UNSYMMETRICAL FAULTS IN POWER SYSTEM

It covers all kind of unsymmetrical faults like LG,LL,LLG faults etc.

1. POWER SYSTEM ENGINEERING CE00603-3 UNIT- II Unbalanced System Analysis

2. Contents Introduction Symmetrical Faults on 3-Phase System Percentage Reactance and Base KVA Steps for Symmetrical Fault Calculations Unsymmetrical Faults on 3-Phase system Significance of Positive, Negative and Zero sequence components Average three phase power in terms of symmetrical components Sequence impedances Single Line-to-Ground Fault Line-to-Line Faults Double Line-to-Ground Faults Slide 2 of 11

3. Introduction Short-Circuit “Whenever a fault occurs on a network such that a large current flows in one or more phases, a short circuit is said to have occurred”. Causes of short-circuit ( i ) Internal effects are caused by breakdown of equipment or transmission lines, from deterioration of insulation in a generator, transformer etc . Such troubles may be due to ageing of insulation, inadequate design or improper installation. ( ii ) External effects causing short circuit include insulation failure due to lightning surges, overloading of equipment causing excessive heating; mechanical damage by public etc . Slide 3 of 11

4. Effects of short-circuit When a short-circuit occurs, the current in the system increases to an abnormally high value while the system voltage decreases to a low value. Short-circuit causes excessive heating which may result in fire or explosion. Sometimes short-circuit takes the form of an arc and causes considerable damage to the system Low voltage created by the fault has a very harmful effect on the service rendered by the power system . If the voltage remains low for even a few seconds, the consumers’ motors may be shut down and generators on the power system may become unstable . Slide 4 of 11

5. Short-Circuit Current Calculations The calculations of the short-circuit currents are important for the following reasons: ( i ) A short-circuit on the power system is cleared by a circuit breaker or a fuse . It is necessary, therefore, to know the maximum possible values of short-circuit current so that switchgear of suitable rating may be installed to interrupt them. ( ii ) The magnitude of short-circuit current determines the setting and sometimes the types and location of protective system. ( iii ) The magnitude of short-circuit current determines the size of the protective reactors which must be inserted in the system so that the circuit breaker is able to withstand the fault current. ( iv ) The calculation of short-circuit currents enables us to make proper selection of the associated apparatus ( e.g . bus-bars, current transformers etc.) so that they can withstand the forces that arise due to the occurrence of short circuits. Slide 5 of 11

6. Faults in a Power System Symmetrical faults : That fault which gives rise to symmetrical fault currents ( i.e . equal faults currents with 120 o displacement) is called a symmetrical fault. Example : when all the three conductors of a 3-phase line are brought together simultaneously into a short-circuit condition. Unsymmetrical faults: Those faults which give rise to unsymmetrical currents ( i.e . unequal line currents with unequal displacement) are called unsymmetrical faults. Single line-to-ground fault Line-to-line fault Double line-to-ground fault Common : a short-circuit from one line to ground Slide 6 of 11

7. Symmetrical Faults on 3-Phase System “That fault on the power system which gives rise to symmetrical fault currents (i.e. equal fault currents in the lines with 120 o displacement) is called a symmetrical fault ” symmetrical fault rarely occurs in practice The symmetrical fault is the most severe and imposes more heavy duty on the circuit breaker Slide 7 of 11

8. Limitation of Fault Current 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 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 Slide 8 of 11

9. Percentage Reactance Slide 9 of 11

10. Conclusion For instance, if the percentage reactance of an element is 20% and the full-load current is 50 A, then short-circuit current will be 50 × 100/20 = 250 A when only that element is in the circuit. It may be worthwhile to mention here the advantage of using percentage reactance instead of ohmic reactance in short-circuit calculations. Percentage reactance values remain unchanged as they are referred through transformers, unlike ohmic reactances which become multiplied or divided by the square of transformation ratio. This makes the procedure simple and permits quick calculations . Slide 10 of 11

11. Percentage Reactance and Base kVA Slide 11 of 11

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15. Short-Circuit kVA Slide 15 of 11

16. Reactor Control of Short-Circuit Currents With the fast expanding power system, the fault level ( i.e. the power available to flow into a fault) is also rising The circuit breakers connected in the power system must be capable of dealing with maximum possible short-circuit currents that can occur at their points of connection Generally, the reactance of the system under fault conditions is low and fault currents may rise to a dangerously high value If no steps are taken to limit the value of these short-circuit currents, not only will the duty required of circuit breakers be excessively heavy , but also damage to lines and other equipment will almost certainly occur Solution: additional reactance's known as reactors are connected in series with the system at suitable points A reactor is a coil of number of turns designed to have a large inductance as compared to its ohmic resistance Slide 16 of 11

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18. Steps for Symmetrical Fault Calculations Slide 18 of 11

19. Example 1 Fig. shows the single line diagram of a 3-phase system. The percentage reactance of each alternator is based on its own capacity. Find the short-circuit current that will flow into a complete 3-phase short-circuit at F Slide 19 of 11

20. Solution Slide 20 of 11

21. Example 2 A 3-phase, 20 MVA, 10 kV alternator has internal reactance of 5% and negligible resistance. Find the external reactance per phase to be connected in series with the alternator so that steady current on short-circuit does not exceed 8 times the full load current. Slide 21 of 11

22. Solution Slide 22 of 11

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27. Operator ‘a’ Slide 27 of 11

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29. Symmetrical Components in Terms of Phase Currents Slide 29 of 11

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33. Example 1 In a 3-phase, 4-wire system, the currents in R, Y and B lines under abnormal conditions of loading are as under: I R = 100 ∠ 30º A ; I Y = 50 ∠ 300º A ; I B = 30 ∠ 180º A Calculate the positive, negative and zero sequence currents in the R-line and return current in the neutral wire.

34. Example 2 The currents in a 3-phase unbalanced system are : I R = (12 + j 6) A ; I Y = (12 − j 12) A ; I B = (−15 + j 10) A The phase sequence is RYB. Calculate the zero, positive and negative sequence components of the currents. Slide 35 of 11

35. Solution Slide 36 of 11

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37. Example 3 The sequence voltages in the red phase are as under : E R0 = 100 V ; E R1 = (200 − j 100) V ; E R2 = − 100 V Find the phase voltages E R ,E Y and E B . Slide 38 of 11

38. Solution Slide 39 of 11

39. Example 4 A balanced star connected load takes 90 A from a balanced 3-phase, 4-wire supply. If the fuses in the Y and B phases are removed, find the symmetrical components of the line currents ( i) before the fuses are removed (ii) after the fuses are removed Slide 40 of 11

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44. Example 5 Slide 45 of 11

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47. Sequence Impedances Each element of power system will offer impedance to different phase sequence components of current which may not be the same. Therefore, in unsymmetrical fault calculations, each piece of equipment will have three values of impedance—one corresponding to each sequence current viz.   ( i ) Positive sequence impedance ( Z 1 ) ( ii ) Negative sequence impedance ( Z 2 ) ( iii ) Zero sequence impedance ( Z ) Slide 48 of 11

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49. Analysis of Unsymmetrical Faults In the analysis of unsymmetrical faults, the following assumptions will be made : ( i) The generated e.m.f. system is of positive sequence only. ( ii) No current flows in the network other than due to fault i.e. load currents are neglected. ( iii) The impedance of the fault is zero. ( iv) Phase R shall be taken as the reference phase. In each case of unsymmetrical fault, e.m.fs’ per phase are denoted by E R , E Y and E B and the terminal p.d. per phase by V R , V Y and V B . Slide 50 of 11

50. Single Line-to-Ground Fault

51. Line-to-Line Fault Slide 55 of 11

52. Double Line-to-Ground Fault

53. Observations on Faults Positive sequence currents are present in all type of faults Negative sequence currents are present in all unsymmetrical faults Zero sequence currents are present when the neutral of the system is grounded and faults also involves the ground and magnitude of the neutral current is equal to 3I R0 Slide 62 of 11

54. Example 1 A 3-phase, 10 MVA, 11 kV generator with a solidly earthed neutral point supplies a feeder. The relevant impedances of the generator and feeder in ohms are as under : Generator feeder Positive sequence impedance j 1·2 j 1·0 Negative sequence impedance j 0·9 j 1·0 Zero sequence impedance j 0·4 j 3·0 If a fault from one phase to earth occurs on the far end of the feeder, calculate ( i ) the magnitude of fault current ( ii ) line to neutral voltage at the generator terminal

55. Solution

56. Example 2 A 3-phase, 11 kV, 25 MVA generator with X = 0·05 p.u., X 1 = 0·2 p.u. and X 2 = 0·2 p.u. is grounded through a reactance of 0·3 Ω . Calculate the fault current for a single line to ground fault.

57. Solution

58. Example 3 Two 11 KV, 12 MVA, 3f, star connected generators operate in parallel. The positive, negative and zero sequence reactance's of each being j0.09, j0.05 and j0.04 pu respectively. A single line to ground fault occurs at the terminals of one of the generators. Estimate ( i ) the fault current (ii) current in grounding resistor (iii) voltage across grounding resistor. Slide 69 of 11

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61. Example 4 A 50 MVA, 11 kV three-phase alternator was subjected to different types of faults. The fault currents are as under : 3-phase fault = 2000 A ; Line-to-Line fault = 2600 A ; Line-to-ground fault = 4200 A The generator neutral is solidly grounded. Find the values of the three sequence reactance's of the alternator. Ignore resistances . Slide 72 of 11

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64. Any Queries? Slide 75 of 11