per unit Reactance of Transformer T: 3 3 2 3 3 10 1000 10 5000 10 11 10 11 05 . 0 ) . ( j u p T X new Total Per Unit Reactance Up to Point of Fault: u p j j j u p Total X . 45 . 0 25 . 0 2 . 0 ) . ( Short Circuit Current ISC Per unit: u p X V u p SC I u p u p . 22 . 2 45 . 0 / 0 . 1 / ) . ( . . ISC Calculation at 5,000 kVA as Base kVA: On this base value, the reactance's of the various elements are calculated as below: New Per unit Reactance of Generator G: 3 3 2 3 3 10 2500 10 5000 10 11 10 11 1 . 0 ) . ( j u p G X new 30 u p j j . 2 . 0 2 1 . 0 u p j j . 25 . 0 5 05 . 0 Solution:
- 31. Actual Short Circuit Current ISC : Hence, it can be observed from Equation 1 and 2 that actual value of short circuit current does not depend upon the base kVA. However, in the interest of simplicity, numerically convenient value for the base kVA should be chosen. Base V kVA Base Base SC I 3 ) ( Base Short Circuit Current ISC : A Base SC I u p SC I Actual SC I 1 . 97 74 . 43 22 . 2 ) ( ) . ( ) ( Equation …2 Answer 31 A 74 . 43 3153 . 114 5000 1000 66 3 1000 5000 66 3 5000 Solution:
- 32. A three phase transmission line operating at 33 kV and having a resistance and reactance of 5 ohms and 20 ohms respectively is connected to a generating station bus bar through a 15 MVA step-up transformer which has a reactance of 0.06 p.u. connected to bus bar are two generators, one 10 MVA having 0.1 p.u reactance and another 5MVA having 0.075 p.u reactance. 32 Calculate the short circuit MVA and the fault current when a three phase short circuit occurs: (a) At the high voltage terminals of transformer (b) At the load end of the transmission line. Select 15MVA and 11kV as the base values. (c) Which fault will have higher value of current (Added by J. A. Laghari) Example:
- 33. We know that the formula for new per unit impedance is given by: given new new given old u p new kVA base kVA base kV base kV base Z u p Z 2 ) . ( ) . ( New Per unit Reactance of Generator G1: 6 6 2 3 3 10 10 10 15 10 11 10 11 1 . 0 ) . ( 1 j u p G X new New Per unit Reactance of Generator G2: 6 6 2 3 3 10 5 10 15 10 11 10 11 075 . 0 ) . ( 2 j u p G X new New Per unit Reactance of Transformer: 6 6 2 3 3 10 15 10 15 10 11 10 11 06 . 0 ) . ( / j u p F T X new 33 u p j j . 15 . 0 5 . 1 1 1 . 0 u p j j . 225 . 0 3 1 075 . 0 u p j j . 06 . 0 1 1 06 . 0 Solution:
- 34. New Per unit Impedance of Transmission Line: It can be observed that for transmission line the base voltage is changed. Hence, new base voltage is determined by: V Voltage Base New 3 3 3 3 10 33 10 11 10 33 10 11 Now, it can be noticed that the impedance of transmission line is given in ohms instead of per unit values. Hence, the formula to find per unit impedance of transmission line is given by: 2 2 . ) ( kV Base kVA Base Z V S Z Z ohms B B ohms u p 2 3 6 ) . ( ) 10 33 ( 10 15 ) 20 5 ( j Z u p Line 34 u p j Z u p Line . ) 2754 . 0 06887 . 0 ( ) . ( 1089 15 ) 20 5 ( j 01377 . 0 ) 0 2 5 ( ) . ( j Z u p Line Solution:
- 35. Solution: The total impedance from the generator side up to the point of fault FA is given by: 06 . 0 ) 225 . 0 ( || ) 15 . 0 ( ) . ( j j j u p Total Z 35 (a) Calculation of Short Circuit Current at Point FA: The equivalent per unit single line diagram with fault at point FA is shown as below: 06 . 0 375 . 0 03375 . 0 ) . ( 2 j j j u p Total Z 06 . 0 225 . 0 15 . 0 225 . 0 15 . 0 j j j j j u p j j u p Total Z . 06 . 0 375 . 0 03375 . 0 ) ( ) . ( 06 . 0 375 . 0 03375 . 0 1 j j u p j u p j j . 15 . 0 . 06 . 0 09 . 0
- 36. Solution: The short circuit MVA fed into the fault at FA: 15 . 0 10 15 6 . j Z S S u p Base Actual The fault current at point FA is given by: Alternate Approach to Calculate Fault current FA: (J.A.Laghari) It should be noted that the here question ask first to determine the short circuit MVA, due to which here, fault current is directly calculated by putting the value of short circuit MVA in the formula. However, if any question does not ask to calculate the short circuit MVA and ask to determine the fault current only. In that case, first per unit short circuit current will be calculated followed by short circuit base current. In the end, short circuit or fault current will obtained as below: u p X V u p SC I u p u p . 6667 . 6 15 . 0 / 0 . 1 / ) . ( . . 33 3 10 100 10 33 3 10 100 3 3 3 6 Actual V kVA Actual SC I 36 100 / 15 10 15 15 . 0 10 15 6 6 MVA 100 15 10 100 15 6 A 5 . 1749 10 7495 . 1 1576 . 57 10 100 3 3
- 37. Solution: 3 6 10 33 3 10 15 3 ) ( Base V kVA Base Base SC I (b) Calculation of Short Circuit Current at Point FB: The equivalent per unit single line diagram with fault at point FB is shown as below: A Base SC I u p SC I Actual SC I 5 . 1749 4 . 262 6667 . 6 ) ( ) . ( ) ( 37 A 4 . 262 10 2624 . 0 3 1576 . 57 10 15 33 3 10 15 3 3
- 38. Solution: The total impedance from the generator side up to the point of fault FB is given by: 2754 . 0 06887 . 0 06 . 0 ) 225 . 0 ( || ) 15 . 0 ( ) . ( j j j j u p Total Z The short circuit MVA fed into the fault at FA: MVA j Z S S u p Base Actual 8 . 34 431 . 0 10 15 431 . 0 10 15 6 6 . The fault current at point FB is given by: Note: To calculate short circuit MVA, impedance in polar form is used. However, alternatively, if impedance in rectangular form is used, then short circuit MVA is obtained by rationalizing the impedance. 33 3 10 8 . 34 10 33 3 10 8 . 34 3 3 3 6 Actual V kVA Actual SC I 38 2754 . 0 06887 . 0 06 . 0 09 . 0 ) . ( j j j u p Total Z u p j u p Total Z . 431 . 0 4254 . 0 06887 . 0 ) . ( 2 2 4254 . 0 06887 . 0 j A 8 . 608 10 6088 . 0 1576 . 57 10 8 . 34 3 3
- 39. Solution: Alternate Approach to Calculate Fault current FB: (J.A.Laghari) The short circuit current can also be determined by first determining the per unit short circuit current followed by short circuit base current. In the end, short circuit or fault current will obtained as below: u p X V u p SC I u p u p . 32 . 2 431 . 0 / 0 . 1 / ) . ( . . 33 3 10 15 10 33 3 10 15 3 ) ( 3 3 6 Base V kVA Base Base SC I (c) Which fault has higher value of current From part A & B, it is evident that fault FA is more sever & has higher value. Note: It should be noticed that without calculating the actual current, we can realize that the fault point at generator side will have highest current than all other points and decreases gradually at other points. A Base SC I u p SC I Actual SC I 8 . 608 4 . 262 32 . 2 ) ( ) . ( ) ( 39 A 4 . 262 10 2624 . 0 1576 . 57 10 15 3 3
- 40. Example For Practice: The single line diagram of a power system is shown in Figure. Find the short circuit current that will flow at point F. select 35 MVA and 11 kV as base values. Solution: Solving the system, we will get the following answer. 40 A Base SC I u p SC I Actual SC I 815 . 4724 02 . 1837 572 . 2 ) ( ) . ( ) ( Answer
- 41. Example: A 10 MVA, 6.6kV, 3-phase star connected alternator having a reactance of 20% is connected through a 5MVA, 6.6/33kV kV transformer of 10% reactance to a transmission line having a resistance and reactance per conductor per kilometre of 0.2 ohms and 1 ohm respectively. Fifty kilometres along the line, a short circuit occurs between the three conductors. Find Solution: The formula for new per unit impedance is given by: (a) Short circuit current fed by the Alternator: 41 This example is taken from Book Principles of Power System V K Mehta, Chapter 17, Example 17.14. (b) The current at the fault point (Added by J.A.Laghari). (a) The current fed to the fault by the alternator.
- 42. given new new given old u p new kVA base kVA base kV base kV base Z u p Z 2 ) . ( ) . ( Solution: Let 10MVA and 6.6 kV be the base values. New Per unit Reactance of Generator G: 6 6 2 3 3 10 10 10 10 10 6 . 6 10 6 . 6 2 . 0 ) . ( j u p G X new New Per unit Reactance of Transformer T: 6 6 2 3 3 10 5 10 10 10 6 . 6 10 6 . 6 1 . 0 ) . ( / j u p F T X new New Per unit Impedance of Transmission Line: It can be observed that for transmission line the base voltage is changed. Hence, new base voltage is determined by: V Voltage Base New 3 3 3 3 10 33 10 6 . 6 10 33 10 6 . 6 42 u p j j . 2 . 0 1 1 2 . 0 u p j j . 2 . 0 2 1 1 . 0
- 43. Solution: 2 2 . ) ( kV Base kVA Base Z V S Z Z ohms B B ohms u p 1089 10 ) 50 10 ( ) 10 33 ( 10 10 ) 0 5 10 ( 2 3 6 ) . ( j j Z u p Line Total New Per unit Impedance up to the Fault Point: 459 . 0 0918 . 0 2 . 0 2 . 0 ) . ( ) . ( / ) . ( ) . ( j j j X X X Z u p Line u p F T u p G u p Toal The short circuit MVA fed into the fault at FA: MVA Z S S u p Base SC 57 . 11 8638 . 0 10 10 6 . Now, it can be noticed that the impedance of transmission line is given in ohms instead of per unit values. Hence, the formula to find per unit impedance of transmission line is given by: 43 u p j Z u p Line . ) 459 . 0 0918 . 0 ( ) . ( 00918 . 0 ) 50 10 ( j u p j j Z u p Toal . 8638 . 0 7463 . 0 ) 859 . 0 ( 0918 . 0 859 . 0 0918 . 0 2 2 ) . (
- 44. Solution: The fault current fed by the alternator is given by: 6 . 6 3 10 57 . 11 10 6 . 6 3 10 57 . 11 3 3 3 6 Actual V kVA Actual SC I (b) Short circuit current at the fault point: 33 3 10 57 . 11 10 33 3 10 57 . 11 3 3 3 6 Actual V kVA Actual SC I Alternate Approach to Calculate ISC: (By J.A.Laghari) u p X V u p SC I u p u p . 157 . 1 8638 . 0 / 0 . 1 / ) . ( . . 6 . 6 3 10 10 10 6 . 6 3 10 10 3 ) ( 3 3 6 Base V kVA Base Base SC I The fault current fed by the alternator is given by: For the fault current fed by the alternator, the short circuit base current is given by: 44 A 1012 43 . 11 10 57 . 11 3 A 4 . 202 157 . 57 10 57 . 11 3 A 77 . 874 431 . 11 10 10 3
- 45. Solution: A Base SC I u p SC I Actual SC I 1012 77 . 874 157 . 1 ) ( ) . ( ) ( The fault current at the fault point is given by: u p X V u p SC I u p u p . 157 . 1 8638 . 0 / 0 . 1 / ) . ( . . Base V kVA Base Base SC I 3 ) ( ) ( ) . ( ) ( Base SC I u p SC I Actual SC I For the fault current at fault point, the short circuit base current is given by: 45 3 6 10 33 3 10 10 A 95 . 174 157 . 57 10 10 33 3 10 10 3 3 A Actual SC I 4 . 202 95 . 174 157 . 1 ) (
- 46. Example For Practice: A 3-phase generating station has two 15,000 kVA generators connected in parallel each with 15% reactance and a third generator of 10,000 kVA with 20% reactance is also added later in parallel with them. Load is taken as shown from the station bus-bars through 6000 kVA, 6% reactance transformers. Determine the maximum fault MVA which the circuit breakers have to interrupt on 46 This example is taken from Book Power System Analysis by P.S.R.Murty. Chapter 6, Example 6.17. (a) LV side and (b) HV side of the system for a symmetrical fault.
- 47. given new new given old u p new kVA base kVA base kV base kV base Z u p Z 2 ) . ( ) . ( Solution: Let 15,000 kVA as base kVA (a) Short Circuit MVA at LV side of Transformer: The formula for new per unit impedance is given by: 47 The short circuit MVA is given by: MVA Z S S u p Base SC 250 06 . 0 10 000 , 15 3 . The short circuit MVA is given by: MVA Z S S u p Base SC 43 . 71 21 . 0 10 000 , 15 3 . (b) Short Circuit MVA at HV side of Transformer:
- 48. Example: For the radial network shown in Figure, a three phase fault occurs at F. determine the fault current. 48 This example is taken from Book Modern Power System Analysis by D.P.Kothari. Chapter 9, Example 9.1.
- 49. given new new given old u p new kVA base kVA base kV base kV base Z u p Z 2 ) . ( ) . ( Solution: Let 100 MVA and 11 kV as chosen as base values. New Per unit Reactance of Generator G1: 6 6 2 3 3 1 10 10 10 100 10 11 10 11 15 . 0 ) . ( j u p G X new (a) Short Circuit Current at End of Feeder: The formula for new per unit impedance is given by: New Per unit Reactance of Generator G2: 6 6 2 3 3 2 10 10 10 100 10 11 10 11 125 . 0 ) . ( j u p G X new New Per unit Impedance of Transformer T1: 6 6 2 3 3 10 10 10 100 10 11 10 11 1 . 0 ) . ( 1 j u p T Z new 49 u p j j . 5 . 1 10 15 . 0 u p j j . 25 . 1 10 1 125 . 0 u p j j . 0 . 1 10 1 1 . 0
- 50. New Per unit Impedance of Transmission Line: It can be observed that for transmission line the base voltage is changed. Hence, first new base voltage is required to determined. Then its per unit impedance can be calculated. The formula for finding new base voltage is given by: kV E E Voltage Base Old Voltage Base New 33 10 11 10 33 10 11 3 3 3 1 2 The formula to find per unit impedance of transmission line is given by: Solution: 2 2 . ) ( kV Base kVA Base Z V S Z Z ohms B B ohms u p 2 3 6 ) . ( ) 10 33 ( 10 0 0 1 ) 36 . 0 27 . 0 ( 30 j Z u p Line The values are given for per kilometre. Hence, for 30 kilometre, we have to multiply it with 30. 50 u p j Z u p Line . 99 . 0 744 . 0 ) . ( 1089 0 0 1 ) 36 . 0 27 . 0 ( 30 j 09183 . 0 ) 8 . 10 1 . 8 ( 09183 . 0 ) 36 . 0 27 . 0 ( 30 ) . ( j j Z u p Line
- 51. New Per unit Impedance of Transformer T2: 6 6 2 3 3 10 5 10 100 10 33 10 33 08 . 0 ) . ( 2 j u p T Z new Solution: New Per unit Impedance of Cable: It can be observed that for transmission line the base voltage is changed. The new base voltage is determined as below: kV E E Voltage Base Old Voltage Base New 6 . 6 10 33 10 6 . 6 10 33 3 3 3 1 2 The impedance value of cable is given for per kilometre. Hence, for 3 kilometre, we have to multiply it with 3 as below: 2 3 6 ) . ( ) 10 6 . 6 ( 10 0 0 1 ) 08 . 0 135 . 0 ( 3 j Z u p Cable 51 u p j j . 6 . 1 20 1 08 . 0 u p j Z u p Cable . 55 . 0 93 . 0 ) . ( 56 . 43 0 0 1 ) 08 . 0 135 . 0 ( 3 j 295 . 2 ) 24 . 0 405 . 0 ( 295 . 2 ) 08 . 0 135 . 0 ( 3 ) . ( j j Z u p Cable
- 52. Solution: It can be observed that since the system is on no load prior to occurrence of the fault, the voltages of the two generators are identical (in phase and magnitude) and are equal to 1.0 p.u. the generator circuit can thus be replaced by a single voltage source in series with the parallel combination of generator reactance's as shown above. The equivalent per unit single line diagram with fault at point F is shown as below: 52
- 53. Solution: The total impedance up to the fault point is given as: 55 . 0 93 . 0 6 . 1 99 . 0 744 . 0 0 . 1 ) 25 . 1 ( || ) 5 . 1 ( ) . ( j j j j j j u p Total Z The per unit short circuit current is given by: u p Z V u p SC I u p u p . 8 . 70 196 . 0 8 . 70 1 . 5 / 0 . 1 / ) . ( . . 53 55 . 0 93 . 0 6 . 1 99 . 0 744 . 0 0 . 1 25 . 1 5 . 1 25 . 1 5 . 1 ) . ( j j j j j j j j u p Total Z 55 . 0 93 . 0 6 . 1 99 . 0 744 . 0 0 . 1 75 . 2 875 . 1 ) . ( 2 j j j j j j u p Total Z 55 . 0 93 . 0 6 . 1 99 . 0 744 . 0 0 . 1 681 . 0 ) ( ) 1 ( ) . ( j j j j j u p Total Z 55 . 0 93 . 0 6 . 1 99 . 0 744 . 0 0 . 1 681 . 0 ) . ( j j j j j u p Total Z u p j u p j u p Total Z . 8 . 70 1 . 5 82 . 4 674 . 1 . 82 . 4 674 . 1 ) . ( 2 2
- 54. Solution: Now, the short circuit base current is given by: 3 6 10 6 . 6 3 10 100 3 ) ( Base V kVA Base Base SC I 54 A 8750 43 . 11 10 100 6 . 6 3 10 100 3 3 A Base SC I u p SC I Actual SC I 1715 8750 196 . 0 ) ( ) . ( ) (
- 55. Example: Three 6.6 kV generators A, B and C, each of 10% leakage reactance and MVA ratings 40, 50 and 25 respectively are interconnected electrically as shown in Figure, by a tie bar through current limiting reactors, each of 12% reactance based upon the rating of the machine to which it is connected. A three phase feeder is supplied from the bus bar of generator A at a line voltage of 6.6 kV. The feeder has a resistance of 0.06 Ω/phase and an inductive reactance of 0.12 Ω/phase. Estimate (a) The maximum MVA that can be fed into a symmetrical short circuit at the far end of the feeder. (b)The maximum short circuit current at the feeder. (added by J.A.Laghari) 55 This example is taken from Book Modern Power System Analysis by D.P.Kothari. Chapter 9, Example 9.4.
- 56. given new new given old u p new kVA base kVA base kV base kV base Z u p Z 2 ) . ( ) . ( Solution: Let 50 MVA and 6.6 kV as chosen as base values. New Per unit Reactance of Generator G1: u p j j j j u p G X new . 125 . 0 4 5 . 0 4 5 1 . 0 10 40 10 50 10 6 . 6 10 6 . 6 1 . 0 ) . ( 6 6 2 3 3 1 (a) Short Circuit MVA at Far End of Feeder: The formula for new per unit impedance is given by: New Per unit Reactance of Generator G2: New Per unit Reactance of Generator G3: u p j j j u p G X new . 1 . 0 1 1 1 . 0 10 50 10 50 10 6 . 6 10 6 . 6 1 . 0 ) . ( 6 6 2 3 3 2 u p j j j j u p G X new . 2 . 0 2 1 1 . 0 25 50 1 1 . 0 10 25 10 50 10 6 . 6 10 6 . 6 1 . 0 ) . ( 6 6 2 3 3 3 56
- 57. Solution: New Per unit Reactance of Reactor A: u p j j j j u p A X new . 15 . 0 4 6 . 0 4 5 12 . 0 10 40 10 50 10 6 . 6 10 6 . 6 12 . 0 ) . ( 6 6 2 3 3 u p j j j u p C X new . 24 . 0 2 1 12 . 0 10 25 10 50 10 6 . 6 10 6 . 6 12 . 0 ) . ( 6 6 2 3 3 New Per unit Reactance of Reactor B: u p j j j u p B X new . 12 . 0 1 12 . 0 10 50 10 50 10 6 . 6 10 6 . 6 12 . 0 ) . ( 6 6 2 3 3 New Per unit Reactance of Reactor C: New Per unit Reactance of Feeder: Now, it can be noticed that the impedance of transmission line is given in ohms instead of per unit values. Hence, the formula to find per unit impedance of transmission line is given by: 2 2 . ) ( kV Base kVA Base Z V S Z Z ohms B B ohms u p 57
- 58. Solution: u p j j Z j j Z u p Line u p Line . ) 138 . 0 069 . 0 ( 1478 . 1 ) 12 . 0 06 . 0 ( 56 . 43 50 ) 12 . 0 06 . 0 ( ) 10 6 . 6 ( 10 50 ) 12 . 0 06 . 0 ( ) . ( 2 3 6 ) . ( The equivalent per unit single line diagram with fault at point F is shown as below: 58
- 59. Solution: It can be noticed from the equivalent circuit that j0.1 and j0.12 are in series, similarly, j0.2 and j0.24 are in series. Hence, these can be combined to get one value as j0.22 and j0.44 p.u respectively. The modified equivalent per unit single line diagram with fault at point F is shown as below: The total impedance from the generator to the fault point is given as: ) 44 . 0 22 . 0 44 . 0 22 . 0 15 . 0 ( || 125 . 0 ) 138 . 0 069 . 0 ( ) . ( ) 44 . 0 || 22 . 0 15 . 0 ( || 125 . 0 ) 138 . 0 069 . 0 ( ) . ( j j j j j j j u p Total Z j j j j j u p Total Z 59
- 61. Solution: The short circuit MVA is given by: MVA Z S S u p Base SC 86 . 211 236 . 0 10 50 6 . (b) Short circuit current at the fault point: kA V kVA Actual SC I Actual 532 . 18 43 . 11 10 86 . 211 6 . 6 3 10 86 . 211 10 6 . 6 3 10 86 . 211 3 3 3 3 6 Alternate Approach to Calculate ISC: (By J.A.Laghari) In order to avoid the repetition of steps, we just write the steps to calculate the short circuit current. To do this, first, per unit short circuit current is calculated followed by base current is calculated. Finally, total short circuit current is calculated. The formulae for the currents can be used from the previous examples. 61
- 62. Example: Three 6.6 kV generators A, B and C, each of 10% leakage reactance and MVA ratings 40, 50 and 25 respectively are interconnected electrically as shown in Figure. A three phase feeder is supplied from the bus bar of generator A at a line voltage of 6.6 kV. The feeder has a resistance of 0.06 Ω/phase and an inductive reactance of 0.12 Ω/phase. Estimate: (a) The maximum MVA that can be fed into a symmetrical short circuit at the feeder. (b)The maximum short circuit current at the feeder. 62 This example is the modification of previous example. The aim is to analyse the advantage of reactor. Hence, in this example, reactors are not used. This is modified by J.A.Laghari
- 63. given new new given old u p new kVA base kVA base kV base kV base Z u p Z 2 ) . ( ) . ( Solution: Let 50 MVA and 6.6 kV as chosen as base values. New Per unit Reactance of Generator G1: u p j j j j u p G X new . 125 . 0 4 5 . 0 4 5 1 . 0 10 40 10 50 10 6 . 6 10 6 . 6 1 . 0 ) . ( 6 6 2 3 3 1 (a) Short Circuit MVA at Far End of Feeder: The formula for new per unit impedance is given by: New Per unit Reactance of Generator G2: New Per unit Reactance of Generator G3: u p j j j u p G X new . 1 . 0 1 1 1 . 0 10 50 10 50 10 6 . 6 10 6 . 6 1 . 0 ) . ( 6 6 2 3 3 2 u p j j j j u p G X new . 2 . 0 2 1 1 . 0 25 50 1 1 . 0 10 25 10 50 10 6 . 6 10 6 . 6 1 . 0 ) . ( 6 6 2 3 3 3 63
- 64. Solution: New Per unit Reactance of Feeder: The formula to find per unit impedance of feeder is given by: 2 2 . ) ( kV Base kVA Base Z V S Z Z ohms B B ohms u p u p j j Z j j Z u p Line u p Line . ) 138 . 0 069 . 0 ( 1478 . 1 ) 12 . 0 06 . 0 ( 56 . 43 50 ) 12 . 0 06 . 0 ( ) 10 6 . 6 ( 10 50 ) 12 . 0 06 . 0 ( ) . ( 2 3 6 ) . ( The equivalent per unit single line diagram with fault at point F is shown as below: 64
- 65. Solution: The total impedance from the generator to the fault point is given as: u p j j u p Total Z j j j j u p Total Z j j j j j j j j u p Total Z j j j u p Total Z j j j u p Total Z j j j j u p Total Z j j j j j j u p Total Z j j j j u p Total Z . 18 . 69 194 . 0 18146 . 0 069 . 0 18146 . 0 069 . 0 ) . ( 138 . 0 069 . 0 04346 . 0 138 . 0 069 . 0 04346 . 0 ) ( ) 1 ( ) . ( 138 . 0 069 . 0 1916 . 0 00833 . 0 138 . 0 069 . 0 125 . 0 0667 . 0 125 . 0 0667 . 0 ) . ( 138 . 0 069 . 0 ) 125 . 0 ( || 0667 . 0 ) . ( 138 . 0 069 . 0 ) 125 . 0 ( || 0667 . 0 ) ( ) 1 ( ) . ( 138 . 0 069 . 0 ) 125 . 0 ( || 3 . 0 02 . 0 ) . ( 138 . 0 069 . 0 ) 125 . 0 ( || 1 . 0 2 . 0 1 . 0 2 . 0 ) . ( 138 . 0 069 . 0 ) 125 . 0 ( || ) 1 . 0 ( || ) 2 . 0 ( ) . ( 2 2 2 2 65
- 66. Solution: The short circuit MVA is given by: MVA Z S S u p Base SC 73 . 257 194 . 0 10 50 6 . (b) Short circuit current at the fault point: kA V kVA Actual SC I Actual 548 . 22 43 . 11 10 73 . 257 6 . 6 3 10 73 . 257 10 6 . 6 3 10 73 . 257 3 3 3 3 6 Answer It can be observed that without using the reactors, the short circuit MVA is quite large compared to the short circuit MVA using the reactors. Hence, it is proved that use of reactors will highly limit the short circuit current and will reduce the size of the breakers required. This will result in overall cost reduction of the system. 66 THANK YOU