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Cable resistance and voltage drop
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<blockquote data-quote="TonyMitchell" data-source="post: 490973" data-attributes="member: 32191"><p>Let me have a go at this one, although I’m a little confused by the 12kW load and specified with a load current of 40A, since as U0=230V, I would have thought Ib would be 52A, not 40A ?</p><p></p><p>Using Ib=40A</p><p> (Table 4D2B) 10mm2 Two core single phase AC voltage drop = 4.4mV/A/m</p><p> Voltage Drop = (4.4*40*70.25)/1000=12.4V</p><p></p><p>Power Loss = (12.4*40)=496W</p><p></p><p>(GN3 Table B1) 10mm2 R1 1.83mΩ/m</p><p> R1 = (1.83*70.25)/1000=0.13Ω </p><p></p><p>Reduce voltage drop to 5A overall by using a parallel conductor:</p><p> 13V/5V=2.6 ratio</p><p> (10*2.6=26)-10=16mm2 parallel conductor required</p><p></p><p>16mm2 conductor resistance 1.15mΩ/m</p><p></p><p>Parallel conductor resistance (1.15*70.25)/1000=0.08Ω</p></blockquote><p></p>
[QUOTE="TonyMitchell, post: 490973, member: 32191"] Let me have a go at this one, although I’m a little confused by the 12kW load and specified with a load current of 40A, since as U0=230V, I would have thought Ib would be 52A, not 40A ? Using Ib=40A (Table 4D2B) 10mm2 Two core single phase AC voltage drop = 4.4mV/A/m Voltage Drop = (4.4*40*70.25)/1000=12.4V Power Loss = (12.4*40)=496W (GN3 Table B1) 10mm2 R1 1.83mΩ/m R1 = (1.83*70.25)/1000=0.13Ω Reduce voltage drop to 5A overall by using a parallel conductor: 13V/5V=2.6 ratio (10*2.6=26)-10=16mm2 parallel conductor required 16mm2 conductor resistance 1.15mΩ/m Parallel conductor resistance (1.15*70.25)/1000=0.08Ω [/QUOTE]
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Cable resistance and voltage drop
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