Maximum and Minimum fault currents

Question

Bit confused, I was reading this in a book called modern wiring practices 
 Calculation minimum fault current 3P 3 wire. Ipp = VL / 2zp 
 Ipp - phase / Phase PSSC 
 VL - Line voltage 
 2zp inpedance of one phase only So using 400V In the design guide its. 
 Ief = Cmin Uo / Zx + CrZd + Cr'Z1 +Cr"Z2 + Cr'"Zpen 
 So using 230V EIDG tells us A fault across three phases is considered the worst case.... and we get Cmax Uo / Zx +Zd Again voltage to earth is used. But basically a bit unsure why 230V is used to calculate 3 phase fault current and not 400V Any help really appreciated. Thanks

gkenyon · Accepted Answer

Moderators, please could this be re-located to Wiring and the Regulations ? It relates to the Electrical Installation Design Guide, which is all about calculations to BS 7671.

mapj1 · Answer

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.

gkenyon · Answer

Apologies for not noticing earlier. This should have been posted in the Wiring and the Regulations forum as it relates directly to BS 7671 and its guidance publications. 
 I think you have mis-interpreted the phrase "A fault across three phases is considered worst-case" ... this is a three-phase bolted fault across all phases, not a line-to-line fault. 
 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. 
 The maximum line-to-line fault current must take into account the impedance of each line and each phase winding, and therefore using the same symbols as present in Section 6.3.2 of the IET Electrical Installation Design Guide , 5th Edition, would be: I pf(L-L ) = U OC(L-L) /[2&times;(Z x +Z d )] 
 
 As you point out, U OC(L-L) =&radic;(3)&times;U OC , and therefore we get: 
 I pf(L-L ) = &radic;(3)&times;U OC /[2&times;(Z x +Z d )] &cong; &radic;(3)&times;C max U O/ [2&times;(Z x +Z d )] 
 
 Hence: 
 I pf(L-L ) &cong; [&radic;(3)/2]&times;C max U 0 /(Z x +Z d ) = [&radic;(3)/2]&times;I pf where I pf is the bolted fault current across all three phases and neutral or PEN as above. 
 
 Quite simply, because [&radic;(3)/2] < 1, mathematically I pf(L-L) < I pf . Or in plain English, the line-line prospective fault current is less than the three-phase-to-neutral bolted fault current. 
 Hopefully, this demonstrates why the maximum prospective fault current occurs not line-to-line, but between all lines and Neutral or PEN, and why U 0 is used rather than U 0 &times;&radic;(3).

Maximum and Minimum fault currents

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