# Balancing the Breaker Box

Thanks Don Dave Foreman
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There are some electrical engineering books on Ebay. I'd have no idea which is worthwhile.
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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)
>> The meter simply measures what you actually use (+ losses on your side of

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.
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Don Kelly
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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)
>> The meter simply measures what you actually use (+ losses on your side of

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.
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Don Kelly
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krw wrote:

I disagree. Analog computers were around long before digital (the speedometer in a vintage car is an obvious one).
The three coils create a torque on the disk proportional to real power. The arrangement of the coils and inductance of the voltage sensing coil are such that the power factor *is* implicit in the development of the torque. A load current that is 90 degrees out of phase with the applied voltage results in no torque developed. (which is why many claims that 'power factor correction' will save you big money, are bogus)
If that were the only torque applied to the disk, it would spin rather rapidly because the mechanical friction is pretty small. And it would be hard to calibrate since the torque needed to drive all the gearing is somewhat variable. So another permanent magnet called a "drag magnet" is positioned next to the disk creates a counter-torque that is proportional to the disk's speed. Obviously the disk stops accelerating/decelerating when the torque produced by the sensing coils, friction and the drag magnet balance. The result is that the disk speed is proportional to the power in the sensing circuits.
(As an aside, some criminals have tried to cheat the power company by positioning additional 'drag magnets' above/below the disk. This tends to increase the counter-torque developed when the disk is in motion, slowing the speed of the disk for a given power level. The law does provide sanctions for such meter tampering.)
All the gears and wheels form an integrator (as in 'calculates the integral of the disk speed with respect to time'). Because the disk speed is 'revolutions / time', counting the revolutions is the calculus operation of integration with respect to time.
It's a meter that calculates the energy usage by sensing voltage, current and the phase relationship between them (i.e. power factor) and integrating the results over time.
daestrom
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wrote:

Certainly, but this doesn't change the fact that PF is not the measured quantity.

Yes.
Yes, others have clarified this better than I. I have no clue why he's still pursing this.
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RogerN wrote:

Balancing it is a good idea because it makes the most of the incoming capacity, but it will have no effect on your electric bill, the meter reads both sides and records usage accordingly.
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RogerN wrote:

The kWh meters used in typical residential service in the US (i.e. the 'Edison connection') sense the current in *both* hot legs. As such, it effectively 'averages' the total current of all the loads to figure out the kWh used (along with the voltage and power-factor).
You can test this out yourself easily. Open all the load breakers except for one on each leg. Then turn on a 150 watt light on one leg that is still powered and watch the meter (count the disk rotations per minute). Go back inside and turn on a 75 watt light on the other leg that is still powered. If your 'person at work' was right, the meter should still be spinning at the same speed (the first leg is still the highest load). But they're wrong and you'll find the meter now spinning faster than before.
daestrom
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If what the 'person at work' claimed was right, they wouldn't be giving electricity away but would be getting paid for a small amount of electricity they weren't providing. Could you not understand that from the original post? Your test sounds like a good idea though, I may try using 1500 Watt heaters instead of light bulbs though.
Seems more likely though that if what he said wasn't BS, then the household he was speaking of had a bad meter.
RogerN
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: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.
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: ::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. http://www.weschler.com/_upload/sitepdfs/misc/MD3000_data.pdf http://www.elsterelectricity.com/downloads/A3_ALPHA_meter_data_sheet.pdf
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wrote:

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
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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".
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wrote:

/ /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
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wrote:

It will be a ground unless it's a four pin outlet. It's still not safe. A better, but still not great, solution would be to hardwire the two heaters in series with a single 240V plug.
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snipped-for-privacy@gmail.com wrote:

If the heaters are the same wattage and are run at the same time, it will work. It's not really advisable, but it isn't likely to cause any real problems. Some clothes dryer receptacles have a neutral so that's a better idea. Yeah ground and neutral go to the same place, but the ground really shouldn't be used to carry current.
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Yeah, I wouldn't use the ground as a neutral. About the only way I would proceed that way is a complete rewiring of the heaters (oil filled radiators) so that both 600W elements were in series and both 900W elements were in series, and all with proper fusing / breakers for each pair of elements.
RogerN
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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.
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I'm wrong even though I never claimed what my friend claimed? I had never heard it before and didn't know, that's why I asked here. It is amazing that you can become a professional engineer with such a reading comprehension problem. It's also amazing that you parallel an actual electrical measurement, high leg current, with moon phase and price of soybeans. Maybe you claim your P.E. because you passed physical education class? Giving you the benefit of the doubt, you should know if meters ran based off of the high leg then balancing would let you use all the power you were being charged for. That is completely different than moon phase, price of soybeans, or carrying a black rock in your pocket. I suppose you think fixing a leak in a gas tank to save fuel would be like a 100mpg miracle carburetor too. Go back to the gumball machine and get your doctorate.
RogerN
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wrote:

This thread has been so full of rubbish, some of it true, some of false, most of it by people trying to show what the know, and mostly succeding in proving they know very little.
You are too buried in the crap to see that Charles, like me has got fed up with the trash and is making appropriate sarcastic remarks.
Grow up! You or your friend or whoever was wrong in the begining and all this crap has only gone on to prove that, but will you give up Oh No!.
John G.