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A design query on earthing trafo.
An earthing trafo installed on a 11 kv switchboard is rated for 381 kva , 11 kv/ 110v in a star – broken delta arrangement. A resistor rated 1.14 ohms is connected across the broken limb. …

A design query on earthing trafo.
An earthing trafo installed on a 11 kv switchboard is rated for 381 kva , 11 kv/ 110v in a star – broken delta arrangement. A resistor rated 1.14 ohms is connected across the broken limb.
How can the max earth fault current be calculated using the above data? Can the max earth fault current be assumed to be the full load current of the earthing trafo?

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[Hamid]

With above connection, the secondary current can be calculated by multiplying the transformer primary current times the transformer ratio. This is the current through the grounding resistor, and its value establishes the continuous current rating of the grounding resistor. The voltage across the resistor under ground-fault conditions is 1.73 times the secondary volltage of the grounding transformer bank (190V for 110V rating). Therefore, I think correct answer is 5 A.

[GLAVIZH]

the voltage across 1.14 Ohm is not 190 V . It is 110 V . So the earth fault current is 2.89 Amp. not 5 Amp . 5 Amp. is incorrect . If you consider 167 Amp. in secondary side and 1.67 Amp. in primary side , then ampere-turn equality law will not be satisfied. N1/N2 is not 100 , it is 100/?3 . Belive me . UL1/UL2 is (?3 N1/N2) = 100 . So N1/N2 is 57.73 . In similar riddles my dear friend HAMID has the same mistake .

[Benyamin]

I think final conclusion as 5 Ampere is correct . 110/1.14 =96.5 A (96.5/57.73)*3=5 A Althogh the voltage across 1.14 Ohm is 110 V not 190 V , but conversion ratio is 57.73(not 100)

[Marcio macedo]

5 Amps is correct. Probably someone considered the system capacitive charging current around 1,7 A/per phase. So, considered 03 single phase transformers 11000/110 V (ratio 100). Secondary current during bolted ground fault would be 170A under 190V(3 x 11000 /sq3 x100) It means a 1,12 Ohms resistor Safe values for transformers would be 18kVA and 32kW for resistor. Voltage protection relay if installed at broken delta terminals must consider percentage of 190V for alarm and tripping. Limiting the ground fault current to a value slightly greater than capacitive charging current avoids dangerous overvoltages. Max system ground current would be 3 x 1,7A = 5,1 A(remember zero sequence).

[Fredelito Dela Cruz]

The transformation ratio is calculated as primary winding phase voltage divided by secondary winding (in case of grounding transformer, a tertiary winding) phase voltage. The primary winding phase voltage is 11000V/1.73 and the tertiary winding phase voltage is 110V/3. The transformation ratio for the given system is (11000V/1.73)/(110V/3) which is equal to 173.2. The current on 1.14 ohm resistor is (110V/1.14ohm)=96.5A. The current on primary winding is 96.5A / 173.2 = 0.557A per phase. The total ground fault current is 0.557A x 3 = 1.67A. Unless the capacitive charging current is more than 3.33A, the 5A fault current is too big.

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