Simplicio: Calculation of earth fault loop impedance is shown below:

For a 250m, 50 mm² phase copper conductor:

R(phase) = 0.494*0.25 = 0.1235 Ω
X(phase) = 0.0751*0.25 = 0.0188 Ω

Z(phase) = 0.125 Ω

Z(earth) = unknown, depends on earth cable size

Impedance of single phase to earth fault loop, Z(fault) = Z(phase) + Z(earth)

I(fault) = 80 % of phase to neutral voltage/Z(fault) = 0.8*230/Z(fault) = 184/Z(fault)

[Ref AS/NZS 3008:2007, Cl B5.2.1 (b)]

Touch potential, V(touch) = I(fault)*Z(earth)

From Table 8.1 (AS 3000), for a disconnection time of

0.4 seconds : Z(fault)<0.27 which implies Z(earth)<0.27-0.125=0.145 --------------- condition 1

5 seconds : Z(fault)<0.48 which implies Z(earth)<0.48-0.125=0.355 --------------- condition 2

From Table 5.1 (AS 3000), minimum earth conductor size for 50mm² copper active conductors is 16mm²

Calculating impedance for 250m, 16mm² copper earth conductor gives

Z(earth) ~ 1.47*0.25 = 0.3675 which is unacceptable according due to condition 2

This can also be seen from the prospective fault current graph for the 100A fuse shown below, which shows that for a disconnection time of less than 5s [which is the maximum allowable disconnection time according to AS 3000:2007, Cl 5.7.2 (b)], fault current must be greater than 430A. Using a 16mm² earth conductor in the equation, gives I(fault) = 374A

Using the next available earthing conductor size of 25 mm²

Z(earth) = 0.2326

Z(fault) = 0.3576

I(fault) = 514A

V(touch) = 120V

Disconnection time = 3s

But from Fig 5 (AS 3007.2:2004), for disconnection time of 3s, V(touch) should be less than 60V

Selecting the next conductor size of 35mm²

Z(earth) = 0.1684

Z(fault) = 0.2934

I(fault) = 627A

V(touch) = 106V

Disconnection time ~1.3s (Not acceptable due to same reason as above)

Selecting the next conductor size of 50mm²

Z(earth) = 0.125

Z(fault) = 0.25

I(fault) = 736

V(touch) = 92V

Disconnection time = 0.6s

Acceptable V(touch) ~ 90V

We are just short of meeting our criteria. I don’t think we use earth conductors larger than active conductors, so I am not doing a calculation for a 70mm² conductor, which will definitely meet the criteria. However, we may use the 50mm² if we know that the operating temperature will be lower than 90^oC, in which case, Z(fault) will decrease.



Here are the links for PART 1 & PART 2

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