Deriving the neutral on 120/208

[dbuckley]

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.

Thats the definition of af an RMS voltage, so that is absolutely correct.

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

This is a mathmatical convenience, and as noted, only applies with a sinusoidal waveform.

But..... If you measure at a single point in time, you dont have a sinusoidal waveform (or any waveform): you have a 'point'. So at this 'point' (and at any other point) the formula is not valid. The formula only works when you have a sinusoidal waveform, and that means whole half-cycles.

 

[charlie b]

But..... If you measure at a single point in time, you don?t have a sinusoidal waveform (or any waveform): you have a 'point'. So at this 'point' (and at any other point) the formula is not valid.

Thanks for that statement, dbuckley. I wasn?t able to convince qmavam of that simple fact myself.

To qmavam: If you use a meaningless mathematical process, and you happen to come up with an answer that happens to be the right answer, that does not mean that the process you used was valid. It means only that you got lucky, and that you do not yet understand the nature of the process.

Here is the essence of the situation: The line-to-line voltage is derived by subtracting one waveform from another. You are taking one voltage that can be expressed as V1(t) = Vm sin(wt), and subtracting from it another voltage that can be expressed as V2(t) = Vm sin(wt-120). The result will be a sinusoidal waveform, with a peak value that is higher than the peak value of the original two by a factor of the square root of three. Therefore, the RMS value of the ?new waveform,? the line-to-line waveform, will be higher than that of the original two by the same factor. That is, if you started with a pair of line-to-neutral voltages with an RMS value of 120, the RMS value of the line-to-line will be 208.

The point in time at which the line-to-line waveform will reach its peak value turns out to be at 150 degrees, as seen from the line-to-neutral waveform of what I called V1(t). That is why, when you measure voltages at the single point in time equivalent to 150 degrees, you are seeing the actual peak voltage of the line-to-line waveform. And yes, if you multiply that value by 1.732, you will get 208 as a result. But you got there by accident, not by viewing the actual new waveform, by squaring its value point by point, by averaging these values, and by taking the square root of the result.

 

[winnie]

If you correctly assess that the waveform being measured is a sinusoid, and you correctly determine the peak voltage of that waveform, then it is no accident that {peak value}/sqrt(2) is the RMS value of that sinusoid. For a _fixed_ continuous differentiable waveform shape, there will always be a relationship between the peak value of that waveform and the RMS value of that waveform. I am not prepared to make that claim if the waveform is not differentiable, but it is also true for a square wave.

It is known that the sum or difference of two different sinusoids of the same frequency, but arbitrary amplitude and phase, is yet another sinusoid.

As qmavam noted, his error was in not correctly determining _where_ the peak value was, and thus not correctly determining the actual peak of the sinusoid.

-Jon

 

[qmavam]

I have been chastised in a private email, saying I should apologize to charlie b.
If I offended you charlie b, I'm sorry.

I want to put up a little defense of my position.

But..... If you measure at a single point in time, you don't have a sinusoidal waveform (or any waveform): you have a 'point'. So at this 'point' (and at any other point) the formula is not valid. The formula only works when you have a sinusoidal waveform, and that means whole half-cycles.

I agree with that, here's how I phrased that in post #14.
"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."
I emphasized sinewave and I said "at peak".
My error was, I was not on the peak of the 208 V waveform, which I admitted in both posts 16 and 17. I thought I was, but I made an error and was off by 30 degrees. (This is the very reason I came to the thread, because I knew I had an error and was trying to figure it out. The error again, was looking 30 degrees off peak.)

Thanks for that statement, dbuckley. I wasn?t able to convince qmavam of that simple fact myself.

To qmavam: If you use a meaningless mathematical process, and you happen to come up with an answer that happens to be the right answer, that does not mean that the process you used was valid. It means only that you got lucky, and that you do not yet understand the nature of the process.

Again I say Vrms = 0.707 x Vpeak is not meaningless, and yes anywhere
off peak it is meaningless. One more time, my error, I thought I was on peak but was off peak by 30 degrees.

Here is the essence of the situation: The line-to-line voltage is derived by subtracting one waveform from another. You are taking one voltage that can be expressed as V1(t) = Vm sin(wt), and subtracting from it another voltage that can be expressed as V2(t) = Vm sin(wt-120). The result will be a sinusoidal waveform, with a peak value that is higher than the peak value of the original two by a factor of the square root of three. Therefore, the RMS value of the ?new waveform,? the line-to-line waveform, will be higher than that of the original two by the same factor. That is, if you started with a pair of line-to-neutral voltages with an RMS value of 120, the RMS value of the line-to-line will be 208.

Yes

The point in time at which the line-to-line waveform will reach its peak value turns out to be at 150 degrees, as seen from the line-to-neutral waveform of what I called V1(t). That is why, when you measure voltages at the single point in time equivalent to 150 degrees, you are seeing the actual peak voltage of the line-to-line waveform.

Yes that has been my eureka moment, finding that the 208V peak did not correspond to the 120V peak. I assumed that several days ago when I started looking at 3 Phase graphs. Bad assumption!

And yes, if you multiply that value by 1.732, you will get 208 as a result. But you got there by accident, not by viewing the actual new waveform, by squaring its value point by point, by averaging these values, and by taking the square root of the result.

I have never seen the 1.732 number, don't know if it has any real use
in design work. Could be I just don't know it.

Mike (qmavam)

 

[Smart $]

... For a _fixed_ continuous differentiable waveform shape, there will always be a relationship between the peak value of that waveform and the RMS value of that waveform. I am not prepared to make that claim if the waveform is not differentiable...

There is always a relationship (ratio) of peak to rms values no matter how sporadic the waveform. Differentiating is only yet another shortcut to determining the RMS value. For sporadic value waveforms it is likely the sample method will have to be used to determine the RMS value. I have a digital audio program which does it this way.

 

[Smart $]

I have never seen the 1.732 number, don't know if it has any real use
in design work. Could be I just don't know it.

1.732 is the square root of three (√3). It is commonly used in calculations where 3? systems are utilized.

 

[dbuckley]

I have never seen the 1.732 number, don't know if it has any real use in design work. Could be I just don't know it.

I can say with absolute certainty that 1.73 is a really really important number for anyone who tools with power electrics to know

Remember it well.

 

[charlie b]

I have been chastised in a private email, saying I should apologize to charlie b. If I offended you charlie b, I'm sorry.

I was not offended, and no apology was necessary.

I have never seen the 1.732 number, don't know if it has any real use in design work. Could be I just don't know it.

My turn to apologize. I typed that earlier message too quickly, and lost my own train of thought. While it is true that in three phase systems, the number "the square root of three" comes into play in almost anything we try to calculate, it was not the appropriate number to use in the context of my post. The relationship between the peak and RMS values of a sine wave is "half the square root of two," or about 0.707.

 

[qmavam]

I was not offended, and no apology was necessary.
My turn to apologize. I typed that earlier message too quickly, and lost my own train of thought. While it is true that in three phase systems, the number "the square root of three" comes into play in almost anything we try to calculate, it was not the appropriate number to use in the context of my post. The relationship between the peak and RMS values of a sine wave is "half the square root of two," or about 0.707.

Ok guys, I showed my ignorance, but "ya can't know everything about everything". My background is in electronics, often working with 5v
and 12v and up DC systems. This is generally derived from a single phase 120v line.
I started trying to get an understanding of where 208V comes from in a three phase system when I saw this on a newsgroup I read.

>So on a 3 phase supply, one leg will rise up to 120volts, but the other
>leg will not be at -120 volts. It will be at about -88volts (if i did
>the head math correctly).

I knew that was incorrect and searched for a graph to show the poster
so he could get get it right. Then I found several graphs that I still could not
glean the info I needed. Finally through a long google search of "3 phase power"
Images I found Smart $'s graph and got in to this fine discussion and was educated.
Looking back over the posts, there were a few responses that I probably
should have got my understanding from, but the visual representation did it
best for me.
charlie b, after you told me twice that I didn't understand rms my ego got
in the way and I wasn't paying attention to what you were explaining. bad on me, lol. I see that from your point it looked like I didn't, but
it was that 30 degree error caused a "failure to communicate".
I got the answer to my question, thanks all.
Now if you want to know about Fr=1/2pi(sqrt(LC)), Xl=2piFL, Xc=1/(2piFC)
or Z=sqrt(R^2 +Xl^2) or 3db points, we're in my area.
Thanks Guys