13. Chapter 13 Power Quality

8/13/2019 13. Chapter 13 Power Quality

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CHAPTER 13:

POWER QUALITY

INSTRUCTOR: DINH THANH VIET,DEPARTMENT OF ELECTRICAL ENGINEERING

DANANG UNIVERSITY OF TECHNOLOGY


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POWER QUALITYIS


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A POWER QUALITYPROBLEM IS

When something doesn't work because ofthe electric power supplied to it.


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POWER QUALITY

Our power system were designed for:


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POWER QUALITY

Now the power system servers

Computers


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POWER QUALITY

The power system disturbing sensitive loads


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POWER QUALITY EFFECT

Various odd equipment behavior:

Spring winding counter resets to zero Programmable logic controller (PLC) errors Numerically-controlled machines

malfunction


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VOLTAGE PROBLEMS

Typical medical equipment: CT Scanner and many others

Resets, loses data during series of scans


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VOLTAGE PROBLEMS

Industrial, electronic factories, hospital

need high power quality


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VOLTAGE PROBLEMS


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Voltage problems

Vs

vload

Zline

I2I1

Z1 Z2

Load1 Load2

+

=

+==

Z

Z

VZ

ZZ

VZIV

line

s

line

sload

1


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VOLTAGE PROBLEMS

Fluctuations: Small changes in voltage between 90% and110% of the rated value. If the change is slow, every fewhours, the fluctuations have no effect on people orequipment.

Flickers: fast and cyclic changes in voltage that aredetected by human eyes. Although the magnitude of the

flicker could be within the fluctuation range of the voltage,it is often fast enough allowing human eyes to detect it.


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VOLTAGE PROBLEMS

Sags: voltage below 90% of the rated value for a short

period (up to a few seconds).

Swells: voltage above 110% for a short period (last forless than 1 minute).

Undervoltage (brownout): voltage drop below 90% for atleast several minutes.

Overvoltage: voltage increase above 110% for at leastseveral minutes.

Interruptions:decrease in voltage below 10% for anyperiod.


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VOLTAGE FLICKER


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VOLTAGE FLICKER

Flicker is the fluctuation of voltageat frequencies much lower than the

power frequency. Most irritating effect of voltageflicker is light flickers (fast variations

in light intensity).


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VOLTAGE FLICKER

Light intensity is a measure ofbrightness of a light source

incandescent ~ P ~ V2

ballast ~ V

Incandescent lamps tend to be more

sensitive to changes in voltages thanballasted lamp (arc and fluorescentlamps).


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Vo

ltageflic

ker

Computer freeze up

Jitter of television

pictures

Faulty signals inelectronic circuits

Malfunction of sensitiveequipment

Loss of informationstored in electronic

memories


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VOLTAGE FLICKER

Two types of flickers:Cyclic flicker: is caused by periodheavy load fluctuations.

Noncyclic flicker: is caused byoccasional heavy load switching


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VOLTAGE FLICKER

V

Vmax

Vmin

Vssmax

Loadvoltage

Time

Cyclic flicker


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VOLTAGE FLICKERThe instantaneous value of the flicker voltage vf

where

Vssmax is the peak value of the steady state voltagewithout any flicker

V/2 is the modulation amplitude of the flicker

is the frequency of the supply voltage without theflicker

f is the frequency of the flicker

)sin(2)sin(max t

VtVv fssf

=


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VOLTAGE FLICKER The load voltage:

The flicker factor F:

If F = 0, the voltage waveform without any flicker

The larger F, the more severe is the flicker problem

Voltage fluctuation VF:

)cos())cos(21(max tt

V

Vvvv fssfssload

+=+=

minmaxmax

2/

VV

V

V

VF

ss +

=

=

FV

VVF

ss

2max

=

=


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VOLTAGE FLICKER

2 5 10 20 30 2 5 10 20 30 1 2 5 1011

6

5

4

3

2

1Border of visibility

Border of irritationVF%

Fluctuations per hour Fluctuations per minute Fluctuations per second

Effect of flickers on humans


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VOLTAGE FLICKER Example

A system with a steady-state voltage of 220V

is connected to a cyclic load that fluctuates ata rate of 15 cycles/h. Compute the maximumallowable flicker V

According to Figure above, VF at the irritationlevel for 15 cycles/h is about 3%. In this case

About 9V variation in voltage can irritatepeople if the flicker is repeated 15 times/h.

max 0.03 220 2 9.3ssV VFxV x x V = = =


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VOLTAGE SAG


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VOLTAGE SAGVoltage sag:IEEE defines voltage sag as

a reduction in voltage for a short time.The durationof a voltage sag is less than1 minute but more than 10 milliseconds

(0.5 cycles). The magnitude of thereduction is between 10 percent and 90percent of the normal root mean square

(rms) voltage at 50Hz.

The flicker is repetitive or cyclic while the

sag is not.


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VOLTAGE SAG CAUSED BY Switching on heavy electric loads

(elevators, cranes, pumps, airconditioners, refrigerators, x-rayequipments, heaters)

Short circuits or faults in power systemcomponents.

Motors starting.


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VOLTAGE SAG CAUSED:

Vo

ltage

sag

sInterrupting controller orrelays to malfunction of

sensitive medical equipment

Some types of motors to run

at lower speeds or take alonger time to start

Disruptions to assembly linesand central air-conditioning

systems


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VOLTAGE SAG

Voltage sag


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VOLTAGE SAG

Voltage sag


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Motors continuously start and stop causing

transients in currents (inrush currents):

Where

i -instantaneous value of inrush current

1

- rising time constant

2 - falling time constant

k is the factor that determines the maximum crest

of the inrush current Icrest

INRUSH CURRENT

+

=

21 11)sin(max

tt

ss keetIi


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INRUSH CURRENTIcrest

Steady state

Asymmetrical inrush current

Inrushcur

rent

Typical inrush current waveform at motorstarting or transformer switching(b) Asymmetrical inrush current with a dccomponent


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VOLTAGE SAG

V

Voltage

Vssmax

Time

Voltage sag due to symmetrical inrush current


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VOLTAGE SAG

Vload = Vs XLI

The voltage sag VS:

ss

ssload

ss V

VV

V

VVS

=

=


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HARMONIC PROBLEMS


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HARMONIC PROBLEMS

f(x) = sin(x) f(x) =sin(5x)

5

+

The resulting wave shows a strong distortion from the smoothoriginal waves:

f(x) = sin(x) +sin(5x)

5=


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HARMONIC PROBLEMSSinusoidalwaveform

Periodic andDistortedwaveform

Time

Sinusoidal and distorted

waveform


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HARMONIC PROBLEMS

The kth harmonic component:

Where

( ) ( )tkbtkag kkk sincos +=

( )=

2

0

cos1

tdtkga k

( )=

2

0cos

1tdtkgbk

( ) ( ) ( )1max 2 max 3max( ) sin sin 2 sin 3 ...dci t I I t I t I t = + + + +

Current in frequency domain:

...)4()3()2()( 43210 +++++= tgtgtgtggg

Fourier series:


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HARMONIC PROBLEMS The rms current:

( )[ ]

=

2

0

2

2

1tdtiI

( ) ( ) ( )[ ] tdtItItIII dc

22

0

max3max2max1 ...3sin2sinsin

2

1 ++++=

( ) ( ) 1kif;0sinsin2

1 2

0

=

tdtkt

( )( ) ( )( )

+++=

2

0

2

max2

2

0

2

0

2

max1

2 ...2sinsin2

1tdtItdtItdII dc


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( )( )

=

2

0

2

maxsin2

1tdtkII kk

...232

2

2

1

2 ++++= IIIII dc

The rms of any current component:

The rms current of a distorted waveform:

HARMONIC PROBLEMS


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IH - the harmonic current, rms of all harmoniccomponents except the fundamental

...23

2

2

2 +++= IIIIdcH

1I

IIHD kk =

1I

ITHD H=

Individual harmonic distortion (IHD):

Total harmonic distortion (THD)

HARMONIC PROBLEMS


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HARMONIC PROBLEMS

......

21

2

3

21

2

2

21

2

1

2

3

2

2

2

+++=+++

=I

I

I

I

I

I

I

IIITHD dc

dc

...2

4

2

3

2

2

2 ++++= IHDIHDIHDIHDTHDdc

The THD has several other forms


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HARMONIC PROBLEMS Example:

A distorted current waveform has 4harmonic components represented by theirrsm values: I1= 50A, I3=20A, I5= 5A, I7=2A.Compute:

a) Rms current of the distorted waveform

b) IHD of the 5th harmonicc) THD


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HARMONIC PROBLEMS

a)

b)

c)The harmonic current:

Conclusion: This waveform contain 41.42%of its magnitude as harmonics.

AIIIII 12.54252050 22222

7

2

5

2

3

2

1 =+++=+++=

%1050

5

1

55 ===

I

IIHD

AIH 71.202520 222 =++=

%42.4150

71.20

1

===I

ITHD H


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HARMONIC DISTORTION OFELECTRIC LOADS


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HARMONIC PROBLEMS

The source voltage ispurely sinusoidal

without any harmonicsand the load

impedance is linear

The current of the loadis sinusoidal and free

of harmonics

The load is nonlinear The current is distorted


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HARMONIC PROBLEMS Example:

A 220V (rms) purely sinusoidal voltagesource is feeding a nonlinear resistance.Assume that the value of the resistance

is dependent on the applied voltage andcan be expressed by

Draw the current waveform.

+= 4

2

1010

v

eR


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The current of the load:

( ) ( )max

2 2 4 2 2 4

max

sin( ) 311.13sin( )

10 exp sin /10 10 exp 311.13 sin /10

V tv ti

R V t t

= = =

+ +

C

urrent

Angle

10

8

6

4

2

0

-2

-4

-6

-8

-101 2 3 4 5 6 7


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HARMONIC PROBLEMS

VbusVs

L

i SW

VR

R

Representation of a simple system consisting ofsource, cable, and switching load


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The waveforms of the source voltage, current, andbus voltage, affected by power electronic converters

vs

vbus

vs

i


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Example:A feeder is powering 100 computers. The totalcurrent of all computers can be represented byi=4+ 50sin(250t) + 30sin(2150t) +10sin(2250t)

+5sin(2350t) A

Compute the THD of the feeder. Then, assume thata linear heating load of 100 A (rms) is connected tothe same feeder, compute the new THD of the

feeder.

HARMONIC PROBLEMS


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A.IAI

AIAI

5432

5

721.212

30

07.72

1036.35

2

50

3

51

====

====

The rms value of the harmonic current

AIH 2354.307.721.214 2222

=+++=

Compute the rms values of the harmonic components:i=4+ 50sin(250t) + 30sin(2150t) +10sin(2250t)

+5sin(2350t) AA4

dcI =

HARMONIC PROBLEMS


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%6536.35

23

1

===I

ITHD H

When the heating load is added

AI 36.135100

2

501

=+=

AIH 2354.307.721.214 2222 =+++=

%1736.135

23

1

===I

ITHD H

HARMONIC PROBLEMS

PROBLEM CAUSED BY HARMONIC


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PROBLEM CAUSED BY HARMONICDISTORTIONS

Resonance in power system that could stress the

components beyond their designed ratings

Increasing losses in transmission lines, transformers

and generators

Damaging to various compensation capacitor banks

Creating drag forces inside motors that overheat andreducing their torques

PROBLEM CAUSED BY HARMONIC


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PROBLEM CAUSED BY HARMONICDISTORTIONS

Reducing the overall power factor (pf) of the system and

producing inaccurate power measurements

Increasing the electromagnetic interference to

communication networks.

Premature aging of insulators in equipment and cables

Causing picture jitters, and freezing or rebooting ofcomputers and sensitive equipment


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RESONANCE DUE TOHARMONICS


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HARMONIC PROBLEMS

VsIC

Lcable

IL

I

Zload

L

Compensated load


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The total load impedance Zload

LC

Lj

CjLj

CjLj

Z load2

11

1

=

+

=

The voltage across the load

IZloadload=V

HARMONIC PROBLEMS


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The current of the circuit

cableload

s

ZZ

V

I +=

where cablecable LjZ =

s

cableload

load

load VZZ

ZV

+=

( ) ( ) s

cablecable

load VCLLLL

LV

+=

2

HARMONIC PROBLEMS


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Load voltage reach a theoretical value of infinity ifa frequency 0 makes the denominator equalzero.

( ) ( ) 02

0 =+

CLLLL cablecable

At this condition, the frequency is calledresonance frequency f0 or 0

CLL

LLf

CLL

LL

cable

cable

cable

cable

+=

+=

2

10

0


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The total impedance on the load side is:

( )

( )

( )

( )

4

4

101 10 0.1

1 1010 0.1

load

jR j Ljj C

Z

R j L j

j C j

++ = =

+ + + +

Where2 2 (3 50) 942 / f x rad s = = =

The total impedance at the load side is:

0.159 11.95loadZ j=

HARMONIC PROBLEMS


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Source current of the system:

100

36.56 147.20.159 11.95 11.31

s

load cable

V

I j Aj jZ Z= = = ++

Voltage across the load:

( )( ) 00.159 11.95 36.56 147.2 1813 165.3

loadloadV Z I

j j V

=

= =

HARMONIC PROBLEMS

Notice: load voltage is more than 18 times

the third harmonic source voltage.This high voltage may cause damage toequipment


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EFFECT OF HARMONICSON TRANSMISSION LINES

AND CABLES


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Two main transmission line problems

associated with system harmonics: losses andexcessive voltage.

If the harmonic is close to the resonant

frequency, the voltage across the cable or lineis to be excessive.

The harmonic currents produce line loss Pline

= k kline RIP 2

HARMONIC PROBLEMS


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EFFECT OF HARMONICON CAPACITOR BANKS


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The voltage across the capacitorcould be excessive. This high voltagecauses high current in the capacitor

Icapacitor= CVload

HARMONIC PROBLEMS


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When

high and Vload is also high The capacitor current could beexcessive and could damage the often

expensive capacitor bank

HARMONIC PROBLEMS

Example:


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At the 50 Hz, the load impedance:

Example:

Compute the capacitor current at 50 Hz and at the thirdharmonic voltage. Assume that the fundamental voltage is240 V and the third harmonic voltage is 100V

The 50 Hz current of the source:

4

4

101(10 0.1 314)( )

314Z 101.2 27.3

1 10( ) (10 0.1 314)

314

load

j xR j Ljj C

j

R j L j xj C j

++

= = = + + + +

240 1.8 06.310 31.4 3.77

s

load cable

VI j Aj jZ Z

= = = + ++


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The current of the capacitor at 50 Hz:

The voltage across the load:

The current of the capacitor at 150Hz:

Icapacitor

= CVload

= 314 x 10-4 x 688.2 = 21.6 A

Icapacitor = CVload = (3 x 314) x 10-4 x 1813 = 170.8 A

HARMONIC PROBLEMS

0

688.2 89.2loadloadV Z I V = =

This is almost 8 times the current at 50Hz


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EFFECT OF HARMONICSON ELECTRIC MACHINES


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Harmonic lossesExcessivevoltage

Drag torque(braking torque)

Harmonics

cause avariety of

problem toelectricalmachines


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Suppose that rotation of magnetic field inthe airgap follows the abc sequence:

a= V

maxsin(t)

b= Vmaxsin(t - 120)c= Vmaxsin(t + 120)

HARMONIC PROBLEMS

Fifth harmonic waveforms:


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Hence,

Fifth harmonic waveforms:

a= Vmaxsin5(t)b= Vmaxsin5(t - 120)

c= Vmaxsin5(t + 120)

a= V

maxsin(5t)

b= Vmaxsin(5t + 120)c= Vmaxsin(5t - 120)

Magnetic field rotates in the acbsequence, opposite to the abc sequenceof fundamental frequency voltage.


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EFFECT OF HARMONICSON ELECTRIC POWER


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The instantaneous power:

p(t) = v(t) i(t)The active power:

HARMONIC PROBLEMS

= ==

2

0

)()(2

1

0

2

0

)(2

1)(

1tdtitvtdtpdttpP


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In circuits with harmonics, the pf is defined:

Using the definition of the THD:

THDV is the THD of the voltage waveformTHDI is the THD of the current

...)...)((

...

2

3

2

2

2

1

22

3

2

2

2

1

2321

++++++++

++++==

IIIIVVVV

PPPP

VI

Ppf

dcdc

dc

)1)(1(

...

)1)(1(

...22

1

321

22

11

321

IV

dc

IV

dc

THDTHDP

PPPP

THDTHDIV

PPPPpf

++

++++=

++

++++=

HARMONIC PROBLEMS

ExampleAn industrial load has the following rms harmonic


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gcomponents:

a) Assume that simple devices are used to measurethe power and the pf, the devices monitor only thefundamental components. What are their readings?b) The actual values of power and pf?

c) If the price of electricity is $0.2/kWh, compute thelosses for the utility in 1 year assuming that the load isactive for 10 h daily.

V I

Dc component 10 1

First harmonic 460 10 300

Third harmonic 90 5 00

Fifth harmonic 30 2 100


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Solutiona) The power meter measures the fundamentalfrequency component only

kWIVP 430cos1040cos1111

===

866.030cos11

1

1 ===

IV

Ppf

HARMONIC PROBLEMS


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O C O S


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c) The loss in measurements due to harmonics

The annual energy unmonitored due to the harmonics is

The annual lost revenue of the utility is

Lost revenue= Eharmonics.cost of kWh = 1898.$0.2=$379.6

Conclusion: The loss revenue for this one load is probablyhigh enough to justify the installation of more accurate meters

WPPPharmonics 520400045201 ===

kWhtPE harmonicsharmonic 189836510520 ===

HARMONIC PROBLEMS


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EFFECT OF HARMONICS

ON COMMUNICATIONS

HARMONIC PROBLEMS


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Electrical current produces a magneticfield surrounding the wire.

HARMONIC PROBLEMS

High-frequency harmonic componentsmay not be filtered out and could interferewith communication signals.


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STANDARD OF POWERQUALITY IN VIETNAM

V lt


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Voltage

In normal operation At load side: 5%

At source side: +10% v -5%

In the single failure or the recovery process afterfailure, at load side: +5% v -10%

In serious failure operation: 10%

Operating permissible voltage in comparison withrated voltage:

H i


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Harmonic

Voltage levelTotal harmonic

distortion (THD)

Individual harmonic

distortion (IHD)

110kV 3,0% 1,5%


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Voltage flicker Voltage flicker level of short-term (Pst) -

measured value during the period often minutes by measuring equipment.

Pst95% is the threshold value of Plt so that

about 95% of the measured time (at leastone week) and 95% of the Pstmeasurement position does not exceed this

value.

Voltage flicker


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Voltage flicker Voltage flicker level of long-term (P

lt) is calculated from 12 consecutive

measurements Pst (in 2 hours),

123

3

1

1

12lt stj

jP P==

Plt95% is the threshold value of Plt so that about

95% of the measured time (at least one week) and 95%of the Plt measurement position does not exceed thisvalue.

Voltage flicker


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Voltage flickerVoltage level

Permissible

flicker level

110kVPst95% = 0,80

Plt95% = 0,60

Distribution Network Pst95% = 1,00Plt95% = 0,80

Low voltage NetworkPst95% = 1,00

Plt95% = 0,80


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13. Chapter 13 Power Quality

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Transcript of 13. Chapter 13 Power Quality