oso@mhuxh.UUCP (04/17/87)
I need some help in understanding the operation of the
domestic KW-Hr. meters the electric utility companies
use to calculate my electric bill every month.
It was my understanding that the KiloWatts in each
phase of my 220 Volt service was measured, and integrated
to obtain the KW-Hrs.
A friend of mine, who is an electrician by trade, claims
that the >>DIFFERENCE<< between each phase is integrated.
Anybody have any insight??
References will be helpful
Thanks
Ed Osoliniec
AT&T- Bell Labs
600 Mountain Ave.
Murray Hill, N.J.
(201) 582-3060
mhuxh!oso
.larry@kitty.UUCP (Larry Lippman) (04/19/87)
In article <923@mhuxh.UUCP>, oso@mhuxh.UUCP) (Edward J. Osoliniec) writes: > I need some help in understanding the operation of the > domestic KW-Hr. meters the electric utility companies > use to calculate my electric bill every month. > > It was my understanding that the KiloWatts in each > phase of my 220 Volt service was measured, and integrated > to obtain the KW-Hrs. > > Anybody have any insight?? The most common type of watthour meter for domestic 124/240 volt service is called a "two-element" watthour meter. In this device, the power through each hot wire is measured through an individual current winding, and through an individual potential winding which measures the potential of each hot leg to ground. There are therefore TWO current windings (one for each hot leg), and TWO potential windings (one for each hot leg to ground). The net energy measured is the algebraic sum of the energy in each hot leg. Since the potential of each leg is measured with respect to ground (which is 120 volts), a 240 device which has no ground current flow shows 1/2 its power in each leg - which is, of course, summed to read actual power. The watthour meter is effectively calculating: P = (I1)(E1) + (I2)(E2) where P is total power, I1 is current through leg 1, E1 is voltage from leg 1 to ground, I2 is current through leg 2, and E2 is voltage from leg 2 to ground. P above is actually represented as an AC magnetic field, so the strength of this magnetic field is proportional to the instantaneous power being consumed. Contained within this magnetic field is a metal disc which functions as the rotor of an induction motor; the magnetic field, as created by the stator current and potential coils, induces eddy currents in the metal rotor disc, thereby making it function as a motor. The greater the magnetic field, the faster the rotor disc turns. The rotor disc drives a totalizing gear train to display kilowatt-hours. There are also permanent magnets through which the rotor passes; these magnets provide damping. This magnetic field/rotor arrangement provides a true integration of instantaneous power, which therefore results in an energy measurement as kilowatt-hours. Some older 120/240 volt watthour meters have only one potential coil which measures voltage between the hot legs. Since many loads on a 120/240 volt power system are 120 volts only, this is not as accurate as a meter with two potential coils. > A friend of mine, who is an electrician by trade, claims > that the >>DIFFERENCE<< between each phase is integrated. There could be some confusion in semantics in this statement, but I am inclined to say it is oversimplified and false. See above for actual explanation. > References will be helpful These publications used to be given away free of charge by General Electric: (1) "Manual of Watthour Meters", Publication GET-1840C (2) "Application of Watthour Meters", Publication GET-1905D (3) "Manual of Demand Meters", Publication GET-2327C (4) "Metering Products Buyers' Guide", Publication GEP-550B (5) "Guide for Installing General Electric Watthour Meters", Publication GET-2669H The above materials were in my watthour measurement technical data file. This is NOT a "plug" for General Electric, although they do give away helpful technical publications. :-) <> Larry Lippman @ Recognition Research Corp., Clarence, New York <> UUCP: {allegra|ames|boulder|decvax|rocksanne|watmath}!sunybcs!kitty!larry <> VOICE: 716/688-1231 {hplabs|ihnp4|mtune|seismo|utzoo}!/ <> FAX: 716/741-9635 {G1,G2,G3 modes} "Have you hugged your cat today?"
sdp@omssw1.UUCP (Scott Peterson) (04/20/87)
A while back, someone told me you could make watt-hour meters run slow, or even backwards by messing with the power factor angle. I think it involved putting a huge inductive load on the line. This came up when someone was explaining why large industrial motors have large capacitors in parallel with them. -- Scott Peterson, Intel Corp., Hillsboro, OR, ...!tektronix!ogcvax!omssw1!sdp
ken@rochester.ARPA (Ken Yap) (04/21/87)
|A while back, someone told me you could make watt-hour meters run |slow, or even backwards by messing with the power factor angle. |I think it involved putting a huge inductive load on the line. |This came up when someone was explaining why large industrial motors |have large capacitors in parallel with them. Large motors have capacitors in parallel to improve the power factor, to "cancel out" the inductive load. You can make the watt-hour meter run backwards by feeding power back into the supply. Ken
commgrp@silver.bacs.indiana.edu (04/22/87)
Re: Watt-hour meters Yes, you can make them run "slow" by connecting a reactance (either inductive or capacitive) to the line, but nothing is gained by doing so; reactive power isn't useful for doing work, and you still get billed for the "real" power you use. Someone asked the same question in an EE course I took; the prof, instead of answering the question, launched into a tirade on ethics. Frank Reid reid@gold.bacs.indiana.edu "It's not my fault!" -- St. Andreas
keithe@tekgvs.TEK.COM (Keith Ericson) (04/22/87)
In article <27181@rochester.ARPA> ken@rochester.UUCP (Ken Yap) writes: > >Large motors have capacitors in parallel to improve the power factor, >to "cancel out" the inductive load. > Large industrial sites have more sophisticated power metering equipment installed which tracks the VARs (Volt-Amp Reactive) of the load and can be used to apply a premium (i.e., extra cost) if it exceeds some limit. (Source: my dad worked for the metering dept. of an electric utility company for 33 years.) keith
wtm@neoucom.UUCP (04/23/87)
Not having immediate access to the publications larry mentioned in his helpful article, I thought I'd ask the Net a follow-up question. It seems to me like the process of metering that Larry described: <P> = <I1><E1> + <I2><E2> is the measure of instantaneous *REAL* power consumed by the load, as opposed to *REACTIVE* power in the circuit, whcih would not actually be energy consumed in the load. Eventhough the reactive power load doesn't dissipate energy in the customer's equipment, it would cause (I**2)*R power loss in the power co.'s lines to the customer. In essence real power woud be: p(t)=v(t)i(t)cos(theta), theta being the phase angle time difference between the voltage and power waveforms. Note that if the voltage and current are 90 degress out of phase, no actual power is delivered. This could be the case if you stuck an inductor or capacitor into your friendly wall socket. (Please don't try it, though.) I've heard of corporate customers using things called synchronous capacitors and synchonous motors to apply reactive loads to the power line to bring the phase angle of the current waveform back into step with voltage. Presumably it results in lower bills. I remeber playing with a synchronous motor in a college course and being able to widely vary the phase angle by applying a control voltage to a field coil. What I was wondering if there is any convenient way for the electric co. to measure domestic customers' reactive power load, or do they just estimate the average domestic power factor and bill everyone under the same default assumption? I presume that the electric co. keeps close tabs, and maybe even measures the phase angle at corporate customers' sites by hand. Thanks in advance for any responses, --Bill Bill Mayhew Division of Basic Medical Sciences Northeastern Ohio Universities' College of Medicine Rootstown, OH 44272 USA phone: 216-325-2511 (wtm@neoucom.UUCP ...!cbatt!neoucom!wtm)
dwj@mtx5d.UUCP (04/23/87)
> It seems to me like the process of metering that Larry described: > <P> = <I1><E1> + <I2><E2> > is the measure of instantaneous *REAL* power consumed by the load, > as opposed to *REACTIVE* power in the circuit, whcih would not > actually be energy consumed in the load. > [The reactive load would ] cause (I**2)*R power loss in the > power co.'s lines to the customer. > > In essence real power woud be: > p(t)=v(t)i(t)cos(theta), theta being the phase angle time > --Bill When talking to a 'power' engineer (5-6yrs) ago, he said that the phase angle had been realativly insignificant in residential uses (lights, stove, hot-water and heaters are all resistive). Industrial users, however, had their power factors measured and were billed at some rate, set by the PUC. This put the incentive on the industrial users to get their power factors down by using synchronous moters and capacitor banks. david james mtx5d!dwj 201-957-1652 AT&T - ISL
roy@phri.UUCP (Roy Smith) (04/24/87)
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!"ornitz@kodak.UUCP (barry ornitz) (04/24/87)
In article <24300006@silver> commgrp@silver.bacs.indiana.edu writes: >Re: Watt-hour meters > >Yes, you can make them run "slow" by connecting a reactance (either >inductive or capacitive) to the line, but nothing is gained by doing so; >reactive power isn't useful for doing work, and you still get billed for >the "real" power you use. I once call Sangamo who makes many of these meters about this same question. They replied that the basic design of these meters had changed little since the 1020's when they were first designed. They had more modern materials to work with but the concept was the same. This engineer said the basic meter indicated VIcos@ or real power for reasonable values of cos@, the power factor. When I quizzed him about what was reasonable, he said cos@ greater than 1/10. This is a quite reactive load. He said the biggest problem was people who would place a jumper around the meter. Normally this would be obvious from drastic changes in use from one month to the next. However, he said that if someone wanted to cut his bill by only 20% or so, it would be difficult for the power company to notice the change. If the cheater were caught, however, the courts generally ruled that in addition to the fine and possible jail sentence, the criminal would have to pay the estimated higher rates retroactive to the initiation of his service. ----------------- | ___ ________ | | | / / | | Barry L. Ornitz UUCP:...!rochester!kodak!ornitz | | / / | | Eastman Kodak Company | |< < K O D A K| | Eastman Chemicals Division Research Laboratories | | \ \ | | P. O. Box 1972 | |__\ \________| | Kingsport, TN 37662 615/229-4904 | | -----------------
jeffw@midas.TEK.COM (Jeff Winslow) (04/24/87)
In article <834@mtx5d.UUCP> dwj@mtx5d.UUCP (David W. James) writes: >Industrial users, however, had their power factors measured and were >billed at some rate, set by the PUC. This put the incentive >on the industrial users to get their power factors down by ... >capacitor banks. Or they could buy lots and lots of computer equipment which uses off-line switching supplies for power conversion. How's that for a great reason to buy CAM equipment? 1/2 :-) Jeff Winslow
michael@m-net.UUCP (Michael McClary) (04/25/87)
In article <923@mhuxh.UUCP> oso@mhuxh.UUCP) (Edward J. Osoliniec) asks: >I need some help in understanding the operation of the >domestic KW-Hr. meters the electric utility companies >use to calculate my electric bill every month. You can find descriptions of watthour meters in electrical engineering texts. Check at your local university's library. First, you referred to "two phases". Well, yes, you could think of a single-phase service as two phases 180-degrees apart, but we'd better stick to power company nomenclature and think of it as a single-phase supply, with two hot conductors carrying current in opposite directions. (Otherwise we'd have somewhere between 9 and 12 phases running around by the time we'd dealt with a 3-phase meter.) I will Keep It Simple and Stupid by discussing only an ideal meter, neglecting the dozen or so minor ways they deviate from perfection. A (single-phase) watthour meter is a two-phase electric motor: - The spinning disk is the rotor. - A forward force proportional to the power consumed is applied to it by a pair of electromagnets driven by currents 90 degrees out-of-phase. - A retarding force proportional to the speed of the rotor is applied by a permanent magnet. Thus the rotor rapidly reaches an equilibrium speed proportional to the power consumed, and its rotation (which drives a gear-train to an analog counter) integrates the power. It will also run backward if you feed power to the grid. (A polyphase watthour meter has two or three separate disks on a common shaft, one to receive a forward force proportional to the power delivered by each phase of the service. Under some conditions you can get away with two disks in a 3-phase service.) The two electromagnets that apply the force are a "voltage" magnet and a "current" magnet. Assuming they are driven by sine waves, the force they apply is proportional to the product of their field strengths times the sine of the phase angle between them. Thus if the "voltage" magnet has a field strength proportional to the voltage, but delayed by 90 degrees, and the "current" magnet has a field strength proportional to, and in phase with, the current, the total force will be proportional to the power through the meter. The "voltage" magnet is wound with many turns of thin wire, placing a large resistance in series with its inductance. This causes it to carry a small current, retarded about 90 degrees from the voltage's actual phase. It is connected between the two "hot" wires on the input side of the meter (thus making the assumption that the "neutral" wire is at a voltage exactly midway between the two "hot" wires.) The field strength is proportional to the current times the number of turns, which is what we wanted. (The current is small, but there are a >lot< of turns, so the field is usable.) The "current" magnet has two windings, one for each of the "hot" wires. There are very few (3-ish) turns of very heavy wire, so there is negligible voltage drop through the meter. The windings are connected so that if you pull your current from hot wire to hot wire, the fields will add. Thus if you pull one ampere from hot to hot (220 volts), you'll cause twice as much of a field as if you pull one ampere from one hot wire to neutral (110 volts). Does it all make sense now? =========================================================================== "I've got code in my node." | UUCP: ...!ihnp4!itivax!node!michael | AUDIO: (313) 973-8787 Michael McClary | SNAIL: 2091 Chalmers, Ann Arbor MI 48104 --------------------------------------------------------------------------- Above opinions are the official position of McClary Associates. Customers may have opinions of their own, which are given all the attention paid for. ===========================================================================
michael@m-net.UUCP (Michael McClary) (04/27/87)
In article <243@omssw1.UUCP> sdp@omssw1.UUCP (Scott Peterson) writes: >A while back, someone told me you could make watt-hour meters run >slow, or even backwards by messing with the power factor angle. >I think it involved putting a huge inductive load on the line. >This came up when someone was explaining why large industrial motors >have large capacitors in parallel with them. Well, you can, but that's not why the capacitors are there. (If that were then only reason for the capacitors, the big companies would leave them out, reducing both their motor costs and electric bill.) If you pull your current way out of phase, the meter gets inaccurate, but not seriously. If you pull a LOT of current 90 degrees out-of-phase (but I forget whether it's leading or lagging), you can get the meter to back up just a tad. It's not enough to worry about. The power company must provide current to supply whatever load is presented. This means it must provide both "real" power (in-phase current times voltage) plus "reactive" or "imaginary" power (90 degrees leading current times voltage). If you hang an inductor across the line, it will pull "imaginary" power, and if you hang a capacitor across the line, it will "supply" it. (An inductor is called a "load" and a capacitor a "source" of imaginary power, rather than the other way around, because big loads, such as induction motors, tend to be inductive rather than capacitive.) The power company must feed this "reactive power" to its customers. The amount that isn't "generated" by capacitors somewhere will come from the generators, causing them to run out-of-phase, providing a mix of in-phase and reactive current. The generators are limited by the total current, not the total power, they produce, so this lowers the amount of real power they can generate. Also, the resistive losses in the transmission lines are proportional to the square of the current, regardless of the current's phase, and this is real energy that eventually comes from fuel. Therefore it costs the power company to provide this "reactive" power, and it charges its large industrial customers a surcharge if they run too far out-of-phase. So the customers buy capacitors to generate their own, keeping their bills down and everybody happy. Sometimes they buy motors which have their own power-factor correction capacitors, so the capacitors and the motor will be switched on and off together, but usually they just buy a few big banks of capacitors and turn them on and off to keep the plant under the surcharge limit. (Most cpacitors that you find with a motor are "starting" capacitors, used to provide an out-of-phase current during motor start-up to give it extra torque.) =========================================================================== "I've got code in my node." | UUCP: ...!ihnp4!itivax!node!michael | AUDIO: (313) 973-8787 Michael McClary | SNAIL: 2091 Chalmers, Ann Arbor MI 48104 --------------------------------------------------------------------------- Above opinions are the official position of McClary Associates. Customers may have opinions of their own, which are given all the attention paid for. ===========================================================================
michael@m-net.UUCP (Michael McClary) (04/27/87)
In article <555@neoucom.UUCP> wtm@neoucom.UUCP writes: > >What I was wondering if there is any convenient way for the >electric co. to measure domestic customers' reactive power load, or >do they just estimate the average domestic power factor and bill >everyone under the same default assumption? I presume that the >electric co. keeps close tabs, and maybe even measures the phase >angle at corporate customers' sites by hand. Yes, there is a convenient way. You just run a normal watthour meter with the "voltage" coil 90 degrees out of its normal phase. This integrates reactive power just as the normal connection integrates real power. Read both meters and a simple formula gives you your average power factor. At an auto plant where I worked, the real power was read automatically every 15 minutes by a recorder on the meters, and the reactive power was read by the meter reader when he came to eyeball-check the real power meters and change the recording medium. I assume different utilities track power factor differently, depending on their regulatory agencies. =========================================================================== "I've got code in my node." | UUCP: ...!ihnp4!itivax!node!michael | AUDIO: (313) 973-8787 Michael McClary | SNAIL: 2091 Chalmers, Ann Arbor MI 48104 --------------------------------------------------------------------------- Above opinions are the official position of McClary Associates. Customers may have opinions of their own, which are given all the attention paid for. ===========================================================================
wtm@neoucom.UUCP (04/30/87)
Roy's article got me wondering what your friendly local power company does when they want to put an additional alternator on line. It must be pretty tricky to match the rpm, and then the phase angle of the output waveform too. I've had a mental image of the operators getting the rotation of the machine matched up as close as possible and then taking cover as the machine lurches into phase with the rest of the power grid. It seems like the multipole generators used in hydroelectric plants would be especially tricky since they spin at relatively low rpm. I suppose that in a hydroelectric plant, you could start out by motoring the alternator and then letting the water add useful power when the machine was up to speed. --Bill
commgrp@silver.bacs.indiana.edu (05/02/87)
Re: Putting alternators on line I don't know how the power companies do it, but it's very simple to put a small 115-volt alternator on line: Connect two 115-volt lightbulbs in series and put them between line and alternator. Start the alternator and bring it up to speed. As the alternator frequency approaches that of the line, the bulbs will begin to flash-- there are 230v across the lamps when the two sources are 180 deg. out of phase, that's why you need two bulbs in series. Make the bulbs flash very slowly, and when both are extinguished, close a shorting switch across the bulbs. Your alternator now locks itself to the line, and the more mechanical power you put into it, the more electricity goes into the line. If it's a 3-phase alternator, you will need three sets of lightbulbs. Also, put fuses in the line in case you goof and throw the switch at the wrong time. Frank Reid reid@gold.bacs.indiana.edu
jans@stalker.gwd.tek.com (Jan Steinman) (05/04/87)
In article <561@neoucom.UUCP> wtm@neoucom.UUCP (Bill Mayhew) writes: >I suppose that in a hydroelectric plant, you could start out by >motoring the alternator and then letting the water add useful power >when the machine was up to speed. This is exactly how you start a hydro micropower (under 50kW) plant. Related issue: does anyone know why a generator costs 50% - 100% more than a comparable motor? Or why a turbine costs so much more than a pump? All the microhydro books I've read recommend using centrifugal pumps and synchronous motors to save money, but were short on what one loses in the process. Is is simply supply and demand? The efficiency differences appear to be minimal: 5% - 10%. :::::: Software Productivity Technologies --- Smalltalk Project :::::: :::::: Jan Steinman N7JDB Box 1000, MS 60-405 (w)503/685-2956 :::::: :::::: tektronix!tekecs!jans Wilsonville, OR 97070 (h)503/657-7703 :::::: :::::: Software Productivity Technologies --- Smalltalk Project :::::: :::::: Jan Steinman N7JDB Box 1000, MS 60-405 (w)503/685-2956 :::::: :::::: tektronix!tekecs!jans Wilsonville, OR 97070 (h)503/657-7703 ::::::