Why is residential wiring known as single phase?

[mivey]

Lets enter the ideal world that Rattus wants us to be in, at Time t0 spin your 120/240V generator;
Does a voltage appear between terminals A and N at the same exact time that a voltage occurs between terminals B and N?

Yes, with a 180? phase displacement between Van and Vbn and a 0? displacement between Van and Vnb.

Do we have to wait 8.33-10msec (60 or 50hz) before two voltages can be sensed?

No.

Now take the wye system, you keep wanting to bring up, at Time t0 spin your 208Y/120V generator;
Does a voltage appear between terminals A and N at the same exact time that a voltage occurs between terminals B and N?

No, and there is a 120? displacement between Van and Vbn.

Do we have to wait 2.78-3.33msec (60 or 50hz) before any two voltages can be sensed?

No.

During either of the two examples above: Is the generator ever taking away a (i.e. the equivalent of 'producing negative') voltage between any two points? While mathematically accurate, that does not sound good, so how about; Is the generator ever producing a voltage opposite of what it is designed to produce?

A generator produces what it was designed to produce.

 

[rattus]

That is because one is the inverse of the other. Not because one waveform is time displaced from another.

Select the 'invert' option on the scope. Does it displace the waveform in time, or simply provide its 'opposite polarity'?

Waveforms are not displaced in time simply by swapping reference points, B versus N. There is only a single voltage between terminal B and N, therefore there can only be a single voltage waveform.

I'm guessing you didn't look at the reference I posted concerning phases and polarities.

Jim, it doesn't matter how the waveforms got shifted!

We still describe them as follows:

V1n = 120Vrms*sin(wt)
V2n = 120Vrms*sin(wt + PI)

Their phase constants are DIFFERENT, their phases are DIFFERENT.

Free your mind from the transformer and scope just this once and consider these voltages to be synchronized, independent sources. They are phase opposed; they are out of phase by PI radians. To claim they carry the same phase is to say that wt = wt + PI! Clearly not the case.

 

[mivey]

Wow, -Vm*sin(wt) NE Vm*sin(wt)
Could you provide the reference for this?

Just solve the equation for the voltage (i.e. -X_volts NE X_volts)

I am still trying to figure out why you think:

Other than that it proves a waveform and its inverse have the same phase, because they both occur at the same point in time..

Consider that for an example three-phase source with all three voltages starting at t=0: Van=158v, Vbn=-25v, Vcn=-133v and they all occur at the same time t=1ms. So while one might claim that these three points are the same phase because of what they see on their stopwatch, it is obviously not what we call the same phase.

Again, I challenge you to find where math or physics texts say they are the same phase. There is a little more to comparing phases than using a stopwatch. That is not how we compare phases in the electrical world.

Here is a reference for you, it even includes colored pictures. It discusses audio waves, but I think you might agree it is applicable to electricity also.

The reason this is not applicable to electricity has been covered ad nauseam. I covered it again in #1744 along with the follow-up graphic in #1778 .

A phase shift due to a time shift is not the type phase shift we are concerned with in transformers as I explained here:

A phase shift and a time shift go hand in hand but it really does depend on your reference frame. For example, in a three-phase generator all of the voltages start at the same time so they all have the same t₀. But from a different reference frame, we could consider the time that they peak. In that frame, the time of the positive peak (t_peak) will be different for the voltages. With the three-phase generator, we produce a shift by a physical shift, just like I did for the voltages on the left side of my generator example.

Using the winding voltages in different ways does not produce a real time shift for t₀ in one reference frame. But if we look at the positive peak times (t_peak) for those voltages, they do occur at different times.

The other way to get a shift is to actually delay the waveform. You could do this using delay boxes. I have a delay cascade circuit that I use on my workbench to get a 3-phase set of voltages from a single-phase source. The time shift due to flux lag from the primary to secondary side of the transformer is another example of a time delay.

This actual time delay appears to be the shift you are thinking about, but is not the phase shift we normally talk about when discussing transformer phase shifts. Outside of the shift from the flux delay (which we routinely ignore), the transformers do not have that kind of time shift.

 

[gar]

120310-2133 EST

If I multiply a sine wave with another sine wave of the same frequency and both sine waves are "in-phase" by my definition, then the result is
v_a = A₁*sin (wt) * A₂*sin (wt)
v_a = A₁*A₂*sin² (wt)
v_a = A₁*A₂*(1 - cos (2wt) ) / 2
Extracting the DC component we have
V_a = A₁*A₂/2

Let A₁ be a constant amplitude reference signal, and A₂ be a input variable amplitude proportional to some physical quantity that ranges from 0 to 1, the output varies from 0 to A₁*A₂/2 and has a positive sign.

Next suppose that the phase of the input variable changes by Pi at the zero transition point as the physical input goes negative.

Now the DC output goes negative, and thereby the DC output can indicate the plus or minus condition of the physical input. Thus, we have a phase sensitive detector. There is no useful sense to try to claim that the input variable has the "same phase" under both the negative and positive input conditions. Use of the term "same phase" needs to have a practical useful meaning to assist in communication. Otherwise there is no need for the term. "Same phase" is of no use to me if it equates to include a 180 degree phase shift.

If I do not use a phase sensitive detector but simply full wave rectify the variable, then my output is always positive and I can not distinguish the minus physical input from the plus input.


Now consider a different situation. This time add signals. This is what happens in various wave motion studies. Such as in the study of interference patterns in light.

v_a = A₁*sin (wt) + A₂*sin (wt)
v_a = ( A₁ + A₂ ) *sin (wt)

This result got bigger than either input.

Next change the phase of the A₂ component by Pi and the result is

v_a = ( A₁ - A₂ ) *sin (wt)

and if A₂ = A₁, then the result is ZERO. The dark band in an interference pattern.

Again one does not want to use a useless definition for "same phase" that can not distinguish a half wavelength from a full wavelength difference.

The definition from Physical Optics of same phase does not create this problem.

.

 

[mivey]

The reason this is not applicable to electricity has been covered ad nauseam. I covered it again in #1744 along with the follow-up graphic in #1778 .

I forgot to mention this additional follow-up in #1827

 

[mivey]

120310-2133 EST

If I multiply a sine wave with another sine wave of the same frequency...

Another nice explanation.

 

[Besoeker]

How long you been at it Bes?

Not long. Just a bit over forty years.
Just half a lifetime.

 

[Besoeker]

Are you saying

1. Vmsin(ωt+π) = - Vmsin(ωt)
2. Vmsin(ωt+4π/3) = - Vmsin(ωt+π/3)
3. Vmsin(ωt+5π/3) = - Vmsin(ωt+2π/3)

aren't identities?

Nope.

Just

1. Vmsin(ωt+π) ≠ Vmsin(ωt)
2. Vmsin(ωt+4π/3) ≠ Vmsin(ωt+π/3)
3. Vmsin(ωt+5π/3) ≠ Vmsin(ωt+2π/3)

Thus six different waveforms.

 

[rbalex]

Nope.

Just

1. Vmsin(ωt+π) ≠ Vmsin(ωt)
2. Vmsin(ωt+4π/3) ≠ Vmsin(ωt+π/3)
3. Vmsin(ωt+5π/3) ≠ Vmsin(ωt+2π/3)

Thus six different waveforms.

I guess we agree that

1. Vmsin(ωt+π) = -Vmsin(ωt)
2. Vmsin(ωt+4π/3) = -Vmsin(ωt+π/3)
3. Vmsin(ωt+5π/3) = -Vmsin(ωt+2π/3)

are identities then.

Care to answer the second question?: "Or are you saying every time you can use a different argument for the same function a new "phase" magically appears?"

 

[rattus]

Not long. Just a bit over forty years.
Just half a lifetime.

I've had one year of experience 53 times.

 

[Besoeker]

Care to answer the second question?: "Or are you saying every time you can use a different argument for the same function a new "phase" magically appears?"

I'm saying no such thing.
Not once have I used a different argument for the same function.

 

[rattus]

I guess we agree that

1. Vmsin(ωt+π) = -Vmsin(ωt)
2. Vmsin(ωt+4π/3) = -Vmsin(ωt+π/3)
3. Vmsin(ωt+5π/3) = -Vmsin(ωt+2π/3)

are identities then.

Care to answer the second question?: "Or are you saying every time you can use a different argument for the same function a new "phase" magically appears?"

Nothing magic about it. Six arguments from six sine functions, six phase constants, six phases. Self evident. So what if half of them are inverses?

Your identities merely show that a wave shifted by PI is the inverse of the unshifted wave. We already know that. So why keep harping about it as if you have discovered something new and mysterious?

 

[rattus]

Seems to me that the requirements to be 'of the same phase' or 'in phase' are the same, that is they must of course be of the same frequency and have their negative to positive zero crossings coincide. In fact, with the application of an appropriate scaling factor, the two waves would follow the same locus.

Just look at the plots of sin(wt) and -sin(wt) and tell me if the minus to plus zero crossings coincide.

 

[rbalex]

I'm saying no such thing.
Not once have I used a different argument for the same function.

It wasn't an accusation Bes; it was an inquiry.

So if:

1. Vmsin(ωt+π) = -Vmsin(ωt)
2. Vmsin(ωt+4π/3) = -Vmsin(ωt+π/3)
3. Vmsin(ωt+5π/3) = -Vmsin(ωt+2π/3)

Then

1. Vmsin(ωt)
2. Vmsin(ωt+π/3)
3. Vmsin(ωt+2π/3)
4. -Vmsin(ωt)
5. -Vmsin(ωt+π/3)
6. -Vmsin(ωt+2π/3)

are equally valid expressions

AND

1. Vmsin(ωt), -Vmsin(ωt) [common ωt]
2. Vmsin(ωt+π/3), -Vmsin(ωt+π/3) [common ωt+π/3]
3. Vmsin(ωt+2π/3),-Vmsin(ωt+2π/3) [common ωt+2π/3]

 

[rattus]

It wasn't an accusation Bes; it was an inquiry.

So if:

1. Vmsin(ωt+π) = -Vmsin(ωt)
2. Vmsin(ωt+4π/3) = -Vmsin(ωt+π/3)
3. Vmsin(ωt+5π/3) = -Vmsin(ωt+2π/3)

Then

1. Vmsin(ωt)
2. Vmsin(ωt+π/3)
3. Vmsin(ωt+2π/3)
4. -Vmsin(ωt)
5. -Vmsin(ωt+π/3)
6. -Vmsin(ωt+2π/3)

are equally valid expressions

AND

1. Vmsin(ωt), -Vmsin(ωt) [common ωt]
2. Vmsin(ωt+π/3), -Vmsin(ωt+π/3) [common ωt+π/3]
3. Vmsin(ωt+2π/3),-Vmsin(ωt+2π/3) [common ωt+2π/3]

So what? They are inverses of each other. The minus sign merely inverts the wave. I can't believe you believe your own argument. The negative>positive zero crossings do not coincide; they are out of phase, they are NOT of the same phase,

 

[Besoeker]

It wasn't an accusation Bes; it was an inquiry.

So if:

1. Vmsin(ωt+π) = -Vmsin(ωt)
2. Vmsin(ωt+4π/3) = -Vmsin(ωt+π/3)
3. Vmsin(ωt+5π/3) = -Vmsin(ωt+2π/3)

Then

1. Vmsin(ωt)
2. Vmsin(ωt+π/3)
3. Vmsin(ωt+2π/3)
4. -Vmsin(ωt)
5. -Vmsin(ωt+π/3)
6. -Vmsin(ωt+2π/3)

are equally valid expressions

AND

1. Vmsin(ωt), -Vmsin(ωt) [common ωt]
2. Vmsin(ωt+π/3), -Vmsin(ωt+π/3) [common ωt+π/3]
3. Vmsin(ωt+2π/3),-Vmsin(ωt+2π/3) [common ωt+2π/3]

Your point being?

 

[rbalex]

Your point being?

What I've said all along; your "hexiphase" is a glorified three-phase system.

 

[rattus]

Arguments:

Arguments:

The argument of

sin(wt) is wt, no question.

The argument of

[-sin(wt)] is (wt + PI), no question.

Just reverse the well known trig identity to the expression inside the brackets.

 

[rattus]

What I've said all along; your "hexiphase" is a glorified three-phase system.

The glory is three more phases to provide smoother rectification. No difference in principle from the millions, maybe billions, of single phase, full wave rectifiers in the world.

 

[rbalex]

The argument of

sin(wt) is wt, no question.

The argument of

[-sin(wt)] is (wt + PI), no question.

Just reverse the well known trig identity to the expression inside the brackets.

No the argument of [-sin(wt)] is wt - by your own defintion. Sin(wt) is -sin(wt + PI) by identity.