Calculating Neutral current in unbalanced TP

[Simonslimline]

I have been looking at this today and found it interesting so thought i would post a question for any of you to answer.

Calculate the neutral current on a lv 3 phase system.

L1 = 30Amps

L2 = 60Amps

L3 = 22Amps.

 

[telectrix]

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

and if the answer is not 42, then i'm out. there's beer to be drunk. but a head guess is 24amps odd.

 

[Simonslimline]

My calculation came up with 34.7 amps. hopefully i am doing it right :smiley2:

 

[Silly Sausage]

It's when you lose the neutral it starts getting interesting!

 

[D Skelton]

The answer is 34.7A you are correct.

 

[Simonslimline]

Could you explain this a bit more please. Do you mean where a neutral is not required?

It's when you lose the neutral it starts getting interesting!

 

[telectrix]

34.69870315. so i'll give you 34.7. LOL.

 

[telectrix]

a rough guess ( and i mean rough) to check if you've don the calc. correctly ( i.e. not got way out) is to add the 3 line currents and divide by 3.

 

[Simonslimline]

I first tried the calculation using the same value for each phase to make sure my answer came back as zero.

 

[D Skelton]

Or a little cheat I like is to use a line measurement representing L1 in cm, then draw a line representing L2 from the end point of line 1 at 120 degrees from line 1, then a third line representing L3 from the end point of line 2 at 120 again. The distance between the end point of line 3 and the start point of line 1 will represent the neutral current. It will look something like this:



p.s. Don't let your college lecturers see you doing this! lol

 

[telectrix]

Or a little cheat I like is to use a line measurement representing L1 in cm, then draw a line representing L2 from the end point of line 1 at 120 degrees from line 1, then a third line representing L3 from the end point of line 2 at 120 again. The distance between the end point of line 3 and the start point of line 1 will represent the neutral current. It will look something like this:

View attachment 21606

p.s. Don't let your college lecturers see you doing this! lol

if you did it in inches, would the neutral current be 2.54 x as large? :rofl:

 

[mr7526]

Could you explain this a bit more please. Do you mean where a neutral is not required?

He means in fault conditions where the neutral becomes disconnected. What tends to happen is that the neutral point != 0v.

 

[Silly Sausage]

Could you explain this a bit more please. Do you mean where a neutral is not required?

As the previous poster hinted, things go haywire!!!
On an unbalanced star, if you lose the neutral, depending on the 3 loads, the star point wanders away from zero for starters!
I've got a link to an excellent book which will explain all.
You'll have to remind later though, too tired at the moment.

 

[Simonslimline]

This in an interesting read.Impact of Floating Neutral in Power Distribution | Electrical Notes & Articles

 

[Silly Sausage]

Ey up Simon, this is the book...
Alternating Current Machines - R. K. Rajput - Google Books

You used to be able to read it on line, but the barstewards seem to have taken it down. :-(
Going to have to buy a copy now!

 

[Simonslimline]

Cool thanks for the link, ill be sure to buy a copy, watersones have it and a family member can get me a 50% discount

Ey up Simon, this is the book...
Alternating Current Machines - R. K. Rajput - Google Books

You used to be able to read it on line, but the barstewards seem to have taken it down. :-(
Going to have to buy a copy now!

 

[Silly Sausage]

Cool thanks for the link, ill be sure to buy a copy, watersones have it and a family member can get me a 50% discount

HANDY!!!
I might be getting in touch with you sometime soon.

Better warn you, it is quite heavy/theoretical, but has loads of worked examples, which is always good.