What is the correct amperage for line B in a 3 phase system?

[Awwtumn]

If you have 3 phase power and let's say

line A & B = 10 amperes
line B & C = 10 amperes
line C & A = 10 amperes

For the single line B. Since it's in both in the first and second lines, then B can experience 10 + 10 = 20 Amperes? Or is it 15 A or still 10 Ampere?

Thank you.

 

[berkeman]

What do the phases of those 3 current waveforms look like?

 

[anorlunda]

If you have 3 phase power and let's say

line A & B = 10 amperes
line B & C = 10 amperes
line C & A = 10 amperes

For the single line B. Since it's in both in the first and second lines, then B can experience 10 + 10 = 20 Amperes? Or is it 15 A or still 10 Ampere?

Thank you.

Huh? We measure voltages between two phases AB, BC, CA. But we measure currents, one wire at a time, A, B, C.

 

[Awwtumn]

Huh? We measure voltages between two phases AB, BC, CA. But we measure currents, one wire at a time, A, B, C.

So the B line still get 10 A even if it's used twice because it doesn't occur at same time, correct?

 

[Baluncore]

If you have 3 phase power and let's say

line A & B = 10 amperes
line B & C = 10 amperes
line C & A = 10 amperes

The currents appear to be measured between lines.
Can we assume the load is wired in delta ?
The currents you specify are therefore phase currents in the three phases of the load.

 

[anorlunda]

So the B line still get 10 A even if it's used twice because it doesn't occur at same time, correct?

We are talking about RMS measurements of AC currents. RMS inherently averages over an entire cycle. It is not time variable through the cycle.

 

[Averagesupernova]

I assume you have equal loads between all three legs. This is what you mean by line A, B, and C. You have stumbled on the beauty of 3 phase without realizing it. Yes, each leg supplies two different loads but the loads are in not in phase. So the current that one leg supplies to two different loads is NOT double of one loads current. This property allows us to transmit the same power with less conductor as compared to single phase.

 

[Awwtumn]

I assume you have equal loads between all three legs. This is what you mean by line A, B, and C. You have stumbled on the beauty of 3 phase without realizing it. Yes, each leg supplies two different loads but the loads are in not in phase. So the current that one leg supplies to two different loads is NOT double of one loads current. This property allows us to transmit the same power with less conductor as compared to single phase.

In the building electrical plan. I noticed they multiply the highest current in the A, B or B, C or C, A by square root of 3. For example.

I (T) = square root of (3) x 10A. = 17.3 A.

So does it mean each of the wire of the A, B and C should be sized for 17.3A or for 10 A?

I just want to understand it. Not trying to rewire anything. So no one here will get sued, ok?

 

[Borek]

So the B line still get 10 A even if it's used twice because it doesn't occur at same time, correct?

Close, but still a bit wrong. In a way these currents do occur at the same time, but google a plot of how the currents change during the cycle. Each current is in its own phase, and they are separated by 120 deg angle, so you are looking for a maximum of a sin(x)+sin(x+120) function. Which is definitely more than 1, but also less than 2.

 

[Averagesupernova]

There is another way to think about it. Suppose your source is 120/208 Y. 208 V * 10A = 2080 watts. You KNOW if you have 3 balanced loads across each phase then each transformer has to supply 2080 watts. Soooo, 2080/120 = 17.33 amps from each transformer. The math has to work from all approaches.

 

[russ_watters]

So the B line still get 10 A even if it's used twice because it doesn't occur at same time, correct?

No, it's not getting used twice. It's just 10A. The amperage is literally the flow rate of electrons through the wire. What you're saying is like suggesting that somehow the water flowing through a pipe is "used twice". It just doesn't make sense.

In the building electrical plan. I noticed they multiply the highest current in the A, B or B, C or C, A by square root of 3. For example.

I (T) = square root of (3) x 10A. = 17.3 A.

So does it mean each of the wire of the A, B and C should be sized for 17.3A or for 10 A?

I just want to understand it. Not trying to rewire anything. So no one here will get sued, ok?

No, they wouldn't do that. You're misreading/taking out of context a chopped-up bit of the equation for power. There is a safety factor, but that isn't it. Please provide the full calculation and specific context (like a photo of what you are seeing).

 

[Awwtumn]

I was referring to Delta circuit.

So if
line A & B = 10 amperes
line B & C = 10 amperes
line C & A = 10 amperes

These are phase currents as Baluncore sugguested So since the line current = 1.73 x 10 A or about 17.3 A.
Then the actual wire amperage capacity of each of the wire in the Delta must match the 17.3A and not 10A. Is this right?? (this is the primary question)

 

[Averagesupernova]

Then the actual wire amperage capacity of each of the wire in the Delta must match the 17.3A and not 10A. Is this right?? (this is the primary question)

"In the delta" is not the same as the feeders that go from source to load. The delta connected transformers you show will have a 10 amp load each. Only after they are connected together and the conductors become feeders en route to the load will they carry 17.33 amps.

 

[russ_watters]

Ok, yeah, I didn't realize they would express a "line" amperage...but yeah, not the same as the phase wire's amps. I'm actually not clear why that would be useful and can see why it might be confusing.

...it also implies the diagram is mislabeled?

[Edit]Oh, I see, no, the line current still just is what it is, but the phase current is measured in the delta.
https://www.allaboutcircuits.com/textbook/alternating-current/chpt-10/three-phase-y-delta-configurations/
I can see this causing confusion because I was thinking externally the three wires could be called a "line" or "phase" interchangably.

 

[anorlunda]

I can see this causing confusion because I was thinking externally the three wires could be called a "line" or "phase" interchangably.

The nomenclature is a mess, but there's no authority empowered to clean it up.

 

[Averagesupernova]

Where I come from a 'phase' is always a pair of wires. A 'leg' is a single wire. This wire can be a single lead off a transformer in wye configuration or the node that two separate transformers connect together at on delta configuration. But never the neutral.
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I believe it is common to refer to a hot and neutral pair in wye as a phase also. Not sure if it's technically correct. I've also heard an individual hot called a phase wire. This I assume electricians will refer to in order to differentiate between a hot and neutral when running through a conduit or similar raceway. I just refer to it as hot. 120/208 wye may have almost 100% 120 volt loads so it would be common to run one hot and the neutral in conduits.

 

[Awwtumn]

"In the delta" is not the same as the feeders that go from source to load. The delta connected transformers you show will have a 10 amp load each. Only after they are connected together and the conductors become feeders en route to the load will they carry 17.33 amps.

Let's say your electric dryer requires 230 volts, from P = V I, Let's say the power is 2300 Watts so the ampere is 10 A in our example.

So Line A & B = 10 Amperes

These are phase currents. What seems odd to me is the 10 Amperes will only run inside the power company Delta circuit. While what comes out will be 17.33 A??

If so, won't the electric dryer becomes 17.33A? Then what would change here:

P = 2300 Watts
V = 230 Volts
I = 10 Am??

 

[Averagesupernova]

The electric dryer if it's like any clothes dryer I've ever seen is a single phase load. The motor runs on 120 volts. So one of the winding would need a center tap. You are mixing apples and oranges here. You've left one leg unloaded in the dryer example.

 

[Awwtumn]

The electric dryer if it's like any clothes dryer I've ever seen is a single phase load. The motor runs on 120 volts. So one of the winding would need a center tap. You are mixing apples and oranges here. You've left one leg unloaded in the dryer example.

Oh. To avoid confusion. Let's go back to the pure Delta case and the load A-B = 10 A.

Since line current = 1.73 phase current

Will the load draws 10A or 17.33A?? It's confusing. Will 10A or 17.33A go to the load??

 

[Averagesupernova]

3 identical loads in the shape of a delta will draw 10 amps each. The current in each of the three conductors will be 17.33 amps. Makes no difference if the source is delta or wye configured transformers.

 

[Awwtumn]

3 identical loads in the shape of a delta will draw 10 amps each. The current in each of the three conductors will be 17.33 amps. Makes no difference if the source is delta or wye configured transformers.

But for wye configured transformers. Line current is equal to phase current. So where does the 17.33 come from with 10 amps 3 phase load?

 

[Averagesupernova]

But for wye configured transformers. Line current is equal to phase current. So where does the 17.33 come from with 10 amps 3 phase load?

You didn't pay attention to my previous post did you:
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There is another way to think about it. Suppose your source is 120/208 Y. 208 V * 10A = 2080 watts. You KNOW if you have 3 balanced loads across each phase then each transformer has to supply 2080 watts. Soooo, 2080/120 = 17.33 amps from each transformer. The math has to work from all approaches.

 

[Awwtumn]

You didn't pay attention to my previous post did you:
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I found out the transformer of the commercial building I'm leasing uses Open Delta. So if the load is 10A. Would it draw 17.33A from the lines or 10A?? How do you apply your Wye reasoning to this case? The same 17.33A for all cases??

 

[Averagesupernova]

Ok @Awwtumn I'm going to sound a little edgy because it seems to be my nature but I'm going to try to stay in control. Nothing personal.
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If you read my post about a wye source driving a balanced delta load, how can you not understand it? Always go back to watts if you are having trouble. Three loads dissipating 2080 watts have to source that power from three transformers. The power can come from NOWHERE ELSE! The math says this. Do the math, it shows 17.33 amps. Now if you drive with a delta source it makes no difference in the feeder conductors. They will still have to carry 17.33.

 

[Awwtumn]

Ok @Awwtumn I'm going to sound a little edgy because it seems to be my nature but I'm going to try to stay in control. Nothing personal.
-
If you read my post about a wye source driving a balanced delta load, how can you not understand it? Always go back to watts if you are having trouble. Three loads dissipating 2080 watts have to source that power from three transformers. The power can come from NOWHERE ELSE! The math says this. Do the math, it shows 17.33 amps. Now if you drive with a delta source it makes no difference in the feeder conductors. They will still have to carry 17.33.

In our commercial building. There are no 3 phase loads. Only single phase loads. But the source is 3 phase Open delta just in case there are future needs. So imagine the following are separate single phase loads.

A-B = 10 Amps (approx for sake of example)
B-C = 10 Amps (approx... )
C-A = 10 Amps. (approx... )

I understand now that in pure Delta loads. They would draw 17.33 amps from each of the line conductor A B and C because of the nature of connections of 3 phase loads where they are interconnected, like this:

But if the load is not 3 phase but only 3 single-phase connected to each A,B, B,C, C,A. Does it somehow still draw 17.33 A or 10A? Perhaps still 17.33A because the 3 single phase loads are still connected to one another that would still produce the configuration where the currents split and form like in 3 phase loads?? Just clarifying if I got it. Many thanks for all the explanations.

 

[Averagesupernova]

But if the load is not 3 phase but only 3 single-phase connected to each A,B, B,C, C,A.

Can you explain to me what the difference is?

 

[Awwtumn]

Can you explain to me what the difference is?

I guess 3 single phase connected in delta is same as bonafide delta load. So all would draw the same amount of current like 17.33 from the line conductor of the Delta or Wye.

I think we are clear on that.

But say if the single phase Delta load would be very unbalanced with only 1 single phase being on most of the time. Say the A,B = 30 A and the B,C and C, A = 0 A. Would this damage the Delta source or something or would the maximum current only become 30x1.73= 51.9A in any of the line conductors? So it is only the line conductors that will give away and not the Delta source? I will google it but just give last impression of what would happen to the Delta source, like would it blow up or something for totally unbalanced load? So I can remember you. Thank you.

 

[Baluncore]

But say if the single phase Delta load would be very unbalanced with only 1 single phase being on most of the time.

You have dug yourself into a hole of confusion. I cannot understand why you insist on digging it deeper.
This thread is about an unusual situation. Grounding and neutral get no mention, but the ground must be provided somehow. A delta must be grounded through the lines or the insulation will be destroyed by fire. There must be a Y connection somewhere to provide the grounded neutral. Draw yourself a phasor diagram of a 3PH system, mark the lines and the grounded neutral.

 

[Averagesupernova]

If you were to load A,B with 30 amps as you say, why do you think the conductors that feed a single load would carry current that exceeds the current in the load?
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You need to take a step back and simplify. Draw a single transformer winding driving a single load. Easy right? Now draw in an extra winding as would be in an open delta but no changes in the load. Why would this change anything? Finally, draw in the third winding to form the full delta. Again, no change.

 

[Averagesupernova]

Grounding and neutral get no mention, but the ground must be provided somehow. A delta must be grounded through the lines or the insulation will be destroyed by fire. There must be a Y connection somewhere to provide the grounded neutral.

The drawing in post #23 shows a center tap ground. As far as having to have a neutral, this is not true. There are delta systems out there that are basically floating. In these systems a fault to ground from one of the hot conductors will not cause excessive current and trip a breaker or blow a fuse. However, there is usually detector(s) of some sort that indicate a fault needs to be located and fixed. A second fault can result in a tripping breaker, etc.
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At any rate there is no need to confuse the op even farther with neutral and ground. It makes no difference concerning the questions at hand. I doubt the objective here is to rewire the building by DIY.