Unbalance neutral current in a 3 phase system

[rattus]

Agreed:

Agreed:

Knowing the difference was not my point. How many decimal points do you use when performing electrical power circuit calculations? How many significant digits? At which point do you perform your rounding to get the resulting voltages of 208Y/120.

I am not trying to pick any type of fight. My point was that the circuits that most electricians deal with can be handled with "adequate" equations and simplifications. When dealing with circuit ampacity to determine conductor size I am not often concerned with "1s" amps, I am usually satisfied with "5s" or even "10s" instead.

Sure Jim, two significant figures are usually enough, but with computers and calculators, we can easily carry the result out to three. I suspect that newbies start out striving for more accuracy than is needed until they realize it is a waste of time.

I don't want to fight either--too close to Christmas. Santa is watching! Nothing to fight over anyway.

 

[crossman]

I'm looking at these things more from a "Physics class" standpoint than a "practical applications" standpoint, simply because it is interesting to me. And in my opinion, there is absolutely nothing wrong with that.

I can't think of a single place where finding the neutral current of an unbalanced Y system would be required to put in a code-compliant electrical installation.

Rattus, thank you for reviewing my work on the phasors and for the confirmation!

 

[Smart $]

I am thinking it would be (2A -2C)/3 because dividing the individual parts by √3 and then dividing by √3 again is actually dividing by 3. Check post #45 by charlie b.



I'm thinking it would be (|A?B|+|A?C|+|B?C|)/3 for the same reasons.

Oops! My mistake.

Nevertheless, it's still an invalid method :wink::grin:

 

[crossman]

Now you can go and complete my derivation of the algebraic formula.

Couldn't sleep tonight so figured I may as well have some entertainment:

Note for Rattus: You don't have to go through the whole thing for my sake, but definitely check my last few thoughts at the end of the post please.

Given: Y system, 3ph 4w : Ia @ 30, Ib @ -90, Ic @ -210
Find: Neutral current

Convert to phasors in complex plane

Ia = .866Ia + j.5Ia
Ib = 0 + j(-Ib)
Ic = -.866Ic + j.5Ic

Find resultant by adding real part then adding imaginary part

Ire = (.866Ia - .866Ic) + j(.5Ia - Ib +.5Ic)

So the resultant "tail" is at the origin 0 + j0 and the arrow is at the point Ire. Now to make this easier to type, let me replace each "I" expression with the letter of the phase, "Ia" will just be "A", Ib will be B and Ic will be C.

Ire = (.866A - .866C) + j(.5A - B + .5C)

Now using the distance formula (Pythagorus) let M be the magnitude of the vector:

M? = (.866A - .866C)? + (.5A - B + .5C)?

M? = (.75A? - 1.5CA + .75C?) + (.25A? - .5AB + .25CA - .5AB + B? - .5BC + .25CA - .5BC +.25C?)

Combine terms

M? = (.75A? + .25A?) + (B?) + (.75C? + .25C?) + (-.5AB - .5AB) + (-.5BC - .5BC) + (.25CA + .25CA - 1.5CA)

M? = A? + B? + C? - AB - BC - CA

Taking SQRT of both sides

Magnitude of resultant = √(A? + B? + C? - AB - BC - CA)

So I am good to go on the unbalanced Y neutral current.

However, one issue. To make this work, I had to ignore the "j" in the distance formula. If I left it in and squared it with proper algebra, it made a bunch of the signs come out wrong. So on that note I am unsure of what should be done about that.

Did I miss some algebra that did away with the j in the distance formula?

In the cartesian plane, the distance formula is D? = (x2 - x1)? + (y2 - y1)? so.... does j(.5A - B +.5C) - j0 actually do away with the j term?

And how to find the angle of that resultant is going to take some thought. Obviously something to do with sine and cosine?

 

[crossman]

Angle of resultant from post above:

Angle = invTan (imaginary/real)

So for the 208Y/120 neutral:

Angle = invTan (j(.5A - B +.5C)/(.866A - .866C))

And I figure this may be 180 degrees off from the actual angle of the neutral current due to Kirchoff's current law. Also, the j probably isn't used here either. Also looks like something else should be considered to decide which quadrant the angle is in?

 

[Gary Shumaker]

Simple clarification

Simple clarification

Just a little clarification:
√3?√3=1 where as √3x√3=3. Not that it makes any difference at this point since I still like roger?s post #2 the best so far.
Now for a little background; we teach the apprentices are taught that in many cases the NEC allows derating the neutral by 70% because in these many cases the neutral only carries the unbalanced current. I perform a demonstration in our lab with 208/120 3ф connected to a switch board with three 40 watt bulbs, one on each phase. I ask them what will happen if I switch off the neutral. The demonstration shows that with all three bulbs lit, switching off the neutral has no affect on the output of the bulbs.
Naturally, sooner or later a student will ask how we can determine the amount of current on the neutral wire with an unbalanced load. I was originally given a completely false formula (see my first post) and that is how we got here.
Merry Christmas, happy chucknucker and merry quantumzi
Gary Shumaker

 

[roger]

And a Merry Christmas to you Gary.

Roger

 

[crossman]

Just a little clarification:
√3?√3=1 where as √3x√3=3. Not that it makes any difference at this point

I guess it doesn't make any difference at this point as far as this thread goes, but just for the record:

If you take a number and divide it by the √3, and then you take that answer and divide it by √3 again, the result is the same as dividing the original number by 3.

Example:

21 ? √3 = 12.1387...

12.1387... ? √3 = 7

 

[Gary Shumaker]

I guess it doesn't make any difference at this point as far as this thread goes, but just for the record:

If you take a number and divide it by the √3, and then you take that answer and divide it by √3 again, the result is the same as dividing the original number by 3.

Example:

21 ? √3 = 12.1387...

12.1387... ? √3 = 7

I get what your say now.
Gary

 

[charlie b]

. . . but just for the record: If you take a number and divide it by the √3, and then you take that answer and divide it by √3 again, the result is the same as dividing the original number by 3.

I think I already said that, twice.

 

[crossman]

You sure did, the first time was in post #45. The OP must have missed it, so I was reinforcing the idea.:smile:

 

[rattus]

Crossman,

Generally when doing a derivation, one uses general terms insofar as possible. That is, I would choose Ia, Ib, and Ic for the values of the phase currents This approach prevents a fluke derivation based on a specific set of values.

However, it is appropriate to define the phase angles as you have done because we can prove that the choice of reference has no effect on In. That is the formula will work for any set of angles separated by 120 degrees.

Correct, the operator "j" only identifies the imaginary components. You do not use it with Phythagoras. And, this formula does not provide the phase angle of In. You must use phasors to compute the angle, but that is hardly if ever needed in this business.

 

[crossman]

Thanks for hanging with me, rattus. This has been instructive and enjoyable.:smile:

 

[gpedens]

Unbalance neutral current in a 3 phase system

Okay youngins, I glanced through all the replies and didn't notice anyone using the "symetrical component" method to calculate current, voltage. It basically says you can take 3 unequal vectors and come up with 9 equal vectors that are the equavalent of the unequal. This method is BC before calculators. It says there are 3 equal positive sequence, 3 negative sequence, and 3 zero sequence vectors. It applies to current, voltage, and impedance. Say i1 is positive, i2 negative, and i0 is zero seq. There is a unit vector a that has a values of 1 at an angle of 120 and a square (a2) at or 1 at an angle of 240 deg. i1=[IA+a*IB+a2*IC]/3 and i2=[IA+a2IB+aIC]/3 and i0=[IA+IB+IC]/3. From which IA=i1+i2+i0 and IB=a*IA and IC=a2*IA and IN=3*i0. Remember there will be no neutral current in an ungrounded wye or a delta even if i0 has a value. Then there is nodal analysis which says the sum of the currents flowing into a node is zero. IA+IB+IC+IN=0 or typically since IN is shown into the node or neutral point and IA,IB,IC are shown out, we have IN=IA+IB+IC or Ia<0+. IA,IB,IC are vectors and Ia,Ib,Ic is just the magnitude. IA is at 0 degrees Ia<0, IB at 120 Ib<120 and IC at 240 Ic<240. the sum then becomes IN= Ia*[cos0+jsin0]+Ib*[cos120+jsin120]+Ic*[cos240+sin240]. j is the sine term or usually called the imaginary or square root of -1. In this case just look at it as the vertical part of a right triangle or the sine part. The cosine part is the real or horizontal part of the vector.
cos0=1, sin0=0, cos120=-0.5,sin120=0.866,cos240=-0.5, and sin240=-0.866. adding we get IN= Iacos0+Ibcos120+Iccos240+j*[IAsin0+IBsin120+ICsin240]. Reducing further we get IN=[Ia*1-0.5*Ib-0.5*Ic]+j [Ia*0+Ib*0.866-Ic*0.866]. you will basically end up with IN=X+jY where the magnatude of IN or In=square root of (Xsquare +Ysquare). The angle is inverse cosine of X/In or inverse sine of Y/In or inverse tangent of X/Y

 

[LarryFine]

If you don't mind, this much text is much easier to read if broken into paragraphs, thusly:


Okay youngins, I glanced through all the replies and didn't notice anyone using the "symetrical component" method to calculate current, voltage. It basically says you can take 3 unequal vectors and come up with 9 equal vectors that are the equavalent of the unequal.

This method is BC before calculators. It says there are 3 equal positive sequence, 3 negative sequence, and 3 zero sequence vectors. It applies to current, voltage, and impedance.

Say i1 is positive, i2 negative, and i0 is zero seq. There is a unit vector a that has a values of 1 at an angle of 120 and a square (a2) at or 1 at an angle of 240 deg. i1=[IA+a*IB+a2*IC]/3 and i2=[IA+a2IB+aIC]/3 and i0=[IA+IB+IC]/3. From which IA=i1+i2+i0 and IB=a*IA and IC=a2*IA and IN=3*i0.

Remember there will be no neutral current in an ungrounded wye or a delta even if i0 has a value.

Then there is nodal analysis which says the sum of the currents flowing into a node is zero. IA+IB+IC+IN=0 or typically since IN is shown into the node or neutral point and IA,IB,IC are shown out, we have IN=IA+IB+IC or Ia<0+.

IA,IB,IC are vectors and Ia,Ib,Ic is just the magnitude. IA is at 0 degrees Ia<0, IB at 120 Ib<120 and IC at 240 Ic<240.

the sum then becomes IN= Ia*[cos0+jsin0]+Ib*[cos120+jsin120]+Ic*[cos240+sin240]. j is the sine term or usually called the imaginary or square root of -1. In this case just look at it as the vertical part of a right triangle or the sine part.

The cosine part is the real or horizontal part of the vector. cos0=1, sin0=0, cos120=-0.5,sin120=0.866,cos240=-0.5, and sin240=-0.866. adding we get IN=Iacos0+Ibcos120+Iccos240+j*[IAsin0+IBsin120+ICsin240].

Reducing further we get IN=[Ia*1-0.5*Ib-0.5*Ic]+j [Ia*0+Ib*0.866-Ic*0.866]. you will basically end up with IN=X+jY where the magnatude of IN or In=square root of (Xsquare +Ysquare). The angle is inverse cosine of X/In or inverse sine of Y/In or inverse tangent of X/Y


See what I mean?

 

[rattus]

Too much:

Too much:

Paragraphs or no, this method is overly complicated for such a simple problem. If you are going to use phasors anyway, just apply Kirchoff's current law to the neutral node. This works for any problem and gives the phase angle as well.

Otherwise, just use the algebraic formula which is sufficient for the majority of cases.

 

[jghrist]

Rattus said:

Agreed, the algebraic formula should be taught because it is easily applied and can be utilized in a spreadsheet or programmable computer. However it is no more accurate than the other approaches. It should be adequate for most situations, but if the PFs are different among the phase currents, only the phasor/complex math approach will provide exact results. I would suggest that this point be emphasized.

You don't have to have unequal power factors to get current vectors at angles different from 120 deg. If you have some loads connected phase-to-phase and some connected phase-to-neutral, the angles between currents will be different from 120 degrees.

For instance, if you have balanced voltages VA=120@0?, VB=120@-120?, and VC=120@120?, a 2 ohm resistor between phases A and B, and a 2 ohm resistor between C and neutral, the currents will be:

IA=104@30? (208 volts / 2 ohms)
IB=104@-150?
IC=60@120? (120 volts / 2 ohms)
IN=60@120? (IN defined as current going into the neutral bus)

In this example, all loads are resistive (100% power factor). The phase-to-phase load results in no neutral current. The neutral current equals the single phase-to-neutral connected load.

The neutral current is the vector sum of the phase currents.

 

[rattus]

Yes but:

Yes but:

Rattus said:

You don't have to have unequal power factors to get current vectors at angles different from 120 deg. If you have some loads connected phase-to-phase and some connected phase-to-neutral, the angles between currents will be different from 120 degrees.

For instance, if you have balanced voltages VA=120@0?, VB=120@-120?, and VC=120@120?, a 2 ohm resistor between phases A and B, and a 2 ohm resistor between C and neutral, the currents will be:

IA=104@30? (208 volts / 2 ohms)
IB=104@-150?
IC=60@120? (120 volts / 2 ohms)
IN=60@120? (IN defined as current going into the neutral bus)

In this example, all loads are resistive (100% power factor). The phase-to-phase load results in no neutral current. The neutral current equals the single phase-to-neutral connected load.

The neutral current is the vector sum of the phase currents.

We measure the power factor as a function of the angle between phase current and phase voltage. You have presented an example of an unbalanced system which causes these angles to be other than zero. Matters not that the loads are resistive.

The poco would penalize you for running a load like this!

 

[coulter]

... I glanced through all the replies and didn't notice anyone using the "symetrical component" method to calculate current, voltage. ...

We have some that are vehemetly opposed to symetrical comments, vectors, and rotating magnetic fields. I agree symetrical components will handle a general case better than most other methods. With the advent of calculators that will handle matrices(sp) and complex math, the method is much easier to apply.

carl

 

[gpedens]

Unbalance neutral current in a 3 phase system

Thanks, Larry. I like your way better too. It was late and I was rushing. After reading the posts more it looks like someone is looking for a simple way to calculate the neutral current if three phase currents are not equal. I tend to think of sines, cosines, a trig as simple high school math. Good old Pathagorem. The simplest way I can think of is to graph it out adding the LINE currents head to tail.

Rule 1. The easy solution. Remember there is no neutral current if the load neutral and source neutral are not wired together or both grounded. No path no current. Even if the math shows says there is. There will be a voltage between the two neutral points.

From Kirchoffs law In at source In=A+B+C or In-A-B-C=0 if we assume direction of In is pointing from load to source and A,B,C currents are pointing to load.

I draw a horizontal line and a vertical line at 90 degrees to each other. Then figure out what scale I want to use to represent the currents.

I start out by showing the currents as three arrows A, B, C with all the feathered ends touching where the H and V lines cross and the three tips 120 degrees apart forming the classic wye. I show the A arrow on the zero or horizontal line pointing right. The B at 120 degrees CCW from A and the C 120 degrees CW from A.

Now add B to A by drawing another B with its feathers touching A tip and the new B is parallel to the old B.

Now add a new C with its feathers touching the tip of new B and keep the new C parallel with the old C.

Now measure In neutral current from the center of the graph to the head of the new C. The angle of In will be the angle measured CCW from the A or horizontal line to In.

If A was not at angle zero I draw my graph with A at zero the shift In as needed. Say A was really at 20 degrees above the zero or in the CCW direction, I will just add 20 to the measured In or if A was 62 degrees below the horizion , I subtract 62 degrees from the measured In.

As for the real and imaginary. I just look at it like pulling on a rope with a pulley at the center of the horizontal and vertical playing field. Voltage is a woman with a rope and current is the man.

For the imaginary, a capacitor current is 90 deg ahead of voltage. Or man is on the horizon and woman on the vertical. Man goes horizontal and hits end of rope woman is just drug down and goes back vertical jerking man back to center. Same way with an inductor except man is vertical and woman horizontal. Lot of jerking around and sweating and grunting but they never get off center. What one gives the other takes back, the result being a big current that gets nothing done. One is always on the vertical and one on the horizontal out of phase wearing each other out going nowhere.

In a resistor voltage and current move together back and forth along the horizontal creating lots of heat smoking that rope. Even in the minus direction the heat is positive because -x- is +.