Fault current Symmetrical

The magnitudes of symmetrical and non-symmetrical fault currents, under different conditions of fault and configurations of faulty circuits, can be determined from Table 13.5, where Z] = Positive phase sequence impedance, measured under symmetrical load conditions. The following values may be considered ... 

Therefore, the level of phase-to-phase asymmetrical faults will he generally of the same order as the three-phase symmetrical faults. The ground faults, however, will he higher than the symmetrical faults. Special care therefore needs he taken while grounding a generator, when they are solidly grounded, particularly to limit the ground fault currents See also Section 20.10.1. 

This is a simple calculation to determine the maximum symmetrical fault level of a system, to select the type of equipment, devices and bus system etc. But to decide on a realistic protective scheme, the asymmetrical value of the fault current must be estimated by including all the likely impedances of the circuit. 

A fault current on a power system is normally asymmetrical as discussed next, and is composed of a symmetrical a.c. component /sar.m.s.) nd an asymmetrical sub-transient d.c. component (Figure 14.5). The forces arising out of /jc aie referred to as electromagnetic and... 

The peak value of a fault current will depend upon the content of the d.c. component. The d.c. component will depend upon the p.f. of the faulty circuit and the instant at which the short-circuit commences on the current wave. (Refer to Figure 13.27, illustrating the variation in asymmetry with the p.f. of the faulty circuit. For ease of application, it is represented as a certain multiple of the r.m.s. value of the symmetrical fault current /sc )... 

This test is conducted to verify the suitability of the equipment to withstand a prospective short-circuit current that may develop on a fault. It may also be termed the steady slate symmetrical fault current or the short-time (withstand current) rating of the equipment. When the equipment is an interrupting device, it is referred to as its symmetrical breaking current. 

This is also known as the asymmetrical breaking current and tends to become the symmetrical r.m.s. value of the fault current / c after almost four cycles from the instant of fault initiation, as discussed in Section 13.4.1(8). 

The d.c. component, /j ., at any instant should be a niininiuni of 50% that of the corresponding peak value of the a.c. component of the symmetrical fault current /.,c, /.,ci. /jctri. at any instant, during the period of the short-circuit condition, /jc should bo S 0..S V2 /jc- Otherwise the asymmetry may be ignored, being insignificant. 

This is the steady-state symmetrical fault current, which the faulty circuit may almost achieve in about three or four cycles from commencement of the short-circuit condition at point Oi (Figure 14.5) and which the interrupting device should be able to break successfully. 

The instant (sub-transient) fault current, /jjgf, through a generator in a symmetrical three-phase system, irrespective of the condition of neutral as defined in Table 13.9 will be... 

Generalizing the above, the symmetrical fault current, when n number of machine.s are operating in parallel. 

This is less than the corresponding symmetrical fault current of 34.2u and incidentally equal to the ground fault current when only one machine is grounded at a time. By this method, the ground fault current can be controlled to any desired level. We have considered a resistance with a view to improving the p.f. of the fault current and thus, making it easier to interrupt. 

Iq = jiiaximum ground fault current /,j = symmetrical ground fault current D, = decrement factor... 

The electrodynamic forces may exist for only three or four cycles (Section 13.4.1(7)), but the mechanical system must be designed for these forces. On the other hand, the main current-carrying system is designed for the symmetrical fault current, 1 (Table 13.7) for one or three seconds according to the system design. For more details refer to Section 13.5. 

Calculation of fault current - rms symmetrical values Calculate the Sub-Transient symmetrical RMS Fault Current Contributions 11.6.1 Calculate the sub-transient peak fault current contributions... 

The maximnm valne of 4000 A for low voltage bnsbars roughly corresponds to the secondary cnrrent of a fnlly loaded 2500 kVA transformer. 2500 kVA is often chosen as the hmit for transformers that feed motor control centres because the fault current that they allow through is typically near to the limit that the mannfactnrers can normally supply, e.g. 80 kA symmetrical rms current. A 2500 kVA transformer with a 6% leakage impedance and a 400 V secondary winding will pass approximately 60 kA of fanlt cnrrent. If the MCC feeds mostly motors then they will collectively contribute some fault current in addition to that from the transformer, see IEC60363 clause 4 and IEC60909 clause... 

It is a particular characteristic in the solution of differential equations involving resistances and inductances that a DC component accompanies the symmetrical AC component. The magnitude of the DC component can equal that of the peak AC component since both are determined by X J. The decay of the DC component can be reasonably slow and is determined by which is a function of X J and the armature winding resistance Ra With machines that have significant values of X J and particularly low values of R, the value of Ta can become relatively high. When Ta is high in relation to Tj and rj it is possible that the initial AC decay is faster than the DC decay. When this happens the AC instantaneous current does not reach zero until several cycles have passed. This puts an extra strain on the circuit breaker and can cause problems at the point when it starts to open to clear the fault current. 

For a prospective symmetrical fault current of 100 kA rms the upstream fault source impedance... 

The actual value of let-through current for a given fuse will depend upon the nature and magnitude of the prospective fault current e.g. asynunetrical or synunetrical. This is because a greater current has to be reached in the symmetrical case than in the asynunetrical case to create the same amount of melting energy. This is due to the shape of the current waveform in the first cycle, which can be seen in Figure 8.1. 

In a 60 Hz system the peak of the fault current will occm in 0.0042 sec (symmetrical) or... 

Step 3. Calculate the prospective symmetrical and asymmetrical fault currents. 

For the selection of fuses the prospective symmetrical rms value of the off-set fault current is calculated as,... 

E shows the synchronous emf behind the synchronous reactance Xsd- Used for calculating the steady state fault current, which will then be fully symmetrical, since all the sub-ffansient and transient effects will have decayed to zero. The emf E will be the ceiling voltage of the exciter since the AVR will have seen a severe depression in terminal voltage and will have forced the exciter to give its maximum possible output. See also sub-sections 7.2.8 and 12.2.2.1. 

The same method that is described for transformers in sub-section 6.3 is used to find E", E and E. Simply replace Xse in the equations by X, X or Xsd as appropriate and assume R and Rse to be equal to zero. Now that the driving voltage has been calculated, it is a simple matter to calculate the symmetrical fault currents. 

CALCULATE THE SUB-TRANSIENT SYMMETRICAL RMS FAULT CURRENT CONTRIBUTIONS... 

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