Julian

[Dennis Alwon]

What convention are you using to relate the Phase letter to the wire designator?
I will call the wires L1, L2 and L3 to avoid any other association. And I will call the voltage from L1 to L2 Phase A, the voltage from L2 to L3 Phase B and the voltage from L3 to L1 Phase C.

So when you say a load is connected to A and B only, that means that one load is connected between L1 and L2, the other is connected from L2 to L3. So the current in L1 is 49kW divided by the Phase A voltage. The current in L3 is also 49kW/V. The current in L2, however, is the vector sum of the two currents from the two loads. That one will have a sqrt(3) factor in it, and will not be in phase with either load voltage.

Okay then why did gar say if I connect 2 loads to line one and line 2 only the resulting reading on one line would be additive of the 2 lines. At least that is how I read his comment. Is that because we are not using L1, L2 and L3?

 

[gar]

190105-0859 EST

A big part of the problem here is having a clear understanding of the circuit being discussed.

In post #35 I quoted what Dennis said in post #31. I believe that in post #31 he has defined his question at that time as being only two wires of a three wire source being the source voltage. These being A and B. And he further said that C was not connected. No mention of neutral, and therefore assume there is no neutral.

Thus, the circuit of post #31 is simply a two wire single phase circuit. It can not be anything else.

GoldDigger tried to help, and he did, by labeling the wires or points of connection as L1, L2, and L3. This helps because you don't normally call phases L1, L2, or L3. So post #31 is about using L1 and L2 as a single phase voltage source of 480 V. You can connect as many loads in parallel as you want across this source. Dennis chose to put two equal value loads in parallel across L1 and L2.

Before post #31 a different circuit was being discussed.

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[gar]

190105-1011 EST

If we get back to the three phase circuit, which we should, but not before the single phase circuit is understood. Note: Dennis's two bell transformer question, a different thread, is a single or two phase question that needs understanding.

Two sine waves of identical frequency, but with some relative phase shift, have to be considered as two phases whether you like it or not. A 180 degree phase shift produces a two phase system. So does 1 degree, or 30, or 90, or 120. As soon as there is a phase shift it is not a single phase system.

Basic to much of this is:
1. Steady state sine waves that are of identical frequency and relative phase relationship. Not talking about the relationship of sine waves of different frequencies.
2. Sum of two or more identical frequency sine waves is a sine wave of the same frequency, trig identity, possiby some other phase angle. See trig functions. Handbook of Chemistry and Physics is one source.
3. Analysis of steady state sine waves can be done with algebra by use of phasors, vectors.
4. The phasor concept was invented by Steinmetz of General Electric in the late 1800s. At the time he was not understood.

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[GoldDigger]

Okay then why did gar say if I connect 2 loads to line one and line 2 only the resulting reading on one line would be additive of the 2 lines. At least that is how I read his comment. Is that because we are not using L1, L2 and L3?

If you connect two loads in parallel between L1 and L2, then the current clearly adds.

The opportunity for misinterpretation comes when you talk about two loads on Phases A and B rather than lines A and B. A reasonable interpretation is one load on each distinct line to line phase rather than two loads in parallel on two wires.

We have a history of carelessly talking about the high leg of a 120/240 delta having to be on the B phase rather than having to be on the B bus or the B wire, for example.

 

[Strathead]

Okay then why did gar say if I connect 2 loads to line one and line 2 only the resulting reading on one line would be additive of the 2 lines. At least that is how I read his comment. Is that because we are not using L1, L2 and L3?

Dennis, I think your question is not answered yet, or at least not clearly. Golddigger's explanation confuses me, but I think what he is saying and/or what you are getting hung up on is this. Inside the equipment which is actually three transformers hooked up in a delta configuration not a single transformer as earlier discussed, there is a point where the wire we call phase A coming out of the equipment, is split (coming from transformer secondary 1 and transformer secondary 2) After they are joined together you are going to see the full load of 10.4 amps. from the connection point to the place where they split again to go to the two loads. INSIDE the "transformer, some of the current will travel through the transformer connected to A and B and a smaller portion will travel through the other two transformers in series with each other. As I read hear recently, "path of least resistance" is a misnomer, as current will travel both paths, higher and lower resistance, or impedance in this case just not is equal amounts.

 

[Dennis Alwon]

Dennis, I think your question is not answered yet, or at least not clearly. Golddigger's explanation confuses me, but I think what he is saying and/or what you are getting hung up on is this. Inside the equipment which is actually three transformers hooked up in a delta configuration not a single transformer as earlier discussed, there is a point where the wire we call phase A coming out of the equipment, is split (coming from transformer secondary 1 and transformer secondary 2) After they are joined together you are going to see the full load of 10.4 amps. from the connection point to the place where they split again to go to the two loads. INSIDE the "transformer, some of the current will travel through the transformer connected to A and B and a smaller portion will travel through the other two transformers in series with each other. As I read hear recently, "path of least resistance" is a misnomer, as current will travel both paths, higher and lower resistance, or impedance in this case just not is equal amounts.

Thank you that helps alot..

 

[kwired]

Wouldn't it be 204 amps... 49 kw twice for each phase

If both loads were connected to same two source terminals, yes. Since you have two connections to B and one connection to A, one to C the phase angle will it will draw 1.732 times the 102 amps of the two individual loads on B, which is 176. You will still see 102 amps on the lead to each individual load, but the common B back to the source will measure 176. You will have this value at that point even if you took the third load off line.

ETA: sorry if I mentioned things already brought up, for some reason I thought I was at the end of the thread with only 8-10 posts in when I replied, then after submitting reply I see my post was past the 40th post in the thread.

 

[kwired]

Now that I read the other posts I missed

but with three balanced loads as shown , each supply line sees 1.73 times the individual load current not twice.

even if you disconnected one of the loads you still have 1.73 times individual load current on the line with two identical loads connected to it. Make each load a different value and calculating the current on common leads gets more complex.

So question,
1. Load per phase is 49,000/480V = 102 amps,
2. Since there are three balanced loads, each supply line sees 1.73 times the individual load current (not twice), so 102 amps x 1.732 = 176.81 amps,
3. Now, do you fuse at next higher or lower standard size per 240.6 for this feeder OCP???
Y'all are a big help )

You seem to have gotten the concept of what the current will be on the feeder conductors, but you don't want to go next lower overcurrent device from actual current draw - that puts you into trip curve of the OCPD, and after enough time it may trip, you only want to be in the trip curve during an abnormal condition.

Okay, what am I missing? Why are we only looking at 49kw. Doesn't each phase have 2 loads of 49kw?

Each output lead of the source has has 1.732 loads of 49 kw connected to it. You would get same current values even with a 480 volt wye source and same connected loads.

So if there were only 2 49 kw loads both going to phases A and B - loads are sp 480V--what is the load on A or B?

If both loads in question are on A and B then there is only one 180 degree phase angle involved and current is doubled.

 

[gar]

190107-0856 EST

Dennis Alwon:

If that helps, then I think you are now more confused than before.

An ideal transformer has zero internal impedance. It is simply an absolutely constant voltage source. It will put out its constant voltage no matter what the load is on the transformer secondary. And don't get lost in what happens at infinity.

Yes, real transformers have internal impedance and their terminal voltage is not absolutely constant. But getting lost in that discussion does not contribute to an understanding of what is happening.

Do this -- separate the delta into three separate single phase secondaries, and the delta load into three separate loads connected to their corresponding secondaries. Analyze what happens in this case.

I have to leave right now.

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[Strathead]

190107-0856 EST

Dennis Alwon:

If that helps, then I think you are now more confused than before.

An ideal transformer has zero internal impedance. It is simply an absolutely constant voltage source. It will put out its constant voltage no matter what the load is on the transformer secondary. And don't get lost in what happens at infinity.

Yes, real transformers have internal impedance and their terminal voltage is not absolutely constant. But getting lost in that discussion does not contribute to an understanding of what is happening.

Do this -- separate the delta into three separate single phase secondaries, and the delta load into three separate loads connected to their corresponding secondaries. Analyze what happens in this case.

I have to leave right now.

.

What is this in relation to? It doesn't seem to be related to any post. I haven't seen a reference to internal impedance, or its voltage output.

 

[wwhitney]

The graph below was prepared (by others) to show how voltage waveforms add in a 3 phase system. The waveform labeled Phase 1 is 120V AC RMS relative to the neutral; the waveform labeled Phase 2 is identical but shifted 120 degrees in phase; and the difference is a waveform that is 208V AC RMS.

The math is the same for the case of current with two identical resistive loads in a three phase system, one load connected from leg A to leg B, and one load connected from leg B to leg C. Just relabel the curves to be Load Current instead of Phase; the two Load Currents are 120 degrees out of phase. Then the current in leg B is Load Current 1 - Load Current 2, and the result is a sine wave whose magnitude is sqrt(3) times either load current. Not twice the load current, as the two current waveforms do not peak at the same point in time.

Cheers, Wayne

 

[gar]

190107-1536 EST

Strathead:

When one does circuit analysis assumptions are made to approximate the real world. Sometimes some assumptions are made equal to zero to simplify understanding the circuit, or in some cases even making it possible to solve the problem.

In this case, if we assume the voltage sources, the transformers as viewed looking from the load toward the transformer secondary, have zero internal impedance, then the point we are trying to make is easier to see.

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[Strathead]

190107-1536 EST

Strathead:

When one does circuit analysis assumptions are made to approximate the real world. Sometimes some assumptions are made equal to zero to simplify understanding the circuit, or in some cases even making it possible to solve the problem.

In this case, if we assume the voltage sources, the transformers as viewed looking from the load toward the transformer secondary, have zero internal impedance, then the point we are trying to make is easier to see.

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I understand that in that we also don't take in to account the wire resistance, we assume zero. None the less, Dennis is thinking deep enough to get confused but one on the level of you or say Smart$ (I miss him God rest). Nor do I. A regular person would accept that the transformer amperage between A to B in a Delta is the same as that between A and B is a stand alone. Not so Dennis. So I just explained it without too much depth, or the same way that I understand it.

 

[petersonra]

I'm with augie47: 176.81 amps.

147 kW load. 147 ÷ sqrt(3) ÷ 0.480 = 176.81 amps of load leaving the transformer via the feeder.

I concur.

beyond that the selection of conductor sizes and OCPD sizes is based on what the code allows and/or requires.

 

[gar]

190108-0558 EST


julian padilla, Dennis Alwon, and others:

Do you know what a phasor ( vector ) is, how to use them, and why the invention ( late 1800s ) of the concept by Steinmetz was so profound and important? What is the basis of the invention? What assumptions allow the invention's use?

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[Dennis Alwon]

Gar, as I said earlier I have absolutely no knowledge of theory in electricity. Never had a class in it either. I learned hands on doing both commercial and resi work but never had a need to understand this stuff. I am asking now because of curiosity so no I know of vectors but nothing else you mentioned.

 

[gar]

1901008-1027 EST

Dennis Alwon:

Some basics.

In an AC electrical system we at least assume that a voltage source is a pure sine wave. This is a fundamental requirement for use of the phasor concept. In the real world this is not true, but in many cases it is a sufficiently good assumption to be able to perform useful calculations.

Second, we assume loads are linear. This means their defined electrical characteristics are invariant under any conditions, and that if you double the applied voltage that the current will double. A pure resistor, inductor, or capacitor is a linear load. A diode is not. A transformer or motor is not, but in some regions of operation may be considered linear.

Third, for phasor analysis it is assumed we are working under steady-state, not transient, conditions.

Fourth, that all sources are of identically the same frequency, and synchronized, meaning no drifting phase relationship. But just saying identical frequency should be sufficient to mean no phase drift.

Mathematically sine waves can be precisely defined. Some relationships of sine wave can be precisely defined.

Two good references for these relationships are:
"Handbook of Mathematical Tables and Formulas", by Burington. See Fundamental Identities.
"Handbook of Chemistru and Physics", by Chemical Rubber Publishing Co. See Trigonometric Formulae.

My search for a clear simple proof that the sum of two sine waves of identical frequency is another sine wave of the same frequency, some phase shift, and some amplitude, did not produce a find so you will just have to accept this as being true.

I have to stop this discussion for the moment, other things to do. If you had a scope you could experiment. But there are other experiments that can be done.

The following was a reference I found, but it is probably of no value to you.
https://www.cs.sfu.ca/~tamaras/sinusoids318/Adding_two_sinusoids.html

More later.

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[gar]

190109-1936 EST

Dennis Alwon:

To continue:

I think you need to do some experiments, and inter-related look at theoretical means to compare with the experimental results.

So what facilities, components, and measurement equipment you have will determine what can be done. I have to assume you have some reasonably good DVM, but even a Simpson 260 could be useful, but it can not directly do a useful AC current measurement. Some good experiments could be done with 3 doorbell transformers, some suitable resistors, and two phases of a 3 phase wye.

What do you have available?

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[kwired]

Gar, as I said earlier I have absolutely no knowledge of theory in electricity. Never had a class in it either. I learned hands on doing both commercial and resi work but never had a need to understand this stuff. I am asking now because of curiosity so no I know of vectors but nothing else you mentioned.

You do know that if you have three 10 amp @ 240 volt loads balanced across a three phase supply that you have 10 x 240 = 2400 x 3 = 7200 total VA, right? Then when calculating total system load you do have 7200 / 240 / 1.732 = 17.32 amps right? You just don't fully understand why it is 17.32 and not 20 amps?

The current that enters those three main junctions just doesn't split in same manner as it does with simpler situation of single phase supply and only one 180 degree angle involved, at "B" you have some current that is flowing between A-B, some current that is flowing between B-C, as well as some flowing between A-C. Hard to explain much more than that without drawing out the angles and such, just that current is going multiple directions at the junction point but not an equal amount in every direction, but is a constant factor (the square root of 3) involved, with balanced loads across the three phases, or even with just two identical loads involving all three phases, get non identical loads and it gets more complex.

It doesn't have to be a delta source, to do some of your own experimenting and only have wye source to play with, connect three identical loads line to line to line(loads connected in delta) across a wye system and you get same kind of results.