unsymmetrical faults in power system
TRANSCRIPT
POWER SYSTEM ENGINEERING
CE00603-3
UNIT- IIUnbalanced System Analysis
Signal Processing, CE00039-2
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
Signal Processing, CE00039-2
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
Signal Processing, CE00039-2
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
Signal Processing, CE00039-2
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
Signal Processing, CE00039-2
Faults in a Power System
• Symmetrical faults : That fault which gives rise to symmetrical fault currents (i.e. equal faults currents with 120o 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
Signal Processing, CE00039-2
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 120o
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
Signal Processing, CE00039-2
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
Signal Processing, CE00039-2
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
Signal Processing, CE00039-2
Percentage Reactance and Base kVA
Slide 11 of 11
Signal Processing, CE00039-2 Slide 12 of 11
Signal Processing, CE00039-2 Slide 13 of 11
Signal Processing, CE00039-2 Slide 14 of 11
Signal Processing, CE00039-2
Reactor Control of Short-Circuit Currents
• With the fast expanding power system, the fault level (i.e. the power available to flow into a fault) is also rising
• The circuit breakers connected in the power system must be capable of dealing with maximum possible short-circuit currents that can occur at their points of connection
• Generally, the reactance of the system under fault conditions is low and fault currents may rise to a dangerously high value
• If no steps are taken to limit the value of these short-circuit currents, not only will the duty required of circuit breakers be excessively heavy, but also damage to lines and other equipment will almost certainly occur
• 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
Signal Processing, CE00039-2 Slide 17 of 11
Signal Processing, CE00039-2
Steps for Symmetrical Fault Calculations
Slide 18 of 11
Signal Processing, CE00039-2
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
Signal Processing, CE00039-2
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
Signal Processing, CE00039-2 Slide 23 of 11
Signal Processing, CE00039-2 Slide 24 of 11
Signal Processing, CE00039-2 Slide 25 of 11
Signal Processing, CE00039-2 Slide 26 of 11
Signal Processing, CE00039-2 Slide 28 of 11
Signal Processing, CE00039-2
Symmetrical Components in Terms of Phase Currents
Slide 29 of 11
Signal Processing, CE00039-2 Slide 30 of 11
Signal Processing, CE00039-2 Slide 31 of 11
Signal Processing, CE00039-2 Slide 32 of 11
Signal Processing, CE00039-2
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:
IR = 100 ∠30º A ; IY = 50 ∠300º A ; IB = 30 ∠180º A Calculate the positive, negative and zero sequence
currents in the R-line and return current in the neutral wire.
Signal Processing, CE00039-2
Signal Processing, CE00039-2
Example 2
The currents in a 3-phase unbalanced system are :IR = (12 + j 6) A ; IY = (12 − j 12) A ; IB = (−15 + j 10) AThe phase sequence is RYB. Calculate the zero, positive
and negative sequence components of the currents.
Slide 35 of 11
Signal Processing, CE00039-2 Slide 37 of 11
Signal Processing, CE00039-2
Example 3
The sequence voltages in the red phase are as under :ER0 = 100 V ; ER1 = (200 − j 100) V ; ER2 = − 100 VFind the phase voltages ER ,EY and EB .
Slide 38 of 11
Signal Processing, CE00039-2
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
Signal Processing, CE00039-2 Slide 42 of 11
Signal Processing, CE00039-2 Slide 43 of 11
Signal Processing, CE00039-2 Slide 44 of 11
Signal Processing, CE00039-2Slide 47 of 11
Signal Processing, CE00039-2
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 (Z1)(ii) Negative sequence impedance (Z2)(iii) Zero sequence impedance (Z0)
Slide 48 of 11
Signal Processing, CE00039-2 Slide 49 of 11
Signal Processing, CE00039-2
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 ER, EY and EB and the terminal p.d. per phase by VR, VY and VB.
Slide 50 of 11
Signal Processing, CE00039-2
Signal Processing, CE00039-2
Signal Processing, CE00039-2
Signal Processing, CE00039-2
Signal Processing, CE00039-2
Signal Processing, CE00039-2
Signal Processing, CE00039-2
Signal Processing, CE00039-2
Signal Processing, CE00039-2
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 3IR0
Slide 62 of 11
Signal Processing, CE00039-2
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 feederPositive sequence impedance j 1·2 j 1·0Negative sequence impedance j 0·9 j 1·0Zero 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
Signal Processing, CE00039-2
Signal Processing, CE00039-2
Example 2
A 3-phase, 11 kV, 25 MVA generator with X0 = 0·05 p.u., X1 = 0·2 p.u. and X2= 0·2 p.u. is grounded through a reactance of 0·3 Ω. Calculate the fault current for a single line to ground fault.
Signal Processing, CE00039-2
Signal Processing, CE00039-2
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
Signal Processing, CE00039-2 Slide 70 of 11
Signal Processing, CE00039-2 Slide 71 of 11
Signal Processing, CE00039-2
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
Signal Processing, CE00039-2Slide 74 of 11