Balancing the Breaker Box

Nonsense. Ever see a 'power factor meter'?? I have, on old switchgear systems. These only measured the phase-based power factor (harmonic content wasn't a problem in these old systems). But they simply developed a position based on the phase difference between the applied voltage and current (one input polarized a moving vane in a magnetic field created by the other input). Because the position of the needle actually showed the phase angle, the meter face was marked in a cosine pattern (i.e. the distance between 0.9 lagging position and the 0.8 lagging position was small, but the distance between 0.1 lagging and 0.2 lagging was larger).

The position didn't vary with the magnitude of the voltage or current, only the phase relationship (of course if current was below a certain threshold, the meter wasn't reliable)

daestrom

Reply to
daestrom
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One could argue that the Earth is at the center of the universe too, and that the universe indeed exists inside the shell of the Earth. The math gets annoying and Occam doesn't like the idea, but it could be argued.

Interesting. How do they deal with harmonics?

Reply to
krw

Exactly. They don't deal with harmonics, thus are really phase angle meters.

So what? Without measuring harmonic content, they only measure phase angle not PF. ...unless you assume the world is a sine wave.

Reply to
krw

I know that you're the stupidest hack on the Usenet, and that is pretty damned stupid, Roy.

Cheap shots for dumb asses.

No, you are Roy the Retard.

Reply to
krw

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It appears that you may be confusing capacity and utilization. Substitute utilization for capacity and you'll be OK. You have a 100A, 240V system which is center-tapped so that you can get 100 A at 120 V on each leg. If the load is balanced, the center-tapped or neutral connection has 0 current. The capacity of the system then is 100A at 120/240V which works out at 12KW per leg for a total capacity of 24KW. If you put all the load on one leg, then you are (current) limited to 12KW . The other leg is doing nothing. The capacity of the system is unchanged but is only half used.

Here's a little chart: (120V per leg) Leg1 leg2 total neutral utilization current 100A 0

100A 100A 50% KVA 12 0 12 losses 1 units 0 2 units 1unit

current 50A 50A

50A 0 50% KVA 6 6 12 losses 0.25units 0.25 units 0.5units 0 units

current 100A 100A

100A 0 100% KVA 12 12 24 losses 1 unit 1 unit 2 units

current 75A 25A ---

50A 50% KVA 9 3 12 losses 0.56units 0.06 units 0.88units 0.25units

(unit taken as single wire I^r loss at 100A)

I could quibble with you a bit with regard to the motors- it is the load that determines the torque requirement at any given speed. Yes, speed and voltage are related as are torque and current.-you have that right- but the intersection of the torque speed curves of motor and load, determines the operating point. Note that the watthour meter is a type of induction motor and thus is limited to less than synchronous speed. In fact, it would sepf destruct at that speed and is operating in a region where it is nearly at standstill so the torque is effectively constant at a given power, within the operating range. The drag torque is linearly speed dependent.

However krw may have been referring to the fact that the inherent measurement doesn't deal with power factor as electromechanical wattmeters (dynamometer) and watthour meters inherently measure the instantaneous product of voltage and current and average this to get the true power without any consideration of power factor. So does a digital meter. They don't use power factor because of this-effectively they go back to fundamentals to make measurements. If we are using voltmeters and ammeters then we don't have this multiplication and averaging of the instantaneous values so have to account for phase. Power factor is one way to do this. We also use rms voltage and current- but these don't actually exist - what exists are sinusoidal (hopefully) voltages and currents. and their instantaneous product.

Reply to
<dhky

---------------------

It appears that you may be confusing capacity and utilization. Substitute utilization for capacity and you'll be OK. You have a 100A, 240V system which is center-tapped so that you can get 100 A at 120 V on each leg. If the load is balanced, the center-tapped or neutral connection has 0 current. The capacity of the system then is 100A at 120/240V which works out at 12KW per leg for a total capacity of 24KW. If you put all the load on one leg, then you are (current) limited to 12KW . The other leg is doing nothing. The capacity of the system is unchanged but is only half used.

Here's a little chart: (120V per leg) Leg1 leg2 total neutral utilization current 100A 0

100A 100A 50% KVA 12 0 12 losses 1 units 0 2 units 1unit

current 50A 50A

50A 0 50% KVA 6 6 12 losses 0.25units 0.25 units 0.5units 0 units

current 100A 100A

100A 0 100% KVA 12 12 24 losses 1 unit 1 unit 2 units

current 75A 25A ---

50A 50% KVA 9 3 12 losses 0.56units 0.06 units 0.88units 0.25units

(unit taken as single wire I^r loss at 100A)

I could quibble with you a bit with regard to the motors- it is the load that determines the torque requirement at any given speed. Yes, speed and voltage are related as are torque and current.-you have that right- but the intersection of the torque speed curves of motor and load, determines the operating point. Note that the watthour meter is a type of induction motor and thus is limited to less than synchronous speed. In fact, it would sepf destruct at that speed and is operating in a region where it is nearly at standstill so the torque is effectively constant at a given power, within the operating range. The drag torque is linearly speed dependent.

However krw may have been referring to the fact that the inherent measurement doesn't deal with power factor as electromechanical wattmeters (dynamometer) and watthour meters inherently measure the instantaneous product of voltage and current and average this to get the true power without any consideration of power factor. So does a digital meter. They don't use power factor because of this-effectively they go back to fundamentals to make measurements. If we are using voltmeters and ammeters then we don't have this multiplication and averaging of the instantaneous values so have to account for phase. Power factor is one way to do this. We also use rms voltage and current- but these don't actually exist - what exists are sinusoidal (hopefully) voltages and currents. and their instantaneous product.

Reply to
<dhky

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I made no reference to the current rating, I just picked a power out of the air. It might have been clearer to you if I had said 12KW. as I wasn't considering overload conditions.

Reply to
<dhky

I have to disagree here, the instantaneous torque is proportional to instantaneous va or power. Inertia averages this to get the average power over each cycle (or longer). VA*pf implies that it measures the product rms voltage and current and applies a bugger (oops power) factor. The meter "sees" none of these because they are not physically there. the equivalence is there in that the time varying instantaneous values can be represented by their rms frequency domain equivalents in steady state. Analysis of the meter uses the rms approach as you have done (it avoids a nasty mess of non-linear differential equations) but the meter doesn't.

Reply to
<dhky

In the 30's and 40's, when these were often used, the world of electrical loads was a sine wave :-)

daestrom

Reply to
daestrom

What percent error is typically caused by ignoring harmonic content? Do the harmonics fool the utility meter or how does my power meter deal with harmonics? It just seems logical that if a utility power meter can measure true power, and a meter can measure apparent power (VA), then why should it be so difficult to measure power factor?

RogerN

Reply to
RogerN

I know. I used to have such meters (a mechanically resonant Hz meter too). My father was a power engineer and *collected* stuff (several tons of such stuff in the attic when he passed).

Reply to
krw

30%, 40%, maybe more of the VA is in harmonic content, particularly with electronics. Switching power supplies are often *very* bad.

They deal with it perfectly. We keep telling you that they measure

*POWER*.

You can calculate PF from VA and power, sure, but that's not the same as measuring PF. Your logic is backwards.

Reply to
krw

:A person at work was telling about someone that cut their electric bill :approximately in half by balancing their breaker box. If I understand this :correctly, the power company charges for power based on the current draw of :the highest leg. If you are using 100A on L1 and 50A on L2 then you would :pay based on 100A instead of the average 75A. : :If this is the case, since loads change constantly based on what is ON and :what is OFF, would a person save money by installing an isolation :transformer, wired for 240V in and 240/120 out? : :Thanks! : :RogerN :

If you are talking about having a 2 phase supply plus neutral, I would imagine you would have 2 separate single phase power meters - one for each phase. This means that total power, irrespective of the load on each phase, would be accurately recorded, and calculated by summing the readings of each meter at the end of the recording period. Power consumption is usually calculated in cents/kWh based on the totals recorded by each meter. I doubt very much that any domestic supply utility would have a sliding scale of charges which depended upon the load in each phase.

Reply to
Ross Herbert

:On Sat, 21 Nov 2009 20:25:09 -0600, "RogerN" wrote: : ::A person at work was telling about someone that cut their electric bill ::approximately in half by balancing their breaker box. If I understand this ::correctly, the power company charges for power based on the current draw of ::the highest leg. If you are using 100A on L1 and 50A on L2 then you would ::pay based on 100A instead of the average 75A. :: ::If this is the case, since loads change constantly based on what is ON and ::what is OFF, would a person save money by installing an isolation ::transformer, wired for 240V in and 240/120 out? :: ::Thanks! :: ::RogerN :: : : :If you are talking about having a 2 phase supply plus neutral, I would imagine :you would have 2 separate single phase power meters - one for each phase. This :means that total power, irrespective of the load on each phase, would be :accurately recorded, and calculated by summing the readings of each meter at the :end of the recording period. Power consumption is usually calculated in :cents/kWh based on the totals recorded by each meter. I doubt very much that any :domestic supply utility would have a sliding scale of charges which depended :upon the load in each phase.

Obviously, a polyphase meter will read the power consumption on both L1 and L2 simultaneously and come up with the correct total without any user intervention. The Weschler Instruments MD3000 or the Elster A3 polyphase meters are the way to go.

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Reply to
Ross Herbert

Yeah, it seems the original information (power meter runs based on high leg) was wrong. One of my future projects is to make an adapter to plug in to

240 and have 2, 120V receptacles. I have a bad 220V window AC unit that I would like to use the power for electric heaters in the winter. This way I can do an actual test and give the results to the electrician that was led to believe balancing saves money.

RogerN

Reply to
RogerN

:

This is *NOT* a good idea. The outlets used for window ACs don't generally (ever?) have a neutral. Using a ground as a neutral is really a *bad* idea. Get a 220V heater and use outlets elsewhere that happen to be on opposite legs to do your "experiment".

Reply to
keithw86

But wouldn't you say that the torque developed is a function of the phase angle between current and voltage? If the current lags 90 degrees from the applied voltage, then the magnetic fields of the current coil and voltage coil are almost exactly in-phase (owing to the high inductance in the potential coil). With the two magnetic fields pulsing 'in-phase', there is no torque either forward or reverse developed.

The simple fact that power flow in the opposite direction develops torque in the opposite direction shows that phase-angle between current and applied voltage is 'built-in' to the device.

Yes, of course you're right that the inertia *averages* out the torque, but it's not 'instantaneous va', it's 'instantaneous power'. But 'average power' *is* rms-volt * rms-current * power-factor in a sine-wave only system (i.e. no harmonic content)

daestrom

Reply to
daestrom

/ /This is *NOT* a good idea. The outlets used for window ACs don't /generally (ever?) have a neutral. Using a ground as a neutral is /really a *bad* idea. Get a 220V heater and use outlets elsewhere that /happen to be on opposite legs to do your "experiment". /

Yeah, I will check it out first, there are 3 wires, not sure if third is ground or neutral. If it's ground I'll not make the 240 to 2 X 120V wiring harness. If it's a ground wire then it would work as long as I have the load balanced, assuming to 120V 900W elements is equal to one 240V 1800W element.

RogerN

Reply to
RogerN

--------------- I have no problem with this. The meter doesn't measure phase angle. Note that the induction disc motor is essentially a form of single phase motor. The disc wont start but if already rotating there will be a torque bias in that direction. A single phase motor depends on this bias which is greatest near synchronous speed. However, in the case of the induction disc there will be very little unbalance torque -if any as the torque speed curve is essentially (and desirably) constant torque in the operating region (just about standstill) There will be a pulsating torque.

Yes it is, in the same way it is built into a conventional electromechanical (dynamometer) wattmeter. The torque at any instant depends on the product of instantaneous voltage and instantaneous current. On this basis, it simply averages the instantaneous torque. If the voltage and current are 90 out of phase, the instantaneous voltages and currents result in a double frequency power with 0 average

Mathematically we can say -for sinusoids: p(t)=Vmcos(wt)*Imsin(wt+phi) =(VmIm/2){cos(phi) +a second harmonic power term with 0 average] The average over a period is (VmIm/2)*cos phi +0 =Vrms*Irms* cos(phi)

Physically the meter simply produces a torque which is proportional to the instantaneous power and inertially averages it. Specifically, it doesn't measure phase angles, rms voltages or find pf -that is my point. If the voltage and current coils have high R/X values then the meter will be able to handle distorted waves with reasonable accuracy (and without doing a fourier analysis). A digital KWH meter will simply do the same v(t)*I (t) and averaging as a mechanical meter but with a few bells and whistles can be made to measure KVAH (pf*H is possible but meaningless) as well but such measurements aren't required or necessary for determination of energy.

instantaneous power = v(t)i(t) This happens to be the same as instantaneous va but I should not have used that term. The concepts of VA and VAR's are related to phasor analysis which is a mathematical model which gives us the pertinent information without the labor of solving a mess of differential equations. At your 120V outlet- there is no actual 120V source as can be seen if you examine the voltage with an oscilloscope.

Certainly, use of rms volts(magnitude)*rms current magnitude *power factor will give the same average power. That's part of the reason we use this model- it works. Also, this "model" allows us a reasonable chance of solving not-so-simple circuits. Think of what the situation would be in solving load flows for large systems using differential equations! Phasor analysis essentially replaces these differential equations by algebraic equations. Another convenient model is the use of symmetrical components for fault studies. Another is the use of forward/backward fields to model a single phase machine as two opposing machines. In these cases there are direct relationships between the model quantities and the actual quantities present.

You know all this. The point of this long winded diatribe is that, too often, people think that the model is the actual thing.

Reply to
<dhky

I tried to read this whole thread but gave up. Roger, you are wrong, your friend is wrong. Residential revenue meters measure real power. Not high leg current, not moon phase, not price of soybeans. They don't do it by measuring one current and one voltage. They actually measure two currents and one voltage. The currents on both legs are used, as is the voltage between them. I get really tired of electricians saying they have a friend who had a friend who carried a black rock in his pocket and saved energy. I had one actually tell me that the electric utility put "faster" meters on newer homes!!!! Crazy. If you want to lower your electric bill, use less energy.

Charles Perry P.E.

Reply to
Charles Perry

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