Quick three phase question

[La Poste]

Greetings.

I apologise if this is a bit of a daft question but I don't know much about three phase.

In a three phase system there is only current in the neutral if there is an imbalance of the phases right?

So looking at the transformer that supplies our street it is a three phase transformer that supplies various houses on different phases.

Each house has a line and neutral so does the above principle mean that the neutral cable in my own house will be carrying less than the phase?

If the current in the neutral cable is a result of an imbalance between the phases then does this principle apply to houses that are supplied from the same three phase transformer?

Any opinions appreciated.

Thanks.

 

[JUD]

Neutral curent is given by:

In=√(Ia²+Ib²+Ic²)-((Ia*Ib)+(Ia*Ic)+(Ib*Ic))

I think I get that...

In a perfectly balanced load (Ia² + Ib² + Ic²) will be equal to ((Ia x Ib) + (Ia x Ic) + (Ib x Ic)) so they cancel each other out.

So if you had 10A on each phase you'd get...

In = √(10² + 10² + 10²) - ((10 x 10) + (10 x 10) + (10 x 10))
In = √(100 + 100 + 100) - (100 + 100 + 100)
In = √300 - 300
In = √0
In = 0A

but if Ia was 20A for example you'd get...

In = √(20² + 10² + 10²) - ((20 x 10) + (20 x 10) + (10 x 10))
In = √(400 + 100 + 100) - (200 + 200 + 100)
In = √600 - 500
In = √100
In = 10A

Is that about right??

 

[Knobhead]

Spot on hone:

 

[snowhead]

Back to simplicity;

3 phase / neutral cable down the street.
Each house has only a 3kw load switched on.
Live and Neutral current at each house is equal (give or take a harmonica)

Last house is Red, no not paint, L1.
Neutral in the 3ph cable= same current as L1.

Next house back is Yellow, your lucky I didn't say white, L2
Neutral in the 3ph cable = less than the total of the L1 + L2 so far.

Next back is Blue, lost for words, L3
Neutral current in the 3 phase cable = 0

 

[La Poste]

I think I see it.

Within each house the line and neutral currents are the same but when the neutral cable hits the main drag (the three phase neutral cable) this is when things start to cancel out.

And the overall current within the three phase neutral cable will be equal to Tony's equation above.

 

[Moses]

Neutral curent is given by:

In=√(Ia²+Ib²+Ic²)-((Ia*Ib)+(Ia*Ic)+(Ib*Ic))

I think the brackets are in the wrong place.

should be

In=√((Ia²+Ib²+Ic²)-(Ia*Ib)-(Ia*Ic)-(Ib*Ic))


would someone pls confirm?

 

[Silly Sausage]

Simplistic as it might sound, draw a simple line diagram, think where the currents are flowing, apply Kirchoff's Law, and ALL will be revealed! Await that moment of enlightenment.

You'll be (and I have been many a time) amazed how relatively simple it is in the end.

 

[Knobhead]

I think you should retake maths.
If you want to be pedantic:

In=√((Ia²+Ib²+Ic²)-((Ia*Ib)+(Ia*Ic)+(Ib*Ic)))

Which is how it has to written in Excel!

 

[Moses]

I think you should retake maths.
If you want to be pedantic:

In=√((Ia²+Ib²+Ic²)-((Ia*Ib)+(Ia*Ic)+(Ib*Ic)))

Which is how it has to written in Excel!

Or

In=√((Ia²+Ib²+Ic²)-(Ia*Ib)-(Ia*Ic)-(Ib*Ic))

Tony, not being pedantic.

As you know a 0 makes abig difference to the answer, same as where you put a bracket, or a decimal point or a SQrt sign.

 

[topquark]

I think you should retake maths.
If you want to be pedantic:

In=√((Ia²+Ib²+Ic²)-((Ia*Ib)+(Ia*Ic)+(Ib*Ic)))

Which is how it has to written in Excel!

Why does he need to retake maths? He's quite right!

 

[Darkwood]

Or

In=√((Ia²+Ib²+Ic²)-(Ia*Ib)-(Ia*Ic)-(Ib*Ic))

Tony, not being pedantic.

As you know a 0 makes abig difference to the answer, same as where you put a bracket, or a decimal point or a SQrt sign.

They are both correct, its just 2 different ways of coming to the same result, admittedly Tony's excel version is more complex on the eye but still correct.

 

[Moses]

They are both correct, its just 2 different ways of coming to the same result, admittedly Tony's excel version is more complex on the eye but still correct.

Tony's original post is incorrect.

Thats why I raised it.

He has subsequently corrected the formulae in the last post.

 

[brucelee]

Neutral curent is given by:

In=√(Ia²+Ib²+Ic²)-((Ia*Ib)+(Ia*Ic)+(Ib*Ic))

lol I learned that equation recently

 

[Darkwood]

Tony's original post is incorrect.

Thats why I raised it.

He has subsequently corrected the formulae in the last post.

Yes missed that post thought they all were same... missing outer brackets thus give Sqr root on first bracket result only as appose to total bracket equation.

Not like Tony though i must say but he has corrected it :thinking2:

 

[Moses]

Originally Posted by Tony
Neutral curent is given by:

In=√(Ia²+Ib²+Ic²)-((Ia*Ib)+(Ia*Ic)+(Ib*Ic))



lol I learned that equation recently



Bruce, thats the wrong formulae.

Originally Posted by Moses
Or

In=√((Ia²+Ib²+Ic²)-(Ia*Ib)-(Ia*Ic)-(Ib*Ic))

Tony, not being pedantic.

As you know a 0 makes abig difference to the answer, same as where you put a bracket, or a decimal point or a SQrt sign.



This the correct one.

 

[i=p/u]

just type in the digits and out comes the neutral current to the excel version, mind you i might get moses to check you excel pages tony

only joking:indian_chief: