Earth/neutral

[SparkyTris]

I have 3-phase supply to a building. Incoming are 5 separate conductors: the 3 phases, a black cable (neutral) and an uninsulated earth. Am I correct in assuming that this main supply cable travels as far as my local substation?

Am I also correct in assuming that the local substation is fed by a high-voltage overhead supply consisting of 3 conductors, being only the 3 phases?

if so, what happens at the substation (apart from stepping down voltage) to provide me with neutral and earth into the building?

 

[breezer]

the primary side of the transformer is in a triangle the secondary is a 3 pointed star.

the centre of the star is where both earth and neutral are connected

 

[Paul_C]

In more technical terminology, the transformer is a delta-wye type. The following diagrams should help you visualize the connections more easily.

The primary (high voltage) side is wired like this (named "delta" for the resemblance to the Greek letter):

The secondary (low voltage) side is like this (named star or wye, the latter obviously from the resemblance to the letter "Y"):

Ignore the voltage labels on those diagrams. For the application we're discussing, the primary side would most likely be 11,000 volts between phases. On the secondary side, it will be 240V phase-to-neutral, 415V between phases.

 

[SparkyTris]

Thanks very much for the diagrams! I wonder if you or anyone might be able to post a waveform diagram that illustrates how you can get 415V between phases?

 

[ban-all-sheds]

Your desire to learn is commendable, but I'm beginning to wonder how justified your username is.....

 

[SparkyTris]

Well it doesn't mean that I'm trying to be an electrician, if that's what you're poking at. Strikes me that there's a lot of poke and not much else on this 'ere forum.

 

[plugwash]

sparky is a slang word for electrican so people who have it in thier names and then demonstrate a lack of things electricans should know just worry people a bit (ie that you might be one of those cowboy/incompetent electricans)

now we know that isn't the case lets go ahead



note that the phase colors shown are american but the principle is there.

 

[SparkyTris]

OK plugwash, understand the 415V potential now and thanks for the link. Following on from this: the distribution of electricity to a residential street applies adjacent phases to adjacent properties. (correct me if I'm wrong!) But it must be impossible to ensure that each property always presents the same impedance. Therefore at least occasionally, very large currents (up to 3 x highest phase current) would flow in the neutral line. But the neutral conductor out in the street has the same Cross-sectional area as the phase conductors.

Where is/are my mistake/s?

 

[ban-all-sheds]

Plug's fallen asleep at the wheel - he didn't cover the 415V bit.

The graph he quoted shows a 3-phase supply, which is 3 single-phase supplies, 120° apart. The values on the axes of the graph are actually degrees and the value of the sine of the angle, hence they range from -1 to +1. AC voltage is a sine-wave.

If the Y axis were to show volts for a UK supply, the +/- 1 points would be +/- 325V, since that's what our supply voltage is - the quoted value of 230V is 325V/SQRT(2), as that's the equivalent value of a DC voltage that would provide the dame amount of energy (the area under a sine wave is the same as that under a line which is at the peak value divided by the square root of two:

It's known as the Root Mean Squared value, or RMS. See here for more of the maths: http://en.wikipedia.org/wiki/Alternating-current_electric_power#Mathematics_of_AC_voltages

Anyway - as you can see from the 3-phase chart, the difference between phases can be greater than the difference between any single phase and zero, so more than 325V. The equivalent value turns out to be the single phase RMS value x SQRT(3), or 230 x 1.732... = 398V (415V is what you get when you do the calculation with an RMS value of 240V, but you aren't supposed to do that any more, even if it is the reality)

 

[Paul_C]

Following on from this: the distribution of electricity to a residential street applies adjacent phases to adjacent properties. (correct me if I'm wrong!)

Well, the houses aren't always connected in strict A-B-C-A-B-C sequence along the street, but you have grasped the principle that the services will be as evenly balanced as possible between the three phases.

Therefore at least occasionally, very large currents (up to 3 x highest phase current) would flow in the neutral line. But the neutral conductor out in the street has the same Cross-sectional area as the phase conductors.

You're making the mistake of just adding the individual phase currents together and assuming that the total must be the neutral current. You can't use a direct arithmetical sum of the phase currents because there is a 120 degree difference in phase between them.

If the three live lines were all in phase (i.e. the black, red, and blue waveforms above were shifted so that they were on top of each other), then the neutral current would indeed be the sum of the individual phase currents. But as soon as you move the phases apart, that is no longer the case.

Three-phase can be quite complex to understand at first, but an easier way to start getting an understanding of this is to examine the simpler single-phase 3-wire system. This is widespread in North America, but also found in rural areas of Britain.

The 3-wire arrangement has a neutral plus two live lines, which are 180 degrees out of phase with each other. (Imagine two waveforms like those above, arranged so that the zero-point coincides but such that when one is at peak positive the other is at peak negative. We'll c all these A and B.)

Now, assume a load of 15A connected on the A side to neutral. At this point, with nothing on B, you would have 15 amps flowing on the A line and obviously that 15 amps must also flow along the neutral.

Now let's say you connect a load of 5A between B and neutral. Obviously the current in B is now 5 amps, but what happens at the common neutral point? If A and B were in phase, then the currents would add, resulting in 20A. But A and B are 180 degrees out of phase. When A is positive, B is negative, and vice versa (or if you like, when one side is pushing the other is pulling, if that helps you to visualize it better).

If you have 15 amps flowing toward the neutral point from A, but at the same time you have 5 amps flowing away from it via B, then the neutral current is the difference: In = Ia - Ib = 15 - 5 = 10 amps. If the loads on A and B are identical, then Ia = Ib and therefore the neutral current drops to zero. In this 3-wire system, the neutral current can never exceed A or B, whichever is the greater.

Once you can picture what is going on with this simpler system, you can then make the step up to 3-phase where it is more complex. With a 3-phase system, when the same current flows on A, B, and C, the neutral current drops to zero.

 

[plugwash]

please note that what paul_c has said assumes the loads are linear.

If a load draws most of its current on the third harmonic then the neutral current will indeed be the sum of the phase currents.

lukilly in practice harmonic currents aren't usually very big. But it could be a concern when running certain types of electronic equipment that draws a lot of current at the top of the cycle.

 

[Paul_C]

Agreed PG. I was trying to keep the explanation as simple as possible.

 

[Lectrician]

I think Paul made a great effort at explaining with that.

It is a very common mis-conception that the the phases add together and make the neutral.

I have demonstrated to all my apprentices the following...

3 100watt lamps connected in the star config. (ie, R, Y, B to a lamp each, all neutrals together to the neutral).

urrent will be equal, no neutral current. Disconnect the neutral - the lamps stay lit!

You will find that ( due to lamp intollerance ), the lamps will alter in brightness slightly, as one recieves a little more volts than it should, and another a little less. This is due to the lamps not being 100% perfect in manufactur (maybe 98watt, or 101watt etc!)

 

[andy]

I have demonstrated to all my apprentices the following...

3 100watt lamps connected in the star config. (ie, R, Y, B to a lamp each, all neutrals together to the neutral).

urrent will be equal, no neutral current. Disconnect the neutral - the lamps stay lit!

You will find that ( due to lamp intollerance ), the lamps will alter in brightness slightly, as one recieves a little more volts than it should, and another a little less. This is due to the lamps not being 100% perfect in manufactur (maybe 98watt, or 101watt etc!)

what would happen on the neutral is you had 2 bulbs, one on red and one on yellow? would there be no current on the neutral or some current?

 

[plugwash]

same magnitude as with a bulb between one phase and neutral.