Neutral load on 3 phase 4 wire

[lesmcw]

Hi Guys: Could someone check this please. I am an electrician and was asked this question by another electrician and replied as follows, but want to make sure it's correct.

What is the current on the neutral on a three wire branch circuit, that is connected to a three phase 4 wire service. The three wire branch circuit has 12 amps load on each leg.
What is the formula for calculating the vector sum of these two currents.

It would be a Common Conductor when connected to two phases since it is no longer a Neutral. It would only be a true “Neutral” if connected to all 3 phases or as single phase since it only carries the unbalanced current.

To find the neutral current on a wye system the here is the formula:
(I dont have a square root symbol so the V and line are them)
______________________________________
\/ (AI^2 + BI^2 + CI^2) - (AxB)+(BxC)+(CxA)

written out, that would be: The square root of
(A phase squared + B phase squared + c phase squared) - (A phase x B phase + B phase x C phase + C phase x A phase)

Lets now give all three phases your value of 12 amps and find the Neutral load from the formula.

(12x12 + 12x12 + 12x12) = 432 - (12x12 + 12x12 + 12x12) = 432

432 – 432 = 0. The square root of 0 is 0. Therefore the identified conductor is actually a neutral.

With only 2 phases and a common (single phase) used, then:
(12x12 + 12x12 + 0x0 = 288) - (12x12 + 0x0 + 0x0 = 144) = 144

The square root of 144 is 12, so you see the Grounded Conductor carries the same current as the two phase conductors, and your answer is 12 amps.

 

[jghrist]

The current in phase B is the opposite direction from the current in phase A, so if you use -12 for B in your formula, you get B²=432 or B=20.8. Vector addition yields the same value. Assume A=12@0°, B=-12@-120° (or 12@60°). Then the real part (at 0°) 12 for A and 12·cos(60°)=6 for B, for a total of 18. The imaginary part (at 90°) is 0 for A and 12·sin(60°)=10.4 for B, for a total of 10.4. Magnitude is sqrt(18²+10.4²)=20.8.

 

[jghrist]

Forget about my dumb previous posting. B is not -12@-120°. It is 12@180°. The vector sum is zero and the neutral current is zero.

 

[tinfoil]

lesmcw:

Your formula needs a set of parentheses around the (a*b)+(b*c)+(c*a) part.

The formula you used is good for determining the MAGNITUDE of the resisidual or neutral current, and I expect you do not care about the resultant angle. One caveat is that this formula only holds true of the currents are EXACTLY 120° apart, but this is generally close enough in reality to not matter much .

If the magnitudes of your phase currents are exactly the same (and are exactly 120° apart), then a three-phase four wire (3 hots, one neutral) system should have zero return on neutral.

If two phase currents on a two-phase three wire system are exactly even (2 hots, one return), then the neutral return current WILL have the same magnitude as either phase current. This is easy to confirm if you use exact formulae or a vector diagram.


I am curious about your application for this, as outside of distribution utility practices, I have never encountered the use of two-phase three-wire systems?

 

[busbar]

 
If there are no phase-to phase loads, the answer is 12 amperes. It’s sort of an old “trick” question.

 

[buzzp]

Correct me if I am wrong but is not 3 wire service a hot, neutral, and ground? If so, obviously the neutral is carrying the return current so it would be 12A.

 

[lesmcw]

I think what he was looking for is this
Please excuse the poor drawing, as i'm not much of an artist.

http://www.bcelectrician.com/uglyyellow.html

 

[aolalde]

NO. The current in the neutral is not 12 amperes when a single phase is applied to two wires. Per Kirchoff’s current law; “the algebraic sum of all current entering or leaving a node of a circuit is zero”.
Then; Ia + Ib + In = 0
For the single phase condition Ia = 12 and Ib= -12 (as jghrist said, they are 180° out of phase).
Then; 42-42+In=0 => In=0

For a three phase balanced circuit:

Ia+Ib+Ic+In = 0

but Ia+Ib+Ic = 0; then In=0 ( balanced circuit)

For a three phase unbalanced circuit Ia+Ib+Ic >or< 0

and In = Ia+Ib+Ic (note that these are vectorial quantities)

 

[aolalde]

The above formula should be:

Then; 12-12+In=0 => In=0

 

[skiier]

Tinfoil's is right on the money.
"Neutrals" and "commons" get mixed up all the time.
Both neutrals and common connections serve two points. To facilitate proper polyphase or single phase voltage to a load and to serve as a return point for any unbalanced currents of the loads connected.
Neutrals and commons can exist in polyphase and single phase systems. Generally the only place you are going to find a "true" neutral in the real world is at an electrical distribution panel. Where else are you going to find this?
A nuetral exists in a group of wires, be it 3ph or 1ph, when the potential for the current flow in the neutral can be zero. I don't know how else to explain it.
In a single phase 3w system - derived from a "true" single phase source - 2hots and wht make a nuetral, 1 hot and wht makes a common.
In a single phase 3w system - derived from a 3 phase source such as 2 legs of wye connected transformer where the two voltages are 120 out of phase - 2hots and wht make a common, 1 hot and wht makes a common.
You get the idea yet?

 

[rbulsara]

alodale:

With all due respect..........

The two currents are not 180 deg aprart but 120, as they are part of a 3 phase , 4-wire system. For 12A in only 2 of the phases: the vector sum is

12/_0 + 12 /_120 + 0/_240 =
= 12/_0 + 12 /_120 + 0 = 12/_60, and takig mangnitude only its 12A in In.

The kirchoff's law is also not violated, as at any given 'instant' the sum of instantaneous values of current in two phases "and" N is zero. No different than when there is a balanced load of 12A in all 3 phases and N results in zero N current. By your own post the kirchoff's law will in lot of trouble, trying to explain 12+12+12=0, if you simplify 2 phase loads to 12-12+In=0


Lets look at it otherway: Draw a picture of a generator or tranformer secondary winding of a 208Y/120 source. Ground the source N. Draw a load of 20 ohms. Feed it with two of the phases from the Y source. Do NOT conncet the mid point of the R to source N. In this case the curernt through the 20ohm R will be 208/20 = 10.4A very much in phase with 208V (phase to phase voltagye). No current through the neutral as its not even connected. If you look closely the source is behaving as 3 phase "Open delta" or V connected transformers.

Now connect the midpoint of the R to the source N. And voila, the current through the R will increase to 12A (120V/10 ohm). Becuase now each half of R is seeing 120V , L-N voltage of the 208Y/120V system and they are 120 degree apart And as indicated above the current through N (between the load and the source) will be 12A.

And if you can relate to "V" conncetion theory, 10.4/12=0.866. That discussion for some otherday.

 

[rbulsara]

tinfoil:

Classic example would be connection of a electic cooking range in a house fed with a 208/120V,single phase service taken off a 208Y/120V serivce on the utility poles.

If you are in the USA you can relate to it...


And by the way, I did not see lesmcw's link to picture..it can be used to what i said..

buzzp:

No. A 3 wire circuit is 2 hots and a neutral. Ground is not counted when stating system 'wire' configuration. Like 3 phase 3 wire means 3 phases no neutral but there is always a ground. Similarly 3 phase , 4 wire system is 3 phases and a neutral. Ground is on top of that.

Also not all 3 wire circuits are same. A 3 wire service derived from a mid-point of a tranformer is not same as 3 wire service taken off a 3 phase, 4w Y system.

In circuir fed off the midpooint of the transformer, the current through the N will be zero if both hots are carrying equal amps as they will be 180 deg out of phase.

But not in the case of the original post of this thread.

 

[rbulsara]

One more explanation:

When the N is not connceted to the load , in my example, the each half of the load R sees 208/2=104V and 180 deg out of phase with each other...hence In = 0. But Ia or Ib is 10.4 (not 12)

 

[jghrist]

I originally assumed that the 12A load was connected from phase A to phase B (like a 208V range on a 120/208V system). In this case, I_A=-I_B and the neutral current is zero.

From the sketch, it appears that there are two 120V loads. If they have the same power factor, the currents are 120° apart and the neutral current equals the phase current.

Assume I_A=12@0°=12+j0, then I_B=12@-120°=12·cos(-120°)+j·12·sin(-120°)=-6-j10.39

I_N=I_A+I_B=6-j10.39=12@-60°

 

[aolalde]

Rbulsara:

I assumed a single phase load connected line to line ( Ib = -Ia).
If two loads with the same impedance are connected between those two lines to the neutral, yes you are right the phase angle is offset 120°.
The node current formula is still valid.
Ia+Ib+Ic+In=0
Ia=12<0 , Ib=12<120 , Ic=0 (assuming resistive loads.)

In= -12<0 - 12<120= -12<60 = 12<240

 

[rbulsara]

I am glad we all on the same page now

 

[lesmcw]

Thanks Guys. Looks like it may have dusted off some reference books. It's good to know there is help out here when you're stuck on something! Thanks again!!!
Lesmcw