20
\$\begingroup\$

In most books it's said that the harmonic content in an AC line does not transfer power (just the fundamental frequency does), but no explanation. It seems intuitive, but why is that true?

Edit: In the context of electrical power, so I'm imagining a distorted current waveform and a sine wave voltage.

\$\endgroup\$
  • \$\begingroup\$ harmonic content? do you mean harmonic current? \$\endgroup\$ – Jasen Jun 4 at 2:24
  • 4
    \$\begingroup\$ The book is wrong, simple as that. It's even more wrong today because most devices rectify the AC input to DC before using it. \$\endgroup\$ – Navin Jun 4 at 3:32
  • 1
    \$\begingroup\$ harmonic current on an AC line with single frequency sinusoidal voltage does provide usable power to the load. a rectifier is not going to change that. \$\endgroup\$ – Jasen Jun 4 at 4:37
4
\$\begingroup\$

It's true because the voltage is sinusoidal and

sin(a).sin(b) = 1/2(cos(a-b)+cos(a+b))

And only in the case where a=b does the result have a mean value that is not zero.

so all harmonics give a result that has no effect on the real power

\$\endgroup\$
  • 1
    \$\begingroup\$ Yes you got the point, in simpler words harmonics are orthogonal each other \$\endgroup\$ – carloc Jun 4 at 5:43
  • \$\begingroup\$ yeah is is just a somewhat hand-wavy proof of that. \$\endgroup\$ – Jasen Jun 4 at 8:31
28
\$\begingroup\$

It's an overly-general statement. Obviously, with a resistive load, all frequencies transfer power.

It's really a statement about rotating machinery specifically (motors and generators). For these devices, energy at frequencies other than the fundamental is just as likely to oppose the work being done as aid it. Also, the energy of high frequencies is often wasted in the form of unwanted eddy currents, etc.

\$\endgroup\$
  • \$\begingroup\$ Also, in normal situations (other than using an inverter), there's comparatively very little energy in the harmonics to begin with, yes? \$\endgroup\$ – Hearth Jun 2 at 14:18
  • 2
    \$\begingroup\$ @Hearth: Well, yes, but the question is whether that energy is useful or not. \$\endgroup\$ – Dave Tweed Jun 2 at 15:45
16
\$\begingroup\$

That is true only if the current is being distorted by the load and not due to distortion of the voltage waveform of the AC line.

If you multiply the instantaneous values, point by point, of two sine waves of different frequencies, you get a waveform that has an average of zero. You have positive power during some intervals and negative power in other intervals. That shows that the energy is passing back and forth rather than being transferred from the source to the load.

enter image description here

If the AC line voltage is distorted, harmonic power is transferred, but it may not be useful power. In AC motors, the harmonic current would be attempting to make the motor operate at a higher speed in conflict with the fundamental. Some of the harmonics would be attempting to drive the motor in the reverse direction. As a result, all of the net harmonic power transferred is lost as heat, noise and vibration. Some harmonic power would be circulated between the source and load just as reactive poser is circulated.

Harmonic power would be useful to the extent it causes heating where heating is the desired use of power. There is some possibility that harmonic power could be useful in a universal motor. It might also be useful when rectified and filtered. While some of the power transferred might be said to be useful, the undesirable effects would outweigh the usefulness.

\$\endgroup\$
  • 1
    \$\begingroup\$ That's a pretty image, but I think it could use a more detailed explanation. At least, it took me a while to figure out what I think you're trying to illustrate with it. At first I thought you were trying to argue that "a waveform that has an average of zero" cannot transfer useful power, which seems kind of reasonable... until one realizes that a any unbiased periodic waveform, including a pure sine wave, also has an average of zero, yet AC power transmission obviously still works. \$\endgroup\$ – Ilmari Karonen Jun 4 at 9:26
  • \$\begingroup\$ ... I guess what you're actually trying to say is that if the current drawn by the load has some frequency components that are orthogonal to the line voltage waveform (or vice versa), then those frequencies cannot transfer useful power (unless, of course, there's some rectification or other nonlinear stuff going on). But I only figured that out when I noticed the "V" and "I" in the image legend. And I'm still not entirely sure that's what the OP is actually asking about. \$\endgroup\$ – Ilmari Karonen Jun 4 at 9:27
  • \$\begingroup\$ @Ilmari Karonen The OP is asking about what is said "in most books." I am trying to explain what can be true about that. \$\endgroup\$ – Charles Cowie Jun 4 at 12:40
  • \$\begingroup\$ OP here, I was asking about electrical power, so I assumed the current was distorted and the line voltage was clean. Sorry about the lack of clarity. \$\endgroup\$ – k_orolev Jun 6 at 11:10
6
\$\begingroup\$

From a sinewave grid, this is true since the harmonics are due to any "nonlinear device.

For example, the partial magnetic core saturation controlled by the peak excitation voltage.

However, that statement contradicts inexpensive inverters with square wave sources rated in V-rms. The harmonics in this voltage waveform can generate the same power in resistive loads. But then harmonics can increase eddy current losses in motors so it is less efficient.

So if you understand the source of the harmonics and load impedance, you can understand the exceptions to the rule. An electrical heater, being mainly resistive can use either sine or square wave sources.

enter image description here

The only difference between real (used) power and reactive (stored) power is the 90 deg phase shift component of current relative to voltage, regardless if it is fundamental or harmonics.

\$\endgroup\$
  • 5
    \$\begingroup\$ An electrical heater can use DC, even. Since it has no memory, it only cares about the instantaneous voltage, not any previous voltage - except for heat capacity. The temperature of the heater obviously does depend on previous voltage. \$\endgroup\$ – MSalters Jun 3 at 9:13
  • \$\begingroup\$ Adding to this, if your power source is full of distortion and you are feeding power to a switched-mode power supply (SMPS), such as a laptop power brick, it WILL transfer that extra power. If the SMPS has a passive filter on the input, most of that non-fundamental energy will be converted into heat, way more than a pure-sine inverter. Worst case scenario, transients make it through the filters and can even damage circuitry. But more sophisticated supplies (e.g. buck-buck-boost) can tolerate and even clean up the power factor. So yes, power is definitely transferring. \$\endgroup\$ – DeusXMachina Jun 4 at 20:45
  • \$\begingroup\$ Capacitive load in bricks are more lossy to square wave inverters as are inductive in motors with eddy current losses but now due to high peak currents I²*ESR and adds noise ripple to audio. So a good active PFC front end is also better to present a resistive load. @DeusXMachina so I agree \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 4 at 20:57
0
\$\begingroup\$

In the context of the power grid, AC power is generated and transmitted through transformers in three phases 120 degrees apart. The third harmonic (triple the frequency) is identical for all three phases, which you can verify by plotting the sine waves. When you connect a load between any two phases, the third harmonic voltage is the same on both wires, so your load sees nothing. (However, you will see the third harmonic from any phase to ground.). Since a three phase transformer connects only between phases, is cannot receive third harmonic power. This blocking applies for any 3n harmonic. This only happens in three phase power transmission, never when a single signal line is connected to a load.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.