Symmetrical (L-L-L) fault occurs infrequently, as for example, when a line, which has been made safe for maintenance and/or repairs by clamping all the three phases to earth, is accidently made alive or when, due to slow fault clearance, an earth fault spreads across to the other two phases or when a mechanical excavator cuts quickly through a whole cable. It is an important type of fault in that it results in an easy calculation and generally, a pessimistic answer.
The analysis of symmetrical (L-L-L) faults includes the determination of the voltage at any point (or bus) in the power system network, the current in any branch and value of reactance necessary to limit the fault current to any desired value. Such calculations provide the necessary data for selection of circuit breakers and design of protective scheme.
The circuit breaker MVA breaking capacity is based on 3-phase fault MVA. Since the circuit breakers are manufactured in preferred standard sizes, e.g., 250, 500, 750 MVA, high precision is not required in calculations of 3-phase fault level at a point in a power system. Moreover, the system impedances are also never known accurately.
To make complete and accurate calculations is virtually impossible, and certainly too tedious for operational work.
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It is customary to perform the short circuit analysis under the following simplifying assumptions:
1. Load currents are considered negligible as compared to fault currents.
2. Shunt elements in the transformer model that account for magnetizing current and core loss are neglected. The transformer is represented by a reactance in series, as transformer resistance is quite low in comparison with its reactance.
3. Shunt capacitances of the transmission lines are neglected.
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4. System resistance is neglected and only inductive reactance of the system is taken into account. This assumption cannot be applied in case overhead lines or underground cables of considerable length are included in the network. A transmission line is represented by series reactance (and resistance).
5. The emfs of all the generators are assumed to be equal to 1 ∠0° per unit. This means that the system voltage is at its nominal value and the system is operating on no load at the time of occurrence of fault. The selection of zero phase for one source is arbitrary and convenient. Assuming that all sources are in phase and of the same magnitude is equivalent to neglecting pre-fault load current. When desirable, the load current can be taken into account, at a later stage by superposition.
6. The effect of dc component is accounted for by using correction factors. The correction or multiplying factor for determination of breaking capacity of a circuit breaker depends on the speed of the circuit breaker. For example, a two-cycle circuit breaker might require a factor 1.4 whereas with an eight-cycle breaker a factor 1.0 would be sufficient.
Generator reactances are normally taken as their subtransient values in order to depict the most pessimistic condition. However, if transient current is to be determined, then transient reactances should be used.
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For simple systems, calculations can be made by network reduction technique, which will be discussed here. However, for modern complex systems, ac network analyzers or digital computers are used for fault calculations.
Network Reduction Technique:
Because of the balanced nature of fault and the system, any condition which applies to one phase applies equally to the remaining two phases. Thus the problem is reduced itself to a single phase problem involving a single supply source acting through the equivalent network impedance up to the fault. The equivalent network impedance up to the fault can be obtained by network reduction that involves series- parallel combinations and star/delta or delta/star conversion of reactances.
Various steps involved in the short circuit calculations are given below:
1. Make out a single line diagram of the complete network indicating on each component, its rating, voltage, resistance and reactance.
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2. Choose a common base kVA (or MVA) and convert all the resistances and reactances in per unit values as referred to common base kVA (or MVA).
3. From the single line diagram draw a single line reactance (or impedance) diagram showing one phase and neutral. In this diagram write down the reactances (or impedances) of the elements in per unit values, determined under step 2.
4. Reduce the reactance (or impedance) diagram, by network reduction technique keeping the identity of the fault point intact. Find the reactance of the system as seen from the fault point (Thevenin reactance).
5. Determine the fault current and fault MVA in per unit. Convert these per unit values to actual values.
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6. Retrace the steps of calculations to work out the current and voltage distribution throughout the network.
Fault MVA and Fault Current (Steady-State):
Per unit fault (or short circuit current),
ISC pu = PU voltage at fault point/PU Xequivalent …(4.16)
Per unit fault level (MVA) = √3 per unit fault current x per unit source voltage
or Fault MVA = BAse MVA/PU Xequivalent MVA (lagging) …(4.17)
Fault current ISC = Base MVA x 103/√3 x base kV amperes …(4.18)
Example:
A 3-phase, 10,000 kVA, 11 kV alternator has a sub-transient reactance of 8%. A 3-phase short circuit occurs at its terminals. Determine the fault current and fault MVA.
Solution:
Alternator percentage reactance is based on its own voltage and kVA ratings. Let us choose 10,000 kVA as base kVA and 11 kV as base kV
