Maximum and Minimum fault currents

Although, lyledunn, it did make me stop and think, that the formula Uo*1.1/ Zp only applies when the three-phase supply source has an earthed star point (tehcnically, a "neutral point"). If it's a delta-source, such as a delta-secondary transformer, and therefore corner-earthed, then the line-to-line voltage is equal to the line-to-earth voltage (Uo=U) and the formula would be Uo*1.1*√(3)/(3*Zp)

lyledunn said:

Not sure what you mean Graham, in such a fault one assumes that there will be no current in the neutral. It is irrelevant whether it is present or not, the maximum fault current being Uo*1.1/ Zp. 

Yes, exactly that. Bolted fault across all live conductors (three lines plus neutral) in the circuit at a particular point. This is what is shown in EIDG, and the formula matches, except that in EIDG what you have as Zp is split into (Zx+Zd), or Zp=(Zx+Zd).

I agree that it is irrelevant in a fully balanced system whether the neutral is present or not, but having it there and noting the current is zero perhaps provides an explanation that some might find informative.

very true - you could have a 3 phase supply without neutral and a phase-phase-phase fault would still be the worst case.

Mike

gkenyon said:

In a three-phase bolted fault across all phases, where all live conductors are connected together, and the fault current returns via Neutral or PEN. Hence, it is the line-Neutral voltage that is used.

Not sure what you mean Graham, in such a fault one assumes that there will be no current in the neutral. It is irrelevant whether it is present or not, the maximum fault current being Uo*1.1/ Zp. 

Yes, sorry ... long day yesterday.

"according to ESQCR is 230 V + 10 % / - 4 % (and 5 % is roughly the same as 4 %). C_min = 0.95 applies to both U and U_0"

_{Shouldn't that be 230V + 10% / - 6% ? Thus Cmin should actually be 0.94.}

_{Is 5% nearer 6% than 4%?}

JohnE said:

√3 ÷ 2 x 1.1 =0.95  which is the factor for Cmin ??

In this case, 0.95 from the calculation you provided above is a mere coincidence.

C_min = 0.95 ["for supplies to ESQCR" I believe its how you will find it described in BS 7671 and IET Guidance] simply comes from the fact that in the UK, the public supply voltage variation according to ESQCR is 230 V + 10 % / - 4 % (and 5 % is roughly the same as 4 %). C_min = 0.95 applies to both U and U₀.

ah well the winding impedance is one of the many things that means the resistance of the 2 paths may not be quite divided as I suggested

But this can also be estimated at least if you can read the rating plate - binoculars may help - an upper limit might be 5% volt drop at full load on the winding. (rather less on a modern low loss transformer.)

Unless the TX is in the same building, usually the distribution cable dominates - i.e. after the first few tens of metres.

Mike.

Thank you both for the replies. very much appreciate it..

Just to put some numbers in Zx = 0.02 and Zd = 0.15 

Uoc = Open circuit line-to-neutral voltage

I_pf(L-L) = U_OC(L-L)/[2×(Z_x+Z_d)]   

230 ÷ 0.17 x 2 =   230 ÷ 0.34 =676A

I see this as the impedance of the winding plus R1e (External) multiplied by 2 to account for the neutral ?  but the neutral won't have a winding impedance , so I would have thought 2x Zd + Zx.
Im obviously wrong in my thinking , just trying to grasp it...

Ipf = 676A

As you point out, U_OC(L-L)=√(3)×U_OC, and therefore we get:

I_pf(L-L) = √(3)×U_OC/[2×(Z_x+Z_d)] ≅ √(3)×C_maxU_O/[2×(Z_x+Z_d)]

1.73 x 230 ÷ 0.34 = 1171A    ≈  1.73 x1.1 x 230 ÷ 0.34 =  437 ÷ 0.34 = 1287A
Ipf = 1287

Hence:

I_pf(L-L) ≅ [√(3)/2]×C_maxU₀/(Z_x+Z_d) = [√(3)/2]×I_pf where I_pf is the bolted fault current across all three phases and neutral or PEN as above.

0.86 x 1.1 x 230 ÷ 0.17 = 217.58 ÷ 0.17 = 1279A   = 0.86 x 1287 (?) = 1106A`

√3 ÷ 2 x 1.1 =0.95  which is the factor for Cmin ??

Ipf = 1106

The assumption is as per Grahams explanation above, and if it helps you with the visualization of why, consider that  for any fault current down any one line is set by the voltage at the driving end of the line - 230V to ground 400V to any other phase- and the difference between that and whatever the voltage is at the fault end.

In this case the fault end voltage is the point of 3 wire junction - which, if all 3 fault currents are equal is exactly zero. So the full 230V appears along each line.

In contrast in the single phase fault, the fault end voltage rises by the drop in the return path - and usually L and N are assumed the same cross-section so half the voltage (115V) on the line and the other half on the way back,

Handily then having done a single phase PSSC test, (with 115V along the line) and double it and say 'calculated 3 wire fault current'

But be aware that all this assumes all 3 phases lines and the neutral return path, are all the same impedance, are the same. That is not always quite true but is close enough for most purposes.

And for fun, the fault current between any 2 phases puts the fault point midway along the phase to phase voltage, so 200V along each line, and a fault so the current  is  200 / 115 = 1.7 or so of the single phase test result.

Mike.