# Fault Current Across Transformers

#### EC Dan

##### Member
I don’t have access to power analysis software, so I’m doing manual point-to-point short circuit calculations and I have questions about the propagation of fault current across transformers. I’m using the formula described below, but I never use formulas that I don’t understand so I wanted to make sure I fully understand this: In general, for calculation of downstream fault current (I2) from known upstream fault current (I1), the assumption is the line voltage drop is the same in either case, so the relationship exists: I1 * Z1 = I2 * (Z1+Z2), which simplifies to I2 = I1 * 1 / (1 + Z2/Z1). In the above procedure, 1 / (1 + Z2/Z1) corresponds to the M multiplier and Z2/Z1 corresponds to the f-factor, therefore Z2/Z1 = (I1 * V1 * 1.732 * %Z) / (100,000 * KVA). Z2 in this case is the impedance of the primary winding, which can be calculated from %Z by Z2 = Zabs = %Z/100 * Zbase = %Z/100 * V2 * 1.732 / (1000 * SKVA). Z1 is simply V1/I1. Dividing Z2 by Z1 yields the same formula for the f-factor in the method above. I understand that this method ignores the X/R ratios of the line impedances. My questions:

1. In step C, I1 = Isca(p) and I2 = M * Isca(p). Therefore I2 is the fault current as seen by the primary side of the transformer during a secondary fault, which is then converted to the current on the secondary using the standard transformer conversion formula. Does this mean that I2 would be the value I use to determine clearing time on the primary OCPD for arc flash purposes?

2. Step C does not use any efficiency or power factor for conversion. I can get efficiency and X/R ratios from the transformers here. In fault current situations, are these adjustments typically used?

#### EC Dan

##### Member
Realized a mistake in question 1. I know I2 is not to be used directly for clearing time in arc flash event since that's bolted fault current. I actually need to transfer the reduced arcing current on the secondary back to the primary side, therefore is fault current used to determine the primary OCPD clearing time simply Iarc(s) * Vs / Vp?

#### David Castor

##### Member
I've never seen this method of short circuit calculation before and I've been doing this for a long time. Without the full explanation and referenced tables it's hard to help much.

As a first approximation, there are several free arc-flash calculators available online. The Littlefuse one might work for you.

Also, the relationship between bolted fault current and the expected arcing current is complicated (in IEEE 1584). There are a lot of factors.

#### romex jockey

##### Senior Member
Also, the relationship between bolted fault current and the expected arcing current is complicated (in IEEE 1584). There are a lot of factors.

A brief synopsis might go a long way here David

~RJ~

• Carultch

#### EC Dan

##### Member
That point-to-point method is from this Eaton guide: https://www.eaton.com/content/dam/e...nter/bus-ele-tech-lib-electrical-formulas.pdf. I of course use the system configuration that results in the minimum fault current at each node.

I then use this calculator to determine the arc flash characteristics: https://www.jcalc.net/arc-flash-calculator-ieee

My main question was the best way to convert the resulting calculated arcing current at the secondary of a transformer back to the primary side to determine clearing time of the primary OCPD. Using Iarc(s) * Vs / Vp seems overly simplistic.

#### David Castor

##### Member
If you have the arcing current at the point you're interested in, the primary current can be directly calculated from the turns ratio of the transformer.

So your equation is basically correct. For example if you have a 12.47 kV to 480 V transformer and 10,000 A arcing current at 480 V, the primary current will be 480/12470 x 10,000 or 385 A.