08 Harmonics

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Industrial electrical network design guide T & D 6 883 427/AE

8. Harmonics


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8. HARMONICS

The currents consumed by non-linear electrical equipment such as arc furnaces, lighting

systems, convertors, rectifiers, etc., are non-sinusoidal, and flowing through the impedances of

the power supply network they cause the sinusoidal supply voltage to become distorted. Thewaveform distortion is characterised by the appearance of voltages at harmonic frequencies.

These harmonic voltages can disturb the operation of electrical devices on the network.

This chapter deals with the different sources of harmonic disturbance, as well as the methods

and techniques that can be applied to reduce it to an acceptable level.

8.1. Basic concepts

This paragraph gives the technical and theoretical fundamentals necessary for an investigation

of harmonics.

8.1.1. Fourier series periodic signal analysis

The French mathematician Joseph Fourier showed that a periodic signal ( )s t , of period T, can

be expressed as the sum of sinusoidal signal components plus a d.c. component:

( ) ( )s ta

a p t b p t p

p p= + +=

01

2cos sin

where:

=2

T

( )aT

s t dt T

00

2=

( ) ( )

( ) ( )

aT

s t p t dt

bT

s t p t dt

pT

p

T

=

=

2

2

0

0

cos

sin

where p is a whole number

This analysis may also be expressed as follows:

( ) ( )s ta

c p tp

p p= + +=

01

2sin

where: c a b p p p= +2 2

pp

p

b

a= arctan


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8.1.2. Definitions

A distorted voltage ( )v t of period T (T ms= 20 at f Hz= 50 ) can therefore be expressed in

the following manner:

( ) ( )v t V V p t p

p p= + +=

01

2 sin where

=2

T

V0 : amplitude of the d.c. component, generally zero and taken to be so from now on .

p : the initial phase ofVp , ( )t= 0 .

Similarly, a distorted current ( )i t of period T can be expressed as:

( ) ( )i t I I p t p

p p= + +=

01

2 sin

I0 : amplitude of the d.c. component, generally zero and taken to be so from now on .

p : the initial phase ofIp , ( )t= 0 .

fundamental component, or fundamental

V1 is the fundamental component of the signal ( )v t , that is to say the effective value of thesinusoid whose frequency is the same as that of the supply network.

harmonic component, or harmonic

For values of p 2 , Vp is the harmonic component of order p , of the signal ( )v t ; that is, the

effective value of the sinusoidal voltage whose frequency is p times that of the supply

network.

order of harmonic

The whole number equal to the ratio between the frequency of the harmonic and that of the

fundamental.

p is therefore the order of the harmonic.

For example, V3 is the third harmonic voltage, or simply the 3rd harmonic.


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distortion factor according to IEC standard 1000-2-2

- voltage distortion factor ( )V

p

p

V

V% = =

100

2

2

1

- current distortion factor ( )I

pp

I

I% =

=

100

2

2

1

The distortion factor as defined by the IEC standard represents the ratio between the r.m.s.

value of the harmonics to that of the (undistorted) fundamental, and gives a good indication of

the degree of harmonic disturbance on the network. This definition will be used throughout theremainder of the document.

The DIN factor can be converted to the IEC factor as follows:

1 11

2 2D V V= +

1 11

2 2

D I I

= +

As will be seen, IEC distortion factor values can be greater than 100 %.

For low distortion factors the two definitions give nearly identical values. On the other hand, for

high distortion factors the values turn out to be very different.

For example, for I = 10 % we find DI = 9 95. %

for I = 120% we find DI = 77 %

individual harmonic rate

The rate of the harmonic of order p is given by:

( )VV

Vpp

% = 1001

( )II

Ipp

% = 1001


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power values for non-sinusoidal signals

Definitions of power applying to sinusoidal signals are not appropriate to the treatment of non-

sinusoidal signals.

Consider a non-sinusoidal voltage and current whose Fourier series components are:

( ) ( )i t I p t pp

p= +=

1

2 cos

( ) ( )v t V p t pp

p= +=

21

cos

By definition, active power is equal to mean power:

( ) ( )PT

v t i t d t T

= 1 0

and after calculation, we obtain:

( ) P V I p pp

p p= =

1

cos

By definition, apparent power is equal to:

S V Irms rms=

The definition of reactive power applied to purely sinusoidal waveforms is wholly inappropriate

to the treatment of non-sinusoidal signals.

Certain documents do offer a definition, but since the reactive power of a non-sinusoidal signal

has no practical relevance to an investigation of harmonics there is no need to refer to it in thisdocument.


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power factor

The power factor is defined as the ratio between active power and apparent power:

F PSp

=

displacement factor(for the fundamental)

The displacement factor is the ratio between active power and apparent power for the

fundamental component:

cos 1

1

1

=P

S

It can also be defined as the cosine of the phase displacement between the fundamental

voltage and current components : 1 = phase displacement ( )V I1 1, .

Note: Taking the reference (in time) relative to U1 , we have 1 0= .

harmonic factor

This expresses the relationship between the power factor and the displacement factor:

FF

hp=

cos1

peak factor

This is the ratio of peak current to r.m.s. current:

F

Ipk rms=

interharmonics

Interharmonics are sinusoidal components whose frequencies are not multiples of the

fundamental frequency: 130 Hz, 170 Hz, 220 Hz, etc.


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infraharmonics

These are sinusoidal components whose frequencies are below that of the fundamental:

10 Hz, 20 Hz...

The presence of interharmonics or infraharmonics is due to periodic or random variation in the

power consumed by certain electrical equipment. In this case, the signal is not periodic over

time T (period of fundamental), a factor which explains the appearance of frequency

components supplementary to those expressed by Fourier analysis.

Common sources of the phenomena include:

- arc furnaces- cycloconverters- speed variators.

frequency spectrum

A graphical representation of harmonics showing their individual amplitudes and orders.

Usually, the level of each harmonic is represented by its value as a percentage of the

fundamental (see fig. 8-1).

1 3 5 7

100

n

I1(%)

Ip

Figure 8-1: frequency spectrum of a non-sinusoidal current


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linear and non-linear loads

A load is said to be linear if its impedance is constant: the current input is therefore sinusoidal

when the applied voltage is sinusoidal (see fig. 8-2).

A load is said to be non-linear if its impedance varies over a given period: the current input is

therefore non-sinusoidal for a sinusoidal supply voltage (see fig. 8-3).

t

v t

i t

Figure 8-2: linear load

t

v t

i t

Figure 8-3: non-linear load


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8.1.3. The main sources of harmonics

non-linear loads

In this section we shall determine figures for the individual harmonic currents generated by

common non-linear loads.

The values given are approximate. They vary notably in relation to the upstream impedance

(generally, when the upstream impedance increases, the harmonic current levels decrease).

six-phase rectifier bridge (see fig. 8-4)

This is a device to convert 3-phase alternating current into a single-phase direct current.

T

t

t

t

T

3T

6

i t

i ta

i tb

i tc

i tc

i tb

i ta

Figure 8-4: six-phase rectifier bridge current waveforms.

Theoretically, each of the currents I I I a b c, , has a rectangular waveform.

The Fourier series analysis of the rectangular signal gives us the following harmonic currents

of order p k= 6 1 ( that is: 5, 7, 11, 13, 17, 19...) and of amplitude II

pp= 1

I1 : amplitude of fundamental .


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In reality, the current waveforms are not perfectly rectangular, with the result that the actual

values of the harmonic current are less than their theoretical values. An empirical law exists

(see IEC 146-1-2 3.6.2.1) enabling a calculation to be made of the approximate values of

harmonic currents 5 to 31:

II

pp

p =

11

5.2

for 5 31 p

From it we can obtain the individual values of harmonic current as a percentage of the

fundamental (see table 8-1).

I1 I5 I7 I11 I13 I17 I19 I23 I25 I29 I31

100 % 18.9 % 11.0 % 5.9 % 4.8 % 3.4 % 3 % 2.3 % 2.1 % 1.8 % 1.6 %

Table 8-1: values of the harmonic currents produced by a rectifier bridge

The formula yields approximate values, especially when high thyristor delay angles, , are

utilised. Moreover, the inductance of the d.c. circuit influences the harmonic content. A low

value of inductance is conducive to high d.c. ripple, thereby increasing the 5th. harmonic

(multiplied by 1.3 or greater ), as well as, but to a lesser degree, the 11th, 17th... ( )6 1k . The

levels of harmonics 7, 13... ( )6 1k+ are usually lower. IEC standard 146-1-2, paragraphs 3.6.4

and 3.6.5 gives a (very complicated) method of calculating the precise figure for each

harmonic together with its phase angle.

Generally speaking, a calculation of the precise theoretical value of each harmonic current is

not essential for an investigation, indeed whenever possible, on-site measurement is the

preferred method of obtaining the precise values.


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computer switchmode power supplies (see fig. 8-5)

t

CR

i t

v t

i t

v t

Figure 8-5: switchmode power supplies for computers, current waveform

The harmonic currents produced by a switchmode power supply are more or less high in

relation to the load and the upstream network impedance.

A spread of values, in line with the two assumptions of high and low harmonic content, is given

below (see table 8-2).

I1 I3 I5 I7 I9 I11

high 100 % 130 % 70 % 50 % 30 % 10 %

low 100 % 65 % 35 % 25 % 15 % 5 %

Table 8-2: values of harmonic currents for a computer switchmode power supply

lighting loads (fluorescent tubes, discharge lamps)

The values of harmonic currents supplied by fluorescent tubes and magnetic ballast discharge

lamps are given in table 8-3.

Electronic ballast discharge lamps supply harmonic currents having a value comparable to that

of a computer switchmode power supply (see table 8-2: high).

I1 I3 I5 I7 I9 I11 I13 I15

100 % 35 % 27 % 10 % 2.5 % 3.5 % 1.5 % 1.5 %

Table 8-3: fluorescent tube and magnetic ballast discharge lamp current harmonic values


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uninterruptible power supplies (UPS)

For example, the GALAXY model produces harmonic currents as shown in table 8-4.

I1 I5 I7 I11 I13 I17 I19

100 % 33 % 2.5 % 6.1 % 2.4 % 2.5 % 1.6 %

Table 8-4: values of harmonic currents for the GALAXY model of UPS

Note: the newer generation of "sine wave sampling" inverters do not generate harmonics(for example, the 0 to 4 kVA Pulsar models).

speed variators

Speed variators are used to vary the speed of an asynchronous motor.

The harmonic currents they generate depend to a large extent on:

- the ratio between the network short-circuit powerSsc and the apparent power of the variatorSn .

- The variator load S (apparent power of the motor) as a percentage of the apparent powerof the variatorSn .

Measurement results obtained are summarised in tables 8-5, 8-6, 8-7.

S S sc n= 100% I1 I5 I7 I11 I13 I17 I19 I23 I25

S S sc n= 250 100 % 85 % 72 % 41 % 27 % 8 % 5 % 6 % 5 %

S S sc n= 100 100 % 73 % 52 % 16 % 7 % 7 % 5 % 3 % 3 %

S S sc n= 50 100 % 63 % 35 % 6.2 % 1.3 % / / / /

Table 8-5: speed variator with a load of 100% of its apparent power rating


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S Sn= 50 % I1 I5 I7 I11 I13 I17 I19 I23 I25

S S sc n= 250 100 % 90.5 % 82 % 59.5 % 48 % 25.5 % 16.5 % 6 % 4.5 %

S S sc n= 100 100 % 82% 66.5 % 33 % 19.5 % 7 % 6.5 % 5 % 3.5 %

S S sc n= 50 100 % 74.3 % 53.9 % 18.3 % 7.9 % 1.9 % 2.5 % / /


S Sn= 25 % I1 I5 I7 I11 I13 I17 I19 I23 I25

S S sc n= 250 100 % 94 % 89 % 74 % 66 % 47 % 38 % 22 % 15 %

S S sc n= 100 100 % 89 % 78 % 53 % 40 % 17 % 9 % 5 % 6 %

S S sc n= 50 100 % 84 % 68 % 38 % 24 % 6.1 % 2.1 % / /


arc furnace

Arc furnaces as used in iron and steel making can either be a.c. or d.c. supplied.

the a.c. supplied arc furnace (see fig. 8-6)

HV

transformer

cable

furnace

Figure 8-6: case of the a.c. supplied arc furnace


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The arc produced is non-linear, dissymetric and unstable. It gives rise to odd and even

harmonics, as well as to a continuous spectrum of currents (currents at all frequencies). The

levels of harmonics produced together with the continuous spectrum values will depend on the

type of furnace, its power, the duration of the process under consideration (casting, refining),

etc.

Therefore, the precise levels of harmonics can only be determined by measurement.

Figure 8-7 is an example.

1 3 5 7

100

10

1

0.1

9

4

3.2

1.3

0.5

100

order

continuous spectrum

I1(%)

Ip

Figure 8-7: current spectrum of an a.c. supplied arc furnace

the d.c. supplied arc furnace (see fig. 8-8)

HV

transformer

cable

furnace

cable

rectifier

Figure 8-8: case of the d.c. supplied arc furnace


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The d.c. supply is derived from a rectifier, the arc produced being more stable than for the a.c.

supplied furnace.

The current consumed comprises:

- a spectrum of harmonic currents similar to that of a rectifier

- a continuous spectrum with a lower overall value than for the a.c. supplied furnace.

Figure 8-9 gives the harmonic current spectrum for a 144 MVA d.c. driven arc, fed from a dual

bridge rectifier.

The use of the latter explains the high values of 11th and 13th harmonics (see 8-4-6).

order

I1(%)

Ip

3.8%

4.3% 2.5%

3.4%

1.5%1.2%

2% 1.9%

3.4%

1.9% 1.9%

9.1%

7.7%

3.2%3.5% 3.5%

continuous spectrum

Figure 8-9: harmonic current spectrum of a d.c. supplied arc furnace fed from a dual bridge rectifier

magnetic circuit saturation in machines (transformers, motors, ...)

Electrical machines are designed to work close to their magnetic saturation limits when

operating under nominal supply voltage conditions.

Should the supply voltage become abnormally high (greater than 1.1 times its nominal value),

magnetic circuits will saturate and currents will become distorted. Henceforth, the machine

becomes a source of odd order harmonic currents.


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voltage sources

Voltage sources (utility, alternator, UPS) possess pre-existing harmonic voltages which

manifest even when the equipment is connected to perfectly linear loads.

utilities

Utilities possess pre-existing voltage harmonics due to other consumers (industrial and

domestic) who create voltage harmonics on the distribution and transmission network.

Measurements taken on the French electricity authority EDF's MV distribution network have

given the results provided in table 8-8:

V1 V5 V7

high 100 % 9 % 3 %

medium 100 % 6 % 2 %

low 100 % 3 % 1 %

Table 8-8: harmonic voltages pre-existing on the French electricity authority EDF's M.V. network

Note: the 3rd harmonic currents, and multiples of them, which exist at high levels on the industrialand domestic low-voltage network, are eliminated by the delta-star connection of MV/LVtransformers. This is why 3rd harmonic voltages, and multiples of them, do not appear on theM.V. network.

Uninterruptible power supplies (UPS)

The sinusoidal voltage supplied by an inverter will not be perfect even for a linear load.

A UPS will therefore possess pre-existing harmonic voltages (see table 8-9).

Hardware model EPS 5000 EPS 2000 ALPES 1000 GALAXY

Overall distortion factor 5 % 4 % 5 % 2 %

Table 8-9: pre-existing harmonic voltages of UPS


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alternators

The sinusoidal voltage supplied by an alternator will not be perfect even for a linear load.

An alternator therefore possesses pre-existing harmonic voltages. Taking the example ofLeroy-Somer 10 kVA to 5000 kVA alternators, the voltage distortion factor is around the 4 %

mark, with 2 to 3 % of 5th harmonics and some 3rd harmonics.

V 4 %

I

Ito5

1

2 3= %

Note: the more unbalanced the load, the more 3rd harmonic an alternator will generate.


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8.1.4. Method of determining harmonic currents and voltages on a supply network

non-linear loads model

For calculation purposes, non-linear loads are considered to be harmonic current generators

(for harmonics p 2 ). They are modelled as currents injected into the network (see fig. 8-10).

Note: this model is valid for a discrete spectrum. In the case of a continuous spectrum (arc furnace) amore complex model must be devised. Such models are not dealt with in this document.

source

Vp Ip

Z

Ip : harmonic current of orderp

( )Z : impedance of network at angular frequency = =p 0 0( angular frequency at 50 Hz or 60 Hz).

Figure 8-10: model for current harmonics generated by non-linear loads

Each non-linear load will thus be modelled by its impedance at 50 Hz (or 60 Hz) and by current

sources corresponding to the harmonic currents generated by the load.

A harmonic current Ip , flowing across the network to the source through the impedance

( )Z , gives rise to a harmonic voltage Vp , such that:

( )V Z p I p p= 0

The voltage distortion factor resulting from a spectrum of harmonic currents I I I 2 3 4, , ... is then:

( )

Vp

p Z p I

V=

=

2

02 2

1


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Where a network comprises inductances and capacitances the impedance spectrum ( )Z

may vary significantly, as the curve below shows (see fig. 8-11).

0 2 0 4 0 6 0 8 0 p 0

Z( )

Figure 8-11: impedance spectrum

Harmonic voltages created on the network disturb the operation of electrical equipment. One

element of any investigation into harmonic content must therefore be to limit the values of

( )Z for values ofcorresponding to the harmonic currents with the highest values.


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modelling of sources polluted by pre-existing harmonic voltages

For the sake of calculation, pre-existing harmonic voltages are considered to be harmonic

voltage sources (see fig. 8-12).

Z

Vp load

Vp : p order of harmonic voltage

( )Z : impedance of the supply network at angular frequency = =p 0 0( angular frequency at 50 Hz).

Figure 8-12: pre-existing harmonic voltage model

A voltage source polluted by pre-existing harmonic voltages will therefore be modelled by its

nominal value at 50 Hz plus harmonic voltage sources.

impedance of network elements

utility

The impedance of the utility is generally taken to be a pure reactance. Nevertheless, the

capacitor banks found at HV/MV substations may provoke resonances at undesired

frequencies.

This should be taken into consideration during an investigation of harmonics where the

installation comprises non-linear loads drawing large amounts of power (for example, an arc

furnace).

cables, power lines and transformers

These are considered as resistances of constant value and reactances proportional to the

order of the harmonic:

( ) Z p R jpX 0 = +

p : order of the harmonic

X : reactance at 50 Hz.


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capacitor banks

These are treated as capacitors whose impedance is inversely proportional to the order of the

harmonic:

( )Z pCp

00

1=

alternator

It will be explained further on why the reactance of an alternator is not proportional to the order

of the harmonic. An alternator is considered to have negligible resistance.

at the supply network frequency

The field created by the three stator currents rotates at the speed of the rotor and is therefore

fixed in relation to it.

When the stator current increases, the voltage decreases. The impedance of the alternator is

equal to the synchronous reactance Xd which is of the order of 200 to 300 %.

at frequencies other than that of the network

The field created by the three stator currents does not rotate at the speed of the rotor and is

not therefore fixed in relation to it.

This rotating field causes harmonic currents to flow in the rotor, thereby producing a drop in

voltage.

For harmonic currents of order 3 1k , the alternator's impedance is a function of the sub-

transient impedance:

( ) X k X k d3 1 3 1 = "

Xd"

is of the order of 10 to 20 %.

For 3 k order harmonics, if the neutral is not distributed the impedance is infinite. Indeed, 3rd

order harmonics, and multiples of the 3rd, flowing in the neutral are in phase. However, if the

neutral is not distributed they cannot flow, with the result that the impedance is considered to

be infinite.


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If the neutral is distributed, 3rd order harmonic currents and multiples of them flow in it, the

alternator impedance then being a function of the zero-sequence impedance. Indeed, the field

created by harmonics which are multiples of 3 does not rotate, as in a zero-sequence current

system.

X k X k3 03=

X0 is roughly equal to half Xd"

, or of the order of 5 to 10 %

asynchronous motor

The reactance of an asynchronous motor is not proportional to the order of the harmonic. Its

resistance is considered negligible.

At the supply network frequency the motors impedance ZM depends on the load.

Indeed, ZV

IMn=

Vn : single-phase voltage

I : current consumed by the motor, which depends on its load.

For harmonic voltages of order 3 k, the impedance is infinite because generally the neutral of

the motor is not distributed.

For 3 1k order harmonics, the motor's impedance is a function of the locked rotor impedance

(or starting impedance if there is no starting current limiting system):

( )X kI

IXk

n

bn3 1 3 1 =

( )X kV

Ikn

lr

3 1 3 1 =

Xn : motor impedance at rated load, XV

Inn

n=

Ilr : locked rotor current (or start up current if there is no starting current limiting system).


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inverters

The output impedance is vastly different according to whether the machine is of the older type

(EPS 5000) or one of the newer generation (GALAXY or COMET).

older type inverters (EPS 5000)

The output impedance is equal to impedance of the output filter. The equivalent circuit as seen

from the output is that of figure 8-13.

V

Zs ( )

Figure 8-13: older type inverter output impedance

Output impedance is therefore frequency dependant (see fig. 8-14).

At low frequencies the output impedance is close to L .

At high frequencies the output impedance is close to1

C.

At the resonance frequency fL C

r =1

2 , the filter's impedance is very high, having an order

of magnitude equal to the nominal load impedance, that is ( )Z fout r 2 100 % .

In practice, the resonance frequency is chosen to be far from that of the most common

harmonic currents.

For example, 210 Hz might be selected since 4th order harmonics are virtually non-existent.


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newer generation inverters with phase-neutral voltage regulation (GALAXY, COMET)

This type of equipment uses a very high switching frequency (several kHz) and an extremely

fast voltage regulator, enabling output voltage quality to be maintained even with a high

current distortion factor. As a consequence, the output impedance is very low and virtuallyconstant for all orders of harmonic (see fig. 8-14).

150

100

50

0 50 250 500 750

new type of inverter(GALAXY, COMET)

old type of inverter (EPS 5000)

ratio of the output impedance

to the nominal load impedanceZn

%Zs

f Hz( )

Figure 8-14: inverter output impedance


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particular case of 3rd harmonics and multiples of the 3rd

3rd harmonic currents are in phase, with a current equal to their sum flowing in the neutral

when distributed (see fig. 8-15). If the neutral is not distributed, 3rd harmonics and multiples of

them cannot flow, and thus cannot exist.

I3

I3

I3

3 3I neutral

Figure 8-15: 3rd harmonic currents, and multiples of them, flowing in the neutral

why 3rd harmonics, and multiples of the 3rd, are in phase

Take 3 equal currents I ta ( ) , I tb ( ) and I tc ( ) with phase displacements of 1/3 of a period:

( )

( )

( )

i t t

i t t T

i t t T

a

b

c

=

= +

= +

cos

cos

cos

32

3

where T=2

period of signal

Note !: the currents are phase shifted in time and must therefore be expressed as above.

The 3rd harmonics of these currents are obtained by replacing by 3 :

( )i t t a3 3 3= cos

( ) ( )i t t T t t b3 3 3 33 3 2 3= + = + =cos ( ) cos cos

( ) ( )i t t T t t c3 3 3 33 2 3 4 3= + = + =cos ( ) cos cos

giving i t i t i t t a b c3 3 3 33 3( ) ( ) ( ) cos+ + =


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The in-phase relationship can also be shown graphically (see fig. 8-16). The same is true for

all harmonics that are multiples of the 3rd.

I3

I2

I1

t

t

t

3rd harmonic

t

Ineutral

3rd harmonic

3rd harmonic

Figure 8-16: the 3rd harmonics of a 3-phase system are in phase


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superposition theorem

Consider the case of a network with the following characteristics:

- a voltage source with pre-existing harmonics V V V2 3 4, , , ...

- a transformer whose impedance is R LT T, , feeding a busbar

- various non-linear loads grouped together to form an equivalent load generating the

harmonic currents I I I2 3 4, , , ...

The loads are supplied from the busbar through an impedance R L1 1,

- various linear loads grouped together to form an equivalent load fed from the busbar

through an impedance R L2 2,

- the equivalent capacitorC representing the reactive energy compensating capacitors.

Each voltage and current source is considered to act in independently on the network.

The theorem of superposition states: for a linear network, the current (or voltage) generated in

a branch by several independent sources acting simultaneously, is equal to the currents (or

voltages) produced in the same branch by the different sources acting in isolation.

Figure 8-17 illustrates the application of this theorem.


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R LT T,

V source

non-linearloads

linearloads

CR1 R2

L2L1

equivalent

non-linear

loads

linear

loads

V1

oC

1

R LT T o,

R1 R2L o1 L o2 +

non-linearloads

linearloads

Vp

R pLT T o,

pL o1

R1 R2pL o2

1

pC o +

linearloadsIp

R1pL o1

R2pL o2

1

pC o

R pLT T o,

network at 50 Hz + p harmonic networks of orders

1 2, ,..., p supplied by voltagesources V V Vp1 2, ,...,

+ p harmonic networks of orders

1 2, ,..., p supplied by currentsources I I Ip1 2, ,... ,

Figure 8-17: superposition theorem as applied to harmonics


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addition of equal order harmonics arising from different non-linear loads

The harmonic currents of the same order generated by several non-linear loads are not

necessarily in phase. The resultant current is obtained by vector addition (see fig. 8-18).

r

Ip,1

r

Ip, 2

r

Ip sum,

Figure 8-18: vector addition of two 2 harmonic currents of the same order

Take the two extreme situations.

their phase relationship is perfectly random

The modulus of the sum of two p order harmonics is, on average:

I p mean,2

( ) ( )[ ]= + +12 1 22

22

0

2

I I I d p p p, , ,cos sin

( )= + +12 202

12

22

1 2

I I I I d p p p p, , , , cos

= +I Ip p, ,12 22

I p mean, = +I Ip p, ,12

22

By extension, for a number k of non-linear loads supplying harmonics of order p , the

modulus of their sum is, on average: I I p sum p

k

, ,= 2


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the harmonics are perfectly in phase

In this case, the harmonics add together algebraically, giving :

I I p sum p ii

k

, ,==

1

Practical instances fall somewhere between the two extremes, with phase correlation

depending to a greater or lesser degree on: diversity of harmonic sources, transformer vector

group, etc. Consequently, an empirical law has been defined to estimate the modulus of the

sum of equal order harmonics, expressed by the following relationship:

( )I I p sum p ik

, ,

/

=

1

where 1 2

= 1 : the harmonics are perfectly in phase= 2 : harmonic phases are perfectly random.

It can be seen that I p sum,

decreases as increases,

I p sum,

r eqZ

C,

1.


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670

Figure 8-23 illustrates the variations in Zeq as a function of .

N

M

K

R

r

Zeq

Figure 8-23: variations in Zeq as a function of angular frequency

n definition of amplification factor FA

FA is the ratio of the network impedances with and without capacitors, at the resonant frequency.

FKN

KMA= (see fig. 8-22)

i.e. FR

LA sc r =

hence

F R

C

LA sc=

now Q C Un= 02

and LU

Sscn

sc0

2

=

henceF

Q S

PAsc=


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671

definition of the order of resonance pr

The order of resonance is defined as the ratio pr of the angular frequency of resonance to the angular

frequency of the network:

prr=

0

now rccL C

= 1

hencep

S

Qrcc=

determining the network behaviour in the presence of harmonic current the order of

which is equal to pr

For = =r eqZ R,

hence V R Ir r=

V F L I r A cc r r =

Vr : harmonic voltage of orderr appearing on the busbar

The harmonic voltage Vr on the busbar is therefore multiplied by FA compared to the case

where there are no capacitors (V L Ir cc r r = ).

Let IC r, be the harmonic current of order r flowing in the capacitors,

I C V C L C

R IC r r r cc

r, = =1

I F I C r A r , =

Let IL r, be the harmonic current of order r flowing in the upstream network,

IV

L

Lcc C

LccR IL r

r

cc rr, = =

I F I L r A r , =


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The currents and voltages at resonance are those shown in figure 8-24.

R

Ir

IrC

I F I L, r A r I F IC, r A r

V F L I r A sc r r Lsc

Figure 8-24: currents and voltages appearing at resonance

The following paragraphs will demonstrate that the value of FA could be very high on an

industrial network (in general anything from 2 to 5, and as much as 20).

As the diagram shows, harmonic currents flowing in the capacitors and upstream network are

each amplified by a factor ofFA .

Therefore, compared to the case where there are no capacitors, the harmonic voltage of order

r produced on the busbar is amplified by a factor of FA . Compensating capacitors will then

more than likely pose significant problems when large, non-linear loads are connected to the

network.

example

Let us consider a 20 kV / 400 V transformer of S kVAT = 1250 , and Usc = 5 5. % , supplying :

- day and night, a bridge rectifier whose powerS kVAB = 500 and power factorFp B, .= 0 8 .

- a linear load whose power consumption S kVAld = 600 during the daytime period, andS kVAln = 100 during the night-time period, with a power factorFp l, .= 0 9 .

The upstream network short-circuit powerS MVAup = 130 .

The requirement is to compensate the reactive energy of the fundamental during the day so as

to obtain tan .1 0 4= , where tan 11

1

=P

Q.

P1 : active power of the fundamental component

Q1 : reactive power of the fundamental component.


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determining the reactive power of the capacitors to be installed

Using the manufacturers' assumption (see 8.1.5), we can calculate the fundamental

component values for the bridge rectifier:

P P F S B B p B B1 0 8 500 400, , .= = = =

From the example of paragraph 8.1.5, Q kVAB1 277, = .

The daytime period linear load is S kVAld = 600 where Fp l, .= 0 9 .

Calculating the active and reactive powers of a linear load is straightforward:

P S F kW ld ld p l = =, 540

Q S P k ld ld ld = =2 2

262 var

We can also calculate the installation's tan 1 value:

tan .,

,

11

1

277 262

400 5400 57=

+

+=

++

=Q Q

P P

B ld

B ld

To compensate the reactive energy of the fundamental, the necessary capacitor bank power

rating is thus given by:

( ) ( )Q P P k B ld = + = =1 1 0 940 0 17 159 8, tan .4 . . va

The choice would be to install a standard value capacitor bank with a rated power value ofQ kC = 160 var.


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calculation of the harmonic voltages and distortion factor in the absence of capacitors

Firstly, the equivalent impedance for the transformer and upstream network is calculated.

Using the approximation that the transformer and upstream network are composed of purereactance (see 4.2.1.4 - Protection Guide), we have:

XU

Supn

up=

2

and XU

SUT

n

Tsc=

2

hence ( )X UU

S Sm sc n

sc

T up= +

=

+

=2 2

3 6

1400

0055

1250 10

1

130 108 27

..

X Xup T, : are the reactance of the upstream network and the transformer respectivelyXsc : is the equivalent reactance of the transformer and upstream network (short-circuit impedance).

We can now calculate the harmonic voltages (see table 8-12) and the distortion factor.

Order of harmonic 5 7 11 13 17 19 23 25 29 31

( )Ip % 18.9 11.0 5.9 4.8 3.4 3 2.3 2.1 1.8 1.6

( )I Ap

132.7 77.3 41.4 33.7 23.9 21.1 16.2 14.7 12.6 11.2

( )V pX I V p sc p= 5.49 4.47 3.77 3.62 3.36 3.32 3.08 3.04 3.02 2.87

( )V

Vp

1

%2.39 1.95 1.64 1.58 1.46 1.44 1.34 1.32 1.31 1.25

Table 8-12: harmonic voltages in the absence of capacitors

From here, the voltage distortion factor is:

V = 5 07. %

This is an acceptable value, both for the operation of the bridge rectifier and for the rest of the

installation (see 8.2).


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calculation of the harmonic voltages and distortion factor for the daytime period with

the compensating capacitors in service

Calculating the order of resonance:

pS

Qrsc=

where SU

XMVAsc

n

sc= =

2

19 35.

Q k= 160 var

hence pr

= 1100.

The order of resonance is the same as the order of a high value harmonic current generated

by the bridge rectifier.

Calculating the amplification factor:

FQ S

PAsc=

P is the active power of the linear loads, giving P P kW

ld= =540

hence FA = 3 26.

Calculating the equivalent impedance Zeq :

1 1 10

0Z Rj p C

p Leq ld sc= +

Z

RpC

p L

eq

ld sc

=+

1

1 12 0

0

where RU

Pmld

n

ld= =

2

2963.

CQ

Un0 2 100= = .

L X m sc sc

0

8 27= = .


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Figure 8-25 shows the curve of the Zeq ( ) impedance spectrum.

We can go on to calculate the harmonic voltages (see table 8-13)


( )I Ap 132.7 77.3 41.4 33.7 23.9 21.1 16.2 14.7 12.6 11.2

Z meq ( ) 51.3 92.5 296.3 199.7 95.7 76.5 55.4 48.9 39.9 36.6

( )V Z I V p eq p= 6.81 7.15 12.27 6.73 2.29 1.61 0.90 0.72 0.50 0.41

( )V

Vp

1

%2.96 3.11 5.33 2.93 0.99 0.70 0.39 0.31 0.22 0.18

( )I AC p, 34.1 50.1 135 87.5 38.9 30.6 20.7 18.0 14.5 12.7

Table 8-13: harmonic voltages during the daytime period with capacitors in service

From here, the voltage distortion factor is :

V = 7 5. %

The harmonic currents flowing in the capacitor bank are obtained from:

I pC V C p p, = 0 (see table 8-14).

The r.m.s. current in the capacitor bank is :

I IC rms C , ,=

2

IC, 1 is the fundamental current in the capacitors, that is the nominal current, which gives

I I C V AC C n n, ,1 0

230= = = .

hence I AC rms, .= 2930

and I

I

C rms

C n

,

,

.= 1 27

Next, we can calculate Vrms = 230 64. .


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It can be seen that despite the high values of some of the harmonic currents, the r.m.s. voltage

is virtually identical to the nominal voltage. The manufacturers' assumption is thus justified:

U Urms = 1 .

The r.m.s. current in the capacitor bank is close to the limit of 1.3 IC n, , there is thus a risk of

the capacitors overheating.

The voltage distortion factor is only just acceptable, and certain loads may well be disturbed

(see 8.2). The bridge rectifier will be able to operate normally (see 8.2).

calculation of harmonic voltages and distortion factor for the night-time period with

the compensating capacitors in service

Calculating the amplification factor:

FQ S

PAsc=

P is the active power of the linear loads during the night-time period, giving P P S kW = = =ln ln 0 9 90.

hence FA =160 19 350

90

FA = 19 6.

Z

RpC

p L

eq

sc

=

+

1

1 12 0

0ln

where RU

Pmn

ln ln= =

2

1778.

CQ

Un0 2 1 00= = .

L X m sc sc0 8 27= = .

Figure 8-25 shows the curve of the Zeq ( ) impedance spectrum


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We can go on to calculate the harmonic voltages (see table 8-14)


( )I Ap 132.7 77.3 41.4 33.7 23.9 21.1 16.2 14.7 12.6 11.2

Z meq ( ) 52.1 97.2 1778 267.3 101.1 79.1 56.3 49.6 40.3 36.9

( )V Z I V p eq p= 6.91 7.51 73.61 9.01 2.41 1.67 0.91 0.73 0.51 0.41

( )V

Vp

1

%3.01 3.27 32.0 3.92 1.05 0.73 0.40 0.32 0.22 0.18

( )I AC p, 34.6 52.6 809.7 117.1 41.0 31.7 20.9 18.3 14.8 12.7

Table 8-14: harmonic voltages during night-time period with capacitors in service

From here, the voltage distortion factor is :

V = 32 6. %

This figure is unacceptable to the majority of equipment, which will not operate properly and

may be damaged as a result (see 8.2). The bridge rectifier itself will malfunction, the

distortion factor being well above the permitted threshold (see 8.2).

The harmonic currents flowing in the capacitor bank are given by:

I pC V C p p, = 0 (see table 8-15).

The r.m.s. current in the capacitor bank is :

I IC rms C , ,=

2

IC, 1 is the fundamental current in the capacitors, that is, the nominal current. We have then I I C V AC C n n, ,1 0 230= = = .

hence I AC rms, .4= 854

and I

I

C rms

C n

,

,

.= 3 71


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The r.m.s. current in the capacitor bank is much greater than the permitted maximum, with the

result that the protection system will disconnect them.

Next, we can calculate the r.m.s. voltage: V Vrms = 242 .

Being 5 % above nominal voltage, this is an acceptable value in terms of the capacitors.

conclusion

The installation of reactive energy compensating capacitors increases the voltage distortion

factor, especially when harmonic currents exist close to the resonance frequency. When the

power consumed by any linear loads is reduced, voltage distortion increases resulting in higher

levels of disturbance.

For large, non-linear loads, the installation of capacitors should go hand in hand with an

investigation of the harmonics. This will enable calculations to be made for the equipment

necessary (see 8.4) to overcome the problem cited above.

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31

1700

1500

1300

1100

900

700

500

400

300

200

100

with capacitors, night-time

with capacitors, day-time

frequency

(p x 50 Hz)

without capacitors

Z meq

Figure 8-25: the different impedance spectrum curves


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8.2. Effects of harmonics on electrical apparatus, and working guidelines

Harmonics may well cause electrical equipment to malfunction, or to overheat, or to produce

mechanical vibrations which could lead to irreparable damage.

capacitors

Capacitors should conform to the IEC standard 871-1 for M.V. and to IEC standard 831-1 for

L.V. These standards stipulate:

- the ability to permanently pass a current of 1.3 times rated value, corresponding to a current

distortion factor ofI = 83 % . In fact, I

pp

pp rms

I I I I2

2

2

2

12 2

1

2

2

21= =

+

= +=

=

I1 : rated current of the bank

hence ( )I = 1 3 12

.

- the ability to withstand 1.1 times rated voltage 12 h / day for M.V. use, and 8 h / day in a

L.V. application, allowing their operating voltage to be 10% above rated value.

working guidelines

The above specifications are not cumulative, therefore the following condition should be

observed:

3 1 3 I V Qrms rms n .

Qn : nominal or rated power of the bank.

Should this condition not be met, techniques to limit harmonics could be implemented

(see 8.4). Otherwise, one approach might be to install, if this is adequate, over-insulatedcapacitors (class H) able to permanently pass 1.5 times their rated current, a figurecorresponding to a current distortion factor of I = 112% .


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example

A capacitor bank of standard value 200 400k Vvar, is supplied from a busbar whose voltage

characteristics are:

- daytime period VV

d1400

3=

- night-time period VV

t1420

3=

- measured harmonic voltages, identical for day and night-time periods, of:

V V V5 7 113 5 7V V V1 1 1

= = =%, %, % andV13 4V1

= % .

calculating the r.m.s. current in the capacitors during the daytime period

For the fundamental component (at 50 Hz):

I C V d1 0 1=

where V Vd n1 = : rated voltage of capacitor bankI In1 = : rated current of capacitor bank 0 2 50= : angular frequency of the supply network .

For the harmonics:

I pC V p p= 0

henceI

Ip

V

Vp p

d1 1=

I I pV

Vpp

d

212 2

1

2

=

( ) ( ) ( ) ( )( ) I I I pV

VIrms

p

dp

212

12 2

1

2

12 2 2 2 2 2 2 2 2

2

1 0 03 5 0 05 7 0 07 11 0 04 13= +

= + + + +

=

. . . .

I Irms = 1 1.42

hence I Irms n= 1 42.

The capacitors will overheat and deteriorate prematurely if the protection system does not

disconnect them.


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calculating the r.m.s. current in the capacitors during the night-time period

For the fundamental component (at 50 Hz):

I C V C V VV

C Vt nt

nn1 0 1 0

10

420400

= =

=

hence I In1 1 05= .

For the harmonics :

I pC V p p= 0

henceI

I

pV

V

p p

t1 1

=

I I pV

Vpp

t

212 2

1

2

=

I I I pV

Vrmsp

t

212

12 2

1

= +

=

I I I rms n= = 1 42 1 42 1 051. . .

hence I Irms n= 1 49.

The r.m.s. value of current is higher during the night-time period because the voltage is higher.

As a result the capacitors will overheat all the more.


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transformers

When subjected to harmonic currents a transformer will suffer supplementary losses and

possible disturbance to its magnetic circuit. Moreover, it will tend to generate noise due to the

vibrations caused by the harmonics.

Joule losses

These are proportional to the square of the r.m.s. current : loss = R I rms2

.

Harmonic currents therefore increase Joule losses.

iron loss

This comprises eddy current losses and magnetic induction hysteresis losses.

Eddy current loss is proportional to the square of the frequency, whereas hysteresis loss is

proportional to the frequency. The higher frequency harmonic currents will therefore generate

significant iron losses.

disturbance of the magnetic circuit

Harmonic currents generate an extra magnetic flux proportional to the impedance of the

upstream network. This is superimposed on the flux produced in the transformer by the

fundamental.

The peak value of flux is thereby increased, which can lead to saturation and an increase in

magnetising current and iron losses. The transformer itself may also generate harmonics.


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transformer de-rating formula, and working guidelines

To take account of the extra heat developed, transformers should be de-rated. The formula

commonly used is:

kI

Ip

p

p

=

+

=

1

1 0 11

21

2

..

p : order of harmonic

I1 : transformer rated current.

A transformer of rated powerSn is therefore limited to supplying a load of k Sn .

Another solution consists in obtaining the manufacturer's agreement to construct a transformer

specially designed to feed equipment that produces harmonics.

example 1

It is required to feed a 800 kVA, bridge rectifier from a transformer.

Using table 8-1, the de-rating factor is:

11 0 1

2

k

= + . ( ) ( ) ( ) ( ) ( )[ 0 189 5 0 11 7 0 059 11 0 048 13 0 034 172 1 6 2 1 6 2 1 6 2 1 6 2 1 6. . . . .. . . . . + + + +

( ) ( ) ( ) ( ) ( ) ]+ + + + + 0 03 19 0 023 23 0 021 25 0 018 29 0 016 312 1 6 2 1 6 2 1 6 2 1 6 2 1 6. . . . .. . . . .

hence k= 0931.

The bridge rectifier should therefore be supplied from a transformer whose rated power is at

least equal to 859 kVA.

example 2

It is required to feed a 250 kVA load in the shape of switchmode power supply from atransformer.

Using the high harmonic level assumption of table 8-2, the de-rating factor is:

( ) ( ) ( ) ( ) ( )[ ]1

1 0 1 1 3 3 0 7 5 0 5 7 0 3 9 0 1 112

2 1 6 2 1 6 2 1 6 2 1 6 2 1 6

k= + + + + + . . . . . .. . . . .

hence k= 0532.

A transformer of at least 470 kVA should therefore be installed.


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motors

When harmonic voltages are present, motors are subject to supplementary losses and

pulsating torques.

Joule losses

Joule loss is proportional to the square of the r.m.s. value of current : loss = R I rms2

.

Harmonic currents generated by the harmonic voltages present in the supply will thus increase

Joule losses.

iron loss

This comprises Foucault current (eddy current) losses and magnetic induction hysteresislosses.


proportional to the frequency. The higher frequency harmonic voltages will therefore generate


pulsating torques, opposing or motor

Harmonic voltages produce harmonic currents, creating rotating fields whose speeds differ

from that of the 50 Hz field. Such fields will produce either motor or opposing torques at

frequencies other than 50 Hz. These torques, and the vibrations they cause, may lead to

mechanical damage, extraneous noise, supplementary Joule loss in the rotor and a reduction

in motor efficiency.

working guidelines

IEC standard 34-1 lays down that a.c. current motors must be able to operate correctly with a

supply voltage whose harmonic voltage factor ( HVF ) is as follows:

HVF 2 %

The HVF value is calculated from the following formula:

HVFV

V pp

p

=

=

1

2

2

1

According to the standard only harmonics of order 13 need generally be considered.


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example

Using the example of paragraph 8.6.1:

without capacitors

From table 8-12, calculation gives HVF= 162. % (whereas V = 5 07. % ). This value isacceptable for motors conforming to IEC standard 34-1.

with capacitors during the day time period

From table 8-13, calculation gives HVF= 2 50. % . This higher value is the figure laid down bythe standard. Any motor installation must therefore involve specific calculations, or testing at

the manufacturers to check motor thermal withstand or to establish the de-rating figure that

may need to be applied to the machines.

with capacitors during the night time period

From table 8-14, calculation gives HVF= 9 89. % . This is a value far greater than that definedby IEC standard 34-1, the strong resonance at the 11th harmonic producing intolerably high

harmonic voltages.


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alternators

When supplying non-linear loads, an alternator will suffer supplementary losses and pulsating

torques due to the harmonic currents produced.

iron loss

This comprises Foucault current (eddy current) loss and magnetic induction hysteresis loss.


proportional to the frequency. The higher frequency harmonic currents will therefore generate


pulsating torques, opposing or motor

Harmonic currents create rotating fields whose speeds differ from that of the 50 Hz field. Such

fields will produce either motor or opposing torques at frequencies other than 50 Hz. These

torques, and the vibrations they cause, may lead to mechanical damage, extraneous noise,

supplementary Joule loss in the rotor and a reduction in alternator efficiency.

working guidelines

Alternators supplying less than 20% of non-linear loads will not generally have any difficulties.Between 20 and 30 % certain manufacturers de-rate their machines by 10 %. Above 30%, the

the harmonic current spectrum must be shown to the manufacturer so that he can provide the

derating coefficient. For example, for an uninterruptible power supply manufacturers

recommend choosing an alternator with a power of 1.5 to 1.9 times that of the UPS.


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neutral conductor heat problem

3rd harmonic currents, and multiples of the 3rd, are in phase and flow in the neutral conductor

when it exists, and when the transformer vector group makes it possible (e.g. a delta-star

transformer, which is the most common case).

The r.m.s. current in the neutral conductor is given by:

I Ineutral k k

==

3 321

This current may be greater than that flowing in the phase conductors, and as a result the

neutral conductor must be oversized.

working guideline

The cross-section of the neutral conductor must take into account the existence of 3rd

harmonic currents and multiples of the 3rd.

example 1

Consider a lighting load whose current spectrum is that of table 8-15.

I1 I3 I5 I7 I9 I11 I13 I15

100 % 35 % 27 % 10 % 2.5 % 3.5 % 1.5 % 1.5 %

Table 8-15: lighting load current spectrum

We can derive:

I Irms = 110 1.

( ) ( ) I I I neutral = + =3 0 35 0 025 1 0512 2

1. . .

The neutral current value is more or less equal to that flowing in a phase conductor, and as

such all cables can have similar cross-sections.

.


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example 2

Consider computer type loads that have a spectrum as shown in table 8-16.

I1 I3 I5 I7 I9 I11

100 % 65 % 35 % 25 % 15 % 5 %

Table 8-16: current spectrum for computer loads

We can derive:

I Irms = 1 28 1.

( ) ( ) I I I neutral = + =3 0 65 0 15 2 0012 2

1. . .

andI

Ineutral

rms= 1 56.

The current in the neutral is much larger than that flowing in a phase conductor. This demands

the use of a protection device (circuit-breaker with 4-pole protection) plus a neutral conductor

of sufficient cross-section.

electromagnetic disturbances in a TNC earthing system

Currents due to the 3rd harmonic, and multiples of the 3rd, flow in the neutral conductor. In a

TNC earthing system the neutral conductor and circuit protection conductor are one and the

same.

Now, since the circuit protection conductor connects together all exposed conductive parts,

including elements of the building's infrastructure, 3rd harmonic currents and multiples of them

flowing in these circuits will produce local variations in potential.

This phenomenon may cause problems such as :

- corrosion of metal parts

- overcurrent on a telecommunication link connecting the exposed conductive parts of two

loads (for example, between a microcomputer and a printer)

- electromagnetic radiation affecting computer screens.

Where 3rd harmonics or multiples of them are present the TNC earthing system must be

avoided.


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inverters

The level of harmonic disturbance created by an inverter will depend on the technique used to

generate the sine wave voltage, that is, either PWM (Pulse Width Modulation) or non-PWM.

non-PWM inverters

The output current of non-PWM inverters is limited to its rated value, giving an instantaneous

value of current no higher than 2 In .

Manufacturers apply a de-rating factor that is a function of the peak factor F

Ipk n=

This de-rating factor is given by :

dFpk

= 2

PWM inverters

The output current of Merlin Gerin PWM inverters is limited to 2.35 times the r.m.s. value ofrated current, giving an instantaneous value no higher than 2 35. In .

- if In

2 35. de-rating is not applied- if In> 2 35. the de-rating factor is given by:

dI

n=

2 35.

hence dFpk

=2 35.

example

Table 8-17a gives the de-rating coefficients for inverters feeding computer type loads.

Type of load IBM 4381 IBM AS400 IBM PCXT IBM site *

Peak Factor 2.14 2.53 5.54 2.21

Non-PWM de-rating 0.66 0.56 0.26 0.64

PWM de-rating / 0.93 0.42 /

Table 8-17a: inverter de-rating when feeding computer loads

* Example of an IBM site with a large number of terminals.


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semiconductor convertors

IEC standard 146-1-1, 2.5.4.1 lays down the operating limits, without performance loss, for

convertors, defining them according to the machine's immunity class (see table 8-17b):

- class A: limits are applicable to convertors used on highly (harmonically) disturbed

networks.

- class B: limits are applicable to convertors used on networks where there is an average

level of disturbance

- class C: limits are applicable to convertors used on networks where the disturbance is low.

If the immunity class is not specified, class B is deemed to apply.

Immunity class

A B C

Individual even harmonics (%) 2 2 1

Individual odd harmonics (%) 12.5 5 2.5

Total harmonic distortion (%) 25 10 5

Table 8-17b: operating limits without performance loss, for convertors


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special case of the uninterruptible power supply

The IEC standard 146-4, 5.1 lays down the guaranteed operating limits for a UPS:

- voltage distortion factor less than 10 %

- harmonic voltage components less than those defined by the curve shown in figure 8-26.

5%

1%

0.5%

0.4%

0.3%

3 5 7 9 11 13 25 100

Vp

V1(%)

Figure 8-26: maximum permitted harmonic voltage content on a UPS a.c. supply input


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permanent insulation monitors (PIM)

Such devices apply a d.c. or low-frequency voltage between the protection conductor and the

neutral (or phase if the neutral is not accessible), and measure the current.

PIMs using a d.c. voltage are not affected by harmonics since they measure a highly filtered

d.c. current.

On the other hand, the older type of PIM (before 1993), such as the XM100, will be affected.

These inject a 10 Hz voltage and filter the 50 Hz signal. However, the filter becomes ineffective

from, and including, the 3rd harmonic upwards. Thus, any 3rd harmonic currents or multiples

of them or unbalanced harmonic currents of monophased loads flowing in the neutral may be

"seen" by the PIM causing it to alarm in error.

Where an installation has in-circuit fault location core-balance current transformers on its

outgoing feeders, harmonic currents flowing in the neutral will disrupt the fault location system.

No permitted voltage distortion limits have been defined for PIMs, since any likely disruption

will depend on the characteristics of the network and loads.

The newer generation PIMs (post 1993), which use a 2.5 Hz voltage and filters which are

effective above 4 Hz, are immune to harmonics.

circuit-breakers with electronically tripping relay

The thermal or magnetic detection of older model circuit-breakers (still on sale today) aresensitive to peak values of current.

High current distortion factors can mean high peak currents (see the values for computer

equipment, table 8-18) leading in turn to spurious opening of the circuit-breakers. The thermal

detection threshold must then be increased when this is possible or the harmonics filtered out.

Newer generation devices - compact or masterpact - sample the current and calculate its r.m.s.

value, and are therefore immune.

computer loads

The recommended threshold for reliable operation is:

V

Vp

p 1

2

1

5

21 0.2/ * 0.2/ * 10 0.5 0.2/0.5*

19 1.5 1 12 0.2/0.5* 0.2/0.5*

23 1.5 0.7 14 0.2/0.5* 0.2/0.5*

25 1.5 0.7 16 0.2/0.5* 0.2/0.5*

> 25 0 212 5

..

+n

*

0 112 5

..

+n

*

18 0.2/0.5* 0.2/0.5*

20 0.2/0.5* 0.2/0.5*

22 0.2/0.5* 0.2/0.5*

24 0.2/0.5* 0.2/0.5*

> 24 0.2/ * 0.2/ *

Total harmonic distortion

8 % on MV networks 3 % on HV networks

* : values specific to the EDF "Emeraude" contract : no value has been specified

Table 8-19: compatibility levels for harmonic voltages on medium and high-voltage public networks

(taken from Electra, n123 March 1989, and the EDF's Emeraude contract)


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on industrials networks

IEC standard 1000-2-4 lays down the compatibility levels for harmonic and interharmonic

voltages existing on industrial networks (see tables 8-20 and 8-21).

They apply to low and medium-voltage networks .The compatibility levels are given for different classes of environment:

Class 1:

Environments in this class are essentially protected networks, with compatibility levels lower

than for public networks, where equipment particularly sensitive to supply network disturbance

is used: laboratory instruments, certain automatic and protective installations, certain

computers, etc.

Equipment in this type of environment is, of necessity, generally supplied from an

uninterruptible power supply. Only low-voltage networks are included in this class.

Class 2:

This class concerns the utility's take-over point and the internal network.

The compatibility levels are identical to those applying to public supply networks.

Class 3:

This class applies only to internal networks. Compatibility levels are higher than for class 2

environments.


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Odd harmonics not

multiples of 3

Odd harmonics

multiples of 3

Even harmonics

Order of

harmonicn

Harmonic voltage

%

Order of

harmonicn

Harmonic voltage

%

Order of

harmonicn

Harmonic voltage

%

class

1

class

2

class

3

class

1

class

2

class

3

class

1

class

2

class

3

5 3 6 8 3 3 5 6 2 2 2 3

7 3 5 7 9 1.5 1.5 2;5 4 1 1 1.5

11 3 3.5 5 15 0.3 0.3 2 6 0.5 0.5 1

13 3 3 4.5 21 0.2 0.2 1.7

5

8 0.5 0.5 1

17 2 2 4 > 21 0.2 0.2 1 10 0.5 0.5 1

19 1.5 1.5 4 > 10 0.2 0.2 1

23 1.5 1.5 3.5

25 1.5 1.5 3.5

> 25 212 5

+.

n2

12 5+

.

n 511

n

Total harmonic distortion

5 % for class 1 8 % for class 2 10 % for class 3

Table 8-20: compatibility levels for harmonic voltages existing on industrial networks

Order

h

Class 1

( )Up %Class 2

( )Up %Class 3

( )Up %

< 11 0.2 0.2 2.5

11 to 13 inclusive 0.2 0.2 2.25

13 to 17 inclusive 0.2 0.2 2

17 to 19 inclusive 0.2 0.2 2

19 to 23 inclusive 0.2 0.2 1.75

23 to 25 inclusive 0.2 0.2 1.5

> 25 0.2 0.2 1

Table 8-21: compatibility levels for inter-harmonic voltages on industrial networks


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8.3.2.2. Emission levels

Harmonic currents emitted by customer installations and equipment must be limited in order

avoid harmonic disturbance attaining the maximum levels of compatibility defined previously

(see tables 8-18, 8-19, 8-20 and 8-21).

emission limits applied to low voltage apparatus drawing less than 16 A per phase

It would be extremely difficult to check the harmonic current emissions of each customer

connected to a low-voltage network. Besides, IEC 1000-3-4 was drawn up by standards bodies

to define emission limits for low-voltage apparatus drawing less than 16 A per phase, and

destined to be connected to the public low-voltage distribution networks. Table 8-22 defines

the emission limits for class A equipment: i.e. apparatus other than portable tools, lighting and

equipment drawing an input current with a "special waveform" whose active power

consumption is less than 600 W (for equipment of this type see IEC 1000-3-2, 7).

These limits do not apply to professional equipment with a power rating greater than 1 kW.

Order of harmonic

n

Maximum permissible harmonic current

( )A

Odd harmonics

3 2.30

5 1.14

7 0.77

9 0.40

11 0.33

13 0.21

15 39 n015

15.

n

Even harmonics

2 1.084 0.43

6 0.30

8 40 n0 23

8.

n

table 8-22: harmonic current emission limits for class A, low-voltage apparatus

drawing less than 16 A per phase


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emission limits applied to industrial consumers

Precisely limiting the harmonic voltages generated by industrial users would penalise those

connected to a low short-circuit power network (highly impedant).

An equitable solution consists in permitting the level of disturbance generated to be

proportional to the customer's subscribed demand, as applied to each class of voltage ( LV,

MV, HV)

There is no international standard laying down the emission limits applying to industrial users.

case of France

In France, the EDF's new "Emeraude" contract gives the following tolerances applying to its

MV and HV networks for information only. At present, the limits shown are not obligatory

although customers are advised to conform since values more or less equivalent to them will

no doubt become statutory between now and 1998.

In the contract, the customer is obliged to take steps to limit the individual harmonic currents

introduced onto the EDF network. Such limits are calculated on a pro-rata basis according to

each customer's subscribed demand ( Ssubscribed ).

A limiting coefficient, kp , is applied to each harmonic current of order p . The customer must

limit harmonics to a level given by:

I kS

Up p

subscribed

C=

3

where UC is the contract voltage.

Table 8-23 gives the values for kp as a function of the harmonic order p .

08 Harmonics

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