2 What are harmonics It falls into the Power Quality category
of Power Systems It falls into the Power Quality category of Power
Systems There are three main classes of PQ There are three main
classes of PQ Under Voltages Under Voltages Over Voltages Over
Voltages Waveform Distortions Waveform Distortions Harmonics belong
to Waveform Distortion category. Harmonics belong to Waveform
Distortion category.
Slide 3
3 Remember we generate Sine Waves We dont want waveforms that
are not sine waves either current and/or especially voltage. We
dont want waveforms that are not sine waves either current and/or
especially voltage.
Slide 4
4 Three Classes of Voltage disturbances Voltage disturbances
Under Voltage Over VoltageWaveform Distortion Sags Notches Outages
Impulsive Transient Oscillatory Transient Swells Harmonics A.
Voltage B. Current
Slide 5
5 So what is Harmonics First a distorted waveform on the power
system is a waveform that is not sinusoidal such as a square wave.
First a distorted waveform on the power system is a waveform that
is not sinusoidal such as a square wave. If that waveform is
periodic i.e. it repeats itself every cycle it was shown by a man
named Fourier that it can be made up by summing a fundamental sine
wave with sine waves of integer number multiples of the frequency
of the fundamental sine wave If that waveform is periodic i.e. it
repeats itself every cycle it was shown by a man named Fourier that
it can be made up by summing a fundamental sine wave with sine
waves of integer number multiples of the frequency of the
fundamental sine wave These integer number multiples sine waves are
called harmonics. These integer number multiples sine waves are
called harmonics.
Slide 6
6 What causes Harmonics Since we generate sine waves it is not
the utility, but it is the load, which means it is the current that
is distorted, this in turn can cause distortion to the voltage by
the line impedance. Since we generate sine waves it is not the
utility, but it is the load, which means it is the current that is
distorted, this in turn can cause distortion to the voltage by the
line impedance. What cause the current to be distorted. A load that
has a non linear impedance. What cause the current to be distorted.
A load that has a non linear impedance. What is a non linear
impedance, well lets look a what is a linear impedance. Linear
impedance that we know are resistance, inductance, and capacitance.
What is a non linear impedance, well lets look a what is a linear
impedance. Linear impedance that we know are resistance,
inductance, and capacitance. There may be a phase shift between
current and voltage, but both current and voltage are sinusoidal.
There may be a phase shift between current and voltage, but both
current and voltage are sinusoidal.
Slide 7
7 Load Impedance Line Impedances
Slide 8
8 Lets see how adding sine waves of different frequency causes
waveform distortion Fundamental frequency A 3 rd harmonic 3*fund
frequency Distorted waveform Primary freq is that of fund.
Slide 9
9 Lets take the same fundamental and add the same third
harmonic but shifted 180 degrees. The waveform looks completely
different. So what harmonic frequencies the waveform has, what
amplitude the different harmonic frequencies are and what the phase
shift of them is greatly affect the resultant waveform
Slide 10
10 Now lets see how this non linear impedance causes waveform
distortion.
Slide 11
11 We understand the idea behind harmonics but what I have is a
distorted waveform how do I determine what harmonics can make that
waveform First we have to develop the general harmonic form for a
distorted waveform First we have to develop the general harmonic
form for a distorted waveform X(t)= a 0 + (a n cos n(2 f)t + b n
sin n(2 f)t n=1 8 X(t) is the distorted waveform and now we have to
figure out how to solve for a 0, a n, b n That is a lot of math so
here is the answer a0a0 = X(t)dt 1 T -T 2 T 2 anan = X(t) dt 2 T -T
2 T 2 bnbn = X(t) dt 2 T -T 2 T 2 cos n(2 f)tsin n(2 f)t
Slide 12
12 X(t)= a 0 + (a n cos n(2 f)t + b n sin n(2 f)t n=1 8 X(t)= a
0 + (c n cos [n(2 f)t + On] n=1 8 c n = a n 2 +b n 2 On=tan -1 bnbn
anan But it can be expressed instead of cosine and sine terms as a
cosine(or sine) term with a phase angle. This is the way you will
see from meters. When you hand calculate the harmonics you usually
do it so you end up with this expression, as it is what comes from
the formulas.
Slide 13
13 Also many times it is easier to work with angles instead of
time as really a sine function deals with angles in radians. We
change time to an angle in our sine function by 2 f t = O or wt=O
where w = 2 f and f= 1/T So we can rewrite our formulas as: a0a0 =
X(t)dt 1 T -T 2 T 2 anan = X(t) dt 2 T -T 2 T 2 cos n(2 f)t bnbn =
X(t) dt 2 T -T 2 T 2 sin n(2 f)t Remember T is one period or 2 and
So d =dt 2 f t = O O = X( )dOO 1 2 0 2 = X( ) cos(n ) dOO 1 0 2 O =
X( ) sin (n ) dOO 1 0 2 O
Slide 14
14 Lets work a problem 02 v -v We have already changed time
into angles = X( )dOO 1 2 0 2 a0a0 = 1 2 0 VdO-VdO 2 = 1 2 VVV=0+-2
+ a0a0 So the average value or DC component=0 and looking at the
waveform we can see that is correct
Slide 15
15 02 v -v Solving for a n = X( ) cos(n ) dOO 1 0 2 Oanan anan
1 0 V dO-V dO 2 +cos(n )O O = Vsin(n ) 0 O 1 n O n = =0 2
Slide 16
16 02 v -v Solving for b n = X( ) sin (n ) dOO 1 0 2 Obnbn bnbn
1 0 V dO-V dO 2 + sin(n )O O = Vcos(n ) 0 O 1 n O n = 2 + V = n cos
n+ 1 +cos n2 cos n = 1 for all n 2V = n cos n1 2V n cos n1but = 0
for n=2,4,6,8,.. And 4V n For n=1,3,5,7.
Slide 17
17 X ( )= a 0 + (a n cos n(2 f)t + b n sin n(2 f)t n=1 80 0 O
Putting it all together X (O ) = 4V sin + sin 3 + sin 5 + sin 7 +..
1 3 1 5 OOOO 1 7 02 v -v So this harmonic analysis is actually a
good way to mathematically represent this waveform as there is
really no way else to describe it in an equation form Well lets add
up these harmonics and see if we get a square wave
Slide 18
18 I only added up to the seven harmonic and it is starting to
resemble a square wave and I would actually get a square wave if I
went to infinity If v=1 then 4V =1.273. X (O ) = 4V sin + sin 3 +
sin 5 + sin 7 +.. 1 3 1 5 OOOO 1 7 The other cool thing is if I can
figure out how to get rid of the harmonic terms in a square wave I
can get a sine wave
Slide 19
19 0 2 2v 0 Lets see what happens when we shift the x axis 1 2
0 2VdO0 dO 2 + a0a0 = = 1 2 2VO 0 = V anan 1 0 2V dcos(n )O=O
2Vsin(n ) 0 O 1 n ==0 bnbn 1 0 2V dO sin(n )O= -cos(n ) 0 O 2V n =
n =-cos(n )O+1 For n=1,3,5,7,9, 4V n bnbn = For n=2,4,6,8,bnbn =0 X
(O ) = 4V sin + sin 3 + sin 5 + sin 7 +.. 1 3 1 5 OOOO 1 7 V+
Slide 20
20 0 v -v 2 2 3 2 Now lets move the y axis into the center of
the square wave a0a0 = 0No DC offset bnbn 1 V dO-V dO+ sin(n )O O =
2 2 - 2 2 3 Vcos(n )O 1 n O n =+ 2 - 2 2 2 3 cos(n ) V n = + 22 3 2
+ 2 2 = 0 for all n cos(n ) 2 2 3 = 2 2 =
Slide 21
21 anan 1 V dO-V dO+ cos(n )O O = 2 2 - 2 2 3 V sin(n )O 1 n
-Vsin(n )O n =+ 2 - 2 2 2 3 sin(n ) V n = - 22 3 2 + 2 2 2 3 =
-sin(n ) 2 = 2 2 sin(n ) 2 4V n = X (O ) = 4V cos - cos 3 + cos 5 -
cos 7 +.. 1 3 1 5 OOOO 1 7
Slide 22
22 Again this graft shows that the harmonic components are
correct
Slide 23
23 Lets work a real problem A half wave rectifier VRVR IRIR
Graph of IRIR Voltage here is still sinusoidal Current has
harmonics
Slide 24
24 1 2 0 V sin da0a0 =OO cos( )O 0 V 2 == V Has DC anan 1 0
Vsin dcos(n )O=OO V 0 = 1 2 sin(n+1) - sin(n-1) dOOO For n=1 have
to evaluate integral with this term = 0 = V 2 -cos(n+1) cos(n-1)OO
n+1 n-1 00 + Gotta watch this term as it will give us problems for
n=1 = V 2 1 n+1 -cos(n+1) n+1 + cos(n -1) n-1 + 1 For n=2,3,4 = V 2
1 n+1 -cos(n+1) n+1 + anan anan =0 For n=1 anan n 10 2 3 -2V 3 15
-2V 35 0 0 4 5 6 Solve for Fourier coefficients
Slide 25
25 bnbn 1 0 Vsin d sin(n )O=OO V 0 = 1 2 cos(n-1) - cos(n+1)
dOOO For n=1 have to evaluate integral with this term = 1 = V 2
-sin(n+1) sin(n-1)OO n+1 n-1 00 + Gotta watch this term as it will
give us problems for n=1 = V 2 -sin(n+1) n+1 sin(n -1) n-1 + For
n=2,3,4 = V 2 -sin(n+1) n+1 + bnbn bnbn For n=1 bnbn n 1 2 30 0 4 5
6 Solve for Fourier coefficients = V 2 0 0 0 V 2
Slide 26
26 I (O ) = V sin - cos 2 - cos 4 - cos 6 -.. OO OO V 2 2V 3 15
2V 35 + We have DC, sin, cos, odd and even harmonics, hum!
Slide 27
27 Lets look at this waveform 6 6 - 6 5 7 6 11 6 -5 6 OK why
are we looking at this waveform, does it have any practical value
if so what.
Slide 28
28 A diode Bridge ABC This is a typical front end of a drive.
You can see the waveform for one phase is like what we are going to
analyze.
Slide 29
29 6 6 - 6 5 7 6 11 6 -5 6 Lets start, by now we should start
to see that a 0 =?, yes 0 1 2 VdO-VdO+ a 0 = 6 6 5 7 6 11 6 V V 2 =
6 5 6 7 6 6 + =0 anan 1 V dO-V dO+cos(n )O O = 6 6 5 7 6 11 6
Vsin(n )O 1 n O n = 6 7 6 6 5 11 6
Slide 30
30 6 7 6 6 511 6 V n sin(n ) sin(n ) sin(n ) + sin(n ) = V n =
66 5 66 5 sin(n ) 6 5 6 =0 anan bnbn 1 V dO-V dO+ sin(n )O O = 6 6
5 7 6 11 6 Vcos(n )O 1 n O n = 6 7 6 6 5 11 6
Slide 31
31 6 7 66 511 6 V n -cos(n ) + cos(n ) + cos(n ) - cos(n ) = 6
-5 66 5 V n -cos(n ) + cos(n ) + cos(n ) - cos(n ) = cos(n ) 6 6 6
5 66 5 V n = -2cos(n ) +2cos(n ) cos(n ) 6 5 but = cos(n - n )=
cos(n )cos(n ) + sin(n )sin(n ) 666 0 2V n = cos(n ) 6 1-cos(n
)bnbn For n even, b n =0
Slide 32
32 For n odd, bnbn n 1 3 5 0 0 7 9 11 4V cos(n ) 6 bnbn = 4V
cos( ) 6 = 1.102V n b1b1 5 b1b1 7 b1b1 11 Notice no 3 rd or 9 th or
15 or odd multiples of 3s harmonics, that is a good thing
Slide 33
33 So now we know where IEEE 519 gets the formulas
Slide 34
34 So what have we seen so far We generate a 60 hertz voltage
sine wave We generate a 60 hertz voltage sine wave Waveform
distortion is caused by the load having non linear impedance,
therefore causing the current to be distorted Waveform distortion
is caused by the load having non linear impedance, therefore
causing the current to be distorted Constant cycle to cycle
waveform distortion can be represented by integer number harmonics
through Fourier Series Constant cycle to cycle waveform distortion
can be represented by integer number harmonics through Fourier
Series We have learned that this representation can lead well to
filtering out harmonics to get back to a fundamental sine wave We
have learned that this representation can lead well to filtering
out harmonics to get back to a fundamental sine wave We have
learned the math behind the harmonic analysis Ick!! We have learned
the math behind the harmonic analysis Ick!!
Slide 35
35 Lets talk about symmetry As we were going over the harmonics
we saw some had only sine terms some had only cosine terms. Some
had both. As we were going over the harmonics we saw some had only
sine terms some had only cosine terms. Some had both. Some had only
odd harmonics and some had both odd and even harmonics. Some had
only odd harmonics and some had both odd and even harmonics. Is
there any pattern? Is there any pattern? Yes, we have three types
of symmetry Yes, we have three types of symmetryEvenOdd wave
Slide 36
36 Even Symmetry Odd Symmetry sin(- )= - sin( )OO Sine wave has
odd symmetry cos(- ) = cos( )OO Cosine wave has even symmetry
Harmonics only have cosine terms Harmonics only have sine terms Has
both sine and cosine terms as it does not have either odd or even
symmetry Axis placement is everything X(t)=X(-t) X(t)= -X(-t)
Slide 37
37 wave symmetry This wave is the same As here When you have
symmetry you only odd harmonics This wave is not the same as this
Has both even and odd harmonics as it does not have symmetry
X(t)=-X(t+T/2)
Slide 38
38 What components do these waveforms have? (Sine, Cosine,
both, Odd, both odd and even, DC) DC, Sine, Odd Sine, Cosine, Odd,
Even Sine, Odd, Even DC, Cosine, Odd Sine, Cosine, Odd
Slide 39
39 So what does all this mean? Y Axis placement will greatly
simplify hand solution of harmonics by only having either sine or
cosine terms if odd or even symmetry exists. But for meters you
dont get to choose the axis placement everything is usually
referenced off of A phase voltage. So all other harmonics will have
phase angles associated with them, but I dont care to much as I am
not calculating it, but I have to be aware of it. Y Axis placement
will greatly simplify hand solution of harmonics by only having
either sine or cosine terms if odd or even symmetry exists. But for
meters you dont get to choose the axis placement everything is
usually referenced off of A phase voltage. So all other harmonics
will have phase angles associated with them, but I dont care to
much as I am not calculating it, but I have to be aware of it. If I
know the device I am concerned about and realize this device should
produce wave shapes with wave symmetry and I see both even and odd
harmonics I most likely have a problem with the device. If I know
the device I am concerned about and realize this device should
produce wave shapes with wave symmetry and I see both even and odd
harmonics I most likely have a problem with the device. If I use a
regular clamp on CT to measure current I am already eliminating DC
from my picture. This may give me an incomplete picture so I must
be aware of it. If I use a regular clamp on CT to measure current I
am already eliminating DC from my picture. This may give me an
incomplete picture so I must be aware of it.
Slide 40
40 Well we showed the harmonic component of a waveform but now
we would like to plot them instead of in the time domain lets use
the frequency domain 02 v -v X (O ) = 4V sin + sin 3 + sin 5 + sin
7 +.. 1 3 1 5 OOOO 1 7 X (O ) = 4V Cos( - ) + cos(3 - )+ cos(5 - )+
cos(7 - ) 2222 OOOO 1 3 1 5 1 7 h1234567 h 1234567 Magnitude Phase
angle
Slide 41
41 X (O ) = 4V Cos( - ) + cos(3 - )+ cos(5 - )+ cos(7 - ) 2222
OOOO 1 3 1 5 1 7 h1234567 h 1234567 Magnitude Phase angle It is
easy to see that for a period waveform the harmonics are discrete
integers and what has to be removed to get back to a fundamental
sine wave. But what happens to something that is not periodic. This
would be something like a transient waveform. Now these waveforms
create harmonics that are integer and non integer harmonics and
there can be a lot of them. However, because it is a transient it
is short lived and usually doesnt cause a problem unless resonance
is involved, such as voltage magnification we learned about in
transients class. That means we dont usually worry about harmonics
associated with transients.
Slide 42
42 It would suit us well to learn however what the transient
spectrum for a transient might look like. This is the job of the
Fourier Transform The Fourier Transform encompasses the Fourier
series. Remember the Fourier Series used for calculating the
harmonics for a periodic function. And that is discrete integer
harmonics. But a transient waveform is not periodic and exist for
only a short period of time. The Fourier Transform encompasses the
Fourier series. Remember the Fourier Series used for calculating
the harmonics for a periodic function. And that is discrete integer
harmonics. But a transient waveform is not periodic and exist for
only a short period of time. Lets look at a single square wave to
illustrate. Lets look at a single square wave to illustrate.
Slide 43
43 F (W) = f(t) e -jwt dt (2) 8 8 T 2 -T 2 V v e -jwt dt= -T 2
= -V jw e -jwT/2 -e jwT/2 e jx -e -jx 2j =sin x = 2V sin(wT/2) w =
-V jw e -jwT/2 -e jwT/2 = 2V sin(wT/2) WT/2 *T/2 VTsin(x) x Which
is a sinc function wT/2= W=2 T =2 f For f = 60 w=377 or 1 st
harmonic
Slide 44
44 OK what does this mean? DC component Amplitude of some of
the harmonics and there are infinite harmonics between each of them
X(t)= a 0 + ( b n sin n(2 f)t n=1 8 1 2 3 0 The unit step function
has infinitely many harmonics that are non integer, just so happens
that the integer harmonics are zero for this function
Slide 45
45 What Coopers Harmonic book says about transients
Slide 46
46
Slide 47
47
Slide 48
48 Again transient conditions which produce all these non
integer harmonics dont last too long and are usually not a problem
unless resonance exist so we wont deal with them again!
Slide 49
49 Lets look at the symmetrical component expressions of
harmonics as to whether they are positive, negative, or zero
sequence Why is this important. Well it is a good way to determine
whether a motor will vibrate more with different harmonics and
whether neutrals will overload Why is this important. Well it is a
good way to determine whether a motor will vibrate more with
different harmonics and whether neutrals will overload From
Symmetrical Components A B C A B C PositiveNegative Zero B A C
Slide 50
50 A B C Positive A=Vcos(wt) B=Vcos(wt-120) C=Vcos(wt+120) This
is the fundamental The second Harmonic A=Vcos2(wt)=Vcos2(wt)
B=Vcos2(wt-120)=Vcos(2wt-240) =Vcos(2wt+120)
C=Vcos2(wt+120)=Vcos(2wt+240) =Vcos(2wt-120) A B C Negative So
second harmonic is negative sequence. It rotates opposite in a
motor than the positive
Slide 51
51 The 3 rd Harmonic A=Vcos3(wt)=Vcos3(wt)
B=Vcos3(wt-120)=Vcos(3wt-360) =Vcos(3wt)
C=Vcos3(wt+120)=Vcos(3wt+360) =Vcos(3wt) Zero B A C So third
harmonic is zero sequence. It has no rotation these currents will
add in the neutral 4 th is positive 5 th is negative 6 th is zero 7
th is positive .
Slide 52
52 A=cos(wt)+1/3cos(3wt) B=cos(wt-120)+1/3cos[3(wt-120)]
C=cos(wt+120)+1/3cos[3(wt+120)] N=cos(3wt) Third harmonic adds in
the neutral
Slide 53
53 360/5=72 120 o -72 o =48 o = 2/3 of the 5 th harmonic wave
If I take one cycle of 5 th to be 360 o the 5 th from phase B
starts at the 240 o offset on phase A. So it is indeed Negative
Sequence Lets show another way the 5 th harmonic is negative
sequence
Slide 54
54 L ets talk about THD, RMS, True Power Factor. Displacement
Power Factor The industry has standardized on a term call THD or
Total Harmonic Distortion. The industry has standardized on a term
call THD or Total Harmonic Distortion. You have to specify if THD
is current or voltage You have to specify if THD is current or
voltage THDv V DC 2 +V 2 2 +V 3 2 +V 4 2 +V 5 2 +..+V N 2 V1V1 =
THD I I DC 2 +I 2 2 +I 3 2 +I 4 2 +I 5 2 +..+I N 2 I1I1 = All
quantities are RMS
Slide 55
55 RMSv V DC 2 +V 1 2 +V 2 2 +V 3 2 +V 4 2 +V 5 2 +..+V N 2 =
Same type of formula for current Again all quantities are RMS
Slide 56
56 Orthogonality Principle sin (w 1 t+ ) dtsin(w 2 t+O )O 0 T =
1/2cos (w 1 t+ )O 0 T (w 2 t+O ) 1/2cos (w 1 t+ ) dtO(w 2 t+O )+-
cos ((w 1 -w 2 ) t + )O 0 T -1/2O cos ((w 1 +w 2 ) t + )OO+dt sin
((w 1 -w 2 ) t + )OO w 1 -w 2 0 T 1 2 sin((w 1 +w 2 ) t + )OO+ - w
1 +w 2 0 T = 2 f 1 = w 1 2 = w 1 T1T1 2 = T 1 w 1 let w 2 =nw 1
Where n is an integer so T=T 1 as it has the longer period than T
2
Slide 57
57 sin ((w 1 -w 2 ) t + )OO w 1 -w 2 0 1 2 sin((w 1 +w 2 ) t +
)OO+ - w 1 +w 2 0 w1 w1 2 w1 w1 2 sin ((w 1 -nw 1 ) t+ )OO w 1 -w 2
0 1 2 sin((w 1 +nw 1 ) t+ )OO + - w 1 +w 2 0 w1 w1 2 w1 w1 2
sin((1-n)2 + ) -OO w 1 -w 2 1 2 sin((1+n)2 + )OO + - w 1 +w 2 sin(
)OO OO+- OO OO+ =0 for w 1 =w 2 for w 1 =w 2 cos ((w 1 -w 2 ) t +
)O 0 T -1/2O cos ((w 1 +w 2 ) t + )OO+dt=
Slide 58
58 cos ( )O 0 -1/2O cos ((2w 1 ) t + )OO+dt= w1 w1 2 sin(4 +
)OO + - 2w 1 sin( )OO+- OO+ OOcos ( ) T1T1 2 OO T1T1 2 = Summary:
If you integrate or want the average of two functions multiplied
together of different integer frequencies over one period- it is
0
Slide 59
59 0 T Lets see how to use this Orthogonality Principle RMS of
a function If f = f 0 + f 1 cos O + f 2 cos 2 +f 3 cos 3 + . + f n
cos nOOO 1 T f 2 dt= f 2 RMS f 2 RMS = 0 T 1 T (f 0 + f 1 cos O + f
2 cos 2 +f 3 cos 3 + . + f n cos n ) 2 OOO dt f 2 RMS = 0 T 1 T (f
0 2 + 2f 0 f 1 cos O + 2f 0 f 2 cos 2 + . + 2f 0 f n cosnOO O +f 1
2 cos 2 +2f 1 f 2 cosOcos 2 +2f 1 f 3 cos cos 3 + .+2f 1 f n cos
cosnOOOOO O +f 2 2 cos 2 2 +2f 2 f 3 cos2Ocos3 +2f 2 f 4 cos2 cos4
+ .+2f 2 f n cos2 cosnOOOOO +..+f n 2 cos 2 n )Odt Let T=2
Slide 60
60 f 2 RMS = 0 T 1 T f 0 2 + 2f 0 f 1 cos O + 2f 0 f 2 cos 2 +
. + 2f 0 f n cosnOO O + f 1 2 cos 2 + 2f 1 f 2 cosOcos 2 + 2f 1 f 3
cos cos 3OOO OOO + f 2 2 cos 2 2 + 2f 2 f 3 cos2Ocos3 O O OOO +..+
f n 2 cos 2 nOdt 0 00 0 0 0 00 0 f12f12 2 f22f22 2 fn2fn2 2 f02f02
f12f12 2 f22f22 2 f32f32 2 fn2fn2 2 f02f02 f 2 RMS = + ++++ f 1 2 2
f 2 2 f 3 2 f n 2 f02f02 f RMS = + ++++ 222 0 T 0 T 0 T 0 T + 2f 2
f 4 cos2 cos4 +.+ 2f 2 f n cos2 cosn 0 T 0 T 0 T 0 T + .+ 2f 1 f n
cos cosn 0 T 0 T 0 T 0 T dt
Slide 61
61 Power with waveform distortion S=V*I If V=V 1 sin(wt+O 1 ) +
V 2 sin(2wt+O 2 ) + V 3 sin(3wt+O 3 ) I=I 1 sin(wt+O 1 ) + I 2
sin(2wt+O 2 ) + I 3 sin(3wt+O 3 )and Then instantaneous power =
V*I=V 1 I 1 sin(wt+O 1 )sin(wt+O 1 )+V 1 I 2 sin(wt+O 1 )sin(2wt+O
2 )+ V 1 I 3 sin(wt+O 1 )sin(3wt+O 3 )+V 2 I 1 sin(2wt+O 2
)sin(wt+O 1 )+ V 2 I 2 sin(2wt+O 2 )sin(2wt+O 2 )+V 2 I 3 sin(2wt+O
2 )sin(3wt+O 3 )+ V 3 I 1 sin(3wt+O 3 )sin(wt+O 1 )+V 3 I 2
sin(3wt+O 3 )sin(2wt+O 2 )+ V 3 I 3 sin(3wt+O 3 )sin(3wt+O 3 )
Slide 62
62 V 1 I 1 sin(wt+O 1 )sin(wt+O 1 ) +V 1 I 2 sin(wt+O 1
)sin(2wt+O 2 ) + V 1 I 3 sin(wt+O 1 )sin(3wt+O 3 ) +V 2 I 1
sin(2wt+O 2 )sin(wt+O 1 ) + V 2 I 2 sin(2wt+O 2 )sin(2wt+O 2 ) +V 2
I 3 sin(2wt+O 2 )sin(3wt+O 3 ) + V 3 I 1 sin(3wt+O 3 )sin(wt+O 1 )
+V 3 I 2 sin(3wt+O 3 )sin(2wt+O 2 ) + V 3 I 3 sin(3wt+O 3
)sin(3wt+O 3 ) = 0 T 1 T 0 T 0 T 0 T 0 T 0 T 0 T 0 T 0 T 0 T 1 T dt
Average PowerV*I dt P=
Slide 63
63 V 1 I 1 sin(wt+O 1 )sin(wt+O 1 ) +V 1 I 2 sin(wt+O 1
)sin(2wt+O 2 ) + V 1 I 3 sin(wt+O 1 )sin(3wt+O 3 ) +V 2 I 1
sin(2wt+O 2 )sin(wt+O 1 ) + V 2 I 2 sin(2wt+O 2 )sin(2wt+O 2 ) +V 2
I 3 sin(2wt+O 2 )sin(3wt+O 3 ) + V 3 I 1 sin(3wt+O 3 )sin(wt+O 1 )
+V 3 I 2 sin(3wt+O 3 )sin(2wt+O 2 ) + V 3 I 3 sin(3wt+O 3
)sin(3wt+O 3 ) 1 T 0 T 0 T 0 T 0 T 0 T 0 T 0 T 0 T 0 T dt 0 0 0 0
00 T*V 1 I 1 cos(O 1 -O 1 ) T*V 2 I 2 cos(O 2 -O 2 ) T*V 3 I 3
cos(O 3 -O 3 ) 2 2 2 P=
Slide 64
64 V 1 I 1 cos(O 1 -O 1 ) 2 V 2 I 2 cos(O 2 -O 2 ) 2 V 3 I 3
cos(O 3 -O 3 ) 2 ++ P= 2 V1I1V1I1 V 1 I 1 22 V RMS1 *I RMS1 == Same
for the other terms
Slide 65
65 Lets see how Power Factor, and VARS are affected by
Harmonics First Power Factor can be corrected for only 60 hertz
using a capacitor to unity, but if the waveform is harmonically
distorted you cant correct it First Power Factor can be corrected
for only 60 hertz using a capacitor to unity, but if the waveform
is harmonically distorted you cant correct it Things we learn about
the Distortion Factor, True Power Factor and Harmonic VARS really
dont mean much as long I realize a Cap cant correct it. Things we
learn about the Distortion Factor, True Power Factor and Harmonic
VARS really dont mean much as long I realize a Cap cant correct it.
But we have to learn about these terms as they are talked
about(though not really used) in industry But we have to learn
about these terms as they are talked about(though not really used)
in industry
Slide 66
66 Voltage I-inductor I-resistor I-capacitor VARS Watts V I
Power Factor Angle VA=V*I Watts=V*Icos(angle) V I
VARS=V*Isin(angle) Load is resistive and reactive Power Triangle
for single frequency case Voltage and current are RMS
quantities
Slide 67
67 Use handwriting notes
Slide 68
68 Why is power electronics so prevalent today Power
electronics allows you to control voltage/current/power to the load
without introducing losses. It also allows for easier frequency
conversion to control the speed of an induction motor. Power
electronics allows you to control voltage/current/power to the load
without introducing losses. It also allows for easier frequency
conversion to control the speed of an induction motor. Lets look at
voltage control Lets look at voltage control
Slide 69
69 V out I R1R1 R2R2 V =VR 2 R 1 +R 2 P=V out II= V R 1 +R 2 P=
V 2 R 2 (R 1 +R 2 ) 2 For R 1 =0 P=V 2 R2R2 For R 1 = 8 P= 0 V out
=0 I=0 For minimum Power to R 2 0=dP =d dR 2 V 2 R 2 (R 1 +R 2 ) 2
= V2V2 -2V 2 (R 1 +R 2 ) 2 R2R2 (R 1 +R 2 ) (R 1 +R 2 ) 4 (R 1 +R 2
)-2R 2 =0R 1 =R 2 P min =V 2 4R 2 The old way
Slide 70
70 V out I R2R2 V The new way P= 1 2 0 V2V2 R2R2 Cos 2 OdO 1
(1+cos2O) 2 =V 2 (O+sin2O) R2R2 42 0 = V2V2 4R 2 No loss in an
ideal diode
Slide 71
71 Lets Examine how Equipment produces Harmonics Electronic
Equipment is becoming the big offender so we will concentrate on
it. Electronic Equipment is becoming the big offender so we will
concentrate on it. ABC Rectifier- This is the prime building block
for most of the power electronic equipment being used DC
Slide 72
72 A diode Bridge ABC A B C 135 246 1-41-63-63-25-25-4 Load The
load can be resistive or inductive or a source when in regen We are
going to represent the load as a constant current as it makes the
understanding easier, but it is only a constant current in a
current source drive This is a fairly orderly transfer from one
diode to the next as the voltages are equal at that time so
commutation notching is not a problem here
Slide 73
73 Now if you take the rectifier of the DC drive and then build
an inverter off of the DC link you can build an AC wave and can
vary the frequency which means you can control the speed of an
induction motor ABC ABC
Slide 74
74 ABC Phase back SCR by angle 30 deg, you can see DC voltage
is reduced. Now lets use SCRs instead of Diodes
Slide 75
75 Commutation notches at 30 firing angle A-phase to ground A-B
phase to phase voltage All the phase to ground voltages Shows the
currents so you know when transfer occurs
Slide 76
76 ABC Phase back SCR by angle 60 deg, you can see DC voltage
is farther reduced.
Slide 77
77 Commutation notches at 60 firing angle All the phase to
ground voltages A-phase to ground A-B phase to phase voltage Shows
the currents so you know when transfer occurs
Slide 78
78 ABC Phase back SCR by angle 90 deg, you can see you get
incomplete conduction for a resistive load This shows a resistive
load + -
Slide 79
79 ABC Phase back SCR by angle 90 deg, you can see DC voltage
is zero. This shows an inductive load + + - -
Slide 80
80 ABC Phase back SCR by angle 120 deg, you can see DC voltage
is in regen. A V A I
Slide 81
81 ABC Phase back SCR by angle 180 deg, you can see DC voltage
is in regen. A V A I
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82 Clocks Run Fast
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83
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84
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86
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87
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88
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89 Switch Mode Power Supplies Diode & Capacitor found in
all types of office automation equipment and controls: PCs,
printers, FAX, copiers, controllers, etc. Diode & Capacitor
found in all types of office automation equipment and controls:
PCs, printers, FAX, copiers, controllers, etc. Current THD in 70%
to 90% range Current THD in 70% to 90% range Displacement power
factor near unity, true power factor is low Displacement power
factor near unity, true power factor is low
Slide 90
90 In all cases the current waveform looked the same it was
just shifted with respect to the voltage So the harmonics were the
same for the current independent of the phase angle So the
harmonics were the same for the current independent of the phase
angle For this case the harmonics are defined as For this case the
harmonics are defined as h=K*P+1 where P is the pulse number and K
is an integer from 1,2,3,, h=K*P+1 where P is the pulse number and
K is an integer from 1,2,3,, Magnitude is I 1 /h Magnitude is I 1
/h _ 8 This is a 6 pulse drive so using the formula the harmonics
are 5,7,11,13,17,19,23,25, Magnitudes are I 1 /5,I 1 /7.I 1
/11
Slide 91
91 Lets see some regen schemes
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92
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93 The malfunctioning X-Ray Machine
Slide 94
94 Lets look at a 12 pulse device ABC 135 246 ABC 135 246 + -
We have already looked at this one and know the harmonics this is a
6 pulse Lets see what the primary current harmonics are for this
section a a b b B c c C A A C B
Slide 95
95 a b B cC A I A =I a -I b IbIb IbIb IaIa IaIa O=0 O=60
O=0
Slide 96
96 B C A I A =I a -I b O=0 a b c Adding the two together After
adjusting the turns ratio Of the delta wye by to get the same
output voltage 3 ab c
Slide 97
97 33 33 2 45 Odd symmetry no DC bn= sin(nO)dO 0 3 2 3 2 3 4 3
4 3 5 3 5 2 3 3 +2+ sin(nO)dO - + +2 sin(nO)dO- - + I = 1+cos(n ) 4
I n 3 For n odd Need the harmonic components for this waveform
Slide 98
98 33 33 2 45 6 6 5 6 7 6 11 = 1+cos(n ) 4 I n 3 cos(n ) 6 bnbn
= n bnbn 3 4I cos(n ) 6 bnbn = n 1+cos(n ) 3 3 +
Slide 99
99 4I cos(n ) 6 bnbn = n 1+cos(n ) 3 3 + bnbn n 1 3 5 0 07 9 11
4I = 2.205 b1b1 11 0 0 b1b1 13 h=K*P+1 where P is the pulse number
and K is an integer from 1,2,3,, h=K*P+1 where P is the pulse
number and K is an integer from 1,2,3,, Magnitude is I 1 /h
Magnitude is I 1 /h This is a 12 pulse drive so using the formula
the harmonics are 11,13,23,25,35,37, Magnitudes are I 1 /11,I 1
/13,I 1 /23 -
Slide 100
100 a b B c C A I A =I a -I b a b c What happens if I buy a
delta wye transformer as shown? The resultant waveform is screwed
up and you wont get harmonic cancellation. This is a problem for
people who make a poor mans 12 pulse
Slide 101
101 Other typical harmonic spectrums
Slide 102
102 Voltage Distortion Harmonic producing devices require
sinusoidal voltage and draw non sinusoidal currents. Therefore it
is the current that has harmonic terms, so where do we get harmonic
voltages? Harmonic producing devices require sinusoidal voltage and
draw non sinusoidal currents. Therefore it is the current that has
harmonic terms, so where do we get harmonic voltages? Harmonic
Voltage Distortion is the result of harmonic currents flowing
through line impedance. This produces harmonic voltage drop.
Harmonic Voltage Distortion is the result of harmonic currents
flowing through line impedance. This produces harmonic voltage
drop. Harmonic voltage distortion is actually worst on the power
system than harmonic current flow because this voltage is what we
give to other customers. And if it is not sinusoidal then even if
the load is linear it will draw harmonic currents. Harmonic voltage
distortion is actually worst on the power system than harmonic
current flow because this voltage is what we give to other
customers. And if it is not sinusoidal then even if the load is
linear it will draw harmonic currents.
Slide 103
103 Harmonic Load Harmonic currents flow back to the substation
as it has a lot lower impedance than the other connected loads.
Voltage distortion comes from this harmonic current flowing through
the line impedance
Slide 104
104
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105 Vr V 60Hz Vr = V-I*Z This is the typical way we model
harmonic flow and the resultant voltage distortion. Additional
loads can be modeled either as harmonic producing or linear just by
adding it to the model. The harmonics are represented as current
sources.
Slide 106
106 For a system with a line impedance to the load of 5kM of #2
and with the installation of a 6 pulse converter 2MVA. Find the
voltage distortion and current distortion for out to the 25
harmonic. At 12kV unity Power factor Only need the positive
sequence impedance as this is a balanced three phase load. For #2 R
1 =1.0501 /kM X 1 = 0.5288/kM Z b = 12 2 /100 I b
=100000/(1.732*12) 57111317192325 3.646 j1.836
109 How to determine if a load is a harmonic sink or a harmonic
source Just like in directional relaying you need to have a
reference and that is the harmonic voltage. Then you can determine
direction Just like in directional relaying you need to have a
reference and that is the harmonic voltage. Then you can determine
direction
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119 Frequency Scan Analysis Hempfield Substation Cool Valley
12kV
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124
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125 Prior to any component failure, the circuit breaker
protecting the capacitors tripped on ground fault on several
occasions. This was indicated by a latching ground fault. Prior to
any component failure, the circuit breaker protecting the
capacitors tripped on ground fault on several occasions. This was
indicated by a latching ground fault. The initial failure of the
unit occurred either during the week or on the first shift (12am-8
am) on Saturday. A mine electrician stopped at the site on Saturday
and observed smoke coming out of the station. Visible damage to the
unit was as follows: three banks of fuses were blown and destroyed,
several fuseholders were destroyed, and power cables feeding the
capacitors were damaged. The damage was repaired and power
reapplied to the unit at approximately 4 a.m. on Sunday. At mine
start-up (belt drives starting) at 10 p.m. on Sunday, the unit
failed a second time. During the failure, all four sets of fuses
protecting the capacitors failed. Again several of the fuses and
fuseholders were destroyed. A ground fault indication was present
for the circuit breaker, but the breaker had not tripped.
Additional investigation of the failure revealed that one of the 15
kV oil switches and one of the 600 kvac capacitors had failed.
These had not been tested on Saturday and may have failed in the
initial incident. The initial failure of the unit occurred either
during the week or on the first shift (12am-8 am) on Saturday. A
mine electrician stopped at the site on Saturday and observed smoke
coming out of the station. Visible damage to the unit was as
follows: three banks of fuses were blown and destroyed, several
fuseholders were destroyed, and power cables feeding the capacitors
were damaged. The damage was repaired and power reapplied to the
unit at approximately 4 a.m. on Sunday. At mine start-up (belt
drives starting) at 10 p.m. on Sunday, the unit failed a second
time. During the failure, all four sets of fuses protecting the
capacitors failed. Again several of the fuses and fuseholders were
destroyed. A ground fault indication was present for the circuit
breaker, but the breaker had not tripped. Additional investigation
of the failure revealed that one of the 15 kV oil switches and one
of the 600 kvac capacitors had failed. These had not been tested on
Saturday and may have failed in the initial incident. Following the
second failure, the damaged capacitor bank was removed from service
and the remaining capacitors put in service. One week later one of
the fuses protecting the 450 kvac capacitor bank blew. The fuse was
replaced and power restored. Following the second failure, the
damaged capacitor bank was removed from service and the remaining
capacitors put in service. One week later one of the fuses
protecting the 450 kvac capacitor bank blew. The fuse was replaced
and power restored. Two weeks after the second failure, the
substation manufacturer replaced the damaged oil switch and
capacitor. AT that time, the fuses were replaced with fuses rated
at 50 amperes for the 450 kvac bank and 60 amperes for the 600 kvac
banks. Two weeks after the second failure, the substation
manufacturer replaced the damaged oil switch and capacitor. AT that
time, the fuses were replaced with fuses rated at 50 amperes for
the 450 kvac bank and 60 amperes for the 600 kvac banks. On
December 20, the unit failed a third time. The failure occurred
during mine start-up at approximately 10 p.m. Again, three banks of
fuses were destroyed along with fuseholders and wiring. The ground
fault relay tripped, but the circuit breaker did not open. The
fault was eventually cleared by fuses on the primary of the West
Penn Power transformer. On December 20, the unit failed a third
time. The failure occurred during mine start-up at approximately 10
p.m. Again, three banks of fuses were destroyed along with
fuseholders and wiring. The ground fault relay tripped, but the
circuit breaker did not open. The fault was eventually cleared by
fuses on the primary of the West Penn Power transformer. During
subsequent testing, the capacitor circuit breaker operated
property. The capacitors and oil switches were Hi- Pot tested and
found to be within manufacturers tolerances. The damage was again
repaired and the fuses replaced with a McGraw Edison Fuse typically
found on capacitors supplied by the capacitor manufacturer. During
subsequent testing, the capacitor circuit breaker operated
property. The capacitors and oil switches were Hi- Pot tested and
found to be within manufacturers tolerances. The damage was again
repaired and the fuses replaced with a McGraw Edison Fuse typically
found on capacitors supplied by the capacitor manufacturer.
Following the latest failure, two fo the capacitor banks were
removed from service. The belts have been started several times
since that time without further failures. Following the latest
failure, two fo the capacitor banks were removed from service. The
belts have been started several times since that time without
further failures. The only indication of the amount of capacitance
on the system at failure would be the number of blown fuses. The
controller for the capacitor bank would have continued to call for
additional capacitors to correct the power factor even after the
failure. If one or more of the banks were de-energized at the time
of the failure, they would have been turned on by the controller.
The only indication of the amount of capacitance on the system at
failure would be the number of blown fuses. The controller for the
capacitor bank would have continued to call for additional
capacitors to correct the power factor even after the failure. If
one or more of the banks were de-energized at the time of the
failure, they would have been turned on by the controller. A coal
mine problem
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133
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134
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135
Slide 136
136 Harmonic Distortion Limits As a Utility, harmonics are
going to affect other customers and therefore it is prudent to
establish harmonic limits. As a Utility, harmonics are going to
affect other customers and therefore it is prudent to establish
harmonic limits. Basically these limits come from IEEE 519
Basically these limits come from IEEE 519 Primarily Current
harmonics produced by a customers non linear load dont affect other
customers, but they add additional stress to transformers and wires
that are in series with the equipment which is usually utility
owned. Most of the time our faculties are sized with some excess
capacity that this does not cause a problem. Primarily Current
harmonics produced by a customers non linear load dont affect other
customers, but they add additional stress to transformers and wires
that are in series with the equipment which is usually utility
owned. Most of the time our faculties are sized with some excess
capacity that this does not cause a problem. However, if to much
current harmonics are injected into the system it could cause
excessive voltage distortion, which will affect other customers
loads. However, if to much current harmonics are injected into the
system it could cause excessive voltage distortion, which will
affect other customers loads.
Slide 137
137
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139
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Slide 144
144 RMS Voltage and Current RMS is the equivalent heat through
a resistor from a DC voltage. RMS is the equivalent heat through a
resistor from a DC voltage. For a pure sine wave RMS = Peak Voltage
divided by Square root of 2. For a pure sine wave RMS = Peak
Voltage divided by Square root of 2. Waveform distortion causes RMS
voltage changes. Can no longer divide by Square root of 2 Waveform
distortion causes RMS voltage changes. Can no longer divide by
Square root of 2
Slide 145
145 Measuring Voltage and Current Not all meters are created
equal Not all meters are created equal Fluke 77 and Fluke 87 will
not provide the same voltage reading if the waveform is distorted.
Fluke 77 and Fluke 87 will not provide the same voltage reading if
the waveform is distorted. The Fluke 87 is a true RMS voltmeter The
Fluke 87 is a true RMS voltmeter
Slide 146
146 Meter Comparisons
Slide 147
147 Affects of Harmonics on Equipment Capacitors Capacitors
Transformers Transformers Neutrals Neutrals
Slide 148
148
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149
Slide 150
150 Derating a Transformer due to Harmonics Because of Eddy
current Losses in a transformer. Because of Eddy current Losses in
a transformer.
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151
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154
Slide 155
155 Filter Building Control of harmonics is sometimes necessary
because of excessive current injection Control of harmonics is
sometimes necessary because of excessive current injection Or
because of excessive Voltage Distortion Or because of resonance
Therefore a Harmonic Filter is sometimes necessary
Slide 156
156 Lets look at a situation where harmonics are causing
excessive voltage distortion Is it because of high current
injection? Is it because of high current injection? Or is it
because of Resonance? Or is it because of Resonance?