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09-03-2013 09:34 PM

Three phase short circuit current

I have three phase short circut current now i want to do setting for earth fault relay. I want to find Single phase to earth curretn,

Is there any approximation that can i made by using three phase short circuit current?
09-03-2013 09:35 PM
Top #2
09-03-2013 09:35 PM
If the system is solidly grounded, a single-phase-to-earth fault current would be equal to the three phase short circuit current.
09-03-2013 09:35 PM
Top #3
09-03-2013 09:35 PM
First, the Earth Fault Current ( between phase & PE ) is depending on the used Earthing System, as we know all, there are 4 types that : TT, TN-S, TN-C, IT, where in the most of industrial applications we use either " TN-S or TN-C or both ", so, for these types, you have 2 possibilities :
1- If the PE cable's section is equal of Phase cable's section, in this case, the Earth Fault Current between one phase & PE is " 1/2 of 3 phases short circuit current ".
2- Is the cables are different, you should calculate the values of " R & X " for the cables of Phase and the cable of PE from the source to the fault's point, then " Rtotal & Xtotal ", then Ztotal, after that you can calculate the Earth Fault current by using the formula :
Ief = 0.8 x V / Ztotal

By the way, sorry Mr. Sivakumar that's not right. Whatever the sections of cables, the Earth Fault Current is smaller than the 3 Phases short circuit, because the formula of " 3 Phases short circuit current " is :
Isc3max = U / 1.732 x Z or Isc3max = V / Z
where " Z " is the cables' impedance of " 1 phase " from the source till short circuit point.
but the Earth Fault current formula is :
Isf = V / Ztotal
where " Ztotal " is cables' impedance of " 1 phase and PE " from the source till earth fault point.
09-03-2013 09:36 PM
Top #4
09-03-2013 09:36 PM
You need to do a short circuit calculation to determine your earth fault current. Your three phase fault current was determined utilizing the positive and negative sequence impedance of the system in the faulted zone. In your earth fault current, you need to do an additional task / effort to determine your zero sequence impedances which is mainly the most essential parameter in the earth fault calculation. You can do it by manual calculation for a smaller scope or you can use a software program for a large system or network.
09-04-2013 10:18 PM
Top #5
09-04-2013 10:18 PM
But for short circuit calculation i need system data.I want to find current in auto transformer during single phase to earth fault for relay setting .I have symmetrical and asymmetrical short circuit current and impedance of ATR .The system impedance,where High and medium voltage winding is connected ,is not with me.

Can i make a approximation of impedance ???If yes how much impedance i need to consider??
09-04-2013 10:18 PM
Top #6
09-04-2013 10:18 PM
What do you mean when you say Earth Fault, do you then mean actual Earth Fault (active phase to earth), or do you mean single-phase short circuit to Neutral?You have to consider how your system is earthed in order to calculate the (worst case) fault current, that you want to protect against.
09-04-2013 10:20 PM
Top #7
09-04-2013 10:20 PM
I beg to differ with you!

The formula for single-phase-to-earth fault current is:
3 Vph
I(L-G) = ------------------------------------------
Z1 + Z2 + Z0 = 3 (Zn + ZGP)


I (L-G) = Single-line-to-earth fault current, in Amperes
Vph = Phase-to-Neutral Voltage, in Volts
Z1 = System Positive Sequence Impedance, upto the point of fault, in Ohms
Z2 = System Negative Sequence Impedance, upto the point of fault,in Ohms
Z0 = System Zero Sequence Impedance, upto the point of fault,in Ohms
Zn = Neutral-to-Ground Path Impedance, in Ohms
ZGP = Ground Return Path Impedance, in Ohms

Now, in solidly earthed systems (i.e.) in TN (TN-C or TN-S) networks, Zn & ZGP are negligible and can be ignored. And, in transformer fed industrial networks, Z1 = Z2 = Z0; thus, the formula becomes:
3 Vph
I(L-G) = ------------
3 Z1
= --------

= Root 3 x VL

Which, incidentally, is the formula for three phase symmetrical short circuit current, vide IEC 60909.

Hence, I standby my statement that in solidly earthed systems, the singe-line-to-earth fault current is equal to the three phase symmetrical short circuit current.
09-04-2013 10:21 PM
Top #8
09-04-2013 10:21 PM
A very conservative approach would be to take the full load amp and dived by the percent impedance, that gives you the transformer capability to deliver fault current on a per phase bases. This value is higher than standard fault calculation assuming 1 per unit voltage at the primary terminal. Additional system reactance and ground resistance and /or neutral reactance will reduce this current.

This method can give you an approximate ball park figure as to what symmetrical KA rating equipment downstream of the transformer should be rated for.
09-05-2013 09:27 PM
Top #9
09-05-2013 09:27 PM
First, that's right if you mean exactly at the Transformer's output, because we have at this point only the impedances of MV and that of transformer, but after even a few meters where we have cables and the phases cables' section is certainly different than PE cable's section, in this case, the impedance " Z " in the formula is composed of the initial impedances ( MV + Tr ) + the cables' impedances, and because the cables' impedances are, as value, more important, that will drive us to have different values of " 3 phases short circuit " and " Earth Fault current ". in the same time, you know well that when we calculate the minimum short circuit current and earth fault current we should take into consideration the worth case, that means the state of cables is " Hot ", but when we calculate the maximum short circuit current we suppose the state of cable is " Cold ", that means the resistivity of cable will be different, so the impedance " R + jX " will be also different.

By the way, please let me just re-write your formula :
I(L-G) = Vph / Z = VL ( U ) / root 3 x Z
and not ( root 3 x VL / Z ) because Vph = VL ( U ) / root 3 ( 1.732 )

I jus calculated the " 3 phases short circuit " and " Earth Fault Current " of the following network :
- MV : 20kV, 500MVA
- Earthing System : TN-S
-Transformer : 1600 kVA, 20/0.4kV , Ucc : 6%, Copper Losses : around 30000 watts, Dyn11
- Cables : Single Core, PVC, Phase : 6 x 240mm2 , N : 3 x 240mm2 , PE : 1 x 240mm2 , L : 10meters

The results at the end of cables ( only 10 m ) are :
Isc3max : 35,65 kA
Ief : 27.70 kA
09-05-2013 09:27 PM
Top #10
09-05-2013 09:27 PM
So Actually it should be by using ur example, Trans current I = P/V* 1.732*cos, Short circuit current Isc = I / %Z (0.06) + total cable Z = this is just a "rule of thumb". Your figures are very much correct.
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