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What's Ssc?

AJJewsbury

I've just read in the installation instructions for a heat pump...

This unit complies with IEC 61000-3-12 under the prerequis-
ite that the short circuit line Ssc at the connection point for
the customer's installation to the public grid is greater than or
equal to 33. It is the responsibility of the installer or end user
of the unit to ensure, if necessary after consultation with the
network operator, that this unit is only connected to one con-
nection point with an Ssc value that is greater than or equal
to 33.

which got me wondering what on earth Ssc was (I've seen a few other things over they years that have specified a minimum L-N loop impedance for the supply, but at 33 this looked somewhat different).

A bit of research (e.g. Cahier technique no. 158 and this WPD document) seem to suggest that Ssc can be calculated as U²/Zsource - but plugging in some vaguely plausible numbers into that e.g. 400²/0.15Ω gives me values many orders of magnitude away from 33. Working it backwards 33 would imply a Z of around 150kΩ which feels equally implausible.

Obviously there's something I'm not understanding here. Or did they perhaps mean Rsce = 33?

   any ideas?

   - Andy.

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  • AJJewsbury said:
    Or did they perhaps mean Rsce = 33?

    I think that it is a typo. BS EN ISO-3-12:2011+A1:2024 refers to a R_sce limit of 33. (Article 5.2)

    For single phase equipment, R_sce = S_sc/3S_equ

    S_sc is the short-circuit power, which is equal to (U_nominal)²/Z where U_nominal is the interphase system voltage and Z is the line impedance at the PCC. The PCC is the point of common coupling, i.e. the point in the public system which is closest to the customer concerned and to which other customers are or may be connected.

    S_ecu is the rated apparent power of the equipment, i.e. (for single phase) the product of the rated voltage and the rated current.

    So let's put in some figures. Take a 32 A heat pump. S_ecu = 32 x 0.23 = 7.36 kW.

    I assume that Z would be the line impedance where the service cable joins the street mains, so let's say 0.2 Ω. So S_sc = 400²/0.2 = 800 kW.

    Now R_sce = 800k/(3 x 7.36k) = 36.

    So, Andy, I think that your original hunch was correct.

  • in reply to Chris Pearson

    thanks Chris!

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