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An analogy for three phase current flow
 

 

Subject - An analogy for three phase current flow

June 7, 2007
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An analogy for three phase current flow

by Eric Stromberg, P.E.

 

An attendee of a seminar that Mike Holt held in Puerto Rico asked the following question:

 

Q:  I don't want to be a bother, but I'm looking for the explanation of an age long notion in my industry, PREPA (Puerto Rico Electric Power Authority). The people in charge of the distribution standards here indicate that the neutral or common line can be smaller than the live lines.  The revelations during your seminar here got me thinking about stuff like that, which I always felt was wrong.  The gauge chosen for the neutral is usually half that of the other lines, and I say this is wrong because at any given moment it may carry the same amount of current as the line.

 

In your first answer you stated that if all phases carry the same current, the neutral current will be zero...how? I understand this if the load is purely 3 phase, there's no use of the neutral. But for line-neutral loads, how? And I don't mean mathematically, I mean, what's happening with the flow of current there?

 

 

A:  Julio. You've given me a significant challenge with your statement "and I don't mean mathematically." Effectively, if the currents are balanced on each phase, they add up to zero on the Neutral. Think of a single phase 120/240 Volt system. If 'phase' A has 20 Amps on it and 'phase' B has 0 Amps on it, the Neutral has 20 Amps on it. If both 'phases' have 20 Amps on them, the Neutral has zero Amps. If 'phase' A has 20 Amps on it and 'phase' B has 10 Amps on it, then the Neutral has 10 Amps on it. The Neutral carries the imbalance current of the ungrounded conductors.

 

 

OK, we're going to try to do this totally non-mathematically. After typing that last sentence, however, I've been sitting here thinking of exactly how to describe this without mathematics. Trouble is, mathematics is the best way to describe it.  The "adding up to zero" is a convenient mathematical construct that has no real world analogy. This is because the voltages are time-shifted. Our brains don't think like this. Try to imagine three garden hoses, each at the point of an equilateral triangle, spraying water into a single point. Now, imagine that each hose is alternately spraying water and then 'sucking' water back, according to a specified frequency. If all the hoses are spraying/sucking water, but are time shifted between them, it turns out that whatever water is being sprayed into the center (at any given time) is also being sucked up by the other hoses. As long as each hose is spraying/sucking the same volume of water, there is no residual water laying on the ground. If, however, the hoses are not spraying/sucking the same volume of water, an amount of water will be deposited on the ground. This is the 'imbalance' water that returns to the source via a different path.

 

 

Julio, you are absolutely correct in that, if only phase A (or B, or C) loads are present, then the full current of phase A will also be present on the Neutral. If full current is present on two phases, then full current is present on the Neutral.

 

 

I'm not sure if any of this helps. It is a difficult concept that takes some time to figure out.

 

Q: OK Eric, I made a drawing and thought about what you explained. My main issue is if the Neutral current might be zero or near zero, then how multiple, unbalanced, line-neutral loads work?

 

If I understand correctly what you explained, there is a sharing of current between the lines, consequently, at a given time, 2 loads might be sharing current from a different phase. For example, there's a load in A-N and another in B-N, at a given time both loads will have an A-B current; depending on the system, it could be A-C and B-C on each load. Is this thinking correct?

 

If so, that makes clear why the Neutral current is usually low or may reach zero, and indeed your analogy is an excellent one. A new question would be, why is that path used instead of the lower resistance path of the neutral?

 

A:  You’re right.  This where my analogy falls short.  Let me change my analogy from hoses spraying out on the ground to hoses connected to a tank.  Imagine three hoses connected to a tank.  We will label the hoses A, B, and C.  The hoses alternately fill and drain the tank.  While A is filling at full volume, B and C are both draining at half volume.  While B is draining at full volume, both A and C are filling at half volume.  While C is neither filling or draining (zero crossing), A is at .866 and B is at -.866 volume (or vice versa).  Point is, no matter where you are, on a balanced circuit, whatever hoses are filling the tank, the other hoses are draining the tank to the same exact degree.  Here is a graphic.

 

 

Image 1

 

You can see on the graphic that no matter where you are in time (represented by a vertical line cutting through all three phases), the rates of “fill” and “drain” equal each other.  You can view the currents as complimentary. 

 

Now, add the condition that the level in the tank must stay exactly the same.  This necessitates all currents to be equal.  If the currents are not equal, the level in the tank will change.  In order to account for this, connect a fourth hose to the tank and call it neutral.  The fluid imbalance will flow in the neutral hose.

 

Now, imagine two tanks.  Hoses A and B are connected to the first tank and Hose C is connected to the second tank.  There is a neutral hose between tanks and back to the source.  If A, B, and C all have the same volume, then the rates of fill and drain are again the same.  In this case there will be current flow back and forth between the tanks, but no current flow in the neutral hose back to the source.  In the case where there is a slight imbalance, there will be equalization flow between the tanks and a small imbalance flow back to the source.

 

As for how the current from two phases results in a neutral that is the equivalent of one phase, look at this:

 

Image 2

 

The addition of Phase A current, Red, and Phase B current, Blue, results in an equivalent (time-shifted) sinusoid (black).

 

Looking at the graph a little more closely, you will notice that the imbalance sinusoid (black) is exactly opposite that of the missing Phase C sinusoid.  Therefore, when the Phase C sinusoid is added, the imbalance sinusoid goes to zero.

 

 

Eric Stromberg, P.E.

Stromberg Engineering, Inc.

www.strombergengineering.com

eric@strombergengineering.com

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Comments
  • The reason the neutral conductor in an electrical distribution system is sometimes found to be a smaller size than the phase conductors is when a multi-grounded neutral distribution system is used.

    Unfortunately the majority of the electrical distribution systems in NORTH AMERICA are multi-grounded neutral distribution systems, which are dangerous and hazardous. The Electric Power Research Institute (EPRI) in one of their documents state that they expect to find only 40 percent of the return current flowing over the electrical neutral conductor while 60 percent will return to the substation over and / or through the earth.

    Thus, if only 40 percent will flow over the neutral conductor why bother using the same size as the phase conductor when installing a multi-grounded neutral distribution system? The reduced neutral size saves costs.

    We have measured as high as 80 percent of the phase current flowing back to the substation over the earth. This stray current, the correct term, NOT stray voltage, cannot be controlled in how much flows over the earth and cannot be controlled where over or in the earth the stray current flows. It can and has flowed through humans and cows and pigs causing them harm.

    An Enlightned Electrical Engineer, P.E.
    Reply to this comment

  • To Eric Stromberg, P.E.

    Mr. Puerto Rico stated “The people in charge of the distribution standards . . .” not the electrical inspector inspecting an office building or industrial complex where 5 wire systems are used. Your answer is correct for everything except multi-grounded neutral distribution systems used in the majority of North America electrical utility systems.

    So, Mike, Mr. Puerto Rico question was not answered. The answer to Mr. Puerto Rico’s question is:

    The reason the neutral conductor in an electrical distribution system is sometimes found to be a smaller size than the phase conductors is when a multi-grounded neutral distribution system is used. As I recall the NESC at one time permitted a reduced neutral conductor and I have not researched this statement.

    Unfortunately the majority of the electrical distribution systems in NORTH AMERICA are multi-grounded neutral distribution systems, which are dangerous and hazardous. The Electric Power Research Institute (EPRI) in one of their documents state that they expect to find only 40 percent of the return current flowing over the electrical neutral conductor while 60 percent will return to the substation over and / or through the earth.

    Thus, if only 40 percent will flow over the neutral conductor why bother using the same size as the phase conductor when installing a multi-grounded neutral distribution system? The reduced neutral size saves costs.

    We have measured as high as 80 percent of the phase current flowing back to the substation over the earth. This stray current, the correct term, NOT stray voltage, cannot be controlled in how much flows over the earth and cannot be controlled where over or in the earth the stray current flows. It can and has flowed through humans and cows and pigs causing them harm.

    An Enlightned Electrical Engineer, P.E.
    Reply to this comment

  • I'll try once more. First of all a balanced neutral as a basic concept does not depend on 3 phase current or any two legs thereof. Imagine you had a SINGLE phase transformer with a center tap on the secondary. Wire this to the most simple balanced load possible: A fixed resister with a central tap. The central tap of the transformer is wired directly to the central tap of the resistor and the legs to each end of the load. Under normal AC service conditions there would be no current in the central-or neutral leg. The reason why is because of the application of Kirkoff's law for current (based on the law of energy conservation) which says the total current from any node of an electrical circuit must have an algerbraic sum of 0: Entering an leaving currents have opposite signs. This is an application of the most fundamental law in physics: The conservation of energy, which translates- you can't have any more than what you got. There's another law (entropy) that says-and not even that.

    Bob
    Reply to this comment

  • ... I almost had it ...in my mind, ....seeing it but a few of the calculations presented have mudded up the water a bit. ,..

    So I'm up for an experiment... what do ya say we take three of the milkmans cows run hoses between them .... throw the switch and see what happens

    I'll bring the Oreos ;)

    .......................... thanks again for the time and resources discussing the topic.

    Joe
    Reply to this comment

  • Joe , I think we need more oreos ! That's a lot of milk !

    Javier
    Reply to this comment

  • I always use a bicycle analogy when explaining the benefit of three-phase. With two legs, and your two pedals, you get a couple of dead spots in your stroke just before the pedal on either side gets to the top. But if you had three legs, you could have a third pedal aligned so it's on its power stroke while the others are in the dead spot. Thus a three-legged man could whip Lance Armstrong in a race.

    That's just for explaining the evening out of power delivery part vs. "two-phases." Anything beyond three phases is diminishing returns.

    The water is a very good analogy for balancing and I may "borrow" it next time it comes up.

    Matt

    Matt
    Reply to this comment

  • The algebraic sum of all currents in an electrical circuit is always zero. If 20 amps are flowing in hot leg A and -20 amps in B than that comes to zero. There is no current available for the neutral.The current we measure is a statistical average (RMS) over time. The vector approach is quite correct but overkill to explain linear neutral loads. We use the RMS because it cancels out opposing signs. The square of both a minus value and a plus value is always plus.

    Bob
    Reply to this comment

  • Hmmm.... Great expalnations....but I was unable to open the grahs and pictures that you provided... :( Thank you for your E-Mail. Allias, Safeguy.

    Safeguy
    Reply to this comment

  • Three phase loads Since this is not a perfect world the impedances in a 3 phase device is never perfect. and without a neutral return path the imbalance is turned into heat.

    rwbeaster
    Reply to this comment

  • Imagine that I had a single resister with a center tap as the only load on two legs of a 3 phase transformer. The current in both legs is the same: Call it C. Then attach the center tap to a wire running to the neutral leg of the transformer. The current in each of the hot legs is still C. The same power as before is flowing through both hot legs. That’s logical is it not? Then how can the neutral be carrying any power? We'd be creating power from nothing if it too had power (hey, that's not a bad idea though) No power means no current. If you go back to the water analogy this might be clearer. Same amount of water flowing in both pipes we add a third than we must add more water if it has flow and the flow in the other two remain the same. The lack of current in the neutral of a balanced load is simply an example of the conservation of energy. There is an important catch. The current must be pure sinusoidal and we are measuring RMS. In a non linear load conservation laws still apply but our measuring instruments must be more refined.

    Bob
    Reply to this comment

  • good analogy.

    Scott "C"
    Reply to this comment

  • Refrencing to you question on Balanced Three Phase Crkts. If you had 3 loads weather they are lights or motors, it doesn't matter.. Example each of light bulbs consume 100 watts one is connected on phase A the other is on Phase B and the other is on Phase C, then the return on the neutral is zero... Because the load is consuming all the current and there is nothing left. If one lamp was replaced by a lower wattage the what is left over will be the unbalance on the neutral. I think this is a simpler explanation...

    Gerald A Gioia
    Reply to this comment

  • Excellent analogy, Eric. It usually always helps to use the water analogy to explain with touchie-feelies what is invisible and only mathematical otherwise. Even though it is mathematical, a graphical analysis using vector lengths and angular relationships might also be of benefit. It DOES however mean that the reader has at least some geometry under their belt. But, if done to scale, it is easy to see with a pocket ruler how the lengths balance out to zero.

    Dan Lawrence, PE dan@djlawrence.eng.pro

    Dan Lawrence, PE
    Reply to this comment

  • In today's environment, there are many electronic power supplies whcih create a large amount of harmonic currents. To the worse, these harmonic currents can even overload the neutral wire. So full size neutral is preferrable in nowadays installation. In addition, we sould use true RMS ammeter to measure the exact current flowing through the phase and neutral wires.

    Michael
    Reply to this comment

  • I may be taking this whole question out of context please excuse me if this is the case.

    Julio's original question as I read it was "how can the "neutral", (N), conductor be a smaller gage than the "current carrying" conductor (A, B, C)."

    In a "blanced" system (whatever that may be),as you explained, all is fine the (N) is zero.

    My question is what happens to that smaller (N) conductor should the "balanced" system become "un-balanced" would not the smaller (N) conductor become exposed to an excessive load that the (N) conductor is not necesarrily rated for?

    Would it not be better to have the (N) conductor sized for the maximum load the circuit is designed for, ie the same as the (A, B, or C) conductor?

    Neil
    Reply to this comment

  • Thank you Eric Your detail analogy remind me electrical engineering school days when my professor explaining how 3-phase current flow - excellent explanation thru graphic Thanks Mike - in today's world these type of questions answer are necessary to keep brain working.

    Please update my e-mail address.

    Suresh Shah
    Reply to this comment

  • great explanation, I have been trying to explain the same thing without math to some helpers and now I can

    James Blankenburg
    Reply to this comment

  • If you can get ahold of a ripple tank. You can actually \"see\" how it works.

    Randy Buck
    Reply to this comment

  • Great analigy... But the last question is..... wait for it ...... WHY ????

    Why must all three phases eual out. I get it.... Phase A=20a phase B=15a phase C-10a nutral= A-0+B-5a+C10A = 15a flowing tward netral.

    AND, as I read it, the nutral must be the same size as the load wires...NOT smaller

    Randy
    Reply to this comment

  • I have been trying to explain this to the mechanical group that I work with and have never been able to it as effectivly as you did here. Thank you very much.

    Dale Kessler
    Reply to this comment

  • The graphs caused me to remember an "ah ha!" moment that occured many years ago in an electrical power engineering class.

    We all know the three phases are "time shifted" 120 degrees with respect to each other. This leads to zero neutral current if the loads are all equal phase-to-phase, which is typical in a three-phase motor and the main reason Nikola Tesla invented the concept in the first place.

    Now suppose each pair of phases is connected to an electromagnet, the two poles of each electromagnet are aligned along a radius of a common circle, and the electromagnets are equally spaced around the circle, 120 degrees apart. What is the resultant magnetic field in the center of the circle? I almost fell out of my chair when (with the aid of graphs) it was shown that the magnetic field is constant and rotating! Try to come up with a water hose analogy for THAT.

    Howard B. Evans, Jr.
    Reply to this comment

  • Very good analogy, but I still am not sure if the neutral can be smaller than the other conductors? I'm a mechanical engineer, I need some heilp with electricity.

    Jack Kagan
    Reply to this comment

  • Eric

    It is great to see you back writing for the forum!!

    In reality the water tank analogy may not completely illustrate the issue and please forgive this old dairyman for commenting on your answer.

    If I am not correct on the following-- PLEASE let me know where I am going wrong because I am also finding it difficult to explain this extremely complex yet basic electrical building block.

    In your example, if you have two water hoses (phases) filling at the same pressure (volts) and rate of flow (amps) and only one drawing the level down, you will be raising the level in the water tank, at that moment.

    When you have two hoses drawing down and only one filling, you are lowering the tank level, at that moment.

    I believe that example would create a surging action in the water tank (neutral conductor current).

    If I may offer the following (please note this example is for a 3 piston water pump).

    Each water hose (piston or phase) is alternating from pressure (filling) to suction (drawing down) sixty times (cycles) per second.

    And each water hose (piston or phase) is separated by 120 degrees (rotation position of the 3 cylinder piston pump).

    This means that at any given moment with equal pressure (volts) and volume (amps) you would have each of the 3 hoses filling and drawing down at three different times (120 degrees of rotation) during one single complete pump rotation (120, degrees, 240 degrees and 360 degrees).

    In the above description, you will have equalized the phase duration and flow (1/3, 1/3 and 1/3 of current flow) in each water hose (phase) filling and drawing out in each cycle (rotation) of the pump (generator) or stated as a shift in phase of 120 degrees between each phase.

    Does the above help clarify the explanation or just "muddy" the waters??

    Just a dumb dairy farmer, Chuck Untiedt

    Chuck Untiedt
    Reply to this comment

  • I'm thirsty!

    john villafane
    Reply to this comment

  • Thank you Julio for asking the question, one of my electricians asked me the same question the other day. I can use Eric's analogy to answer his question. Thanks Guys

    Lou
    Reply to this comment

  • Thank you Eric for the analogy.

    I thought it was insightful as to why good connections of the neutral in branch circuits .are so important.

    A TV repairman told me sometime ago.... \" a good many bad electricians have made me a bundle\"

    Joe
    Reply to this comment

  • The loads shown appear to be linear, what is the resulting effect of the nuetral current in a non-linear load situation?

    Greg
    Reply to this comment

  • An excellent analogy that should help clarify current flow for anyone who will ponder it. I was just explaining how a transformer works to my son this afternoon (for the tenth time). I had to remind him to think in terms of how water flows in order to get a mental picture of what electricity does. Thanks again for a very good analogy. Bob

    Bob Boan
    Reply to this comment


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