PE conductor size for TT system

[kk_bb]

Hello Everyone,

I'm using ecostruxure power design tool for my system. The problem is it always size the PE conductor following the rule of (half size of main conductor), although the earth fault current doesn't exceed 11A. As the earthing system is a TT system with a fault loop impedance of 20 ohm. So, isn't sizing as per the adiabatic equation available within the software.

Thanks in advance for your help

[Genghiz · Accepted solution]

Hello,

Thanks for reaching the EPD FORUM.

The cable (conductor sizing) on the EPD Ecodial is based on the IEC standard (IEC 60364-5-52 & 54).

The sizing of protective earth (PE) conductor is specific and half of the ‘live conductor.’

Thanks & Regards,

Team - Ecodial Expert Support

[kk_bb]

Thanks for your reply, and sorry for my late response. 

Thanks for the attached excerpt above, but it seems that it is extracted from an old version of IEC. As the updated IEC 60364-5-54 states that you can use the above table in case you do not use the adiabatic equation. The IEC even states that the protective conductor cross section shall not exceed 25mm2 for copper conductor and 35mm2 for aluminum conductor

Thanks and Best Regards

[Genghiz]

Thanks for your comments.

Can you please provide the extract from the IEC standard on this point.

The expectation is not yet clear...

The size of the PE conductor provided by the software is higher or smaller than the expected.? kindly please comment.

-Team Ecodial Expert Support.

[kk_bb]

Thanks for your reply.

I already shared an extract above in my comment (IEC 60384-5-54:2011- clause 543.1) which states that you can use the adiabatic equation in 543.1.2 for sizing the protective conductor. It even states that in TT system (which is my case) the protective conductor size shall not exceed 25mm2 for copper and 35mm2 for aluminum (refer to 543.1.1 shown above).

Back to my case, I have a TT system where the fault current does not exceed 11.5A, hence it is enough to have a protective conductor of 4mm2 ( using the adiabatic equation), however the software also propose half size of the main conductor which way more than enough. Refer to the attached photos in my original post.

Thanks and best regards.