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IEC 60364-5-52Electrical installations of buildings –Part 5-52:Selection and erection of electrical equipment – Wiring systems
Standard prescriptions
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Neutral conductor sizing
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Standard prescriptions
IEC 60364-5-52 Art. 524.3For polyphase circuits where each phase conductor has a cross-sectional area greater than 16 mm2 in copper or 25 mm2 in aluminium, the neutral conductor may have a smaller cross-sectional area than that of the line conductors if the following conditions are simultaneously fulfilled:
– the expected maximum current including harmonics, if any, in the neutral conductor during normal service is not greater than the current-carrying capacity of the reduced cross sectional area of the neutral conductor;
– the neutral conductor is protected against overcurrentsaccording to the rules of 431.2 of IEC 60364-4-43;
– the size of the neutral conductor is at least equal to 16 mm2 in copper or 25 mm2 in aluminium.
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Il mondo è cambiato per sempreThe world …
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… changed forever!
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Theorethical background
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Theorethical backgroundBalanced linear loads
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Theorethical backgroundUnbalanced linear loads
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Non linear unbalanced loads
Theorethical background
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Theorethical background3rd order harmonics
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Sizing criteria
IEC 60364-5-52 Annex DReduction factors
for harmonic currents in four-core and five-core cables
Third harmonic content of phase
current (%)
Size selection is based on phase
current
Size selection is based on neutral
current
0-15 1,00 - 15-33 0,86 - 33-45 - 0,86
45 - 1,00
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installation method C
Example
Phases 3
design load 37,5 A
Number of cores 4
Insulation EPR
Installation method Clipped to a wall
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Example
Harmonics free environment
– Cable size related to phase currents
– ⇒ 4 mm2
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Office # 1
3 h = 22%
− IN = 37,5 · 0,22 · 3 = 24,7 A − IF = 37.5 A − IN < IF size related to phase
current− Equivalent phase current:
− ⇒ 6 mm2
(IEC 60364-5-52 tab. A.52-5 )
Example
6,4386,0
5,37=
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Office # 2
3 h = 40%
− IN = 37,5 · 0,4 · 3 = 45 A − IF = 37.5 A − IN > IF size related to neutral
current− Equivalent neutral current:
− ⇒ 10 mm2
(IEC 60364-5-52 tab. A.52-5 )
Example
3,5286,0
45=
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Lighting equipment FLC
3 h = 72% + 9 h = 33,9% + …
− IN = 37,5 · 1,30 · 3 = 146,2 A − IF = 37.5 A − IN > IF size related to neutral
current− Equivalent neutral current:
− ⇒ 35 mm2
(IEC 60364-5-52 tab. A.52-5 )
2,1461
2,146=
Example
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The End
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