The Star Point

[La Poste]

Greetings.

I am hoping someone could help me understand the star point in a three phase star configured transformer.
It's something that is mentioned a lot in electrical discussions.

I know the star point is attached to the ground and held at 0 Volts.
I also know that the star point is always at a difference of 230 Volts RMS from the line conductor but I am not sure why.

The star point is right in the middle of the three lines that are 120 degrees out of phase in a star wound transformer.
Does the fact it is right in the middle account for the reason it is always at 0 volts, is it some sort of balancing act that is done by the three phases that results in it being at 0 volts?

My knowledge is very patchy on this subject, I mean I have seen the diagrams, I learned to do the calculations at college but that was a long time ago.

I also know that when electricity is flowing in the line it is always trying to get back to the star point or does the return part of the electrical current actually cross the star point and then flow in one of the other phases sort of treating the star point like a cross roads for traffic between the three phases?

So could someone just explain what exactly is the star point and why it is at 0 volts.

I sure would appreciate it.

Thanks very much.

 

[plugsandsparks]

Will try and answer your questions in the quote body. Not really sure what you are struggling with but here goes

Greetings.

I am hoping someone could help me understand the star point in a three phase star configured transformer.
It's something that is mentioned a lot in electrical discussions.

I know the star point is attached to the ground and held at 0 Volts.
I also know that the star point is always at a difference of 230 Volts RMS from the line conductor but I am not sure why. It is specifically designed that way to get the correct voltage, remember this is the secondary windings of which the primary can be 11000V

The star point is right in the middle of the three lines that are 120 degrees out of phase in a star wound transformer.
Does the fact it is right in the middle account for the reason it is always at 0 volts, is it some sort of balancing act that is done by the three phases that results in it being at 0 volts? It is at Zero because it is bonded to earth which is our reference point for 0 volts. Not sure why you say it is in the middle, i may be just the middle of the drawing you are looking at. The phase angle are created by the generators, not the transformer. the transformer will transform whatever gets presented to it. (as long as its AC)

My knowledge is very patchy on this subject, I mean I have seen the diagrams, I learned to do the calculations at college but that was a long time ago.

I also know that when electricity is flowing in the line it is always trying to get back to the star point or does the return part of the electrical current actually cross the star point and then flow in one of the other phases sort of treating the star point like a cross roads for traffic between the three phases? The route that is taken by the current really depends how the load is connected. Single phase loads use Neutral, i.e. the star point to get the 230V it needs, hence the current is traveling up and down one Phase Line and neutral or star point. Three phase loads do not require the star point to pull current because they can use the voltage difference across the Phase lines - so in this case the current is running up and down the different phase line conductors and the star point is largely un-used.
This is a bit general as there are always exceptions but this hopefully will give you the gist

So could someone just explain what exactly is the star point and why it is at 0 volts.

I sure would appreciate it.

Thanks very much.

 

[Silly Sausage]

Some scans from Hughes Electrical Technology, good book.

 

[La Poste]

Thanks.

I think I'm going to read those books Marvo recommended.

What I was wondering is if for instance we have a single phase supply from one of the star windings let's call it L1.
This 230V supply goes to a house then returns back through the neutral to the star point of the transformer then through the windings of L1 and then back along the line L1 to the house again (I'm imagining the current going one way just to keep things simple)

Now is that all there is to it or does the current go through the windings of L1, supply the house return via the neutral and then at the star point a little bit of current goes into L2 and bit into L3, and most into L1.

The star point to me seems to be a place where three currents from three different phases mix and I am wondering of the simple idea of current simply going along one line say L1 and back through the neutral to again return along L1 is a bit of a simplified approach.

I don't go out of my way to confuse myself but am just wondering if there is more to it.

 

[Marvo]

A 3-phase transformer has three primary windings and 3 secondary windings. The secondary windings can be connected in star of delta configuration depending what voltage you want from them. If you configure them in star then the center point can be used as a neutral and this point may or may not be earthed depending on the application.

The load can be connected across two or three of the phases or it could be connected across one of the phases and the center point of the star if it's a single phase load.

 

[yellowvanman]

Kirkhoffs law state that the sum of current at a node = 0, so with the star point of a 3 phase system current going to the star point = current going away from the star point. So in effect the current from L1 will go back thro L2 and L3, there is only current in the neutral conductor if the phases are unbalanced.

If you look at a diagram with the sinusoidal waveforms of the 3 phases, you see how this is.

Take one point in time , say L1 = maximum, then look at L2 and L3 and you should see they each will be @ -0.5 of maximum, so 1 + ( 2* -0.5)=0! Obviously this would be a balanced system.

I hope that explains it in as few as words as possible!

 

[Knobhead]

I’ve got some drawings of the two main types of distribution transformers (DYn11 and DYn1) and the vector diagrams, earthing arrangements, etc. Also the effect of an unearthed neutral, which can be quite devastating. It blew every PC in an office where I worked.
Give me an hour to find them in my wonderful filing system, it’s a total shambles!

 

[i=p/u]

tongues out:biggrin:

 

[brman]

It is probably all been said but I find it helps to treat all the phases as separate circuits and visualise how all the circuits add up.

So if you have say 10A going out of L1 that same 10A will return through the neutral to the star point and hence back to the L1 winding.
You might also have say 5A doing the L2->neutral route.
So Neutral then has the vector sum of the 10A from L1 and the 5A from L2. Vector rather than arithmetic sum because the two currents are of course not in phase with each other.

If you have a load from L1 to L2 (for example) then the current loop is out from the L1 winding, through the load, back through the L2 winding to the star point and back through the L1 winding. This current isn't going down the neutral but it is still going through the star point to get back to its origin.

As has already been said, the fact that the star is at 0V is purely because that is the point that has been connected to earth/neutral.

I think that might be how you are thinking anyway and yes, it really is that simple as long as you are happy doing vector sums to find the magnitude/phase of the currents.

 

[Knobhead]

So much for my filing system!

To start the two types of main distribution transformer. I will add I’ve only come across a DYn1 once in 40 years and that was due to a cock up when we bought a 2[SUP]nd[/SUP] hand transformer. (No wonder it was cheap)! Nearly every transformer will be DYn11.

View attachment 11596
View attachment 11597

If the neutral earth link is removed things happen.
With a balanced load you will never notice.

View attachment 11598

Un balanced and the fun and games start.

View attachment 11599

As I said I’ve had bad experiences with this.

From another board I’m on


One to add to the experience pile. Got a call to a plant, all the office equipment had failed. On arrival I tested the supply to the office and all seemed OK. Called the lads from office services to come and sort their junk out. (It was junk, it's a wonder you didn't need to shovel coal in to the computers). All power supplies smoked! I'm puzzled now, how could 4 go wrong together. Got talking to the plant manager and he tells me that the lighting had "gone weird" for a while. Went back and checked the Ph -N voltages again and all OK.
It wasn't until lunchtime when I got talking to some of the other electricians that things became clear. Two of them had installed a 150mm 3½C cable in to the plant main switch board that morning and had disconnected the neutral - earth cables to make it easier to get their cable in, when finished they reconnected them. To quote one of them "I thought there was a bit of a spark!" What galled me was I'd fitted an engraved warning label to the earth bar when I installed it!

 

[ipf]

Good stuff Tony, brings back memories. We'll never know why idiots ignore warnings, cause problems and, in most cases, let lucky.

 

[La Poste]

Great stuff.

When the star point is connected to the ground via an earth cable is there much current flowing through the cable to the ground?

If you were to measure the potential gradient around where the cable enters the soil would there be much of a volt drop the further you moved away from the cable?

I could walk up the road with my clamp meter and clamp around the earth cable that goes from our local pole mounted TX into the ground but I don't know if this is a good idea.

I have seen cows lying around on the ground right where the earth cable enters the soil so there can't be much current going through it.

 

[brman]

you have to think about current loops. Current doesn't go anywhere without coming back to the origin.

So for current to flow through the earth cable there must be a path for it to return from earth to one of the live conductors. Unless you have an earth fault then this will not happen.

As someone has said, all the currents at any particular node equal zero. This just means that all the current going out through a transformer winding will return to the same point via one of the live conductors (either lines or neutral depending on the load configuration). The earth isn't involved unless there is actually a fault path via earth.

This a bit simplified as you can get earth currents for reasons not directly relating to load currents but that is probably another discussion.

 

[ackbarthestar]

Three phase generating technology was established as a compromised between cost, complexity, and max power transfer. There were experiments with 6, 9 and 12 phase systems....

So in a three phase Delta system the sum of the instantaneous currents will be zero, In a Star system it is also zero, but now includes the neutral.

If single phase circuits are used and are unbalanced then there will be significantly more current flowing through the neutral.

The star point is formed by the linking of the transformer's coils to allow for the use of single phase circuits and appliances.

 

[La Poste]

I guess the next logical question will be what would happen if you removed the earth connection from the star point, what voltage would the star point rise to?

I have had a look at Tony's vectors but unfortunately they are a bit beyond me, I will have to do some further reading to understand them.

But let's say you were standing on the ground underneath the transformer and you thought it would be a good idea to cut through the earth cable ( I am no copper thief by the way), would you receive an electric shock from touching one end of this cable with one hand (the end still attached to the star point) whilst you were standing on the ground, what sort of voltage would you be subject too?

Thanks.

 

[ackbarthestar]

Theoretically, with a perfectly balanced load/loading it would be equal to the sum of the instantaneous currents which would be zero.
It is very unlikely that this would be the case since it would rise and fall according to the imbalances pulled by all the various single/three phase loads.

 

[SirKit Breaker]

I’ve got some drawings of the two main types of distribution transformers (DYn11 and DYn1) and the vector diagrams, earthing arrangements, etc. Also the effect of an unearthed neutral, which can be quite devastating. It blew every PC in an office where I worked.
Give me an hour to find them in my wonderful filing system, it’s a total shambles!

Tony is the man, i will keep my gob shut, so as to not complicate it.:wink_smile:

 

[La Poste]

I am thinking if the earth connection is removed from the star point the voltage at the star point will rise above 0 volts, I also think the line voltage will be 230 volts with respect to the star point but this 230 volts may be higher with respect to the ground.

I am not sure how much the star point voltage will float about though.

 

[Knobhead]

The amount the voltage floats depends on the current being drawn. See the vector diagrams in my post.