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40    Chapter Three
























                                  FIGURE 3.9 Diagrammatic representation of Class II equipment in bonded
                                  conductive enclosure.


                                  supplementary insulation (S SI ). Eq. (3.3) applies:


                                  S II (t) = 1 − [1 − S BI (t)][1 − S SI (t)] = e −  BI t  + e −  SI t  − e −(  BI +  SI )t  (3.13)

                                     Let us compare S II and S BI . Since   SI <   BGCPD , we obtain:

                                                          S II (t) > S I (t)           (3.14)


                                     In addition, in the absence of a conductive enclosure, the prob-
                                  ability that persons can touch a fault potential caused by the failure
                                  of both insulation layers is much lower than in the case of Class I
                                  equipment (i.e., k II < k I ), hence, r II (t) < r I (t).
                                     What would happen if Class II equipment were installed in a metal
                                  enclosure that is bonded (Fig. 3.9)?
                                     We can reasonably assume that the enclosure is more likely to be
                                  energized due to voltages transferred by the bonding connection than
                                  due to failure of its own double insulation. In fact, the probability of
                                  failure of Class II equipment is considered very low when compared
                                  to the probability F TP (t) that the enclosure becomes live due to trans-
                                  ferred voltages. Thus, international standards prohibit the bonding of
                                  Class II equipment. 8
                                     We can express safety of Class II equipment in bonded enclosure
                                  S IIBE (t)as

                                          S IIBE (t) = 1 − [1 − S I (t)][1 − S SI (t)] − F TP (t)  (3.15)
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