chapter 7 harmonics part 2

1

Principles of Establishing Harmonic Voltage and Distortion Limits

System Response Characteristics

Principles for Controlling Harmonics

Power system quantities for harmonic analysis

Passive Filters

Chapter 7 –Analysis of Power Systems Harmonics

References: 1. S. Santoso, Fundamentals of Electric Power Quality, Spring 2009 Ed.2. R. C Dugan, M. F McGranaghan, S. Santoso, W. Beaty,

Electrical Power Systems Quality, McGraw Hill 2002. 3. Bin Wu, High-Power Converters and AC Drives, IEEE Press 2006.

EE 362Q/EE394.9 – Fall 2012Power Quality & HarmonicsS.Santoso

Page - 2

Root causes: Current and Impedance

perfectvoltage waveshape

voltage drops {

Customer Bus

Utility

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

111 jnXRtnItZIVV nn

nssload

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

11 tnItItin

n

))((sin

)sin(

1,...7,5,3

1111

jnXRtnIV

tZIV

nn

nn

s

2

Page - 3

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) – v. distortions will be high.

Fundamentals of Electric Power Quality (S. Santoso)

Page - 4

Concept of Point of Common Coupling

Utility System

Customer Under Study

Other UtilityCustomers

PCC

IL

PCC at the transformer primary where multiple customers are served

3

Page - 5

Concept of Point of Common Coupling

Utility System

Customer Under Study

Other UtilityCustomers

PCC

IL

PCC at the transformer secondarywhere multiple customers are served

Page - 6

IEEE 519 Voltage Limit

Bus Voltage at PCC (Vn)

Individual Harmonic Voltage Distortion

(%)

Total Voltage Distortion - THDVn

(%)

V kVn 69 3.0 5.0

69 161kV V kVn 1.5 2.5

V kVn 161 1.0 1.5

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

4

Page - 7

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

A3

1000

kV

MVAISC

kVPF

kWIL

3

L

SC

I

ISCR

Page - 8

Current distortion limits

5

Page - 9

Current distortion limits

Page - 11

Harmonic current flows

NORMAL PATHALTERED PATH

6

Page - 12

System Response Characteristics

• Short-circuit MVA at the substation:

SCSC

SCSCSC

I

kVMVAkV

jXRZ

3

2

.1hXXh

Page - 13

Short-circuit capacity at customer facilities

• For example for a 1500 kVA, 6% transformer, the equivalent impedance on the 480 V side is:

.txSC XX

(%),3

2

txtx ZMVAkV

X

. 0092.006.05.1

480.0(%)

2

3

2

txtx ZMVAkV

X

7

Page - 14

Example 7.1

Page - 15

Example 7.1

8

Page - 16

Capacitor Impedance under Harmonic Frequencies

h

XX

Q

kVX

fCX

chc

ratedcap

LLc

c

2

2

1

Page - 17

Parallel resonance

• When the system resonant frequencies corresponds to one of the harmonic frequencies being produced by the nonlinear load (characteristic frequencies), harmonic resonance can occur.

CLf

L

R

CLf

sp

ssp

1

2

1

1

2

12

2

9

Page - 18

Simplified distribution system

Parallel circuit fromthe nonlinear load perspective

pc

pps

p

s

pc

s

ps

s

pss

pc

pc

pss

pss

pc

pc

pssp

XQXQR

X

R

X

R

jXRjX

XXjR

jXRjX

jXjXRZ

22 )()(

)(

)(

)(

)(

Page - 19

Parallel resonance

• Zp is max.

• Denominator is min, limited to Rs

p p pcap c pV Q X I

s

pc

s

psp

R

X

R

XQ p

ss XR

pc

pps

p

s

pc

s

ps

s

pss

pc

pc

pss

pss

pc

pc

pssp

XQXQR

X

R

X

R

jXRjX

XXjR

jXRjX

jXjXRZ

22 )()(

)(

)(

)(

)(

10

Page - 20

Parallel Resonance: current behavior

Let’s find magnitudes of currents flowing in the cap. Bank and in thepower system.

Capacitor failure, fuse blowing, transformer overheating.

pps

pps

p

ps

p

ppc

pp

cp

pc

ppcap

QIX

IXQ

X

V

QIX

IXQ

X

VI

Page - 21

Estimating resonant frequencies

• Voltage and current magnifications are determined by the size of the cap. Bank.

ratedcap

sc

s

cp Q

MVA

X

Xh 3

11

Page - 22

Estimating resonant frequencies

Page - 23

12

Page - 26

Frequency/Impedance Scan

RL

C

33

.42

Ica

p

BR

KEs

System voltage = 69 kV

System short-circuit capacity is 1500 MVA

Series resistive damping = 1 ohm

Cap bank is rated 60 Mvar at 69 kV

Z(f)

0.0 -2000 [Hz]

0 200 400 600 800 1000 1200 1400 1600 1800 20000

50

100

150

200

250

300

frequency

|Z|

Frequency Scan

With Capacitor

Without Capacitor

Page - 27

Example case for harmonic resonance (1)

RL

C

33

.42

Ica

p

BR

KEs

5th order harmonic

0 Amps





Voltage and Current

0.080 0.100 0.120 0.140 0.160 0.180 0.200 0.220 0.240

-100 -75 -50 -25

0 25 50 75

100

kV

Es

-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

kA

Icap

No resonance, no harmonic current The system is simply weak

13

Page - 28


Resonance occurs !!Fifth order harmonic current = 80 A

RL

C

33

.42

Ica

p

BR

KEs

5th order harmonic

80 Amps





Voltage and Current

0.080 0.100 0.120 0.140 0.160 0.180 0.200 0.220

-125 -100 -75 -50 -25

0 25 50 75

100

kV

Es

-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

kA

Icap

Page - 29


No resonanceHarmonic current is 80 A but atthe 25th order

RL

C

33

.42

Ica

p

BR

KEs

25th order harmonic

80 Amps





Voltage and Current

0.080 0.100 0.120 0.140 0.160 0.180 0.200 0.220

-100 -75 -50 -25

0 25 50 75

100

kV

Es

-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

kA

Icap

14

Page - 30

Series Resonance

DistributionSubstation Bus

CustomerPower FactorCorrection

HighVoltageDistortion

HighHarmonicCurrents

Page - 31

Series and Parallel Resonance

Harmonic Number h

Z

series resonance:

parallelresonance :

T

Cs X

Xh

sourceT

Cr XX

Xh

sourceXTX

CXhI

15

Page - 32

Principles of controlling harmonics

• Recall these key concepts in harmonic analysis:– Nonlinear loads: producers of harmonic currents. Who 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)

Page - 33

Three common causes of harmonic problems

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

16

Page - 34

Basic options for controlling harmonics

• Large HARMONIC CURRENTS vs Make it smaller, i.e., reduced 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.

• Large HARMONIC CURRENTS vs. 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.

• 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.

Page - 35

Where to control harmonics: utility feeders

• Cap. bank placed on utility feeders usually done without harmonic evaluations.

• When harmonic resonance occurs (in rare cases), change cap. bank size or move to other locations.

• Triplen harmonics on wye-ground cap. 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.

• Widespread harmonic sources along the feeder: distribute a few filters toward the ends of the feeder.

17

Page - 36

Harmonic filter installation on an overhead distribution-feeder

Oil-insulated iron-core reactorsCapacitor banks and switches

Page - 37

Harmonic filters on a substation

18

Page - 38

Harmonic filters on a feeder

Page - 39

Where to control harmonics: End-user facilities

• Harmonic problems at end-user facilities is it due to resonance with p.f capacitors in the facility?

• YES change capacitor size.

• NO use filters.

19

Page - 40

Example – Parallel resonance

Page - 41

20

Page - 42

Page - 43

21

Page - 44

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.

Page - 45

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 ORDERHIGH-PASS

2ND ORDERHIGH-PASS

3RD ORDER HIGH-PASS

22

Page - 46

One-phase of a 3-phase filter

Page - 47

Filters for industrial power system applications

23

Page - 48

A system response of a 5th harmonic filter

Harmonic Number h

0

10

20

30

40

50

1 3 5 7 9 11 13 15 17

(a) Typical low voltage filter configuation.

(b) Equivalent circuit of system with filter.

B

C

A

X

X

XC

F

XF

XC

3

SCHarmonicSource

(c)System

frequencyresponse(Z = 1.0).

hZ

1

capacitoronly

capacitorconverted

tofilter

hNotch

F

Cnotch X

Xh

3

Page - 49

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 power factor 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 kVAtransformer 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

HARMONICSOURCE

EQUIVALENT CIRCUIT

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

R

L Ih

with no reactor

24

Page - 50

A system response of a 5th harmonic filter

Harmonic Number h

0

10

20

30

40

50

1 3 5 7 9 11 13 15 17

(a) Typical low voltage filter configuation.

(b) Equivalent circuit of system with filter.

B

C

A

X

X

XC

F

XF

XC

3

SCHarmonicSource

(c)System

frequencyresponse(Z = 1.0).

hZ

1

capacitoronly

capacitorconverted

tofilter

hNotch

F

Cnotch X

Xh

3

Page - 51

Series and Parallel Resonance

Harmonic Number h

Z

series resonance:

parallelresonance :

T

Cs X

Xh

sourceT

Cr XX

Xh

sourceXTX

CXhI

25

Page - 52

Step 1 - Select a tuned frequency for the filter

• 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 slightly below the harmonic frequency of concern to allow for tolerances in the filter components and variations in system impedance

– This will prevent the filter from acting as a direct short circuit for the offending harmonic current

– Reduce duty on the filter components

Page - 53

Step 1 - Select a tuned frequency for the filter

26

Page - 54

Step 2: Compute capacitor bank size and the resonant frequency

• 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 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 demand for a 0.75 power factor would be:

1600 x0.75= 1200 kW.

• Required compensation Qfilt from the filter:Qfilt =1200[tan(acos(0.75))-tan(acos(0.98))]

= 814.28 kvar

Page - 55

Step 2: Compute capacitor bank size and the resonant frequency (con’t)

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

• 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 higher than Xcap

calculated above)

LCapFilt XXX

2828.01028.814

48.0

]Mvar[ 3

22

filt

LLFilt

Q

kVX

LLCap XXhX 22 7.4

2962.17.4

)7.4(2828.0

1 2

2

2

2

h

hXX Filt

Cap

kvarX

kVkvar

Cap

LL 75.7772962.

)1000(48.0)1000( 22

XCap = 0.3072

27

Page - 56

Page - 57

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

0.307222

)()( h

XX wyeCap

fundL

mH. 0.0369602

)(

fundLXL

,C2

1

(wye)

hL

f

Hz. 282607.4 hf

28

Page - 58

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.

Page - 59

Step 5: Computation of fundamental duty requirements

• Determine a fundamental frequency operating voltage across the capacitor bank.

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

• B) The fundamental frequency filter current is:

• C) The fundamental frequency operating voltage across the capacitor bank is

• D) 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)( wyeCapLfiltfund XXX

A 86.4490.2933

3480

3,

fund

kVfilt

fundduty XI

actual

V. 76.5023 )(,, wyeCapfilt

funddutycap

fundduty XIV

kvar. 822.81 3 ,, capfundduty

filtfundduty

capduty,fund VIQ

29

Page - 60

Step 6: Computation of harmonic duty requirements.

• Compute the maximum harmonic current expected in the filter

• A) 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:

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

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

– the utility impedance will be neglected

• Fundamental frequency impedance of the service transformer:

• The fifth harmonic impedance of the service transformer

A. 9.48348.03

160020.0

3.).()(

actual

hampshkV

kVAupII

. 0.00580.2

48.005.0(%)

22

)( Xfmr

actualTfundT MVA

kVZX

. 0.02880.00585)(5, fundTT hXX

Page - 61

Step 6: Computation of harmonic duty requirements (cont’d)

• The harmonic impedance of the capacitor bank is:

• The 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:

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

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

. 0614.05

3072.0)(5),(

h

XX wyeCap

wyeCap

. 06951.00139.05)(5, fundLL hXX

5,5),(5,

)()(

3

)(

LwyeCapT

actualutilityhutilityh

XXX

kVpuVI

A. 23.150

06951.00614.00288.03

48002.0

A. 13.535.23501 384.9)( totalhI

h

XIV wyeCap

totalhcap

duty)(

)(5, 3

V. 95.565

0.307213.3553

30

Page - 62

Step 7: Evaluate total rms current and peak voltage requirements

• A) Total rms current passing through the filter:

• This is the total rms current rating required for the filter reactor

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

• C) The RMS voltage across the capacitor is:

• D) The total kvar seen by the capacitor:

A. 89.108513.53586.944 222)(

2, utilityhfundtotalRMS III

capduty

capfundduty

cappeakLLduty VVV 5,,, 22

.V 55.791295.562502.76 peak

25,

2,, )()( cap

dutycap

funddutycap

rmsLLduty VVV

V. 505.9795.56502.76 22

caprmsLLduty

filttotalduty

captotalduty VIQ ,,, 3

kvar. 614508.06983

Page - 63

Step 8: Evaluate capacitor rating limits

• Comparison 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

kvar1172480

600750

2

2

31

Page - 64

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.

• Beware of transformer energization events

.95.30139.00058.0

3072.0

)()(

)('0

fundLfundT

wyeCap

XX

Xh

Page - 65

Harmonic control: load side

• Most common – use a pair of transformers configured as delta-delta and delta-wye, or wye-wye and wye-delta

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

32

Page - 66

Harmonic control: Six-pulse converters

• Three-phase six-pulse converters

iDC

Constant DC current source to representa current source inverter

Va

1 3

64

vDCLink

iSAA

B

C

R=0iSB

iSC

2

Vb

Vc

5

Page - 67

Voltage and currents of 6-pulse convertersiDC


Va

1 3

64

vDCLink

iSAA

B

C

R=0iSB

iSC

2

Vb

Vc

5Voltage and currents

0.1750 0.1800 0.1850 0.1900 0.1950 0.2000 ...

-0.40

-0.20

0.00

0.20

0.40

0.60

0.80

kV

Va Vb Vc vDC

-250 -200 -150 -100 -50

0 50

100 150 200 250

y

iSourceA

33

Page - 68

Phase shifts in transformers

• ANSI/IEEE C57.12.00-2000: Std General Requirements for Liquid Immersed Distribution, Power, and Regulating Transformers.

• The angular displacement between HIGH- and LOW-voltages of 3-phase transformers:

– Delta-delta OR Wye-Wye connections shall be zero degrees

– Wye-delta OR delta-Wye connections shall be 30 degrees with the low voltage lagging the high voltage.

Page - 69

Phase shifts in transformers

• ANSI/IEEE C57.12.00-2000: Std General Requirements for Liquid Immersed Distribution, Power, and Regulating Transformers.

34

Page - 70

12-pulse operation using d-d, and d-y


Va

1 3

64

vDCLink

A

B

C

R=0

2

Vb

Vc

5

1 3

64 2

5

A

B

C

A

B

C0.48

#2#1

12.47

1.5 [MVA]

A

B

C

A

B

C0.48

#2#1

12.47

1.5 [MVA] iT_Xa

iB_XaiB_Ha

ISa

ISb

ISc

iT_Ha

Voltage and currents

0.1300 0.1350 0.1400 0.1450 0.1500 0.1550 0.1600 0.1650 0.1700 0.1750

-12.5 -10.0 -7.5 -5.0 -2.5 0.0 2.5 5.0 7.5

10.0 12.5

kV

Va Vb Vc vDC

-20.0 -15.0 -10.0 -5.0 0.0 5.0

10.0 15.0 20.0

A

iSa

0.00

0.50

1.00

1.50

KV

vDC

Page - 71


35

Page - 72


Page - 73


36

Page - 74


Page - 75

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%.

• 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

Line input choke (% on Drive Base)

Inpu

t C

urr

ent

Dis

tort

ion

(%

)

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)

37

Page - 76

Effect of transformer size

30%

50%

70%

90%

110%

130%

0% 5% 10% 15% 20% 25% 30%

No Choke Choke

ASD kVA

Transformer kVA

I(t) at 33% kVA ratio

I(t) at 5% kVA ratio

Transformer5% Z

(on xfmr kVA)Choke3% Z

(on ASD kVA)ASD

I THD

chapter 7 harmonics part 2

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