Tech Te ch Tips T ips by Jef Jowett, Megger
A Look at Fault Currents
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n adequate grounding system is able to deal successfully with fault conditions. Put crudely crudel y, an electrical fault is a breakdown in insulation that sends current curren t in an unwanted direction. direct ion. Te current must return from the fault point to its source. For personnel safety, protection of equipment, and continued functioning of the electrical system, such current must be quickly diverted and normal operation restored. Return to source may be accommodated through the earth, through metallic conducting paths, or both. When soil is the means of return, a voltage gradient develops in the vicinity vicini ty of the groun grounding ding system.Magnit Magnitude ude of this gradient is directly proportional to soil resistivity and inversely to distance. Safe and eective grid design depends on a knowledge of current distribution and what the grid must accommodate under various conditions. A so-called “ground” may may occur by a fallen conductor contacting earth, or the grounding system may be directly involved, as in a ash-over. Single-circuit transmission lines and feeders tend to have faults of low magnitude. Tese can be dicult to clear and also make shock hazards of long duration. On double or multicircuit lines, fault currents are of greater magnitude but also clear more quickly. For grid design, in order to mitigate ground potential rise, touch potentials, and step potentials, it is necessary to begin with the ability to calculate total symmetrical ground fault current, It:
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It = 3Io =
3E ———————————————— 3R f + R 1 + R 2 + R 0 + j(X 1 + X 2 + X 0)
where I0 is total zero sequence fault current at the location, in rms amperes (note that 3I0 is total ground fault current) E is phase-neutral prefault voltage, in rms volts R f is estimated minimum resistance, in ohms (R f = 0 may be assumed) R 1 + jX 1 is positive sequence equivalent fault impedance, in ohms R 2 + jX 2 is negative sequence equivalent fault impedance, in ohms (usually R 1 + jX 1 = R 2 +jX 2 is assumed) R 0 + jX 0 is the zero sequence equivalent fault impedance at the fault location, in ohms For double-line faults, between two phases and ground, the equation becomes: It = 3I0 =
3E(R 2 + jX 2) ———————————————————————————— (R 1 + jX 1) (3R f + R 2 + R 0 + j(X 2 + X 0)) + (R 2 + jX 2) (3R f + R 0 + jX 0)
otal otal symmetrical grid current (I g), the current owing from the grid into surrounding soil, is therefore a component of It with its magnitude dependent upon specic conditions. Te location of the fault has a major impact on current distribution. Consider rst a fault within a substation on the secondary side of a typical delta/wye-grounded transformer. Tere already exists a low impedance metallic path through the grid from the point of the fault and the source, the transformer. transformer. Current need only travel as far as the grid and back to the transformer. In this case, negligible current ows into surrounding soil, so overall grid current and ground potential rise (GPR) are negligible (Ig = 0). However, suppose the fault were on the primary side. Te on-site transformer does not now contribute to the fault current. It all originates from a remote source, beyond the substation. In this case, there is no direct metallic path as previously described; all of the fault current must return through the soil. Symmetrical grid current, that owing from the grid in a radial pattern, comprises 100% of total current (Ig = It). GPR may be high. Spring 201 2010 0 NETA WORLD
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Tese cases demonstrate the extremes of current distribution. What if ground fault current originated from both local and remote sources? An example would be a substation grounded-wye/delta/grounded-wye transformer and a remote source with a grounded-wye secondary. If a fault occurred within the substation, fault current is contributed by both local and remote sources. Tis situation represents a combination of the rst two examples. Current from the local source circulates through the grid while that from the remote source must return through the soil. Symmetrical grid current is now a fraction of total current and is equal to the percent of total ground fault current contributed by the remote source. Finally, imagine that the fault occurs, not within the substation, but at a distant point on a transmission line. Now, current divides and returns through soil to both the substation and the remote source which could be a main station or generating plant. Te substation grounding grid is not involved in the latter of these tow paths. Again, the symmetrical grid current of the substation in question is some fraction of total current. In this case, it equals the percent contributed by the local source. Te preceding examples represent clear-cut cases where no shield wires or feeder neutrals are present. However, this simplied condition often does not apply. Overhead shield wires and feeder neutrals provide convenient parallel paths for fault current. Feeder neutrals in wye-grounded systems are also grounded and connected to the transformer neutral bushing. Shield wires are normally grounded at intervals along the line’s length and at the substations at each end. In addition, other alternate paths may exist through convenient, continuous metal objects like fences, railroad tracks, and grounded pipes. Tis phenomenon of parallel current paths is known as current division, and the IEEE has dubbed this “split factor.”.Terefore, in such a system there are two current divisions to be computed: the one previously described between local and remote sources, and the division between the grounding system and the alternate current paths. Consider a fault on the primary side of the substation transformer: whereas in the simplied example grid current equaled total current, the multiple grounds downstream of the grounded wye now carry the major portion of fault current. Te grid current may now be reduced to something like a third of total current. A fault on the secondary side, however, will circulate current directly through the grid between the fault point and the grounded wye. As in the simplied situation, symmetrical grid current to remote earth is negligible, but there will be a small amount due to induction from live conductors onto the overhead static wire. Because of the short path and low impedance, total fault current will be greater than in the case of a primary fault, and so this needs to be considered when sizing the conductor. Lastly, there is the case of a fault occurring remotely at some distance down the line. Most of the current will return to the source through the feeder neutral. Because of the lower impedance, total current will be greater than for a fault on the primary side. Te farther the fault occurs down the line, the added impedance decreases total current and increases the amount returning through the soil by parallel 2
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pole grounding electrodes. In this case, grid current is typically only a small fraction of total current, perhaps less than ten percent, and a feeder fault is rarely a worst case for the substation grounding system. Fault clearance, then, is not a simple case of current going to ground via the on-site grid. Current divides in known and understood ways, but the details and calculation of magnitude can be complex. A thorough knowledge of current distribution is critical in grid and grounding system design. Te mathematics of some typical examples will be examined in a later edition. Source of information: ennessee Valley Public Power Ass’n; Allen & Hoshall Architects Engineers, Nashville, N; Alexander Publications, Newport Beach, CA. Jerey R. Jowett is Senior Applications Engineer for Megger in Valley Forge, Pennsylvania, serving the manufacturing lines of Biddle, Megger, and Multi-Amp for electrical test and measurement instrumentation. He holds a BS in Biology and Chemistry from Ursinus College. He was employed for 22 years with James G. Biddle Co. which became Biddle Instruments and is now Megger.
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