Last Week of October Question...

[Silly Sausage]

400V 3P Star, NO Neutral.
Loads of 100, 120 & 150 Ohm, purely Resistive.
Star point of supply is Earthed at 0V.


What are...

Current through each Load.
Voltage across each Load.
Voltage to Earth of the Star point.




May add more sub Qs later when I've attempted the answer myself.


The only prize I can offer is Kudos!!!

 

[pennychew]

Is that an aftershave or something?

 

[Silly Sausage]

Is that an aftershave or something?

OK, cut the 'amusing' quips and knuckle down to the question!

 

[pennychew]

OK, cut the 'amusing' quips and knuckle down to the question!

Sorry mate, will do!

 

[pennychew]

Might be barking up completely the wrong tree here but is this what you mean?

Only touched on this a little last year so i know im not strong on it

 

[pennychew]

I even rotated that before i posted it to stop it ending up sideways!

 

[Silly Sausage]

No.

The supply Star point is Earthed (as normal), NOT the Load. Read the Q!

Clue...Floating Neutral

 

[pennychew]

I'm going to have to get back to the books on this one

 

[Simonslimline]

The phase voltages are going to vary due to the unbalanced loading. I need to refresh on this and have a good read first before trying to answer it though.

 

[Silly Sausage]

To be fair, it's not as trivial as it may appear!

Points will be awarded for researching the method to solve it and having a go at the Maths.

 

[happysteve]

I had a think about this after the pub last night, but I didn't get very far. I'd have a punt at the voltage to earth of the star point: I reckon it has a phase angle of 29.8 degrees (assuming 100 ohm load is on L1, and 120 ohm load is on L2). The voltage I'm less sure about: 66V? I'm too embarrassed to show my workings.

As a hint to everyone: is the order in which you ask for bits of information (current through each load, voltage across each load, voltage of star point) related to the order in which you solve the problem, or not?

 

[pennychew]

If Simon and Steve are struggling with this one it will be way beyond me! if you work it out put the workings up so i can get my head around it as well, cheers

 

[ElectroChem]

Ooh, time to look up the lesson on unbalanced star points.

Archy you're an evil man, making me dig into the pit of despair that is my pile of school notes.

 

[Bobster]

Christ, taken me back to working on bitumen heating tanks. The heating elements were all connected in this manner, caused loads of fun when elements failed and it became imbalanced.

 

[Bobster]

Christ, taken me back to working on bitumen heating tanks. The heating elements were all connected in this manner, caused loads of fun when elements failed and it became imbalanced.

I know there is a thread on here, regarding how it's worked out. I think Archy was involved.

That's my little nudge in the right direction.

 

[happysteve]

Well, I'm trying to post the answer to this, but I keep getting the anti-spam thing. Seems a bit harsh, in the trainee section. Maybe I'll try splitting it up (if it lets me post anything!)

 

[happysteve]

Part 1:

Right, y'bugger.

Instead of L1, L2, L3 let's call the 3 points a, b, c.
So Ra = 100, Rb = 120, Rc = 150.

Ia+Ib+Ic = 0 (Kirchoff's current law) (eqn 1).

Expressing the resistances as conductances (or, more generally, admittances):

Ya = 1/Ra, Yb = 1/Rb, Yc = 1/Rc.

Let "0" be the floating neutral point, so "Va0" is the voltage between Va and the floating neutral, i.e. the voltage across resistance Ra (or, to put it another way, across admittance Ya)

Ia = Va0 x Ya, Ib = Vb0 x Yb, Ic = Vc0 x Yc

So substituting those into the KCL equation:

(Va0 x Ya) + (Vb0 x Yb) + (Vc0 x Yc) = 0 (eqn 2)

Let "n" be the neutral point (if neutral was still connected to Earth).

Va0 = Van + Vn0
Vb0 = Vbn + Vn0
Vc0 = Vcn + Vn0 (equations 3a-3b)

Substituting these into (eqn 2):

(Van+Vn0)Ya + (Vbn+Vn0)Yb + (Vcn+Vn0)Yc = 0

Therefore:

V0n = ((Van*Ya) + (Vbn*Yb) + (Vcn*Yc)) / (Ya + Yb + Yc) (equation 4)

(... to be continued...)

 

[happysteve]

But we have to be careful, because we have to take phase into account. I do this using complex numbers, but you can think of this as

resolving your voltage plus phase angle into "x" (the "real part") and "y" (the "imaginary part")

 

[happysteve]

Let's define "a" as zero phase. So Van is easy:

Van = 230

In complex form, Vbn = -115 + 199j

 

[happysteve]

Sorry, won't let me post any more.

 

[happysteve]

Oh, apparently it will let me post inanities, but not maths. Great.

 

[happysteve]

I'll resort to taking a screen shot then:


Sorry about that. Best I could do.

 

[happysteve]

Practical application of all this: If you find yourself the victim of a floating neutral (as I once did... well, one with a high resistance to Earth), put the kettle on (and any other loads you can). You will reduce your voltage so there's less chance of stuff blowing up or a fire... but your neighbours won't be so lucky.

 

[pennychew]

Damn! i was just about to post all that and you beat me to it!

Seriously though, top work fella :coolgleamA:

 

[Silly Sausage]

I knew you'd be first to answer this Steve!
I was considering barring you.

Haven't checked your answer yet, I still need to work it myself, got 7 days left.

What was that battery thread, I've got to own up on that one!

 

[happysteve]

I knew you'd be first to answer this Steve!
I was considering barring you.

Haven't checked your answer yet, I still need to work it myself, got 7 days left.

What was that battery thread, I've got to own up on that one!

I don't tend to pipe up unless it's a question I struggle with myself. There is a lot to be gained by struggling on and working stuff out for yourself, so if I reckon it's a fair question I'll generally stay out of it (unless no-one's volunteered an answer for ages).

But this was on another level. That's not a Level 3 question (well, not any more, dunno if it used to be taught). I'd be interested to hear from our overseas trainees what they thought of that!

And the battery thread is this one: Tonight. I noticed Tony went quiet on that one, too...:angel_smile:

Cheers for posting it up... I certainly learnt summat! I also learnt about wye-delta and delta-wye transformations in my research (though it turned out ultimately to not be useful for this question). And it's good to be able to put numbers in.

 

[Baker1988]

i honestly do not remember doing this in college i could be wrong but i do not remember ant of that

 

[Simonslimline]

If this was taught it would come under unit 309 on the 2357 C&G. We never did anything regarding floating neutrals. There was a question on the 309 written exam asking for the neutral current of unbalanced loads on a 3 phase 4 wire system.

 

[happysteve]

I didn't come across it on my 2365 unit 305, either (like Simon, we also did neutral currents in unbalanced star).

I looked back through my old uni notes, I found this from the first year (Level 4):


"Calculation of the line currents can be done using mesh analysis but we shall not enter that here."
Couldn't find it in my Level 5 notes either.

Saying all that, it's an interesting problem! Summat that looks so simple, a problem that's so easily stated and understood, yet takes a page of maths to sort out.

So cheers