Is it Single or Two Phase?

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jim dungar said:
Rattus,
The consistency I am looking for has to do with the methodology of solving networks, regardless the point of reference. The base formulae should not change if the reference point is chosen as L1 or L2 instead of N. So if two voltage sources are in series the resultant voltage is the summation of the two individual sources even if it means having to add a negative value. With this basic concept it is much easier to be consistent with the application of Kirchoff's Law for solving 120/240V multi-wire and 2-wire combination circuits.
Click to expand...

Sure Jim,

You can simply add the magnitudes in a 3-wire ckt to obtain V12. The phase angle is unimportant here. But how would you compute line voltages and phase angles in a 3-ph wye? There is probably a formula for that somewhere.

But, somewhere in the derivation of that formula, subtraction is involved. It is fundamental that a voltage difference is obtained by subtracting one voltage from another! Of course we use a common reference point, and the neutral is the logical reference point. I would never use any other point if a neutral is available.

I think the crux of this matter is that most in the trade use formulas, and that is OK, but the more general approach is to use phasors, and this is not a trivial endeavor. In a sense, you must derive your own formulas.
 
Draw three phasors: ?A? at 0 degrees, ?B? at 240 degrees, ?C? at 120 degrees.
(?tails? at origin with arrows all pointing out at their corresponding direction)
Now put a plus sign at all the arrows, and negative signs where the ?tails? meet in the center)

Now calculate a phase-to-phase voltage (follow the tip of one arrow, to the origin, to another tip.

How did the +/- signs go along the way? + - + - or + - - + ?

That is why one phasor is subtracted to get the L-L voltage?
 
But is it right?

But is it right?

steve66 said:
Yes, but you only posted it last night while every one else was sleeping:p
Click to expand...

But is it right now? Do you add or subtract?

I maintain that even if you mislabel the phasor as -120@0 instead of 120@180, you still subtract!

BC,

If you flew through your circuits courses, you should be able to show the math involved in computing line voltages from wye phase voltages.
 
(120@ -30) - (120@ -150) = 208 @ 0 is absolutely correct! at least according to hp. I don't have a slide rule to check it on.

I thought we already talked about absolute values of magnitude?
 
According to my text books Kirchoff's Laws are:
1: The algebraic sum of the currents toward any junction point is zero.
2: The algebraic sum of the voltages around any closed path in the network is zero.

So in 2-wire circuit the resulting voltage drop V12 = V2n + Vn1
and in a 3-wire circuit the voltage drops are V1n+Vn2 = V2n+Vn1.

Just because V1n = -Vn1 does not mean Kirchoff's Law should be restated as the "algebraic subtraction of voltages".
 
"Just because V1n = -Vn1 does not mean Kirchoff's Law should be restated as the "algebraic subtraction of voltages"."

Nobody is trying to re-state any law.
Assume A is 120V at 0 degrees, and C is 120V at 120 degrees.
Remember A and C are ?joined at the hip? at the origin. (neutral)
If you want the vector CA, you start at ?Point A?, travel magnitude A at 180 degrees, then magnitude C at 120 degrees.

You can say you are traveling magnitude A at 180 degrees or ?(A at 0 degrees), either way you are headed the same direction.

If you just do A + C you will get 120V, but we know it is 208V.
 
vectors

vectors

Is this correct?

If you want the vector CA, you start at ?Point A?, travel magnitude A at 180 degrees, then magnitude C at 120 degrees.


If you want CA, you should start at A (120 @ 0), then travel a magnitude of C at 120 degrees. Next connect this new point to the origin and you have CA.
 
jim dungar said:
According to my text books Kirchoff's Laws are:
1: The algebraic sum of the currents toward any junction point is zero.
2: The algebraic sum of the voltages around any closed path in the network is zero.

So in 2-wire circuit the resulting voltage drop V12 = V2n + Vn1
and in a 3-wire circuit the voltage drops are V1n+Vn2 = V2n+Vn1.

Just because V1n = -Vn1 does not mean Kirchoff's Law should be restated as the "algebraic subtraction of voltages".
Click to expand...

Jim, no argument with Kirchoff! Your method works just fine, but when you reverse the subscripts, you have in effect reversed the polarity of the sinusoid. That is you are subtracting without realizing it.

Algebraic addition includes subtraction at times. Don't get stuck on the word "sum".

With all due respect, I gather that you are not comfortable with phasor manipulation, and that is OK. But, if you are not comfortable with phasors, you will have trouble grasping what we are saying.
 
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I fully grasp phasors.

I refuse to say that I analyze a balanced three phase circuit by adding voltages and currents per Kirchoff but that I subtract them when working with single phase.
 
mpross said:
Is this correct?

If you want the vector CA, you start at ?Point A?, travel magnitude A at 180 degrees, then magnitude C at 120 degrees.


If you want CA, you should start at A (120 @ 0), then travel a magnitude of C at 120 degrees. Next connect this new point to the origin and you have CA.
Click to expand...

With your method CA would be 120V
How do you then rationalize L-L voltage of 208V?
 
jim dungar said:
I fully grasp phasors.

I refuse to say that I analyze a balanced three phase circuit by adding voltages and currents per Kirchoff but that I subtract them when working with single phase.
Click to expand...

Even using circuit analysis you add or subtract voltages around the loop, depending on the components. Same with currents - in/out. It depends on how you set up the mesh loops.

Thats why when you calculate the CA vector I recently described, you need to keep track of your +'s and -'s....
 
Your (the one you are talking about) phasor is a pure sine wave approximation of a Fourier Series. The approximation involves measuring the RMS value and adding a phase angle, so don?t tell me your so-called ?complex waves? can?t be given a phase angle. We give everyday electrical signals a phase angle, and they are ?complex.? While you find comfort in equating an inversion with a 180 degree phase shift, it is incorrect. Only when your phasor is not an approximation (only happens in theory), will you be correct. I?d rather be correct more often than that.

Here?s how it works:

Let?s say we have a function:

y = x^(1/2)

What Rattus is saying is that

y^2 = x

is an equivalent equation because it gives the same answers.

I mean, (1,1), (4,2),(9,3) all satisfy both equations. However, just put in a negative value for y and the whole system breaks down. One equation will allow it, while the other will not. (It actually will, but you will obtain a different answer than the other equation delivers.) One might correctly infer that the second equation is an approximation of the other for some instances?..usually it works, but sometimes not. At the end of anycalculations, one always needs to go back to the start and see if the answer is compatible with the source data.
 
Here is some clarification for whoever wants it. I know some will try to dispute it and pick it apart. That's ok, it's difficult for some to accept defeat. But the following is based on sound engineering and math principles.

For the record, I did not breeze through my circuits classes, I used them as an opportunity to learn everything I could and get a good foundation. Maybe it was the quality school I attended. My analogy to "breezing" through anything, is that it may get your boat out into deep water, but the sound principles is what will make the return journey possible when the hot air runs out.

The IEEE dictionary defines a phasor as a complex number. The term phasor is understood to be used only in the context of steady state alternating linear systems. The absolute value of the complex number corresponds to either the peak amplitude, or root-mean-square value, and the phase corresponds to the phase angle at zero time reference.

Further, according to J. Lewis Blackburn, who was the guru of protective relaying, in normal practice the magnitude, unless specifically stated otherwise, is understood to be the rms value of the positive half cycle of the corresponding sinusoid.

Mathematically averaging the values of all the points on a sine wave to a single, aggregate number by considering their sign, either positive or negative, the average value for most waveforms is technically zero, because all the positive points cancel out all the negative points over a full cycle.
02412.png



This, of course, will be true for any waveform having equal-area portions above and below the ?zero? line of a plot. However, as a practical measure of a waveform's aggregate value, ?average? is usually defined as the mathematical mean of all the points' absolute values over a cycle. In other words, we calculate the practical average value of the waveform by considering all points on the wave as positive quantities, as if the waveform looked like this:
02413.png

It should be understood that the RMS value (practical average) is for pure waves only. Meters are calibrated for pure sine waves and therefore false readings will occur if the wave shape is distorted.

Now for vectors-
Vectors have magnitude and direction. The length of the vector represents the magnitude (or amplitude) of the waveform, and therfore the MAGNITUDE OF A VECTOR IS ALWAYS POSITIVE!!!!! For those of you who breezed through their basics, maybe you were sleeping that day from boredom, and missed this rule.
The angle of the vector, however, represents the phase shift in degrees between the waveform in question and another waveform acting as a ?reference? in time. Usually, when the phase of a waveform in a circuit is expressed, it is referenced to the power supply voltage waveform (arbitrarily stated to be ?at? 0 deg). The phase is always a relative measurement between two waveforms rather than an absolute property.
The reference point is not unlike that of the ?ground? point for the benefit of voltage reference. With a clearly defined point in the circuit declared to be ?ground,? it becomes possible to talk about voltage ?on? or ?at? single points in a circuit, being understood that those voltages (always relative between two points) are referenced to ?ground.? Correspondingly, with a clearly defined point of reference for phase it becomes possible to speak of voltages and currents in an AC circuit having definite phase angles.
In accordance with the rules for complex vector addition, the magnitudes of vectors in phase will add like scalers when they have additive polarity, and subtract with subtractive polarity. This is similar to simple DC. So if you take the 120/240V service, and we assume they are in phase (as some have proposed) with the secondary transformer windings connected as +120- -120+, is subtractive (or opposing) and the vectors would subtract like scalers: 120 -120 = 0. But, we know we have 240 volts between them. therefore the theory that they are in phase doesn't hold true.
With AC, two voltages can be aiding or opposing one another to any degree between fully-aiding and fully-opposing, inclusive. Without the use of vector (complex number) notation to describe AC quantities, it would be very difficult to perform mathematical calculations for AC circuit analysis.
Vectors do not directly add for angles for any degree out of phase. For the same 120/240V connection,
+120 - -120+, to get 240V, you have to subtract the opposing voltage vectors, i.e. (120@ 0) - (120 @ 180) = 240 @ 0.

Conclusion
It would not be possible to combine subtractive polarities together and get 240 volts if indeed the vectors were in phase. Therefore, they must be subtractive and fully opposing, and thats why you get 240 @ 0.
 
Typically, I don?t ?calculate? line voltages electrically, I tend to view these things geometrically, just Law of Sines?.whatever. But I can crunch vectors.

Take for example: (remember, per Rattus? request, these are perfect sinusoids  )

Va-b = Voltage from a to b

L1: 120@0 V
L2: 120@120 V
L2: 120@240 V


VL1-L2 = { ( [L2(cosΘL2)-L1(cosΘL1)]2 + [L2(sinΘL2)-L1(sinΘL1)]2 )1/2 }

@ tan-1 { [L2(sinΘL2)-L1(sinΘL1)]/ [L2(cosΘL2)-L1(cosΘL1)] }


This is just the general form for adding or subtracting two vectors and gives the result in phasor form.
 
BC, turn off the smudge pot. I said we commonly assign phase angles to complex waves although this is not correct because the phase angle applies to the fundamental only.

AC analysis assumes a pure sinusoid, so any talk of the Fourier series or complex waveforms is just so much more smoke.

Now back to the root of the discussion. You insist on attaching a negative sign to a phasor so you can claim no phase shift. But, we all know that phasors are described by MAGNITUDES and phase angles. You have yet to explain or justify your position on this, so I must presume you cannot.

Explain, if you can, why it is permissible to describe 3-ph voltages with magnitudes and angles, but it is not permissible to describe 1-ph voltages the same way.

Don't start harping on complex waves either, that was never an issue!
 
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