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Calculating Earth Fault Loop Impedance in three phase + Neutral circuits

Nick Parker

I would like to calculate the Zs (external impedance). Would it be fine just calculating the impedance of "L1" and "N" in series from the secondary of the transformer till the final load and completing the loop.

Just concerned the "Ze" zone being the three phase circuit and have a thinking that just adding the impedance would yield inaccurate "Zs".

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    Er, for starters - your Earth fault loop should consist of the L plus c.p.c. impedances, not N.

    If it's a 'large' system (e.g. conductors over 25mm²) for accuracy, you might need to add the resistive and reactive components separately rather than just adding the overall impedances.

        - Andy.

  • 0 in reply to AJJewsbury
    AJJewsbury said:
    If it's a 'large' system (e.g. conductors over 25mm²) for accuracy, you might need to add the resistive and reactive components separately rather than just adding the overall impedances.

    If armoured cable is used, even relatively small csa, the armour csa is usually > 16 sq mm, and therefore the relevant method in PD CLC/TR 50480 should be used due to the reactance of the armour, or the armour/external cpc combination. Note that often - but especially if the cable is buried - the armour (perhaps in parallel with a copper conductor) has to be capable of acting as a cpc, whether you use it downstream or not, because a fault to it is possible, and it's often connected to exposed-conductive-parts.

    The 5th Edition (2022) of the IET Electrical Installation Design Guide, and the 9th Edition of Guidance Note 6 Protection Against Overcurrent, discuss the relevant calculation methods according to PD CLC/TR 50480.

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  • 0 in reply to AJJewsbury
    AJJewsbury said:
    If it's a 'large' system (e.g. conductors over 25mm²) for accuracy, you might need to add the resistive and reactive components separately rather than just adding the overall impedances.

    If armoured cable is used, even relatively small csa, the armour csa is usually > 16 sq mm, and therefore the relevant method in PD CLC/TR 50480 should be used due to the reactance of the armour, or the armour/external cpc combination. Note that often - but especially if the cable is buried - the armour (perhaps in parallel with a copper conductor) has to be capable of acting as a cpc, whether you use it downstream or not, because a fault to it is possible, and it's often connected to exposed-conductive-parts.

    The 5th Edition (2022) of the IET Electrical Installation Design Guide, and the 9th Edition of Guidance Note 6 Protection Against Overcurrent, discuss the relevant calculation methods according to PD CLC/TR 50480.

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