Short Circuit Calculations

[dvcraven0522!]

I made a request from a local utility co. for available SSC at utility pole for pad mounted transformer. Below is the data that was provided.


The data provided is from pole but they did not model the transformer or UG wire. I do not completely understand the x/r ratio so, my question is can use the max fault (LLLG) 1865A as starting point for my pad mounted transformer?

Dan Craven

 

[JoeStillman]

X/R is the ratio of reactance to resistance upstream of the point of fault (the utility pole, in your case.) It affects the DC offset in the initial fault waveform. 4.14 is a pretty low X/R ratio and so you're not at a whole lot of risk to just add impedances instead of doing vector algebra. But if you want to calculate impedances of the source, cables and transformer in vector form, adding reactance to reactance and resistance to resistance for a total X/R at the point of fault. That's why it's best to use software. It makes it easy.

 

[dvcraven0522]

X/R is the ratio of reactance to resistance upstream of the point of fault (the utility pole, in your case.) It affects the DC offset in the initial fault waveform. 4.14 is a pretty low X/R ratio and so you're not at a whole lot of risk to just add impedances instead of doing vector algebra. But if you want to calculate impedances of the source, cables and transformer in vector form, adding reactance to reactance and resistance to resistance for a total X/R at the point of fault. That's why it's best to use software. It makes it easy.

I do have a short circuit program in excel and requires the transformer KVA rating and impedance to start the calculation. I try to use the utility SSC valve if available. If not, I will use an transformer KVA rating suitable for the service and use the Bussmann book for transformer impedance valves listed for that size transformer.

If I understand your response I could take the LG fault of 1406A at the pole an estimate the conductor impedances from the pole to my pad mounted transformer and use that value to start my calculation down stream of the distribution.

 

[jim dungar]

I do have a short circuit program in excel and requires the transformer KVA rating and impedance to start the calculation. I try to use the utility SSC valve if available. If not, I will use an transformer KVA rating suitable for the service and use the Bussmann book for transformer impedance valves listed for that size transformer.

If I understand your response I could take the LG fault of 1406A at the pole an estimate the conductor impedances from the pole to my pad mounted transformer and use that value to start my calculation down stream of the distribution.

Do not use L-G values when performing 3-phase fault analysis.
Most free apps do not perform 1-phase short circuit calculations.

 

[ron]

The info they gave you was on the primary side of the XFMR. To go further downstream, you will need to know the kVA of the XFMR and the impedance of the XFMR and any low side conductor impedance that they may add before the meter.

 

[Julius Right]

According to IEC 60909-0 standard
ILLLG [noted here as I"kE2E line-to-line short circuit with earth connection]
I"kE2E=sqrt(3)*c*Un/(Z1+2*Zo)
where Un is line-to-line voltage [rms],
Z1=Positive-sequence short-circuit impedance,
Z2=Negative-sequence short-circuit impedance
If the short-circuit it is far from generators, then Z1=Z2
Zo=Zero-sequence short-circuit impedance
and c it is voltage factor-maximum= 1.1
Let's put ZeqA=Z1+2*Zo
ILG[noted I"k1]=sqrt(3)*c*Un/(2*Z1+Zo)
Let's put ZeqB=2*Z1+Zo
ILLLG=1860 A
ILG=1465 A
From here we get:
ZeqA=Z1+2*Zo=sqrt(3)*1.1*25000/ILLLG=25.53962 ohm
ZeqB=2Z1+Zo=sqrt(3)*1.1*25000/ILG=32.62424 ohm
We don't know yet for which impedance the value of X/R has to be applied. In my opinion it is referred to the equivalent impedance in both cases.
Let’s take
ZeqA=ReqA+iXeqA
ZeqB=ReqB+iXeqB
XA/RA=3.86
XeqA= ZeqA/sqrt[1+1/(Xa/Ra)^2])=24.723 ohm
ReqA=XeqA/(XA/RA)= 6.405033 ohm
XB/RB=4.14
XeqB=ZeqB/SQRT(1+1/(XB/RB)^2)= 31.71224
ReqB=XeqB/(XB/RB)= 7.659962
X1=(2*XeqB-XeqA)/3=12.90035
R1=(2*ReqB-ReqA)/3=2.97163
For the pole grounding purpose ILLLG is the considered current.
For short-circuit at low voltage transformer terminal, Z1 it is the system impedance.
For protection action minimum current at high-voltage terminal the minimum current it is as indicated [328 A].

 

[Julius Right]

I forgot Zo:
Xo=XeqB-2*X1=5.911538
Ro=ReqB-2*R1=1.716701