Generator L-G Fault

[wbd]

Is there a quick back of the envelope type calculation for a generator line to ground fault at the terminals?
Thank You in advance

 

[Cooda76]

It depends upon, what type of grounding it has, As SLG fault is supposed to be less that 3 ph fault
so I=E/(X1+X2+X0+3XN)

this is calculation for fault

 

[wbd]

Thank you. I believe that a fault close in to the generator can have a SLG higher than the 3ph fault. I'm trying to figure out how Cat has on one of their generator data sheets a 3ph fault of 30,926 A and a line to neutral fault of 45,293. My software program has a line to grn of 23540 A. Am I missing something here?

 

[alehman]

If you look at the formula posted by Cooda, you will see the terms X1, X2 and X0. These represent the generator's internal postitive, negative and zero sequence reactances. XN is the external neutral impedance. For this example we'll assume XN=0 (meaning the neutral is solidly grounded).

For a 3-phase fault, I = E / 3X1. If E=1 and X1=1,
then I = 1 / 3 = 0.33.

For generators X1 is normally about the same as X2, but X0 is typically very much less.

If X1=X2, and say X0 = 0.1 X1,
then I = 1 / 2.1 = 0.48

I suspect your program doesn't have the correct X1, X2 and X0 values (you can't assume default values are correct). Also, this could be explained by a relatively high XN.

 

[davidbeach]

A couple of '3's are out of place. The SLG fault is I-fault=3*E/(X1+X2+X0+3*XN) and the 3-phase fault is just I-fault=E/X1. This leaves the relationship between the two the same, but increases the fault current by a factor of 3 in both cases.

 

[wbd]

Thank you. Wouldn't the x1, x2,&x0 values be used for steady state conditions? I'm looking at the first 1/2 cycle so I would think that sub transient and transient reactances would drive the result.

 

[alehman]

david is correct about the misplaced 3's. To be more precise, the values acutally need to be direct-axis subtransient reactances.

 

[Cooda76]

Sorry guys, I missed three.
cheers !

 

[davidbeach]

steady state, with balanced load, is all X1, which for a round rotor machine is Xd. For faults, X1 starts off as Xd'', then becomes Xd', and, if the fault lasts long enough, becomes Xd. Salient pole rotors add the complexity of dealing with Xq. X2 (at all times) is approximately equal to Xd''. X0, which really ought be Z0, or even R0, is all most entirely resistive, and is basically just the resistance of the stator. That's why it is so much lower than the other two.

Because of the way the sequence networks parallel, a decent sized generator (say 2MW) in parallel with a comparable utility service produces an SLG fault value much higher than the sum of the two sources taken individually.

 

[wbd]

Thank you all. I found that on my model I had selected the wrong type of generator and once I corrected that ( to a Synchronous Salient Pole with Armortisseur winding) it matched the manufacturer's data. Does anyone have any good resources for information on the different types of generators and how to recognize which is which?
As always, thanks to everyone and this is a great forum.