Why is residential wiring known as single phase?

[mivey]

...Amplitude and polarity are just irrelevant to defining phase
...
even though the examples are, in fact, perfectly consistent with my definition of phase

Pure spin. From my "examples":

the sine waves representing them are zero at the same time and positive and negative together
...
both waves must be written with positive amplitudes
...
If the phase difference δ is Π or an interger times Π, then the systems are said to be 180? out of phase.

Keep dreaming.

 

[T.M.Haja Sahib]

Pure spin. From my "examples":

Keep dreaming.

You are correct.But what are you trying to prove?Please note that in science any thing that can not be measured does not exist.

 

[mivey]

You are correct.But what are you trying to prove?

That a 180? phase displacement does not mean "in phase".

The waveforms are symmetrical. This symmetrical characteristic is typical of systems that have multiple phases being labeled with a lower phase order number (the single-phase label even though we can have two phases, the two-phase label even though we can have four phases, etc).

Please note that in science any thing that can not be measured does not exist.

Then I guess it is a good thing we have only been discussing what we are able to measure.

 

[T.M.Haja Sahib]

That a 180? phase displacement does not mean "in phase".

The waveforms are symmetrical. This symmetrical characteristic is typical of systems that have multiple phases being labeled with a lower phase order number (the single-phase label even though we can have two phases, the two-phase label even though we can have four phases, etc).

Then I guess it is a good thing we have only been discussing what we are able to measure.

Brilliant.:thumbsup:
Have you measured any load voltages in any home with 180 degree phase displacement?

 

[mivey]

Have you measured any load voltages in any home with 180 degree phase displacement?

Countless times. In homes, businesses, and in utility circuits.

I don't think anyone here disputes that we can measure these. Are you raising that dispute?

 

[T.M.Haja Sahib]

Countless times. In homes, businesses, and in utility circuits.

If there were 180 degree phase displacement,the currents in the hot conductors should add in the neutral for resistive loads.But what is measured is the difference of two hot conductors currents in the neutral.So 180 degree phase displacement does not exist here.

 

[T.M.Haja Sahib]

Further to post #946:
If V1n is acting at one moment towards neutral in one half of 120v/240v supply and if there were 180 degree phase displacement for the voltage in the other half, this voltage in the other half would also be acting towards the neutral at that moment so that the current in the neutral would be sum of the two hot conductors currents for resistive loads.But this is never measured and hence does not exist.

 

[mivey]

If there were 180 degree phase displacement,the currents in the hot conductors should add in the neutral for resistive loads.But what is measured is the difference of two hot conductors currents in the neutral.So 180 degree phase displacement does not exist here.

The currents do add. Do you understand what it means for one wave to peak above the axis at 0? and for a different wave to peak above the axis 180? later?

Draw it out and maybe you will understand.

 

[T.M.Haja Sahib]

The currents do add.

In the neutral?

 

[mivey]

Further to post #946:
If V1n is acting at one moment towards neutral in one half of 120v/240v supply and if there were 180 degree phase displacement for the voltage in the other half, this voltage in the other half would also be acting towards the neutral at that moment

Then you do not understand what a 180? phase shift means because that is not correct. I'll ask you again to just work it out with a calculator. Take the equations and write down the voltages for Van and Vbn for every 30? or something so you can get a clearer picture.

You will find that the positives of one wave are in synch with the negative of the other wave.

 

[mivey]

In the neutral?

What do you think? Make the calculations I asked and then if it is still not clear I will help you with the calcs.

 

[T.M.Haja Sahib]

Then you do not understand what a 180? phase shift means because that is not correct. I'll ask you again to just work it out with a calculator. Take the equations and write down the voltages for Van and Vbn for every 30? or something so you can get a clearer picture.

You will find that the positives of one wave are in synch with the negative of the other wave.

It would be very nice if you do it yourself for yourself......

 

[T.M.Haja Sahib]

Make the calculations

Please see post#952.

 

[mivey]

It would be very nice if you do it yourself for yourself......

Please see post#952.

I already have in many prior posts. Do your own lab work and learn.

 

[T.M.Haja Sahib]

I already have in many prior posts. Do your own lab work and learn.

Please mivey produce your calculation in a nutshell here.
One more point I want to bring to your notice is the following. Let ,according to you, V1n=V*sin(wt) is the voltage across one half and V2n=V*sin(wt+180) is the voltage across the other half of 120/240v supply.In any case,their sum of instantaneous values should equal 240*sin(wt).But
V1n+V2n=V*sin(wt)+V*sin(wt+180)=V*sin(wt)+[-V*sin(wt)]=V*sin(wt)-V*sin(wt)=0 at any instant.A contradiction.So,the voltages across each of the two halves should be V*sin(wt) so that their sum is 2*V*sin(wt)=2*120*sin(wt)=240*sin(wt) at any instant.

 

[david luchini]

One more point I want to bring to your notice is the following. Let ,according to you, V1n=V*sin(wt) is the voltage across one half and V2n=V*sin(wt+180) is the voltage across the other half of 120/240v supply.In any case,their sum of instantaneous values should equal 240*sin(wt).

No, this is incorrect. The DIFFERENCE, not the SUM of V1n and V2n should equal 240. V12 = V1n - V2n. It is really that simple.

 

[T.M.Haja Sahib]

No, this is incorrect. The DIFFERENCE, not the SUM of V1n and V2n should equal 240. V12 = V1n - V2n. It is really that simple.

What is the basis for your taking the difference and not the sum of V1n and V2n?

 

[jim dungar]

No dispute about that. Normally, X2 gets connected to X3 giving the equivalent of a centre-tapped arrangement. To hots, say V1 and V2, and one neutral N.
Neutral in this arrangement is the common. Resulting in V_1n and V_2n being mutually displaced by 180 degrees. They cross zero at the same time. They are in anti-phase. Not in phase. If they are not in phase it is difficult to construe that as single phase. It's pretty much the same logic that gets a hexaphase circuit from three phase.

Nope you changed the discussion. I agree V2n=-Vn2

But that is not what I asked about, I really and truly do not care how something appears, based on some absolutely irrelevant change in an arbitrary reference point.

I want a simple answer based on the actual construction of a specific transformer.
If the windings can be paralleled are they 'in-phase'? YES or NO

 

[Besoeker]

Why can't you make an attempt in interpreting the voltage values under the heads AC and DC and share your marvelous knowledge on it with us?

What's to interpret?
120Vdc is usually taken to mean mean.
120Vac is usually taken to mean root mean square.

Just common knowledge. At least for those in the electrical field.

 

[david luchini]

What is the basis for your taking the difference and not the sum of V1n and V2n?

It is simple. Here's two methods to visualize this.

For the first method, we have source voltages of V1n and V2n and connect a load between 1 and 2 and reference the load voltage as being Vload=V12. Using KVL around the loop created by the circuit we would see that Vload+V2n-V1n=0, or V12+V2n-V1n=0. This rearranges to V12=V1n-V2n.

The second method uses vector math. At your source draw a vector from 1 to N and then draw a vector from N to 2. You can see that these vectors are in the same direction. Using vector math, you would add these to vectors together to get the vector from 1 to 2. So V12 = V1n + Vn2.

Now draw vectors from 1 to N and from 2 to N. You can see that these vectors are in opposite directions. Using vector math, you would subtract these vectors to get the vector from 1 to 2. So V12 = V1n - V2n.