jumper
Senior Member
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- 3 Hr 2 Min from Winged Horses
Only in context. Only works as a comparison under ideal conditions, balanced resistive load.
Will not work under normal/real conditions.
May I ask a question about the single vs two phase stuff
- Thread starter Russs57
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Only in context. Only works as a comparison under ideal conditions, balanced resistive load.
Will not work under normal/real conditions.
jaggedben
Senior Member
- Location
- Northern California
- Occupation
- Solar and Energy Storage Installer
Are you kidding? This whole discussion is a tangent!
Lol. Fair enough.
At some point I had hopes that some conclusions could be drawn about which mathematical functions better describe certain real world aspects of the system. But it seems no one wants to have that discussion.
jumper
Senior Member
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- 3 Hr 2 Min from Winged Horses
I would like to talk about that aspect. But we never get that far because it always goes back to the basics being confused.
There are at least four different sub topics in this thread if not more. Totally muddled again.
wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
OK, let's try that.
In the idealized system with sine wave voltages of frequency f, we have two relations that both hold: (A) VL1-N(t) = - VL2-N(t) and (B) VL1-N(t) = VL2-N(t - 1/2f). So what are the ways in which the real world differs from the idealized? Here's the start of a list:
1) Primary voltage is not a pure sine wave.
2) Primary voltage source is not infinitely stiff.
3) Secondary windings are not identical.
4) Secondary conductors have non-zero impedance
5) Secondary loads are not purely resistive.
6) Secondary loads are unbalanced.
7) Secondary loads are non-linear.
Probably there are a bunch more. Under which of those realities (taken individually) will (A) or (B) still hold?
Cheers, Wayne
And if you consider the vectors/waveforms as L1-N and the other as N-L2 rather the conventional......
jumper
Senior Member
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- 3 Hr 2 Min from Winged Horses
It is just one of the teaching progressions from the tutorial.
Problem is that someone sees that out of context and it can easily be misunderstood.
Like the tranny pic I posted earlier, people misread stuff.
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
. . . which I apparently do, because it means that the center tap does not change how the secondary functions as a whole.
jaggedben
Senior Member
- Location
- Northern California
- Occupation
- Solar and Energy Storage Installer
Yeah, good. I'll hold off bringing back in generators and inverters until we've settled on answers for a single phase-primary transformer.
120-0-120.
Call it what you will, select whichever common reference you wish.
I'm just a pragmatic old phart.
Ideal series load.
Even some math.
Doesn’t change a thing.
the signing convention is wrong
if at common you have a + and - the associated v's must be opposite signs
as is you have a line shorted to ground
also the currents are both wrong as are the loads
jumper
Senior Member
- Location
- 3 Hr 2 Min from Winged Horses
It was a simple point. The neutral is just an extra wire as defined by the pic. It is a really 240V series circuit.
The + and - just show series supply.
It is from the OP tutorial. Same tutorial Mivey and I have used for years.
https://www.allaboutcircuits.com/textbook/alternating-current/chpt-10/single-phase-power-systems/
It makes sense it the context it is used for.
jumper
Senior Member
- Location
- 3 Hr 2 Min from Winged Horses
Prolly best. Or if you really want a decent discussion start a new thread with just that as a focus.
We are never gonna get past tranny 101 here......
jumper
Senior Member
- Location
- 3 Hr 2 Min from Winged Horses
Iggy please glance at the tutorial, that section was just part of the progression that leads to conventional analysis.
It leads to this: - and + are reference points for scope.
Split phase 120/240 Vac source is equivalent to two series aiding 120 Vac sources.
To mathematically calculate voltage between “hot” wires, we must subtract voltages, because their polarity marks show them to be opposed to each other:
If we mark the two sources’ common connection point (the neutral wire) with the same polarity mark (-), we must express their relative phase shifts as being 180o apart. Otherwise, we’d be denoting two voltage sources in direct opposition with each other, which would give 0 volts between the two “hot” conductors.
jumper
Senior Member
- Location
- 3 Hr 2 Min from Winged Horses
So,
“If we mark the two sources’ common connection point (the neutral wire) with the same polarity mark (-), we must express their relative phase shifts as being 180o apart. Otherwise, we’d be denoting two voltage sources in direct opposition with each other, which would give 0 volts between the two “hot” conductors.”
Leads to this:
Two 120 V signals and and a 240V signal. 120V signals referenced to the noodle. 240V L-L.
“If we mark the two sources’ common connection point (the neutral wire) with the same polarity mark (-), we must express their relative phase shifts as being 180o apart. Otherwise, we’d be denoting two voltage sources in direct opposition with each other, which would give 0 volts between the two “hot” conductors.”
Leads to this:
Two 120 V signals and and a 240V signal. 120V signals referenced to the noodle. 240V L-L.
So,
“If we mark the two sources’ common connection point (the neutral wire) with the same polarity mark (-), we must express their relative phase shifts as being 180o apart. Otherwise, we’d be denoting two voltage sources in direct opposition with each other, which would give 0 volts between the two “hot” conductors.”
Leads to this:
Two 120 V signals and and a 240V signal. 120V signals referenced to the noodle. 240V L-L.
thanks
got it
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
180803-1618 EDT
Ingenieur:
In your post #732 in the circuit diagram you have corrected some errors of other drawings of this circuit, but now have added an error by reversing the + and - s on the loads.
The sum of the voltages around a loop must equal 0. Thus, the polarity signing on source and load must be the same for the same wire. With the top end of the source being labeled + this means the top end of the load must be +.
A source has a rise in voltage in going from - to + thru the source, and a load has a drop in voltage in going from + to - thru the load.
For those unhappy + and - being used in an AC, then view the drawing as labeled being for one instant of time.
.
Ingenieur:
In your post #732 in the circuit diagram you have corrected some errors of other drawings of this circuit, but now have added an error by reversing the + and - s on the loads.
The sum of the voltages around a loop must equal 0. Thus, the polarity signing on source and load must be the same for the same wire. With the top end of the source being labeled + this means the top end of the load must be +.
A source has a rise in voltage in going from - to + thru the source, and a load has a drop in voltage in going from + to - thru the load.
For those unhappy + and - being used in an AC, then view the drawing as labeled being for one instant of time.
.
jumper
Senior Member
- Location
- 3 Hr 2 Min from Winged Horses
Your drawing looked fine to me.
I gotta and read gar again. I think we got a small mix up going.
You both are saying the same thing I think.
I think I see what he is saying, the load polarities should be reversed
kinda iffy, it could go either way
he thinks the load neg should be on the system common/ground
my way, at load
v = -v
i =+i since flow is - to +
p = +v x -i = -vi
negative to me implies load is absorbing power
positive would be supplying
???
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