Harmonics

[Grouch1980]

Charlie has a great explanation on harmonics in Post #9 in the following thread:

Delta and 3rd harmonics

Remembering that you used to remember... I recall that there was a derivation of the benefits of using a delta connection to reduce/eliminate 3rd harmonics. It's probably next to my car keys and a couple of unpaid bills... Does anyone recall or can name the reference? Thanks. Matt

forums.mikeholt.com

All the even harmonics disappear. The triplen harmonics only include the 3rd, 9th, 15th harmonic, and so on. My question is, what happened to the 5th harmonic, 7th harmonic, 11th harmonic, etc?

 

[wwhitney]

All the even harmonics disappear. The triplen harmonics only include the 3rd, 9th, 15th harmonic, and so on. My question is, what happened to the 5th harmonic, 7th harmonic, 11th harmonic, etc?

Those harmonics will also tend to cancel in the neutral of a 3 phase wye system.

E.g. if you have 3 identical non-linear loads supplied A-N, B-N, and C-N, then 5th harmonics of the current would add in the neutral to be proportional to something like sin(5*t) + sin(5*(t+1/3 cycle)) + sin(5*(t+2/3 cycle)) [where something like means I left out the factor of 2*pi/60 for 60 Hz, and "cycle" means 2*pi radians]. And if you try graphing that sum [after adding the factors I left out], you find that it is identically zero. Which happens whenever the order of the harmonic n (5 in this case) is relatively prime to the number of phases P (3 in this case).

Cheers, Wayne

 

[Grouch1980]

[where something like means I left out the factor of 2*pi/60 for 60 Hz, and "cycle" means 2*pi radians].

where something like what was left out? I could not follow this.

 

[Hv&Lv]

The harmonics multiply by the phase shift.
30 degrees times the fifth harmonic.
30--150 is 180.
The 7th is a 210 degree shift.
The shift is determined by the sequence.

 

[winnie]

The harmonics multiply by the phase shift.

This.

You probably learned in early schooling (in a music class) that harmonics are a multiple of the frequency. But the phase shift of the harmonic components gets multiplied by the same amount.

This key point is better understood by thinking of time shifts first. Consider some non-linear load, say a triac dimmer running some lamps. There is a sharp change in current at a fixed _time_ after the 60 Hz zero crossing. This sharp change in current is associated with a huge number of harmonics, all tied to the timing of the 60 Hz zero crossing.

Take that same load and put it on a different phase (on a 3 phase system). The zero crossing, the sudden current change, and all of the harmonics are shifted by 5.556ms, exactly 1/3 of the 60Hz AC cycle.

Now express that 5.556ms as degrees of a cycle. It is 120 degrees of a 60Hz AC cycle, 240 degrees of a 120Hz harmonic cycle, and 360 degrees of a 180Hz harmonic cycle. In a cyclic system, 360 degrees is the same as 0 degrees, 'in phase'. Thus the problem with any triplen harmonic in a three phase system.

Note that the above analysis depends on the loads having similar waveform shapes and thus the same relation between harmonic timing and fundamental timing. There is no rule that says that this relationship must be fixed, but many common non-linear loads have similar harmonic timing.

By a similar analysis, even harmonics cause problems on split single phase systems. However most large non-linear loads don't generate much even harmonic content, so practically this is not a problem.

The 5th, 7th, etc harmonics are balanced with respect to the neutral and simply balance, but they can cause problems in 3 phase motors. Each harmonic creates its own rotating magnetic field, superimposed upon the desired fundamental field. These can reduce motor efficiency.

 

[wwhitney]

However most large non-linear loads don't generate much even harmonic content,

Can you elaborate on why this is the case?

Thanks, Wayne

 

[Grouch1980]

Those harmonics will also tend to cancel in the neutral of a 3 phase wye system.

E.g. if you have 3 identical non-linear loads supplied A-N, B-N, and C-N, then 5th harmonics of the current would add in the neutral to be proportional to something like sin(5*t) + sin(5*(t+1/3 cycle)) + sin(5*(t+2/3 cycle)) [where something like means I left out the factor of 2*pi/60 for 60 Hz, and "cycle" means 2*pi radians]. And if you try graphing that sum [after adding the factors I left out], you find that it is identically zero. Which happens whenever the order of the harmonic n (5 in this case) is relatively prime to the number of phases P (3 in this case).

Cheers, Wayne

So it's all the harmonics that are prime numbers that cancel out on the neutral? Similar to the even harmonics?

 

[wwhitney]

where something like what was left out? I could not follow this.

OK, let me try again and fix an error:

sin(t) has period 2*pi, as its argument is in radians. For simplicity, let's use sin1(t) to mean a 1 Hz sinewave, so sin1(t) = sin(2*pi*t). Likewise sin60(t) means a 60 Hz sinewave, so sin60(t) = sin1(60t) = sin(2*pi*60*t). [That's the error in my earlier post, I had 2*pi/60.]

For the actual application we would be working with sin60(t), but the presentation will be clearer if we just work with sin1(t), 1 Hz AC instead of 60 Hz AC. [Or if you prefer, we can just measure time t in terms of cycles, so t = 1 means 1/60 seconds.] So for 3 phase power, the voltage waveforms would be (up to a constant scaling) sin1(t), sin1(t+1/3), and sin1(t+2/3). I.e. the second function reaches its maximum 1/3 cycle before the first function, and the third function reaches its maximum 2/3 cycle before (equivalently 1/3 cycle after) the first function.

The 5th harmonic of sin1(t) is sin1(5t) (which is sin5(t), but that's not helpful for this point). The 5th harmonic of sin1(t+1/3) is sin1(5*(t+1/3)) = sin1(5t+5/3). So the phase shift has changed from 1/3 to 5/3, which is what is meant in posts 4 and 5 by "the harmonics multiply the phase shift." Of course, phase shift is only defined up to an integer number of cycles, so a 5/3 cycle phase shift is the same as a 2/3 cycle phase shift, but we can see that it arrives by multiplication by 5.

Likewise the 5th harmonic of sin1(t+2/3) is sin1(5*(t+2/3)) = sin1(5t+10/3), a phase shift of 10/3 cycles, which is the same as 1/3 cycles. So the act of taking the 5th harmonic has transformed the phase shifts from the starting collection of (0,1/3,2/3) to the collection (0,5/3,10/3) = (0,2/3,1/3). Adding up sinewaves with a given even set of phase shifts around one cycle like this always gives you zero.

You can also see that the 3rd harmonic behaves differently. The phase shifts starting as (0,1/3,2/3) becomes (0,1,2) = (0,0,0). Adding these up just gives you a factor of 3, there is no cancellation. The same will happen for harmonics of order any multiple of 3.

Cheers, Wayne

 

[wwhitney]

So it's all the harmonics that are prime numbers that cancel out on the neutral? Similar to the even harmonics?

Harmonics whose order is relatively prime (largest common divisor is 1) to the number of phases. So for 3 phase, it would be any order other than a multiple of 3. 6th harmonics, if present, would add in the neutral. But for reasons I'm uncertain of, apparently the loads usually don't generate even harmonics.

If you had a crazy power system using 15 phase power, then both 3rd and 5th harmonics would be a problem in terms of current adding in the neutral.

Cheers, Wayne

 

[winnie]

Can you elaborate on why this is the case?

I think it pretty much boils down to the waveform symmetry of common non-linear loads.

If a waveform like a balanced square wave then it has lots of odd harmonics. If it looks like a sawtooth wave then it has both even and odd harmonics.

If you have a full bridge rectifier feeding a capacitor filter, then you get current pulses symmetrically at the peak of the fundamental sine wave, and lots of odd harmonics. This pretty much describes the majority of non-linear loads.

If you have a 'half wave' rectifier, then you get current pulses only on one side of the fundamental sine wave, and lots of even harmonics.

I'd expect even harmonics in a triac dimmer. But I don't think that half wave rectifiers show up in large loads. I don't have information on large banks of triac dimmers; but if the different dimmers have _different_ dimming then each dimmer will have a different waveform and different harmonics phase angles.

Anyone have info on the harmonic content of large SCR phase angle controlled systems?

-Jonathan

 

[winnie]

Harmonics whose order is relatively prime (largest common divisor is 1) to the number of phases. So for 3 phase, it would be any order other than a multiple of 3. 6th harmonics, if present, would add in the neutral. But for reasons I'm uncertain of, apparently the loads usually don't generate even harmonics.

If you had a crazy power system using 15 phase power, then both 3rd and 5th harmonics would be a problem in terms of current adding in the neutral.

Cheers, Wayne

I've actually run high phase order motors. (9, 17, and 18 phase, ahh research...)

If you have a 15 phase power system, 3rd and 5 harmonics would balance/cancel on the neutral. For a _symmetric_ system, you only lose cancellation when the harmonic is equal to a multiple of the phase count.

-Jonathan

 

[wwhitney]

If you have a 15 phase power system, 3rd and 5 harmonics would balance/cancel on the neutral. For a _symmetric_ system, you only lose cancellation when the harmonic is equal to a multiple of the phase count.

Ah, thank you for the correction.

Let me switch from 15 = 3*5 to 6 = 2*3 so I don't have to write as much. Then for the 2nd order harmonics on a six-phase system, the starting phase shifts of (0, 1/6, 2/6, 3/6, 4/6, 5/6) would become (0, 2/6, 4/6, 6/6, 8/6, 10/6) = (0, 1/3, 2/3, 0, 1/3, 2/3). Which is still (two copies of) a set of evenly spaced phase shifts around the circle, and so the sum will still be zero. Like wise for 3rd order harmonics, which gives you final phase shifts of (0, 1/2, 0, 1/2, 0, 1/2).

Cheers, Wayne

 

[Grouch1980]

Harmonics whose order is relatively prime (largest common divisor is 1) to the number of phases. So for 3 phase, it would be any order other than a multiple of 3. 6th harmonics, if present, would add in the neutral. But for reasons I'm uncertain of, apparently the loads usually don't generate even harmonics.

I thought 6th order harmonics, since even, would cancel out and not add up in the neutral. Since the even harmonics are symmetrical but opposite waves.

I'm going to have to read more into this.