309 written exam. Neutral current in an unbalanced 3 phase system.

[Simonslimline]

Had the 309 science and principles exam today at college and had a question on calculating the neutral current flow in a 3 phase 4 wire system with unbalanced loads. The formula is below, and definitely one to memorise.

N=√((A²+B²+C²)-((A×B)+(A×C)+(B×C)))

 

[happysteve]

Cheers for that Nice handy formula. Will probably stick to:

x = A cos 0 + B cos 120 + C cos 240 (or, x = A - 0.5B - 0.5C)
y = A sin 0 + B sin 120 + C sin 240 (or, y = √0.75B - √0.75C)
N = √(x²+y²)

... as I can picture in me head where it comes from.

You way's neater, though!

Got 2365 302 (which I think is pretty much the same as 2357 309?) in a few weeks' time. It's been a looooong unit....

 

[Sintra]

Had the 309 science and principles exam today at college and had a question on calculating the neutral current flow in a 3 phase 4 wire system with unbalanced loads. The formula is below, and definitely one to memorise.

N=√(A²+B²+C²) - (A×B+A×C+B×C)

You need a few more sets of brackets in that formula to make it work.

N=√((A²+B²+C²)-(A×B)+(A×C)+(B×C)))

Try working it out both ways with made up values and you'll see what I mean.

 

[happysteve]

You need a few more sets of brackets in that formula to make it work.

N=√((A²+B²+C²)-(A×B)+(A×C)+(B×C))

Try working it out both ways with made up values and you'll see what I mean.

One more set of brackets for ya:

N=√((A²+B²+C²)-((A×B)+(A×C)+(B×C)))

... otherwise you end up adding the (AxC) and the (BxC); they need to be subtracted from the squared terms just like the (AxB).

Between us, we'll get there eventually!

 

[Simonslimline]

Cool thanks for the corrections on that. I wrote it down on paper in the exam so the answer was ok. I will see what answers i get with the calculator using the methods you have both posted.

 

[Sintra]

Yes Steve that's correct was still sleeping when I put the brackets in lol. Going to edit now as per my notes.
In=√((Ia²+Ib²+Ic²)-((Ia*Ib)+(Ia*Ic)+(Ib*Ic)))

 

[Simonslimline]

Ok so i tried it on my calculator. What i found was the sum is the same regardless of having any of the 3 brackets on the end or not, but If you remove any of the others the value changes.

 

[ElectroChem]

Applying this formula to the neutral current calculations from my AC theory class today suggests that it only applies on resistive loads, have you covered reactive loads and vector addition yet Simonslimline? Happysteve's variant could be made to work, but only if you changed the Sin/Cos 120/240 to the correct phase angle of the load.

 

[Simonslimline]

I took the formula from the thread below a while back. I am going to do some reading about what you have posted, Nice 1



http://www.electriciansforums.co.uk/commercial-industrial-electrical/39855-electrical-theory.html

Applying this formula to the neutral current calculations from my AC theory class today suggests that it only applies on resistive loads, have you covered reactive loads and vector addition yet Simonslimline? Happysteve's variant could be made to work, but only if you changed the Sin/Cos 120/240 to the correct phase angle of the load.

 

[Simonslimline]

The thread below is from another forum and goes for 15pages discussing formulae for neutral. It is an interesting read if you have the time


Unbalance neutral current in a 3 phase system

 

[Baker1988]

i got a question about the neutral current in my 309 exam 2 weeks ago but we wasn't allowed to do it mathematically it asked us to do it graphically so we had to draw the vector diagram and that but i thought it was a easy question really haha

 

[Ste69]

simple way we have been taught, dont know how to put math symbols up so ill do my best.

3 phase and neutral, Ph1 50, Ph2 40, Ph3 10 and N ?

Take the smallest number away from the other numbers, so 50-10 and 40-10, gives you 40 and 30 respectively.

N= square root of 40 squared-30 squared so N= square root of 1600-900 so N= square root of 700 so N=26.5

sorry about how its written out, hope you can follow it

 

[Simonslimline]

I make that 36.055A Neutral current. I did it this way

N=√((A²+B²+C²)-((A×B)+(A×C)+(B×C)))

 

[Ste69]

N=√((A²+B²+C²)-((A×B)+(A×C)+(B×C)))

yep, i get 36 when i do it that way aswell!........i never thought to check against what the tutor told us.

bit ****ed off, i have my 309 written exam on tuesday and no more classes to clarify this.

cheers

 

[Simonslimline]

Try this link for typing maths symbols then copy and paste it in mate. Type mathematical symbols - online keyboard

Don't sweat the 309 too much it was not as bad i thought it was going to be. Still worried over it though to be honest.. That formulae is worth remembering though for sure.