Why is residential wiring known as single phase?

[rattus]

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.

dan, we are talking about phasor representation of voltages. No circuitry is involved. You need to read up on phasors.

 

[rattus]

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.

There is no neutral in a pure delta, let's concentrate on the wye.

In a wye, how do you get 208V between phases?

Here is what I get.

Va + Vb = 62V @ 60 degrees

Va - Vb = 208V @ 30 degrees

Which is correct?

 

[__dan]

dan, we are talking about phasor representation of voltages. No circuitry is involved. You need to read up on phasors.

So it's a ploy to further confuse the subject description. There are two indentical windings in the circuit, not much else.

 

[rattus]

So it's a ploy to further confuse the subject description. There are two indentical windings in the circuit, not much else.

dan, we are talking about voltages represented as phasors. There is no mention of a transformer. Just voltages. Read up on phasors.

 

[rattus]

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.

Jim, my phasors are connected tail to tail, my diagram indicates that. When you reverse one of the phasors, that is tantamount to subtraction.

The arrows in a wye are connected tail to tail, why not in a split phase system?

The reason we connect the arrows tail to tail is we want to know the voltage and phase at L1 and L2, not on the neutral, and the neutral is a good place to do that.

 

[__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))
Arrow points right

The complex term jsin(0) will produce a phase shift when you substitute wt in for 0.

V2n = -120Vrms(cos(0) + jsin(0))
Arrow tries to point right, but the negative sign won't let it.

The negative sign is added by you by reversing the polarity of the leads relative to the winding turn direction. If you maintain consistency of measuring method with winding turn direction, there is no multiplication by (-1).

 

[rattus]

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

The complex term jsin(0) will produce a phase shift when you substitute wt in for 0.

V2n = -120Vrms(cos(0) + jsin(0))
Arrow tries to point right, but the negative sign won't let it.

The negative sign is added by you by reversing the polarity of the leads relative to the winding turn direction. If you maintain consistency of measuring method with winding turn direction, there is no multiplication by (-1).

dan, you are missing the point--entirely!

 

[rbalex]

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.



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?

In that case V1n = Vm*sin(wt) = V2n and you're only describing a 120V system, not a 120/240V system. Yes they are "in phase" and have the same phase - SO?

If you are describing a 120/240V system then:

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

They have the same phase, but are not "in phase"

 

[rbalex]

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.

No - actually you didn't. You fought their application from the beginning. You had absolutely no idea how they were applied properly - and from what I can tell still don't.

 

[rattus]

No - actually you didn't. You fought their application from the beginning. You had absolutely no idea how they were applied properly - and from what I can tell still don't.

Are you a mind reader as well? All these insults indicate to me that you are feeling the heat, that you know you are wrong and refuse to admit it. Very unprofessional.

 

[rattus]

In that case V1n = Vm*sin(wt) = V2n and you're only describing a 120V system, not a 120/240V system. Yes they are "in phase" and have the same phase - SO?

If you are describing a 120/240V system then:

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

They have the same phase, but are not "in phase"

That is my point, you would not use a split phase transformer in that way. When you say V1 and V2 have the same phase, you are saying,

V2n = Vm*sin(wt).

Simple put, the phase should be taken from the POSITIVE sine function, not the NEGATIVE. That way you do not discard the phase constant,

V2n = Vm*sin(wt + PI)

So which 'phase' do you use?

How do you decide? Where did you learn this trick? Is it in a book of magic somewhere, or is it just in your head?

 

[rattus]

Case Closed:

Case Closed:

Two functions exhibit the same phase if and only if their start points coincide. That is (wt + phi1) = (wt + phi2), that is phi1 = phi2.

For sines, the start point is the negative to positive transition.

But for the split phase system, V1n and V2n start PI radians apart.

Therefore the two voltages exhibit two phases.

Case closed.

 

[T.M.Haja Sahib]

But for the split phase system, V1n and V2n start PI radians apart.

Therefore the two voltages exhibit two phases.

Case closed.

But for the split phase system, V1n and Vn2 start 0 radians apart.

Therefore the two voltages exhibit in phases.

Case re-opened.

 

[pfalcon]

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.

In other words, YOU must justify not using mathematics. Rbalex nor I have to justify it.

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!

Incorrect equations.
V_1n = V_m*sin(wt)
V_n2 = V_m*sin(wt)
therefore
V_2n = -V_m*sin(wt+180)

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.

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

-120Vrms @ 0 ------------>0---------->120Vrms @ 0

 

[david luchini]

Incorrect equations.
V_1n = V_m*sin(wt)
V_n2 = V_m*sin(wt)
therefore
V_2n = -V_m*sin(wt+180)

I'm afraid you're math is wrong. If V_1n = V_m*sin(wt) and V_n2 = V_m*sin(wt), then V_2n = +V_m*sin(wt+180)

That made your vectors wrong as well. Using vector addition on the following will result in zero volts, not 240 volts.

-120Vrms @ 0 ------------>0---------->120Vrms @ 0

-120@0 + 120@0 = 0

 

[rattus]

In other words, YOU must justify not using mathematics. Rbalex nor I have to justify it.


Incorrect equations.
V_1n = V_m*sin(wt)
V_n2 = V_m*sin(wt)
therefore
V_2n = -V_m*sin(wt+180)



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

-120Vrms @ 0 ------------>0---------->120Vrms @ 0

We are discussing V1n and V2n which are clearly out of phase. Even rbalex admits that. My phasor diagram is correct as I drew it. No editing required or requested.

 

[rattus]

But for the split phase system, V1n and Vn2 start 0 radians apart.

Therefore the two voltages exhibit in phases.

Case re-opened.

The voltages in question are V1n and V2n.

 

[rbalex]

That is my point, you would not use a split phase transformer in that way. When you say V1 and V2 have the same phase, you are saying,

V2n = Vm*sin(wt).

Simple put, the phase should be taken from the POSITIVE sine function, not the NEGATIVE. That way you do not discard the phase constant,

V2n = Vm*sin(wt + PI)

So which 'phase' do you use?

How do you decide? Where did you learn this trick? Is it in a book of magic somewhere, or is it just in your head?

Flip a coin. As arguments for the Sine function, either wt OR wt + PI may be used - but not both.

 

[jim dungar]

There is no neutral in a pure delta, let's concentrate on the wye.

In a wye, how do you get 208V between phases?

Maybe you need to read up on phasors just like your advice to anyone who disagrees with you.

In a tail-to-point configuration we ADD the phasors together.
In a tail-to-tail or a point-point configuration we SUBTRACT them.

Given two identical transformer windings;
V12=V34:two phasors pointed in the same direction, this is the actual physical connection of the transformer
V1n+Vn4= V14: two phasors connected tail-to-point, when terminals 2&3 are connected to form a neutral
Vn4=-V4n: two phasors, where one equals the opposite of a phasor that has been rotated. NOTE: two opposing actions occur in this equation.
V14=V1n+(-V4n)=V1n-V4n: two phasors connected same end-to-same end, this is a mathematical model of the connection of the transformer (we know if the one winding was actually physically rotated the result would be V12-V43=V13=0).

 

[stockexpert]

no time!

no time!

Would love to get into this discussion..just no enough time!