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? For instance, you could have 60kA short circuit available then simply put a 30 kVA transformer and now short circuit on secondary just dropped to about 3kA without any regard of how high it was on the primary. Is this magic?
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The transformer is only capable of producing it's rated amperage divided by it % impedance. So, if a transformer can produce, say 100A, at its full KVA rating...and if its impedence is 5%, then it can only produce a maximum of 100/.05 or 2000 amps. This is a maximum number regardless of its lineside configuration. In reality, the 2000A number could be very slightly reduced by line losses in the supply side....even if the supply side available fault current is a very VERY high number. It's almost exclusively the transformer itself that governs what will be available on the secondary.Facilities Electrical Engineer
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Originally posted by cppoly View Post. . . 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?
Charles E. Beck, P.E., Seattle
Comments based on 2017 NEC unless otherwise noted.
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Originally posted by RD35 View PostThe transformer is only capable of producing it's rated amperage divided by it % impedance. So, if a transformer can produce, say 100A, at its full KVA rating...and if its impedence is 5%, then it can only produce a maximum of 100/.05 or 2000 amps. This is a maximum number regardless of its lineside configuration. In reality, the 2000A number could be very slightly reduced by line losses in the supply side....even if the supply side available fault current is a very VERY high number. It's almost exclusively the transformer itself that governs what will be available on the secondary.
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Originally posted by acrwc10 View Postwhere would you find the impedance of the transformer?
Charles E. Beck, P.E., Seattle
Comments based on 2017 NEC unless otherwise noted.
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Impedance values are all over the place.
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).
Without data you’re just another person with an opinion – Edwards Deming
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Originally posted by cppoly View PostWhen 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
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.
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
(edit to add) It has been a few years since I have had to do this. Anyone wants to jump in and correct is welcome.
Last edited by iceworm; 092019, 03:19 PM.Without data you’re just another person with an opinion – Edwards Deming
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Originally posted by cppoly View PostWhen 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? For instance, you could have 60kA short circuit available then simply put a 30 kVA transformer and now short circuit on secondary just dropped to about 3kA without any regard of how high it was on the primary. Is this magic?
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 3phase:
ffactor:
f = (√3 * I_{SCA(pri) }*V_{pri *} %Z) / (100,000 * KVA)
Mmultiplier:
M = 1/(1+f)
Fault current on secondary:
I_{SCA(}_{sec) =} I_{SCA(}_{pri)} * M * (V_{pri}/V_{sec})
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_{SCA(}_{sec) =} X * M/V_{sec}
Define C as a constant, to combine the impedance, KVA and sqrt(3) factor:
C = √3 * %Z/(100,000*KVA)
Rewrite 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 Xvalues 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 phasetophase. 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 utilityside fault current may be, as utilities upgrade substations or bring new generating facilities online.
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Originally posted by Carultch View Post
Notice that 100,000*KVA/(√3 * Vsec) is the operating secondary current associated with the KVA rating, when Vsec is phasetophase. 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 phasetophase.
It is still a matter of dividing by impedance. Or rather dividing by (%Z/100). 100 of course comes from the definition of percent.
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