Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Path: utzoo!mnetor!seismo!rutgers!ames!hc!beta!cmcl2!phri!roy From: roy@phri.UUCP (Roy Smith) Newsgroups: sci.electronics Subject: Re: Domestic KiloWatt Hour Meters Message-ID: <2642@phri.UUCP> Date: Thu, 23-Apr-87 17:36:04 EST Article-I.D.: phri.2642 Posted: Thu Apr 23 17:36:04 1987 Date-Received: Sat, 25-Apr-87 19:01:21 EST References: <923@mhuxh.UUCP> <1715@kitty.UUCP> <243@omssw1.UUCP> Reply-To: roy@phri.UUCP (Roy Smith) Organization: Public Health Research Inst. (NY, NY) Lines: 145 Summary: A brief tutorial on power factor et al. and pointers to texts In article <243@omssw1.UUCP> sdp@omssw1.UUCP (Scott Peterson) raises some questions concerning power factors, capacitors on industrial motors, and making watt-hour meters run backwards. In this article I hope to explain a bit about the above, and delve a bit into the theory behind how it all works. Everybody who knows anything about electricity knows that you compute power as P=VI (eqn. 1) The problem is that eqn. 1 only holds for DC. For AC, I and V are vectors (phasors, actually) and the equation is P = V dot I (eqn 2.) which can also be written as P = |V| |I| cos(phi) (eqn. 3) where phi is the phase difference between the V and I vectors. The factor of cos(phi) is called the power factor (pf) and ranges from +1.0 to -1.0. Let's look at some limiting cases. 1) If you have a purely resistive load (like a toaster) phi = 0 and pf = 1.0. From the point of view of the electric utility, you are being a nice customer for reasons which will be explained shortly. 2) If you have a purely inductive load (like a big lossless transformer with no load on the secondary) phi = -90 and pf = 0.0 lagging. This means that while you may be drawing a substantial amount of current you are drawing zero net power! I say net because during half the cycle you are drawing power to pump potential energy into the inductor's magnetic field and during the other half of the cycle, the magnetic field collapses and returns all the energy to the power source. Assuming your watt-hour meter is designed properly, it will register 0. 3) If you have a purely capacitive load (a big lossless capacitor) phi = +90 and pf = 0.0 leading. Just as in the inductive case, you are drawing current but no net power. A few notes are in order here. I've always had trouble keeping track of the sign of phi. It's possible that according to proper engineering practice I've got the -90/+90 and leading/lagging mixed up (i.e. does current lag voltage or does voltage lead current?). If that's the case, I've just described the dual of the real situation; no big deal. Since cos is an even function, the power factor is the same regardless of the sign of phi. Also, all quantities are taken to be RMS values. Of course, none of the three cases above actually happen in real life. Most loads are a mixture of inductive (motors, fluorescent light ballasts) and resistive (incandescent lights, heaters). I can't think of a single item you might commonly find in a home or industrial plant which is a capacitive load. This adds up to a phi somewhere between 0 and -90 or a pf of somewhere between 0.0 lagging and 1.0. Now, just why does the power utility give a damn about your pf? The lower your pf (be it leading or lagging), the more current you draw for the same amount of power. To handle the greater current, the utility has to run fatter wires (which means more copper or aluminum which means more capital dollars invested), but since your watt-hour meter reads true power, you don't pay any more for it. Power companies usually slap a (hefty) surcharge on industrial customers if they don't keep their pf higher than some given value (I think pf's above about 0.8 are considered reasonable; can anybody confirm that?). OK, now let's say you have a factory with lots of motors and arc welders and fluorescent lights. You've got some disgusting pf and Con Ed if charging you an arm and a leg in surcharges. What can you do? A couple of things. One is to get some big capacitors and hook them up across your power feed. Simple and effective. If you want to be fancy, you can replace some of your motors with ones that have separately exitable windings which you can run at *leading* pf by diddling with the rotor current (I forget the details of how this works but it's pretty nifty, especially after you've had "motor = inductive load = lagging pf" drilled into your head all semester). A motor running like this is not as efficient as a normal motor, but it may be more cost effective than adding a capacitor bank. What you can't do is adjust your pf so the meter runs backwards. Well, that's not *quite* true. If you take those big three-phase synchronous motors and drive the shafts backwards, you've just turned them into generators. A synchronous motor and a synchronous generator are the same machine. If you're putting in electrical energy at the terminals and taking out mechanical energy at the shaft, you say you're motoring. If you put in mechanical energy and take out electrical energy, you say you're generating. In this case, your phase angle is now in the range of -90 to -180 (or +90 to +180), you have a *negative* power factor, and you are feeding power back to the utility company and sure enough, your meter will run backwards. I'm not sure if my negative power factor analysis is the way a power engineer would phrase it, but the end result is the same; you are now selling power to Con Ed instead of buying it. Just a few problems. First is that to turn those shafts backwards you need to get mechanical energy from somewhere. Unless you've invented a perpetual motion machine, this will cost you money (maybe not as much as Con Ed charges for electricity, but money none the less). Second, since you probably don't have all sorts of fancy equipment to keep your generators strictly in phase with the rest of the power grid, the utility is going to be mighty pissed at you. In reality, since you don't have that fancy phase regulating equipment, what will really happen is you'll blow a fuse or (if you put a big enough penny in the fuse box) break a shaft and trash your factory. I've gone on long enough, so I'll leave you with some reading to do if you are really interested in this. The first book is the easiest but still assumes you know something about basic electricity along with differential and integral calculus, linear algebra, and complex variables. Sorry folks, but to really understand this, you gotta know the math. Most apropos to this discussion would be chapter 11, "Average Power and RMS Values", especially section 11-5, "Apparent Power and Power Factor". The second and third books assume in addition that you are well versed in circuit analysis (which you learn from the first book, that's why it's first). If you can't find any of these, don't fret; there are lots of good circuits and motors books out there which should cover this. Try your local engineering library. %A William H. Hayt, Jr. %A Jack E. Kemmerly %T Engineering Circuit Analysis, 3rd edition %D 1978 %I McGraw-Hill %X Library of Congress TK454.H4, ISBN 0-07-027393-6 %A Charles A. Gross %T Power System Analysis %D 1979 %I John Wiley & Sons %X Library of Congress TK1005.G76, ISBN 0-471-01899-6 %A A. E. Fitzgerald %A Charles Kingsley, Jr. %A Alexander Kusko %T Electric Machinery, 3rd edition %D 1971 %I McGraw-Hill %X Library of Congress 70-137126 [this doesn't look like a LoC # to me but that's what it says --RHS] -- Roy Smith, {allegra,cmcl2,philabs}!phri!roy System Administrator, Public Health Research Institute 455 First Avenue, New York, NY 10016 "you can't spell deoxyribonucleic without unix!"