Why is residential wiring known as single phase?

[__dan]

Perhaps you can provide an official definition of "in phase". Meanwhile, I believe the statement from Tang's textbook will suffice.

I repeat, positive peaks can never coincide in a sinusoidal wave and its inverse, therefore, the two waves can never be in phase.

Let us look at it another way:

Let the expressions for the phases of the two waves be:

phi1 = (wt + phi0) for V1
phi2 = (wt + theta0) for V2

The waves are in phase if and only if phi0 = theta0

New let phi0 be zero and let theta0 = PI.

0 NE PI

Therefore

phi1 NE phi2

Are those waves in phase??

Now don't answer with another question. Just tell us how they can be in phase with phi0 NE theta0.

Simple. At the transformer, the winding's turn direction matches and phi 0 = theta 0. This should be obvious because 120(wt + phi) + 120(wt + theta) = 240(wt + phi).

To get phi 0 != theta 0 you have to reverse the polarity of the connection leads relative to the winding's turn direction. The transformer natively offers you two windings that are wound in the same direction.

 

[Besoeker]

First, thanks for asking in the interrogatory. (Sincerely)

Let me ask this one more time with maybe a little better grammar so that I can clearly understand what you are claiming.
Are you saying that, in a single phase circuit with a 0.8 pf lagging load, the current and voltage are in phase?

 

[rbalex]

Perhaps you can provide an official definition of "in phase". Meanwhile, I believe the statement from Tang's textbook will suffice.

I repeat, positive peaks can never coincide in a sinusoidal wave and its inverse, therefore, the two waves can never be in phase.

Let us look at it another way:

Let the expressions for the phases of the two waves be:

phi1 = (wt + phi0) for V1
phi2 = (wt + theta0) for V2

The waves are in phase if and only if phi0 = theta0

New let phi0 be zero and let theta0 = PI.

0 NE PI

Therefore

phi1 NE phi2

Are those waves in phase??

Now don't answer with another question. Just tell us how they can be in phase with phi0 NE theta0.

I don't need a definition of "in-phase." Your's is fine, but it is what is irrelevant. Only the period is necessary to explain "Why is residential wiring known as single phase?"

I don’t have much time to spend today. I haven’t tried to follow the manufacturers’ branch of the discussion, so if what I say here has already been covered, I apologize.

Using gar’s “black box,” assume three terminals labeled left to right “E-F-G” that randomly originate from a conventional 120/240V residential system.

Assuming the measuring convention was “left to right,” it would take ten seconds or less to determine the neutral and which pair of terminals had 240V between them with a simple voltmeter. Again assuming the labeling convention is “left to right,” you would quickly identify “L1” and “L2.” “L1” would be to the left of “L2.”

Using an oscilloscope, and the same “left to right” measuring convention you would conclude “L1 to neutral” was “in-phase” with the 240V set and the “L2 to neutral” was “out of phase.”

If the original “E-F-G” labels remained left to right but the measuring and labeling conventions were reversed to “right to left” you would still conclude “L1 to neutral” was “in-phase” with the 240V set and the “L2 to neutral” was “out of phase.” However; now “L1” would now be to the right of “L2.”

That means for a conventional 120/240V residential system which “live to neutral” terminal is “in phase” or “out-of-phase” with the 240V set would be entirely based on how it is measured.

However, with respect to a conventional 120/240V residential system the definition of phase is not direction sensitive nor dependent on the means of measurement. It only requires that the period P and the effective or equivalent t₀ for a periodic voltage function to be validly determined. Trig identities are what enables us to determine the "effective or equivalent t₀ for a periodic voltage function."

If two periodic functions have the same period P and t/P is effectively or equivalently the same throughout the period, they have the same phase. In fact, for a conventional 120/240V residential system every relevant voltage function has an identical phase. So there is only one phase in the system; i.e., a single-phase. Which is why residential wiring is known as "single-phase."

 

[rbalex]

Let me ask this one more time with maybe a little better grammar so that I can clearly understand what you are claiming.
Are you saying that, in a single phase circuit with a 0.8 pf lagging load, the current and voltage are in phase?

No I'm saying they have the same phase, assuming they use a common t₀. See my explanation above.

I must leave for most of the day; I may be able to check in occasionally, but I can't count on it.

 

[Besoeker]

Only the period is necessary to explain "Why is residential wiring known as single phase?"

All three phases of a three phase system have the same period.
By your definition they would all be in phase.
I suspect that more than a few of us here, including some from the single phase lobby, just might not concur.

 

[gar]

120228-1154 EST

rbalex:

This is in response to your post 1516.

I will choose AtoN and BtoN.

You have written a ratio t/P where t is a variable time in some units, I will choose the units as seconds, and P is the period of the waveform in time, and again I will choose seconds. Thus, t/P is a fractional part of a period. To allow working over multiple periods I would write this as t mod P / P where t and P are real numbers. I do not really understand your reason for using this numerical ratio because I believe I could just as well work in degrees or radians. The t/P has a range of 0 thru 1, or maybe 0 up to 1. Multiplied by 100 and it is a percentage of a period. But I do not know why this is important to the discussion.

If I am comparing two waves for their timing relationship to each other, then I need to work from the same time base. So when t=0 one wave might have a voltage of 0, but the other one might be 5 V as a result of their being a 30 degree phase difference between the two waves. Assumed here is a peak voltage of 10 V, and sin 30 = 0.5 .

If I want to see if two wave shapes are identical in form (shape, or whatever), then I may want different time base origins so that I can slide one waveform over the other to do a correlation of one to the other. If this is what you are doing, then I believe your approach would would classify all the waveforms in a three phase system as being in-phase.

.

 

[rattus]

Dodged the question yet again!

Dodged the question yet again!

I don't need a definition of "in-phase." Your's is fine, but it is what is irrelevant. Only the period is necessary to explain "Why is residential wiring known as single phase?"

I don’t have much time to spend today. I haven’t tried to follow the manufacturers’ branch of the discussion, so if what I say here has already been covered, I apologize.

Using gar’s “black box,” assume three terminals labeled left to right “E-F-G” that randomly originate from a conventional 120/240V residential system.

Assuming the measuring convention was “left to right,” it would take ten seconds or less to determine the neutral and which pair of terminals had 240V between them with a simple voltmeter. Again assuming the labeling convention is “left to right,” you would quickly identify “L1” and “L2.” “L1” would be to the left of “L2.”

Using an oscilloscope, and the same “left to right” measuring convention you would conclude “L1 to neutral” was “in-phase” with the 240V set and the “L2 to neutral” was “out of phase.”

If the original “E-F-G” labels remained left to right but the measuring and labeling conventions were reversed to “right to left” you would still conclude “L1 to neutral” was “in-phase” with the 240V set and the “L2 to neutral” was “out of phase.” However; now “L1” would now be to the right of “L2.”

That means for a conventional 120/240V residential system which “live to neutral” terminal is “in phase” or “out-of-phase” with the 240V set would be entirely based on how it is measured.

However, with respect to a conventional 120/240V residential system the definition of phase is not direction sensitive nor dependent on the means of measurement. It only requires that the period P and the effective or equivalent t₀ for a periodic voltage function to be validly determined. Trig identities are what enables us to determine the "effective or equivalent t₀ for a periodic voltage function."

If two periodic functions have the same period P and t/P is effectively or equivalently the same throughout the period, they have the same phase. In fact, for a conventional 120/240V residential system every relevant voltage function has an identical phase. So there is only one phase in the system; i.e., a single-phase. Which is why residential wiring is known as "single-phase."

No, the phase is wt + phi0. Then we are concerned with

(wt + phi0)/2PI

So, unless phi0 is identical for all waveforms, they cannot all have the same phase.

BTW, phase is sometimes described as the argument of the sinusoid which describes the wave, each wave carries a phase angle or phase constant. Certainly is true with 3-phase systems For example, the 3-phase wye carries 6 phase constants. Can't all be in phase or we couldn't draw a phasor diagram. wouln't work right either.

 

[gar]

120228-1340 EST

pfalcon:

LoL. They go in the same direction. It's your leads that go in opposite directions. As always. You keep forgetting to invert the magnitude when you invert the leads.

When one uses an instrument to obtain information it is necessary understand how the instrument operates and how to properly interpret that information.

So consider the oscilloscope. Normally in the English world left to right is increasing time and upward is more positive. There may be places where these are inverted. With single ended inputs and not inverting the channel the chassis of the scope is common and may be connected to the power system EGC. Relative to scope common a positive voltage on the Y-axis causes the beam to move upward on the screen.

I shall pick the voltage v_AN and its positive zero crossing to trigger the horizontal sweep so that a positive zero crossing of the v_AN signal occurs at the left end of the horizontal sweep. This means that scope common is connected to neutral (N), and the probe tip of scope channel A connects to phase A terminal. Also connected to phase A is the horizontal trigger input. None of these leads are going to be moved.

Scope channel B is also in non-inverting mode. Note channel B's common also goes to neutral because it is common to channel A.

Now put channel B's probe tip on the phase A wire. There are now two superimposed sine waves precisely overlapping each other. These two waves are in phase, because they are one and the same signal.

Next put channel B's probe tip on phase B's wire. Now channel B is an inverted image of channel A. What we expect. v_BN is 180 degrees out-of-phase with v_AN. It would be wrong to invert channel B to make its display look like that of channel A.

Back to my statement that you can parallel two voltages sources if they are the same voltage, frequency and are in-phase. If they are 180 degree out of phase then big sparks. Can you connect my above phase A and phase B and not get big sparks? No you can not.

If you look at "Basic Electrical Measurements", by Melville B. Stout, Professor of Electrical Engineering, University of Michigan, Prentice-Hall, Copyright 1950, Second Printing 1951 on pages 312 and 313 you will find a discussion of a phase-sensitive bridge detector. The circuit is on p 312 and shows a center tapped secondary. On page 313 is stated "..... The plates of this tube are supplied with voltage at bridge frequency but 180 degree apart because of the transformer connection. ...."

.

 

[rbalex]

All three phases of a three phase system have the same period.
By your definition they would all be in phase.
I suspect that more than a few of us here, including some from the single phase lobby, just might not concur.

Lunch break

I?ve been waiting for this. As it applies to the OP, by my application - not by my definition. I'm genuinely glad you noticed.

They certainly have the same period; but they don?t have the same phase.

For a single-phase system:

V sin([ωt+φ₀]) = -V sin([ωt+φ₀]?180?) or, choosing the inverse:
- V sin([ωt+φ₀]) = V sin([ωt+φ₀]?180?)
​

V sin([ωt+φ₀]) and - V sin([ωt+φ₀]) are the system?s basic general functions or their equivalents. They aren?t equal to each other, but they have an identical phase characteristic, [ωt+φ₀]; i.e., they have the same phase.

For the special case, φ₀=0:

V sin([ωt]) and - V sin([ωt]) are the basic system functions, with the identical phase characteristic, [ωt]

Similarly, for a three-phase system; through trig identities and without loss of generalization, assuming a common or equivalent t₀, the system?s basic functions can only be resolved to:

V sin([ωt+φ₀]+0),
V sin([ωt+φ₀]+120?) and
V sin([ωt+φ₀]+240?)

​

Or their inverse equivalents which can still be written in terms of the phase characteristics ([ωt+φ₀]+0), ([ωt+φ₀]+120?) and ([ωt+φ₀]+240?)

Phase is not dependent on magnitude or polarity; only the period P and a common t₀, which doesn?t even have to be zero, are required to establish the phase of a system.

 

[rbalex]

No, the phase is wt + phi0. Then we are concerned with

(wt + phi0)/2PI

So, unless phi0 is identical for all waveforms, they cannot all have the same phase.

BTW, phase is sometimes described as the argument of the sinusoid which describes the wave, each wave carries a phase angle or phase constant. Certainly is true with 3-phase systems For example, the 3-phase wye carries 6 phase constants. Can't all be in phase or we couldn't draw a phasor diagram. wouln't work right either.

See my last reply to Bes.

 

[__dan]

120228-1340 EST

pfalcon:

When one uses an instrument to obtain information it is necessary understand how the instrument operates and how to properly interpret that information.

So consider the oscilloscope. Normally in the English world left to right is increasing time and upward is more positive. There may be places where these are inverted. With single ended inputs and not inverting the channel the chassis of the scope is common and may be connected to the power system EGC. Relative to scope common a positive voltage on the Y-axis causes the beam to move upward on the screen.

I shall pick the voltage v_AN and its positive zero crossing to trigger the horizontal sweep so that a positive zero crossing of the v_AN signal occurs at the left end of the horizontal sweep. This means that scope common is connected to neutral (N), and the probe tip of scope channel A connects to phase A terminal. Also connected to phase A is the horizontal trigger input. None of these leads are going to be moved.

Scope channel B is also in non-inverting mode. Note channel B's common also goes to neutral because it is common to channel A.

Now put channel B's probe tip on the phase A wire. There are now two superimposed sine waves precisely overlapping each other. These two waves are in phase, because they are one and the same signal.

Next put channel B's probe tip on phase B's wire. Now channel B is an inverted image of channel A. What we expect. v_BN is 180 degrees out-of-phase with v_AN. It would be wrong to invert channel B to make its display look like that of channel A.

Back to my statement that you can parallel two voltages sources if they are the same voltage, frequency and are in-phase. If they are 180 degree out of phase then big sparks. Can you connect my above phase A and phase B and not get big sparks? No you can not.

If you look at "Basic Electrical Measurements", by Melville B. Stout, Professor of Electrical Engineering, University of Michigan, Prentice-Hall, Copyright 1950, Second Printing 1951 on pages 312 and 313 you will find a discussion of a phase-sensitive bridge detector. The circuit is on p 312 and shows a center tapped secondary. On page 313 is stated "..... The plates of this tube are supplied with voltage at bridge frequency but 180 degree apart because of the transformer connection. ...."

.

Hi gar

Your scope display shows you two waveforms that are A, 120sin(wt+0), and B, 120sin(wt+180). On that, I believe we agree.

Now if you vector sum A + B, 120sin(wt) + 120sin(wt+180), the sum = 0 volts because the waveforms cancel each other out. You would have to carry out the vector subtraction (-) to get the difference and 240 volts.

However, the transformer under test also offers you this vector sum for measuring, the two windings in series are an underlying physical reality and the subject of the repeated inquiries. By observation, placing the probes A to B shows you a 240 volt sinewave, 240(sin(wt+0) volts, not zero volts.

This link from post 1468 shows this. Dial the second sinewave for 180 deg phase difference and the sum = 0.

http://www.acoustics.salford.ac.uk/feschools/waves/super.htm#phase

The first test yields the premise or conclusion that, 120sin(wt) + 120sin(wt+180) = 0. Testing this we find 120sin + 120sin = 240sin, proving the first premise or conclusion defective. Logic specifies, when arriving at a defective conclusion, it is time to check the premises for defects.

The transfomer windings are in fact in phase and sum to 240 volt in series. However, one of the windings can be connected to the load in the reversed direction relative to the other winding's turn direction. The reversed connection to the transformer has the effect of multiplying the vector by (-1) and rotating it 180 deg. The phase reversal on the scope is an artifact of the measuring protocol and not a result of the underlying physical reality.

 

[rbalex]

120228-1154 EST

rbalex:

This is in response to your post 1516.

...
.

Thanks. The reason I specify t/P is it is used in the IEEE Std. 100 definition of phase and is compatible with the Kerchner and Corcoran, Alternating-Current Circuits, definition rattus offered.

Off to my meetings again.

 

[rattus]

Hi gar

Your scope display shows you two waveforms that are A, 120sin(wt+0), and B, 120sin(wt+180). On that, I believe we agree.

Now if you vector sum A + B, 120sin(wt) + 120sin(wt+180), the sum = 0 volts because the waveforms cancel each other out. You would have to carry out the vector subtraction (-) to get the difference and 240 volts.

However, the transformer under test also offers you this vector sum for measuring, the two windings in series are an underlying physical reality and the subject of the repeated inquiries. By observation, placing the probes A to B shows you a 240 volt sinewave, 240(sin(wt+0) volts, not zero volts.

This link from post 1468 shows this. Dial the second sinewave for 180 deg phase difference and the sum = 0.

http://www.acoustics.salford.ac.uk/feschools/waves/super.htm#phase

The first test yields the premise or conclusion that, 120sin(wt) + 120sin(wt+180) = 0. Testing this we find 120sin + 120sin = 240sin, proving the first premise or conclusion defective. Logic specifies, when arriving at a defective conclusion, it is time to check the premises for defects.

The transfomer windings are in fact in phase and sum to 240 volt in series. However, one of the windings can be connected to the load in the reversed direction relative to the other winding's turn direction. The reversed connection to the transformer has the effect of multiplying the vector by (-1) and rotating it 180 deg. The phase reversal on the scope is an artifact of the measuring protocol and not a result of the underlying physical reality.

No dan, if the phasor arrows connect tail to tail, you subtract to get the potential DIFFERENCE as in a wye. If the phasor arrows connect head to tail,you add to get the phasor sum.

 

[Besoeker]

Lunch break

I?ve been waiting for this. As it applies to the OP, by my application - not by my definition. I'm genuinely glad you noticed.

They certainly have the same period; but they don?t have the same phase.

For a single-phase system:
V sin([ωt+φ_{0[/SUB) = -V sin(ωt+φ0?180?)}

<sub>You seem to have a propensity to attempt to obscure the simple by obfuscation.
Of course Vmsin ωt = -Vmsin (ωt+π)
It's a very simple trigonometrical relationship.
Sin π/2 =1
Sin 3 π/2 =-1
Thus Sin π/2 = -Sin 3 π/2
That's fine for a two wire system

But not terribly illuminating in the context of a three wire system where Van = Vm Sin(ωt) and Vbn = Vm Sin(ωt+π)
NOT - Vmsin (ωt+π)</sub>

 

[__dan]

No dan, if the phasor arrows connect tail to tail, you subtract to get the potential DIFFERENCE as in a wye. If the phasor arrows connect head to tail,you add to get the phasor sum.

Yes, you are making my point for me.

If you multiply one of the vectors on the scope by (-1), then it will sum to the correct answer (V1 + V2 = 240 volt), not zero volts. You have been emphatic that the voltages are displaced by 180 deg, which sums to zero. Measuring the actual, the voltages sum to 240.

 

[Labrat]

You seem to have a propensity to attempt to obscure the simple by obfuscation.
Of course V_msin ωt = -V_msin (ωt+π)
It's a very simple trigonometrical relationship.
Sin π/2 =1
Sin 3 π/2 =-1
Thus Sin π/2 = -Sin 3 π/2
That's fine for a two wire system

But not terribly illuminating in the context of a three wire system where V_an = V_m Sin(ωt) and V_bn = V_m Sin(ωt+π)
NOT - V_msin (ωt+π)

It works for three wire systems too unless you demand that polarity is a critical element of phase. If you do demand it, it doesn’t show up in any definition offered so far:
From IEEE Std 100 The IEEE Standard Dictionary of Electrical and Electronic Terms:

phase (of a periodic phenomenon ƒ(t), for a particular value of t) The fractional part t/P of the period P through which ƒ has advanced relative to an arbitrary origin.

Note: The origin is usually taken at the last previous passage through zero from the negative to the positive direction. See also: control system, feedback; simple sine-wave quantity. (IM) [120]

Phase: Phase is the fractional part of a period through which time or the associated time angle wt has advanced from an arbitrary reference...........

[Kerchner and Corcoran, Alternating-Current Circuits, Wiley, 1951]

 

[Besoeker]

It works for three wire systems too unless you demand that polarity is a critical element of phase. If you do demand it, it doesn’t show up in any definition offered so far:
From IEEE Std 100 The IEEE Standard Dictionary of Electrical and Electronic Terms:

Got it in one!
Thank you.

phase (of a periodic phenomenon ƒ(t), for a particular value of t) The fractional part t/P of the period P through which ƒ has advanced relative to an arbitrary origin.

The fractional part, in this case, is Pi.

 

[rbalex]

Got it in one!
Thank you.


The fractional part, in this case, is Pi.

I confess I?ve been tiptoeing around the fact that you haven?t actually accepted or offered an alternate definition of phase. But I guess you have now tacitly accepted the IEEE definition.

I believe most people reading the definition for comprehension , rather than defending the Myth, would recognize the period in radians for any relevant voltage function in a single-phase residential system is 2π and t/P isn?t a fixed number. You don't get to pick it arbitrarily, like you do t_0.

 

[Besoeker]

I confess I?ve been tiptoeing around the fact that you haven?t actually accepted or offered an alternate definition of phase. But I guess you have now tacitly accepted the IEEE definition.

Nothing tacit about it.
It matches exactly what I have been saying.

phase (of a periodic phenomenon ?(t), for a particular value of t) The fractional part t/P of the period P through which ? has advanced relative to an arbitrary origin.

I_a and I_b are a fraction of the period apart.

 

[rbalex]

Nothing tacit about it.
It matches exactly what I have been saying.

are a fraction of the period apart.

Do you believe the periods for the driving voltages aren't 2π? Because if they aren't Mr. Ohm needs a serious talking to.