Why is residential wiring known as single phase?

[rattus]

Well then I think you should review 'mathematical substitutions', so that you can use them correctly.
Simplified.
-Y[wt] = Y[wt-PI]

If Vbn = -Vnb and if Van=Vnb then -Vbn=Van. Their waveforms must start at the same point in time: t0. A simple inversion does not create a time delay, Vbn 'is created' at the exact same time as is Vnb.

Your PI is not a direct part of the phase constant: rather it is simply a modifier to it.

Yes,

-Vbn does start at the same time as Van because they are equivalents.

But,

Van starts at 0; phase is wt
Vbn starts at PI; phase is wt + PI

Phase has commonly taken to be the argument of the POSITIVE sine function. You can't slip a negative in front of it.

The phase constant in the expression for phase is the sum of all constants; in this case the constant it is PI.

 

[rbalex]

Nope.
Post #1004 was an attempt to provide a cogent, almost self-evident, explanation of why more more than one phase exists in a 120-0-120 system.
It shows exactly why it is more than one phase.
I specifically chose the SCR arrangement.
It requires firing pulses at 180 degrees intervals. That is two per cycle.
The circuit wouldn't require phase displaced pulses if there was just one phase.

I've yet to see a cogent explanation that reasonably refutes this.

I thought it was pretty simple - we went through it in several different iterations. But it boiled down to down stream circuits don't determine the characteristics of the voltage system. In other words, disconnect the down stream and the voltage system you have is the voltage system you have - the down stream is irrelevant.

 

[rattus]

All engineering is applied physics. All physics is applied mathematics. There's never any justification required to use physics or math to describe an engineering problem. The burden is on YOU to prove there isn't a justification for it. Without the math there are no phases. Phases are a mathematical construct. As such, math is ALWAYS a valid method of description.


Any and all math books that include trigonometry. I'm sure you can access some.

In other words, you cannot justify the reasoning behind rbalex's claim that we have only one phase? Don't refer me to trig books. Tell me why his claim is valid.

 

[jim dungar]

Why can't I write simply V2n = Vm*sin(wt + PI)? Everyone else does.

Go ahead and do it. Just remember the relationship is -V2n = Vn2

-V2n = Vm*sin(wt + PI-PI)= Vm*sin(wt)=Vn2

 

[rattus]

Consider the split phase system.

If it only one phase is present, say (wt), then

V1n = Vm*sin(wt)
V2n = Vm*sin(wt)

Then V1n and V2n are equal and in-phase. But rbalex has already said they are out of phase although they carry the same phase. I am so confused!

Something wrong here. I couldn't run my AC!

 

[rattus]

Go ahead and do it. Just remember the relationship is -V2n = Vn2

-V2n = Vm*sin(wt + PI-PI)= Vm*sin(wt)=Vn2

That's silly Jim.

I want the cleanest expression possible for V2n.

 

[jim dungar]

Yes,

-Vbn does start at the same time as Van because they are equivalents.

But,

Van starts at 0; phase is wt
Vbn starts at PI; phase is wt + PI

You seem to be forgetting a minus sign that you inserted along with your phasor rotation.
According to you V12= V1n-V2n, this means you do not have +V2n you actually have -V2n. Didn't we all agree that V1n+V2n=0?
-Vbn=Vnb

 

[gar]

120308-1653 EST

rbalex:

On p 24 of Robertson's book the following statement is made:

At any instant, two particles whose times of vibration differ by an integral multiple of the periodic time are said to be in the same phase. Two consecutive particles in the same phase are separated by a wave-length.

This whole chapter is on wave motion, and at one point gets to Lissajous figures and a phase difference of 180. Much earlier in this thread I mentioned the difference in the Lissajous pattern between a 0 degree phase difference and one of 180 degrees.

.

 

[rbalex]

Consider the split phase system.

If it only one phase is present, say (wt), then

V1n = Vm*sin(wt)
V2n = Vm*sin(wt)

Then V1n and V2n are equal and in-phase. But rbalex has already said they are out of phase although they carry the same phase. I am so confused!

Something wrong here. I couldn't run my AC!

Don't speak for me again. I have no idea what you are misinterpreting. I've said to be "in phase" is dependent on phase; i.e.,
the same phase is required to be "in phase". However phase is not dependent on "in phase";i.e., two functions may have the same phase and not be "in phase."

You didn't understand phase in 1956 - you still don't seem to have caught on.

 

[rbalex]

120308-1653 EST

rbalex:

On p 24 of Robertson's book the following statement is made:

At any instant, two particles whose times of vibration differ by an integral multiple of the periodic time are said to be in the same phase. Two consecutive particles in the same phase are separated by a wave-length.

This whole chapter is on wave motion, and at one point gets to Lissajous figures and a phase difference of 180. Much earlier in this thread I mentioned the difference in the Lissajous pattern between a 0 degree phase difference and one of 180 degrees.

.

Thanks - you're a good man.

I confess I really have no idea how Lissajous figures apply to your field. Even more confession - I had no idea what Lissajous figures were until I did a brief google search.

It does appear the definition may allow at least some marginal validity to my concept of "same phase"; i.e., "At any instant, two particles whose times of vibration differ by an integral multiple of the periodic time are said to be in the same phase." But we're talking particle physics so I sure wouldn't attempt to push it past "appearance."

I haven't paid much attention to the "manufacturers" branch of the thread (almost none, in fact). I've intentionally tried to stay out of it until I was specifically addressed with respect to the math applications since manufacturing/testing isn't my field of expertise.

 

[rattus]

You seem to be forgetting a minus sign that you inserted along with your phasor rotation.
According to you V12= V1n-V2n, this means you do not have +V2n you actually have -V2n. Didn't we all agree that V1n+V2n=0?
-Vbn=Vnb

No Jim,

V1n + V2n = 0V

V1n - V2n = 240V

We subtract in this case, just like we do with a wye.

 

[jim dungar]

No Jim,

V1n + V2n = 0V

V1n - V2n = 240V

We subtract in this case, just like we do with a wye.

You absolutely did not read what I posted.

And by the way we add the voltage like we do in a delta.

If you are going to insist on using a negative voltage, then you will have to have your waveform start on the negative zero crossing.

 

[jim dungar]

That's silly Jim.

Yes it is. But it is not wrong.

I want the cleanest expression possible for V2n.

Then use -Vm*sin(wt+0).
V1*sin(wt+0) -(-V2*sin(wt+0) = (V1+V2)*sin(wt+0).
But that seems pretty cluttered to me, so why not just use V1+V2=V12?

 

[rattus]

rattus: If it only one phase is present, say (wt), then

V1n = Vm*sin(wt)
V2n = Vm*sin(wt)

Then V1n and V2n are equal and in-phase.

Don't speak for me again. I have no idea what you are misinterpreting. I've said to be "in phase" is dependent on phase; i.e.,
the same phase is required to be "in phase". However phase is not dependent on "in phase";i.e., two functions may have the same phase and not be "in phase."

You didn't understand phase in 1956 - you still don't seem to have caught on.

But these two functions are equal, have the same argument, how can they NOT be in phase??

If two functions have the same phase, their starting points are the same--phi0 is the same, in this case 0, their peaks and troughs coincide. I still don't understand why they are NOT in phase?

 

[rattus]

Yes it is. But it is not wrong.

Then use -Vm*sin(wt+0).
V1*sin(wt+0) -(-V2*sin(wt+0) = (V1+V2)*sin(wt+0).
But that seems pretty cluttered to me, so why not just use V1+V2=V12?

Because Jim, phasors are complex numbers. They are often expressed in polar form with a magnitude and phase angle. The magnitude is an absolute value, always POSITIVE. In general the voltage is expressed as,

Vrms(cos(wt + phi) + jsin(wt + phi))

To obtain the rectangular form, we merely evalute the trig functions. If we turn up with negative values, so be it. A negative sign in front would cause the arguments to change and greatly confuse the issue.

Besides, that is the way it is done!

 

[rattus]

You absolutely did not read what I posted.

And by the way we add the voltage like we do in a delta.

If you are going to insist on using a negative voltage, then you will have to have your waveform start on the negative zero crossing.

Well Jim, would you please demonstrate how we get 208V by adding. Let the voltages be,

Va = 120V @ 0
Vb = 120V @ -120
Vc = 120V @ -240

Or, whatever you wish.

 

[rattus]

Trig Identites:

Trig Identites:

Let's end this silly talk about simple trig identities. We knew from day zero, even before that:

sin(wt + PI) = -sin(wt)

Nothing hard about that.

 

[rattus]

A Phasor Diagrm:

A Phasor Diagrm:

What say we make a phasor diagram for this problem?

Let

V1n = 120Vrms(cos(0) + jsin(0))
Arrow points right

V2n = -120Vrms(cos(0) + jsin(0))
Arrow tries to point right, but the negative sign won't let it.
May as well let it point left and let the argument of V2n be PI.

120Vrms @ PI <----------0---------->120Vrms @ 0

Looks like the phase constants or phase angles are 0 and PI whatever we do, and the waves are separated by PI radians.

We easily convert this to rotating phasors, but that would not change the phase constants.

 

[__dan]

What say we make a phasor diagram for this problem?

Let

V1n = 120Vrms(cos(0) + jsin(0))
Arrow points right

V2n = -120Vrms(cos(0) + jsin(0))
Arrow tries to point right, but the negative sign won't let it.
May as well let it point left and let the argument of V2n be PI.

120Vrms @ PI <----------0---------->120Vrms @ 0

Looks like the phase constants or phase angles are 0 and PI whatever we do, and the waves are separated by PI radians.

We easily convert this to rotating phasors, but that would not change the phase constants.

V1n = 120Vrms(cos(0) + jsin(0))

The complex component, jsin(), usually represents an energy storage element, capacitance or inductance, in the circuit. Basically, the second, complex, term is a second voltage source that is 90 deg out of phase with the first voltage. j is a rotation operator, by 90 deg, the same way (-1) is a rotation operator, by 180 deg.

Is that what you want to do, pop some inductive or capacitive elements into the circuit to resolve your dilemma? Capacitance or inductance in the circuit will cause a phase shift. However, no energy storage elements were proposed in the OP inquiry.

Is this a ploy to resolve your adding in the 180 deg rotation by swapping the leads at the time of measuring, or is there something else going on that you see. Do you wish to clarify what you wish to convey or wish to make the description of the underlying physical subject even more cloudy.

 

[jim dungar]

Well Jim, would you please demonstrate how we get 208V by adding.

Again you did not read what I posted.
The phasors for V1n and Vn2 are connected point to tail so they add just like in a delta connection.

120V L-N does not equal 208V L-L if we have a delta. Lets stay away from the 3-phase because it does not appear to be helping the discussion.