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| ==== The same calculation using the simplified method recommended in this guide ====
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| *Dimensioning circuit C1
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| The MV/LV 630 kVA transformer has a rated no-load voltage of 420 V. Circuit C1 must be suitable for a current of:
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| <br>Two single-core PVC-insulated copper cables in parallel will be used for each phase.These cables will be laid on cable trays according to method F.<br>Each conductor will therefore carry 433A. Figure G21a indicates that for 3 loaded conductors with PVC isolation, the required c.s.a. is 240mm².<br>The resistance and the inductive reactance, for the two conductors in parallel, and for a length of 5 metres, are:
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| (cable resistance: 22.5 mΩ.mm2/m)
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| X = 0,08 x 5 = 0,4 mΩ (cable reactance: 0.08 mΩ/m)
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| *Dimensioning circuit C3
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| Circuit C3 supplies two 150kW loads with cos φ = 0.85, so the total load current is:
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| Two single-core PVC-insulated copper cables in parallel will be used for each phase. These cables will be laid on cable trays according to method F.<br>Each conductor will therefore carry 255A. Figure G21a indicates that for 3 loaded conductors with PVC isolation, the required c.s.a. is 95mm².<br>The resistance and the inductive reactance, for the two conductors in parallel, and for a length of 20 metres, are:
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| *Dimensioning circuit C7
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| Circuit C7 supplies one 150kW load with cos φ = 0.85, so the total load current is:
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| One single-core PVC-insulated copper cable will be used for each phase. The cables will be laid on cable trays according to method F.<br>Each conductor will therefore carry 255A. Figure G21a indicates that for 3 loaded conductors with PVC isolation, the required c.s.a. is 95mm².<br>The resistance and the inductive reactance for a length of 20 metres is:<br>
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| (cable resistance: 22.5 mΩ.mm2/m)
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| (cable reactance: 0.08 mΩ/m)
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| *Calculation of short-circuit currents for the selection of circuit-breakers Q1, Q3, Q7 (see'''Fig. G67''')
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| {| style="width: 798px; height: 243px" cellspacing="1" cellpadding="1" width="798" border="1"
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| '''''Fig. G67:'''''<i>Example of short-circuit current evaluation</i>
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| *'''The protective conductor'''
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| When using the adiabatic method, the minimum c.s.a. for the protective earth conductor (PE) can be calculated by the formula given in Figure G58:
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| For circuit C1, I = 20.2kA and k = 143.<br>t is the maximum operating time of the MV protection, e.g. 0.5s<br>This gives:
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| A single 120 mm<sup>2</sup> conductor is therefore largely sufficient, provided that it also satisfies the requirements for indirect contact protection (i.e. that its impedance is sufficiently low).<br>Generally, for circuits with phase conductor c.s.a. Sph ≥ 50 mm2, the PE conductor minimum c.s.a. will be Sph / 2. Then, for circuit C3, the PE conductor will be 95mm<sup>2</sup>, and for circuit C7, the PE conductor will be 50mm2.
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| *'''Protection against indirect-contact hazards'''
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| For circuit C3 of '''Figure G65''', '''Figures F41 '''and'''F40''', or the formula given '''page F25 '''may be used for a 3-phase 4-wire circuit.<br>The maximum permitted length of the circuit is given by:
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| (The value in the denominator 630 x 11 is the maximum current level at which the instantaneous short-circuit magnetic trip of the 630 A circuit-breaker operates).<br>The length of 20 metres is therefore fully protected by “instantaneous” over-current devices.
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| *'''Voltage drop'''
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| The voltage drop is calculated using the data given in'''Figure G28''', for balanced three-phase circuits, motor power normal service (cos φ = 0.8).<br>The results are summarized on '''figure G68:'''<br>
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| <br>'''''Fig. G68:'''''<i>Voltage drop introduced by the different cables</i>
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| The total voltage drop at the end of cable C7 is then: 0.77%.
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