High Resistance grounding on Medium Voltage System

[m.smith]

I am working at a plant that has a 6.9kV medium voltage system. The auxiliary transformer is three winding. The primary is 70MVA and the secondary and tertiary are 35MVA in wye configuration with both windings high resistance grounded through a transformer, refer to attached for clarification. I calculated the maximum Ground fault current as 5.67A.

Neutral Grounding Transformer
Vprimary = 4000V
Vsecondary = 240V
Rngr = 1.59 ohms

Transformer Ratio = Vprimary / Vsecondary = 4000V/240V = 16.7
Rprimary resistance = (Transformer Ratio^2)(Rngr^2) = (16.7^2)(1.59^2) = 705 ohms
Ig = Vprimary / Rprimary resistance = 4000V / 705 ohms = 5.67A

5.67A is consistent with a high resistance ground system, but if I use the rule of thumb for the capacitive charging current of 1.5A/MVA the capacitive charging current is approximately 52A. Have I calculated the maximum ground fault current incorrectly?

Note recently a medium voltage motor had a ground fault. All feeders on the medium voltage line-up indicated a ground fault alarms. Once the motor was removed from service all ground fault alarms cleared.

 

[waross]

I get:
R = 1.59 Ohms
E = 240 Volts
I = 151 Amps @ 240 Volts.
151 Amps / 16.7 Ratio = 9.06 Amps @ 4000 Volts.
Why are you squaring everything. Am I losing it in my dotage?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[mgtrp]

Squaring the ratio is right; squaring the resistance is not. It comes about because R=V/I, so when you convert to primary values by the turns ratio you end up with R = (V*n)/(I/n) - the turns ratio is in both the V and the I terms. Expanding becomes R = V/I * n^2. Actual reflected resistance seen on the primary is 442 Ohms in this case.

Bill's calculation method is good, but the final bit of math is off: 151A / 16.7 = 9.05A.

With regards to charging current, I suspect that the actual charging current is nowhere near 52A which is fairly extreme. Perhaps the 1A/MVA value is for a LV system, but doesn't scale well--as you increase the voltage, the insulation thickness goes up so that capacitance goes down. 5-10A of charging current is fairly typical.

Cheers,
mgtrp

 

[waross]

Thanks. Correction made.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[Stephen Nuchia]

How are your feeder relays deriving their ground fault indications? Do you know how the relay scheme was tested at commissioning? I could tell stories ...

Were the lightning arrestors speced for 4kv or 7kv? That's one of the stories that comes to mind. Old plant added new utility feed with resistance grounding, still had arrestors in their substations sized for solid grounding. When they took a phase-to-ground fault it blew them all out, which would be seen as a momentary ground fault at all the feeder breakers.