Does 3 phases transmit exactly 3 times more power than single phase ?

[gf]

A fully rectified 3-phase signal will have 6 overlapping peaks in a full cycle so the difference between the peak and the valley between peaks will be cos(pi/6) = 0.8660 and that's about 13.3%.

So we have "continuous voltage", i.e. at no time it drops to zero.
Should be actually sufficient in order that the large smoothing capacitor after the rectifier can be renounced in a SMPS?

[bdunham7]

It results in a higher power factor, as well as a smoother output voltage.

Yes, and as previously unstated, the power factor of a 3-phase 'full-wave' rectifier to a single DC link is not 1.0 even in the case of a purely resistive load, unlike a single-phase full-wave bridge where it is 1.0 not accounting for diode voltage drop.

[Zero999]

To answer the original question: yes three phase can transmit three times the power, than single phase. It means less copper can be used, for the same power level.

[ledtester]

Just 2 cents. That is about comparing maximum power that can be transferred in each scenario. In all this discussion no one forgot that in three-phase scenario not only voltages go up, but in calculations you no longer divide by the resistances from the one-phase version, right? Right?

So one thing I just realized is that in the OP's formulation of the problem the three phases are represented by sin(t), sin(t+2pi/3) and sin(t+4pi/3). But then the amplitude of the difference between the phases will be sqrt(3) which will increase the power by a factor of 3. So perhaps that's why the OP calculated a factor of 9 instead of 3.

[Simon]

It is 3x the power, the problem here is the phase difference. So take 3x 230V phases 120 degrees apart and connect them together. You will no longer measure 230V between phases but a higher one, 400V I believe. The problem here is understanding that the 400V comes from 2 of the phases so part of the current comes from one of the other phases, so you cannot have it both ways, it is either the current of one phase multiplied by 3 times the single phase voltage or it is the phase to phase voltage that will be higher than the phase to neutral multiplied by a lower current as any one phase is adding to the voltage of the other two with it's current split between them. This is where the square root of 3 comes in.

It's a right mess and I got myself completely confused when talking to the local DNO for work about how much power we could have as they give you numbers but you have to understand what context they refer to.

[gf]

It is 3x the power, the problem here is the phase difference. So take 3x 230V phases 120 degrees apart and connect them together. You will no longer measure 230V between phases but a higher one, 400V I believe. The problem here is understanding that the 400V comes from 2 of the phases so part of the current comes from one of the other phases, so you cannot have it both ways, it is either the current of one phase multiplied by 3 times the single phase voltage or it is the phase to phase voltage that will be higher than the phase to neutral multiplied by a lower current as any one phase is adding to the voltage of the other two with it's current split between them. This is where the square root of 3 comes in.

It's a right mess and I got myself completely confused when talking to the local DNO for work about how much power we could have as they give you numbers but you have to understand what context they refer to.

Phase-to-phase voltage is 230V * sqrt(3) = 400V. But the maximum current that can be drawn phase-to-phase with a delta load connection must not exceed the maximum current allowed by a single phase, divided by sqrt(3), because two currents with amplitude 1/sqrt(3) and 120° phase differnce add up to amplutide 1. At the end, the max. possible phase-to-phase power in a delta connection is the same as the phase to neutral power in a star configuration. In both configurations the current through N is zero. With delta connection, N is not connected, so there can't be any current, and in star configuration, the three currents through N add up to zero, and cancel out.

[Simon]

Yes I think a lot of confusion is in that the current referred to is the current of each phase, while the voltage given is the voltage between phases, this can lead to the incorrect calculation of 3xV*I when it is (1/1.7)*V*I

For example our new feed at work was referred to as 3x100A at 230V this gave the illusion of 69kW when in fact the current is shared out so it is (1/1.7)*300*230 which is 40kW. I'm not sure what the arrangement is and I don't know if there is one that is usually used.

[langwadt]

Yes I think a lot of confusion is in that the current referred to is the current of each phase, while the voltage given is the voltage between phases, this can lead to the incorrect calculation of 3xV*I when it is (1/1.7)*V*I

For example our new feed at work was referred to as 3x100A at 230V this gave the illusion of 69kW when in fact the current is shared out so it is (1/1.7)*300*230 which is 40kW. I'm not sure what the arrangement is and I don't know if there is one that is usually used.

if it 230V phase to neutral and 400V phase to phase and you have 100A on each phase you have 69kW

[Simon]

This is the problem, I don't know, all I know is that we don't have the 69kW I assumed it to be. Maybe it is a star at 230V phase to phase, this makes each phase less than 230V.

[gf]

Yes I think a lot of confusion is in that the current referred to is the current of each phase, while the voltage given is the voltage between phases, this can lead to the incorrect calculation of 3xV*I when it is (1/1.7)*V*I

For example our new feed at work was referred to as 3x100A at 230V this gave the illusion of 69kW when in fact the current is shared out so it is (1/1.7)*300*230 which is 40kW. I'm not sure what the arrangement is and I don't know if there is one that is usually used.

if it 230V phase to neutral and 400V phase to phase and you have 100A on each phase you have 69kW

Exactly.

With star connection the voltage across the each load resistor is 230V, and the current through each load resistor is 100A. So the total power is 3 * 230 * 100 = 69kW.

With delta connection, the voltage across each load resistor is 398V, but the current through each load resistor can only be 58A, in order that the current through each phase does not exceed 100A. Total power is 3 * 398 * 58, i.e still 69kW.
The resistance of the load resistors must be 3x higher with delta connection than with star connection.

[ Btw, I'm considering only purely resistive load here, and only equal load on each phase. ]

[IanB]

This is the problem, I don't know, all I know is that we don't have the 69kW I assumed it to be. Maybe it is a star at 230V phase to phase, this makes each phase less than 230V.

A usual three phase consumer supply in the UK is 415 V phase to phase, 240 V phase to neutral. At the distribution panel there would be four wires: blue (neutral), brown (L1), black (L2), grey (L3).

So unless it is some special kind of installation, you would have 240 V x 100 A x 3 phases = 72 kW.

Who is saying you don't have that, and for what reason?

[Simon]

This is the problem, I don't know, all I know is that we don't have the 69kW I assumed it to be. Maybe it is a star at 230V phase to phase, this makes each phase less than 230V.

i don't know, maybe it is not 100A supply and there is confusion about the fusing of the supply and the actual rating.

A usual three phase consumer supply in the UK is 415 V phase to phase, 240 V phase to neutral. At the distribution panel there would be four wires: blue (neutral), brown (L1), black (L2), grey (L3).

So unless it is some special kind of installation, you would have 240 V x 100 A x 3 phases = 72 kW.

Who is saying you don't have that, and for what reason?

[Zero999]

In the UK, most residential properties only have a single phase feed. The neighbours will most likely be on a different phase. This means the currents roughly balance, so the neutral doesn't have to carry the full load current. A single distribution transformer often powers a whole housing estate. I believe the transformer often has a delta secondary, with the neutral provided by an auxiliary winding, to save copper, as it doesn't have to carry as much current has the phases.

[Simon]

delta makes sense, I would assume there would be less voltage fluctuation than with star as the current from one house is not being forced through other households or the lack of demand from another house. In this case it's an industrial supply but I would assume it would be the same.

[IanB]

In the UK, most residential properties only have a single phase feed. The neighbours will most likely be on a different phase. This means the currents roughly balance, so the neutral doesn't have to carry the full load current. A single distribution transformer often powers a whole housing estate. I believe the transformer often has a delta secondary, with the neutral provided by an auxiliary winding, to save copper, as it doesn't have to carry as much current has the phases.

Can you draw, or describe, how that could work? The neutral has to be connected to each of the phases to complete a circuit. A winding has two ends. If one end of an auxiliary winding provides the neutral, to what is the other end of the winding connected?

I have looked at the rating plate of distribution transformers in the UK, and I do recall that it would say something like "Primary 11 kV Delta, Secondary 240 V Wye".

[gf]

I believe the transformer often has a delta secondary, with the neutral provided by an auxiliary winding, to save copper, as it doesn't have to carry as much current has the phases.

Delta must be wound for higher voltage (more turns) which actually leads to a more expensive coil than star¹⁾. I rather think that delta-star (i.e. delta on the primary side, and star on the secondary side) is more common, as it provides a neutral for the secondary side (which is required for unbalanced line-to-neutral loads, and for single-phase loads), while a neutral can still be renounced on the feed line connected to the primary side. Any current on the neutral is eventually transformed to line-to-line current imbanaces on the primary side.

¹⁾ https://www.electronics-tutorials.ws/transformer/three-phase-transformer.html

[IanB]

i don't know, maybe it is not 100A supply and there is confusion about the fusing of the supply and the actual rating.

There could be a difference between peak load and continuous sustained load? It is common for a transformer to provide an output much higher than its continuous rating for a short period of time. The fuses or breakers might then be set according to the continuous rating rather than the peak rating?

I guess it would need someone familiar with the situation to provide an answer.

[Simon]

This is the local supplier network I was talking to, they are only interested in the maximum, after that they want you to have a different meter etc. At the end of the day I am not bothered. I don't know why I was put in charge of trying to deal with it, well I do, my MD assumed electronics engineer should deal with electrical thing and as he has an electric car must know all about car charging. Well I knew more than enough about car charging but trying to coheres 3 different companies of very untechnical people to talk sense and work together needed a professional project manager.

[Zero999]

In the UK, most residential properties only have a single phase feed. The neighbours will most likely be on a different phase. This means the currents roughly balance, so the neutral doesn't have to carry the full load current. A single distribution transformer often powers a whole housing estate. I believe the transformer often has a delta secondary, with the neutral provided by an auxiliary winding, to save copper, as it doesn't have to carry as much current has the phases.

Can you draw, or describe, how that could work? The neutral has to be connected to each of the phases to complete a circuit. A winding has two ends. If one end of an auxiliary winding provides the neutral, to what is the other end of the winding connected?

I have looked at the rating plate of distribution transformers in the UK, and I do recall that it would say something like "Primary 11 kV Delta, Secondary 240 V Wye".

I'm confusing an earthing transformer, which is a separate transformer, not another winding.
https://www.sciencedirect.com/topics/engineering/earthing-transformer

I believe the transformer often has a delta secondary, with the neutral provided by an auxiliary winding, to save copper, as it doesn't have to carry as much current has the phases.

Delta must be wound for higher voltage (more turns) which actually leads to a more expensive coil than star¹⁾. I rather think that delta-star (i.e. delta on the primary side, and star on the secondary side) is more common, as it provides a neutral for the secondary side (which is required for unbalanced line-to-neutral loads, and for single-phase loads), while a neutral can still be renounced on the feed line connected to the primary side. Any current on the neutral is eventually transformed to line-to-line current imbanaces on the primary side.

¹⁾ https://www.electronics-tutorials.ws/transformer/three-phase-transformer.html

Thinking about this again, I was wrong. The amount of copper used is the same. Three 230V 460A secondary windings will give 400V phase to phase 460A connected in star, or 230V, 800A connected in delta. In both cases the maximum power will be 317kW.

[Terry Bites]

P = 3 x VL x ( IL/√3) x CosФ …… (IPH = IL / /√3)

P = √3 x√3 x VL x ( IL/√3) x CosФ   (3 = √3x√3 )

P = √3 x VLx IL x CosФ   in a balanced star or delta. It makes no difference.
A neutral cannot add power to the output of a delta>star transformer. That's the normal transformer wiring at your substation or pole pig.
3-Phase increases the utilisation of the conductors, it saves tons of copper or aluminium and thats more economically efficient than 3 single phase. Becacue there is no neutral it has lower I2⋅R losses.

For unbalanced systems you have to work out the power in each phase and add the vectors.