1 harmonics. 2 what are harmonics it falls into the power quality category of power systems it falls...

1 Harmonics

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.

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.

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

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.

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.

7 Load Impedance Line Impedances

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.

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

10 Now lets see how this non linear impedance causes waveform distortion.

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

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.

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

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

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

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.

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

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

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+

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 =

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

22 Again this graft shows that the harmonic components are correct

23 Lets work a real problem A half wave rectifier VRVR IRIR Graph of IRIR Voltage here is still sinusoidal Current has harmonics

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

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

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!

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.

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.

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

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

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

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

33 So now we know where IEEE 519 gets the formulas

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!!

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

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)

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)

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

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.

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

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.

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.

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

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

45 What Coopers Harmonic book says about transients

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!

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

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

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 .

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

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

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

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

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

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=

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

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

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

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 )

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=

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=

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

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

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

67 Use handwriting notes

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

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

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

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

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

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

74 ABC Phase back SCR by angle 30 deg, you can see DC voltage is reduced. Now lets use SCRs instead of Diodes

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

76 ABC Phase back SCR by angle 60 deg, you can see DC voltage is farther reduced.

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

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 + -

79 ABC Phase back SCR by angle 90 deg, you can see DC voltage is zero. This shows an inductive load + + - -

80 ABC Phase back SCR by angle 120 deg, you can see DC voltage is in regen. A V A I

81 ABC Phase back SCR by angle 180 deg, you can see DC voltage is in regen. A V A I

82 Clocks Run Fast

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

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

91 Lets see some regen schemes

93 The malfunctioning X-Ray Machine

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

95 a b B cC A I A =I a -I b IbIb IbIb IaIa IaIa O=0 O=60 O=0

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

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

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 +

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 -

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

101 Other typical harmonic spectrums

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.

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

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.

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

107 1 st harmonic= 2000000/3 12kV/1.732 = 96 amp =0.019pu From the 6 pulse spectrum 5 th = 96/5=19.24A=.004pu 7 th =96/7=13.71A=.00285pu 11 th =96/11=8.727A=.00181pu 13 th =96/13=7.384A=.00153pu 17 th =96/17=5.647A=.00117pu 19 th =96/19=5.052A=.00105pu 23 rd =96/23=4.17A=.00086pu 25 th =96/25=3.85A=.00079pu.004 2 +.00285 2 +.00181 2 +.00153 2 +.00117 2 +.00105 2 +.00086 2 +.00079 2.019 I THD = =.00579/.019 =.304=30.4%

108 Vr 0 = 1 O-[(3.646+j1.836)(0.019)]=.930 +(3.646+j5*1.836) +(3.646+j7*1.836) +(3.646+j11*1.836) +(3.646+j13*1.836) +(3.646+j17*1.836) +(3.646+j19*1.836) +(3.646+j23*1.836) +(3.646+j25*1.836) (0.004)= (0.00285)= (0.00181)= (0.00153)= (0.00117)= (0.00105)= (0.00086)= (0.00079)=.0395.038.037.036 1.0pu.930puI 1 O-[(3.646+j1.836)(0.019)] Vr=cosO-.069 Vr= 0.03488=sinO O=.03489rad Vr=.930 Vr rms =. 930 2 -.0395 2 -.038 2 *7=.924 Vr THD =.0395 2 +.038 2 *7.930 =.113=11.3%

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

119 Frequency Scan Analysis Hempfield Substation Cool Valley 12kV

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

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.

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

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

146 Meter Comparisons

147 Affects of Harmonics on Equipment Capacitors Capacitors Transformers Transformers Neutrals Neutrals

150 Derating a Transformer due to Harmonics Because of Eddy current Losses in a transformer. Because of Eddy current Losses in a transformer.

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

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?

1 harmonics. 2 what are harmonics it falls into the power quality category of power systems it falls...

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