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Line impedance query

me1899
Hi 


I am looking at a job where a supply to a replacement CT scanner has to be calculated.


The current scanner is fed by a cable with line impedance 0.06 at source end of the sub panel that feeds it.


The new scanner must have a line impedance no more than 0.06 so basically I can't use the way on the current panel.


The client engineer has said to run from the intake and provide a cable calc to prove the line impedance is OK for the new scanner feed (this is D and B).


I am using AMTECH and was wondering if there was a way to output line impedance or if I need to show a calculation another way?


The client engaged a firm to do line impedance tests, and their report shows values measured between L1-L2, L2-L3 and L1-L3, but looks like they did testing at the sub panel and not at the intake.


This is basically like PSCC right? So I can get PSCC from AMTECH, divide by root 3 and work out impedance from there?


Am I on the right track? Assistance would be much appreciated.


Regards


Anisur
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  • 0
    To get a closer actual measurement in the field measure the fault current phase to phase with a decent loop tester and divide the reading by 0.87.

    Agreed. Noting L-L not L-N.


    L-L PFC would typically be √3 times the L-N result - as usually there would be similar impedances (presuming N is similar to L) but driven by √3 times the voltage (433 rather than 250, or 400 rather than 230).


    0.87 I think comes from √3 / 2 - so dividing the L-L PFC result by 0.87 is the same as multiplying it by 2/√3. If the L-L result is √3 times the L-N result then you get L-N result * √3 * 2/√3 - the √3s cancel out so you end up with just L-N result *2.


    I think the rule of thumb doesn't actually give much of a margin for error at all. (Indeed might under-estimate if the N has a higher impedance than L - say old cables with a reduced N).


       - Andy.
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  • 0
    To get a closer actual measurement in the field measure the fault current phase to phase with a decent loop tester and divide the reading by 0.87.

    Agreed. Noting L-L not L-N.


    L-L PFC would typically be √3 times the L-N result - as usually there would be similar impedances (presuming N is similar to L) but driven by √3 times the voltage (433 rather than 250, or 400 rather than 230).


    0.87 I think comes from √3 / 2 - so dividing the L-L PFC result by 0.87 is the same as multiplying it by 2/√3. If the L-L result is √3 times the L-N result then you get L-N result * √3 * 2/√3 - the √3s cancel out so you end up with just L-N result *2.


    I think the rule of thumb doesn't actually give much of a margin for error at all. (Indeed might under-estimate if the N has a higher impedance than L - say old cables with a reduced N).


       - Andy.
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