POWER SYSTEM ENGINEERING

1. POWER SYSTEM ENGINEERING EET301/4 EET301 POWER SYSTEM ENGINEERING

2. CHAPTER 6 SYMMETRICAL FAULT IN POWER SYSTEM EET301 POWER SYSTEM ENGINEERING

3. On completion of this lesson, a student should be able to: Ability to explain and evaluate symmetrical fault and protection in power system EET301 POWER SYSTEM ENGINEERING

4. TOPIC OUTLINE 6.1 Introduction 6.2 Cause of Faults 6.3 Type of Faults 6.4 Effect of Faults 6.5 Fault Current Transient in Synchronous Machines 6.6 Balance Three-phase Fault in Power System EET301 POWER SYSTEM ENGINEERING

5. 6.1 INTRODUCTION • A faultin a circuit is any failure that interferes with the normal flow of current to the load. • A fault takes place when two or more conductors come in contact with each other when normally they operate with a potential difference between them. • In most faults, a current path forms between two or more phases, or between one or more phases and the neutral (ground). • A fault represents a structural network change equivalent with that caused by the addition of an impedance at the place of fault. • If the fault impedance is zero, the fault is referred as the bolted fault or the solid fault. • Since the impedance of a new path is usually low, an excessive current may flow. EET301 POWER SYSTEM ENGINEERING

6. 6.2 Cause of Faults A fault may occur on a power system due to a number of reasons. • Generator failure is caused by insulation breakdown between turns in the same slot or between the winding and the steel structure of the machines. The same can take place in transformer. The breakdown is due to insulation deterioration combined with switching and over voltages. • Overhead lines are constructed of bare conductors. Wind, sleets, trees, cranes, kites, airplanes, birds, or damage of supporting structure are causes for accidental faults on overhead line. EET301 POWER SYSTEM ENGINEERING

7. Cause of Faults (cont..) • Contamination of insulators and lightning over voltages will in general results in short-circuit faults. • Deterioration of insulation in underground cables results in short circuit faults. This is mainly attributed to ageing combined with overloading. EET301 POWER SYSTEM ENGINEERING

8. 6.3 TYPE OF FAULTS • Fault may occur at different points in a power system. • Faults that occur on power system are broadly classified as: • Symmetrical fault • Unsymmetrical faults EET301 POWER SYSTEM ENGINEERING

9. TYPE OF FAULTS Symmetrical fault • All three phases of a transmission line are shorted together Unsymmetrical Faults • Single line-to-ground faults can also occur if one phase of the line breaks and comes into contact with the ground or if insulators break. • Two phases of a line may touch, or flashover may occur between two phases – a line-to-line fault. • When two lines touch each other and also touch the ground, the fault is called a double line-to-ground fault. EET301 POWER SYSTEM ENGINEERING

10. TYPE OF FAULTS Three-phase fault Single line-to-ground fault Line-to-line fault Three-phase to ground fault Zf Double line-to-ground fault Three-phase to ground fault with fault impedance EET301 POWER SYSTEM ENGINEERING

11. 6.4 EFFECT OF FAULTS Faults can damage or disrupt power system in several ways. • Faults give rise to abnormal operating conditions, usually excessive voltages and currents at certain points on the system. Large voltage stress insulation beyond their breakdown value while large currents results in overheating of power system component. • Faults can cause the three-phase system to become unbalance with the result that the three-phase equipment operates improperly. • Sometimes, faults block the flow of power. • Faults can cause the system become unstable. Upon the occurrence of the fault, the faulty section should be disconnected as rapidly as possible to in order that the normal operation of the rest of the system is nor affected. If this is no done, the equipment may be damaged and the power supply is disrupted. EET301 POWER SYSTEM ENGINEERING

12. FAULT ANALYSIS Fault analysis is also known as short-circuit study or short-circuit analysis. The important of fault analysis includes: • To determine the value of voltages and currents at different points of the system during the fault. • Determination of the ratings of the required circuit breaker. • Selection of appropriate schemes of protective relaying. Thus, the purpose of fault analysis is to save the system from abnormal conditions within minimum time. EET301 POWER SYSTEM ENGINEERING

13. 6.5 FAULT CURRENT TRANSIENTS IN SYNCHRONOUS MACHINES • Consider the series RL circuit below. • The closing of switch at t = 0 represents to a first approximation a three phase short circuit at the terminals of an unloaded synchronous machine • For simplicity, assume zero fault impedance, so the short circuit is a solid or bolted fault • Writing a KVL equation of the circuit EET301 POWER SYSTEM ENGINEERING

14. FAULT CURRENT TRANSIENTS IN SYNCHRONOUS MACHINES • The total fault current, called the asymmetrical fault current can be written as • Where the ac fault current (also called symmetrical or steady state current fault current) • And dc offset current EET301 POWER SYSTEM ENGINEERING

15. FAULT CURRENT TRANSIENTS IN SYNCHRONOUS MACHINES • When a symmetrical 3-phase fault occurs at the terminals of a synchronous generator, the resulting current flow in the phases of the generator can appear as shown. • The current can be represented as a transient DC component added on top of a symmetrical AC component. • Therefore, while before the fault, only AC voltages and currents were present within the generator, immediately after the fault, both AC and DC currents are present. EET301 POWER SYSTEM ENGINEERING

16. FAULT CURRENT TRANSIENTS IN SYNCHRONOUS MACHINES Symmetrical AC component of the fault current: • Subtransient: first cycle or so after the fault – AC current is very large and falls rapidly; • Transient: current falls at a slower rate; • Steady-state: current gets back to normal. EET301 POWER SYSTEM ENGINEERING

17. FAULT CURRENT TRANSIENTS IN SYNCHRONOUS MACHINES • It is possible to observe the three periods of fault current if the r.m.s magnitude of the AC component current is plotted as a function of time on a semilogarithmic scale. • It is possible to determine the time constants for the three period. EET301 POWER SYSTEM ENGINEERING

18. FAULT CURRENT TRANSIENTS IN SYNCHRONOUS MACHINES • The time varying reactance are typically approximated using three different value, each valid for a different time period: Xd” = direct axis subtransient reactance Xd’ = direct axis transient reactance Xd = direct axis synchrounous reactance • The instantaneous a.c. fault current can be written as • With Eg is the rms line to neutral prefault terminal voltage of the unloaded synchronous machine EET301 POWER SYSTEM ENGINEERING

19. FAULT CURRENT TRANSIENTS IN SYNCHRONOUS MACHINES • While the ac rms fault current • The duration of current is determine by the following time constant: Td” = direct axis short circuit subtransient time constant Td’ = direct axis short circuit transient time constant TA = armature time constant EET301 POWER SYSTEM ENGINEERING

20. FAULT CURRENT TRANSIENTS IN SYNCHRONOUS MACHINES • At t = 0, when the fault occurs, the rms value of iac(t) which is called the rms subtransient fault current, I” is • The rms transient fault current, I’ is given by • And the steady state fault current, I is given by EET301 POWER SYSTEM ENGINEERING

21. FAULT CURRENT TRANSIENTS IN SYNCHRONOUS MACHINES • In addition to the ac fault current, each phase has a different dc offset. • The maximum dc offset in any one phase, which occurs when α = 0 is • The rms asymmetrical fault current with maximum dc offset EET301 POWER SYSTEM ENGINEERING

22. EXAMPLE 6.1 A 500 MVA, 20 kV, 60 Hz synchronous generator with reactancesXd” = 0.15, Xd’ = 0.24, Xd = 1.1 p.u. and time constant Td” = 0.035, Td’ = 2.0, TA = 0.2 seconds is connected to a circuit breaker. The generator is operating at 5% above rated voltage and at no load when a bolted three-phase short circuit occurs on the load side of the breaker. The breaker interrupts the fault 3 cycles after fault inception. Determine • The subtransient fault current in per unit and kA rms • Maximum dc offset as a function of time • Rms asymmetrical fault current EET301 POWER SYSTEM ENGINEERING

23. SOLUTION • The no load voltage before fault occurs in Eg = 1.05 p.u. The subtransient fault current that occurs in each of three phases is The generator base current is The rmssubtransient fault current EET301 POWER SYSTEM ENGINEERING

24. SOLUTION • The maximum dc offset that may occur in any one phase • The rms ac fault current at 3 cycles = 0.05 s is EET301 POWER SYSTEM ENGINEERING

25. SOLUTION The rmsassymetrical fault current EET301 POWER SYSTEM ENGINEERING

26. 6.6 BALANCE THREE-PHASE FAULT IN POWER SYSTEM In order to calculate the subtransient fault current for a three-phase short circuit in a power system, we make the following assumptions: • Transformers are represented by their leakage reactances. Winding resistances, shunt admittances and Δ-Y phase shifts are neglected. • Transmission lines are represented by their equivalent series reactances. • Synchronous machines are represented by constant-voltage sources behind subtransientreactances. Armature resistance, saliency and saturation are neglected. • All non-rotating impedance loads are neglected. • Induction motor either neglected (especially for small motors rated less than 50hp) or represented in the same manner as synchronous machines. EET301 POWER SYSTEM ENGINEERING

27. BALANCE THREE-PHASE FAULT IN POWER SYSTEM • A fault represents a structural network change equivalent with that caused by the addition of an impedance at the place of fault. If the fault impedance is zero, the fault is referred to as the bolted fault or the solid fault. • The faulted network can be solved conveniently by the Thevenin’s method. The following figure shows a single-phase representation of a power system with a fault. EET301 POWER SYSTEM ENGINEERING

28. BALANCE THREE-PHASE FAULT IN POWER SYSTEM • The supply source, a generator, or in power system, many generators operating in parallel in many different power stations. • The impedance, ZTh is made up of total per-unit impedance from the source to the fault point by circuit analysis. Hence, base current; fault current (p.u); fault current (Ampere); fault level; EET301 POWER SYSTEM ENGINEERING

29. PROCEDURE FOR CALCULATING FAULT CURRENT AND FAULT LEVEL Procedure for calculating fault current, IF and fault capacity, SF in power system: • Draw a single-line diagram of the complete network. On this diagram, indicate the rating, voltage, resistance and reactance of all generators, transformers, transmission lines, loads, etc. • Select a common base SB (kVA or MVA) and convert all impedances to per-unit values on the same voltamperes base SB. • Corresponding to the single-line diagram of the network, draw the reactance (or impedance) diagram showing one phase of the system and the neutral. On this diagram, indicate all the per-unit resistances and per-unit reactances of the components calculated in step 2. EET301 POWER SYSTEM ENGINEERING

30. PROCEDURE FOR CALCULATING FAULT CURRENT AND FAULT LEVEL • Calculate the total per-unit impedance from the source to the fault point by circuit analysis. This may involves series-parallel combination, star-delta or delta-star transformations. It should be noted that this total per-unit impedance is the Thevenin impedance of the networks as seen from the fault point. • Determine the fault current and the fault level. EET301 POWER SYSTEM ENGINEERING

31. EXAMPLE 6.2 The synchronous generator in Figure 6.2 is operating at rated MVA, 0.95 p.f. lagging and at 5% above rated voltage when a bolted three-phase short circuit occurs at bus 1. Calculate the per-unit values of • subtransient fault current • subtransient generator and motor currents, neglecting prefault current • subtransient generator and motor currents including prefault current G 1 2 M T1 T2 j20 Ω 100 MVA 138/13.8 kV X = 0.1 p.u. 100 MVA 13.8 kV X” = 0.2 p.u. 100 MVA 13.8 kV X” = 0.15 p.u. 100 MVA 13.8/138 kV X = 0.1 p.u. Figure 6.2 EET301 POWER SYSTEM ENGINEERING

32. SOLUTION (a) Using 100 MVA, the base impedance in the zone of the transmission line is The reactance of the generator, motor and transformer is already given on the system base, so it will not have to change. and EET301 POWER SYSTEM ENGINEERING

33. SOLUTION • Single line diagram EET301 POWER SYSTEM ENGINEERING

34. SOLUTION • Calculate total per-unit impedance from source to the fault point EET301 POWER SYSTEM ENGINEERING

35. SOLUTION • the prefault voltage at the generator terminals is • The subtransient fault current is then j0.11565 EET301 POWER SYSTEM ENGINEERING

36. SOLUTION (b) using current division method, the subtransient generator and motor currents (neglecting prefault current) Generator: IG” IM” IF” – VF + Motor: EET301 POWER SYSTEM ENGINEERING

37. SOLUTION (c) The generator base current is And the prefault generator current is EET301 POWER SYSTEM ENGINEERING

38. SOLUTION • The subtransient generator and motor currents including prefault current Current from the generator: Current from the motor: EET301 POWER SYSTEM ENGINEERING

39. EXAMPLE 6.3 The one line diagram of a simple three-bus power system is shown in the Figure 6.3. Each generator is represented by an emf behind the transient reactance. All impedances are expressed in per unit in a common 100 MVA base, and for simplicity, resistances are neglected. The following assumption are made: • shunt capacitances are neglected and the system is considered on no load. • all generators are running at their rated voltage and rated frequency with their emf in phase • Determine the fault current when a balanced three-phase fault occurs on bus 3 • Repeat (a) with a fault impedance of Zf = j0.16 p.u. EET301 POWER SYSTEM ENGINEERING

40. EXAMPLE 6.3 20 kV, j0.1 p.u. 20 kV j0.2 p.u. G2 G1 20/220 kV j0.1 p.u. 20/220 kV j0.2 p.u. T1 T2 j0.8 1 2 j0.4 j0.4 3 EET301 POWER SYSTEM ENGINEERING Figure 6.3

41. SOLUTION • a) single line diagram j0.8 j0.1 j0.2 j0.4 j0.4 j0.1 j0.2 F EET301 POWER SYSTEM ENGINEERING

42. SOLUTION • Using delta to star transformation j0.8 j0.2 j0.4 j0.4 j0.4 F EET301 POWER SYSTEM ENGINEERING

43. SOLUTION j0.2 j0.2 j0.2 j0.4 j0.1 F the total per-unit impedance EET301 POWER SYSTEM ENGINEERING

44. SOLUTION • The base current on the fault point • And the fault current is j0.34 p.u. EET301 POWER SYSTEM ENGINEERING

45. SOLUTION • b) with the fault impedance of j0.16Ω, the fault current is j0.5 p.u. EET301 POWER SYSTEM ENGINEERING

46. EXAMPLE 6.4 A 25 MVA, 11 kV generator with Xd” = 20% is connected through a transformer to a bus that supplies three identical motors as shown in Figure 6.4. Each motor has Xd” = 25% and Xd’ = 30% on a base of 5 MVA, 6.6 kV. The three-phase rating of the step up transformer is 25 MVA 11/66 kV with a leakage reactance of 10% and that of the step down transformer is 25 MVA, 66/6.6 kV with a leakage reactance of 10%. The bus voltage of the motors is 6.6 kV when a three-phase fault occurs at the point F. For the specified fault, calculate the subtransient current at the fault point. Given the reactance of transmission line is 15% on a base 25 MVA, 66 kV. Choose a system base of 25 MVA and 11 kV at generator side and assume that the system is operating on no load. Figure 6.4 EET301 POWER SYSTEM ENGINEERING F

47. SOLUTION • The base voltage for generator, transmission line and motor side are 11 kV, 66 kV and 6.6 kV respectively. • Subtransient reactance for each motor, • For generator XG = j0.2 p.u. • For transformer XT1 = XT2 = j0.1 p.u. • For transmission line Xline = j0.15 p.u. EET301 POWER SYSTEM ENGINEERING

48. SOLUTION • Reactance diagram EET301 POWER SYSTEM ENGINEERING

49. SOLUTION • Reduce the circuit to obtain the Thevenin impedance EET301 POWER SYSTEM ENGINEERING

50. SOLUTION • Reduce the circuit to obtain the Thevenin impedance j0.237 p.u. EET301 POWER SYSTEM ENGINEERING