David Littlewood | 09/03/2013 19:58:44 |
533 forum posts | I always understood the reason why larger customers are charged for "reactive power" (by whatever name - I believe it may be called VARs, for Volts-Amps-Reactive) is that the out-of-phase current still has to be trasmitted from the generator to the customer, and back again; this gives rise to transmission losses just as does all other current. Also, if there is too much of it, it can destabilise the system, and require the purchase and fitting of expensive power factor correction equipment in the distribution system. David |
Russell Eberhardt | 09/03/2013 20:21:56 |
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Posted by David Littlewood on 09/03/2013 19:58:44:
I always understood the reason why larger customers are charged for "reactive power" (by whatever name - I believe it may be called VARs, for Volts-Amps-Reactive) is that the out-of-phase current still has to be trasmitted from the generator to the customer, and back again; this gives rise to transmission losses just as does all other current. Also, if there is too much of it, it can destabilise the system, and require the purchase and fitting of expensive power factor correction equipment in the distribution system. David The out of phase current doesn't go back again but, yes, you're right. That current flowing through the cable resistance will develop a voltage drop in phase with that current and therefore will absorb power. Russell. |
David Littlewood | 09/03/2013 23:07:52 |
533 forum posts | Russel, Sorry to disagree with you, but the out-of-phase current flows in both directions, so it does go backwards and forwards along the transmission cables, causing losses. David |
martin perman | 09/03/2013 23:29:17 |
![]() 2095 forum posts 75 photos | Gentlemen, Something else I have just spotted in one of your threads is Power Factor Correction, where I served my Apprenticeship as a macine tool fitter we had two factory's close by each other, the older factory had tucked away in its roof space a very large electric motor with no extended shaft just running constantly and near by was a meter which had a zero in the centre of the dial with -1 and +1. I was told by the works engineer that the motor and meter were there to keep the power factor as near to zero as possible, is this anything to do with what we are discussing here.
Martin Perman
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Stub Mandrel | 10/03/2013 10:01:32 |
![]() 4318 forum posts 291 photos 1 articles | Hi Martin, Yes it probably was. Another way is a 'capacitor farm'. Like inductors, capiocitors are reactive loads, but they store a voltage even when current flow is zero, the opposite of inductors that keep pushing current when voltage is zero. This means a big inductive load (motors and other wound components) can be balanced by a big capacitative load. Wikipedia has a good article on power factor from which the picture below is linked. Neil
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Russell Eberhardt | 10/03/2013 10:04:07 |
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Posted by David Littlewood on 09/03/2013 23:07:52
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Yes, of course it does - doh! In exactly the same way that in phase current does - it's AC! Russell. |
David Littlewood | 10/03/2013 12:00:02 |
533 forum posts | Russell, Well, you were the one who said that "the out of phase current doesn't go back again". David |
David Littlewood | 10/03/2013 15:20:10 |
533 forum posts | Just out of interest, I got out my power meter and plugged my larger VFD through it. The latter is a Newton Tesla one which drives the 3 HP (2.25 kW) motor on my M300 lathe. At rest, i.e. with the motor off, the quiescent state showed a PF of about 0.4. At slow speed it drew 300 W resistive, and showed a PF of 0.6; mid-speed was 1000 W and 0.66 (total VA of 1500 W), and at top speed (2500) it drew about 2000 W and a PF of 0.70. The lathe was unloaded at each setting, but of course the headstock gears do take up a lot of power, especially at the high speeds. (Incidentally, if you've never run a large lathe at 2500 rpm or more, it's pretty scary!). Not sure what this proves, just interesting. It does rather show that the previous suggestion (that inverters do not draw much reactive flow - VARs - they do. Incidentally, my textbook on Electric Power Systems (B M Weedy) speaks of reactive power quite consistently when discussing the subject. High voltage power transmission systems are a complex subject which I don't pretend to understand, but it appears that the voltage drop in the lines is largely determined by the reactive power; this causes increased line current, and hence much increased I^2R losses, as I suggested above. David Edited By David Littlewood on 10/03/2013 15:22:53 |
william johnson | 10/03/2013 15:53:08 |
2 forum posts | In the factory where I worked the Electricity Co. said the power-factor was bad so supplied a correction capacitor which I fitted. When switched on the meter would slow down,nearly stop ,and on occasion turn slowly backwards. The heavy cables that supplied it heated up. The supply was liable to interuption so we had a generator,only 40KW so much juggling which machines were on at the same time. I thought the capacitor may help so wired it so it could be connected to the generator. The fluorescent lamps seemed rather bright then I saw the generators 600 volt meter needle hard over. Fortunately no machines were on and only a few lamps suffered but the office typewriter never wrote another word nor did the calculator ffigure out another sum. |
Mike Poole | 10/03/2013 18:14:53 |
![]() 3676 forum posts 82 photos | I think this explains why electricity suppliers do not like poor power factor and charge major users for poor power factor. From E. Hughes Electrical Technology. 11.22 The practical importance of power factor If an alternator is rated to give, say, 2000 A at a voltage of 400 V, it means that these are the highest current and voltage values the machine can give without the temperature exceeding a safe value. Consequently the rating of the alternator is given as 400 X 2000/ 1000 = 800 kVA. The phase difference between the voltage and the current depends upon the nature of the load and not upon the generator. Thus if the power factor of the load is unity, the 800 kVA are also 800 kW; and the engine driving the generator has to be capable of developing this power together with the losses in the generator. But if the power factor of the load is, say, 0.5, the power is only 400 kW; so that the engine is developing only about one-half of the power of which it is capable, though the alternator is supplying its rated output of 800 kVA. Similarly, the conductors connecting the alternator to the load have to be capable of carrying 2000 A without excessive temperature rise. Consequently they can transmit 800 kW if the power factor is unity, but only 400 kW at 0.5 power factor, for the same rise of temperature. It is therefore evident that the higher the power factor of the load, the greater is the power that can be generated by a given alternator and transmitted by a given conductor. The matter may be put another way by saying that, for a given power, the lower the power factor, the larger must be the size of the alternator to generate that power and the greater must be the cross-sectional area of the conductor to transmit it; in other words, the greater is the cost of generation and transmission of the electrical energy. This is the reason why supply authorities do all they can to improve the power factor of their loads either by the installation of capacitors or special machines or by the use of tariffs which encourage consumers to do so. Mike |
Swarf, Mostly! | 10/03/2013 21:06:35 |
753 forum posts 80 photos | Hi there, all, I'm not going to write about phase angle and power factor - instead I'd like to pick up a point from Andrew's post early in this thread. I make no apologies if the following is too electronic! Just look at the title page of Model Engineer issues from the Percival Marshall days. Andrew referred to the conversion of incoming single-phase AC to DC using rectifier diodes feeding a (reservoir) capacitor. With such a circuit configuration, current only flows through each diode when/while the instantaneous input voltage exceeds the voltage on the capacitor (apologies for expressing that in non-rigorous language). The result of this is that the current drawn from the AC supply consists of two brief but large pulses per cycle (assuming full-wave rectification). Not only does this give the electricity generator a bad time, it also requires rectifier diodes with a very high forward current capability (which is used for only a small percentage of the time!). The reservoir capacitor is also required to have a high 'ripple current' rating An alternative configuration, the 'choke input filter', follows the rectifier diodes NOT with a reservoir capacitor but with a suitably valued inductor (aka 'choke' ) which is, in turn, followed by a filter capacitor (and a 'bleeder' resistor to set the minimum output current). With a properly designed choke input filter each rectifier diode conducts continously for its complete half-cycle but at a much much reduced current. Rectifier diodes with lower forward current capability can be used and the filter capacitor can be rated at a lower ripple current and hence reduced physical size. This promises reduced component stress and hence better reliablity for the electronics and a much better input current waveform to the benefit of the electricity supplier. It might also reduce the filtering burden that seems to cause nuisance tripping of RCCBs. The differences between capacitor input and choke input rectifier schemes are very well explained and illustrated in a Texas Instruments Application Note on power supply design which may, just may, be available on-line. The cost of these benefits is the inclusion of the choke or inductor, a wound component of non-trivial but non-outrageous cost. However, during my career as an electronics engineer, it was notable that many of my contemporaries were uneasy with inductors! I've no idea whether or not the designers of workshop inverters employ choke input filters - I do suggest that if they don't then they should have a good think about it. Best regards, Swarf, Mostly!
Edited By Swarf, Mostly! on 10/03/2013 21:07:22 Edited By Swarf, Mostly! on 10/03/2013 21:08:35 Edited By Swarf, Mostly! on 10/03/2013 21:09:38 Edited By Swarf, Mostly! on 10/03/2013 21:10:05 Edited By Swarf, Mostly! on 10/03/2013 21:10:54 |
Russell Eberhardt | 11/03/2013 09:19:52 |
![]() 2785 forum posts 87 photos | Hi Swarf, Mostly, A modern, less bulky and costly, alternative to using a large choke (inductor) is active power factor conversion which uses something akin to a switched mode psu to make the current drawn more sinusoidal. Russell. |
Andrew Johnston | 11/03/2013 11:58:19 |
![]() 7061 forum posts 719 photos | Swarf Mostly: Many thanks for the excellent electronics exposition. It's always interesting to revisit older, but not necessarily obsolete, techniques. Another way of looking at the choke/capacitor combination is as a filter. The rectifier output is a full wave rectified sine wave, with a fundamental (in the UK) of 100Hz. The choke and capacitor act as a low pass filter, removing the fundamental and harmonics while leaving the DC component. Herein lies a problem if using an LC filter in a VFD. The output of the LC filter will be the average of the input voltage, not near the peak of the incoming sine wave as for the diodes and capacitor. If we look at the Fourier series for a full wave rectified sine wave the DC component is 2A/pi, where A is the amplitude. So, assuming we start with a 240Vrms incoming sine wave, for a diode/capacitor we get a DC voltage of around 339V minus diode drops, and less some ripple voltage depending on the load current. For an LC filter that removes all AC components we get an output voltage of 216V. It would not be possible to generate a 240Vrms, between phases, 3 phase output from 216V. If I remember correctly resevoir capacitors were often fairly large, let's say 470uF. The nominal cutoff frequency of a two pole LC filter is f=1/2pi*sqrt(LC) and the filter rolls off at 12dB/octave. So let's set the filter cut off at about 12.5Hz to give 36dB of attenuation at 100Hz. With 470uF that gives an inductance of 0.345H. That's quite a large value, which equals a large component, especially one which needs to carry quite a few amps. It is often forgotten that engineering is as much about money as science. Given that the VFD market is pretty competitive I doubt manufacturers would include large inductors and capacitors requiring manual assembly. Incidentally I suspect that the 'bleed' resistor was there to critically damp the LC circuit at resonance. I agree that many people shy away from inductors. Of all the passive components they are the ones least likely to approach the ideal. For power applications they nearly always needed to be custom designed. That has changed a bit with the proliferation of switch-mode power supply ICs but you still need to design your own for anything out of the ordinary. I would agree with Russell in that I suspect that most VFDs incorporate an active PFC at the front end, if anything. The PFC inductor, and any front end AC filtering, would involve small components that could be auto assembled on a PCB. I'd like to say that none of the above is intended to denigrate the post by Swarf, which I enjoyed, but just to highlight some issues that may preclude the use of LC filters in VFDs. Regards, Andrew
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jason udall | 11/03/2013 13:40:36 |
2032 forum posts 41 photos | Further to Andrews above.. SMPS for laptops say ..oftern made for US market ...so far so good..because US line sockets are line and not_line the filtering on the frount end is designed for this.. and the output is refferenced to the midpoint of these two phases ...due to the nature of europe wide line being line and neutral ( sort of 0 V) the filter can express around 200 V onto the 0V of your laptop...which being not earth bonded sits at around that voltage waiting for you to complete the circuit...what about PAT testing I hear you cry ...Computers are exempt from that part of PAT because they fail due to earth "leakage" in the filter capacitors..( exactly what the testing is supposed to do!) ,,, You couldn't make it up....
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Swarf, Mostly! | 11/03/2013 14:37:49 |
753 forum posts 80 photos | Hi there, all, Thank you, Russel, Andrew and Jason for your responses to my post. Russel, I have to confess that I have been retired from electronics for several years and am not familiar with the technique you mention. Please PM me with some more information on it. Andrew, your analysis differs from what I remember of the TI application note, my copy of which is buried three house-moves deep in my 'filing system'! I'd like time to disinterr it and compare, and then come back to you. I do agree that my own post neglected to take account of the absence of an input mains transformer - I'm conditioned to relying on the mains transformer to account for the reduced DC output from the choke input filter. Jason, your point is interesting, are you referring to the 'brick on a string' usually employed to power laptop computers? Best regards, Swarf, Mostly!
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jason udall | 11/03/2013 16:32:33 |
2032 forum posts 41 photos | 'brick on a string' ...and ones used in desktops and skyboxes etc. it comes from design for US supply and use elsewhere... computer equipement is exempt from some elements of PAT .. |
Stub Mandrel | 11/03/2013 17:07:09 |
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'brick on a string' as distinct from "wall wart"
Electronics is such a technical subject Neil |
Ken Fox | 11/03/2013 18:27:23 |
13 forum posts | A lot of interesting reading here and a lot of mythe and folklore. Here are my comments: 1/ 115/230 volt power is definitely not 2 phase. It is single phase. 2/ In polyphase power, that is, true 2 phase, 3 phase, 6 phase, 12 phase (yes it is used in special cases) the power is supplied continuously, not at 120 (or 100) pulses per second as it is in single phase. This is what distinguishes single phase from polyphase. 3/ In 2 phase the two voltages are 90 degrees apart, 3 phase 120 degrees, 6 phase 60 degrees, etc. 4/ Strictly speaking, inversion is the conversion to from DC to AC. The common inverters first generate DC then chop it into AC. Rectification goes the other way; AC to DC. 5/ The cons of low pf lagging are low HP per amp drawn, as noted elsewhere, increased voltage drop in power lines, and on the large scale loss of sychronization of the power line which means that the transmitting and receiving ends fall out of step. Also low pf lagging demagnetizes the generator so can make it fall out of step with the power system. I don't think anyone reading this should worry about it except the low HP per amp. 6/ Low pf leading can be just as bad for the power utility co. as it can cause a rise in voltage in the power line and over magnetization of the generator. Again no one reading this need be concerned 7/ The power utility co. is happy to get leading power factor because it offsets the lagging pf of most loads, but in principle at least it can be carried too far. 8/ Mention was made above of a motor just spinning, not connected to a load. This is a "sychronous condenser" and is an overexcited synchronous generator which runs at a leading pf. It is a common strategy used by power utilities to offset low pf lagging loads. 9/ To answer an earlier question, if the a full wave rectifier is used on the input to an inverter package feeding into a capacitor I would expect the power factor seen by the power system to be fairly high. Perhaps someone it a pf meter could measure this to see how far off I am. 10/ For those wanting 3 phase power an MG set is another option to the inverter. This would use a single phase motor spinning a 3 phase generator. To make it practical you would need some source of second hand, inexpensive machines.
Ken |
Muzzer | 11/03/2013 20:36:51 |
![]() 2904 forum posts 448 photos | Ken 1/ American house have "2-phase" in the sense that you get 2 x 120V withneutral centere tap. You can connect a load across the "ends" to get 240V. Call it what you want but 2-phase seems fairly sensible? 2/ Not sure what you mean. The voltages in all cases are sinusoidal. In fact, "single phase" in a typical home or factory is simply one of the 3 phases. 3/ You mean 180 degrees apart. 9/ The PF (angle) may be good but the harmonics (distortion) is pretty poor. Which means you are taking a higher RMS current than you would for the same outcome. And which you pay for literally. 10/ Isn't this scheme called a rotary phase converter? To answer an earlier point, reactive (imaginary) power involves a real quantity of real energy flowing through the circuit. A purely reactive (ie capacitive or inductive) load can generate a very significant current. The current and voltage are 90 degrees out of phase. You pay for the current but get no benefit from it in terms of heat or shaft power (which is what "real" power results in). "Real" and "imaginary" are the formal words you'll find in the text books for describing resistive and reactive impedences and currents. "Imaginary" doesn't mean it doesn't exist.... Merry |
Stub Mandrel | 11/03/2013 21:02:34 |
![]() 4318 forum posts 291 photos 1 articles | Merry, I think Ken may be right. The American '2 phase' would be two opposed 120V supplies 180 degrees out of phase, but really that's just single phase 240V with the neutral pegged in the middle instead on one line. A two phase that was 90-degrees could bring the same benefit as three-phase in deleivering a constant power through all 360 degrees of the cycle. Two phase at 180 degrees would be just as lumpy (twice as lumpy?) as single phase. Indeed, it seems I am right in my guess about 2-phase! the american system if apparently rightly descrivbed as split phase. Neil Edited By Stub Mandrel on 11/03/2013 21:05:24 |
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