Is it Single or Two Phase?

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That it what I said!

That it what I said!

bcorbin said:
I encourage everyone to go here and read.

http://www.moultonlabs.com/more/about_comb_filtering_phase_shift_and_polarity_reversal/P1/
Click to expand...

I think he said that it is incorrect to attach a phase angle to a complex wave, inverted or not. He did imply that a polarity reversal of a sinusoid amounts to a 180 degree phase change.

And why are you so hung up on this? We get the right solution by using the 180 degree phase difference. If if you were right, which you are not, the outcome would be the same.

Furthermore, your assumption of complex waves precludes you from attaching a phase angle to any of the waveforms..
 
Same same:

Same same:

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.
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Same same Jim,

Draw the phasor diagrams.

3-ph wye:

Vab - Van + Vbn = 0

Vab = Van - Vbn

1-ph:

V1n + V2n = 0

V12 - V1n + V2n = 0

V12 = V1n - V2n

This is the classic way to do it with phasors.

Now you can do it by the seat of your pants with the same results--perhaps quicker. You are swapping subscripts in order to add. You are in effect subtracting even you won't admit it.
 
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bcorbin said:
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.
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BC,

This looks like it should work except that you are multiplying by 2 instead of squaring, and you are multiplying by 1/2 instead of taking the square root. Furthermore, there is no magnitude. Still it is not phasor math. Do it with phasor math and plug in the numbers. It only takes a few minutes.
 
rattus said:
You are swapping subscripts in order to add. You are in effect subtracting even you won't admit it.
Click to expand...

Simply stated: One of the key reasons for always using "addition" when solving networks using Kirchoff's laws is the ability to determine the "direction of flow" of the unknown variable by checking for a "negative" magnitude in the answer.

This is why I ask that all networks be solved in a consistent manner, whether they be a parallel combination of single phase 2 and 3-wire circuits or an unbalanced open-delta with a high leg.

I have asked you before: How do you account for your "phase shift" when solving for the current flow through the center tapped transformer windings when the secondary load is an unbalanced 120/240V 3-wire circuit in parallel with a 240V 2-wire load?
 
Please, someone explain to me why a web site for sound and recording engineers is being used as a reference to try and explain power system voltages, phases and angles?

This topic has gotten way off coarse.
 
kingpb said:
Please, someone explain to me why a web site for sound and recording engineers is being used as a reference to try and explain power system voltages, phases and angles?
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I will if you tell us why you think this is "a web site for sound and recording engineers."
 
Rattus...Those 2's and 1/2's were powers (superscript). I guess this forum doesn't support those, so it just translated them as plain text. Sorry about that, at least you knew what they were supposed to be. I'll use the little "^" symbol from now on to denote a power.

I'm not really sure what you mean by this is not phasor math. You can't technically add phasors without separating them into their real and complex components, which is what I did. Multiplying phasors is easy; magnitude times magnitude at an angle that equals the sum of their phase angles; division would be the dividend of the magnitudes with the phase angle being the difference between the minuend and the subtrahend. I'm not sure if you were referring to that, but it's a shot.

As to the use of a sound engineering reference, sound waves, being pressure waves, are exactly like electrical signals. They are longitudinal, meaning electrons move back and forth along the path of travel, not up and down, so to speak. Oh yeah....I'm not trying to prickle anyone here, but this talk just reminded me of a something to consider when you're arguing about whether a vector can have a negative value. I know some of you state that a magnitude can not be negative. That is correct. However, one can have a negative sign in front of that magnitude. The rule is that if you have a vector "r" at some phase angle, the magnitude of that vector is the absolute value of r. r can be negative; the magnitude of r can not.

This makes sense when considering compression and rarefaction waves (also longitudinal). If you consider a compression wave of magnitude "r" moving down a pipe (picture blowing on a pipe very quickly, and repeatedly), and specifying that compression (increased pressure) is positive, you would then have a vector of magnitude "r" at a phase angle of let's just arbitrarily say 90 degrees.

Now, if along side this pipe, there were an identical pipe, except something was sucking on it at exactly the same frequency and at the same strength, the waves would still travel in the same direction as if you were blowing into it. This wave would have a negative value at the same phase angle. The magnitude, or strength of the pulse would be identical, but in different directions, call it an inversion, a polarity reversal, whatever.
 
Your kiddin, right?

Your kiddin, right?

LarryFine said:
I will if you tell us why you think this is "a web site for sound and recording engineers."
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Ah, lets see, how about : "The Art and Science of Sound", as stated on their home page; for starters anyway.

I particularly liked the article entitled; Making Loudspeakers and Control Rooms That Make Music "Sound Good". It was refreshing and enlightening at the same time. I'm on my way home now to make the necessary mod's.

Ok, Larry, start splane'in, I got all afternoon!!
 
LarryFine said:
Oh, I see: you were referring to the website linked to above, not this one.

Never mind. :eek:
Click to expand...

Nice Out :rolleyes:

I thought I was going to have something to read while I was drinking my beer. Now I'll have to get drunk to kill the pain of disappointment!
 
Neutral current:

Neutral current:

jim dungar said:
Simply stated: One of the key reasons for always using "addition" when solving networks using Kirchoff's laws is the ability to determine the "direction of flow" of the unknown variable by checking for a "negative" magnitude in the answer.

This is why I ask that all networks be solved in a consistent manner, whether they be a parallel combination of single phase 2 and 3-wire circuits or an unbalanced open-delta with a high leg.

I have asked you before: How do you account for your "phase shift" when solving for the current flow through the center tapped transformer windings when the secondary load is an unbalanced 120/240V 3-wire circuit in parallel with a 240V 2-wire load?
Click to expand...

Jim, first off, the 240V load does not contribute to neutral current. Now let the voltages on L1 and L2 be,

V1n = 120Vrms@0
V2n = 120Vrms@180

Now let the load impedances be Z1 and Z2, then the currents into the neutral are,

I1 = 120V@0/Z1
I2 = 120V@180/Z2

In = I1 + I2

If Z1 and Z2 are equal, I1 and I2 are of equal magnitude but are 180 degrees out of phase. Therefore they ADD to zero.

If they are not of equal magnitude and not of opposite phase, the addition will yield the magnitude and phase of In.

You would compute In in a wye the same way. That is consistency.
 
Rattus,

I asked for the currents in the center tapped transformer windings, not the neutral current.

The correct solution to the problem will have values (and direction) for the following currents:
Winding #1 (N-1)
Winding #2 (2-N)
Load 1-N
Load 1-2
Load N-2

And I did notice that in one problem you subtracted voltages but in a different one you added currents.
 
jim dungar said:
Rattus,

I asked for the currents in the center tapped transformer windings, not the neutral current.

The correct solution to the problem will have values (and direction) for the following currents:
Winding #1 (N-1)
Winding #2 (2-N)
Load 1-N
Load 1-2
Load N-2

And I did notice that in one problem you subtracted voltages but in a different one you added currents.
Click to expand...

You are right. You add or subtract as required; is that wrong?

For the problem at hand let's make it easy and assign some resistive load values:

R1 = 10
R2 = 8
R12 = 12

Now define:

I1 to flow out of L1 thru R1 to N
I2 to flow out of L2 thru R2 to N
I12 to flow out of L1 thru R12 to L2

V1n = 120V@0
V2n = 120V@180
V12 = 240V@0

Then,

I1 = 120V@0/10 = 12A@0
I2 = 120V@180/8 = 15A@180
I12 = 240V@0/12 = 20A@0

Then,

Iv1 = 20A@0 + 12A@0 = 32A@0 [N to L1]
Iv2 = 20A@0 - 15A@180 = 20A@0 + 15A@0 = 35A@0 [L2 to N]
In = 12A@0 + 15A@180 = 15A@180 - 12A@180 = 3A@180
 
The Rotating Phasor:

The Rotating Phasor:

The rotating phasor is of the form:

1) v(t) = Vpk[cos(wt + phi) +jsin(wt + phi)]

where (wt) is the independent variable in radians and phi is the phase shift in radians.

Vpk is the magnitude of the peak value of the sinusoid

The real part of this expression provides the instantaneous value.

By definition the rotating phasor describes a pure sinusoid. Any argument to the contrary is baseless.
 
rattus said:
V1n = 120V@0
V2n = 120V@180
V12 = 240V@0

Iv1 = 20A@0 + 12A@0 = 32A@0 [N to L1]
Iv2 = 20A@0 - 15A@180 = 20A@0 + 15A@0 = 35A@0 [L2 to N]
In = 12A@0 + 15A@180 = 15A@180 - 12A@180 = 3A@ 180
Click to expand...

So now are you saying, in a purely resistive circuit, the current flow from L2-N is @0 but the voltage from L2-N is @180?
 
jim dungar said:
So now are you saying, in a purely resistive circuit, the current flow from L2-N is @0 but the voltage from L2-N is @180?
Click to expand...

Yes, that is right! I2 and I12 are defined to flow in opposite directions, therefore they must be subracted to yield the total current. If I had chosen to compute current flow from N-L2 (in the coil), the current would be 35A@180. Just depends on how you look at it.

This is seeming anomaly is common when writing loop equations. One is free to assign a current as CW or CCW. It is common for two current loops to oppose each other. It is common to find that your assumption on current flow to be backwards so to speak even if the solution is correct.

I could have chosen to represent V2 as -V1 and assume all currents to be CW, but one would not do that in a 3-ph analysis, so I don't do it here. Consistency!
 
But, for the sake of general electrcial theory consistency, why would you say that the current caused by a purely resistive circuit is 180 degrees out of phase with the voltage?

I still find it easier to describe a 120/240V system as two 120V circuits connected +-+- rather than as +--+.
 
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