#### Engineering_Is_Fun

##### Member

- Location
- Chicago

- Occupation
- Electrical Engineer

1. Why is my SLG fault higher than my 3ph fault?

2. What value would you assume for the X/R of an infinite bus?

- Thread starter Engineering_Is_Fun
- Start date

- Location
- Chicago

- Occupation
- Electrical Engineer

1. Why is my SLG fault higher than my 3ph fault?

2. What value would you assume for the X/R of an infinite bus?

- Location
- United States

Bump

- Location
- New York, 40.7514,-73.9925

We typically set the X/R the same for 3ph and SLG

- Location
- Wisconsin

- Occupation
- Engineer

This can happen when the conductors on the secondary side are relatively short.1. Why is my SLG fault higher than my 3ph fault?

This is not an uncommon occurrence on unit substations where the LV switchgear is bussed directly to the transformer output.

Types of faults

So you aren't necessarily doing anything wrong.

- Location
- Oklahoma City

- Occupation
- Engineer

1. If you're close to a ground fault current source, such as the wye side of a delta-wye transformer, then it's conceivable that SLG fault current is higher.

1. Why is my SLG fault higher than my 3ph fault?

2. What value would you assume for the X/R of an infinite bus?

2. A sufficiently high X/R will do. 9999 is good. I was going to say 100, but those fault-current angles are both close enough to 90 degrees to essentially be the same.

- Location
- Chicago

- Occupation
- Electrical Engineer

I contacted SKM tech support and they said that the X//R of 9999 is assuming a very small X and an infinitely small R.

I.E. X = 0.001 and R = 0.00000000001

2. A sufficiently high X/R will do. 9999 is good. I was going to say 100, but those fault-current angles are both close enough to 90 degrees to essentially be the same.

Thank you! This was a big help.

- Location
- Springfield, MA, USA

That makes sense. The 'infinite bus' approximation is the assumption that there is _no_ impedance on the primary side of the transformer.I contacted SKM tech support and they said that the X//R of 9999 is assuming a very small X and an infinitely small R.

I.E. X = 0.001 and R = 0.00000000001

X=0 R=0 would give an undefined X/R (0/0), so they pick some really small numbers that are small enough to be effectively zero but large enough to let the equations converge.

-Jon

- Location
- Vancouver, WA