1

HARMONICS IN POWER SYSTEM

A harmonic is any voltage or current whose frequencies are integral multiples of f. For example a set of sine waves whose frequencies are 50, 150, 250, 450 Hz is said to possess the following components:

Fundamental frequency 50 Hz (the lowest frequency)Third harmonic: 150 Hz (3 x 50 Hz)Fifth harmonic: 250 Hz (5 x 50 Hz)Ninth harmonic: 450 Hz (9 x 50 Hz)

The distortion of a voltage or current can be traced to the harmonics it contains. This distortion can be produced by magnetic saturation in the core of transformers or by the switching of thyristors or IGBTs in electronics drive.

2

All periodic signals of frequency “f" can be represented in the form of a composite sum:

1. of a sinusoidal term at frequency “f": the FUNDAMENTAL (H1).

2. of sinusoidal terms of which frequencies are integer multiples of fundamental H1: the HARMONICS (Hn).

3. of a possible continuous component (DC component)

y(t) = h1(t) + h3(t)

3

Harmonics :Order and Spectrum

Order:The order of the harmonic is the value of the integer which determines its frequency.Example: harmonic of order 5, frequency = 250 Hz(when fundamental f is 50 Hz)

Spectrum:The spectrum of a signal is the graph representing amplitudes of the harmonics as a function of their frequency.

4

To summarize: the harmonics are nothing less than the components of a distorted waveform and their use allows us to analyse any periodic non-sinusoidal waveform through different sinusoidal waveform components.Figure below shows a graphical representation of this concept.

Non-sinusoidal waveform

First harmonic (fundamental)

Third harmonic

Fifth harmonic

5

fundamental and third harmonic

-80

-60

-40

-20

0

20

40

60

80

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37

fundamental

third harmonic

total

Two sinusoidal sources connected in series

U1max = 60 V @ 50 Hz

U2max = 20 V @ 150 HzU1 U2

U3

6

EXAMPLE OF THE HARMONICS

harmonics analyzis

-25

-20

-15

-10

-5

0

5

10

15

20

25

Am

pe

rs

Basic harmonic

3th harmonic

5th harmonic

7th harmonic

9th harmonic

11th harmonic

total curent

7

fundamental + harmonics

-25-20-15-10-505

10152025

1 22 43 64 85 106 127 148 169 190 211 232 253 274 295 316 337 358

fundamental all harmonics total current

8

How harmonics are generated?

Harmonics are generated by nonlinear loads. When we apply a sinusoidal voltage to a load of this type, we shall obtain a current with non-sinusoidal waveform.

Nonlinear loadLinear load

u

i i

u

t

t t

t

9

Linear and non-linear loads

A load is said to be linear when there is a linear relationship between current and voltage. In simpler terms, a linear load absorbs a sinusoidal current when it is supplied by a sinusoidal voltage: this current may be displaced by an angle compared with voltage.

When this linear relationship is not verified, the load is termed non-linear. It absorbs a nonsinusoidal current and thus harmonic currents, even when it is supplied by a purely sinusoidal voltage current absorbed by

a non-linear load.

10

GENERATING HARMONICS IN TRANSFORMER MAGNETIZING CURRENT

The excitation current Ie is split into two components: the magnetizing current I µ and IFe, proportional to the core power losses. These currents are displaced from each other by an angle /2. This displacement can be explained by means of excitation current waveform. If the coil is supplied with sinusoidal voltage the flux Φ must be sinusoidal too. Since the magnetizing characteristic B-H is nonlinear, and has a hysteresis loop, the current waveform obtained from magnetizing curve is far from sinusoidal.

11

12

Definition and characteristic quantities related to harmonics

Joseph FOURIER proved that all non-sinusoidal periodic functions can be represented by a sum of sinusoidal terms, the first one of which, at the recurrence frequency of the function, is said to be fundamental, and the others, at multiple frequencies of the fundamental, are said to be harmonic. A DC component may complete these purely sinusoidal terms.

Fourier's formula: y (t) = Yo + Yn √2 sin (nt – n)n = 1

n =

where:- Yo: DC component value, generally nil and would not be

considered - Yn: rms value of the nth harmonic component,

- : angular frequency of the fundamental,- n: displacement of the nth harmonic component.

13

Calculation of effective current absorbed by single-phase load:

I fund. = 56.2A ; Ih3 = 27.2A ; Ih5 = 2.7A ; Ih7 = 9.2A ; Ih9 = 7.8A

Irms = 56.22 + 27.2

2 + 2.7

2 + 9.2

2 + 7.8

2 = 63.6 A

Harmonics: Effective (RMS - Root Mean Square) Value

The effective value of a non-sinusoidal periodic value is equal to:

Effective value = Y2

1 + Y2

2 + Y2

3 + Y2

4 +…..+Y2

n

Y1 = fundamental component; Y2,..,Yn = harmonic components.

Yrms =1T

T

0y

2(t)dt = Y

2nn=1

n =

14

Total harmonic distortion

Total harmonic distortion is a parameter globally defining distortion of the alternating quantity.

There is another definition which replaces the fundamental Y1 with the total rms value Yrms.

This definition is used by some measuring instruments.

This is the ratio of the RMS value of the harmonics over the RMS value of the fundamental:

Y2

2 + Y2

3 + Y2

4 + Y2

5 +…..+Y2

n

Y1

THD = 100

15

(distortion factor) = 1

1 + (THD)2

Distortion factor vs. THD

16

Individual harmonic ratio

This quantity represents the ratio of the value of an harmonic over the value of the fundamental (Y1), according to the

standard definition or over the value of the alternating quantity (Yrms).

Representation of harmonic amplitude as a function of their order: harmonics value is normally expressed as a percentage of the fundamental.

(Frequency) spectrum

17

Power factor (PF) and Displacement Power Factor (DPF)

It is important not to confuse these two terms when harmonics are present, as they are equivalent only when currents and voltages are completely sinusoidal.

The power factor () is the ratio between active power P and apparent power S:

= P / S

The displacement power factor (cos1) relates to

fundamental quantities, thus:

cos 1 = P1 / S1

In pure sinusoidal waveform: cos1 = cos =

Distortion factor

The IEC 146-1-1 defines this factor as the ratio between the power factor and the displacement power factor cos 1 :

= / cos 1

18

Peak factor

The ratio of peak value over rms value of a periodic quantity.

Fc = Ypeak / Yrms

Some peak factor examples:• Linear load Fc = 1.41

• IT load Fc = 2 to 2.5

• Micro computing load Fc = 2.2 to 3

19

The current drawn by non-linear loads passes through all of the impedance between the system source and load. This current produces harmonic voltages for each harmonic as it flows through the system impedance. These harmonic voltages sum and produce a distorted voltage when combined with the fundamental. The voltage distortion magnitude is dependent on the source impedance and the harmonic voltages produced.

20

A non linear load is effectively drawing current from the power source at the fundamental frequency, and generating current back at higher frequencies. This results in a distorted current waveform as shown previous. Current harmonics disturb the supply voltage and this also results in a distorted voltage at the point of common coupling. Example: Consumer A and B are fed from the same line. The non linear loads of consumer A will distort the voltage of consumer B even if the latter has only linear loads.

A B

Point of common coupling

systemimpedance

21

Voltage and current total harmonic distortion

A non-linear load generates harmonic voltage drops in the circuits supplying it. In actual fact all upstream impedances need to be taken into consideration right through to the sinusoidal voltage source.

Consequently a load absorbing harmonic currents always has a non-sinusoidal voltage at its terminals. This is characterized by the voltage total harmonic distortion:

where Zn is the total source impedance at the frequency of

harmonic n, and In the rms value of harmonic n.

22

Output impedance of the various sources as a function of frequency.

23

Sources of Harmonics

There are many sources of power system harmonics. Some examples of harmonic producing devices are:Transformers: Third harmonic currents are present in the magnetizing current (a small portion of the transformer full load current). If the transformer saturates (due to over-voltage), the harmonic distortion level of the current increases substantially.Fluorescent Lamps: These devices produce a predominantly third order harmonic currenton the order of 20% to 30% of the fundamental current. Electronic ballasts have slightlydifferent characteristics but exhibit similar levels of harmonics.Pulse-Width Modulated Converters: These devices use an external controller for switching the input transistors allowing the current waveform to be shaped more desirably. However, these converters are limited in power and typically used inapplications less than a few hundred kilowatts.

24

Switched Mode Power Supplies: Typically found in single-phase electronic devices such as computers and other business and consumer electronics, these devices use a switching regulator to precisely control the DC voltage. The input of these power supplies normally consists of a full-wave bridge rectifier and a DC filter capacitor which producesan alternating pulse current waveform rich in third harmonic. Though they are not used in large power applications, the cumulative effects of many devices may create concerns,particularly for 400/230 Volt Y systems.

25

Light dimmer or heating regulator

Switch mode power supply rectifier

Wave shape of current absorbed by some non-linear loads.

H3 H5 H7 H9 H11 H13 H15 H17 H1954 18 18 11 11 8 8 6 6

H3 H5 H7 H9 H11 H13 H15 H17 H1975 45 15 7 6 3 3 3 2

26

Three-phase rectifier with front end capacitor

Three-phase rectifier with DC filtering reactor

H3 H5 H7 H9 H11 H13 H15 H17 H190 80 75 0 40 35 0 10 5

H3 H5 H7 H9 H11 H13 H15 H17 H19 0 25 7 0 9 4 0 5 3

27

Harmonic currents in three phase systems

Harmonics get more complicated in three phase applications. Here not only do we have to deal with phase conductors, but also the neutral conductor, triplen (odd multiples of 3 i.e. 3rd,

9th, 15th etc,) harmonics, and sequence harmonics. The triplen harmonics are the major cause of heat because they add together in the neutral conductor. The magnitude of the harmonic current produced by the triplens can approach twice the phase current. This causes the neutral conductor to overheat because neutral conductors were historically designed with the same ampacity as the phase conductors.

Neutral conductor

28

With, for example, an harmonic 3 of 75%, the current flowing in the neutral is 2.25 times the fundamental. The current in each phase is only SQR (1+ 0.752 ) = 1.25 times the fundamental.

neutral current

-150

-100

-50

0

50

100

150

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37

Ia Ib Ic Ia 3rd Ib 3rd Ic 3rd neutral current

29

A situation that produces abnormal amounts of heat inmotors is the combination of positive and negative sequenced harmonics. The positive sequenced harmonics are the fundamental, 7th, 13th, 19th, etc. They tend to apply an additional forward force in the direction of the motor rotation. The negative sequenced harmonics are the 5th,

11th, 17th, etc. They present a force that opposes the motorrotation and tries to make the motor rotate in the oppositedirection. The force of these harmonics acting upon each other creates heat which leads to premature failure. Harmonic voltage distortion causes increased eddy current losses in the motors, in the same way as seen for transformers.

Induction motor

30

Transformers

The effects of harmonics inside the transformers involve mainly three aspects:

• a) increase of iron losses (or no-load losses)• b) increase of copper losses• c) presence of harmonics circulating in the windings

a) The iron losses are due to the hysteresis phenomenon and to the losses caused by eddy currents; the losses due to hysteresis are proportional to the frequency, whereas the losses due to eddy currents depend on the squareof the frequency.

31

b) The copper losses correspond to the power dissipated by Joule effect in the transformer windings. As the frequency rises (starting from 350 Hz) the current tends to thicken on the surface of the conductors (skin effect); under thesecircumstances, the conductors offer a smaller cross section to the current flow, since the losses by Joule effect increase.

These two first aspects affect the overheating which sometimes causes a de-rating of the transformer

32

c) The third aspect is relevant to the effects of the triple-N harmonics (homopolar harmonics) on the transformer windings. In case of delta windings, the harmonics flow through the windings and do not propagate upstream towards the network since they are all in phase; the delta windings therefore represent a barrier for triple-N harmonics, but it is necessary to pay particular attention to this type of harmonic components for a correct dimensioning of the transformer. The triplen harmonics are trapped and circulate in the delta primary of the transformer. Since most loads produce high levels of the 3rd harmonic (one of the triplens), the harmonic content reflected back to the source is reduced. The circulating harmonics in the primary of the transformer creates heat because of their higher frequencies. For this reason, a transformer that can handle the excess heat is needed. This transformer is called a K-rated transformer

33

A situation that produces abnormal amounts of heat inmotors is the combination of positive and negative sequenced harmonics. The positive sequenced harmonics are the fundamental, 7th, 13th, 19th, etc. They tend to apply an additional forward force in the direction of the motor rotation. The negative sequenced harmonics are the 5th,

11th, 17th, etc. They present a force that opposes the motorrotation and tries to make the motor rotate in the oppositedirection. The force of these harmonics acting upon each other creates heat which leads to premature failure. Harmonic voltage distortion causes increased eddy current losses in the motors, in the same way as seen for transformers.

Induction motor

34

Skin effect losses

The transmission distribution grid was designed to carry thefundamental 50 Hertz frequency. A problem exists with higherfrequencies (harmonics), that is, they do not fully penetrate the conductor. They travel on the outer edge of the conductor. This is called skin effect. When skin effect occurs, the effective cross sectional area of the conductor decreases; increasing the resistance and the I2R losses, which in turn heats up the conductors and anything connected to them. This heating effect can cause circuit breakers to trip, neutral and phase conductors to heat up to critical flash over temperatures, and premature failure of motors and transformers.This is costly in terms of downtime, loss of production, repair,and possible reconstruction.

35

In the presence of high-order harmonics, it is necessary to take skin effect into account, because it affects the life of cables. In order to overcome this problem, it is possible to use multiple conductor cables or busbar systems formed by more elementary isolated conductors.

36

Harmonic currents in three-phase four-wire networks

ia(t) = Ian cos (nt – an)n=1

n =

ib(t) = Ibn cos [n(t – 120) - bn]n=1

n =

ic(t) = Icn cos [n(t + 120) - cn]n=1

n =

uan(t) = Um cos t

ubn(t) = Um cos (t-120)

ucn(t) = Um cos (t-120)

37

Neutral current

iN(t) = { Ian cos (nt – an) + Ibn cos [n(t – 120) – bn] + Icn

cos [n(t + 120) – cn] }n=1

n =

If the load is unbalanced, then the neutral connection may contain currents having spectrum similar to the line currents.

In the balanced case, Ian = Ibn = Icn = In and an = bn = cn =

n , for all n; i.e., the harmonics of the three phases all have

equal amplitudes and phase shifts. The neutral current is then

iN(t) = 3 In cos (nt – n)n=3.6.9..

n =

38

iN(t) = 3 In cos (nt – n)n=3.6.9..

n =

Fundamental and most harmonics cancel out

Triplen (triple-n, or 3, 6, 9, ...) harmonics do not cancel out, but add.

rms neutral current is INrms = 3 n=3,6,9..

n = I2n

2

39

40

Example

A balanced nonlinear load produces line currents containing fundamental and 20% third harmonic: ian(t) = I1 cos(t – 1) +

0.2 I1 cos(3t – 3). Find the rms neutral current, and compare

its amplitude to the rms line current amplitude.

INrms = 60% of I1rms

The triplen harmonics in the three phases add, such that 20% third harmonic leads to 60% third harmonic neutral current. Significant unexpected neutral current flows.

= 3 ( 0.2 I1 )

2

2= 0.6

I1

2INrms = 3

n=3,6,9..

n = I2n

2

41

Generally the rms current = n=1

n = I2n

2

=( I1 )

2

2I1rms =

n=1

n = I2n

2

( I3 )2

2+

=( I1 )

2

2I1rms =

(0.2 I1 )2

2+

I1

21 + 0.04

I1

2

Yet the presence of the third harmonic has very little effect on the rms value of the line current.

42

Y-connected nonlinear load, no neutral connection:

If the load is balanced, then it is still true that

iN(t) = 3 In cos (nt – n)n=3.6.9..

n =

43

But iN (t) = 0, since there is no neutral connection and the ac

line currents cannot contain triplen harmonics.

What happens?

A voltage is induced at the load neutral point, that causes the line current triplen harmonics to become zero.The load neutral point voltage contains triplen harmonics.

With an unbalanced load, the line currents can still contain triplen harmonics

44

Delta-connected load

There is again no neutral connection, so the ac line currents contain no triplen harmonics.The load (phase) currents may contain triplen harmonics: with a balanced nonlinear load, these circulate around the delta.

45

Harmonic currents in power factor correction capacitors

PFC capacitors are usually not intended to conduct significant harmonic currents.Heating in capacitors is a function of capacitor equivalent series resistance (esr) and rms current. The maximum allowable rms current then leads to the capacitor rating:

rated rms voltage Urms = Irms

2 f C

rated reactive power QC = I2

rms

2 f C

46

47

i (t) = In cos (nt – n)n=1

Voltage and current as Fourier series:

u (t) = Un cos (nt – n)n=1

Power per cycle

Pcycle = v(t) i(t) dt

0

T

This is related to average power as follows:

Pav = Pcycle

T=

1

Tv(t) i(t) dt

0

Tinfluence of harmonics on average power:

Pav = 1

T0

T

Un cos (nt – n)] [n=1

[ In cos (nt – n)] dtn=1

Average power

48

Integrals of cross-product terms are zero

0

T

Un cos (nt – n)] [[ In cos (mt – m)] dt

{0 if n = m

Un Incos (n – m) if n = m

2

=

Expression for average power becomes

Pav = n=1

Un Incos (n – n)2

So net energy is transmitted to the load only when the Fourier series of u(t) and i(t) contain terms at the same frequency. For example, if the voltage and current both contain third harmonic, then they lead to the average power:

U3 I3cos (3 – 3)2

49

Example 1

Voltage: fundamental onlyCurrent: third harmonic only

u (t) i (t)

Power: zero average

50

Example 2u (t), i (t)

Voltage: third harmonic onlyCurrent: third harmonic only,in phase with voltage

Power: nonzero average

51

Example 3

Fourier series:

u(t) = 1.2 cos (t) + 0.33 cos (3t) + 0.2 cos (5t)

i(t) = 0.6 cos (t + 30°) + 0.1 cos (5t + 45°) + 0.1 cos (7t + 60°)

Average power calculation:

Pav =(1.2)(0.6)

2cos (30°) +

(0.2)(0.1)

2cos (45°) = 0.32 W

52

Voltage: 1st, 3rd, 5thCurrent: 1st, 5th, 7th

Power: net energy is transmitted at fundamental and fifth harmonic frequencies

53

In AC circuits the fundamental current and fundamental voltage together produce fundamental power. This fundamental power is the useful power that cause motor to rotate and deliver work on the rotor’s shaft or to make electrical heater to deliver heat. The product of a harmonic voltage times the corresponding harmonic current also produces a harmonic power. That one is usually dissipated as a heat and does not do useful work. Harmonic currents and voltages should be kept as small as possible.

The product of a fundamental voltage and a harmonic current yields zero net power.

54

GENERATING HARMONICS WITH SWITCH

1000 V60 Hz

Synchronous switch R

10

1410 V

141 A

I = 1000/10 = 100 A

Closed switch

P = I2R = 1002x10 = 100 kW

Operational switch(half time opened)

Dissipated power = 50 kW

I2 = P/R = 50000/10 = 5000I = 70.7 A

P = I2 x R = 70.72 x 10

P = 50 kW

Irms = 70.7 A

55

1410 V

141 A

Chopped current The chopped current can be decomposed to fundamental and harmonics component.

84 A

32.50

Fundamentalcomponent

Apparent fundamental power supplied by sourceS = U x I = 1000 x 84/1.414 = 59.3 kVA

Active fundamental power supplied by sourceP = S x cos = 59.3 x cos 32.5 = 50 kW

Reactive fundamental power supplied by source

Q = V S2 - P2 = V 59.32 – 502 = 31.9 kVAr

The 10 resistor absorbs a fundamental active power

P = I2 x R = 59.32 x 10 = 35.2 kW

The difference of 50 – 35.2 = 14.8 kW goes to the harmonic power absorbed by resistor

10

IF = 59.3 A

1 kV

∠-32.5

50kW

31.9kVAr14.8kW

35.2kW

14.8kW

56

The switch carries a fundamental current of 59.3 A and it absorbs 14.8 kW and 31.9 kVAr, it can be represented by resistance and inductive reactance connected in series.

R = P / I2 = 14800 / 59.32 = 4.21 X = Q / I2 = 31900 / 59.32 = 9.07

Effective value of the harmonic current is

IH = V I2 - IF2 = V 70.72 – 59.32

= 38.5 A

Consequently the voltage across the 10 resistor is U = I x R = 38.5 x 10 = 385 V.

4.21 j9.07

101kVIF = 59.3A

Equivalent circuit for the fundamental component

10IH = 38.5A

385 V

Equivalent circuit for all harmonic components

57

58

Low-power harmonic limits

In a city environment such as a large building, a large fraction of the total power system load can be nonlinear• Example: a major portion of the electrical load in a building is comprised of fluorescent lights, which present a very nonlinear characteristic to the utility system.• A modern office may also contain a large number of personal computers, printers, copiers, etc., each of which may employ peak detection rectifiers.• Although each individual load is a negligible fraction of the total local load, these loads can collectively become significant.

59

Short Term Effects

Over consumption of RMS currentUnwanted tripping of protectionsMalfunction of sensitive applicationsInterference of remote control and telecommunication systemsAbnormal vibration and noise (LV panels, motors, transformers)

Long Term Effects-Overheating

Overheating of capacitor banksOverheating of transformers, alternatorsOverheating of phases, particularly neutral

60

Harmful Effects on Receivers

Cables:Overheating of cablesAdditional losses due to skin effectIncrease in dielectric losses of insulation

Induction motors:Increase in core (stator) and Joule lossesPulsating torques causing efficiency reduction, abnormal vibration, rotor overheating

61

General Solutions

Limit injected harmonic currents:Install limitation induction coils for speed drivesInstall specific rectifiers called active front end

Install anti-harmonics induction coils

Install filters to trap harmonics:Passive filtersActive filtersHybrid filters

Oversize equipment

62

With linear loads, the neutral can be the same size as the phase conductors because the neutral current cannot be larger than the largest phase current, even when the load is completely unbalanced.

The 120° phase shift between linear load currents will result in their balanced portions instantaneously canceling in the neutral.

Harmonic Currents add in the Neutral

63

With non-linear loads, the neutral current generally exceeds the largest phase current, even when the loads are in perfect RMS current balance.

When the load is non-linear however, the current pulse on one phase will not have a pulse on either of the other phases for which to cancel. The pulses are additive which often leads to heavier current on the neutral conductor than on any phase conductor. The frequency of this neutral current is primarily 150 Hz (3rd harmonic).

64

The presence of harmonic currents can also lead to some special problems in three-phase systems:• In a four-wire three-phase system, harmonic currents can lead to large currents in the neutral conductors, which may easily exceed the conductor rms current rating• Power factor correction capacitors may experience significantly increased rms currents, causing them to fail

65

K factor

Harmonic currents are generated whenever a non-linear load is connected to the mains supply. The problems caused by harmonic currents include overheating of cables, especially the neutral conductor, overheating and vibration in induction motors and increased losses in transformers.Where power factor capacitors are fitted, harmonic currents can damage them and care must be taken to avoid resonance with the supply inductance.Losses in transformers are due to stray magnetic losses in the core, and eddy current and resistive losses in the windings. Of these, eddy current losses are of most concern when harmonics are present, because they increase approximately with the square of the frequency. Before the excess losses can be determined, the harmonic spectrum of the load current must be known.

66

The eddy current loss at a particular harmonic is given by:

Pn = Pf In2 n2

wherePn is the eddy current loss at harmonic number n

Pf is the eddy current loss at the fundamental frequency f

In is the fraction of total rms load current at harmonic number nThe total eddy current loss is given by summing the losses for the individual harmonics and the fundamental:

Pt = Pf In2 n2

n=1

n = max

where Pt is the total eddy current loss.

67

There are two distinct approaches to accounting for this increased eddy current loss in selecting a transformer. The first, devised by transformer manufacturers in conjunction with Underwriters Laboratories in the United States, is to calculate the factor increase in eddy current loss and specifya transformer designed to cope; this is known as ‘K-Factor’.

The second method, used in Europe, is to estimate by how much a standard transformer should be de-rated so that the total loss on harmonic load does not exceed the fundamental design loss; this is known as ‘factor K’. The figures produced by each method are numerically different; ‘factor K’ is a total rating factor while ‘K-factor’ is a multiplier (although a de-rating factor can be derived from it). The fact that both methods useK as a designation can lead to confusion when talking to suppliers.

68

K-Factor

In US practice, where dry-type transformers are often used, the K-factor is the ratio of eddy current losses when driving non-linear and linear loads:

K = Pt

Pf

= In2 n2

n=1

n = max

This K-factor is read directly by many power meters (e.g. Fluke 41 & 43). Once the K-Factor of the load has been determined, it is a simple matter to specify a transformer with a higher K-rating from the standard range of 4, 9, 13, 20, 30, 40, 50.

69

Factor K

In Europe, the transformer de-rating factor is calculated according to the formulae in BS 7821 Part 4. The factor K is given by:

K = { 1+e

1+e[

I1I

]2 n = N

n = 2[ nq (

In

I1)

2]}

0.5

where

e is the eddy current loss at the fundamental frequency divided by the loss due to a dc current equal to the RMS value of the sinusoidal current, both at reference temperature.

n is the harmonic orderI is the rms value of the sinusoidal current including all harmonics given by:

70

I = [ n = N

n = 1

(In

I1)

2]

0.5( In)

2]0.5

= I1[ n = N

n = 1

In is the magnitude of the nth harmonic

I1 is the magnitude of the fundamental current

q is an exponential constant that is dependent on the type of winding and frequency. Typical values are 1.7 for transformers with round or rectangular cross section conductors in both windings and 1.5 for those with foil low voltage windings.

71

A standardized empirical formula (NFC 52-114) is used to calculate the de-rating factor k to be applied to a transformer.

For example whereH5 = 25% ; H7 = 14% ; H11 = 9% ; H13 = 8%,

the factor k is 0.91.

72

K-Rated or De-Rated?

The great advantage of a ‘K-rated’ transformer is that it will have been designed with harmonic loads in mind and care will have been taken to keep losses low. For example, eddy current losses will have been reduced by the use of stranded conductors and magnetic losses will have been reduced by the use of low loss steels. The neutral point connections are usually brought out individually, so that the star point has a 300% current rating.

73

On the other hand, de-rating a standard transformer has a number of disadvantages. Because the transformer is oversized, the primary over-current protection level may be too high to protect the secondary, but if the protection level is reduced, the inrush current may cause tripping. A de-ratedtransformer is less efficient; the excess losses are still being generated and dissipated within the transformer, rather than being designed out, and a larger core than necessary, with larger losses, is being magnetized.

There is also a potential maintenance problem – long after installation, changes in the needs of the facility may result in additional load being added without reference back to the initial de-rating. This may lead to overloading and consequent failure.

74

Typical calculation according to BS 7821 Part 4

(taking q as 1.7 and assuming that eddy current loss at fundamental is 10% of resistive loss i.e. e = 0.1).

Harmonic

No.

RMS

current (In)In/I1 (In/I1)

2nq nq (In/I1)

2

1 1 1 1

3 0.82 0.82 0.6724 6.473 4.3525

5 0.58 0.58 0.3364 15.426 5.1893

7 0.38 0.38 0.1444 27.332 3.9467

9 0.18 0.18 0.0324 41.900 1.3576

11 0.045 0.045 0.0020 58.934 0.1193

=2.1876 =14.9653

Total rms (I)=1.479

(I1/I)2 =

0.457

e/(1+e) = 0.091

K2=1+(0.091x0.457x14.9653); K=1.27

de-rate to

78.52%

75

Typical calculation according to Underwriters’ Laboratories method:

Harmonic

No.

RMS

current (In)In/I1 (In/I1)

2 In/I (In/I)2 (In/I)

2 x

n2

1 1 1 1 0.6761 0.4571 0.4571

3 0.82 0.82 0.6724 0.5544 0.3073 2.7663

5 0.58 0.58 0.3364 0.3921 0.1538 3.8444

7 0.38 0.38 0.1444 0.2569 0.0660 3.2344

9 0.18 0.18 0.0324 0.1217 0.0148 1.2000

11 0.045 0.045 0.0020 0.0304 0.0009 0.1120

=2.1876 11.6138

Total rms (I) = 1.479

K-factor 11.6138

76

Failure of power factor capacitors

The impedance of a circuit dictates the current flow in that circuit. As the supply impedance is generally considered to be inductive, the network impedance increases with frequency while the impedance of a capacitor decreases.

XL = 2 f L ()

Inductive reactance

XC = ()

Capacitive reactance

2 f C

1

f (hz)

X() XL

XC

XL = XC

resonantfrequency

77

XL = XC = X = 2 f L = 2 f C

1fr =

2 L C

1

If this condition occurs on, or close to, one of the harmonics generated by non linear load, then large harmonic currents can circulate between the supply network and the capacitor equipment. These currents are limited only by the damping resistance in the circuit. Such currents will add to the harmonic voltage disturbance in the network causing an increased voltage distortion. This results in a higher voltage across the capacitor and excessive current through all capacitor components. Resonance can occur on any frequency, but in general, the resonance we are concerned with is on, or close to, the 5th, 7th, 11th and 13th harmonics for 6 pulse systems.

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Example of the resonant frequency calculation

3 phase 11/0.4 kV distribution transformer of 1.6 MVA, with(uk) 5.82 % impedance, would have a short circuit current at

main panel of

ISC = S

3 U2 uk

= 1600

1.732 x 0.4 x 0.0582= 39681 A

and short circuit power of

SSC = 3 ISC U2 = 1.732 x 39681 x 0.4 = 27491 kVA

Assuming the size PFC capacitors of 230 kVAr

System harmonic calculation: n = SSC

QC

=27491

230= 10.9

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This calculation indicates that the resonant frequency can develop if the distribution system contains the 11th harmonic in any significant amount, causing the capacitor to overheat and possibly, explode.

Using detuning reactors

This solution consists of protecting the capacitors, designed to improve the displacement power factor by installing a series reactor. This reactor is calculated so that resonancefrequency matches none of the harmonics present.Typical tuning frequencies are for a 50 Hz fundamental: 135 Hz (order 2.7), 190 Hz (order 3.8) and 255 Hz (order 4.5).

Thus for the fundamental, the battery can perform its displacement power factor improvement function, while the high impedance of the reactor limits amplitude of the harmonic currents.

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PFC capacitors are usually not intended to conduct significant harmonic currents.Heating in capacitors is a function of capacitor equivalent series resistance (esr) and rms current. The maximum allowable rms current then leads to the capacitor rating:

rated rms voltage Urms = Irms

2 f C

rated reactive power QC = I2

rms

2 f C

81

Evaluating System Harmonics

In order to prevent or correct harmonic problems that could occur within an industrial facility, an evaluation of system harmonics should be performed if the facility conditionsmeet one or more of the criteria below.1.The application of capacitor banks in systems where 20% or more of the load includes other harmonic generating equipment or where background distortion exceeds 2%.2.The facility has a history of harmonic related problems, including excessive capacitor fuse operation.3.Large single non-linear loads are being added greater than about 10% of the transformer rating.4.Many small identical non-linear loads are being added that operate together.In facilities where restrictive power company requirements limit the harmonic injection back into their system to very small magnitudes.

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Mitigation of Harmonics

There are many ways to reduce harmonics, ranging from variable frequency drive designs to the addition of auxiliary equipment. Following are some of the more common methods used today for controlling power system harmonics.

Addition of a downstream reactor or reduction in upstream source impedance reduces voltage THD at the point considered.

83

Carefully choosing the installation structure

Sensitive loads should not be parallel-connected with non-linear loads

Solution to avoid Solution to recommended

Very powerful non-linear loads should preferably be supplied by another MV/LV transformer.

84

Passive harmonic filters

issue, power factor correction is required, and specific harmonics such as 5th or 7th are creating the problem. Passive filters are ideal for systems that have a high percentage of 6 pulse drives and other linear loads. However, the filters may need to be retuned for changes in the power system. Filters can be designed for several nonlinear loads or for an individual load,

Passive or ‘trap’ filters employ ‘passive’ elements (capacitors and inductors) to ‘trap’ or absorb harmonics. Passive filters are generally configured to remove only one or two specific harmonics. Passive filters are generally regarded as unsuitable for filtering 3rd harmonics. For this reason, they are

best suited for applications in which 3rd harmonics are not an

Non linear load

Harmonic trap filter

85

86

Principle of compensation of harmonic components by “shunt-type” active harmonic conditioner

The device should able to inject at any time a current where each harmonic current has the same amplitude as that of the current in the load and is in opposition of phases, then Kirchoff’s law at point A guarantees that the current supplied by the source is purelysinusoidal

87

Isolation Transformers: An isolation transformer provides several advantages. First and foremost, it provides impedance to the drive, which reduces current distortion. Itobviously resolves voltage mismatch between the supply and the load. If the secondary is grounded, it isolates ground faults and reduces common mode noise.

88

The Engineering Recommendation G5/3 has been in place for decade and a half and has provided certain guidelines. This include limits of harmonic currents that can be fed into the system by customers, limits for harmonic distortion caused, suggestion for whether a load can be connected and procedures for measurement and assessment of new loads. The new G5/4 is more stringent than G5/3, requiring more careful assessment of new loads and measurements to be taken at different voltage levels. It also extends the range of harmonics covered and includes inter harmonics and notching currents.

Engineering recommendation G5/4

The original edition of G5/4 included certain limits which were to be applied from1st January 2005. However, after discussion with industry it was agreed to modify this requirement, and this has resulted in G5/4-1:2005

89

The standard G5/4 published in March 2001,seeks to limit harmonic voltage distortion levels on public networks at the time of connection of new non-linear loads to ensure compatibility of all connected equipment.

It does this by seeking data from the customer and then making an assessment to see whether the planning limits are likely to exceed at the time of connection

The enforcement of this is via the Electricity Supply Regulations, the Grid and Distribution Codes, and the connection agreements between NOCs and customers.

90

Features of G5/4

G5/4 defines planning levels and introduces compatibility levels for individual harmonics and THD over the voltage range from 400V to 400 kV.

Emphasis placed on voltage distortion levels in Stage 2 and 3 assessment (compared with other standard)

The three stage assessment process of G5/3 retained

Information on harmonic impedance for use in network modeling has been updated.

Description material and examples moved to the Application Guide ETR 122.

91

Uses IEC standards wherever possible

Introduces specific emission requirements for number of aggregated low voltage equipment

THD assessment required up to 50th harmonic

5th harmonic current emissions levels reduced

Harmonic combination rules clarified to account for harmonic phase angles likely to be

Harmonic emissions modifiable relative to fault level

Introduces a flow chart to help users through the assessment process

92

What are the G5/4 limits?

Total harmonic distortion limits are recommended based upon the type of installation. It is the defacto standard for power utilities and is often required of large consumers and medium voltage systems. This specification has also become increasingly common in low voltage system.

Total voltage distortion not to exceed 5% in 400V systems and 4% in 6.6kV, 11kV and 20 kV systems, and 3% for the other systems with voltage level above 20kV.

93

Gaps between compatibility levels and planning levels for THD harmonics level

Voltage level (kV)

Compatibility level

Planning level

0.4 8% 5%

6.6 – 20 8% 4%

>20 – 36.5 8% 3%

66 - 145 5% 3%

275 - 400 3.5 % 3%

Compatibility levels in voltage systems are based on the immunity of capacitors as they are susceptible to harmonic voltage distortion and are common in use.

94

Total current distortion is limited on harmonic by harmonic basis measured in absolute RMS amps.

Table 7 is based on a system fault level of 10MVA at 400V. If the fault levels varies from this base level, the figures in Table 7 may be varied prorate, as will the powers that can be connected.

95

Voltage distortion limits

96

97

98

99

100

101

102