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Power System Harmonics

Tuesday, June 11, 20138:30AM – 12:30PM

Florida Electric Cooperatives AssociationSand Pearl ResortClearwater, Florida

Ralph Fehr, Ph.D., P.E.Senior Member, IEEE

[email protected]

Topics

Fundamentals of Harmonics

Causes of Harmonics

Effects of Harmonics

Power Factor

Harmonic Mitigation Techniques

IEEE/FECA Harmonics Seminar Jun. 2013 Ralph Fehr, Ph.D., P.E. 2




Jean Baptiste Joseph Fourier1768‐1830

Any periodic waveform W ( t )can be represented by a series of sinusoidal waveforms with frequencies of integer multiples of the periodic waveform.

)(sin)(cos2

)(1

0 ntbntaa

tW nn

n

The coefficients are

0,)(cos)(1

ndtnttWan

1,)(sin)(1

ndtnttWbn



The coefficients an and bn are of most interest.

How do we go about finding them?

)(sin)(cos2

)(1

0 ntbntaa

tW nn

n

0,)(cos)(1

ndtnttWan

1,)(sin)(

1

ndtnttWbn



Analytically:

Empirically:

http://www.integral-calculator.com/#



Power system harmonics typically have half‐wave symmetry.

Half‐wave rectification destroys this symmetry.

With half‐wave symmetry, a0 = 0 and only odd‐ordered harmonics are present.

Let’s play with some harmonics!

)(sin)(cos2

)(1

0 ntbntaa

tW nn

n

0,)(cos)(1

ndtnttWan

1,)(sin)(

1

ndtnttWbn



For a square wave,

1 12

122sin4)(

nsquare k

tfktW

...10sin

5

16sin

3

12sin

4tftftf

first term first two terms first three terms



Start with a complex waveform



Superimpose the first harmonic (fundamental frequency)



Add the third harmonic



Add the fifth harmonic



Add the seventh harmonic



Add the ninth harmonic



Add the eleventh harmonic



Change your perspective a bit



Separate the harmonics



Measure the amplitude of each harmonic



The harmonic amplitude spectrum representsthe Fourier transform of the original waveform

1 3 5 7 9 11



Harmonic spectrum (frequency domain)

Original waveform (time domain)

1 3 5 7 9 11



Relationship between harmonic voltages and harmonic currents:

Vharmonic = (Iharmonic) (Zsystem)

Weak systems (with high Zsystem) will have greater voltage distortion than stiff systems (with low Zsystem).

So voltage distortion may be acceptable when on the utility system, but becomes problematic when running off a diesel generator.


Causes of Harmonics


Causes of Harmonics

Harmonics are the result of non‐linear loads.

A purely resistive load is linear, since the graph of voltage versus current is a straight line.

V

I

slope = resistance (R)I V

+

_

R


Causes of Harmonics

Transformer iron has a non‐linear hysteresis characteristic.

The sinusoidal flux results in a distorted excitation current.

Transformer excitation current is rich in third harmonic, and also contains some second harmonic.


Causes of Harmonics

But most harmonics come from loads such as induction and arc furnaces, and power electronics.

Power electronics are the biggest culprit when it comes to producing harmonics. Static VAR compensators, variable‐frequency motor drives, and switching power supplies are the greatest concern.


Causes of Harmonics

Rectifiers are essentially harmonic current generators. The number of pulses of the rectifier determine the harmonic orders that will be dominant.

6‐Pulse RectifierDominate harmonics: 5th and 7th

12‐Pulse RectifierDominate harmonics: 11th and 13th


Causes of Harmonics

Variable‐frequency drives can be a major source of harmonics. Proper DC link design and good filtering help mitigate harmonics.





Conductor Overheating

Increases as the square of the rms current per unit volume of the conductor. Harmonic currents are subject to “skin effect”, which increases with frequency and effectively reduces the conductor cross‐sectional area.

Skin Effect

Direct Current High FrequencyLow Frequency



The skin depth () of a conductor is defined as the boundary of the region below the surface of the conductor that contains 63% of the charge carriers.

Skin Depth

63.0e

11

63% of thechargecarriers



Skin depth of a cylindrical conductor is given by

Skin Depth

f

s

where

= resistivity of conductor in ‐mf = frequency in hertz = permeability of free space (4 10‐7 H/m)


Skin Depth Example

Find the skin depth of copper and aluminum at 60 Hz and at 180 Hz. (Cu = 1.72 10‐8 ‐m and Al = 2.82 10‐8 ‐m)

in335.0cm852.010460

1072.17

8

60Cu

in194.0cm492.0104180

1072.17

8

180Cu

in430.0cm091.110460

1082.27

8

60Al

in248.0cm630.0104180

1082.27

8

180Al

42% decreasefrom fundamentalto 3rd harmonic


Effects of HarmonicsElectronic Device Misoperation

Various electronic circuits can misoperate due to multiple zero crossings of the voltage waveform.



Capacitor Bank Problems

Increased heating in the capacitor units can lead to reduced life.

Resonance can be a concern if the capacitor is tuned near a critical harmonic. Resonance causes overvoltage which can lead to dielectric failure.

Capacitor banks can be used to mitigate the effects of harmonics by being configured as a harmonic filter. More on this later...



Protection Problems

Harmonics can cause nuisance tripping of circuit breakers, and fuses to blow for no apparent reason.

Since harmonic levels at a particular location of the power system can vary significantly from one moment to the next, protection problems due to harmonics can be very difficult to diagnose.



Transformer Overheating

Like conductors, harmonics cause transformer windings to experience additional heating due to skin effect.

Additional core heating also occurs due to increased eddy currents and stray flux losses.

For these reasons, k‐factor transformers should be used with non‐linear loads. More on this later...



Motor Overheating

Motors experience the same types of problems due to harmonics as transformers.

Additionally, motors can also overheat due to unbalanced voltages due to waveform distortion.



Metering Errors

Metering devices can improperly measure electrical quantities due to harmonic distortion.

Pure Sinusoid Highly‐distorted Sinusoid

2

II peakrms

22

II peakrms 50% difference


Effects of HarmonicsWhat does “rms” mean?

Root‐mean‐square (rms) averaging is done with AC waveforms to express an effective DC value.

I V

+

_

R

If 1A DC is passed through a resistor of resistance R, the power dissipated is (1)2 R = R watts.

What AC current must the resistor carry to dissipate the same power?

)usoidsinpureif(2

AA1 peak

rms

Average

SomethingElse

Peak



Root‐mean‐square Calculation

dt)t(iTT

1I

2

1

T

T

2

12

rms



Metering

Inexpensive meters sense the peak of a waveform, then adjust it to “rms” by scaling by .

This is fine if the waveform is close to sinusoidal.

Distortion can cause significant error.

“True rms” meters sample the waveform, calculate the rms value, and display accurate results regardless of the level of distortion.

2


Effects of Harmonics“True rms” Metering

0.0‐2.3‐7.8‐0.19.418.626.243.762.158.361.570.771.482.188.683.576.172.877.683.3


91.498.195.594.793.194.795.598.191.483.377.672.876.183.588.682.171.470.761.558.3

62.143.726.218.69.4‐0.1‐7.8‐2.3‐1.8‐0.3‐0.10.12.37.80.1‐9.4

‐18.6‐26.2‐43.7‐62.1

‐58.3‐61.5‐70.7‐71.4‐82.1‐88.6‐83.5‐76.1‐72.8‐77.6‐83.3‐91.4‐98.1‐95.5‐94.7‐93.1‐94.7‐95.5‐98.1‐91.4

‐83.3‐77.6‐72.8‐76.1‐83.5‐88.6‐82.1‐71.4‐70.7‐61.5‐58.3‐62.1‐43.7‐26.2‐18.6‐9.40.17.82.30.0

Effects of Harmonics“True rms” Metering


0.0‐2.3‐7.8‐0.19.4

18.626.243.762.158.361.570.771.482.188.683.576.172.877.683.3

91.498.195.594.793.194.795.598.191.483.377.672.876.183.588.682.171.470.761.558.3

62.143.726.218.69.4‐0.1‐7.8‐2.3‐1.8‐0.3‐0.10.12.37.80.1‐9.4

‐18.6‐26.2‐43.7‐62.1

‐58.3‐61.5‐70.7‐71.4‐82.1‐88.6‐83.5‐76.1‐72.8‐77.6‐83.3‐91.4‐98.1‐95.5‐94.7‐93.1‐94.7‐95.5‐98.1‐91.4

‐83.3‐77.6‐72.8‐76.1‐83.5‐88.6‐82.1‐71.4‐70.7‐61.5‐58.3‐62.1‐43.7‐26.2‐18.6‐9.40.17.82.30.0

2222222222rms 3.581.627.432.266.184.9)1.0()8.7()3.2(0.0

10

1i

~

2222222222 3.836.778.721.765.836.881.824.717.705.61

~~

2222222222 3.834.911.985.957.941.937.945.951.984.91

~~

2222222222 3.585.617.704.711.826.885.831.768.726.77

~~2222222222 )3.0()8.1()3.2()8.7()1.0(4.96.182.267.431.62

~~

2222222222 )1.62()7.43()2.26()6.18()4.9(1.08.73.21.0)1.0(

~~

2222222222 )6.77()8.72()1.76()5.83()6.88()1.82()4.71()7.70()5.61()3.58(

~~

2222222222 )4.91()1.98()5.95()7.94()1.93()7.94()5.95()1.98()4.91()3.83(

~~

2222222222 )5.61()7.70()4.71()1.82()6.88()5.83()1.76()8.72()6.77()3.83(

~~

2222222222 0.03.28.71.0)4.9()6.18()2.26()7.43()1.62()3.58(

~ = 63.7 Arms

A4.692

ipeak

Power Factor


AC Power

v(t) = Vm cos (t + v) i(t) = Im cos (t + i)

= Vm Im cos (t + v) cos (t + i)

s(t) = (Vm Im)/2 [cos (v i) + cos (2ωt + v + i)]

s(t) = v(t) x i(t)

coscos2

1coscos


AC Power

s(t) = (Vm Im)/2 [cos (v i) + cos (2ωt + v + i)]

ivvivmm t2coscos2

I

2

V)t(s

ivvivvivrmsrms sint2sincost2coscosIV)t(s

2ωt + v + i = 2(ωt + v) – (v – i)

cos ( – ) = cos cos + sin sin


AC Power

ivvivvivrmsrms sint2sincost2coscosIV)t(s

Constant component of real power

Oscillating component of real power

Reactive power

s(t) = P + P cos [2(t + v)] + Q sin [2(t + v)]

Let P = Vrms Irms cos (v – i)and Q = Vrms Irms sin (v – i)


AC PowerAC Power


AC Power

s(t) = P + P cos [2(t + v)] + Q sin [2(t + v)]

P = Vrms Irms cos (v – i)

Q = Vrms Irms sin (v – i)

= Vrms Irms [cos (v – i) + j sin (v – i)]S = P + j Q


P = P + P cos [2(t + v)]

Q = Q sin [2(t + v)]

AC Power

= Vrms Irms [cos (v – i) + j sin (v – i)]

jesinjcos

= Vrms Irms ej(v) e j(– i)

= Vrms v Irms –i

S = P + j Q

S = V I*

S = Vrms Irms ej(v – i)

S = Vrms ej(v) Irms ej(– i)


Complex Conjugate

Power Triangle

Total or Apparent PowerS

kVA

Real PowerPkW

Reactive PowerQ

kVAR

cos Power Factor

This is the only “useful” work done.

This component of the power furnishes the energy required to establish and maintain electric and magnetic fields.

Our electrical system must handle this component.

THEREFORE, VARS REDUCE THE “USEFULNESS” OF THE TOTAL ELECTRIC POWER.


Not quite...

Power Factor

cos is called “displacement power factor” because it is due to the displacement between the voltage and current phasors

complexplane

voltage

current


watts

vars

Q

S

P

Power Factor

Just like the current being out of phase from the voltage reduces the “usefulness” of the power, distortion of the current waveform has a similar effect.

This is because a harmonic current times a fundamental frequency voltage produces zero real power (watts).

Distortion of Current Waveform


Effects of HarmonicsHarmonic Power

Alex McEachern’sPower Quality Teaching Toy

http://www.powerstandards.com/PQTeachingToyIndex.php


Power Factor

watts

vars

distortion


complex plane

Total Effective Power

NOT this

Power Factor

Distortion power factor

rms,effective

rms,lfundamenta

distortionI

Ipf

1n

2

neffective II


Distortion Power Factor Example

A power quality analysis revealed the following current components:

%74.969674.0A399

A386pfdistortion

A39948121627425066386I 222222222effective

I1 = 386 AI3 = 66 AI5 = 50 AI7 = 42 AI9 = 27 AI11 = 16 A

I13 = 12 AI15 = 8 AI17 = 4 AI19 = 3 AI21 = 1 AI23 = 1 A

Find the distortion power factor.


Ignore harmonics if |Ix| < 1% of I1

Total Power Factor(Power Factor)

The total power factor, considering both displacement and distortion effects, is expressed as:


powerapparent

powerrealpf

ntdisplacemedistortion pfpf

cosI

I

eff

1

effLL

1LL

IV3

cosIV3

Harmonic MitigationTechniques


Neutral Conductor Sizing

Harmonic currents behave like sequence currents.


HarmonicOrder

Sequencing

35791113

zeronegativepositivezero

negativepositive

(see Alex McEachern’s Power Quality Teaching Toy)

zero‐sequenceharmonics are called“triplen harmonics”

Triplen harmonics do not cancel – they are additive on the neutral.

3, 9, 15, 21, ...

Neutral Conductor Sizing

Neutral conductors supplying non‐linear loads should not be downsized.

The neutral on such circuits shall be at least equal in size to the phase conductors.

In situations where triplen harmonic content is very high, one neutral conductor equal in size to the phase conductors should be provided for each phase.

Switching power supplies are notorious triplenharmonic producers.


k‐Factor Transformers

Special transformers can be designed to better tolerate the heating effects due to non‐linear loads.

These transformers do not reduce current distortion –they simply perform better in the presence of harmonics.

Multiple small winding conductors are paralleled to reduce skin effect.

Additional iron is used in the core to reduce core heating.

These transformers are called k‐factor transformers.



The distortion of the current flowing to a non‐linear load can be described using the k‐factor.

The k‐factor is calculated as


2eff

1n

2n

2

factorI

In

k


Calculate the k‐factor for the following distorted current waveform:


HarmonicOrder

Magnitude(Amperes)

1 385

3 120

5 85

7 40

9 15

11 12

13 8

15 6

17 4

19 3

Amperes415

346812154085120385I 2222222222eff

48.3

415

319417615813121115940785512033851k

2

22222222222222222222


Various load types have different typical k‐factors:

HID and fluorescent lighting systems k‐4

Classrooms, healthcare facilitiesProduction lines with VFDs k‐13

Office buildings, computer facilities k‐20and up


Twelve‐Pulse Rectifiers

Producing a six‐phase systemfrom a three‐phase system.


I I

B

A

B

I

H2

x b

X1

I

c

CI

A

C

H1

I

H3

z yI

X2

I

X3

b

c

aI

I y

xI

zIc

IcC

BI

B

c

b

I

X2 c

aA

A

I

C

I Ia

H1 I

I

I b

I c

aI

Ib InX3

X0

X1

b

n

a

H2

H3

,

,

,

A B C

Six-Phase

a

b

c

a’

b’

c’

a

b

c

Th

ree-

Ph

ase

Twelve‐Pulse Rectifiers Supplying rectifiers with six‐phase input substantially reduces harmonics.


Six‐phase input Twelve‐pulse

rectifier

Dominant harmonics:11th

13th

Harmonic Filters

Once a harmonic analysis is done to find the problematic harmonic(s), a harmonic filter (capacitor and reactor in series with each other, connected in shunt with the load) can be designed to mitigate the effects of the problematic harmonic(s).


Harmonic Filters

The reactor impedance at the fundamental frequency to achieve resonance at the nth harmonic frequency is


2C

23

2LL

Ln

X

n)MVAR(

kVX

A common practice is to tune the harmonic filter slightly below (3% to 10%) the harmonic of concern. Doing so reduces the duty on the filter, since the path to neutral will not be a short circuit at the harmonic frequency.

Harmonic Filters

Consider the following example, which converts an existing 600 kVAR capacitor bank to a fifth harmonic filter.

The harmonic analysis indicated that the fifth harmonic current comes from a 3000 kVA load with a 5% fifth harmonic component.


Harmonic Filters

The line‐to‐line voltage is 12.47 kV, so the impedance of the reactor at 60 hertz needed for the harmonic filter is


4.105)600.0(

47.12X

2

2

L

The rated current of the capacitor at the fundamental frequency is

amperes8.2747.123

600IC

5th harmonic

Harmonic Filters

The impedance of the capacitor at the fundamental frequency must then be


2598.27

3/470,12XC

Since the algebraic signs of XL and XC are opposite, their impedances in series are subtractive. Therefore, the shunt impedance at 60 hertz of the harmonic filter is

6.2484.10259XXX LCf

Harmonic Filters

The current through the harmonic filter at the fundamental frequency is


amperes8.296.248

3/470,12I )1(f

Note that this current is slightly higher than the current drawn by the shunt capacitor alone.

Harmonic Filters

The fifth harmonic current to which the harmonic filter will be subjected is


amperes9.647.123

300005.0I )5(f

Therefore, the effective current flowing through the harmonic filter is

amperes8.29III 2)5(f

2)1(feff

Harmonic Filters

The next step is to verify that the voltage across the capacitor does not exceed the capacitor's maximum voltage rating. The fundamental frequency voltage across the capacitor is


volts7511)259()29(V )1(C

This means that the fundamental frequency voltage across the reactor is

volts31175117200V )1(R

Harmonic Filters

The fifth harmonic voltage across the capacitor is


volts3575

259)9.6(V )5(C

Assuming that VC(1) and VC(5) are in phase, which produces a worst case scenario,

volts75193577511V 22)total(C

This is less than 110% of the nominal voltage, so is within the capacitor rating.

Harmonic Filters

The last step is to verify that the total kVARproduction of the capacitor is less than 135% of nominal, which is an ANSI‐defined limit.


/kVAR1.224IVkVAR efftotal1

The three‐phase kVAR production is

kVAR2.672kVAR3kVAR 13

This value exceeds the nominal kVA rating by only 12%, so is within the acceptable range. Therefore, simply placing a 10.4 (at 60 hertz) reactor in series with the existing 600 kVAR shunt capacitor creates an acceptable fifth harmonic filter.

Thank you!

Power System Harmonics

Tuesday, June 11, 20138:30AM – 12:30PM

Florida Electric Cooperatives AssociationSand Pearl ResortClearwater, Florida

Ralph Fehr, Ph.D., P.E.Senior Member, IEEE

[email protected]

power system harmonicsfeca.com/harmonics.pdfpower system harmonics tuesday, june 11, 2013 8:30am –...

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