Nothing on the box, so I thought new starters would like an idea how the above is obtained.
R ohms X ohms
1 x 500 kVA transformer 0.005 0.017
100 m of 2 x185 mm2 CNE 0.008 0.004
aluminium (Ø)
100 m x 2 x 185 mm2 CNE
aluminium (neutral/earth) 0.008 0.001
Total 0.021 0.022
Earth fault loop impedance
Zs(A) = V(R2 + X2)
Zs(A) = V(0.021sq + 0.022sq)
= 0.030 ohms (at distribution board A)
A 630A fuse(supply intake) requires an impedance of less than 0,054 ohms.
50 m x 185 mm2 swa
aluminium (Ø) 0.008 0.004
aluminium (armouring) 0.032
Total 0.061 0.026
Earth fault loop impedance
Zs(B) =V(R2 + X2)
Zs(B) =V(0.061sq + 0.026sq)
= 0.066 (at B)
A 250A fuse( at distribution board B) requires an impedance of less than 0.16 ohms.
5 m 6mmsq pvc single cores
in 25mm conduit 0.023
5 m of 25mm conduit (light gauge) 0.007 0.008
Total 0.091 0.034
Earth fault loop impedance
Zs(C) = V(R2 + X2)
Zs(C) = V(0.091sq + 0.034sq)
= 0.097 ohms (at C)
A 32A fuse requires an impedance of less than 1.8 ohms.
Regards UKPN
