Deriving the neutral on 120/208

[K2X]

I think I understand the grounded conductor on single pase 120/240 but now I'm working on a new, large, data storage facility. Our job is just piping for hvac controls but i see alot of big Brown, Orange, Yellow all over the place and the 480 to 120/208 transformers, and the 120/208 panels. Like I said, I think I understand the center tapped grounded conductor on single phase but i can't seem to get it in my head how they are taking the 3 phases and comming up with 120 volts?

Thanks for any help and the help in the past.

 

[augie47]

Here is a diagram of a Wye Transformer that could provide you with 208 betwen phases and 120 to neutral on each phase. In a situation like you describe, the primary would be 480, the secondary 208/Y120.


3 Phase Transformer Winding Combination
As can be seen, the three-phase transformer actually has 6 windings (or coils) 3 primary and 3 secondary. These 6 windings will be pre-connected at the factory in one of two configurations:

Configuration 1. Three primary Windings in Delta and Three Secondary Windings in Wye

 

[LarryFine]

To add to Gus's pix with words, a 3ph wye supply is made up of three 120v transformer secondaries with one end of each all tied together to create the neutral, which is almost always grounded.

The reason you receive 208v between hots instead of 2409v is because, unlike the two 120v halves of a 240v secondary winding, the hots are not 180 degrees apart, timing-wise, but 120 degrees.

The difference in timing causes opposing hots to not reach their opposing voltage peaks at the same time; when one phase peaks, one of the others is on the way up, and the other is on the way down.

That means that 120v + 120v does not = 240v. Added: The center and right diagrams in Gus's pic are the same configuration shown two different ways; they're electrically identical.

 

[K2X]

Got it ! Thanks! So if I looked at a sine wave graph of 3 phase I might see the waves intersecting at 208 ?? Or something like that? I got the idea anyway. Tnx.:grin:

 

[K2X]

Quick question on the timing.
Is the timing set from the generators back at the power plant??
Does it take 3 generators to make 3 phase??

 

[LarryFine]

Got it ! Thanks! So if I looked at a sine wave graph of 3 phase I might see the waves intersecting at 208 ?? Or something like that? I got the idea anyway. Tnx.:grin:

Not intersecting, exactly, but measured from the "top" (positive peak) of one sine wave to the opposing point on the other two waves (both of which will be negative (below zero)).

Remember, each wave's positive peak occurs when the other two waves are between their respective negative waves, one of which just passed 60 degrees ago, and the other of which will occur in the next 60 degrees.

So, in that sense, the positive peak of one phase occurs when the other two phases' negative waves are intersecting, and vice versa.

Who has a good 3-phase wave pic?

 

[LarryFine]

Quick question on the timing.
Is the timing set from the generators back at the power plant??

The timing is "set" by the physical positioning of the three sets of windings. You could literally make an alternator any number of phases you want, but you have to transmit that power over that number of conductors, using that number of transformers.

Does it take 3 generators to make 3 phase??

No, it takes a single three-phase alternator (AC generator). If you have more than one, they're wired and run in parallel, synchronously.

Three-phase transformers, on the other hand, are three single-phase transformers whose three cores are bolted together in a frame.

Three individual transformers can also be used in a three-phase bank. That's what you see when you look at three on a utility pole.

 

[Smart $]

...

Who has a good 3-phase wave pic?

How's this?

 

[qmavam]

How's this?

Hi Smart $,
In the last few days I've been on a quest to find a proper graph of three phase power. I have found a few graphs and they don't agree with each other or with my thinking. I may need to be educated or I still need to find an accurate graph.
So you have posted a graph, to simplify the discussion about your graph I'll be talking about the 3 phases of 120 volts to ground only.
First, do we agree that between any two legs of of 3 phase power you will measure 208 Volts RMS? I'll assume for my discussion you agree, if not let me know.
Let's start with the red line, the peak at 120 degrees is at 169.71 volts (120 x 1.414) Agreed. Now the black line at the same 120* point it is at -84.85. So the difference between the two phases would be 169.71 + 84.85 = 254.56 volts, but this is at a peak so 254.56 x .707 =179.97 Vrms. This is not the
208 volts I think we agreed on.
My thinking (right or wrong) is that the neg. voltage at 120 degrees should be 124.43. Then 169.71 + 124.43 = 294.11 and 294.11 x .707 = 207.93 Volts
I hope I described this so it's understandable. If you can help me understand
where I'm wrong or if I persuaded you something needs to be changed, please let me know.
Thanks, Mike

 

[charlie b]

First, do we agree that between any two legs of of 3 phase power you will measure 208 Volts RMS?

Agreed. But it is apparent that you do not understand what RMS means. Let's look at the rest of your post:

Let's start with the red line, the peak at 120 degrees is at 169.71 volts (120 x 1.414) Agreed. Now the black line at the same 120* point it is at -84.85. So the difference between the two phases would be 169.71 + 84.85 = 254.56 volts, but this is at a peak so 254.56 x .707 =179.97 Vrms. This is not the 208 volts I think we agreed on.

The reason is that you are looking at a single point in time. But the RMS (Root Mean Square) value is a "sort of average" value that represents the entire time between one peak and the next. So you are not going to be able to compare an RMS value with a measurement taken at a single point in time.

To obtain the RMS value of a wave form (it need not be a sine wave, any repetitive pattern will have an RMS value), you begin by squaring the value at each point in time, from the beginning of a cycle to the end of the cycle. This gives you a different shaped wave form. Then you take the average value over the duration of the cycle. This gives you a single number - the average of the squared version of the original wave form. Finally, you take the square root of that single number, and that gives you the RMS value of the original wave form.

 

[Smart $]

.... If you can help me understand
where I'm wrong or if I persuaded you something needs to be changed, please let me know.
...

As Charlie noted you are picking out an instantaneous value. It so happens that the 208V Line-to-Line waveform peaks occur either plus (+) or minus (–) 30? to the 120V waveform peaks, depending on how you are measuring them (i.e. the polarity of your test leads) and which waveform is the reference. The diagram shows the 208V waveforms as "lightened" dashed lines, with the black (–) lead connected to the Line having the longer-dash color, and obviuosly the red (+) lead connected to the Line having th shorter-dash color.

Knowing this, move your instantaneous reference to 150?. Line B "red" is at +146.97V, while Line A "black" is at –146.97V. Absolute difference is 146.97V ? 2 = 293.94V... dividing this value by sqrt(2) equals 207.85V.

 

[winnie]

My thinking (right or wrong) is that the neg. voltage at 120 degrees should be 124.43. Then 169.71 + 124.43 = 294.11 and 294.11 x .707 = 207.93 Volts

Note that the _peak_ of the red-blue voltage does not correspond to the peak of the red voltage.

If the peak of the red voltage is at 120 degrees, and the negative peak of the blue voltage is at 60 degrees, then the peak of the red-blue voltage will be at 90 degrees.

As charlie said, RMS is calculated by taking a sort of average of the entire waveform. It simply happens that for a sine wave there is a known fixed relation between the peak value and the rms value; the use of the .707 multiplier is only valid for sine waves when you find the actual peak.

-Jon

 

[LawnGuyLandSparky]

I think I understand the grounded conductor on single pase 120/240 but now I'm working on a new, large, data storage facility. Our job is just piping for hvac controls but i see alot of big Brown, Orange, Yellow all over the place and the 480 to 120/208 transformers, and the 120/208 panels. Like I said, I think I understand the center tapped grounded conductor on single phase but i can't seem to get it in my head how they are taking the 3 phases and comming up with 120 volts?

Thanks for any help and the help in the past.

For you to completely understand how a neutral is derived, you have to understand that it is only a potential difference referenced to the other phases, and not because the neutral is some magical "ground / zero."

Once a separately derived system is created, the electrons flowing through the secondary windings are completely unrellated to the current powering the transformer. Those secondary electrons do not care about anything other than returning to where they started from. Without a neutral reference on the load side, neither side of that SDS would want to short to ground.

 

[qmavam]

Hi and thanks for your interest,

Agreed. But it is apparent that you do not understand what RMS means.

I think I do understand RMS. In simplistic terms, A 10 Vrms AC voltage would produce the same amount of heat in a resistor that a 10 DC voltage would. It could be a square wave, sawtooth, or a pulse, but if the rms value equaled 10 Volts it would produce the same amount of heat in a resistor as 10 Vdc.

The reason is that you are looking at a single point in time. But the RMS (Root Mean Square) value is a "sort of average" value that represents the entire time between one peak and the next. So you are not going to be able to compare an RMS value with a measurement taken at a single point in time.

I'll agree with your statement for now However with a sinewave the formula, Vrms = peak x 0.0707 is correct to calculate the RMS value if you know the peak value. I did the calculations at the peak of the waveform.

Now here is a 3 phase power graph, if you look at the (black line) first full peak (169.71 Vpeak) and the red line that is negative, it is at 0.6 units. So 1.6 units x 169.71 = 271.53 peak volts, then .707 x 271.53 = 192Vrms not 208 Vrms.
http://www.ukslc.org/images/articles/threephase/fig30001.png

Here is a second 3 phase power graph, in this graph when the first line peaks the other negative line is at -70.7V. So in this case, 169.71 + 70.7 = 240.41
Vpeak. then .707 x 240.41 = 169.96 Vrms not 208 Vrms.
http://www.eetasia.com/IMAGE/EEOL_2008MAR15_POW_NT_01_fig1.jpg

The graph Smart $ posted has 169.71 + 84.85 = 254.56 and .707 x 254.56 =
179.79Vrms not 208 Vrms.
Somewhere I have a fourth graph that has a voltage different than these three.
So three differing graphs and my opinion, there can only be one correct answer.
Thanks again for your interest.
Mike

 

[charlie b]

I think I do understand RMS.

No, I am sorry, but you don?t.

In simplistic terms, A 10 Vrms AC voltage would produce the same amount of heat in a resistor that a 10 DC voltage would.

That is true. That is the physical significance of RMS values, as we use them in the electrical world.

However with a sinewave the formula, Vrms = peak x 0.0707 is correct to calculate the RMS value if you know the peak value.

Again, this is correct.

I did the calculations at the peak of the waveform.

That is where you went wrong. You did the calculations at a single point in time, the point at which one of the values reached its peak. As the sine wave goes up and down with the passage of time, the RMS value does not go up and down with it. You don?t take the value of the sine wave at any given point in time, multiply that value by 0.707, and get the RMS value at that point in time. RMS is a single number, a constant value.

When you take the positive peak value of 1.0, and add it to the negative value of 0.6, then multiply that by 169.71, you will get a voltage difference at that moment in time. When you multiply that by .707, and call it an RMS value, you have in fact arrived at a meaningless number.

 

[qmavam]

It so happens that the 208V Line-to-Line waveform peaks occur either plus (+) or minus (?) 30? to the 120V waveform peaks,

That's it! I wrongly assumed the peak would be where one of the 120V lines peaked.
Clearly it's not, as you say peak voltage is 30 degrees from the 120V waveform peak.

Knowing this, move your instantaneous reference to 150?. Line B "red" is at +146.97V, while Line A "black" is at ?146.97V. Absolute difference is 146.97V ? 2 = 293.94V... dividing this value by sqrt(2) equals 207.85V.

Yes, and the sqrt(2) = 1.414, So 293.94 Vpeak/1.414 = 207.87 Vrms
Or another way .707 x 293.94 Vpeak = 207.82 Vrms
Thanks Smart $ for the clear explanation!
Mike

 

[qmavam]

I think I understand RMS.

No, I am sorry, but you don?t.

That's not enough explanation, especially when I give a couple of descriptions of RMS
and you say.

That is true.

And

Again, this is correct.

When you multiply that by .707, and call it an RMS value, you have in fact arrived at a meaningless number.

Yes it is meaningless if I miss the peak, and I did.
It all boiled down to what Smart $ said about where the true peak is.
I was missing true peak by 30 degrees. With the proper numbers as shown
in Smart $ graph, the equation of Vpeak x .707 = Vrms or as Smart $ puts it
Vpeak / 1.414 = Vrms.
Finding where the true peak of the 208 V gave me the understanding I was looking for. But I think I still have three graphs that do not have a properly
shaped sinewaves, in that they gave differing voltages at the same point in the cycle.
Thanks, Everybody

 

[qmavam]

I just want to thank Smart $ again.
I looked at his graph again, this time I looked at the dotted red/green,
blue/red and green/blue lines and I see they are the additive value of two lines that their colors represent.
This is beautiful.
Thanks, Mike

 

[Smart $]

I just want to thank Smart $ again.
I looked at his graph again, this time I looked at the dotted red/green,
blue/red and green/blue lines and I see they are the additive value of two lines that their colors represent.
This is beautiful.
Thanks, Mike

Glad you are getting it... but i'm feeling a bit picky...

What you called green is in fact semi-transparent BLACK... the dashes are of the same color as the Lines the voltage is measured between, only made semi-transparent (computer-graphics-wise)

 

[qmavam]

Glad you are getting it... but i'm feeling a bit picky...

What you called green is in fact semi-transparent BLACK... the dashes are of the same color as the Lines the voltage is measured between, only made semi-transparent (computer-graphics-wise)

ok, ok, lol

Mike