Does 3 phases transmit exactly 3 times more power than single phase ? - Page 1
Forum section: electronics / beginners

author: tigrou
author_rank: Regular Contributor
timestamp: June 10, 2022, 09:13:26 am
content: I read somewhere that 3 phases (eg: 220V) transmit 3 times more power than single phase, using only one more wire.Is this really the case or is this a rough approximation ?I have tried to manually calculate how much power it transmit (it might be totally wrong as I am a beginner).Here is how I did it.Phase to phase voltage:V1 = sin(t) - sin(t + 2*PI/3)V2 = sin(t + 2*PI/3) - sin(t + 4*PI/3)V3 = sin(t + 4*PI/3) - sin(t)P = P1 + P2 + P3P = V1 * I1 + V2 * I2 + V3 * I3I consider a resistive load, current and voltage are in phase (V = I)At the end I end up with P = 4.5If I integrate it over [0;2*PI] = 2 * PI * 4.5If I do the same with single phase :V = sin(t)P = V * I = sin(t) * sin(t)If I integrate this over [0;2*PI] = PIRatio = (2 * PI * 4.5) / PI = 9 (instead of 3)There must be something wrong. I have probably to take the square root of the final result.

author: BeBuLamar
author_rank: Super Contributor
timestamp: June 10, 2022, 11:41:56 am
content: Sorry I do not know the math but I know it's square root of 3 times that is 1.732 times the single phase with same voltage and current.

author: langwadt
author_rank: Super Contributor
timestamp: June 10, 2022, 01:11:34 pm
content: the handwaving explanation:3 lives and 3 neutrals, can obviously transmit 3x the power of 1 live and 1 neutral,combine the 3 neutrals to one fat neutral = 3+1 wiresrealise that with 3 equally loaded phases the currents in the neutral adds up to zero so you can remove it = 3 wires

author: CaptDon
author_rank: Super Contributor
timestamp: June 10, 2022, 01:20:56 pm
content: Well Lang, The answer ISN'T 3X!! Because of the phase relationship. The peaks do not occur at the same time. The poster who gave the answer 1.73 wins the prize for the correct answer. If the load is delta connected there is no 'neutral' so to speak although there may be a safety ground. If the load is star connected a neutral is usually used because any imbalance in the phases will need a return path and the neutral is also how here in the states we get 3 X 120vac (Any phase to neutral) while at the same time getting 208 3 phase for the larger loads.

author: bdunham7
author_rank: Super Contributor
timestamp: June 10, 2022, 01:59:17 pm
content: Quote from: CaptDon on June 10, 2022, 01:20:56 pmI see no holes inlangwadt'sexplanation. The problem is that you don't agree on the fundamental assumptions of the question, namely what the voltage of the system 'is'.Take an old-school 208/120 4-wire system and do the math. Assume the current is limited to 20A on your circuit and that you have a linear load, no complex harmonic issues.The power available on any one P-N circuit is 120V x 20A = 2400W.The power available on any one P-P circuit is 208V x 20A = 4160W. Note that 4160/2400 is about 1.732 or √3.The power available from the entire circuit is 7200W, which is 3X the individual P-N maximum and 1.732 or √3 x the individual P-P maximum. And absent harmonics, the neutral is not needed thus you can reduce the system to 3 wires.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: June 10, 2022, 02:07:04 pm
content: Quote from: CaptDon on June 10, 2022, 01:20:56 pmYes it is.Quote from: CaptDon on June 10, 2022, 01:20:56 pmNope.So where do those other guys go wrong?There is a factor sqrt(3) involved, but that is the voltage relation ship between a phase and neutral on one side, and between two phases otherwise. 120V * sqrt(3) = 208, or as we say here in Europe, 230V*sqrt(3) = 400V. But if you have a symmetric 3-phase load (no matter whether it's connected in star or delta) you multiply sqrt(3) with 3 again and you get back to 3. It is one of the main reasons that a 3-phase net is used for power distribution. You get 3 times the power for only 50% more wiring. Another reason is that if you sum the three phases together, you get a continuous and constant power flow, and because of this 3-phase motors run smoother then single phase motors, and this is important for very big motors and the big generators.The answer langwadt gave is watertight and easiest to follow, with the addition that the sum of the neutral currents is 0 because of the symmetric phase shifts in the 3-phase system.Here in Europe Neutral and "safety ground" is pretty much the same. (almost?) any (or group of) residence (s) has a long earth electrode which is literally put in the dirt. There is a difference between those wires inside residences because the "neutral" is fed though ground leakage switches, while the safety earth is lead around that switch. Another difference is that inside residences we normally do not distinguish between "live" and "neutral". Both are considered "live" and all 230Vac outlets are non-polarized (Except maybe in France and Italy) Here in Europe the high power trasmission lines usually have 8 wires. Two groups of 3-phase power and two extra leads which do not even have insulators. These extra leads are more used for lingning arrestors then for the neutral wire.CaptDon is right in that the Neutral / Safety ground wire is usually present, but this is only true for an equalizing current due to assymetry for the many 230Vac single phase users.

author: langwadt
author_rank: Super Contributor
timestamp: June 10, 2022, 02:09:53 pm
content: Quote from: CaptDon on June 10, 2022, 01:20:56 pmyou are comparing apples and oranges and mixing voltages and powerif your wire is rated for 10A you can supply 10A to a load with 1 live and 1 neutral => 10A in two wiresif you put three 10A loads in star you need 3*live, neutral is redundant because the currents cancel, => 3*10A in three wires

author: Doctorandus_P
author_rank: Super Contributor
timestamp: June 10, 2022, 02:22:25 pm
content: Quote from: tigrou on June 10, 2022, 09:13:26 amYou're making it very complicated. Do not do that. do not use the phase to phase voltages, because then you use different references for each of the voltages, and this makes it very easy to make mistakes, and hard to correct those mistakes.The answer langwadt gave, by first looking at each voltage compared to the neutral wire, then combining and eliminating the neutral wire is easy to understand, and it is also correct. The only difficulty is that if we talk about a 3-phase net (here in Europe) then we normally assume the phase-to-phase voltage (which is 400Vac) instead of the phase to GND voltage (which is 230Vac)Also note these are nominal voltages, and have certain tolerances. In large area's of the Netherlands we have a bit of a problem on sunny days because of lots of solar installations pushing power into the net. Solar installations even turn themselves off because the local voltage reaches 252Vac and this is the top of the allowed tolerance on the low voltage distribution net.

author: jonovid
author_rank: Super Contributor
timestamp: June 10, 2022, 03:14:54 pm
content: my question if you had a very large quadcopter = 4 x 30kw motors with variable pitch propellers a VTOL fixed wing aircraft.powered by an onboard 150 kilowatt gas turbine alternator generator power plant. with variable pitch propeller providing the forward thrusthowmany phaseswould best transmit the power needed to the 4 outboard motors at maximum efficiency with the minimum of cable & motor/ alternator weight.at the maximum Hz and voltage the system can reliably tolerate. its not like I'm into rocket science or hard core aerospace engineering.

author: TimFox
author_rank: Super Contributor
timestamp: June 10, 2022, 03:41:21 pm
content: When faced with the simple question, "How many kVA is 10 A/phase at 208 V line-line 3-phase?", I used the following mnemonic to calculate it (V and A are rms, power is mean):1. 208 V line-line = 120 V line-neutral.2. Consider three equal light bulbs, 12 ohms each, connected from the three lines to neutral.3. There is no current in the neutral line, but each bulb draws 10 A at 120 V.4. Therefore, the total power drawn by the load(s) is 3 x 10 A x 120 V = 3.6 kW.

author: capt bullshot
author_rank: Super Contributor
timestamp: June 10, 2022, 03:47:02 pm
content: Quote from: jonovid on June 10, 2022, 03:14:54 pm3 Phase w/o neutral.It's the result of long term (in engineering long term units) evolution, and evolution most of the time settles at the optimum solution.Regarding three times the power in a 3 phase system:If you look at the voltages of phase to neutral and current in each phase, its three times the power than single phaseIf you look at the voltages phase to phase, it's sqrt(3) times the power of single phase (but this voltage is sqrt(3) times the single phase voltage, so in terms of power the same result as above)

author: Doctorandus_P
author_rank: Super Contributor
timestamp: June 10, 2022, 03:49:14 pm
content: Edit: Remove: @TimFox: (Oops, sorry).Edit: Add @Jonovid:There is no benefit of using more then 3 phases.In a 3-phase system the sum of the powers delivered though the 3 wires is constant (Unlike a single phase system, where power fluctuates between 0 (at zero crossing of either voltage or current) and the peaks).There is some silly youtuber which has millions of views of a video about 6, 12, or more phase systems, but it's utter shite.In general, size and weight reduces when higher frequencies are used. See it like this: A chunk of iron or ferrite (motor, transformer, inductor or whatever it's used in) can store a limited (and fixed) amount of energy before it saturates. So if you "charge" and "discharge" it at a frequency that is twice as high, then you can also transfer twice as much energy in the same time frame.If you only look at the wiring, then a DC voltage and current with just two wires is optimal, and a higher bus voltage is better, because it needs less current, and thus can use thinner wires for the same amount of energy. The most extreme example of this are HVDC lines (High Voltage DC). These transmission lines can be more then a Mega Volt and span several thousand kilometers. There are two big real uses of HVDC lines. One is for very long power lines, for examply from a hydro plant deep in the amazon rain forest to industrialized areas far away. The other use is for deep sea cables where the capacitive loading of AC cables is a serious limitation.

author: ledtester
author_rank: Super Contributor
timestamp: June 10, 2022, 04:35:55 pm
content: Quote from: tigrou on June 10, 2022, 09:13:26 amIs this possibly the answer?It seems you are considering a delta configuration. But if you don't power one of the corners of the delta then the resistance between the other nodes is a parallel combination of R and 2R = 2/3 R so the single-phase power would be V^2/(2R/3) = 3/2 V^2/R.As you calculated, the 3-phase power is 9/2 V^2/R so the ratio would be (9/2) / (3/2) = 3.

author: TimFox
author_rank: Super Contributor
timestamp: June 10, 2022, 05:06:46 pm
content: Quote from: Doctorandus_P on June 10, 2022, 03:49:14 pmThe voltage and current in my example are all rms, and the power is mean. In your discussion, the voltage, current, and power are instantaneous values.

author: Ground_Loop
author_rank: Frequent Contributor
timestamp: June 10, 2022, 05:33:25 pm
content: Power 3-phase = (Phase-Phase voltage) * (Phase current) * SQRT(3) * (power factor)

author: CaptDon
author_rank: Super Contributor
timestamp: June 10, 2022, 07:03:55 pm
content: I suppose we hugely overbuilt our locomotive systems then. We used the 1.73 number for scaling finding at any one instant calculating the voltage/current and phase relationship the total power was 1.73 X the power in any one phase to phase leg as when L1 to L2 were at the peak then L1 to L3 and likewise L2 to L3 would not be at their peak and therefore producing less power in those phases. However, I can't argue against the fact that with 3 resistors each drawing 10 amps phase to neutral or even phase to phase you would seemingly see 3X the individual power. So is there a term that shows at any given instant calculating currents/voltages and phase relationships you would apparently be seeing only 1.73 of the provable 3.0?

author: langwadt
author_rank: Super Contributor
timestamp: June 10, 2022, 09:27:17 pm
content: Quote from: CaptDon on June 10, 2022, 07:03:55 pmthe voltage from phase to phase is sqrt(3) times the voltage from phase to neutral, that's it

author: CaptDon
author_rank: Super Contributor
timestamp: June 11, 2022, 12:18:17 am
content: I wonder if I have been jaded by the numbers we use on the locomotives. The diesel engine drives a 3 phase alternator which drives a six diode full wave bridge rectifier producing the link voltage (Apprx 1400vdc max). When we calculate the available electrical horsepower on the link we use the phase to phase voltage and current (for example L1 to L2) E times I then multiply that by 1.73 and call that the available link voltage horsepower. Is it because of the diodes and link voltage that we only obtain 'horsepower' from the alternator on the crests? Our link voltage is stored in monstrous capacitors. I think each is 1700uf at 2000vdc and there are 12 of them. At full throttle with 4500 horsepower being pushed into the D.C. link and the energy storage of the capacitor bank when one of those capacitors decides to dead short it has been known to blow the door completely off the auxiliary cab with a huge hot oil bleve fireball. Rather spectacular indeed!

author: Doctorandus_P
author_rank: Super Contributor
timestamp: June 11, 2022, 01:42:22 am
content: What is a "horse power" isn't that some form of animal abuse?

author: IanB
author_rank: Super Contributor
timestamp: June 11, 2022, 01:53:31 am
content: Quote from: CaptDon on June 11, 2022, 12:18:17 amNo, 1.73 is the square root of 3.If you had a single phase the power would be V times I. When you have three phases you get three times the power compared to one phase, hence 3 times V times I. However, that is only true if you take V as the voltage between line and neutral. If instead you take the voltage between L1 and L2 it is already 1.73 times too high, so you can just multiply by another 1.73 to get to 3 (since 1.73 times 1.73 equals 3).This is nothing to do with crests. Actually, if you full-wave rectify a three-phase supply you get perfectly flat and smooth DC, with no ripple. It is really nice.

author: Monkeh
author_rank: Super Contributor
timestamp: June 11, 2022, 02:04:19 am
content: Quote from: IanB on June 11, 2022, 01:53:31 amNo, not quite..

author: Doctorandus_P
author_rank: Super Contributor
timestamp: June 11, 2022, 02:36:56 am
content: Quote from: Monkeh on June 11, 2022, 02:04:19 amIndeed, there will be some ripple left,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%. However, power delivery is not done in pure sines and the peaks of the power net tend to be flattened, and probably the induction of the long transmission lines may also smooth it out a little bit.In high power applications sometimes a phase shift generator is used to create an offset and then 12 peaks per cycle are rectified, but that is old technology. Maybe they do it these days with some power factor correction circuit.

author: golden_labels
author_rank: Super Contributor
timestamp: June 11, 2022, 07:47:57 am
content: 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?

author: Zero999
author_rank: Super Contributor
timestamp: June 11, 2022, 03:05:37 pm
content: Quote from: Doctorandus_P on June 11, 2022, 02:36:56 amIt's normally done with transformers and is still used, as far as I'm aware. It results in a higher power factor, as well as a smoother output voltage.

author: TimFox
author_rank: Super Contributor
timestamp: June 11, 2022, 03:46:58 pm
content: Quote from: Zero999 on June 11, 2022, 03:05:37 pmSome quantitative values for (unfiltered) ripple voltage waveforms for different multi-phase rectifier circuits from F E Terman, "Radio Engineers' Handbook" First Edition, McGraw-Hill, 1943, p 598.Single-phase full-wave bridge: +0.363, -0.637 at 2x line frequency (with respect to DC mean value = 1.0)Three-phase: +0.209, -0.395 at 3x line frequencySix-phase: +0.0472, -0.0930 at 6x line frequencyHere, six-phase can mean "Six-phase single Y", "Double three-phase with balance coil", "Double three-phase, series connection", or "Six-phase star".I once used a high-power (250 V, 250 A) DC current supply from Danfysik for an electromagnet. A DCCT measured the output current. The output linear regulator used a huge number of 2N3772s on a water-cooled plate. There were two three-phase rectifiers, one with a delta and the other with a wye secondary, where one was driven from a three-phase "Slidak" variable transformer, servoed to keep the CE voltage of the 2N3772s constant. The ripple voltages from the two rectifiers (no filter capacitor) were out of phase, but not equal due to the one transformer's variable input. The output ripple current was almost totally removed (< 1 ppm) by the linear regulator and the inductance of the magnet.

author: gf
author_rank: Super Contributor
timestamp: June 11, 2022, 03:48:32 pm
content: Quote from: Doctorandus_P on June 11, 2022, 02:36:56 amSo 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?

author: bdunham7
author_rank: Super Contributor
timestamp: June 11, 2022, 03:55:28 pm
content: Quote from: Zero999 on June 11, 2022, 03:05:37 pmYes, 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.

author: Zero999
author_rank: Super Contributor
timestamp: June 11, 2022, 04:25:25 pm
content: 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.

author: ledtester
author_rank: Super Contributor
timestamp: June 11, 2022, 05:04:11 pm
content: Quote from: golden_labels on June 11, 2022, 07:47:57 amSo 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.

author: Simon
author_rank: Unknown
timestamp: June 11, 2022, 06:43:27 pm
content: 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.

author: gf
author_rank: Super Contributor
timestamp: June 11, 2022, 07:20:20 pm
content: Quote from: Simon on June 11, 2022, 06:43:27 pmPhase-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.

author: Simon
author_rank: Unknown
timestamp: June 11, 2022, 08:00:46 pm
content: 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*IFor 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.

author: langwadt
author_rank: Super Contributor
timestamp: June 11, 2022, 08:11:15 pm
content: Quote from: Simon on June 11, 2022, 08:00:46 pmif it 230V phase to neutral and 400V phase to phase and you have 100A on each phase you have 69kW

author: Simon
author_rank: Unknown
timestamp: June 11, 2022, 08:27:31 pm
content: 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.

author: gf
author_rank: Super Contributor
timestamp: June 11, 2022, 08:35:15 pm
content: Quote from: langwadt on June 11, 2022, 08:11:15 pmExactly.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. ]

author: IanB
author_rank: Super Contributor
timestamp: June 11, 2022, 08:36:29 pm
content: Quote from: Simon on June 11, 2022, 08:27:31 pmA 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?

author: Simon
author_rank: Unknown
timestamp: June 11, 2022, 08:59:01 pm
content: Quote from: IanB on June 11, 2022, 08:36:29 pm

author: Zero999
author_rank: Super Contributor
timestamp: June 11, 2022, 09:08:02 pm
content: 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.

author: Simon
author_rank: Unknown
timestamp: June 11, 2022, 10:01:09 pm
content: 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.

author: IanB
author_rank: Super Contributor
timestamp: June 11, 2022, 10:59:56 pm
content: Quote from: Zero999 on June 11, 2022, 09:08:02 pmCan 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".

author: gf
author_rank: Super Contributor
timestamp: June 11, 2022, 11:01:34 pm
content: Quote from: Zero999 on June 11, 2022, 09:08:02 pmDelta 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

author: IanB
author_rank: Super Contributor
timestamp: June 11, 2022, 11:11:38 pm
content: Quote from: Simon on June 11, 2022, 08:59:01 pmThere 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.

author: Simon
author_rank: Unknown
timestamp: June 12, 2022, 06:29:31 am
content: 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.

author: Zero999
author_rank: Super Contributor
timestamp: June 12, 2022, 08:18:40 am
content: Quote from: IanB on June 11, 2022, 10:59:56 pmI'm confusing an earthing transformer, which is a separate transformer, not another winding.https://www.sciencedirect.com/topics/engineering/earthing-transformerQuote from: gf on June 11, 2022, 11:01:34 pmThinking 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.

author: Terry Bites
author_rank: Super Contributor
timestamp: June 13, 2022, 05:06:45 pm
content: 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.