The IT earthing system (unearthed neutral) in LV

3 The 2 nd insulation fault with the IT earthing systemAs we have already seen in the previouschapter, the advantage of using the IT system innetwork operation lies in the possibility ofcontinuity of supply even though an insulationfault has occurred on a circuit.This message has been received loud and clearby standard drafters who, in order to maintain ahigh level of availability, stipulate in installationstandards indication and tracking of this first faultso as not to fear a second fault. Protectiondevices are also provided for this second fault inorder to guarantee the same level of safety ofpersons as for the TN and TT earthing systems.The two sections below study the fault currentsand contact voltage which depend on how theframes are earthed. There are two possibilities,namely:c The load frames are all interconnected by aPE protective conductor: this is the generalcase.c The frames are not interconnected and areconnected to separate earth connections(configuration to be avoided due to EMC:see “Cahier Technique” no. 187).3.1 Analysis of the double insulation faultIn this section, fault currents and contact voltageare calculated by considering two full insulationfaults on two different live conductors (on onephase and the neutral if the neutral is distributed,or on two different phase conductors if theneutral is not distributed) of two circuits ofidentical cross-section and length.This assumption, which results in a minimumfault current, is the one normally chosen tocalculate the maximum lengths protected by theshort-circuit protection devices.Contact voltage and double fault currentwhen the frames are interconnectedWhen a fault current occurs between two faultyframes, a current flows in the phase conductorsand the PE protective conductor ensuringinterconnection of frames (see fig. 3b).This current is only limited by the impedance ofthe fault loop equal to the sum of theimpedances of the live conductors concernedand the circuit of the equipotential links (PE).There are a number of methods for calculatingfault currents for an electrical installation (see“Cahier Technique” no. 158).In this case, the conventional method has beenchosen, as it enables calculation of fault currentand contact voltage values without making toomany assumptions on installation characteristics.It will thus be used from now on in this “CahierTechnique” to give an idea of the value of thecurrents and voltages involved on a double faultin the IT system.It is based on the simplified assumption thatconsiders that, during the duration of the fault,the voltage at the origin of the feeder consideredis equal to 80% of installation nominal voltage.This assumes that the impedance of the feederin question accounts for 80% of total impedanceof the faulty loop, and that upstream impedanceaccounts for 20%.For the following calculations:U’ = phase to neutral voltage, (= U 0 if one of thetwo faults is on the distributed neutral),orU’ = phase-to-phase voltage, (= e U 0 if theneutral is not distributed).R a = ρ L = resistance of the live conductorS a(phase or neutral) of the circuit on which the faultoccurred.R pe = ρ L = resistance of the circuit protectiveS peconductor.S a = cross-section of the live conductor.S pe = cross-section of the protective conductor.L = length of the faulty circuits.m = Sa= ratio of live conductor cross-sectionSpeover protective conductor cross-section(normally i 1).c If we consider that the live and PE conductorsof the two faulty feeders have the same crosssectionand length and if we ignore their reactance:v if one of the faults is on the neutral0,8 UI0Sd =, i.e. Ia2R ( a + R pe )d =0,8U 02ρ ( 1+mL ),v if the double fault concerns two phaseSconductors Iad =0,8eU0 .2 ρ 1 + mL( )Cahier Technique Schneider Electric no. 178 / p.15