Tech Tips
by Jeff Jowett, Megger
A Look at Fault Currents
A
n adequate grounding system is
able to deal successfully with fault
conditions. Put crudely, an electrical fault is a breakdown in insulation that
sends current in an unwanted direction. The
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 of the grounding system. Magnitude
of this gradient is directly proportional to
soil resistivity and inversely to distance.
Safe and effective grid design depends on a
knowledge of current distribution and what
the grid must accommodate under various
conditions. A so-called “ground” may occur
by a fallen conductor contacting earth, or the
grounding system may be directly involved,
as in a flash-over. Single-circuit transmission lines and feeders tend to have faults
of low magnitude. These can be difficult 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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I = 3Io =
t
3E
————————————————
3Rf + R1 + R2 + R0 + j(X1 + X2 + X0)
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
Rf is estimated minimum resistance, in ohms
(Rf = 0 may be assumed)
R1 + jX1 is positive sequence equivalent fault impedance, in ohms
R2 + jX2 is negative sequence equivalent fault impedance, in ohms
(usually R1 + jX1 = R2 +jX2 is assumed)
R0 + jX0 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(R2 + jX2)
————————————————————————————
(R1 + jX1) (3Rf + R2 + R0 + j(X2 + X0)) + (R2 + jX2) (3Rf + R0 + jX0)
Total symmetrical grid current (Ig), the current flowing from the grid
into surrounding soil, is therefore a component of It with its magnitude
dependent upon specific conditions. The location of the fault has a major
impact on current distribution. Consider first a fault within a substation
on the secondary side of a typical delta/wye-grounded transformer. There
already exists a low impedance metallic path through the grid from the
point of the fault and the source, the transformer. Current need only travel
as far as the grid and back to the transformer. In this case, negligible current
flows 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. The 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
flowing from the grid in a radial pattern, comprises 100% of total current
(Ig = It). GPR may be high.
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These 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. This situation represents
a combination of the first 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. The 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.
The preceding examples represent clear-cut cases where
no shield wires or feeder neutrals are present. However, this
simplified 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. This phenomenon of parallel current paths
is known as current division, and the IEEE has dubbed this
“split factor.”.Therefore, 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 simplified example grid current
equaled total current, the multiple grounds downstream
of the grounded wye now carry the major portion of fault
current. The 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
simplified 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. The farther the fault occurs down
the line, the added impedance decreases total current and
increases the amount returning through the soil by parallel
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NETA WORLD Spring 2010
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. The mathematics of some typical examples will be
examined in a later edition.
Source of information: Tennessee Valley Public Power
Ass’n; Allen & Hoshall Architects Engineers, Nashville,
TN; Alexander Publications, Newport Beach, CA.
Jeffrey 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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