The Eects of Harmonics in
Power Systems and Methods to
Reduce or Eliminate Them
Wilbur N. Dale, Ph.D., P.E.
wilburd@rstVA.com
March 15, 2012
Overview
•
Introduction
•
Analysis of Harmonics
•
Sources of Harmonics
•
Tolerating Harmonics
•
Eliminating Harmonics
2
Introduction
•
Fourier series properties
•
Harmonic relationships
•
Fundamental Theorem of Engineering
3
Fourier Series Properties
v(t) =
i(t) =
∞ √
X
k=1
∞ √
X
2Vk cos (kωt + θk)
2Ik cos (kωt + φk)
k=1
Typically, we do not see a DC term.
We see only odd harmonics (half-wave symmetry).
4
Fourier Series Properties
v
u∞
uX
u
VRMS = t
Vk2
k=1
v
u∞
uX
u
I2k
IRMS = t
k=1
∞
X
VkIk cos (θk − φk)
P=
k=1
5
Harmonic relationships
The Fourier coecients for a 1piecewise
continuous signal decrease at the rate of O k .
Theorem 2 The Fourier coecients for a continuous
signal decrease at the rate of at least O k1 .
Theorem 3 Narrow current pulses are rich in harmonics.
Theorem 1
6
Fundamental Theorem of
Engineering
Follow the money.
Corollary 1 Revenue is good.
Corollary 2 Costs are bad.
Theorem 4
7
Analysis of Harmonics
+
√
−
2VRMS cos ωt
Non-linear
load
8
Analysis of Harmonics
+
√
−
2VRMS cos ωt
Non-linear
load
9
Analysis of Harmonics
√
p(t) = v(t)i(t) =
=
∞
X
2V1 cos(ωt)
∞ √
X
2Ik cos(kωt + φk)
k=1
V1Ik · [cos((k + 1)ωt + φk)
k=1
+ cos((k − 1)ωt + φk)]
= V1I1 cos(φ1) + V1I2 cos(ωt + φ2)
∞
X
+
V1Ik−1 cos(kωt + φk−1)
+
k=2
∞
X
V1Ik+1 cos(kωt + φk+1) = T (t) ω
k=2
10
Analysis of Harmonics
•
•
•
•
Average P = V1I1 cos(φ1).
Harmonic currents do not transfer power from the
generator to the load.
Harmonic currents do generate losses in the distribution system.
Harmonic currents cause torsional vibrations in the
generator shaft. There is a danger of fatigue fracture.
11
Analysis of Harmonics
•
•
In three phase systems, a balanced non-linear load
will have generator vibrations only at the triplen
frequencies.
If the balanced load also has odd half-wave symmetry, the generator vibration will be only at the
sextuplen frequencies.
12
Analysis of Harmonics
+
√
−
2VRMS cos ωt
Linear
load
Non-linear
load
13
Analysis of Harmonics
Transmission
line
+
√
−
2VRMS cos ωt
Linear
load
Fundamental
current
Harmonic
currents
14
Analysis of Harmonics
Transmission
line
Linear
load
Harmonic
currents
15
Analysis of Harmonics
•
•
Harmonic currents cause losses throughout the network.
Harmonic currents cause larger RMS voltage drops
in the lines.
16
Analysis of Harmonics
For inductors:
v
v
u∞
u∞
uX
uX
u
u
2
VRMS = t
Vk = t
I2kX2L
k=1
k=1
v
v
u∞
u∞
uX
uX
u
u
2
2
Ik (kωL) = t
I2kk2ω2L2
=t
k=1
k=1
v
u∞
uX
u
I2kk2
= ωLt
k=1
v
u ∞ uX
1 2 2
u
→ ωLt
O
k →∞
k=1
k
17
Analysis of Harmonics
For 3-phase neutral currents:
√
2IRMS cos (kωt + φk)
√
T
+ 2IRMS cos kω t −
+ φk
3
√
T
+ 2IRMS cos kω t +
+ φk
3
√
2πk
= 2IRMS cos (kωt + φk) 1 + 2 cos
3
√
= 3 2IRMS cos (kωt + φk) if k mod 3 = 0
IA + IB + I C =
=0
otherwise
18
Sources of Harmonics
√
+
2VRMS cos ωt
−
19
Sources of Harmonics
√
+
2VRMS cos ωt
−
20
Tolerating Harmonics
Previously:
v
u∞
uX
u
VRMS = ωLt
I2kk2
k=1
Can we reduce L?
Simple case of two parallel wires.
d
µ0
0
2 ln
+ 2 ln 2 + 1
L =
4π
δ
where d is the center to center distance between the
wires and δ is the diameter of the wires.
21
Tolerating Harmonics
•
Smallest realizable inductance occurs with d = δ.
µ0
(2 ln 2 + 1)
• L0min = 4π
• L0min
•
is not aected by wire size.
Use several smaller cables and connect in parallel.
22
Tolerating Harmonics
1015™
IEEE Recommended Practice for
Applying Low
Voltage Circuit
Breakers Used in
Industrial and
Commercial Power
Systems
Published by the
Institute of Electrical and
Electronics Engineers, Inc.
IEEE Std 1015™-2006
(Revision of
IEEE Std 1015-1997)
Authorized licensed use limited to: Wilbur Dale. Downloaded on June 30,2010 at 04:20:08 UTC from IEEE Xplore. Restrictions apply.
23
Tolerating Harmonics
•
Circuit breakers vs. fuses:
•
Capacitor banks are a low impedance path.
•
Resistance welding machines.
•
Modern microprocessor controlled breakers.
I2t.
24
Tolerating Harmonics
1100™
IEEE Recommended Practice for
Powering and
Grounding
Electronic
Equipment
Published by the
Institute of Electrical and
Electronics Engineers, Inc.
IEEE Std 1100™-2005
(Revision of
IEEE Std 1100-1999)
Authorized licensed use limited to: Wilbur Dale. Downloaded on June 30,2010 at 04:09:21 UTC from IEEE Xplore. Restrictions apply.
25
Tolerating Harmonics
K=
kX
max
(Ikk)2
15 < kmax < 34
k=1
Ik
is in per-unit of transformer ratings.
Non-linear load transformers with larger neutral conductors and temperature tolerance.
26
Eliminating Harmonics
•
•
Delta-Wye eliminates triplen harmonics by absorption.
Wye-Wye zigzag eliminates triplen harmonics and
the 5th and 7th.
27
Eliminating Harmonics
Change the load.
Power factor correction power supplies are switching
power supplies with near unit power factor and low harmonic
content. In addition, the harmonics fall o at
O 12 .
k
28