harmonics
TRANSCRIPT
New Loads onOld SystemsStefan FassbinderDKIGerman Copper InstituteAm Bonneshof 5D-40474 DüsseldorfTel.: +49 211 4796-323Fax: +49 211 [email protected]
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Standards and Power Quality – a major problem in publicand commercial buildings
The substantial growth in the number of electronic devices being used in recent years has resulted in a significant change in the types of load being driven by today’s power distribution systems.
Because these devices are equipped with rectifiers and smoothing capacitors, the current drawn from the power system is significantly distorted from the sinusoidal waveform provided by the utility companies.
This has serious consequences for power quality...
Power quality problems
Turn off the mixer love, the monitor’s
flickering again!
are usually of terrestrial
origin.
The good old days:
Ideal
three-phase system voltage
Three balanced ohmic-inductive single-phase loads on the three phase mains
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gleichgerichteteNetzspannung
Kondensator-spannung
gleichgerichteterNetzstrom
The situation today:
Computed current and voltage profiles when a230 V, 58 W fluorescent lamp with an old-style electronic ballast is driven on the phase wire of a typical power distribution system...
System voltage: 230 VSystem frequency: 50 HzSource resistance: 500 mLongitudinal induction: 904 µHSystem impedance: 575 mMean d.c. current: 180 mASmoothing capacitance: 220 µF
And what about electronic ballasts that are rated over 25W? Introduce electronic power factor correction (PFC)
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Though today’s ballasts are better if rated >25 W
Only the small ones continue to pollute, just like PCs, TV sets, ... do
This leads to a number of previously unknown effects:
1.Deformed voltage curves
2.Huge inrush current peaks
3. Instrument dependent measured parameters
4. Increased loading of phase conductors
5.Loading and overloading of the neutral conductor
6.Overheating and start-up problems of induction motors
7.Additional stray losses in transformers8.Repercussions of generators upon the mains9. Influence of capacitors, repercussions upon the mains
... and four additional complications in TN-C systems:
10.Stray currents in equipotential bonding system: magnetic fields
11. Contamination of data streams by stray currents
12.Stray currents in the earthing system: corrosion damage
13.Lightning currents in devices and equipment
1. Deformed voltage curvesin theory ...
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System voltage profile when driving a single-phase load comprising
twenty 58 W energy-saver lamps in parallel on a typical 230 V power
system
... and in practice ...
An actual current profile
and
a voltage profile
recorded in a large storage, dispatch and
administrative building
... at home or in industrial environments
Television, 40 W
200 V/div., 2 A/div.,5 ms/div.
Harmonic spectrum
Frequencyconverter
50 A/div.Phase currents:I1=I2=I3=16.0 A
IN=29.5 A5 ms/div.
Every periodic waveform can be writtenas the infinite sum of sinusoidal waves ...
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Synthesis of a triangular current profile 100mAContent fundamental: I = 99 mA I² = 9855 mA²Sum of squares: I² = 9855 mA²Root of sum = RMS value: I = 99 mA
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Synthesis of a triangular current profile 100mAContent fundamental: I = 99 mA I² = 9855 mA²Cont. 3 rd harmonic: I = -11 mA I² = 122 mA²Sum of squares: I² = 9977 mA²Root of sum = RMS value: I = 100 mA
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Synthesis of a triangular current profile 100mAContent fundamental: I = 99 mA I² = 9855 mA²Cont. 3 rd harmonic: I = -11 mA I² = 122 mA²Cont. 5 th harmonic: I = 4 mA I² = 16 mA²Sum of squares: I² = 9993 mA²Root of sum = RMS value: I = 100 mA
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Synthesis of a triangular current profile 100mAContent fundamental: I = 99 mA I² = 9855 mA²Cont. 3 rd harmonic: I = -11 mA I² = 122 mA²Cont. 5 th harmonic: I = 4 mA I² = 16 mA²Cont. 7 th harmonic: I = -2 mA I² = 4 mA²Sum of squares: I² = 9997 mA²Root of sum = RMS value: I = 100 mA
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itot i1
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Synthesis of a triangular current profile 100mAContent fundamental: I = 99 mA I² = 9855 mA²Cont. 3 rd harmonic: I = -11 mA I² = 122 mA²Cont. 5 th harmonic: I = 4 mA I² = 16 mA²Cont. 7 th harmonic: I = -2 mA I² = 4 mA²Cont. 9 th harmonic: I = 1 mA I² = 2 mA²Sum of squares: I² = 9998 mA²Root of sum = RMS value: I = 100 mA
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Synthesis of a triangular current profile 100mAContent fundamental: I = 99 mA I² = 9855 mA²Cont. 3 rd harmonic: I = -11 mA I² = 122 mA²Cont. 5 th harmonic: I = 4 mA I² = 16 mA²Cont. 7 th harmonic: I = -2 mA I² = 4 mA²Cont. 9 th harmonic: I = 1 mA I² = 2 mA²Cont. 11 th harmonic: I = -1 mA I² = 1 mA²Sum of squares: I² = 9999 mA²Root of sum = RMS value: I = 100 mA
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itot i1
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Synthesis of a triangular current profile 100mAContent fundamental: I = 99 mA I² = 9855 mA²Cont. 3 rd harmonic: I = -11 mA I² = 122 mA²Cont. 5 th harmonic: I = 4 mA I² = 16 mA²Cont. 7 th harmonic: I = -2 mA I² = 4 mA²Cont. 9 th harmonic: I = 1 mA I² = 2 mA²Cont. 11 th harmonic: I = -1 mA I² = 1 mA²Cont. 13 th harmonic: I = 1 mA I² = 0 mA²Sum of squares: I² = 9999 mA²Root of sum = RMS value: I = 100 mA
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itot i1
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Synthesis of a triangular current profile 100mAContent fundamental: I = 99 mA I² = 9855 mA²Cont. 3 rd harmonic: I = -11 mA I² = 122 mA²Cont. 5 th harmonic: I = 4 mA I² = 16 mA²Cont. 7 th harmonic: I = -2 mA I² = 4 mA²Cont. 9 th harmonic: I = 1 mA I² = 2 mA²Cont. 11 th harmonic: I = -1 mA I² = 1 mA²Cont. 13 th harmonic: I = 1 mA I² = 0 mA²Cont. 15 th harmonic: I = 0 mA I² = 0 mA²Sum of squares: I² = 10000 mA²Root of sum = RMS value: I = 100 mA
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itot i1
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Synthesis of a triangular current profile 100mAContent fundamental: I = 99 mA I² = 9855 mA²Cont. 3 rd harmonic: I = -11 mA I² = 122 mA²Cont. 5 th harmonic: I = 4 mA I² = 16 mA²Cont. 7 th harmonic: I = -2 mA I² = 4 mA²Cont. 9 th harmonic: I = 1 mA I² = 2 mA²Cont. 11 th harmonic: I = -1 mA I² = 1 mA²Cont. 13 th harmonic: I = 1 mA I² = 0 mA²Cont. 15 th harmonic: I = 0 mA I² = 0 mA²Cont. 17 th harmonic: I = 0 mA I² = 0 mA²Sum of squares: I² = 10000 mA²Root of sum = RMS value: I = 100 mA
... whose frequencies, called harmonics, are integer multiples of the fundamental frequency.
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Synthesis of a rectangular current profile 100mAContent fundamental: I = 90 mA I² = 8106 mA²Sum of squares: I² = 8106 mA²Root of sum = RMS value: I = 90 mA
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Synthesis of a rectangular current profile 100mAContent fundamental: I = 90 mA I² = 8106 mA²Cont. 3 rd harmonic: I = 30 mA I² = 901 mA²Sum of squares: I² = 9006 mA²Root of sum = RMS value: I = 95 mA
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Synthesis of a rectangular current profile 100mAContent fundamental: I = 90 mA I² = 8106 mA²Cont. 3 rd harmonic: I = 30 mA I² = 901 mA²Cont. 5 th harmonic: I = 18 mA I² = 324 mA²Sum of squares: I² = 9331 mA²Root of sum = RMS value: I = 97 mA
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Synthesis of a rectangular current profile 100mAContent fundamental: I = 90 mA I² = 8106 mA²Cont. 3 rd harmonic: I = 30 mA I² = 901 mA²Cont. 5 th harmonic: I = 18 mA I² = 324 mA²Cont. 7 th harmonic: I = 13 mA I² = 165 mA²Sum of squares: I² = 9496 mA²Root of sum = RMS value: I = 97 mA
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Synthesis of a rectangular current profile 100mAContent fundamental: I = 90 mA I² = 8106 mA²Cont. 3 rd harmonic: I = 30 mA I² = 901 mA²Cont. 5 th harmonic: I = 18 mA I² = 324 mA²Cont. 7 th harmonic: I = 13 mA I² = 165 mA²Cont. 9 th harmonic: I = 10 mA I² = 100 mA²Sum of squares: I² = 9596 mA²Root of sum = RMS value: I = 98 mA
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itot i1
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Synthesis of a rectangular current profile 100mAContent fundamental: I = 90 mA I² = 8106 mA²Cont. 3 rd harmonic: I = 30 mA I² = 901 mA²Cont. 5 th harmonic: I = 18 mA I² = 324 mA²Cont. 7 th harmonic: I = 13 mA I² = 165 mA²Cont. 9 th harmonic: I = 10 mA I² = 100 mA²Cont. 11 th harmonic: I = 8 mA I² = 67 mA²Sum of squares: I² = 9663 mA²Root of sum = RMS value: I = 98 mA
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itot i1
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Synthesis of a rectangular current profile 100mAContent fundamental: I = 90 mA I² = 8106 mA²Cont. 3 rd harmonic: I = 30 mA I² = 901 mA²Cont. 5 th harmonic: I = 18 mA I² = 324 mA²Cont. 7 th harmonic: I = 13 mA I² = 165 mA²Cont. 9 th harmonic: I = 10 mA I² = 100 mA²Cont. 11 th harmonic: I = 8 mA I² = 67 mA²Cont. 13 th harmonic: I = 7 mA I² = 48 mA²Sum of squares: I² = 9711 mA²Root of sum = RMS value: I = 99 mA
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Synthesis of a rectangular current profile 100mAContent fundamental: I = 90 mA I² = 8106 mA²Cont. 3 rd harmonic: I = 30 mA I² = 901 mA²Cont. 5 th harmonic: I = 18 mA I² = 324 mA²Cont. 7 th harmonic: I = 13 mA I² = 165 mA²Cont. 9 th harmonic: I = 10 mA I² = 100 mA²Cont. 11 th harmonic: I = 8 mA I² = 67 mA²Cont. 13 th harmonic: I = 7 mA I² = 48 mA²Cont. 15 th harmonic: I = 6 mA I² = 36 mA²Sum of squares: I² = 9747 mA²Root of sum = RMS value: I = 99 mA
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Synthesis of a rectangular current profile 100mAContent fundamental: I = 90 mA I² = 8106 mA²Cont. 3 rd harmonic: I = 30 mA I² = 901 mA²Cont. 5 th harmonic: I = 18 mA I² = 324 mA²Cont. 7 th harmonic: I = 13 mA I² = 165 mA²Cont. 9 th harmonic: I = 10 mA I² = 100 mA²Cont. 11 th harmonic: I = 8 mA I² = 67 mA²Cont. 13 th harmonic: I = 7 mA I² = 48 mA²Cont. 15 th harmonic: I = 6 mA I² = 36 mA²Cont. 17 th harmonic: I = 5 mA I² = 28 mA²Sum of squares: I² = 9775 mA²Root of sum = RMS value: I = 99 mA
Even far better simulations are available free of charge at:
www.powerstandards.com/McEachern
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Applying the Fourier analysis to a real PC
current does not work directly ...
... but it can be modelled by a similar triangular curve in which not all of the higher harmonics have been included
Triangular current profile of the same amplitude and a
pulse duty cycle of 1:7
Input current of a PC with monitor
Analysis of the model triangular waveform
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M3
Harmonics
Energy
(active
power)
L1
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N
PE
Important:Harmonics are created within the loads themselves and can flow “upstream” to contaminate the power system!
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Squared values Without
Fundamental: 502 mA 251986 mA² the fundamental3rd harmonic: -479 mA 229000 mA² 229000 mA²5th harmonic : 434 mA 188595 mA² 188595 mA²7th harmonic : -374 mA 139903 mA² 139903 mA²9th harmonic : 304 mA 92527 mA² 92527 mA²11th harmonic : -232 mA 53671 mA² 53671 mA²13th harmonic : 163 mA 26600 mA² 26600 mA²15th harmonic : -104 mA 10780 mA² 10780 mA²17th harmonic : 57 mA 3299 mA² 3299 mA²Sum of the squares: 996362 mA² 744377 mA²Root thereof: 998 mAeff THD = 863 mA
What actually is THD?Example shown here: triangular current profile with an r.m.s. current of 1000 mA and a pulse duty cycle of 1:7
THDr (root mean square) = 863mA/1000mA = 86 %THDf (fundamental) = 863mA/502mA = 172 %
Effect on phase-to-neutraland phase-to-phase voltages
Recorded on June 30, 2002,2:30 p. m.
What was the matter then?
The soccer final: Germany vs. Brazil!
The triple-N harmonics drive a circulating current in
the delta winding of a distribution transformer ...
... but thevoltage harmonics propagate into the next low-voltage power distribution system!
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Harmonics in the course of a week
Harmonics on Saturday
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U13 650Hz U15 750Hz
Harmonics on Sunday
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Switching on just as supply voltage passes through zero
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Switching on just as supply voltage is at its peak
2. Huge inrush current peaks– in theory ...
... and in practice
Inrush current of a compact energy-saver lamp and its effect on the system voltage (recorded at the Technical University of Budapest)
3. Measured parameters are instrument dependentThe root mean square value of an alternating or pulsating current is the value that a smooth (pure) direct current would have to have in order to cause the same heating effect in a fixed resistive load.
Analogue measurement systems: No significant difference in price between average-reading and rms-reading a.c. meters – but no longer in common use.Digital meter:Much more expensive if true rms value is really displayed!
Moving-iron meter:Displays rms value of current.
Moving-coil meter:Displays average magnitude of current if used in conjunction with a bridge rectifier.
4. Increased conductor loadingin theory – (e. g. from older style electronic ballasts):
Line current – average magnitude 179 mALine current – rms value 615 mALine current – peak value 2712 mAApparent power 141 VAD.c. power 58 WLine current – form factor 3.436Line current – crest factor 4.410
Comparison with values for a sinusoidal current:Form factor 1.1107Crest factor 1.4142
...and in practice– e. g. power supplyfor a laptop PC
Now let us compare an RMS to a True RMS meter:
blindblind
All currents are equal ...
î =1A
R=1
ûR=R*î=1V
pR= ûR*î= ûR²/R=1W
q=2*10ms*1A=20mAs
WR=20ms*1W=20mWs
UR=1V, I=1A
î =2A
R=1
ûR=R*î=2V
pR= ûR*î= ûR²/R=4W
q=2*5ms*2A=20mAs
WR=2*5ms*4W=40mJ
UR=1.414V, I=1.414A
...but RMS currents are less equal than others!
... enables us to infer what happens when run by a normal three-phase supply
The same fluorescent lamp with a magnetic ballast:
The behaviour measured when connected to
a d.c. supply ...
Behaviour of a 58 W fluorescent lamp connected to a d.c. supply
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Measurement
Behaviour of a 58 W fluorescent lamp connected to a d.c. supply
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MeasurementCalculation
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-0,6
-0,4
-0,2
0,0
0,2
0,4
0,6
0,8
1,0
i /
A
Systems voltage
Current
-350
-300
-250
-200
-150
-100
-50
0
50
100
150
200
250
300
350
0 5 10 15 20
t / ms
u /
V
-1,0
-0,8
-0,6
-0,4
-0,2
0,0
0,2
0,4
0,6
0,8
1,0
i /
A
Systems voltage
Lamp voltage
Current
Warning: “Compensation” is not always what you think it is!Distinguish between:
Reactive power compensationa.k.a. power factor correction
Remedial measure:
Parallel (or sometimes series) compensation using capacitors
Filter circuits tuned to the individual harmonic frequencies
Harmoniccompensation
5. (Over)loading the neutral conductorMagnetic ballast: Old-style electronic ballast:
-2,5A
-2,0A
-1,5A
-1,0A
-0,5A
0,0A
0,5A
1,0A
1,5A
2,0A
2,5A
0 5 10 15 20
t / ms i
i(t) N
-2,5A
-2,0A
-1,5A
-1,0A
-0,5A
0,0A
0,5A
1,0A
1,5A
2,0A
2,5A
0 5 10 15 20
t / ms
i
i(t) N
-2,5A
-2,0A
-1,5A
-1,0A
-0,5A
0,0A
0,5A
1,0A
1,5A
2,0A
2,5A
0 5 10 15 20
t / ms
i
i(t) L1 i(t) L2 i(t) L3
-2,5A
-2,0A
-1,5A
-1,0A
-0,5A
0,0A
0,5A
1,0A
1,5A
2,0A
2,5A
0 5 10 15 20
t / ms
i
i(t) L1
i(t) L2
i(t) L3
Adding the 3rd harmonics
in the neutral wire
-150-100-50
050
100150
0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360°
L1
-150-100-50
050
100150
0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360°
L2
-150-100-50
050
100150
0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360°
L3
-450-400-350-300-250-200-150-100-50
050
100150200250300350400450
0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360°
f
i/î /
%
N
Physics dictates that at any moment in time the phase and neutral currents must sum to zero
Effect on a three-phase a.c. motor appears almost identical:
The same motor is being driven ...
... simultaneously in these three modes!
homopolar system,zero-sequence system
direct system,positive-sequence system
inverse system,negative-sequence system
The transformerinfluences the load...
... but the load also effects the transformer!
Total transformer loss is:2
)()(
nomnomCunomFeLoss I
IPPP
While total transformer loss really is:
2
)(
2
)(
2
)( *
nomnomnomad
nomnomCu
nomnomFeLoss I
I
f
fP
I
IP
U
UPP
6. “supplementary” additional losses in transformers
can be calculated rapidly using the following two simple formulae:
5,0
2
2
1
2
11
Nn
n
nqh
I
In
I
I
e
eK
5,0
1
2
11
5,0
1
2
Nn
n
nNn
nn I
IIII
where
Oh well,perhaps a practical example is clearer:
1000 compact 11W (15VA) energy-saver lamps powered by a 15kVA transformer, uSC=4%, Pa=0.1PCu
Harmonics of an Osram Dulux 11W CFL and a serial impedance of
R =29.1Ω & X L=113ΩU U ² I L I L² P a/P Cu
n V V² mA mA²
1 230.2 52992.0 48.5 2352.3 5.6%3 8.3 68.9 37.1 1376.4 29.5%5 10.7 114.5 20.3 412.1 24.5%7 4.3 18.5 5.3 28.1 3.3%9 1.1 1.2 3.0 9.0 1.7%
11 2.3 5.3 3.8 14.4 4.2%13 1.0 1.0 1.5 2.3 0.9%15 0.6 0.4 1.5 2.3 1.2%17 1.1 1.2 1.5 2.3 1.5%19 0.5 0.3 0.9 0.8 0.7%21 0.5 0.3 1.3 1.7 1.8%23 0.6 0.4 0.8 0.6 0.8%25 0.4 0.2 0.6 0.4 0.5%27 0.6 0.4 0.8 0.6 1.1%29 0.4 0.2 0.5 0.3 0.5%31 0.3 0.1 0.5 0.3 0.6%33 0.3 0.1 0.5 0.3 0.6%35 0.3 0.1 0.4 0.2 0.5%37 0.3 0.1 0.4 0.2 0.5%39 0.3 0.1 0.3 0.1 0.3%41 0.1 0.0 0.3 0.1 0.4%43 0.2 0.0 0.2 0.0 0.2%45 0.1 0.0 0.2 0.0 0.2%47 0.1 0.0 0.2 0.0 0.2%49 0.1 0.0 0.1 0.0 0.1%51 0.1 0.0 0.1 0.0 0.1%
P a/P Cu = 81.4%81.4%
etc.
etc.
Harmonics measurement on an Osram Dulux 11W compact
fluorescent lamp
Harmonics of an Osram Dulux 11W CFL and a serial impedance of
R =29.1Ω & X L=113ΩU U ² I L I L² P add/P Cu
n V V² mA mA²
1 232.7 54149.3 48.9 2391.2 3.7%3 0.6 0.4 39.1 1528.8 21.5%5 4.4 19.4 26.4 697.0 27.3%7 2.3 5.3 20.0 400.0 30.7%9 0.1 0.0 19.2 368.6 46.7%
11 0.1 0.0 16.6 275.6 52.2%13 0.1 0.0 12.7 161.3 42.7%15 0.1 0.0 11.0 121.0 42.6%17 0.1 0.0 10.2 104.0 47.1%19 0.1 0.0 8.7 75.7 42.8%21 0.1 0.0 7.7 59.3 40.9%23 0.1 0.0 7.3 53.3 44.1%25 0.1 0.0 6.1 37.2 36.4%27 0.1 0.0 4.9 24.0 27.4%29 0.1 0.0 4.2 17.6 23.2%31 0.1 0.0 3.6 13.0 19.5%33 0.1 0.0 3.0 9.0 15.3%35 0.1 0.0 3.3 10.9 20.9%37 0.1 0.0 3.1 9.6 20.6%39 0.1 0.0 2.5 6.3 14.9%41 0.1 0.0 2.5 6.3 16.4%43 0.1 0.0 2.5 6.3 18.1%45 0.1 0.0 1.9 3.6 11.4%47 0.1 0.0 1.8 3.2 11.2%49 0.1 0.0 1.9 3.6 13.6%51 0.1 0.0 1.6 2.6 10.4%
P a/P Cu = 701.7%
To some extent the transformer protects itself...
Always remember to consider it!
If the influence of the transformer upon the load did not exist, then the influence of the load upon the transformer would be nearly 9 times as high!
701.7%
etc.
etc.
For this, too, a tool is provided:
K Factor Calculator by
www.cda.org.uk
www.cda.org.uk/frontend/pubs.htm#ELECTRICAL/ENERGY%20EFFICIENCY
Rule of thumb: Select transformers 35% larger than specification by apparent power would require!
This suffices to err on the safe side even in the worst case ... and can hardly ever be wrong, since maximum efficiency always lies between 25% and 50% of rated load.
Efficiency of a 1 MVA transformer plotted against loading
98,5%
98,6%
98,7%
98,8%
98,9%
99,0%
99,1%
99,2%
99,3%
99,4%
0% 25% 50% 75% 100% 125%
Relative load h
Design with max. Cu loss and min. Fe lossDesign with min. Cu loss and max. Fe loss1 MVA oil transformer according to HD 428
Generator:uSC≈15%...40%Extreme examplebicycle dynamo:uSC≈500%!
The generator also impacts the load:
Transformer:uSC=4% / 6%
Measured response of an 11 W fluorescent lamp operated with a magnetic ballast
We are dealing with a complex problem
These effects are not isolated but are mutually inter-dependent.
Effects 2, 3 and 5 can be demonstrated on DKI’s display panel
A conventional, approximately resistive-inductive load
mA
NL1 L2 L3
Two conventional, approximately resistive-inductive loadsThree conventional, approximately resistive-inductive loads
A modern electronic load
mA
N
Two modern electronic loadsThree modern electronic loads
L1 L2 L3
Discoveries on a refurbished junction box
System voltage
(as trigger signal)
Current in the earthing
conductor between the
consumer unit
and the bonding
busbar
The TN-C system, that was perfectly adequate some years ago, is unable to meet present-day requirements
-6
-4
-2
0
2
4
6
0 5 10 15 20
t / ms
i / A
-350-300-250-200-150-100-50
050
100150200250300350
0 5 10 15 20
t / ms
u /
V
“THF”(third harmonic filter) made in
Finland
In certain situations, this affordable filter can be of help
German version “THX”
0Ω
5Ω
10Ω
15Ω
20Ω
25Ω
30Ω
35Ω
40Ω
0Hz 50Hz 100Hz 150Hz 200Hz 250Hz 300Hz
f
Z
-90°-75°-60°-45°-30°-15°0°15°30°45°60°75°90°
φ
Reactorreactance
Capacitorreactance
Parallelimpedace
Phase angle
Data and frequency response of the filter
1320 µF
875 14 m µH
The filter in use:
In a nursery running sodium vapour lamps (500 kW)In a typical office and administrative building
The EMC Transformer from Switzerland:7. The effects of stray magnetic fieldsin TN-C systems“When we build transformers, the first thing we focus on is complying with our
customers’ requirements and only then on complying with standards. Why? Because it is customer needs and not standards that offer real scope for product innovation.”
8.. In a TN-C system, problems arise from data streams and working currents mixing!
Others who have drawn attention to this source of data transmission errors include engineering ...
... and insurance companies, experts, consultants and many others besides
9. Increased corrosive damageThis (once) galvanized steel strip – the earth electrode of This (once) galvanized steel strip – the earth electrode of
a TN-C-S system – was located close a TN-C-S system – was located close to the transformer stationto the transformer station
Side facing transformer
station
Side facing transformer
station
Side facing away from transformer
station
Side facing away from transformer
station
WRONG!
RIGHT!
Working currents have no place in earthing systems and protective conductors
10. Lightning currents
Yet another skimpers’ network:The TT system...
...easily turns into an explosive »TNT system!«
Storage room
Storage room
e. g. for tri nitro tolulene
Some experts already claim for the »TN-S-S system«
Experience from a radio station:
N current 150 Hz:150 A PE current:32 AN current 450 Hz:14 A PE current:12 A
And this even in a clean TN-S systemwith a CEP!?
Some experts already claim for the »TN-S-S system«
L1
L2
L3
Fi
N
FE
PE
There is onlyone earthon earth
Don’t you believe it!
Protective earth
Operational earth
Functional earth
Power-system earth
IT earth
My earth
Your earth
So what actually is a TN-S system?As shown here, this is in the best case a»TN-S system h. c.«!
But now it is an »academic« one.
Which bridge to cross, which bridge to burn?
Sometimes you just don‘t believe what you see!
This one will soon look rather charred!
IEC 60364-5-54:543.4.3 If, from any point of the installation, the neutral and protective functions are provided by separate conductors, it is not permitted to connect the neutral conductor to any other earthed part of the installation.
Well, then the rest is no longer a Well, then the rest is no longer a miracle!miracle!
This has, at last, been addressed in EN 50174-2
Disturbingequipment
Sensitiveequipment
Disturbingequipment
Sensitiveequipment
Disturbingequipment
Sensitiveequipment
Not recommended
Transformer
Better Excellent
in the most recent edition from Sept. 2001
– the only problem is ... it’s wrong!
Though they’d already got it right in version 2:2000!
“The PEN conductor plays a dualrole in TN-C systems. Its primary role is as a protective conductor, its secondary function is that of a neutral or return wire.”
(Volker Schulze:“Vorgefertigte Klemmenblöcke für PEN-Leiter-Verlegung”[“Prefabricated terminal blocks for PEN conductor installation work”], in “de” 13/1999, p. 1050)
Well, we wouldn’t mind if a PEN con-ductor generally just looked like this...
Yesterday’s truths ...
Cables appropriate for modern electrical
installation work are available
Still too much of this around:
Standard four-core for the miserly and the short-sighted
Contact International Cablemakers FederationGraben 30A-1014 WienPhone: + 43 1 532 9640Fax: +43 1 532 9769http://www.icf.at
And has already integrated it into its Logo:4-core good, 5-core better!
One fabricator says it quite clearly right from the start:
Turn your brains on!
Take a look at EN 50174 fromFeb 2000, subclause 6.4.3:
“…it shall be considered that a PEN conductor through which the unbalanced currents and the accumulating of harmonic currents and other disturbances are transmitted cannot provide an appropriate earthing. It shall also be considered that the TT and IT mains systems need more corrective measures in particular against over voltage; therefore:
there should be no PEN within the building, i.e. the respective option in 546.2.1 of HD 384.5.54 S1:1988 should not be used.
wherever possible, the TN-S system should be used.”
Have a look at CENELEC Guide R064-004:1999-10, where it says:
For buildings which have, or are likely to have, sigificant information technology equip-ment installed, consideration shall be given to the use of separate protective conductors (PE) and neutral conductors (N) beyond the incoming supply point in order to minimize the possibility of electromagnetic problems due to the diversion of neutral current through signal cables causing damage or interference
Or at EN 50160:
Under normal operating conditions rapid voltage changes usually remain below 5% of the rated voltage, but deviations of up to 10% may under certain circumstances occur several times a day.
Under normal operating conditions the number of voltage dips lies between some tens and several thousands per year.
They usually last less than 1 s and have a retained voltage of over 40%.
Short interruptions of up to 3 minutes occur some tens up to several hundred times a year. Up to 70% of these may last for less than 1 s.
Unbalance: 95% of all 10-minute mean values must have an inverse system of less than 2% the direct system. But where there are many singe- and two-phase loads in operation, it may as well give rise up to 3%.
Switching transients usually do not exceed 1.5 kV, other transients commonly stay below 6 kV. In individual cases, however, they may be higher than that.
The voltage magnitude on the low voltage level is 230 V ±10%,measured between phase and neutral conductor in 4-conductor systemsand between phase conductors in 3-conductor systems.
5-conductor systems obviously do not exist!
Or at EN 50160:
The frequency should be between 49.5 Hz and 50.5 Hz for at least 99.5% of a given year.
Island networks not running synchronous to the UCTE mains are exempted.
So this already starts with the British isles, doesn‘t it?
Or at EN 50160:
Indeed...
...But let us have a look at a »real« island – here it comes:
Observations on Malta: Frequencies between 49.80 Hz and 50.13 Hz
While all of this discussion (and the measurement) apparently proves to be a waste, since:
Or at EN 50160:
These limits refer to normal operating conditions only, not to fault conditions.
So does the responsibility for supply drop out when power supply drops out?
Hopefully we will never ever get a supply according to EN 50160!
May be you better have a look at VDE 0100 section 100 of August 2002 first of all:
“The features given in DIN EN 50160:2000-03 represent extreme situations but do not describe the usual situation in the mains. For planning electrical installations with a normal usage it is sufficient to consider the most likely situation in the mains at the point of common coupling.”
Or how aboutIEC 60364-4-44 from 2001:
“For buildings which have, or are likely to have, significant information technology equipment installed, consideration shall be given to the use of separate protective conductors (PE) and neutral conductors (N) beyond the incoming supply point, in order to minimize the possibility of electromagnetic problems due to the diversion of neutral current through signal cables causing damage or interference.”
Fortunately, the 2002 edition now emphasises doing something rather than just thinking about doing it.
Unfortunately, the word “shall” has been replaced with “should”.
The recent amendment to EN 50310, section 6.3from Sept. 2000 appears to have got exactly the right approach at last:
“The AC distribution system inside a building shall conform to the requirements of the TN-S system. This requires that there shall be no PEN conductor inside the building, i.e. the option in 546.2.1 of HD 384.5.54 S1:1980 shall not be used.”
So why all the fuss about interference suppression? After all, the EN 61000-3-2 standard has been in force since 1 Jan. 2001 – so interference is a thing of the past!Right ...?
Class B: Portable power tools
Class A: Balanced three-phase equipment and all other equipment not classified below
Class C: Lighting equipment including lighting controls (except dimmers up to 1000 W)
Class D: Personal computers, PC monitors and televisions with an input power range from 75 to 600 W.
We now have limits on harmonic emissions for ...
The tolerances that the mains voltage has to meet are very tight:
2.0% max. permissible deviation from rated value,
0.9% max. permissible content of 3rd harmonic,
0.4% max. permissible content of 5th harmonic,
0.3% max. permissible content of 7th harmonic,
0.2% max. permissible content of 9th harmonic,
0.2% max. permissible content of even-order harmonics
0.1% max. content of harmonics of the orders 11 to 40during testing!
It’s important that these tolerances are so tight because the effect of voltage distortions is huge ...
And what exactly are the specified limits?
e.g. for Class D equipment
“For the 3rd harmonic, a Class D unit must draw no more than 3.4 mA per watt of input power.”
And how do we ensure compliance ...
No problem ...
But ... are we talking about the rated or measured input power ?
And what about the power system parameters (e.g. resistance, reactance, voltage profile) ?
... in the case of, say, a conventional PC ?
Emission limit (3rd harm.): 116 W x 3.4 mA/W =395 mA
Measured emission (3rd harmonic): 411 mA
Solution 1: Specify rated power 4 % above the measured value
Solution 2: Consider each device separately(PC: 40 W, Monitor: 60 W, Peripherals: 16 W)
Solution 3: Add an ohmic resistance in series
Difference to be “measured away”: 16 mA
Class D: Personal computers, PC monitors and tele-visions with an input power range from 75 W to 600 W
Solution 4:
The perfect PCfor all editorial staff...
Jack the power up to over 600 W by connecting a large resistive load in parallel.
Class D:Personal computers, PC monitors and televisions with an input power rangefrom 75 W to 600 W
Another means of deforming the current profile:
The phase-control dimmer
L1-N L2-N L3-N
φ 0° 0° 0°
u AV 207,1 207,1 207,1 V
U Eff 230,6 230,6 230,6 V
L1 L2 L3 N
i AV 0,235 0,235 0,235 0,000 A
IEff 0,261 0,261 0,261 0,000 A
IEff / i AV 1,114 1,114 1,114 ---
î / i Eff 1,410 1,410 1,410 --- -0,4A
-0,3A
-0,2A
-0,1A
0,0A
0,1A
0,2A
0,3A
0,4A
0 5 10 15 20
t / ms i
-350V
-250V
-150V
-50V
50V
150V
250V
350V
0 5 10 15 20
t / ms
U
L1-N L2-N L3-N
φ 60° 60° 60°
u AV 153,7 153,7 153,7 V
U Eff 205,8 205,8 205,8 V
L1 L2 L3 N
i AV 0,174 0,174 0,174 0,181 A
IEff 0,233 0,233 0,233 0,204 A
IEff / i AV 1,339 1,339 1,339 1,128
î / i Eff 1,581 1,581 1,581 1,562 -0,4A
-0,3A
-0,2A
-0,1A
0,0A
0,1A
0,2A
0,3A
0,4A
0 5 10 15 20
t / ms i
-350V
-250V
-150V
-50V
50V
150V
250V
350V
0 5 10 15 20
t / ms
U
L1-N L2-N L3-N
φ 45° 90° 135°
u AV 176,7 101,7 30,3 V
U Eff 219,9 161,3 69,5 V
L1 L2 L3 N
i AV 0,200 0,115 0,034 0,155 A
IEff 0,249 0,183 0,079 0,188 A
IEff / i AV 1,244 1,585 2,293 1,213
î / i Eff 1,479 2,016 3,251 1,702 -0,4A
-0,3A
-0,2A
-0,1A
0,0A
0,1A
0,2A
0,3A
0,4A
0 5 10 15 20
t / ms i
-350V
-250V
-150V
-50V
50V
150V
250V
350V
0 5 10 15 20
t / ms
U
No-load currents in transformers:a further – though overrated – source of current distortion
No-load current in a 630 kVA distribution transformer, excitated from the 420 V LV side!
Another means of (over)loading the neutral conductor:d.c. currents ...
Original “Fake”
... and their effects
U = 228.5 VI = 9.2 mAP = 1.82 WS = 2.09 VAQ = 1.03 VArLF = 0.87cos φ = 0.99
U = 224.3 VI = 1.26 AP = 38.0 WS = 286.0 VAQ = 283.0 VAr
No-load current in atoroidal-coretransformer
200 VA
-350-300-250-200-150-100
-500
50100150200250300350
0 10 20 30 40
t / ms
u /
V
-12-10-8-6-4-2024681012
i /
mA
VoltageCurrent
No-load current in a toroidal-coretransformer 200 VA when
running a 1500 Whairdryer athalf-powerin parallel
-350-300-250-200-150-100
-500
50100150200250300350
0 5 10 15 20 25 30 35 40
t / ms
u /
V
-6-5-4-3-2-10123456
i /
A
VoltageCurrent
The later models don‘t do it any more
The EN 61000-3-2 has at least managedto put an end to this approach.
Domestic appliances: Max. 1.05 A of 2nd harmonic
An electrical engineering wonder:d.c. current generates a.c. voltage
Spicing things up with a pinch of HF: electronic halogen lamp transformers
Decorative HF reactors
Using simplematerials to ...
... model the electrical installation in a building
A power system that doesn’t get dirty doesn’t need to be cleaned:
0
50
100
150
200
250
300
350
400
450
500
20 30 40 50 60 70 80 90 100 110 120
f / Hz
Z /
-90°
-60°
-30°
0°
30°
60°
90°
φ
XL
XC
ZS
Current and power loss as a function of voltage
0,0
0,2
0,4
0,6
0,8
1,0
1,2
1,4
1,6
1,8
0 40 80 120 160 200 240
U / V
I /
A
0
20
40
60
80
100
120
PV
/ W
I [A]Pv [W]
Power loss, capacitor voltageand total voltage as a function of current
0
50
100
150
200
250
300
350
400
450
0,00 0,25 0,50 0,75 1,00 1,25 1,50
I / A
U /
V
0
5
10
15
20
25
30
35
40
45
50
PV
/ W
Utot [V] Eff
UR [V] Eff
Pv [W]
The passive harmonic filter
made of readily available
components ...
... is both reliable ...
... and effective
before
after
Half the solution has already been implemented:
Professor Manfred Fender Wiesbaden University of Applied Sciences:
We should use more two- and three-phase rectifier loads!
L3
L1
L2
N
iu
L1
t
N
i
prU
t
u
L3
L2
t
Usek
t
ui
B2B6 B2 B2
i
t
ui
uBild d Bild c Bild b Bild a
Bild e
We should use transformers with different vector groups!
usc for zero-seq. system:
60% of rated value
usc for zero-seq. system:
5...10% of rated value
And if that’s not enough ...
Passive filter circuits will do it
So what do we need to do?In low-voltage power distribution systems: Do not use cables in which the neutral or protective earth conductor
has a reduced cross-section. Use only 5-core cable. Do not install TN-C- or TN-C-S systems. Be generous when dimensioning conductor cross-sections. This helps
to reduce line voltage drops and thus reduces the effect of current distortions on the voltage, as well as lowering energy losses (see VDE 0298 Part 100). Generously dimensioned conductors will have no problem coping with any future increase in demand.
Only use measuring instruments that display the true root mean square value (TRMS meters).
In medium-voltage power distribution systems: Use a varied mix of distribution transformers with different vector
groups.
the European Union has been providing a total of three million euros over a three year period to enable experts from across Europe to co-operate in the development of the definitive internet site covering all aspects of power quality!
To follow the latest developments visit
www.lpqi.org
and take a look at the growing body of information that has been made available by the Leonardo Power Quality Initiative.
Our aim is to develop and disseminate teaching materials in 13 languages dealing with the detection, mitigation and management of EMC problems.
Target groups include electrical technicians, engineers, those in the skilled trades, building system engineers, architects, planners as well as apprentice technicians and students and their teachers.
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