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Technical description
3 Supporting functions
3.16 Earth-fault location
VM259.EN007
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183
3.16.
Earth-fault location
The device includes a sophisticated stand-alone earth-fault
location algorithm. The algorithm can locate an earth-fault
accurately in radically operated compensated earthed
networks.
The function can locate a fault only if the fault resistance is
low, say less than 50 ohms. The fault location is given in
reactance value. This value can then be exported, for example,
with event to a DMS (Distribution Management System). The
system can then localize the fault and display it on a map.
The fault location must be used in the incoming bay of the
substation. Therefore, the fault location is obtained for the
whole network with just one device. This is very cost-effective
upgrade of an existing system.
Please note also that the earth-fault location function requires
a change during an earth-fault. This change is done by
switching the secondary resistor of the compensation coil on or
off. The fault should be allowed to be on at least 200 ms, of
which 100 ms without the resistor. The resistor change can be
done by using the logic functionality of the device.
The reactance value is converted to distance in the DMS. The
following formula is used:
2
1
*
3
X
X
Xo
X
s
Where,
s = distance in km
X = reactance calculated by the device
X
0
= zero sequence reactance per kilometre of the line
X
1
= positive sequence reactance per kilometre of the line
X
2
= negative sequence reactance per kilometre of the line
The algorithm functions in the following order:
1.
The needed measurements (phase currents and voltages)
are continuously available.
2.
The fault distance calculation can be triggered in two ways:
by switching ON or OFF the secondary resistor (that is, by
using a digital input) or the calculation can be triggered if
there is a change in earth fault or negative sequence current
3.
The fault phase is identified by that the voltage of the
faulted phase is decreased at least by half.
4.
The fault distance is calculated by dividing the change of
the voltage by the change of the negative sequence current.
5.
Only the imaginary part is used, so then the reactance is
solved.