Fault current calculations for transformers with Sub-600V on both sides

[Carultch]

I'm familiar with calculating KAIC from the fault current calculation spreadsheet available on this website. However, I'm surprised to notice no intake for the voltage on the primary side of the transformer. I know this is geared toward applications where this would be medium voltage and utility service transformers, anywhere from 10 kV thru 25 kV.

What is the strategy for calculating how KAIC is affected by customer-owned, utilization voltage-to-utilization voltage transformers? Such as a 480V primary and a 208V secondary. Or other combinations? Does frequency play a role? I often have applications at 50 Hz.

Given KAIC on the primary, and the voltages of the transformer, the KVA, and the percent impedance, how does one calculate KAIC on the secondary?

Example:
225 kVA
2.5% impedance
480V primary
208V secondary
Fault current available at the primary is 20 KA.

What KAIC do I need on the secondary equipment?

 

[Fitzdrew516]

Ok, this will be a long answer so here it goes...

You need to determine the "f" factor for the xfmr first. That formula is as follows -

f = (√3 * I_SCA(pri) *V_{pri *} %Z) / (100,000 * KVA)

So in your example this would be -

f = (√3 * 20,000* 480 * 2.5) / (100,000 * 225)

f = 1.85

Now that we have the "f" factor we can find the multiplier (M) for the transformer. That formula is as follows:

M = ((1)/(1+f)

So to continue the example the formula would be:

M = ((1)/(1+1.85)

M = .351

So the final formula to actually find the short circuit rating on the secondary would be as follows:

I_{SCA(sec) =} I_SCA(pri) * M * (V_pri/V_sec)Finally we get the following :

I_{SCA(sec) =} 20,000 * .351 * (480/208) = 16,200


Someone should look this over because it doesn't make sense to me that the secondary is more than the primary. I'm pretty sure I got my formulas right, but when I got to the end it really confused me. I think it may be because your impedance is low. I've had this happen before once and I just rated everything downstream of the primary the same rating as the primary. I thought I had a good answer here, but now I've just created another question haha.

Thanks,
-Drew

 

[Carultch]

Ok, this will be a long answer so here it goes...

You need to determine the "f" factor for the xfmr first. That formula is as follows -

f = (√3 * I_SCA(pri) *V_{pri *} %Z) / (100,000 * KVA)

So in your example this would be -

f = (√3 * 20,000* 480 * 2.5) / (100,000 * 225)

f = 1.85

Now that we have the "f" factor we can find the multiplier (M) for the transformer. That formula is as follows:

M = ((1)/(1+f)

So to continue the example the formula would be:

M = ((1)/(1+1.85)

M = .351

So the final formula to actually find the short circuit rating on the secondary would be as follows:

I_{SCA(sec) =} I_SCA(pri) * M * (V_pri/V_sec)Finally we get the following :

I_{SCA(sec) =} 20,000 * .351 * (480/208) = 16,200


Someone should look this over because it doesn't make sense to me that the secondary is more than the primary. I'm pretty sure I got my formulas right, but when I got to the end it really confused me. I think it may be because your impedance is low. I've had this happen before once and I just rated everything downstream of the primary the same rating as the primary. I thought I had a good answer here, but now I've just created another question haha.

Thanks,
-Drew

To me, it does make sense that the secondary would be more than the primary, when the secondary is lower voltage. Because transformers trade voltage for current.

With a gear analogy, imagine a reduction gear that reduces the speed. And a sudden extreme impulse torque is applied on the input shaft. The gear system amplifies this fault torque to the output shaft, causing the potential for greater failure of the output shaft than the input shaft.

I'll look this over and see if it makes sense. Thank you for the help.

 

[Phil Corso]

Carlutch... here is my Gi.Fi.E.S. method (pronounced Jiffy’s) for solution, where: Gi represents "Given"; Fi for "Find"; E for "Equation"; and S for "Solution!" Following is the solution;

Example 1
Given:
A 3-phase transformer having the following design or rated capacity parameters:
o 225kVA; kVpri = 0.48; kVsec = 0.208; Imp Z = 2.5%; connected to an infinite source.
Find:
Short-circuit current on secondary, Isc.
Equation(s):
Infinite means no impedance between the transformer primary and the source voltage, then the short-circuit current magnitude is the transformer's rated kVA, divided by its impedance in per-unit,
o kVAsc = kVA / (Z / 100), and then,
o Isc = kVAsc / (1.732 * kVsec)
Solution:
o kVAsec = 225 / 0.025 = 9,000 kVA.
o Isc = 9,000 / (1.732 * 0.208) = 24,981A ~ 25.0 kA

Example 2
Given:
Parameters above except that SC fault-current level at xfmr primary is given in Amperes = 20,000 A
Then, primary SC Duty, kVApri = 1.732 * Isc * kVpri = 16,627 kVA ~ 16,600 kVA
Find:
Short-circuit current on secondary, Isc.
Equation(s):
Because only reactance’s are involved, that is, resistances are ignored, I will use a short-cut procedure called the kVA Value Method.
Solution:
Combine the pri and sec fault kVA’s together as if they were two resistances in parallel, that is,
o kVAequiv = (kVApri * kVAsec) / (kVApri + kVAsec).
Solve equations as required to find equivalent SC-Duty on 208V bus:
o kVAequiv = (16,600 * 9,000) / (16,600 + 9,000) = 5,840 kVA
o Isc = 5,840 / (1.732 * 0.208) = 16,210A = 16.2 kA

Please let me know if additional detail is required.

Regards, Phil Corso

 

[Phil Corso]

Carultch...

You only ratio V and I when determining Xfmr pri & sec currents related to load!

Phil

 

[Carultch]

Carlutch... here is my Gi.Fi.E.S. method (pronounced Jiffy’s) for solution, where: Gi represents "Given"; Fi for "Find"; E for "Equation"; and S for "Solution!" Following is the solution;

Example 1
Given:
A 3-phase transformer having the following design or rated capacity parameters:
o 225kVA; kVpri = 0.48; kVsec = 0.208; Imp Z = 2.5%; connected to an infinite source.
Find:
Short-circuit current on secondary, Isc.
Equation(s):
Infinite means no impedance between the transformer primary and the source voltage, then the short-circuit current magnitude is the transformer's rated kVA, divided by its impedance in per-unit,
o kVAsc = kVA / (Z / 100), and then,
o Isc = kVAsc / (1.732 * kVsec)
Solution:
o kVAsec = 225 / 0.025 = 9,000 kVA.
o Isc = 9,000 / (1.732 * 0.208) = 24,981A ~ 25.0 kA

Example 2
Given:
Parameters above except that SC fault-current level at xfmr primary is given in Amperes = 20,000 A
Then, primary SC Duty, kVApri = 1.732 * Isc * kVpri = 16,627 kVA ~ 16,600 kVA
Find:
Short-circuit current on secondary, Isc.
Equation(s):
Because only reactance’s are involved, that is, resistances are ignored, I will use a short-cut procedure called the kVA Value Method.
Solution:
Combine the pri and sec fault kVA’s together as if they were two resistances in parallel, that is,
o kVAequiv = (kVApri * kVAsec) / (kVApri + kVAsec).
Solve equations as required to find equivalent SC-Duty on 208V bus:
o kVAequiv = (16,600 * 9,000) / (16,600 + 9,000) = 5,840 kVA
o Isc = 5,840 / (1.732 * 0.208) = 16,210A = 16.2 kA

Please let me know if additional detail is required.

Regards, Phil Corso

That's a nice acronymn, thanks for the help. I'll have to use this as a template to comment my calculation files.

 

[kingpb]

MVA Method-

Line: 20kA x 480V x sqrt3 = 16.63MVA

XFMR: 225KVA/0.025 = 9MVA


Combine as impedance in parallel: 1/[(1/16.63MVA) + (1/9MVA)] = 5.84MVA

This is the available MVA on the LV side of the transformer.

KAIC: 5.84MVA/(208 x sqrt 3) = 16.21KA


To assume infinite bus, would simply be XFMR 9MVA/(208V x sqrt3) = 25KA; which frankly is what I would specify the equipment to be rated for.

 

[mull982]

MVA Method-

Line: 20kA x 480V x sqrt3 = 16.63MVA

XFMR: 225KVA/0.025 = 9MVA


Combine as impedance in parallel: 1/[(1/16.63MVA) + (1/9MVA)] = 5.84MVA

This is the available MVA on the LV side of the transformer.

KAIC: 5.84MVA/(208 x sqrt 3) = 16.21KA


To assume infinite bus, would simply be XFMR 9MVA/(208V x sqrt3) = 25KA; which frankly is what I would specify the equipment to be rated for.

I understand the math but cant seem to recall why we treat these two impedances as being in parallel? Aren't the source impedance and transformer impedance in series? Am I missing something?

 

[kingpb]

I understand the math but cant seem to recall why we treat these two impedances as being in parallel? Aren't the source impedance and transformer impedance in series? Am I missing something?

Good question, Yes the impedance is in series, but it has to do with the fact that the foundation for the MVA method was derived from a modification of the ohmic method; and uses the admittance; which by definition is the reciprocal of impedance.

The method for combining admittance is reciprocal of combining ohms; therefore admittance in series combines in the same way as ohms in parallel, and vice versa.