# Bolted Fault Calculation

• posted
We have a situation where a customer needs to put three single-phase,
> 50 KVA pole mounted transformers into a three-phase bank. The voltage
> of each transformer is 13800-277/480Y. Two of the units have an
> impedance of 2.9%. The other unit has a 1.6% impedance.
>
> Here's the problem. One transformer design engineer is telling me that
> the Wye connection will prevent any significant load and/or voltage
> imbalances. Another transformer design engineer is telling me that the
> Wye connection will not correct for the impedance difference.
>
> Are there any applications or systems engineers out there who can give
> me their thoughts on this? Thanks.
No connection will compensate for the difference. Voltage imbalance
(voltage at each phase divided by the average) will be minimal under
normal balanced load, but as was pointed out it could be worse with
There is a bigger problem, however. Maximum fault current on the low
impedance phase could be up to 80% higher than on the other phases.
There is a danger that someone will analyze this using the higher
impedance, not realizing that they differ, which will give too low a
value and result in inadequately rated load equipment.
Recommended practice is to use three identical transformers. Why ask
for problems?
Ben Miller
• posted
Only under no-load conditions. Voltage drop under load IS affected by the impedance.
The original post mentioned a Wye connection, so I assume that each transformer is 277V, 50kVA, single-phase (150kVA total three-phase). This gives 181A full load. For the 2.9% impedance, fault current at the xfmr secondary will be 6241A, assuming an infinite primary bus. For the 1.6% unit, it will be 11,313A.
Ben Miller
• posted
| |>[Secondary voltage is determined by the turns ratio NOT the impedence | of the windings.] | | Only under no-load conditions. Voltage drop under load IS affected by the | impedance. | | |> Anyway: |> Your transformers would be I-max = 60 amps at 480 volts |> Your mod factors would be 34.48 for the 2.9% xfmrs and 62.5 for the 1.6% |> xfmr. |> So your max SCA will be 2068 amps for the 2.9% and 3750 amps for the 1.6% |> xfmr |> |> ALL electrical equipment must be rated a minimum of 5000 amps AIC rating |> so you don't have enough short circuit current to worry about. If these |> are the right numbers. |> | | The original post mentioned a Wye connection, so I assume that each | transformer is 277V, 50kVA, single-phase (150kVA total three-phase). This | gives 181A full load. For the 2.9% impedance, fault current at the xfmr | secondary will be 6241A, assuming an infinite primary bus. For the 1.6% | unit, it will be 11,313A.
How much for a L-L fault between a 2.9% unit and the 1.6% unit? I get 6967A (but I've never done a calculation like this before).
This is a definite issue if someone assumes 2.9% overall from checking just one transformer and selects ratings based on that. Could a licensed PE ever make such an assumption. I sure hope not.
• posted
Ah, if only life were that simple. My numbers apply to a full three-phase+ground fault. You really need to use symmetrical components to analyze anything else. Google it and have fun!
Ben Miller
• posted
I would have thought this would be way easier in a,b,c land than 0,1,2 land. In a,b,c it is just a circuit diagram that has, essentially, two voltage sources, and a couple of impedances between them. One KVL might solve it. For sequences, not only do you have to find the sequence impedances, and remember to asign the correct sequence sources, but you will then have to figure (or find) what the fault looks like in each sequence circuit.
• posted
----------------------------
------------------------------------- | I would suggest about 6940A for a line to line fault. I get a base impedance of 1.54 ohms so the transformer impedance combined is 0.0691 ohms. The current then is 480/0.0691=6944 A (magnitude) The same answer can be obtained by taking the current 180A and dividing it by (0,029+0,016) and multiplying by root (3) The difference between your answer and mine is that I started with 480V and you started with 277.000....V (something we have argued about before :)).
This is actually easier in this case than the symmetrical component approach as operator Jay suggests. In this case it is not necessary to use symmetrical components as all impedances except the tranformer are ignored.
To use symmetrical components in this case it is easiest to consider that the transformer impedances in the sequence models are 0 and the sum of the impedances is then treated as a fault impedance in the connection between the positive and negative sequences. This avoids messing around with unbalanced transformer impedances (which is messy and where that is necessary, the advantage of symmetrical components is lost). If this is done Ip=-In =277/j0.0691 =-j4010A leading to a fault current magnitude of 6940A
Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer
• posted
| I would suggest about 6940A for a line to line fault. I get a base | impedance of 1.54 ohms so the transformer impedance combined is 0.0691 ohms. | The current then is 480/0.0691=6944 A (magnitude) | The same answer can be obtained by taking the current 180A and dividing it | by (0,029+0,016) and multiplying by root (3) The difference between your | answer and mine is that I started with 480V and you started with | 277.000....V (something we have argued about before :)).
Actually, I started with 277.128129 volts to figure the impedance. Then I summed the impedance and divided 480.000000 by that. But I believe I used 181A rather than 180A. I can double check later but it doesn't quite account for the whole difference. As you know, I don't round my intermediate numbers, and defer rounding results until I need a rounded result.
• posted
whatever, you are in the same ballpark considering that 480V isn't 480V in practice.

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