[dwstrpkt-170712-01.tif, 1, en_US]
Figure 6-281
Application with Neutral Reactor
As side rated value, the following results:
140 MVA/(√3 · 34.4 kV) = 2343 A
You can thus define the lower limit for the threshold value:
[foschwe5-170712-01.tif, 1, en_US]
If the fault is in the middle of the winding, the minimum ground current will arise, as shown in Figure 6-281.
The driving voltage is:
V
Gnd min
= V
rated,S2
/(2 √3) = 34.5 kV/(2 √3) = 9.96 kV
The following minimum ground current results:
I
Gnd min
= V
Gnd min
/R
Gnd
= 9.96 kV/19.05 Ω = 523 A
With reference to the side rated current, the relationship is:
I
Gnd min
/I
rated,S
= 523 A/2343 A = 0.223
With a safety margin of 2, 0.223/2 = 0.1115 results. Select this value as threshold value (rounded: 0.12 I/
Irated,S).
•
Recommended setting value (_:103) Threshold = 0.12 I/Irated,S
As gradient, the following results:
Protection and Automation Functions
6.43 Restricted Ground-Fault Protection
SIPROTEC 5, Overcurrent Protection, Manual 877
C53000-G5040-C017-8, Edition 07.2017
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