wye connected loads

[LarryFine]

The current through the neutral conductor is actually the current which results from moving the floating neutral point to its grounded position. (Did that make any sense?)

It does, and I've described it like that many times. The same is true for the neutral of a 120/240v 1ph system: its current is driven by pulling the neutral to 0v.

Of course, theoretical discussions like this ignore voltage drops, but, in real life, we treat neutrals as CCC's and don't rely on them for EGC'ing beyond the main.

 

[Cold Fusion]

Originally Posted by Rick Christopherson
The current through the neutral conductor is actually the current which results from moving the floating neutral point to its grounded position. (Did that make any sense?)

---Of course, theoretical discussions like this ignore voltage drops, but, in real life, we treat neutrals as CCC's and don't rely on them for EGC'ing beyond the main.

The term "grounded" has shown up a few times. I've been translating as, "connecting the load wye point to the supply wye point."

I'm pretty sure chris knows that connecting the load wye point to earth isn't going to make much difference.

cf

 

[Smart $]

What you say is correct if the definition of balanced load includes both magnitude and vector. If we use only current magnitude [and assume a phase displacement of 120? for each current], then your statement is incorrect.

Without magnitude and direction, how would the currents cancel? Without direction, even when perfectly balanced, the currents would just be additive.

Phasing was implied as revised in the quote of my comment. I was of the impression anyone intelligent enough to read and comprehend the scope of this thread topic would have grasped the implication... but apparently my presumption was off-base :smile:

Your next quote below helps me understand why you made this statement, but it doesn't make is any less incorrect.

The formula I referred to is as follows:

My statement holds true (i.e. you are incorrect :smile because the formula assumes a 120? phase difference between currents and only the current magnitudes are entered into the formula.

You are entitled to your opinion, but my statement is 100% correct. You will never convince me otherwise. Your opinion just does not carry that much weight with me.

First I have to admit that this gave me a good chuckle. :grin:

You seem to find amusement in a most peculiar way :grin:

How could you say "you are incorrect" if this is the first time the "alleged" equation was presented. :grin:

Because I knew what the equation was when I made the satement. What you are saying makes you look even worse in the fact you said my statement was incorrect not even knowing the equation to which I referred. :grin:

It really isn't an equation per se, ...

You can view it however you like, but it is an equation, nonetheless. I am uncertain of how well documented this equation is, but it is fairly well known among persons knowledgeable in this area. Click here for an entire thread on this formula. One publication that I know it is included in the American Electrician's Handbook, Thirteenth Edition. It is shown in Section 1-154, Equation 59, on page 1.64. FWIW: Division One in this publication is titled "Fundamentals".

...it is an intermediate step in the solution of neutral current with assumptions made. Those assumptions are so limiting that the equation have very little value, and that is why I had never heard of it before.

Getting back to the topic of magnitude-only currents; based on the stock you put into this equation, I can understand why you made the statements you did, but that doesn't make it right. The mere existence of this equation is based on both magnitude and direction. Even though you are only entering the magnitudes into the equation, the prerequisite conditions for using the equation are where the directions, and hence the vector analysis, come into play.

This is all just more rhetoric on your opinion. I personally give it no regard. I understand the fact the equation is rendered invalid under many conditions. But the fact it is published in the American Electricians' Handbook refutes your opinion it has very little value, while it mentions nothing about conditions which make the equation's result valid or not.

Yes, I know that you understand how to solve a 3-phase circuit analysis, but I have never heard of anyone suggesting or discussing it from a magnitude standpoint.

Welcome to the real world... :grin:

 

[LarryFine]

It really isn't an equation per se . . .

Sure it is, with values on each side of an = .

 

[Cold Fusion]

---But the fact it is published in the American Electricians' Handbook refutes your opinion it has very little value, while it mentions nothing about conditions which make the equation's result valid or not.---

Smart -
The American Electricians' Handbook, "magnitude only equation", has very little to do with this discussion. The OP is asking about some pretty strange unbalanced loading that require the phase currents to not be inline with the voltages. As you noted:

---Example:
Line A, 5A, pf=1
Line B, 10A, pf=1
Line C, 8.67A, pf=0.867

I didn't check your math, but if it balances as you say it does, then it won't fit the AEHB equation. I looked up the magnitude only equation in my 1981 AEHB and it clearly says, "All are effective values and balance is assumed" Probably the first time I ever bothered to look at that section in 25 years I've had it.

This discussion is pretty much vector math without any constrainsts on phase angle displacement.

From your responses in this and other posts, I'm thinking you already knew this.

cf

 

[Rick Christopherson]

Phasing was implied as revised in the quote of my comment. I was of the impression anyone intelligent enough to read and comprehend the scope of this thread topic would have grasped the implication... but apparently my presumption was off-base :smile:

Put your tongue back in your mouth. When I responded to your previous discussion I mistakenly assumed that you were an engineer and gave you the leeway I assumed you deserved....my mistake.

When you referenced that equation, I assumed you understood where it came from....again, my mistake.

If the best source of your knowledge comes from a ?hand book?, and you want to argue mathematics with an engineer, well, that is your mistake. If you want to believe that your equation has any merit, then you go right ahead. Your limited knowledge leads to limited answers.

 

[Smart $]

Smart -
The American Electricians' Handbook, "magnitude only equation", has very little to do with this discussion.

I agree. But when I made the comment the topic hadn't fully digressed into a "phasor" level discussion. I thought commenting on the formula was noteworthy at that juncture, especially since it appeared PhaseShift hadn't yet fully grasped the concept of unbalanced current returning on the neutral.

The OP is asking about some pretty strange unbalanced loading that require the phase currents to not be inline with the voltages. As you noted:



I didn't check your math, but if it balances as you say it does, then it won't fit the AEHB equation.

...and that is why I brought it up in the first place.

... I looked up the magnitude only equation in my 1981 AEHB and it clearly says, "All are effective values and balance is assumed" Probably the first time I ever bothered to look at that section in 25 years I've had it.

You are correct. It's been years since I read that section, too. I only glanced over the text when I looked it up to cite the reference.

This discussion is pretty much vector math without any constrainsts on phase angle displacement.

From your responses in this and other posts, I'm thinking you already knew this.

Your thinking is correct...

 

[Smart $]

Put your tongue back in your mouth. When I responded to your previous discussion I mistakenly assumed that you were an engineer and gave you the leeway I assumed you deserved....my mistake.

When you referenced that equation, I assumed you understood where it came from....again, my mistake.

If the best source of your knowledge comes from a “hand book”, and you want to argue mathematics with an engineer, well, that is your mistake. If you want to believe that your equation has any merit, then you go right ahead. Your limited knowledge leads to limited answers.

Actually I am a former professional engineer... and while not the only reason I no longer pursue it as a carrer, one of the reasons I do not is because I can't tolerate working with so-called professionals with dispositions such as yours.

And BTW, the AEHB is more than just a handbook that the title implies. I don't know the exact number of pages but it is large enough to be an excellent paperweight for your desk. The first edition was published in 1913, so it has been around since before you were born (ohhh... excuse me, another presumption on my part). Additionally and as a matter of fact, it is on a short list of reference material which are permitted to have on hand while taking some of the proctored journeyman tests I've encountered in my limited capacity.

 

[PhaseShift]

Finally got a chance to read back through this thread. Sorry if I cause any confusion by not being clear on what I was asking.

I guess to clear things up, lets use a 3-phase wye connected motor as an example. If there is no neutral connected in the motor then all current weather it is balanced or unbalanced will return on the other two phases. If there is unbalanced current then the neutral point will float to a value.

If a neutral is then connected to the wye point of this motor then the motor is no longer technically a three phase motor but rather has (3) single phase windings. The wye point will be held to ground and any unbalanced current will flow through the neutral. There will be no flowing of current through the other two phases due to the fact that the neutral is now there. If the phase currents are balanced then there will be no return flow in the neutral or other phases.

That is my understanding up until this point.

 

[PhaseShift]

I know 3 CT's in a residual configuration will sum to zero even with unbalanced current, QUOTE]

Not quite correct. Three CT's in a residual connection have one side of each CT connected together (the neutral or return) and the other side running to the meter or relay for each phase. After the currents run through the phase meters and relays, the three phase wires are joined together to make the residual connection and return to the common point of the CT's, possibly running through a residual current relay (device 50N) on the way.

If the loads are balanced, you will see current in all three phase wires and none in the return lead. Any unbalance will cause current to flow in the return or neutral lead.

I have seen this residual current scheme used for ground fault monitoring. Could this lead to false trips since you are saying that unbalanced current could cause the residual current relay to pickup? If unbalanced currents go through CT's and dont cancel then they will go through the neutral of the CT's and cause relay to pickup?

 

[Smart $]

... There will be no flowing of current through the other two phases due to the fact that the neutral is now there. If the phase currents are balanced then there will be no return flow in the neutral or other phases.

...

This part is not correct. As long as all three lines are conducting current, some or all of the current from each line is returning on the other two. It is only when the current of one or two lines is too much to return on their complementing lines that current flows on the neutral. This is the unbalanced portion of the current

 

[LarryFine]

I guess to clear things up, lets use a 3-phase wye connected motor as an example. If there is no neutral connected in the motor then all current weather it is balanced or unbalanced will return on the other two phases. If there is unbalanced current then the neutral point will float to a value.

Correct.

If a neutral is then connected to the wye point of this motor then the motor is no longer technically a three phase motor but rather has (3) single phase windings. The wye point will be held to ground and any unbalanced current will flow through the neutral.

I'm with you so far.

There will be no flowing of current through the other two phases due to the fact that the neutral is now there.

If we were talking about three individual 120v line-to-neutral loads, and only one phase's load was energized, I'd agree.

A 3ph motor will have current in all three lines, neutral connection or not.

If the phase currents are balanced then there will be no return flow in the neutral or other phases.

For perfect balance, neutral no, phases yes.

 

[PhaseShift]

This part is not correct. As long as all three lines are conducting current, some or all of the current from each line is returning on the other two. It is only when the current of one or two lines is too much to return on their complementing lines that current flows on the neutral. This is the unbalanced portion of the current

Thanks for sticking with me guys.

I guess the one part I am having a hard time visualizing is that for balanced current why is there still current still returning on the other two phases. Wouldn't the balanced current try to return on the neutral and thus cancel to zero. Why would the balanced current try to return on the other two phases that would have more impedence than the neutral connection. Doesn't current want to take the path of least resistance?

Same for unbalanced current. Wouldn't all the current try to return non the neutral (less resistance) but only the unbalanced portion actually returning on the neutral. I cant see why some of the current will still return on other phases.

 

[PhaseShift]

I have seen this residual current scheme used for ground fault monitoring. Could this lead to false trips since you are saying that unbalanced current could cause the residual current relay to pickup? If unbalanced currents go through CT's and dont cancel then they will go through the neutral of the CT's and cause relay to pickup?

I thought about this one some more. I guess depending on how the CT inputs were wired you could use the residual scheme to detect unbalance or ground.

To detect unbalance you would wire the CT's as described above with only the unbalanced portion returning on the neutral of the CT connection.

However if you wire it into a relay in such a way where it adds them similar to a zero sequence CT then I guess it could be used for ground detection. I guess the relay would have to somehow add the currents in each CT so that they looked for balanced current

 

[Smart $]

Thanks for sticking with me guys.

Glad to...

I guess the one part I am having a hard time visualizing is that for balanced current why is there still current still returning on the other two phases. Wouldn't the balanced current try to return on the neutral and thus cancel to zero.

Visuallizing as current "returning" through the other two loads may be the problem. What current passes through one load and returns through the other two isn't per se actually returning. It is simply a portion of the current which passes through the other two. It is the same current, not two different currents.

When you only have one load energized, the current which passes through it "returns" through the neutral conductor. When two other equal loads whose current's phase angle is plus and minus 120? from the first load, the first load's current splits into two out of phase by that same plus and minus 120? to become half of the other loads current. This is true for each of the three loads when equal in both magnitude and have 120? phase angle displacement between them.

When parameters are other than equal as just mentioned, the division into out-of-phase component current is not equal, but they balance out in similar manner. For truly unbalanced loads, the loads divide into their matching vectorial components, but it is similar to arithmetically dividing by 2 and having a remainder.

...Why would the balanced current try to return on the other two phases that would have more impedence than the neutral connection. Doesn't current want to take the path of least resistance?

Because the other two phases have a voltage pushing the current along. The load neutral point sees no impedance associated with the load currents on the other two phases.

Same for unbalanced current. Wouldn't all the current try to return non the neutral (less resistance) but only the unbalanced portion actually returning on the neutral. I cant see why some of the current will still return on other phases.

Well you have to grasp what I already wrote before you understand this part.

 

[resistance]

The formula I referred to is as follows:

My statement holds true (i.e. you are incorrect :smile because the formula assumes a 120? phase difference between currents and only the current magnitudes are entered into the formula.

If you can demonstrate, either by your own working example or by that of another's, that the formula is accurate for currents having different power factors (i.e. not 120? apart), I will concede.

Absolutely! I agree!!

 

[PhaseShift]

When you only have one load energized, the current which passes through it "returns" through the neutral conductor. When two other equal loads whose current's phase angle is plus and minus 120? from the first load, the first load's current splits into two out of phase by that same plus and minus 120? to become half of the other loads current. This is true for each of the three loads when equal in both magnitude and have 120? phase angle displacement between them.

When parameters are other than equal as just mentioned, the division into out-of-phase component current is not equal, but they balance out in similar manner. For truly unbalanced loads, the loads divide into their matching vectorial components, but it is similar to arithmetically dividing by 2 and having a remainder.
.

If I have a wye connected load with a neutral and have the following currents on each phase a=10A b=10A c=10A, then the 10A from phase a will return on phases b and c splitting so that 5A returned on each b and c. No current will flow in neutral

If I have an unbalanced case then currents will not divide eveny among other two phases and there will be some remaining current flowing on the neutral. This is similar to the dividing by two scenario with a remainder as mentioned above. Without the presence of the neutral for the "remainder current" the currents will be forced to divide unevenly and therfore cause the wye point to float to a voltage.

Because the other two phases have a voltage pushing the current along. The load neutral point sees no impedance associated with the load currents on the other two phases.

Can you explain this part further?

 

[Smart $]

Without the presence of the neutral for the "remainder current" the currents will be forced to divide unevenly and therfore cause the wye point to float to a voltage.



Can you explain this part further?

You already know the answer :grin:

Until there is "remainder" current, the load neutral point remains at zero volts. With zero volts between load end and source end of the neutral conductor, no current flows on this path. You have to have a voltage difference for current to flow. Zero volts and zero current could mean mean zero impedance... but it could just as easily mean infinite impedance. It is only by this reasoning that I said "the other two phases have a voltage pushing the current along. The load neutral point sees no impedance associated with the load currents on the other two phases."

When there is a voltage between ends of the neutral conductor, the current sees the neutral conductor as having some impedance. It also sees the other two phases as having some impedance. So under this condition you have an impedance network and out-of-phase voltages and currents in play.

 

[bob]

PhaseShift

If I have a wye connected load with a neutral and have the following currents on each phase a=10A b=10A c=10A, then the 10A from phase a will return on phases b and c splitting so that 5A returned on each b and c. No current will flow in neutral

Ohm law says that Ia + Ib + Ic = 0. If you take the phaseor quantities and add them you get 10@0 + 10@120 + 10@240 = 0. They will cancel. No need to return on the other phases.

 

[Rick Christopherson]

PhaseShift

Ohm law says that Ia + Ib + Ic = 0.

Not to nit-pick, but that is Nodal-analysis via Kirchhoff's Current Law, not Ohm's Law.