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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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