Fault current Peak symmetrical

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 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 )... 

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. 

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. 

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

Hence the peak value of the symmetrical fault current... 

These reactances are measured by creating a fault, similar to the method discussed in Section 14.3.6. The only difference now is that the fault is created in any of the phases at an instant, when the applied voltage in that phase is at its peak, i.e. at Vni- so that the d.c. component of the short-circuit current is zero and the waveform is symmetrical about its axis, as shown in Figure 13.19,... 

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