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## Introduction

* Sizing (c.s.a.)* and the

*of the neutral conductor, apart from its current-carrying requirement, depend on several factors, namely:*

**protection**- The type of earthing system, TT, TN, etc.
- The harmonic currents
- The method of protection against indirect contact hazards according to the methods described below

throughout its length with, in addition, light blue markings at the terminations, or**Green-and-yellow**throughout its length with, in addition, green-and-yellow markings at the terminations**Light blue**

## Sizing the neutral conductor

### Influence of the type of earthing system TT and TN-S schemes

Single-phase circuits or those of c.s.a. ≤ 16 mm^{2} (copper) 25 mm^{2} (aluminium): the c.s.a. of the neutral conductor must be equal to that of the phases.

*or*

**> 16 mm**^{2}copper*: the c.s.a. of the neutral may be chosen to be:*

**25 mm**^{2}aluminium- Equal to that of the phase conductors, or
- Smaller, on condition that:
- The current likely to flow through the neutral in normal conditions is less than thepermitted value Iz. The influence of triplen (Harmonics of order 3 and multiple of 3) harmonics must be given particular consideration or
- The neutral conductor is protected against short-circuit, in accordance with the following Sub-clause G-7.2
- The size of the neutral conductor is
or**at least equal to 16 mm**^{2}in copper**25 mm**^{2}in aluminium

#### TN-C scheme

The same conditions apply in theory as those mentioned above, but in practice, the neutral conductor must not be open-circuited under any circumstances since it constitutes a PE as well as a neutral conductor (*see IEC 60364-5-54 “c.s.a. of PEN conductor” column*).

#### IT scheme

In general, it is not recommended to distribute the neutral conductor, i.e. a 3-phase 3-wire scheme is preferred. When a 3-phase 4-wire installation is necessary, however, the conditions described above for TT and TN-S schemes are applicable.

### Influence of harmonic currents

#### Effects of triplen harmonics

Harmonics are generated by the * non-linear loads* of the installation (computers, fluorescent lighting, rectifiers, power electronic choppers) and can produce high currents in the Neutral.

In particular triplen harmonics of the three Phases have a tendency to cumulate in the Neutral as:

1. Fundamental currents are out-of-phase by 2π/3 so that their sum is zero

2. On the other hand, triplen harmonics of the three Phases are always positioned in the same manner with respect to their own fundamental, and are in phase with each other (* see Figure 1*).

* Figure 2* shows the

*of the neutral conductor as a function of the percentage of 3rd harmonic. In practice, this maximum load factor cannot exceed 3.*

**load factor**Reduction factors for harmonic currents in four-core and five-core cables with four cores carrying current

The basic calculation of a cable concerns only cables with three loaded conductors i.e there is no current in the neutral conductor. Because of the third harmonic current, there is a current in the neutral. As a result, this neutral current creates an hot environment for the 3 phase conductors and for this reason, a reduction factor for phase conductors is necessary (*see Figure 3 below*).

* Figure 3*Reduction factors for harmonic currents in four-core and five-core cables (

*according to IEC 60364-5-52*)

Third harmonic content of phase current (%) |
Reduction factor | |

Size selection is based on phase current | Size selection is based on neutral current | |

0 – 15 | 1.0 | - |

15 – 33 | 0.86 | - |

33 – 45 | - | 0.86 |

> 45 | - | 1 |

Reduction factors, applied to the current-carrying capacity of a cable with three loaded conductors, give the current-carrying capacity of a cable with four loaded conductors, where the current in the fourth conductor is due to harmonics. The reduction factors also take the heating effect of the harmonic current in the phase conductors into account.

Where the neutral current is expected to be higher than the phase current, then the cable size should be selected on the basis of the neutral current.

Where the cable size selection is based on a neutral current which is not significantly higher than the phase current, it is necessary to reduce the tabulated current carrying capacity for three loaded conductors.

In order to protect cables, the fuse or circuit-breaker has to be sized taking into account the greatest of the values of the line currents (phase or neutral). However, there are special devices (for example the Compact NSX circuit breaker equipped with the OSN tripping unit), that allow the use of a c.s.a. of the phase conductors smaller than the c.s.a. of the neutral conductor.

*A big economic gain can thus be made.*

### Examples

Consider a three-phase circuit with a design load of 37 A to be installed using fourcore PVC insulated cable clipped to a wall, installation method C.

From * Figure 4*, a 6 mm

^{2}cable with copper conductors has a current-carrying capacity of 40 A and hence is suitable if harmonics are not present in the circuit.

* Figure 4* – Clustering coefficient according to the number of current consumers

Application | Number of current consumers | Ks Coefficient |

Lighting, Heating | 1 | |

Distribution (engineering |
2…3 | 0.9 |

4…5 | 0.8 | |

6…9 | 0.7 | |

10…40 | 0.6 | |

40 and over | 0.5 |

**Note :** for industrial installations, remember to take account of upgrading of the machine equipment base. As for a switchboard, a 20 % margin is recommended: In ≤ IB x ks x 1.2.

#### If 20 % third harmonic is present:

Then a reduction factor of 0,86 is applied and the design load becomes: **37/0.86 = 43 A**.

For this load a 10 mm2 cable is necessary. In this case, the use of a special protective device (Compact NSX equipped with the OSN trip unit for instance) would allow the use of a 6 mm^{2} cable for the phases and of 10 mm^{2} for the neutral.

#### If 40 % third harmonic is present:

The cable size selection is based on the neutral current which is: **37 x 0,4 x 3 = 44,4 A** and a reduction factor of 0,86 is applied, leading to a design load of: 44.4/0.86 = 51.6 A.

For this load a 10 mm^{2} cable is suitable.

#### If 50 % third harmonic is present:

The cable size is again selected on the basis of the neutral current, which is: **37 x 0,5 x 3 = 55,5 A** .In this case the rating factor is 1 and a 16 mm^{2} cable is required.

In this case, the use of a special protective device (Compact NSX equipped with the OSN trip for instance) would allow the use of a 6 mm^{2} cable for the phases and of 10 mm^{2} for the neutral.

**Resource:** Electrical Installation Guide 2010 – Schneider Electric

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