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D

dave-gardiner

Hi
we have a 3 phase 63A isolator.
i have been asked to supply from this 3 x single phase 63A sockets. I have said this cannot be done because the neutral is only sized for the 63A supply. But now i am questioning myself. Help please
 
sorry must disagree, as the op states he wants 3 x 63A Single phase supplies. Fully loaded you can have a current of 180 Amps flowing down the neutral. 3 phase machines are different, because any unbalanced currents just flow down the neutral.


If you mean the supply neutral then providing the three phases are balanced there will actually be ZERO amps

If the phases are imbalanced the max in the supply neutral will be 63A

If you are adding the phase amperages to determine supply neutral size then you are wasting a lot of copper - don't you care about the Earth's resources?? :)


This is 236 pt 2 or NVQ level 3 stuff


Remember that 3 single phase loads that are balanced are exactly the same as a 3 phase load wired in star
 
Last edited by a moderator:
I wanted to expand on effect of harmonics and why the neutral carries three times the zero phase sequence current but I've been laughing my head off since page 2 and post 43 flipped me completely. I am now incapable of any further rational thought today.
 
I wanted to expand on effect of harmonics and why the neutral carries three times the zero phase sequence current but I've been laughing my head off since page 2 and post 43 flipped me completely. I am now incapable of any further rational thought today.


It's widely googleable (is that a word?) - even TLC have a few words on it:

http://www.tlc-direct.co.uk/Book/4.3.12.htm
 
Here is how to work out the current in the neutral, thanks to tel

SQRT I²A + I²B + I²C - (IA x IB) - (IB x IC) - (IC x IA)

That would be: SQRT ( (IA² + IB² + IC²) - ( (IA x IB) + (IB x IC) + (IC x IA) ) )

I'm guessing Tel just forgot to press his shift key :)
 
That's the same, you've snuck in some more parentheses. A-(B+C+D) = A-B-C-D.
Although, there were some missing from around the whole thing the first time, for the argument of sqrt.
 
but if i'd put the brackets in properly....

SQRT ( I²A + I²B + I²C - (IA x IB) - (IB x IC) - (IC x IA)) it would be the same. n'est pas?
 
mmmm - Bishop's Finger....

...nice, but what a whammer headache in the morning lol

you must down a lot of it then. i've never had a hangover on less than 9 pints. i only ever get hangovers with lager, wine and spirits.
 

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