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.

 

[brucelee]

Hi mate didnt read the whole formulae just recognised it as the one for working Neutral curent in a 3 phase 4 wire system

Iread tonys post didnt check replied then when scrolling down realised
as TONY USUALLY CORRECT 100% lol suppose no one is perfect hehe
dont worry have correct formulae in me notes so would always check first
or draw a phasor diagram lol

 

[Knobhead]

In Excel, Moses's version would not work.

Using Jud's figures of 20 10 10 I get 10A neutral, your version Moses it comes up with 30A!

I must be one of those crazy people that likes Excel.

 

[i=p/u]

you are , you are, you are

 

[telectrix]

Hi mate didnt read the whole formulae just recognised it as the one for working Neutral curent in a 3 phase 4 wire system

Iread tonys post didnt check replied then when scrolling down realised
as TONY USUALLY CORRECT 100% lol suppose no one is perfect hehe
dont worry have correct formulae in me notes so would always check first
or draw a phasor diagram lol

so now star trek comes into the equation!

 

[topquark]

When drawn correctly with the square root symbols horizontal line over the top of the whole equations both versions will work as expected:

 

[Knobhead]

The minus's in the 2nd half should be plus

Lets drop this now

 

[topquark]

Not the way it's expressed it shouldn't! Dropped.

 

[JUD]

When drawn correctly with the square root symbols horizontal line over the top of the whole equations both versions will work as expected:

This is how I interpreted it which is why I broke the calculations down...

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??

Try this...

_____________________________________________
√(Ia² + Ib² + Ic²) - ((Ia x Ib) + (Ia x Ic) + (Ib x Ic))

 

[Silly Sausage]

Forget all this nonsense, convert your currents into complex notation and simply add them together.

 

[Moses]

lol I learned that equation recently

The minus's in the 2nd half should be plus

Lets drop this now

You originally posted a formulea with the brackets in the wrong place, making the formulea invalid.

The formulae that I posted with minuses is quite correct,
its a different matter that you can't recognise a formulae when transposed,
and will work in excel correctly, not as you say in post 32, that it would not work.

You need to practice excel a bit more.

 

[Moses]

This is how I interpreted it which is why I broke the calculations down...




Try this...

_____________________________________________
√(Ia² + Ib² + Ic²) - ((Ia x Ib) + (Ia x Ic) + (Ib x Ic))

This formulae is correct,

but what was posted originally was

√(Ia² + Ib² + Ic²) - ((Ia x Ib) + (Ia x Ic) + (Ib x Ic))

Which is not correct and Which means

( SQRT(Ia² + Ib² + Ic²) ) - ((Ia x Ib) + (Ia x Ic) + (Ib x Ic))

 

[JUD]

This formulae is correct,

but what was posted originally was

√(Ia² + Ib² + Ic²) - ((Ia x Ib) + (Ia x Ic) + (Ib x Ic))

Which is not correct and Which means

( SQRT(Ia² + Ib² + Ic²) ) - ((Ia x Ib) + (Ia x Ic) + (Ib x Ic))

That all depends how you interpret it.

As I said, I interpreted it the correct way as I posted above (with the line across the top).

 

[Moses]

That all depends how you interpret it.

As I said, I interpreted it the correct way as I posted above (with the line across the top).

No it does not depend how 'one' interprets it. That way every body can interpret it the way they want!

It depends how the laws of maths interpret it.

Calculate brackets first,

Then multiply and divide

Then and functions, Square root, Sine, Cosine, Log, Power etc.

Last of all plus and minus.


Order of Operations

 

[JUD]

No it does not depend how 'one' interprets it. That way every body can interpret it the way they want!

This is not what I said.

I said that I interpreted it the way it should have been expressed (which is as you posted) because I understood the connection between the calculations on either side of the subtraction operator i.e. one cancels the other out (in a balanced load).

When entered into a calculator it would look like...

((20² + 10² + 10²) - ((20 x 10) + (20 x 10) + (10 x 10))) and then you would press the √ button which would give you the Square Root of the result of the calculation.

 

[Knobhead]

Go back to school!

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


Put your version of the formula in to Excel and 20A 10A 10A ends up with a neutral current of 470A+