single vs. 3 phase

[crossman]

Well, I think to use the back to back arrows to depict 1ph 120/240, in an educational setting, is needlessly limiting.

I can agree with that. But I'm thinking we would be teaching more with currents than votlages when showing what is going on with the system.

currents that reverse as they pass the magical "N" point to be, well, silly. Can you do it? Sure.

Did I miss something here? Currents that do silly reversals past the neutral point? I don't think they do that! They simply flow, inside the source, from a more positive point to a less positive point.

I find this BTBA model to fall apart when one extends to 3ph D.

But what about a wye? That really bothers me. All the arrows point out from the neutral? Can't be for the currents. That is some crazy! But what else are ya gonna do?

What about 2 sources that were 145 degrees out? What about two sources that were 170 degrees out? What about 179 degrees out? Then all of a sudden 180 out of phase ceases to exist?

Yesterday I would have called that concept wrong. Today; it's silly, limiting, and it bothers the secret engineering moral code.

Engineers moral code?? I never knew of such a thing! Oh... it's secret. Honestly, I would suspect that some physics pHd guy would whole-heartedly embrace the 180 out of phase idea and just love it.

 

[crossman]

A flat line.

Hey Mivey, does your scope have isolated channels with no common lead?

 

[mivey]

Hey Mivey, does your scope have isolated channels with no common lead?

Common lead tied to the chassis.

 

[Don Randall]

A flat line.

Thanks, Don

 

[crossman]

Common lead tied to the chassis.

So can you actually provide a dual channel graph of the voltages showing both of them in-phase?

 

[mivey]

So can you actually provide a dual channel graph of the voltages showing both of them in-phase?

Yes. Commons to L1, probe 1 to n, probe 2 to L2. This gives me 120 & 240 in phase.

I can't do common 1 to L1, probe 1 to n, common 2 to N, probe 2 to L2 to get two 120 volt signals in phase. For that, I can connect common to N, probe 1 to L1, probe 2 to L2, then invert channel 2.

 

[rattus]

A flat line.

sin(x) +(-sin(x)) = 0

sin(x) -(-sin(x)) = 2sin(x)

 

[crossman]

Yes. Commons to L1, probe 1 to n, probe 2 to L2. This gives me 120 & 240 in phase.

And the "add" function gives 360 volts?

I can't do common 1 to L1, probe 1 to n, common 2 to N, probe 2 to L2 to get two 120 volt signals in phase. For that, I can connect common to N, probe 1 to L1, probe 2 to L2, then invert channel 2.

That is what I thought.

So according to your scope, the two sources must be out of phase by 180 degrees to read them properly? To make them in-phase, you have to invert one of the channels? Sounds like a trick to make two 180 outies equal an innie!

For Don, what I am saying is that the "add" function doesn't necessarily give us anything proper.

I thought this thread had about petered out, but I've been enjoying it again.:smile:

 

[rattus]

Big typo, sorry!

 

[Don Randall]

What do you think?

What do you think?

View attachment 1624
In this illustration, L1 and L2 of opposite polarities, are in phase with each other, because the ac pulses are in time, and with the lines of opposite polarities they operate in concert to move current pulses in the same direction when connected to a load. That is, in both lines, the current will move to the left simultaneously on one pulse and to the right on the next. You know the indicated polarities are merely at one instant in time, (but somebody reading this may not know). Don

 

[coulter]

View attachment 1624
In this illustration, L1 and L2 of opposite polarities, are in phase with each other, ...

My translation is, "I obviously don't understand." The terms, "L1 and L2 of opposite polarities" and "are in phase with each other" don't make any sense at all.

To my thinking:
1. One point doesn't have a voltage. A voltage requires measuring between two points
2. Two points don't have opposite polarities. They just have a voltage between them. By convention we have a method to assign a gradient direction.
3. A phase angle measurement requires a reference source to measure the displacement against. To say "L1 and L2 are in phase" doesn't have a meaning

Educate me a little on what I am missing.

carl

 

[mivey]

Yes. Commons to L1, probe 1 to n, probe 2 to L2. This gives me 120 & 240 in phase.

I can't do common 1 to L1, probe 1 to n, common 2 to N, probe 2 to L2 to get two 120 volt signals in phase. For that, I can connect common to N, probe 1 to L1, probe 2 to L2, then invert channel 2.

And the "add" function gives 360 volts?

Correct

So according to your scope, the two sources must be out of phase by 180 degrees to read them properly? To make them in-phase, you have to invert one of the channels?

Also correct if you are saying they are both 120 volts and you want to display them at the same time.

 

[rattus]

View attachment 1624
In this illustration, L1 and L2 of opposite polarities, are in phase with each other, because the ac pulses are in time, and with the lines of opposite polarities they operate in concert to move current pulses in the same direction when connected to a load. That is, in both lines, the current will move to the left simultaneously on one pulse and to the right on the next. You know the indicated polarities are merely at one instant in time, (but somebody reading this may not know). Don

First, assuming L1 and L2 to be the hots in a split-phase service, we must assume a neutral although it is not shown in the diagram.

Second, we should say the voltages--not L1 and L2--are inverses of each other, and that means they are 180 degrees out of phase--not in phase. If they are in phase, there would be zero current.

Third: Assuming V1 and V2 to be 120Vrms, the algebraic difference in V1 and V2 is 240Vrms; then the current is,

Irms = 240Vrrs/R

 

[coulter]

...Irms = 240Vrrs/R

Interesting. I would have thought Irms = 0

carl

 

[coulter]

...Engineers moral code?? ... Oh... it's secret. ...

Shhhh ... And as you surmised, that's why you have never heard of it:grin:

carl

 

[rattus]

Huh?

Huh?

Interesting. I would have thought Irms = 0

carl

Why? Did I make another typo?

 

[quogueelectric]

This post calls for a convention to settle it with multi trace oscilloscopes and arm wrestling tables. Pole pigs and lots of so cord Vegas sounds good to me just need some lead time to convince my wife it is legit.

 

[Don Randall]

Actually, half the time one conductor is considered positive and the other half of the time it is negative. At a single point in time, one conductor would be positive while the other is negative, precisely the same as DC voltage. 120 pulses per second. You actually have DC voltage for 120th of a second, it increases from 0v to max., then decreases back to 0v. Then you have another DC pulse occur for the next 120th of a second, but in the opposite direction. So surely, you can select a point in time and say that one conductor is positive while the other is negative. Is this incorrect? If so please explain as I would like to know. Don

Carl, Above is a previous post you could comment on also, if you like. I was a little sparse on details in the post with the drawing, I meant L1 and L2 to be an ac voltage source. Any line source, I don't care, call it a wall socket if you like. Since L1 and L2 are an ac voltage source, there is a negative charge on one, (excess of electrons), and a positive charge on the other, (deficit of electrons), at any instant except 0?. I'm a little surprised that you don't think there are opposing polarities. Unless, I'm mistaken, (you will correct me if I am), every point in the circuit would be in phase with every other point, with respect to the timing and direction of the pulses. I was getting off topic, but just wondering what you guys think, (Note the heading of the post. I am aware that a voltage is always measured between two conductors.) Don

 

[crossman]

Okay Don, let's go with your example of L1 and L2 being opposite polarities, and assume that we both realize that there must be two points to measure voltages.

If L1 and L2 are opposite polarity, that means that if one of them is positive, the other one is negative. This infers that when one of them is at a positive peak, the other is at a negative peak. Now, if we graph the 2 sine waves based on the condition of L1 at positive peak while L2 is simultaneously at negative peak, then we will see 2 sine waves which are 180 degrees out of phase.

 

[crossman]

Originally Posted by Don Randall
Actually, half the time one conductor is considered positive and the other half of the time it is negative. At a single point in time, one conductor would be positive while the other is negative, precisely the same as DC voltage. 120 pulses per second. You actually have DC voltage for 120th of a second, it increases from 0v to max., then decreases back to 0v. Then you have another DC pulse occur for the next 120th of a second, but in the opposite direction. So surely, you can select a point in time and say that one conductor is positive while the other is negative. Is this incorrect? If so please explain as I would like to know. Don

Don, on all of the above, I am in 100% agreement with you.