Max LG Fault Current on Autotransformer

[dlmartin]

Does anyone know a quick check for obtaining the maximum LG fault current on the low side of a 14.4/7.2 kV (24.94/12.47 kV L-L) autotransformer?

I'm running a fault current study that shows 2723A LG fault on the 14.4 kV side and 674A LG fault on the 7.2kV side. I know the fault currents are based on several impedances (ie: station transformer, line, auto, etc), but 674A seems low.

 

[zkz]

hello,
Can you please provide transformer MVA rating, percentage impedance and fault level on HV side?
regards,
appunni

 

[dlmartin]

1000 kVA, 5% impedance, 2723A LG on the high side of the autotransformer

 

[zkz]

Transformer MVA = 1 MVA
Percentage impedance = 5
HV side L-G Fault current = 2723 A
If we assume L-G fault level = L-L-L fault level,
Per unit impedance = 5/100 = 0.05
HV side fault level = (1.732*24.94KV*2723A)/1000 = 117.62MVA
LV side fault level = 117.62/(1+(117.62/1)*.05)=17.09 MVA
LV side fault current = 17.09*1000/(1.732*12.47) = 791.41 A

Please note that three phase fault current on LV side is 791.41 A.
674A = 85% of 791.41A
You know that
L-L-L fault current = Vph-N/(Zs1+Zt1) and
L-G fault current = 3Vph-N/(2Zs1+Zs0+2Zt1+Zt0)
From the above equations, there is chance for becoming ground fault current less than three phase fault current if total zero sequence impedances of source (Zs0) and of transformer (Zt0) is greater than sum of Zs1+Zt1. So the value you got may be correct.
Regards,
appunni

 

[kh2]

Hi zkz,

Please explain the formula you used to calculate the low voltage (LV) side MVA fault level. Which is given as = " HV side MVA / (1+(HV side MVA/1)*per unit impedance". Why did not use the available HV side fault current to calculate the LV side fault current?

Thank you,

Kh2

 

[zkz]

dear Kh2,
When a fault occurs on LV side, fault path includes transformer impdeance in addition to source impedance and other impedances upto transformer.
But for HV side fault, the transformer impedance is excluded.
Please note that impedance of a transformer is normally high compared to line etc.
Hope this will help you,
regards,
appunni

 

[kh2]

Hi zkz,

Thank you for your response. If the transformer impedance is the only impedance to calculate the transformer LV side fault current, it will be = I(secondary)/per unit impedance = 46.3/.05 =926 Amps. This value is 0.727 of the value that dimartin obtained and is 0.85 of the value you calculated. This is a quick check but not an accurate one. Also, the auto transformer is different from the two winding transformer when it comes to calculating the current and voltage distribution on the windings.

Could you please provide a reference book or a standard that shows the formulas to be used for such calculation.

Appreciate you input.

Regards,

Kh2

 

[kh2]

Hi zkz,

Please note the following correction to my last calculation of 926 amps for the fault current on the secondary side of the 1000 kVA, 24.94/12.47 kV( L-L), 3-phase, 5% Z, auto transformer:

The 926 amps is 1.37 times the 674 amps that dimartin obtained, and is 1.17 the 791 amps that you calculated. I reversed the relationship. Apologize for the confusion

Regards,

Kh2

 

[saladhawks]

There really is no need to calculate the secondary fault current with an infinite bus if the source impedance is known. The original post indicates a SLG source contribution of 2,723 Amps on the high side of the autotransformer. zkz went ahead and made the very reasonable assumption that the SLG source contribution Amps = 3-phase source contribution Amps. From that, it is simple to calculate the equivalent source impedance and simply add that to the autotransformer impedance in order to calculate secondary fault currents.

 

[kh2]

Agree with all you wrote saladhawks. My original question to zhz was related to the formula that he used to calculate the LV side fault. I was trying to understand how the formula was developed. I am familiar with a simple approach which gives the same result as follows:
1) Use MVA base = 1MVA (the transformer MVA).
2) Source p.u. impedance = MVA base/SC MVA system = 1/117.62 =0.0085. (the 117.62 SC MVA was taken from zkz calculation).
3) Total Zp.u. = Zp.u. system + Zp.u. transformer = 0.0085 + 0.05 = 0.0585.
4) LV Fault current at the transformer secondary = I(sec)/total Zp.u.
5) I(secondary) = 1000/(1.732 * 12.47) = 46.3 amps.
6) LV Fault Current = 46.3/0.0585 = 791.45 amps.

Please note that this is the same value that zkz has calculated.zkz value is 791.41 to be exact.

Thank you for your comments.

Kh2