# Effects of harmonics - Overload of equipment

## Contents

## Generators

Generators supplying non-linear loads must be derated due to the additional losses caused by harmonic currents.

The level of derating is approximately 10% for a generator where the overall load is made up of 30% of non-linear loads. It is therefore necessary to oversize the generator, in order to supply the same active power to loads.

## Uninterruptible power systems (UPS)

The current drawn by computer systems has a very high crest factor. A UPS sized taking into account exclusively the r.m.s. current may not be capable of supplying the necessary peak current and may be overloaded.

## Transformers

- The curve presented below (see
**Fig.**M19) shows the typical derating required for a transformer supplying electronic loads

**Example:** If the transformer supplies an overall load comprising 40% of electronic loads, it must be derated by 40%.

- Standard UTE C15-112 provides a derating factor for transformers as a function of the harmonic currents.

## Asynchronous machines

Standard IEC60034-1 ("Rotating electrical machines – Rating and performance ") defines a weighted harmonic factor (Harmonic voltage factor) for which the equation and maximum value are provided below.

### Example

A supply voltage has a fundamental voltage U_{1} and harmonic voltages u_{3}= 2% of U_{1}, U_{5}, = 3%, U_{7}, = 1%. The THDu is 3.7% and the HVF is 0.018. The HVF value is very close to the maximum value above which the machine must be derated.

Practically speaking, asynchronous machines must be supplied with a voltage having a THDu not exceeding 10%.

## Capacitors

According to IEC 60831-1 standard ("Shunt power capacitors of the self-healing type for a.c. systems having a rated voltage up to and including 1 000 V – Part 1: General – Performance, testing and rating – Safety requirements – Guide for installation"), the r.m.s. current flowing in the capacitors must not exceed 1.3 times the rated current.

Using the example mentioned above, the fundamental voltage U_{1}, harmonic voltages u5 = 8% (of U_{1}), U_{7} = 5%, U_{11} = 3%, U_{13}, = 1%, i.e. total harmonic distortion THDu equal to 10%, the result is

I_{r.m.s.}/I_{1} = 1.19, at the rated voltage. For a voltage equal to 1.1 times the rated voltage,the current limit

I_{r.m.s.}/I_{1} = 1.3 is reached and it is necessary to resize the capacitors.

## Neutral conductors

Consider a system made up of a balanced three-phase source and three identical single-phase loads connected between the phases and the neutral (see **Fig.** M20).

**Figure** M21 shows an example of the currents flowing in the three phases and the resulting current in the neutral conductor.

In this example, the current in the neutral conductor has a rms value that is higher than the rms value of the current in a phase by a factor equal to the square root of 3.

The neutral conductor must therefore be sized accordingly.

The current in the neutral may therefore exceed the current in each phase in installation such as those with a large number of single-phase devices (IT equipment, fluorescent lighting). This is the case in office buildings, computer centers, Internet Data Centers, call centers, banks, shopping centers, retail lighting zones, etc.

This is not a general situation, due to the fact that power is being supplied simultaneously to linear and/or three-phase loads (heating, ventilation, incandescent lighting, etc.), which do not generate third order harmonic currents. However, particular care must be taken when dimensioning the cross-sectional areas of neutral conductors when designing new installations or when modifying them in the event of a change in the loads being supplied with power.

A simplified approach can be used to estimate the loading of the neutral conductor.

For balanced loads, the current in the neutral I_{N} is very close to 3 times the 3rd harmonic current of the phase current (I_{3}), i.e.: I_{N} ≈ 3.I_{3}

This can be expressed as: I_{N} ≈ 3. i_{3} . I_{1}

For low distortion factor values, the r.m.s. value of the current is similar to the r.m.s. value of the fundamental, therefore: I_{N} ≈ 3 . i_{3} I_{L}

And: I_{N} /I_{L} ≈ 3 . i_{3} (%)

This equation simply links the overloading of the neutral (I_{N} /I_{L}) to the third harmonic current ratio.

In particular, it shows that when this ratio reaches 33%, the current in the neutral conductor is equal to the current in the phases. Whatever the distortion value, it has been possible to use simulations to obtain a more precise law, which is illustrated in **Figure** M22

The third harmonic ratio has an impact on the current in the neutral and therefore on the capacity of all components in an installation:

- Distribution panels
- Protection and distribution devices
- Cables and trunking systems

According to the estimated third harmonic ratio, there are three possible scenarios: ratio below 15%, between 15 and 33% or above 33%.

### Third harmonic ratio below 15% (i_{3} ≤ 15%):

The neutral conductor is considered not to be carrying current. The cross-sectional area of the phase conductors is determined solely by the current in the phases. The cross-sectional area of the neutral conductor may be smaller than the cross-sectional
area of the phases if the cross sectional area is greater than 16 mm^{2} (copper) or 25 mm^{2} (aluminum).

Protection of the neutral is not obligatory, unless its cross-sectional area is smaller than that of the phases.

### Third harmonic ratio between 15 and 33% (15 < i_{3} ≤ 33%), or in the absence of any information about harmonic ratios:

The neutral conductor is considered to be carrying current.

The operating current of the multi-pole trunking must be reduced by a factor of 0.84 (or, conversely, select trunking with an operating current equal to the current calculated, divided by 0.84).

The cross-sectional area of the neutral MUST be equal to the cross-sectional area of the phases.

Protection of the neutral is not necessary.

### Third harmonic ratio greater than 33% (i_{3} > 33%)

This rare case represents a particularly high harmonic ratio, generating the circulation of a current in the neutral, which is greater than the current in the phases.

Precautions therefore have to be taken when dimensioning the neutral conductor.

Generally, the operating current of the phase conductors must be reduced by a factor of 0.84 (or, conversely, select trunking with an operating current equal to the current calculated, divided by 0.84). In addition, the operating current of the neutral conductor must be equal to 1.45 times the operating current of the phase conductors (i.e. 1.45/0.84 times the phase current calculated, therefore approximately 1.73 times the phase current calculated).

The recommended method is to use multi-pole trunking in which the cross-sectional area of the neutral is equal to the cross-sectional area of the phases. The current in the neutral conductor is therefore a key factor in determining the cross sectional area of the conductors. Protection of the neutral is not necessary, although it should be protected if there is any doubt in terms of the loading of the neutral conductor.

This approach is common in final distribution, where multi-pole cables have identical cross sectional areas for the phases and for neutral.

With busbar trunking systems, precise knowledge of the temperature rises caused by harmonic currents enables a less conservative approach to be adopted. The rating of a busbar trunking system can be selected directly as a function of the neutral current calculated.

For more details, see:

- Harmonic currents in the selection of busbar trunking systems (busways)
- Cahier Technique ECT212: "The neutral: A live and unique conductor"