In the standard 60cycle 240V 3 wire feed from the pole to a house or small business I've noticed that the neutral wire is sometimes smaller than the
two hot wires. Why is that? Has the assumption been made that 1/2 or more of the load will be from 240V devices that don't use the neutral wire? Is that always a reasonable assumption?
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Neutral load is part of the load calculation but it should also be noted that the neutral in a standard triplex cable is not insulated so it will always look smaller from the ground. The utility side is usually smaller than the customer side too. They use the "free air" column of the ampacity chart and they are on the NESC, not the NEC.
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So it's a power dissipation issue and not just circular mils/voltage drop per foot issue? The neutral wire should be twice the circular mils(area) from a voltage drop standpoint if the load is all 120 gadgets, right?
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| |> Neutral load is part of the load calculation but it should also be noted | that |> the neutral in a standard triplex cable is not insulated so it will always | look |> smaller from the ground. The utility side is usually smaller than the | customer |> side too. They use the "free air" column of the ampacity chart and they | are on |> the NESC, not the NEC. | | So it's a power dissipation issue and not just circular mils/voltage drop | per foot issue? The neutral wire should be twice the circular mils(area) | from a voltage drop standpoint if the load is all 120 gadgets, right?
Unless you have connected both hot wires to the same hot coming from the transformer, the neutral wire need be no larger then the larger of the two hot wires ... they should all three be the same. The neutral wire carries only the imbalance between the two sides. Consider putting a 100 watt bulb on one hot-to-neutral, and a 100 watt bulb on the other hot-to-neutral as the only load. Disconnect the neutral in this case and both lights remain lit at the same brightness. The neutral is lame in this condition and is carrying no current. Shut one of those lights off, and now the neutral is carrying the current for the remaining light (unless you left it open).
An open neutral is dangerous because in cases where the load is way out of balance, one side (the one with less load) will get more voltage than the other side. Just having a motor start up can do that. Or as in a case I personally saw, a clock on one side and a stove burner on the other side can heat up the clock more than the burner when the neutral is open.
Personally, I'd rather not have a center tapped neutral system like this. But many devices, such as electric stoves, depend on it by design.
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| Phil Howard KA9WGN | http://linuxhomepage.com/ http://ham.org/ |
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Ron Reaugh wrote:

Nope. The neutral only has to carry the DIFFERENCE between the currents drawn off the 2 hot wires. When the branch circuits are connected to the panel, the electrician SHOULD distribute the loads equally between the 2 sides of the panel, minimizing neutral current.
Bob Weiss N2IXK
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drop
mils(area)
DUH, thanks.
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The neutral only has to carry the unbalanced difference between the other phases. So if each hot wire carries 40 amps, the neutral has nothing.
In a more realistic situation of 50 on one and 30 on the other, the neutral only (if you're talking a single phase 120/240 set of circuits) carries 20 amps.
(If you're talking about a neutral as part of a three phase circuit, the math is a bit trickier, but the concept is the same).
Note that this use of smaller neutrals is frowned upon by many current local codes, and may have been dumped in the latest National ones. (I don't have a copy at hand to double check).
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danny burstein wrote:

Generally, you have to size the neutral for the maximum unbalanced loads. But keep in mind that utilities aren't always governed by the NEC. Depending on the jurisdiction, they are allowed to use their own 'engineering judgement' in calculating load diversity and equipment sizing to a greater degree than allowed by the NEC.
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(remove NS to use the address) 614.937.0463 voice 208.975.1011 fax
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In alt.engineering.electrical Don Phillips
|> Generally, you have to size the neutral for the maximum unbalanced |> loads. But keep in mind that utilities aren't always governed by the |> NEC. Depending on the jurisdiction, they are allowed to use their own |> 'engineering judgement' in calculating load diversity and equipment |> sizing to a greater degree than allowed by the NEC. | | And I would add, depending on the application, harmonics may require the | neutral be increased.
I see a lot of equipment (transfomers, switch panels) rated for 200% neutral current. But I can see scenarios where it could reach as high as 300%. Take, for example, an AC to DC power supply that uses three phases and a full wave rectification. Wouldn't these rectifiers only be conducting when the respective phase is the highest voltage? If this were happening and it were the only (very large) load, you'd have to make sure everything is well overrated for it (3x on neutral for wye, and 2x on each leg for delta). Does anyone even use stuff like that?
I understand the common switching power supply for computers switches on the current when the voltage is high enough to replenish the tank it keeps. So it would see the smaller the computer load relative to the power supply capacity, the narrower the current pulses would be. Once they are smaller than 50%, you could be going past the 200% mark on the neutral, if all your load is these power supplies. For a few computers in an office, maybe that would never happen. But for building a large data center, I suppose it could get serious.
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So you maintain that, as the load drawn from the computer power supply DECREASES, the neutral load INCREASES.
Lets go with that. I'm open minded.
What is magic about the 50% current pulses? Why would a <50% pulse result in 200% neutral current?
Not arguing your point, just looking for an understanding of the theory behind it.
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The increased harmonic current can only happen in 3 phase line to neutral loads and not split phase 120/240 line to neutral loads. What happens is that the current is increasing as the voltage is decreasing along the waveform. At some point, the current really drops. If they all drew current proportional to their voltage, then the neutral currents would cancel. But the definition of non-linear load is that the current drawn is not proportional to the voltage at any given time. Therefore, you can get two harmonic waveforms adding current and the third waveform that normally subtracts current happens to be 0 at that time.
From what I read, the worst case possible current overdraw was 1.717 times the rated load current of one phase.
Search for "3 phase harmonic current" and you'll get a better explanation than the one I just gave.
-- Mark Kent, WA
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| The increased harmonic current can only happen in 3 phase line to neutral loads and not split phase | 120/240 line to neutral loads. What happens is that the current is increasing as the voltage is | decreasing along the waveform. At some point, the current really drops. If they all drew current | proportional to their voltage, then the neutral currents would cancel. But the definition of | non-linear load is that the current drawn is not proportional to the voltage at any given time. | Therefore, you can get two harmonic waveforms adding current and the third waveform that normally | subtracts current happens to be 0 at that time. | | From what I read, the worst case possible current overdraw was 1.717 times the rated load current of | one phase.
I've seen other figures from 1.5 to 2.0 times the rated load current. But these seem to be making assumptions of full cycle square waves which will have a uniform declining distribution of odd-order harmonics. Many loads are more radical than that, and can have narrow spikes in their current waveform, leading to an uneven distribution of harmonics that will more favor the odd multiple of 3 harmonics (3rd, 9th, 15th, 21st, 27th, 33rd, known as "triplens"). More of the energy will be in those harmonics that don't cancel out than in the non-multiple of 3 harmonics.
| Search for "3 phase harmonic current" and you'll get a better explanation than the one I just gave.
Or for "triplens".
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| | |> I understand the common switching power supply for computers switches |> on the current when the voltage is high enough to replenish the tank it |> keeps. So it would see the smaller the computer load relative to the |> power supply capacity, the narrower the current pulses would be. Once |> they are smaller than 50%, you could be going past the 200% mark on the |> neutral, if all your load is these power supplies. For a few computers |> in an office, maybe that would never happen. But for building a large |> data center, I suppose it could get serious. | | So you maintain that, as the load drawn from the computer power supply | DECREASES, the neutral load INCREASES.
No.
As the computer power supply CAPACITY increases over its ACTUAL LOAD, the period of time needed to refill capacitors back to full charge is a SMALLER portion of each AC cycle.
| Lets go with that. I'm open minded. | | What is magic about the 50% current pulses? Why would a <50% pulse result in | 200% neutral current? | | Not arguing your point, just looking for an understanding of the theory | behind it.
It is easier to illustrate with 33.33% current pulses. Given 3 computers, each on a different 120 volt leg (phase) of a 208Y/120 volt three phase service, And assuming identical characterists of these 3 computers (for simplicity of illustration), the current waveforms might look like:
hot leg A: __ __ __ / \________ ________/ \________ ________/ \________ ________ \__/ \__/ \__/
hot leg B: __ __ __ ________/ \________ ________/ \________ ________/ \________ \__/ \__/ \__/
hot leg C: __ __ __ ____ ________/ \________ ________/ \________ ________/ \____ \__/ \__/ \__/
neutral leg (opposite polarity of hot legs): __ __ __ __ __ __ __ __ __ / \ / \ / \ / \ / \ / \ / \ / \ / \ \__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/
If there are sufficient number of computers on each phase to draw an average of 30 amps on each phase, then at 33.33% duty cycle, the pulses would peak at 90 amps. But the average is still 30 amps, so the load is effectively 30 amps in terms of energy usage, heating, etc. However, the neutral wire has all 3 sets of 90 amp pulses overlaid on it, so its overall load is that 90 amps. And it is at 180 Hz (150 Hz in Europe).
Were these loads linear and had sine wave currents, then where the meet at a common neutral, there would at any instantaneous time be a balance of all 3 currents, adding positive and negative values, to equal zero. The worse case in linear loads is when only one hot leg has a load and so the neutral would have the same load.
But with the non-linear load, the timings don't allow summing to zero at instantaneous times in the worst case of 33.33% pulse times or less.
I did say earlier that 33.33% pulse times would be easier to illustrate. If the times were instead 50%, there would be SOME overlap, and thus SOME amount of current summation to zero (or current between various taps on the neutral bus).
The severity depends on just how much of the load is this non-linear load and just how non-linear it is.
A computer power supply that is lightly loaded (for example a 400 watt P/S unit serving a board that uses 50 watts) will have very little draw from it's reserve. Thus it needs less time during the cycle peaks to replenish that reserve.
The concern isn't that the load is low, but rather, that the P/S capacity being higher exacerbates the non-linearity at that (constant) load level. The comparison here would be between a computer board using 50 watts from a 400 watt capacity power supply, vs, the same computer board using the same 50 watts, but from a 100 watt capacity power supply.
When enough of those 50 watt loads are piled up on a wye configured circuit that approachs the configured limit, say for example an average of 100 amps (300 amp peaks for 33.33% of the time), the neutral has the other polarity for all 3 circuits for a total of 300 amps (300 amp peaks, but filled in at 99.99% of the time). If the neutral is rated for 100 amps, it's now in a 300% overload condition.
This is the worst case I'm referring to.
Real life cases perhaps don't typically involve such radical departure of power supply capacity from computer actual usage. Real life cases don't involve all the load being computers (unless it is a data center).
But lots of these power supplies readily operate up to 240 volts and I am wondering if that extra voltage capacity means even shorter current times, making the problem worse for those with such standard voltage.
Note, if running on single phase, 2 computers at opposite poles will have their current pulses in timed opposition (to the extent that they are identical) and their summation will be zero on single phase. It is three phase power systems that have this problem.
200% rated neutrals probably handles most every situation adequately. But there could be situations approaching 300% in rare cases.
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| Phil Howard KA9WGN | http://linuxhomepage.com/ http://ham.org/ |
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I read in sci.engr.electrical.compliance that snipped-for-privacy@ipal.net
Question', on Wed, 25 Feb 2004:

Yes. You have done well to work out these scenarios independently. But the situation isn't quite as bad a pure arithmetic addition of the current pulses in the neutral. The upper limit is about 200 %, not 300%.

But there is a balancing effect; these short pulses are associated with less than maximum loads on the power supplies. In other words, a 300 W supply would do this if loaded to say 100 W, but the currents involved would be related to 100 W, not 300 W.

Yes, it does. Smoke from the neutral cable.
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wrote: | I read in sci.engr.electrical.compliance that snipped-for-privacy@ipal.net
| Question', on Wed, 25 Feb 2004: | |>I see a lot of equipment (transfomers, switch panels) rated for 200% |>neutral current. But I can see scenarios where it could reach as high |>as 300%. Take, for example, an AC to DC power supply that uses three |>phases and a full wave rectification. Wouldn't these rectifiers only be |>conducting when the respective phase is the highest voltage? If this |>were happening and it were the only (very large) load, you'd have to |>make sure everything is well overrated for it (3x on neutral for wye, |>and 2x on each leg for delta). Does anyone even use stuff like that? | | | Yes. You have done well to work out these scenarios independently. But | the situation isn't quite as bad a pure arithmetic addition of the | current pulses in the neutral. The upper limit is about 200 %, not 300%. |> |>I understand the common switching power supply for computers switches on |>the current when the voltage is high enough to replenish the tank it |>keeps. So it would see the smaller the computer load relative to the |>power supply capacity, the narrower the current pulses would be. Once |>they are smaller than 50%, you could be going past the 200% mark on the |>neutral, | | But there is a balancing effect; these short pulses are associated with | less than maximum loads on the power supplies. In other words, a 300 W | supply would do this if loaded to say 100 W, but the currents involved | would be related to 100 W, not 300 W.
But circuit sizing is not usually done for computers based on the power supply capacity, but rather, the actual DC load itself. So a computer with a 300 watt p/s and a 100 watt load could be considered a 100 watt computer. Now put 100 of those computers on one leg of a 208Y/120, and 100 more on the 2nd leg, and 100 more on the 3rd leg. If those computers are drawing power with a narrow enough pulse, you could see 250 amps on the neutral, but only 83.34 amps on each leg.
Keep in mind I'm only describing worst case scenarios which include both very narrow current pulses (and I'm not sure that happens very often) and the bulk of the load being this type (generally not an issue even in an office environment). Real life probably doesn't even see 150% very often.
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| Phil Howard KA9WGN | http://linuxhomepage.com/ http://ham.org/ |
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John Woodgate wrote:

Look at this problem the other way around. Assume that the 100 W per system is fixed. The situation becomes worse as the capacities of the typical system's power supplies are increasingly overrated.
The question is: If one rates the circuit neutral based upon the connected power supply ratings rather than the estimated system loads, will this overrating compensate adequately for the proportional increase (relative to real load) of the neutral harmonics?
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note to spammers: a Washington State resident
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I read in sci.engr.electrical.compliance that Paul Hovnanian P.E.
'Wire Size/Load Question', on Wed, 25 Feb 2004:

The same effect occurs to some extent when increased demands are made for 'hold-up' when the supply voltage decreases or is interrupted, even though there is no increase in system capacity (as opposed to 'capacitance'!); the conduction angle of the rectifiers is still decreased.

Yes. AFAIK, there is no condition in which the ratio of the r.m.s. triplen components to the **full load** fundamental current increases with decreasing conduction angle, at least down to 36 degrees, below which the inrush current becomes a big problem and it is, in any case, difficult to get a low enough total resistance in the rectifier circuit economically.
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wrote:
| Look at this problem the other way around. Assume that the 100 W per | system is fixed. The situation becomes worse as the capacities of the | typical system's power supplies are increasingly overrated.
Good way to look at it.
| The question is: If one rates the circuit neutral based upon the | connected power supply ratings rather than the estimated system loads, | will this overrating compensate adequately for the proportional increase | (relative to real load) of the neutral harmonics?
If the current approached linear as the DC load approached power supply capacity, that might work. But I doubt that happens. Maybe there is a virtual point where it could. What someone would need to do is measure the characteristics of a wide sampling of power supplies to see if such a thing could be done. I suspect there is a lot of variation in quality and design that could affect this.
Fortunately, only the radical cases (like a data center with hundreds or more computers) would need to worry about it. There is a "safe harbor" that can be used if these measurements cannot be made, and that is to assume the 300% worst case, and derate the stuff with 200% neutral to 66% of rated capacity, and the rest to 33% (rather expensive).
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| Phil Howard KA9WGN | http://linuxhomepage.com/ http://ham.org/ |
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I read in sci.engr.electrical.compliance that snipped-for-privacy@ipal.net
Question', on Wed, 25 Feb 2004:

The matter has been thoroughly investigated many times, largely in connection with the provisions of IEC/EN 61000-3-2 on limitation of harmonic current emissions into the public low-voltage supply. Practical ranges of rectifier conduction angle are approximately 36 degrees (below which inrush current becomes a problem) and 72 degrees (above which efficiency is much too poor).

Instead of derating, new neutral cable can be installed. Still not cheap, but at least the outcome of the expense is that the system has been uprated rather than downrated.
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The good news is that nothing is compulsory.