Connecting a neutral to a 3-Phase Wye Load

[Cold Fusion]

(cut)Can you plaese give me the first step in arriving at this solution. I'm stuck and hopefully all I need is a hint to head in the right direction.

See if this helps. For me, the model is the tough part - once you have that set up, the rest is just algebra (okay, complex numbers and matricies - minorly messy, but still just algebra)

There are lots of ways to set up the model. All are valid, some are easier than others. I picked this one since it appeared to minimize the complex algebra.

Your example is not too bad - the impedances are not complex, that helps a lot.

For your example, the common point is way offset from Vn. so I called the common point Vy to differentiate.

Vn is sort of mythical - this example does not have a neutral. But that is okay, Va, Vb, Vc fix Vn - since they are really Van, Vbn, Vcn (all referenced to Vn)

As I mentioned, I didn't check my math, so nothing says I didn't slip a minus sign or a decimal point. But the voltages across the impedances and currents look plausable - so maybe it is okay.

I likely went over a lot of stuff you already know, but if this doesn't make any sense or I left something out - I'll give it another try.

cf

 

[Smart $]

Vn' = 232Vrms @ 125 degrees.

Someone check me please.

I got 237V, >125. But I didn't spend much time at it, didn't check my numbers. Your's could well be right.

Close enough for rough work.

I'll concur... provided A, B, and C phasors are 0?, -120?, and -240? (or 120?) respectively.

 

[rattus]

Did you use the example voltages and impedances that I posted to arrive at this answer?

Can you plaese give me the first step in arriving at this solution. I'm stuck and hopefully all I need is a hint to head in the right direction.

Yes, I used your example. First thing to do is solve for Vn' which is simply a matter of getting a common denominator and collecting terms. I get

Vn' = Va[cos(0) + jsin(0)]/111 + 10*Vb[cos(-120) + jsin(-120)]/111 +100*Vc[cos(120) + jsin(120)]/111

Where Va, Vb, and Vc are the magnitudes of the wye phase voltages (277Vrms) relative to the system neutral and Vn' is the complex value of the floating neutral--note the prime!

You can now plot each of the phasors head to tail to get their sum, or you can evaluate them with a calculator or spreadsheet.

 

[Smart $]

See if this helps...

Pretty good :grin:

Perhaps I can make it a little simpler to understand. Refer to attached drawing. I didn't annotate it so I'll have to explain...

View attachment 2722

Start with the three voltage phasors.

Draw lines between end points. This represents ~480V across the two loads as if the other was disconnected.

Divide this line proportional to the impedances (in this case, pure resistance) between them. For example, Za and Zb are 100 and 10 ohms respectively. ~480 ? 100 ? 110 for the segment closest "A", and ~480 ? 10 ? 110 for "B" end.

At the common node of these two segments, draw a line to the end node of the other voltage phase. This line represents the direction the common node is "pulled" when the third load's impedance varies from infinity (disconnected) to 0 ohms (shorted). Given the third load is neither, "Vy" phasor's endpoint lies somewhere along this line.

Repeat the above process for the other two load pairings (actuall only one more pairing is required, but doing all three shows the inaccuracy— evident if you zoom in on the drawing).

The intersection [area] of the three "pull" lines represents the end of phasor "Vy" (which originates at Vn).

 

[mull982]

Pretty good :grin:
Perhaps I can make it a little simpler to understand. Refer to attached drawing.

I understand your drawing and how you drew the phasors. I guess I'm more of a visual learner.

See if this helps. For me, the model is the tough part - once you have that set up, the rest is just algebra (okay, complex numbers and matricies - minorly messy, but still just algebra)

I follow your model and how you set up your four equations. On the sheet with your matrix's I understand that the first matrix is a representation of the 4 equations on the prior sheet. I see that the second matrix you are strictly dividing out a common multiple to achieve values of 1 for your currents. I cannot see what you are doing in your third matrix labeled eq4 or how you get from the one above it to it.

How do you then get from the third matrix (eq4) to the matrix's at the bottom. I understand how the R and J terms in the bottom matrix's give the magnitude and phase angle of the values but i cant see how you got to the terms on the left. Can you explain?

Yes, I used your example. First thing to do is solve for Vn' which is simply a matter of getting a common denominator and collecting terms. I get

Vn' = Va[cos(0) + jsin(0)]/111 + 10*Vb[cos(-120) + jsin(-120)]/111 +100*Vc[cos(120) + jsin(120)]/111

I guess this is just another way of solving the problem than you posted earlier. You found a commond demonator of 111 and then caried out the math?

Either way the results show that Vn' is 232V@125deg. Does this mean that if you were to measure this "neutral point" to ground or the neutral point on the system supply you would read 232V?

 

[rattus]

I guess this is just another way of solving the problem than you posted earlier. You found a commond demonator of 111 and then caried out the math?

Either way the results show that Vn' is 232V@125deg. Does this mean that if you were to measure this "neutral point" to ground or the neutral point on the system supply you would read 232V?

Yes, and yes. My solution might be a little off because I had to convert between radians and degrees in Excel, and I only used 3 significant figures for Pi.

As expected, Vn' is close to Vc because Zc is the smallest of the loads.

Now that you know Vn', you can compute the currents and see that their sum is zero.

 

[mull982]

Yes, and yes. My solution might be a little off because I had to convert between radians and degrees in Excel, and I only used 3 significant figures for Pi.

As expected, Vn' is close to Vc because Zc is the smallest of the loads.
.

Ok I'm starting to see it now. Still having a hard time trying to figure out the last couple of matrix's though.

Now that you know Vn', you can compute the currents and see that their sum is zero.

Assuming all current in a three phase system sums to zero. Is there an easy way to see what current would now flow through the neutral if we added a neutral to this example. I would think that it would be the current differences between the three legs.

 

[Cold Fusion]

Still having a hard time trying to figure out the last couple of matrix's though.

Yeah, I left a lot out of the matrix math. -I'll be able to put something together tomorrow

Assuming all current in a three phase system sums to zero. Is there an easy way to see what current would now flow through the neutral if we added a neutral to this example. I would think that it would be the current differences between the three legs.

Yes, the solution with the neutral added is fairly easy.

With a neutral connected, Vn is stuck to zero. so the voltage across Za is Van, across Zb is Vbn, across Zc is Vcn,

So:
iA = Va/Za = (277 +j0)/100

iB = Vb/Zb = (-139 -j240)/10

iC = Vc/Zc = (-139 + j240)/1

iA + iB + iC + iN = 0 (iN is referenced going into node at Vn)

iN = -iA -iB -iC
= -2.77 + 13.9 +j24 +139 - j240
= 156 -j216
= 266A >54

cf

 

[rattus]

Clever:

Clever:

Pretty good :grin:

Perhaps I can make it a little simpler to understand. Refer to attached drawing. I didn't annotate it so I'll have to explain...

View attachment 2722

Start with the three voltage phasors.

Draw lines between end points. This represents ~480V across the two loads as if the other was disconnected.

Divide this line proportional to the impedances (in this case, pure resistance) between them. For example, Za and Zb are 100 and 10 ohms respectively. ~480 ? 100 ? 110 for the segment closest "A", and ~480 ? 10 ? 110 for "B" end.

At the common node of these two segments, draw a line to the end node of the other voltage phase. This line represents the direction the common node is "pulled" when the third load's impedance varies from infinity (disconnected) to 0 ohms (shorted). Given the third load is neither, "Vy" phasor's endpoint lies somewhere along this line.

Repeat the above process for the other two load pairings (actuall only one more pairing is required, but doing all three shows the inaccuracy— evident if you zoom in on the drawing).

The intersection [area] of the three "pull" lines represents the end of phasor "Vy" (which originates at Vn).

Rather slick, Smart.

What Smart has done is represent two of the phase voltages, say Va and Vb, as a Thevenin equivalent circuit of value "Vab". This being a linear circuit, he knows that "Vy" lies along a line connecting Vab and Vc. He then repeats this process for Vb and Vc and draws another line to Va. These two lines intersect at Vy. He could repeat the process one more time just to verify the process.

This would be easier to see without the severe imbalance.

Right Smart?

 

[Cold Fusion]

Per PM request from mull:

Note to the math whizes: Feel free to jump in and straighten me out - I'm sure I missed something.

I am still curious in seeing the sheet on how you solved the four equations if you dont mind posting it as an attachment.

mull ?
As I recall, the method I used is called ?elimination? ? but it has been 30+ years since I took a math class, maybe it is called something else. I did it that way because the example posed is relatively simple ? the impedances are real, not complex. The notation I used isn?t really a matrix. It looks like one, but it is just a convenient method to keep track of the coefficients.

I rewrote the sheet on solving the four equations ? included all of the steps. See the attachment.

Hopefully one day I'll be able to explain it to someone else.

That?s my normal fee.:smile:

Oh yeah ? I missed a minus sign in post 28. Should have been 266A >-54

No telling how many more I missed

cf

 

[rattus]

With all due respect:

With all due respect:

Per PM request from mull:

Note to the math whizes: Feel free to jump in and straighten me out - I'm sure I missed something.

CF, I don't see the need for four equations when we have only one unknown.

 

[Smart $]

Rather slick, Smart.

What Smart has done is represent two of the phase voltages, say Va and Vb, as a Thevenin equivalent circuit of value "Vab". This being a linear circuit, he knows that "Vy" lies along a line connecting Vab and Vc. He then repeats this process for Vb and Vc and draws another line to Va. These two lines intersect at Vy. He could repeat the process one more time just to verify the process.

This would be easier to see without the severe imbalance.

Right Smart?

Right! (I think )

 

[glene77is]

If the loads were equal nothing would change. If they were not equal the voltages would float dependant on the load.

QQ,
Good thinking.
The coils would also float on each other, as balanced loads.
This is the nature of a three phase system.

 

[rattus]

Superposition:

Superposition:

Another way to solve for the floating neutral voltage, Vnf, is to apply the Superposition Theorem.

Works like this:

Set all voltage sources to zero except one, say Va.

Compute the currents in each load for this one source.

Repeat the procedure for Vb and Vc.

Sum the currents in one load, say Za.

Then,

Vnf = Va - IaZa

This is a perfect example of currents flowing in opposite directions in the same conductor.

No, Ronald they don't really do that.

 

[Cold Fusion]

CF, I don't see the need for four equations when we have only one unknown.

Interesting, only one unknown. I had not thought of that. I couldn't tell the values of iA, iB, iC, Vy (as noted on my model) by inspection. That makes them unknown to me - dependent, but still unknown.

I believe you used four to come with the one you posted - you just had a batch of substition that was apparent to you by inspection. But, since we don't have access to your model, I can only guess.

So, I'm translating your comment to, "If one can transpose elements by inspection, then those elements are no longer unknowns"

okay with me

cf

 

[rattus]

Interesting, only one unknown. I had not thought of that. I couldn't tell the values of iA, iB, iC, Vy (as noted on my model) by inspection. That makes them unknown to me - dependent, but still unknown.

I believe you used four to come with the one you posted - you just had a batch of substition that was apparent to you by inspection. But, since we don't have access to your model, I can only guess.

So, I'm translating your comment to, "If one can transpose elements by inspection, then those elements are no longer unknowns"

okay with me

cf

CF, there is one equation,

(Va - Vnf)/Za + (Vb - Vnf)/Zb + (Vc -Vnf)/Zc = 0

which contains one unknown, Vnf. Ia, Ib, and Ic are not part of the equation. We know the values of everything else.

After we solve for Vnf, we can compute the phase currents one at a time if we wish--no matrices required.

 

[glene77is]

CF, there is one equation,

(Va - Vnf)/Za + (Vb - Vnf)/Zb + (Vc -Vnf)/Zc = 0

Thanks for exposing the big picture.
It was hidden by all the problematic details .

I made a copy of this thread
and enjoyed reading it again at lunch time.

 

[Cold Fusion]

CF, there is one equation,

(Va - Vnf)/Za + (Vb - Vnf)/Zb + (Vc -Vnf)/Zc = 0

which contains one unknown, Vnf. Ia, Ib, and Ic are not part of the equation.

I really missed that. After figuring out your model (iA, iB, iC all pointing in - same as the one I sketched and attached), I thought your basic equation was KCL,
iA + iB + iC = 0

Then you substituted:
(Va - Vnf)/Za = iA
(Vb - Vnf)/Zb = iB
(Vc - Vnf)/Zc = iC

And your substitution was by inspection.

Sure looks like four equations to me.

--no matrices required.

You're right, this one is pretty simple - not needed

cf
got to go to work for a while

 

[rattus]

I really missed that. After figuring out your model (iA, iB, iC all pointing in - same as the one I sketched and attached), I thought your basic equation was KCL,
iA + iB + iC = 0

Then you substituted:
(Va - Vnf)/Za = iA
(Vb - Vnf)/Zb = iB
(Vc - Vnf)/Zc = iC

And your substitution was by inspection.

Sure looks like four equations to me.


You're right, this one is pretty simple - not needed

cf
got to go to work for a while

CF, there is one unknown in one equation to be solved--one equality sign. Intermediate steps do not count.

Do a search on "Simultaneous Equations" which are often encountered in electrical engineering--but not in this problem.

 

[mull982]

OK thanks alot for the help guys it has made it much clearer what is going on, and I also see there is more than one way to skin this cat.

I wanted to apply this application to real system components to see where it applies. For example:

If I have a wye secondary of a transormer that feeds several single phase loads (L-L) and has no neutral connected then will the neutral of the secondary float? I would think not in this case because all loads are L-L and would be the same as if I did have a neutral on the wye that nothing would flow through the neutral because all loads are L-L?

It seems like we will only have the floating neutral condition talked about in this thread in three phase "loads" and not sources. These loads could be an unbalanced heater arrangement as mentioned, or a (3) phase motor in wye that had three unequal impedances?