Why is residential wiring known as single phase?
- Thread starter TimWA
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K8MHZ
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- Electrician
Pythagoras strikes again.........
Yeah, well.....if Isosceles was a person he would strike right back.
Assuming that he had the right angle to do so.........Yeah, well.....if Isosceles was a person he would strike right back.
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I haven't said anything is "out of phase." This is just basic math. I don't see what's confusing about this. If Van=120<0 and Vbn=120<180 and Ia=10<0 and Ib=10<180, then:
Vab=Van-Vbn = 120<0 - 120<180 = 240<0
In=Ia+Ib = 10<0 + 10<180 = 0
OK, if Van=120<0 and Vbn=120<180, then Vab= Van-Vbn = 120<0 - 120<180 = 240<0. Likewise, Vba= Vbn-Van = 120<180 -120<0 = 240<180.
In your first example you had 1A load from A-N, a 30A load from A-B and a 10A load from B-N. Assuming these are all resistive loads, the first load is 120 ohms, the second load is 8 ohms, and the third load is 12 ohms. I'll call the load currents IL1, IL2, and IL3 respectively.
The first load is 120 ohms connected from A to N, so the load current is IL1=Van/120 ohms = 120<0 volts/120 ohms = 1<0 amps.
The second load is 8 ohms connected from A to N, so the load current is IL2=Vab/8 ohms = 240<0 volts/8 ohms = 30<0 amps.
The third load is 12 ohms connect from B-N, so the load current is IL3=Vbn/12 ohms = 120<180 volts/12 ohms = 10<180 amps.
Using KCL, Ia=IL1+IL2 = 1<0 + 30<0 = 31<0 amps. Ib=IL3-IL2 = 10<180 - 30<0 = 40<180 amps. In=IL1+IL2 = 1<0 + 10<180 = 9<180 amps. The load voltages in this case are the same as the Line-Neutral and Line-Line voltages, so VL1=120<0, VL2=240<0 and VL3=120<180.
Using KVL to check the voltages around the closed loop formed by the 3 resistive loads, we have VL1-VL3-VL2=0 = 120<0 - 120<180 - 240<0 = 0.
Now consider removing the neutral connection. In will will course be zero amps. The load resistors 1 and 3 will now be connected in series across A and B. Using the same current conventions from the first example, IL1 and IL3 will have the same magnitude but will have current angles that are 180 apart. The series connected load resistance will be 132 ohms, so IL1=Vab/132 ohms = 240<0 / 132 ohms = 1.818<0 Amps. IL3=Vba/132 ohms = 240<180 / 132 ohms = 1.818<180 Amps.
The load current (or voltage) for load two has not changed, it is still 240<0 / 8 ohms = 30<0 amps. The load voltages are VL1=1.818<0 * 120 ohms = 218.2<0 volts, VL2=30<0 * 8 ohms = 240<0 volts, and VL3=1.818<180 * 12 ohms = 21.8<180 volts.
The line current are Ia=IL1+IL2= 1.818<0 + 30<0 = 31.818<0 Amps, and Ib=IL3-IL2= 1.818<180 - 30<0 = 31.818<180
Using KVL to check the voltages around the closed loop formed by the 3 resistive loads, we have VL1-VL3-VL2=0 = 218.2<0 - 21.8<180 - 240<0 = 0.
No, none of the math changes. In both cases KCL is used to find the Line current for A and B (Ia=IL1+IL2 & Ib=IL3-IL2) and KVL is used to confirm the loop voltages sum to zero. (VL1-VL3-VL2=0).
K8MHZ
Senior Member
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- Electrician
Assuming that he had the right angle to do so.........
:slaphead:
I should have learned from ham radio to never attempt to beat a Brit at dry humor.
Good one!
Maybe the difference is that we know how to conduct ourselves..............
Joethemechanic
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Maybe the difference is that we know how to conduct ourselves..............
I've always suspected that your rounds are too square :roll:
I think the important point here is not whether you were right or wrong.
Just that you accept that you can learn and are prepared to do so. And that can't be bad.
I've been advocating that for a while. Maybe you have taken a step on that journey.
A journey of 1000 miles starts with a single step.
I have no doubt that the single phase debate will continue here or maybe in other threads. There are some pretty entrenched views expressed that I don't expect any amount of reasoning and constructive discourse will change. That's life.
At least you had the good grace this time to accept that you were in error.
Cathartic, possibly.
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Well, your round next then.
A Lagavulin, if you please.
Make it large.
Or maybe a beer from the centre tap...
Back on topic lest it come to the attention of the mods that we have strayed....
The 120-0-120 arrangement gives equal and opposite voltages. That's where the 240V comes from. And thus the 180deg displaced 120 voltages.
And the cancellation of the anti-phase currents in the neutral.
I don't suppose it matters much what you label it. That doesn't change the practicalities.
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- Wisconsin
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- PE (Retired) - Power Systems
Still waiting to see the comparison between my 2 load and 3 load examples. In particular I am interested in the relationship between the currents and the two sources.Your request is my....etc...
Currents:
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- Wisconsin
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- PE (Retired) - Power Systems
No, my first example was only two loads, one connected L1-N and the other connected L1-L2. I asked for the relationship between the voltages and the currents.
Then I asked for the same information except with 3 loads.
Following your notation at the N point of the sources, a KCL (which effectively says current leaving a node must equal the current leaving a node) equation would be Ia=In+Ib.
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Weisstein, Eric W. "Phase." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/Phase.html:
It is a mathematical concept.
Every proper voltage function for 120/240V (no matter how you measure it) is of the form V = C sin (wt+Φ0) where C is a constant and the “single phase” is wt+Φ0
The real question is not why residential voltages are called “single phase” but why Y-connected secondaries of three winding transformers aren’t described as “six-phase”
Edit add: for some reason only the last link works; but you can get to the other llinks from there.
The problem is phase is not an electrical concept; it has nothing to do with Volts, Amps, polarity, amplitude, Ohm’s Law, Kirchhoff’s Laws, Ampere’s Law, Lenz’s Law, Coulomb’s Law, or any other “law” derivable from Maxwell’s equations.
It is a mathematical concept.
Every proper voltage function for 120/240V (no matter how you measure it) is of the form V = C sin (wt+Φ0) where C is a constant and the “single phase” is wt+Φ0
The real question is not why residential voltages are called “single phase” but why Y-connected secondaries of three winding transformers aren’t described as “six-phase”
Edit add: for some reason only the last link works; but you can get to the other llinks from there.
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T
T.M.Haja Sahib
Guest
Besoeker,please.Do not think your posts are all right Your posts sometimes contradict themselves.See the 'more usual arrangement' circuit diagram in your post #64.You indicated voltages as 120-0-120v.It would mean lines L1 and L2 are in phase and so the current in neutral would double.So it is wrong to designate that way.One correct voltage representation is 120-0-(-120v).With this representation,voltages from L1 to N and from N to L2 are in phase.
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-1 is a vector rotation operator. Multiply a vector by -1 and it rotates 180 deg to point in the opposite direction. The vector continues to lay on the same line so the magnitudes of vectors on the same line add or subtract in full. You can drop the 180 deg notation and just pay attention to the sign, + or -. multiplying the vector by -1 is the same as reversing the polarity of the leads connected at the winding relative to the winding turn direction. The vectors lay on the same line, or axis, or phase.
Square root (-1), or i, is a vector rotation operator. Multiplying a vector by i or SQ root -1 rotates the vector by 90 deg. The vectors no longer lay on the same line. To add the vectors, say 50 + i50, the Pythagorus theorem is necessary. The vector sum is Square root (50^2 + 50^2) = 70.71 The vectors lay on different lines or axis or phase and do not add or subtract linearly. They have a phase difference, a phase angle difference, that places them off of the same line.
Three phase voltages, the sinewaves have a phase angle difference of 120 deg and trig is necessary to add and subtract the vectors.
Nice job rousting the natives!
I find this fascinating.
The way that I always looks at it is that you are supplying the transformer with a single phase supply. As such who can you get more that single phase out? There would have to be a magically created phase shift in the transformer that would have 2 separate secondary windings. Now how can that be? Then, the separate winding would be connected together X2-X3 to make a neutral point. Then it is quite common for the secondary consist of a single winding with a center tap marker as X0. As such either you have 2 winding that end up in a series connect or a single winding. There is no way to get 2 phases.
That?s why I always stay away from using the word phases when discussing a single phase 120/240v connection. As one of the other guys pointed out line to line and line to neutral would be uses L1-L2, L1-N, L2-N.
I have supplied 3ph-2ph and 2ph-3ph as well as 3ph-1ph transformers. I have been also asked for 1ph-3ph transformers on many of occasions. If someone could design won of those that person would be set financially for the rest of their lives
The way that I always looks at it is that you are supplying the transformer with a single phase supply. As such who can you get more that single phase out? There would have to be a magically created phase shift in the transformer that would have 2 separate secondary windings. Now how can that be? Then, the separate winding would be connected together X2-X3 to make a neutral point. Then it is quite common for the secondary consist of a single winding with a center tap marker as X0. As such either you have 2 winding that end up in a series connect or a single winding. There is no way to get 2 phases.
That?s why I always stay away from using the word phases when discussing a single phase 120/240v connection. As one of the other guys pointed out line to line and line to neutral would be uses L1-L2, L1-N, L2-N.
I have supplied 3ph-2ph and 2ph-3ph as well as 3ph-1ph transformers. I have been also asked for 1ph-3ph transformers on many of occasions. If someone could design won of those that person would be set financially for the rest of their lives
T
T.M.Haja Sahib
Guest
Hey,Templdl,What is this 3ph-1ph transformer?If it exists,then 1ph-3ph transformer can also exist.
Rick Christopherson
Senior Member
I really have no interest in engaging in this silly discussion again, but I do wonder why your (our) question keeps going unanswered. :rant: Mesh analysis seems to simply fly out the window for the sake of how scope probes can get connected.
I'm sorry you are unable to understated the two perfectly clear and logical diagrams in post 64. One shows the windings in phase in which case the currents in the neutral would add but you would not get the 240V end to end. The other arrangement is series where the currents in the neutral subtract and you do get 240V with this arrangement. All that is clearly shown on the diagrams and you have already accepted that you blundered on the neutral current issue so I really don't see why you continue to argue the point.
Or maybe I do........
I gave that in post 119.
Was there something I missed?
I posted diagrams in posts 64 and 119.
Do you find fault with them?
If so, can you explain exactly what is wrong with them?
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