Distribution system and power quality

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8_ECE5750_PQ_Harmonics30Apr2020.pdf

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ECE 5750 Distribution System & Power Quality

Part 8: Basic principles, analysis & control of power system harmonics

Instructor: Dr. Ha Le Department of Electrical and Computer Engineering

California State Polytechnic University, Pomona

What will be presented? 1. Linear and nonlinear loads

2. Fourier series of distorted waveforms

3. Sources of harmonics

4. Effects of harmonic distortion on power systems

5. Harmonic voltage and distortion limits

6. Principles for controlling harmonics

Reading: PQ textbook, Chapter 6, 7

Fundamentals of Harmonics

What is harmonics?  Fundamental power system (PS) frequency = f1 = 60 Hz  Harmonic frequency = n f1 where n is integer  Harmonic distortion: typically caused by nonlinear

devices

V(t)

I(t)

Nonlinear Resistor

Linear loads  Linear loads: Those

drawing currents in linear proportion to the applied voltage.

 For a resistive load, the impedance is constant over time as the voltage and current waveforms are sinusoidal but they are also perfectly in phase.

 For an inductive load, the current lags the voltage waveform by 90°. It is linearly proportional to the applied voltage.

Non-linear loads  Nonlinear loads: They

draw currents in nonlinear proportion to the applied voltage.

 Current waveshapes are NOT sinusoidal, i.e. contain fundamental + harmonic frequencies

 Current waveforms are periodic. Hence, they can be decomposed into a weighted sum of sinusoids whose frequencies is an integer multiple of fundamental frequency of the periodic waveform.

Current is discontinuous and periodic

Idealized six-pulse rectifier

Adjustable speed drive v(t)

diode rectifier

transistor inverter

 Total harmonic distortion (THD) current = 80%

 Peak current, approx. 300A  Double hump due to rapid

capacitor charging

100 HP, 6 pulse AC drive

Source: www.schneider-electric.us

ASD with six-pulse PWM VSI

Fourier series Any periodic waveforms can be expressed as a weighted sum of sinusoids, known as the Fourier series.

ꞏ

ꞏꞏ

60 Hz (h = 1)

300 Hz (h = 5)

420 Hz (h = 7)

540 Hz (h = 9)

660 Hz (h = 11)

780 Hz (h = 13)

180 Hz (h = 3)

)sincos()( 1 1

10 tnbtnaati n n

n  









tdtnti T

dtti T

0 1

sin)( 2

cos)( 2

)( 1



i(t) = i(t + T)

Example: Fourier series (1) Example 6.1:

Example: Fourier series (2) Solution for Example 6.1: From Fig. 6.4, it is clear that the current waveform has odd symmetry. Therefore, a0=0 and an=0. Coefficients bn are:

,...7,5,3,1

6 11

cos 6

7 cos

6 5

cos 6

cos

sin 2

sin)( 2

6/11

6/7 01

6/5

6/ 0



 

 

 

 











nwhere

nn n Inn

n I

tdtnI T

tdtnti T

b T

 







The calculation of bn can be simplified since the waveform has both odd and half-wave symmetry. Therefore,

 

 







6 5

cos 6

cos 2

sin 4

sin)( 4

6/5

6/ 0

 





nn n I

tdtnI T

tdtnti T

b T

Example: Fourier series (3) Solution for Example 6.1 (Cont.):

The Fourier series of the current is as follows. Harmonic currents of the rectangular current are simply the individual sinusoids of the Fourier series.



 











  



tnbti n

1 1

25sin 25 1

23sin 23 1

19sin 19 1

17sin 17 1

13sin 13 1

11sin 11 1

7sin 7 1

5sin 5 1

sin 32

sin)(



 



Example: Fourier series (4)  The rectangular current of Fig.

6.2 is decomposed into a series of sinusoids.

 Harmonic content can be quantified.

Harmonic spectrum of a charger

Harmonic Spectrum for EVSE DC Fast Charger is based on lab testing carried out by EPRI.

Parameter Value

Rated Voltage (V) 208

Rated Power (kW) 50

Rated PF 1.0

Specifications

Harmonic Spectrum

Courtesy of A. Maitra, DC Fast Chargers for PEVs – Real and Simulated Results, EPRI PQ and Smart Distribution Conference and Exhibition, San Antonio, Texas, June 5, 2012

Harmonic Magnitude (% Fund)

3 5.93 5 2.13 7 1.52 9 0.83 11 0.45 13 1.07 15 0.37 17 0.40 19 0.57

Harmonics: Causes & effects on PS (1)  Load current becomes distorted

because power electronic devices inside the nonlinear load are switched on and off to achieve desired functions.

 Non-sinusoidal currents interact with PS impedance giving rise to voltage distortion

 The harmonic currents passing through the impedance of the system cause a voltage drop Vn for each harmonic, given by

)(sin 1 ,...7,5,3

n n

nsn tnIZV    



 

  

  





)(sin)sin( 1 ,...7,5,3

111 n n

nsssload tnIZtZIVV 

Fig. 6.6. Harmonic current following through the system impedance results in harmonic voltage at the load

Harmonics: Causes & effects on PS (2) 1) The amount of voltage distortion depends on the impedance and

magnitude of harmonic currents

2) If the impedance Zs at the load bus is very large, the voltage distortion can be significant even though the harmonic currents may be small

3) The worst case is experienced when large harmonic currents are injected into a weak bus. While the load current ultimately causes the voltage distortion, the load has no control over the voltage distortion.

According to IEEE Std. 519-1992

 The control over the amount of harmonic current injected into the system takes place at the end-use application.

 Assuming that the harmonic current injection is within reasonable limits, the control over the voltage distortion is exercised by entity having control over the system impedance, which often is the utility.

Impact of harmonics on RMS current (1) Under the presence of harmonics, the RMS value of i(t) is computed as follows:

)(sin)sin()( 1 ,...7,5,3

11 tnItIti n

n   





Let (6.15)

Impact of harmonics on RMS current (2)  The individual terms is given as,

 The fundamental and harmonic terms can be separated and written as,

where Irms and IrmsH are the RMS of fundamental and harmonic frequency of i(t)

Impact of harmonics on RMS voltage (1)  For a distorted voltage waveform, its instantaneous voltage is

represented as follows:

 The RMS value of the distorted voltage waveform:

where Vrms1 and VrmsH are the RMS of the fundamental and harmonic frequency of v(t).

)(sin)sin()( 1 ,...7,5,3

11 tnVtVtv n

n   





222

531 ...

Hrmsrms

rmsrmsrmsrms

VVVV





Example: RMS value with harmonics Example 6.2: Calculate the RMS value for the current waveform in Fig. 6.2.

Solution: Based on Fourier series of the current i(t) obtained in Example 6.1

22222220

8165.0 3 2

19 1

17 1

13 1

11 1

7 1

5 1

1 1

2 132

IIrms



 

  

  



Power in sinusoidal conditions (1)  Apparent power, S, is given as

 The active power is the rate at which energy is expended, dissipated or consumed by the load and is measured in watts (W). ‘P’ is computed by averaging the product of instantaneous voltage and current.

rmsrms IVS 

)cos(

)cos()cos( 2

)()( 1

1111 11









 

IV IV

dttitv T

rmsrms

)2sin()]2cos(1[)()()( tQtPtitvtp  

 Active power can also be computed from the instantaneous power:

(6.23)

(6.24)

Power in sinusoidal conditions (2)  In a sinusoidal case, the reactive power is simply defined as,

 The relationship between P,Q and S can be summarized in this diagram:

)sin()sin( 2

)sin(

1111 11

11 







rmsrms IV IV

Power in non-sinusoidal conditions (1)

 The instantaneous voltage and current under non-sinusoidal condition is give as,

 Apparent power (S) is given by

)(sin)sin()( 1 ,...7,5,3

11 tnVtVtv n

n   





)(sin)sin()( 1 ,...7,5,3

11 tnItIti n

n   





]][[ 2222

222

11 HH rmsrmsrmsrms

rmsrms

IIVV

IVS





Power in non-sinusoidal conditions (2) Carrying out multiplication for each term, we obtain

Power in non-sinusoidal conditions (3)  Non-fundamental apparent power SN consists of two components, the distortion

power SDP and the harmonic apparent power SHA.  Current distortion power SCDP is defined as the product of the fundamental RMS

voltage and the RMS of the harmonic currents.

 Voltage distortion power SVDP is given as the product of the fundamental RMS current and the RMS of the harmonic voltages.

 Distortion power SDP is

 



 ,...7,5,3

2 11

n rmsrmsrmsrmsCDP nH IVIVS

11 ,...7,5,3

2 rms

n rmsrmsrmsVDP IVIVS nH  





22 VDPCDP SSSDP 

Power in non-sinusoidal conditions (4)  Harmonic Apparent Power SH is the product of the RMS of voltage and

current harmonics. The term can be further resolved into total harmonic active power and total harmonic nonactive power. The harmonic apparent power is given by

Note: There is no such thing as reactive harmonic power. Under non-sinusoidal condition, the part that is in quadrature to the total harmonic power is called as total harmonic nonactive power.

)sin(

)cos(

,...7,5,3

22222

nnrms n

rmsH

nnrms n

rmsH

HHrmsrmsH

IVN

IVP

NPIVS









 







Power factor: Displacement & Total  In a sinusoid condition, power factor is simply given as,

 The power factor in sinusoidal condition is also known as displacement power factor and is written as,

 The power factor that takes into account contribution from all active power both fundamental and harmonic frequencies is known as true or total power factor. It is simply the ratio of all the total active power for all frequencies to the apparent power delivered by the utility:

S P

PF 

)cos( 11 1

1   S P

S PP

PF H 

 1

Practice: I, V, P, THD under harmonics

Fig. 6.21

Solution: I, V, P, THD under harmonics (1)

The Fourier series coefficients are:

Solution: I, V, P, THD under harmonics (2)

For n = 3, 5, 7 … an = 0;

For n = 2, 4, 6 …

For n = 1

Solution: I, V, P, THD under harmonics (3)

For n = 1

Solution: I, V, P, THD under harmonics (4) Hence, the current expression is

Solution: I, V, P, THD under harmonics (5)

Solution: I, V, P, THD under harmonics (6)

Solution: I, V, P, THD under harmonics (7)

Note: Irms1 is the magnitude of fundamental component of load current; Irms2 is the magnitude of the second harmonic component etc.

Solution: I, V, P, THD under harmonics (8)

Part (e):

 By inspection of the voltage waveform (top figure), the load voltage waveform Er is not distorted because there is no system impedance.

 V(t) is an ideal source with infinite short-circuit capacity

Harmonic phase sequences (1)  In determining harmonic phase sequences, the 3-phase power system is

assumed to be balanced. Let phase ‘A’ current, ia(t) be represented as,

 Phase ‘B’ current lags phase ‘A’ current by 120°. The Fourier series of phase ‘B’ current is,

 Phase ‘C’ current leads phase ‘A’ current by 120°. The Fourier series of phase ‘C’ current is

 



 ,...3,2,1

1 )sin()( n

nna tnIti 



 











,...3,2,1 1

,...3,2,1 1 1

) 3

2 sin(

)) 3 2

(sin[)(

n nn

n nnb

ntnI

tnIti

 

  



Harmonic phase sequences (2)

 The first three components of ia(t), ib(t) and ic(t) are:

Harmonic phase sequences (3)

Triplen harmonics  Triplens become an important

issue for grounded-wye system with current flowing in the neutral.

 Two typical problems which arise are overloading the neutral and telephone interference.

 For the system with perfectly balanced 1-phase loads, an assumption is made that fundamental and third harmonic components are present.

 Summing currents at node ‘N’, fundamental current components in the neutral are zero but the 3rd harmonic currents are three times the phase currents because they naturally coincide in phase and time.

Overloading of the neutral Condition: Perfectly balanced single-phase loads in grounded-wye system.

balanced fundamental currents sum to 0, but balanced third harmonic currents coincide

neutral current contains no fundamental, but third harmonic is

300% of phase current

 Neutral has no fundamental

 Neutral has harmonic currents where its values are:

In = 3 I3 + 3 I9 + 3 I15 + …

Triplen harmonics Consider the following balanced currents, ia(t), ib(t) and ic(t).

The neutral current is simply the sum of the line currents, thus,

Impact of triplen harmonics on TRF 1. Transformers with neutral connections: Overheating when serving

single-phase loads on the wye-side.

2. Measuring the current on the delta side of a transformer will NOT show the triplens.

3. The flow of triplen currents can be interrupted by the appropriate isolation transformer connection.

Distribution side

Transmission side

Harmonic indices Two indices: Total Harmonic Distortion (THD) and Total Demand Distortion (TDD)

 THD: A measure of the effective value of the harmonic components Mh of a distorted waveform.

where M1 is fundamental component (50Hz or 60Hz).

 THD is related to the RMS value of the waveform as follows:

TDD: Ratio between RMS value of harmonic components and maximum load (demand) current.

h1 is fundamental and hmax is highest harmonic order to be considered.

Example: Calc. of harmonic indices Example 6.5:

Solution:

Harmonic sources (1) Harmonic sources from commercial loads:

1) Switch-mode power supplies: A very high third harmonic content in the current

2) Fluorescent lighting: Discharge lamps requiring a ballast to provide high initial voltage to initiate discharge between the two electrodes. The current distortion is at a moderate 15%.

3) Three phase power converters: They do not generate third harmonic currents, but they can still be significant sources of harmonics at their characteristic frequencies.

4) Three phase power converters – DC drives: The two largest harmonic currents for the 6-pulse drive are the 5-th and the 7-th.

Harmonic sources (2)

Harmonic sources from industrial loads:

1) Arcing devices: Arc furnaces, arc welders etc. Electric arc is best represented as a source of voltage harmonics.

2) Saturable devices:

 Transformers: Harmonic voltage distortion from TRF over-excitation is generally only apparent under light load conditions.

 Motors: Exhibit some distortion in the current when overexcited, but it has little consequence.

Effect of harmonics on capacitors (1)  A capacitor bank experiences high

voltage distortion during resonance

 The capacitor current is also significantly large and rich in a monotonic harmonic.

 The harmonic current shows up distinctly, resulting in a waveform that is essentially riding on top of the fundamental frequency (e.g. 11th order)

IEEE Std. 18-2002 specifies continuous capacitor ratings as follows:

1) 135% of nameplate kVAR

2) 110% of rated RMS voltage

3) 135% of rated RMS current based on rated kVAR and rated voltage

4) 120% of peak voltage (including harmonics but excluding transients)

 Under presence of harmonics, care must be taken to ensure that capacitors operate within the rating limits.

Effect of harmonics on capacitors (2)

Effect of harmonics on TRF: Losses (1) There are three effects that result in increased transformer heating when the load current includes harmonic components:

1) I2R losses 2) Winding eddy-current losses, PEC 3) Other stray losses, POSL

 The transformer total load loss in watts is,

 When the transformer is supplying a sinusoidal current at power frequency, the total load loss is,

 The total load loss in per unit of PI2R-R is as follows,

][

WPP

WPPPP

ECRI

OSLECRILL





][2 WPPP RECRRIRLL  

][1 puPP RECRLL  

Effect of harmonics on TRF: Losses (2)  When the transformer is supplying a non-sinusoidal current which

consists of the fundamental I1 and harmonic components I3, I5…, I2R losses is

 The winding eddy-current losses is given by,

 Therefore, the transformer total load loss when supplying non-sinusoidal currents is

][ 2

22 W I I

PP R

h RRIRI  



  

 



][2 2

Wh I I

PP R

h RECEC  



  

  

][2 22

Wh I I

P I I

PPP

h REC

h RRI

ECRILL

  

  

 



  

 





Effect of harmonics on TRF: Losses (3)  With a per-unit basis of transformer rated I2R loss and rated full load

current IR, total load loss PLL becomes

][2 22

Wh I I

P I I

PPP

h REC

h RRI

ECRILL

  

  

 



  

 





    



  

 



 





][ ][

][1][

][][][][

22 2

222

puI puIh

puPpuI

puIhpuPpuIpuP

h RECh

hREChLL ][1 puPP RECRLL  

TRF supply Non-sinusoidal current TRF supply sinusoidal current

Derating TRF for non-linear loads (1)

 K-factor has been used by transformer manufacturers to design transformers for nonlinear load applications.

Derating TRF for non-linear loads (2) Example 6.8

Solution:

K-9 transformer would be an appropriate transformer for the given nonlinear loads. Although K-4 transformer may be acceptable at present, when new nonlinear loads are added, K-4 transformer may no longer be sufficient.

Analysis & Control of Harmonics

Root causes: Current & Impedance

Harmonics control: IEEE 519 approach End users:

 Limit the level of harmonic current injection at the point of common coupling (PCC).

 Current injections: Users can control them (think: nonlinear loads).

Utility:  Limit the level of harmonic voltage distortions.  Voltage distortions  utility can control this (think: impedance)

Recall: For the same amount of harmonic current injections

 Strong system (low impedance system)  voltage distortions will be low

 Weak system (high impedance system)  voltage distortions will be high.

Concept of common coupling point (1)

Utility System

Customer Under Study

Other Utility Customers

PCC

PCC at the transformer primary where multiple customers are served

Concept of common coupling point (2)

Utility System

Customer Under Study

Other Utility Customers

PCC

PCC at the transformer secondary where multiple customers are served

IEEE 519 voltage distortion limit Harmonic voltage distortion limits

in % of nominal fundamental frequency voltage. (from IEEE Std. 519-1992, Table 11.1)

Short-circuit ratio 1. Determine the three-phase short-circuit duty, Isc, at the PCC.

2. Find the load average kilowatt demand Pd over the most recent 12 months. This can be found from billing information.

3. Convert the average kilowatt demand to the average demand current in amperes:

4. Short-circuit ratio is

 A 3

1000 kV MVA

ISC 



kVPF kWIL

3 

I I

SCR 

Current distortion limits (1)

Current distortion limits (2)

Current distortion limits (3)

Harmonic current flow

NORMAL PATH ALTERED PATH

System impedance under harmonics  The system Thevenin equivalent impedance is called Short-circuit

capacity (SCC, measured in MVA), which is calculate as

SCSC

SCSCSC

I kV

MVA kV

jXRZ





1,, shs hXX 

ZSC is assume to be purely reactive

 Xs, h is inductive reactance at hth harmonic

 Xs, 1 is inductive reactance at fundamental frequency

 XC, h is capacitive reactance at hth harmonic

 Xc is capacitive reactance at fundamental frequency

h X

X chc ,

(7.1)

(7.8)

SCC at substation

tSC XX 

(%) 2

t t

LL t ZMVA

kV X 



   

 

Equivalent system impedance is often dominated by the service TRF impedance:

Example: Equiv. impedance under harmonics (1)

Example: Equiv. impedance under harmonics (2)

Example: Capacitor impedance under harmonics

h X

X Q kV

X fC

X chcrated cap

LL cc  ,

2 1 

Parallel resonance (1) Parallel circuit from the nonlinear load perspective

p c

pp s

p c

p s

p ss

p c

p ss

p c

p ssp

XQXQ

R X

R jXRjX

XXjR jXRjX

jXjXRZ





 

 





22 )()(

)(

)( )(

)(

Simplified distribution system

Subscript “p” means “parallel” resonance. QP is quality factor that determines the sharpness of the frequency response.

Parallel resonance (2)  When the system resonant frequencies corresponds to one of

the harmonic frequencies being produced by the nonlinear load (characteristic frequencies), harmonic resonance can occur.

CL f

L R

CL f

s p

1 2 1







Rs = Resistance of equiv. sys. impedance (not shown)

Ls = Inductance of equiv. sys. impedance

C = Capacitance of capacitor bank

(7.9)

Parallel resonance (3)  Apparent impedance Zp is maximum; Denominator is minimum, limited to Rs.

p p s

pp cap IXQV 

p c

p sp

R X

Q 

p c

pp s

p c

p s

p c

p ssp

XQXQ

R X

jXjXRZ







22 )()(

)(

p ss XR 

During parallel resonance, a small harmonic current can cause a large voltage drop across Zp. Hence, the voltage near the capacitor bank will be heavily distorted:

(7.11)

Parallel resonance: Current behavior The currents flowing in the capacitor bank and in the power system are:

 The currents are magnified Qp times.

 This may lead to capacitor failure, fuse blowing, transformer overheating.

 The extent of voltage and current magnification depends on the shunt capacitor size.

p p

p s

p p s

p s

p p

p c

p p c

p c

pp cap

IQ X IXQ

X V

IQ X IXQ

X V



Impact of capacitor size on parallel resonance

Estimating resonant frequency  Parallel resonance frequency is computed as follows:

rated cap

c p Q

MVA X X

h 3 (7.13)

Example: Estimating resonant frequency Example 7.4:

Effect of resistive load on parallel resonance

0% Resistive Load

10%

20%

R LC

33 .4

2 Ic

B R

KEs

Sys tem voltage = 69 kV Sys tem s hort-circuit capacity is 1500 MVA

Series res is tive dam ping = 1 ohm

Cap bank is rated 60 Mvar at 69 kV

Z(f)

0.0 - 2000 [Hz]

Frequency - Impedance scan

0 200 400 600 800 1000 1200 1400 1600 1800 2000 0

100

150

200

250

300

frequency

|Z |

Frequency Scan

With Capacitor Without Capacitor

Series resonance During resonance, the PF capacitor forms a series circuit with the TRF and harmonic sources. Resonance frequency is

The voltage at the PF capacitor is magnified and highly distorted:

(7.14)

Vh and Vs,pf is harmonic voltage corresponding to harmonic current Ih and voltage at the PF capacitor, respectively. Reactances are shown in Fig. 7.7

svct

c s X

X h



Series and Parallel resonance

In many systems with potential series resonance, parallel resonance also possible due to circuit topology (Fig. 7.7).

Practice: Capacitor under harmonics & resonance (1)

Practice: Capacitor under harmonics & resonance (2)

Practice: Capacitor under harmonics & resonance (3)

System impedance seen by non-linear load for 0-1500 Hz when the capacitor is ON and OFF line Peak Z =252.4  at 300 Hz

Solution: Capacitor under harmonics… (1)

Solution: Capacitor under harmonics… (2)

Solution: Capacitor under harmonics… (3)

Solution: Capacitor under harmonics… (4)

Solution: Capacitor under harmonics… (5)

Solution: Capacitor under harmonics… (6)

Solution: Capacitor under harmonics… (7)

Solution: Capacitor under harmonics… (8)

Three common causes of harmonics

1. The source of harmonic currents is too great:  Large HARMONIC CURRENTS.

2. The path in which the currents flow is too long (electrically), resulting in either high voltage distortion or telephone interference  HIGH IMPEDANCE (mostly X)  WEAK SYSTEM.

3. The response of the system magnifies one or more harmonics to a greater degree than can be tolerated  HARMONIC RESONANCE

Principles of controlling harmonics (1) Recall these key concepts in harmonic analysis:

Nonlinear loads: Producers of harmonic currents. End-users own these loads.

Utility systems: Utility system impedance may interact adversely with harmonic currents resulting in voltage distortions.

 Strong system: Voltage distortions are likely low and benign.

 Weak system (large reactance): Voltage distortions are likely higher.

 Harmonic resonance: One of system impedance resonant frequencies matches one of harmonic frequencies produced by nonlinear loads (large reactance, small R).

Principles of controlling harmonics (2) 1) Large HARMONIC CURRENTS: Make it smaller, i.e. reducing harmonic

currents.  PWM drives that charge the DC bus capacitor directly from the line  Add inductors, i.e., line reactors.

 Change transformer configurations: Use a pair of delta-delta, and delta- wye.

2) Large HARMONIC CURRENTS: Add filters to siphon currents or block them from entering the system.  Shunt filters  Short-circuiting harmonic currents as close to the source

of distortion as practical.

3) HARMONIC RESONANCE: Modify frequency response by filters, inductors, capacitors.  Change the capacitor size, move a capacitor to a point of the system

with a different short-circuit impedance.  Remove the capacitor and live with it.

Harmonics control: Utility feeders 1) Capacitor bank placed on utility feeders usually done without

harmonic evaluations.

2) When harmonic resonance occurs, change capacitor bank size or move to other locations.

3) Triplen harmonics on wye-ground capacitor banks, change the neutral connection  make it floats or better add a reactor in the neutral to convert the bank into a tuned resonant shunt for a zero-sequence harmonic.

4) Widespread harmonic sources along the feeder: Distribute a few filters toward the ends of the feeder.

Harmonic filter installation on an overhead distribution-feeder

Oil-insulated iron-core reactorsCapacitor banks and switches

Harmonic filters on a substation

Harmonics control: End-user facilities Harmonic problems at end-user facilities

 Is it due to resonance with power factor capacitors in the facility?

 YES  Change capacitor size.

 NO  Use filters.

100

Harmonics control: Load side

Most common method:

 Use a pair of transformers configured as delta- delta and delta-wye, or wye-wye and wye-delta.

 Goal: To reduce characteristic harmonics from six-pulse converters (5, 7, 11,13, 17,19, etc.).

101

Reducing load harmonic current using TRF

 A simple delta-connected TRF can block the flow of triplen harmonics (3rd order).

 By using TRF to shift the phase of half of the 6-pulse power converters in a load by 30 deg. can significantly reduce 5th and 7th harmonics.

 Zig-zag and grounding TRFs can eliminate triplen harmonics off the line.

102

In-line reactors or chokes

A typical 3% input choke can reduce the harmonic current distortion for a PWM-based drives from approximately 80% to 40%.

Line input choke (% on Drive Base)

In pu

t C

ur re

nt D

is to

rt io

n (%

)

0.0%

10.0%

20.0%

30.0%

40.0%

50.0%

60.0%

70.0%

80.0%

0% 1% 2% 3% 4% 5%

Choke (0 to 5) % Z

(on ASD kVA)

 The inductance slows the rate at which the capacitor on the DC bus can be charged and forces the drive to draw current over a longer time period.

 The net effect is a lower-magnitude current with much less harmonic content while still delivering the same energy.

103

Harmonic control: Filters

Passive filters

 These are inductance, capacitance, and resistance elements.

 Tuned to control harmonics.

 Advantage: Economical and straightforward.

 Disadvantage: May interact adversely with the power system.

104

Harmonic filters: Shunt filters

 Known as single-tuned notch filters.  Provide a low impedance path to a particular

harmonic current.  Connected in shunt.

SINGLE-TUNED FIRST ORDER HIGH-PASS

2ND ORDER HIGH-PASS

3RD ORDER HIGH-PASS

105

Response of a 5th harmonic filter

Harmonic Number h

1 3 5 7 9 11 13 15 17

(a) Typical low voltage filter configuation.

(b) Equivalent circuit of system with filter.

X C

XF X C 3

SC Harmonic Source

frequency response (Z = 1.0).

hZ 1

capacitor only

capacitor converted

to filter

hNotch

C notch X

X h

3 

106

Case study: Shunt harmonic filter design A single-tuned notch filter will be designed for an industrial facility and applied at a 480-volt bus. The load where the filter will be installed is approximately 1600 kVA with a relatively poor displacement PF of 0.75 lagging. The total harmonic current produced by this load is dominated by fifth harmonic (20% of fundamental). The facility is supplied by a 2000 kVA transformer with 5.0% of impedance. The fifth harmonic background voltage distortion on the utility side of the transformer is 2.0% of the fundamental when there is no load.

1. A capacitor bank to be installed at 480V bus to improve PF to 98%

2. Single-tuned notch filter to be designed based on capacitor bank size

HARMONIC SOURCE

EQUIVALENT CIRCUIT

h=2, 3, 4, ...C

L I h

with no reactor

107

Step 1: Select a tuned frequency for the filter (1)

The tuned frequency is selected based on the harmonic characteristics of the loads involved.

 Start at the lowest harmonic frequency generated by the load.

 Tune the filter slightly below the harmonic frequency of concern. If the filter is tuned exactly to the harmonic, impedances or other changes would shift the parallel frequency higher into the harmonic being filtered.

 The above situation is worse than without the filter as resonance is very sharp, which require higher duty from filter components.

108

Step 1: Select a tuned frequency for the filter (2)

109

Step 2: Compute capacitor bank size and the resonant frequency (1)

 The filter size is based on the load reactive power requirement for power factor correction.

 Assume that no capacitor is installed.

 The desired power factor is 0.98.

 Thus, the net reactive power from the filter required to correct from 75% to 98% power factor can be computed as follows:

Real power of the load at 0.75 PF lagging is

P = 1600 x 0.75 = 1200 kW

Required reactive compensation Qfilt from the filter:

Qfilt = 1200 * [ tan(acos(0.75)) – tan(acos(0.98)) ]

= 814.28 kVAR

110

Step 2: Compute capacitor bank size and the resonant frequency (2)

 For a nominal 480V system, the net wye-equivalent filter reactance (capacitive), XFilt, is determined by:

 Note that XFilt above is for Y- connected capacitor bank. Also, XFilt is

 For tuning at the 4.7th harmonic,

 Thus, the desired capacitive reactance can be determined by

 At this point, it is not known whether the filter capacitor can be rated the same as the system, 480V, or would have to be rated one step higher at 600V.

 Try a capacitor rated at 480 V

 Standard capacitor 750 kVAR (note that this value is slightly lower than reactive power that Xcap calculated above produces)

LCapFilt XXX 

 

 2828.0 28.814

48.01000 22

filt

LL Filt

Q kV

 

 

 2962.0 17.4

)7.4(2828.0 1 2

h hX

X FiltCap

kVAR X

kV kVAR

Cap

LL 75.777 2962.0

)1000(48.0)1000( 22 

XCap = 0.3072 

XCap = h2XL = 4.72 XL

111

Step 3: Compute filter reactor size

 The filter reactor size can now be selected to tune the capacitor to the desired frequency.

 From Step 1, the desired frequency is at 4.7th harmonic or 282 Hz.

 The filter reactor size is computed from the equivalent wye capacitive reactance above as follows:

 Alternatively, the reactor size can be computed by solving L for in the following equation

 0.0139 7.4

0.3072 22

)( )( h X

X wyeCapfundL

mH 0.0369 602

)(  

 

fundLXL

(wye) h

L f

C2 1

 

Hz 282607.4 hf

112

Step 4: Evaluate filter duty requirements

 Evaluation of filter duty requirements typically involves capacitor bank duties. These duties include

 peak voltage, current, kVAR produced, and RMS voltage.

 IEEE Std 18-2002 is used as the limiting standard to evaluate these duties.

 Computation of the duties are fairly lengthy, therefore, they are divided into three steps, i.e.,

 computation for fundamental duties,

 harmonic duties, and

 RMS current and peak voltage duties.

113

Step 5: Compute fundamental duty requirements

Determine a fundamental frequency operating voltage across the capacitor bank.

1) The apparent reactance of the combined capacitor and reactor at the fundamental frequency is:

2) The fundamental frequency filter current is:

3) The fundamental frequency operating voltage across the capacitor bank is

4) The actual reactive power produced

Evaluation:

 Filter draws more current than capacitor alone, therefore actual reactive power produced by capacitor is greater than its 750 kVAR rating.





2933.03072.00139.0

)(wyeCapL filt fund XXX

A 86.449 0.2933

3 480

3 , 

fund

kV filt

fundduty X I

actual

V 76.5023 )(,,  wyeCap filt

fundduty cap

fundduty XIV

kVAR 822.8 3 ,,  cap

fundduty filt

fundduty cap duty,fund VIQ

114

Step 6: Compute harmonic duty requirements (1)

1) Since the nonlinear load produces 20% harmonic of the fundamental current at the fifth harmonic, the harmonic current in amperes produced by the load would be:

2) Harmonic current contributed to the filter from the source side is estimated as follows:

 It is assumed that the 2% fifth harmonic voltage distortion present on the utility system will be limited only by the impedances of the service TRF and the filter;

 The utility impedance is neglected

Fundamental frequency impedance of the service transformer:

The fifth harmonic impedance of the service transformer:

A 9.483 48.03

1600 20.0

3 .).()(

 



 

actual hampsh

kV kVA

upII

 0.0058 0.2

48.0 05.0(%)

)( Xfmr

actual TfundT MVA

kV ZX

 0.02880.00585)(5, fundTT hXX

115

Step 6: Compute harmonic duty requirements (2)

 The harmonic impedance of the capacitor bank is:

 Harmonic impedance of the reactor is:

 Given that the voltage distortion on the utility system is 0.02 pu, the estimated amount of fifth harmonic current contributed to the filter from the source side would be:

3) The maximum harmonic current is the sum of the harmonic current produced by the load and the harmonic current contributed from the utility side:

4) The harmonic voltage across the capacitor can be computed as follows:

 0614.0 5

3072.0)( 5),( h

X X wyeCapwyeCap

 06951.00139.05)(5, fundLL hXX

 5,5),(5, )(

)( 3

)(

LwyeCapT

actualutilityh utilityh

XXX kVpuV

I 

 

  A 23.150

06951.00614.00288.03 48002.0

 

 

A 13.535.23501 384.9)( totalhI

h X

IV wyeCaptotalh cap duty

)( )(5, 3

V 95.56 5

0.3072 13.3553 

116

Step 7: Evaluate total RMS current and peak voltage requirements

1) Total RMS current passing through the filter:

This is the total RMS current rating required for the filter reactor.

2) Assuming the harmonic and fundamental components add together, the maximum peak voltage across the capacitor is:

3) The RMS voltage across the capacitor is:

4) The total kVAR seen by the capacitor:

A 89.108513.53586.944 22

2 )(

2 ,



 utilityhfundtotalRMS III

cap duty

cap fundduty

cap peakLLduty VVV 5,,, 22 

peakV 55.791295.562502.76 

2 5,

2 ,, )()(

cap duty

cap fundduty

cap rmsLLduty VVV 

V 505.9795.56502.76 22 

cap rmsLLduty

filt totalduty

cap totalduty VIQ ,,, 3 

kVAR 65.51997.50589.10853 

117

Step 8: Evaluate capacitor rating limits

Table for evaluating filter duty limit

 This would be a very marginal application because the capacitor duties are essentially at the maximum limits.

Use a capacitor bank rated at 600 V

 Required kVAR =

 Choose 1200 kVAR

 Repeat all the above.

kVAR1172 480 600

750 2



118

Step 9: Evaluate filter frequency response

 The harmonic at which the parallel resonance below the notch frequency will occur is computed as follows:

 This assumes the service transformer reactance dominates the source impedance. Including the utility system impedance will lower the frequency.

 Hence, the parallel resonance frequency (3.95) is safely away from the harmonic (5th).

 Beware of transformer energization events.

95.3 0139.00058.0

3072.0

)()(

)(' 0 

 

 fundLfundT

wyeCap

XX X

119

References

1. S. Santoso, Fundamentals of Electric Power Quality, 2012.

2. R. C. Dugan, M. F. McGranaghan, S. Santoso, W. Beaty, Electrical Power Systems Quality, McGraw Hill 2012.

3. T. A. Short, Electric Power Distribution Handbook, 2003.

4. J. D. Glover, M. S. Sarma, T. J Overbye, Power System Analysis and Design, 5th Ed., CENGAGE Learning, 2012.

5. T. Gonen, Electric Power Distribution Engineering, 3rd ed., 2014, CRC Press, ISBN 9781482207002.

6. W. H. Kersting, Distribution System Modeling and Analysis, 3rd ed., CRC Press, 2012.

7. IEEE Std. 519-1992.

8. Other sources