3 Phase Question - Page 1
Forum section: electronics / beginners

author: nick.sek
author_rank: Regular Contributor
timestamp: May 14, 2012, 07:01:16 pm
content: If we have a 4 wire wye configuration on a three phase circuit; which has an unbalance load and we add up the phase currents (Ia, Ib, Ic) to find In. And we get a current there must be a voltage; so dictates the wonderful Ohms Law. Is there a way to calculate this voltage? If there is no load on the neutral line?If someone can point me towards a theory I would be really thankful.ThanksNick

author: IanB
author_rank: Super Contributor
timestamp: May 14, 2012, 07:22:57 pm
content: When you think about voltages, you must always think about two measurement points in your system. There is no such thing as a "voltage" as a point property.So when you ask for the voltage on the neutral wire, you must answer the question "Voltage relative to what point?"If you measure the voltage on the neutral wire relative to the neutral wire, then the voltage is always zero by definition.Maybe you can measure the voltage on the load end of the neutral wire relative to the supply end. Now you have a voltage difference along the wire proportional to the current flowing along the wire. So you can use Ohm's law if you know the wire resistance.Perhaps you may ask about the voltage between any of the phase conductors and the neutral wire? Well this will depend on the source characteristics. If you have a perfect voltage source with no source impedance, then the voltage will always be the design voltage.If you have a source with a non-zero impedance, then you can consider how the loads interact with the source impedance and see how the load voltages vary accordingly.

author: Bloch
author_rank: Unknown
timestamp: May 14, 2012, 07:59:14 pm
content: Quote from: nick.sek on May 14, 2012, 07:01:16 pmThere are some things there don't add upIf Ia, Ib, Ic are not the same then there will be a In. That is ohms lawDid you mean that the neutral line is not connected ?

author: IanB
author_rank: Super Contributor
timestamp: May 14, 2012, 08:09:42 pm
content: Quote from: Bloch on May 14, 2012, 07:59:14 pmIsn't that Kirchhoff's current law?

author: ejeffrey
author_rank: Super Contributor
timestamp: May 14, 2012, 08:20:23 pm
content: You have to add the phase currents including their time dependence. If you have linear loads you can add them as phasors. Remember to use the current phase, not the voltage phase.

author: Kremmen
author_rank: Super Contributor
timestamp: May 14, 2012, 09:06:52 pm
content: Quote from: nick.sek on May 14, 2012, 07:01:16 pmThere's lots of theory on multiphase networks but all mine are in books and cannot be bothered to start a net search.Your question has an easy answer though; do as efferjey says and calculate using a phasor diagram. Assuming a resistive load the voltages and curents will be in phase. In case you are just interested in the magnitudes, a static snapshot will do and you don't need to consider the time evolution of the system.So, select one phase (say Ia) and arbitrarily fix its phase angle (theta) to 0. Then the other ones will be (Ib) at 2pi/3 radians (120 degrees) and (Ic) at 4pi/3 radians (240 degrees).The magnitudes in x direction will be Iicos(thetai) and in y direction Iisin(thetai) where i = a,b,c.Next sum all the x magnitudes to get the sum, and same for y magnitudes. The final neutral current will be the vector of these magnitudes, i.e. In = sqrt(Ix^2+Iy^2) at angle atan(Iy/Ix).For example say you have a system with phase currents of 500, 400 and 300 amps respectively. The y axis magnitude would be -86.6 amps and the x magnitude +150 amps. Their vector sum would be 173 amps pointing at -0.52 radians or -30 degrees.This current will pass through the neutral wire. As to the voltage, that would be a voltage loss along some stretch of the neutral conductor and defined by the resistance of said stretch of wire. If you look at a real system, say a star connected heater with connected neutral wire, where some of the elements have e.g. ground faults, then the voltage you see at the star point would most naturally be measured against the local ground wire termination - that providing the closest approximation of true ground potential.

author: FreeThinker
author_rank: Frequent Contributor
timestamp: May 14, 2012, 09:32:56 pm
content: Try reading thishttp://forums.mikeholt.com/showthread.php?t=93575I think this is for non inductive loads as I seem to recall you would need to take into account the reactive components of each phase, but it is a VERY long time since I did this.

author: IanB
author_rank: Super Contributor
timestamp: May 14, 2012, 09:44:35 pm
content: A good process when faced with questions like this is to do a few thought experiments on simpler cases.So for instance, instead of the three phase case consider the single phase case with one phase ("live") wire and one neutral wire forming a circuit.If there is a load on the live wire the current must be returning on the neutral wire (assuming no ground leaks). If there is current in the neutral wire, what then is the voltage on the neutral wire?After answering this question, it is then possible to extend the insight gained to the three phase case.

author: FreeThinker
author_rank: Frequent Contributor
timestamp: May 15, 2012, 12:22:32 am
content: @IanB. That will work for balanced loads but is still not a great method. In a balanced load the neutral can be disconnected and still work as normal (ie there is no neutral current flowing), in an unbalanced load the star point will float around zero with respect to each phase dependant on the load imbalance as the imbalance current shares the return paths of the other two phases, this in turn causes the voltage per phase to vary which can cause significant current fluctuations if loads are dynamic. As an apprentice (Many years ago) we had a neutral cable burn off in a distribution board we lost loads of single phase motors because they were getting over 300 volts across them because of the loading imbalance, not nice!

author: IanB
author_rank: Super Contributor
timestamp: May 15, 2012, 01:32:40 am
content: Quote from: FreeThinker on May 15, 2012, 12:22:32 amI think you misunderstand the idea of looking at a simpler case and thenextendingthe insight gained. Extend being the operative word.As to the unbalanced load--"the star point will float around zero"--what zero, exactly? Isn't the neutral wire the zero reference? How can it float when it is by definition always zero? (If you tie the neutral wire to ground it will be anchored very strongly to ground as well.)I don't doubt the reality of motors burning out, but the cause is not explained in quite the way you propose it to be.

author: Rufus
author_rank: Super Contributor
timestamp: May 15, 2012, 05:29:41 am
content: Quote from: nick.sek on May 14, 2012, 07:01:16 pmOhms law being wonderful tells you that voltage is current multiplied by resistance. Given you know the current all you need is the resistance between where you think 'this' voltage is and where you think the 0v reference for 'this' voltage is.

author: Kremmen
author_rank: Super Contributor
timestamp: May 15, 2012, 05:47:18 am
content: Quote from: IanB on May 15, 2012, 01:32:40 amThe problem here is that a single phase case is not directly extendable to multiphase. For single phase you essentially calculate with magnitudes only (yes, the voltage and current phase may differ but that is not the point). For multiphase you necessarily must consider the system from the vector viewpoint. Anything else is kicking yourself in the foot all the time. Now as to the neutral point, the voltage, as indeed the phase voltages as well, are referenced to "earth", and that usually means the Earth. The spike driven into ground at the power station(s) generating the voltage in the first place. Using that logic it is entirely possible and reasonable that the neutral point of an unearthed star can sail at any voltage subject to the load asymmetry. Only when you bond the star point with earth using a conductor, a neutral current is possible and then the impedance between the star point and earth comes into play. But to find out the magnitude and phase angles of the neutral current and voltage, you must calculate everything using vectorial representations. In single phase systems currents and voltages are just magnitudes that may have a mutual phase angle. In multiphase systems all voltages and all currents are vectors and their combinations are vector sums and products. Calculating by simple magnitudes gets you nowhere.

author: IanB
author_rank: Super Contributor
timestamp: May 15, 2012, 06:45:26 am
content: OK, it seems I must explain better.In the single phase case we have a neutral wire with some current in it. Let's say the originating end of the neutral wire is tied to ground at the transformer/generator and this becomes our zero reference. Now with current flowing in the neutral wire the load end of the neutral wire will float at a voltage caused by the resistance of the wire and Ohm's law.In the three phase case we consider the three loads and do the vector sum as you indicate. When we are done we can consider the magnitude of the resulting current flowing in the neutral wire. The load end of the neutral wire will now float at a voltage caused by the resistance of the wire and Ohm's law. Just the same as the single phase case.On the other hand, we may consider the load end of the neutral wire as our voltage reference, which makes it zero by definition. We can do this in the single phase case and in the three phase case.In the single phase case we can consider the voltage difference between the neutral (zero) and the live conductor. It will be what it is according to the conditions. In the special case where the conductors have negligible resistance and the voltage source has zero impedance then the voltage will remain equal to the source voltage regardless of the load.In the three phase case we can consider the voltage difference between the neutral wire and each of the three phase conductors individually (the phase voltages). If the load is unbalanced, each of these phase voltages may be different, but each voltage will be what it is according to the conditions just as in the single phase case. In the special case where the phase and neutral conductors have negligible resistance and the voltage source has zero impedance, then each of the three phase voltages will remain equal and balanced, regardless of the load imbalance. Any "floating" of the voltages depends only on the current flowing and the resistance (impedance) of the conductors and the impedance of the power source (transformer, generator).Lastly, we may note that we can use phasors and vector sums equally well in both the single phase case and the three phase case when determining the current in the neutral wire. It is trivially redundant to do so for the single phase case, but it still works. So if we look at the problem the right way, we can see no difference in principle between the single phase case and the three phase case.

author: Kremmen
author_rank: Super Contributor
timestamp: May 15, 2012, 07:46:18 am
content: Yes, all true but i don't see what point you are trying to make, exactly.If we are talking about power distribution things then we should use the conventions normally used in that context. For the purposes of the distribution network the load end, whatever node, is not a reference. For the power guys the distribution network is the circuit and that is referenced to planet earth. Only for some special cases is anything else done but this would not be one of those. What we are talking about is a perfectly typical case where you expressly should use the true ground as reference. Othewise you lose sight of the neutral circuit and will miss problems there.Also, for the power guys all conductors are impedances, no such thing as an ideal conductor or source. Additionally, they use their own calculation methods (such as the positive, negative and zero sequence method). A brief intro into that you can find in a technical app note from my previous employer athttp://search-ext.abb.com/library/Download.aspx?DocumentID=1SDC007101G0202&LanguageCode=en&DocumentPartId=&Action=Launchannex B, page 32. All calculations are against the earth.The attachment contains some medium and low voltage distribution network short circuit calculations that i found in my PC archive. Those might be illuminating at least to demonstrate the point that in practice the subject is not without complications.

author: ejeffrey
author_rank: Super Contributor
timestamp: May 15, 2012, 08:27:27 am
content: Quote from: IanB on May 15, 2012, 01:32:40 amZero being the voltage it should be at -- the voltage of the neutral supply you have disconnected from your load or equivalently the average voltage of the three phases. This causes the individual phase-to-neutral voltages to vary anywhere between zero and the phase-to-phase voltage of the supply.

author: FreeThinker
author_rank: Frequent Contributor
timestamp: May 15, 2012, 11:45:31 am
content: Quote from: IanB on May 15, 2012, 01:32:40 amWhen the neutral is floating the ONLY reference you have is local ground (earth) and indeed the neutral conductor is no more than an Extension of the Generators earth point. Think of it this way as a two phase supply (or dual rail supply) with a voltage divider across the supply(say +- 5v) with a 0v tied to the centre.OK you will have 5v dropped across each resistor and the load currents will be determined by ohms law. Remove the 0v and the resistors become in series across a 10v supply and again ohms law applies. As you can see the ONLY way you would get 5v across each resistor was if they were the same value (balanced load).

author: IanB
author_rank: Super Contributor
timestamp: May 15, 2012, 02:19:04 pm
content: Quote from: ejeffrey on May 15, 2012, 08:27:27 amThis is true and logical of course and I get that it's the way engineers think about the problem.But there is nothing to stop you drawing a control envelope around the load for the purpose of analysis and making the star point your reference where the neutral wire connects to the load. In that case each of the phase voltages seen by the load would appear to vary.

author: ejeffrey
author_rank: Super Contributor
timestamp: May 15, 2012, 03:02:24 pm
content: Quote from: IanB on May 15, 2012, 02:19:04 pmTechnically this is true, but it is a weird and potentially misleading way to look at the system. The reason is that the phase-to-phase voltages remain constant. So it isn't that each phase varies. Rather, all three phases vary together relative to the star reference. Once you recognize that, it is much more natural to say that star node is the floating one and the other three remain fixed relative to an imagined reference point which is the mean voltage of the three phases.

author: FreeThinker
author_rank: Frequent Contributor
timestamp: May 15, 2012, 04:06:48 pm
content: Quote from: IanB on May 15, 2012, 02:19:04 pmThats a bit like tying your boat to a jetty and saying the jetty moves as the tide comes in and out, you need a point of reference and that is earth. Whilst you can use the star point as a point of reference any measurements taken from it must be with regard to earth as this is where the current wants to get to, that is its zero potential point and no other.

author: IanB
author_rank: Super Contributor
timestamp: May 15, 2012, 05:17:13 pm
content: It seems like lots of discussion here is going at crossed purposes. I think some people have different configurations pictured in their mind to what others have. I'll make a diagram later to clarify what I am talking about, then it should become clearer.

author: IanB
author_rank: Super Contributor
timestamp: May 16, 2012, 04:33:47 am
content: Quote from: nick.sek on May 14, 2012, 07:01:16 pmOK, here are pictures.Firstly, the 4 wire wye configuration is important. We need to consider the problem as stated.Secondly, I said "consider the single phase case first".So here is the single phase case:Note how the neutral wire from the transformer is grounded near to the transformer. This represents our reference point for measuring voltages. The neutral wire may also be grounded near to the load, but this does not materially change the analysis.We can find out from analyzing the circuit what the current in the neutral wire is. A current in the neutral wire causes a voltage drop in the neutral wire and soVNwill differ from the ground reference point accordingly. If the load is disconnected so that no current flows, thenVNwill be at zero potential.Next, we can extend this to the three phase case. We are told we have a 4 wire wye configuration:Similarly to the single phase case we can find out from analyzing the circuit what the current in the neutral wire is. The necessary extension from the single phase case is to consider phase angles when summing the three load currents, as previously described in the thread. But after doing that, the analysis proceeds in the same way. A current in the neutral wire causes a voltage drop in the neutral wire and soVNwill differ from the ground reference point accordingly. If the three loads are perfectly balanced so that no neutral current flows, thenVNwill be at zero potential.

author: Kremmen
author_rank: Super Contributor
timestamp: May 16, 2012, 05:25:56 am
content: Yes. These pictures are what i have been discussing the whole time.but when you said:Quote from: IanB on May 15, 2012, 06:45:26 amit got confusing. Yes, there are cases when one would make that measurement. Like when one measures the voltage of an ordinary wall socket. But that is not the way one does circuit analysis, esp. multiphase.QuoteThere is the drawback that in this analysis the phase voltages change magically. Heavily loading one phase will mysteriously raise the voltages of the other phases. If the cause of the loading is itself a multiphase load with cos phi less than unity, the behavior of the voltages and currents gets very confusing. There is no benefit in fixing the reference in this way and that is why it is not done either. It is also a somewhat risky method, since fixing the load side neutral point to zero, we lose the neutral current. We know it is there but to find out what it is, we need to consider the earthed generator ground point floating at a nonzero voltage. Someone already compared this to the jetty moving instead of the boat.Shouldn't we agree that the power guys have got it right and true ground is the reference that creates least extra complications?

author: IanB
author_rank: Super Contributor
timestamp: May 16, 2012, 05:32:16 am
content: Quote from: Kremmen on May 16, 2012, 05:25:56 amYes, I see that problem. I stand corrected.QuoteYes, agreed.

author: ejeffrey
author_rank: Super Contributor
timestamp: May 16, 2012, 09:55:16 am
content: Quote from: Kremmen on May 16, 2012, 05:25:56 amNo! Not earth ground. Except in very specialized circumstances earth connections are not supposed to carry any current, not even fault current. The planet earth ground reference is a red herring in 99% of electronics. The resistance of a ground rod is quite large compared with copper wires. The proper zero reference for a 3-phase wye system is the star node of the distribution transformer. That is where the phase currents want to return to by the lowest impedance path. The fact that it is connected to a ground rod is mostly irrelevant.The reason the distribution grid is earthed is for lightning protection and to prevent geomagnetically induced currents from generating large inductive voltages.There are a few exceptions. Single wire earth return power distribution is used for long distance distribution in some places. This requires special grounding rods with low impedance to handle the large currents. I think the old telegraph system used an earth return for the signal currents. These are the exceptions rather than the rule, and by and large referring any electrical signal to the planet earth is at best misleading and at worst wrong.Ground of course matters quite a bit for RF antennas and signal propagation through free space, but not for signals passing through cables.

author: Kremmen
author_rank: Super Contributor
timestamp: May 16, 2012, 11:45:47 am
content: Oh, come on. Could you maybe read through the thread and get the point? This is not about _electronics_, but power distribution.The resistance of a ground rod of some specific installation is a local issue and is largely irrelevant to the general subject.Regarding networks with neutral lines; do the math and you will see in no time why the power companies are not keen to run thousands of kilometers/miles of unnecessary extra conductor. As they do not. The ground is not used as a "return wire" just because the (largely) symmetrical load in the network does not especially need it for that. But it _is_ used as a reference.Grounding does not protect air wires from lightning strikes unlike lightning wires that you see above high tension current carrying lines. Those do. Still, lightning strikes where it wants and one just has to live with the consequences of the resulting surges. In premises the fixed wiring _is_ protected from overvoltages provided a good (preferably Ufer type) grounding and proper surge arrestation is in place.I don't know how things are done in your part of the world, but here the practice is that distribution is 3 phase no ground and the final low voltage leg comes from a 20 kV/400V D/y transformer with neutral wire connected to the star point. The neutral wire is grounded at the transformer, again at the entry to premises and circa every 200m thereafter if there is subdistribution from the entry panel. The neutral and PE (Potential Equalization or ground) are separated at the main entry panel and are separate after that to enable proper operation RCD devices. After this separation the code considers the N (neutral) wire a current carrying one and the PE (ground) not. So the low voltage circuit is definitely grounded, and there definitely are neutral wire currents.Just where the grounding turns into a red herring in this scheme i fail to see.That grounding does not matter for signals (and power) passing through cables is just simply dead wrong. Study the subject before uttering such nonsense, would you please.