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It is because the current flowing to a fault point downstream of the transformer will flow only through the secondary windings of that transformer. There will of course be an impact on the primary side. But the energy source that is feeding the fault is the secondary windings.. . . anyone have a simple explanation why a transformer automatically “resets” the available short circuit current to the value of that particular transformer on the secondary?

where would you find the impedance of the transformer?

It should be on the nameplate. But you are early enough in design or installation process that the transformer has not yet been purchased, I would ask a vendor or two what their typical values would be. My "default" value for rough-order-of-magnitude calculations is 5.75%.where would you find the impedance of the transformer?

For a 1MVA 13.8kv/480Yoil filled, yes, 5.75%IZ is a "normal" number. 21KA SCC - not difficult.

However, I've see one 500KVA, 4160/208Y, oil filled, that was 2%IZ - 69kA SCC - yuck

For the smaller stuff, Square D, EX, is showing 3.5%IZ to 3.8%IZ for 30 KVA, 480/208Y

This one is only looking at 2300A SCC - it doesn't matter much.

Smaller transformers (<500KVA) may not have the impedance on the nameplate. However, mfg data generally shows nominal values and a tolerance

So, if the install could be any where close to where it matters (as in - is the equipment capable of interrupting the SCC?) I'm asking for design specification numbers (not test data).

From your question, I'm thinking you have not been exposed to this. If not, you should consider getting a class covering per unit and changing bases. It comes up quite a bit in Short Circuit studies and Protective Relays Time Current curves.When considering short circuit currents at any point in an electrical one line, anyone have a simple explanation why a transformer automatically “resets” the available short circuit current to the value of that particular transformer on the secondary? ...

Consider a transformer model looks like a voltage source in series with an impedance.

Between your 30KVA transformer and the generation, there are a few transformers. Here is an example:

20 MVA, 138KV/13.8KV 8%Z

1 MVA, 13.8KV/480V 5%Z

30KVA, 480V/208V, 3%Z

1 MVA, 13.8KV/480V 5%Z

30KVA, 480V/208V, 3%Z

The trick is to reflect the impedances from the higher voltage transformers down to the 208V base. As you know, every time the power goes through a transformer:

Voltage changes as the turns ratio

Current changes as the turns ratio

and the series impedance reflects as the square of the turns ratio. I'll use the ratio of the square of the voltages as the square of the turns ratio.

Current changes as the turns ratio

and the series impedance reflects as the square of the turns ratio. I'll use the ratio of the square of the voltages as the square of the turns ratio.

So, looking at the 480V xfm secondary, the 8%Z from the 138/13.8 xfm , looks like .08*(100)*(Vs^2)/(Vp^2) = 8*(13.8^2)/(138^2) = .009% So the series impedance changed from 5% to 5.009%

Then we account for the 480/208 xfm. 5.009 reflected to 208V, %Z = 5.009 * (208^2)/480^2) = .94%. So, the series impedance changes from 3% to 3.9%

Every time the power goes through a stepdown transformer, the upstream impedances have less effect.

At the secondary of the 13.8KV xfm the SCC = 20,000KVA/13.8KV/sqrt(3)/.08 = 10KA

At the secondary of the 480V xfm the SCC = 1,000KVA/.480KV/sqrt(3)/.05009 = 24KA

At the secondary of the 208V xfm the SCC = 30KVA/.208KV/sqrt(3)/.039 = 2.1KA

It isn't that the ASCC goes away, rather the additional transformers add impedance.

I'm hoping this helps and is not just a jumble

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Fault current on the transformer secondary DOES depend on fault current on its primary, but it is a worst-case-scenario calculation you can make as a shortcut to ignore it. This is known as "infinite bus", where you make the assumption that it doesn't matter how high the fault current is on the primary side, the secondary side fault current will never exceed it. You could consider primary fault current in the calculation if desired, and if your fault current is just slightly higher than a standard KAIC rating it might help you. But you don't necessarily need to do so, s a conservative assumption.

It's known as a limit in Mathematics. The function grows ever closer to limiting value, but never exactly gets to it, and never exceeds it. The function that determines transformer secondary fault current as a function of primary fault current, takes off to a "cruising altitude" as primary fault current grows.

In the full form, here are your formulas for how to get fault current, specific to the case of 3-phase:

f-factor:

f = (√3 * I[SUB]SCA(pri) [/SUB]*V[SUB]pri *[/SUB] %Z) / (100,000 * KVA)

M-multiplier:

M = 1/(1+f)

Fault current on secondary:

I[SUB]SCA([/SUB][SUB]sec) =[/SUB] I[SUB]SCA([/SUB][SUB]pri)[/SUB] * M * (V[SUB]pri[/SUB]/V[SUB]sec[/SUB]) [HR][/HR]

Here's the mathematics that show WHY there is an upper limit to the fault current on the secondary:

Notice Isca(pri) always appears multiplied with Vpri? Define X = Isca(pri)*V(pri) and simplify.

f = (√3 * X * %Z) / (100,000 * KVA)

M = 1/(1+f)

I[SUB]SCA([/SUB][SUB]sec) =[/SUB] X * M/V[SUB]sec[/SUB]

Define C as a constant, to combine the impedance, KVA and sqrt(3) factor:

C = √3 * %Z/(100,000*KVA)

Re-write f in terms of X and C:

f = X*C

Substitute in to M multiplier equation:

M = 1/(1 + X*C)

Substitute in to final equation for fault current:

Isca(sec) = X*1/(1 + X*C) * 1/Vsec

Simplify:

Isca(sec) = 1/Vsec * (X/(1 + X*C))

We are interested in taking the limit as X "gets large". In otherwords, as X approaches infinity. "1" becomes insignificant when added to X*C as X approaches infinity. This allows you to disregard this term when calculating the limit.

To show an example that demonstrates this simplification, suppose C = 2, and we try X-values of 5, 10, 100, and 1 million.

5/(1 + 2*5) = 0.4545; 5/(2*5) = 0.5

10/(1 + 2*10) = 0.4762; 10/(2*10) = 0.5

100/(1 + 2*100) = 0.4975; 100/(2*100) = 0.5

1000000/(1 + 2*1000000) = 0.49999975; 1000000/(2*1000000) = 0.5

When neglecting the "1" in the denominator as discussed, you then get:

Isca(sec) = 1/Vsec * X/(X*C)

X then cancels, and we are left with:

Isca(sec) = 1/(C*Vsec)

Recall & replace the definition of C:

Isca(sec) = 1/((√3 * %Z/(100,000*KVA))*Vsec)

Isca(sec) = 100,000*KVA/(√3 * Vsec %Z)

Notice that 100,000*KVA/(√3 * Vsec) is the operating secondary current associated with the KVA rating, when Vsec is phase-to-phase. It is simply a matter of dividing by impedance, if you are already familiar with calculating this. You don't necessarily even need to know the primary voltage of the transformer, just the KVA and impedance. This is useful when you don't know the grid voltage in the neighborhood, and it is not your scope of work to find out. This limiting factor will be an upper limit on your secondary fault current, regardless of if you have a 13.2 kV primary or a 34.5 kV primary. It also is independent of the moving target that utility-side fault current may be, as utilities upgrade substations or bring new generating facilities online.

Correction:Notice that 100,000*KVA/(√3 * Vsec) is the operating secondary current associated with the KVA rating, when Vsec is phase-to-phase. It is simply a matter of dividing by impedance, if you are already familiar with calculating this.

1000*KVA/(√3 * Vsec) is the operating secondary current associated with the KVA rating, when Vsec is phase-to-phase.

It is still a matter of dividing by impedance. Or rather dividing by (%Z/100). 100 of course comes from the definition of percent.