Unbalanced 3-phase wye: Simple way to calculate neutral in your head

[msuchoff]

You can determine neutral current in unbalanced 3-phase wye in your head and estimate's accuracy is pretty good.
This works for loads with PF=1.

1. Examine the phase currents and note the maximum and minimum legs. Minimum can be zero.
2. Estimated neutral current = max - min.

When the remaining phase leg current is zero or same as max phase leg, estimate is exactly same as measured.
Worst estimate is when remaining phase leg is 1/2 value of max - estimate is then 15% high.

Example:

1. Phase leg currents are 40A, 50A, 55A.
2. Max = 55A, min = 40A
3. Estimate = max-min = 15A
4. Measured neutral = 13.23A
5. Estimate's accuracy = +12%

I can give math proof if there is interest.

 

[topgone]

You can determine neutral current in unbalanced 3-phase wye in your head and estimate's accuracy is pretty good.
This works for loads with PF=1.

1. Examine the phase currents and note the maximum and minimum legs. Minimum can be zero.
2. Estimated neutral current = max - min.

When the remaining phase leg current is zero or same as max phase leg, estimate is exactly same as measured.
Worst estimate is when remaining phase leg is 1/2 value of max - estimate is then 15% high.

Example:

1. Phase leg currents are 40A, 50A, 55A.
2. Max = 55A, min = 40A
3. Estimate = max-min = 15A
4. Measured neutral = 13.23A
5. Estimate's accuracy = +12%

I can give math proof if there is interest.

Calculated neutral is indeed 13.23 amperes. I am using resolution of vectors math.

 

[iceworm]

Calculated neutral is indeed 13.23 amperes. I am using resolution of vectors math.

Hummm .... so two of us got nerd-sniped.:blink:

Did you do the generalized case math to see what the limits of the accuracy were?

I started to, that's when I realized I had been sniped and quit.:slaphead:

 

[retirede]

Hummm .... so two of us got nerd-sniped.:blink:

Did you do the generalized case math to see what the limits of the accuracy were?

I started to, that's when I realized I had been sniped and quit.:slaphead:

"Nerd-sniped". LMAO

 

[Carultch]

"Nerd-sniped". LMAO

I'm still trying to figure out the original problem of the nerd sniping comic. "On this infinite grid of 1 ohm resistors, what is the equivalent resistance between the two marked nodes (that are a chess knight's move apart from one another)?"

 

[Carultch]

You can determine neutral current in unbalanced 3-phase wye in your head and estimate's accuracy is pretty good.
This works for loads with PF=1.

1. Examine the phase currents and note the maximum and minimum legs. Minimum can be zero.
2. Estimated neutral current = max - min.

When the remaining phase leg current is zero or same as max phase leg, estimate is exactly same as measured.
Worst estimate is when remaining phase leg is 1/2 value of max - estimate is then 15% high.

Example:

1. Phase leg currents are 40A, 50A, 55A.
2. Max = 55A, min = 40A
3. Estimate = max-min = 15A
4. Measured neutral = 13.23A
5. Estimate's accuracy = +12%

I can give math proof if there is interest.

I've made numeric calculations setting Ia = 1A as the maximum, and Ib and Ic ranging from 0 to 1A, and found similar results. Greatest percent error is 15.47%, which occurs when Ib is halfway between Ic and Ia. This maximum error actually equals (2*sqrt(3) - 3)/3, which the percent error between sqrt(3)/2 and 1.

 

[Smart $]

I've made numeric calculations setting Ia = 1A as the maximum, and Ib and Ic ranging from 0 to 1A, and found similar results. Greatest percent error is 15.47%, which occurs when Ib is halfway between Ic and Ia. This maximum error actually equals (2*sqrt(3) - 3)/3, which the percent error between sqrt(3)/2 and 1.

And this all makes the assumption of equal linear power factors. Yes, it works for not just a power factor of 1. However, it could vary in accuracy a lot more than 15% if unequal and/or has non-linear power factors. Ask yourself, how often do you run into MWBC loads with equal power factors? Used to be commercial lighting circuits one could be reasonably certain, but with the changeover to electronic ballasts and now going to LED, the power factors are nonlinear.

Not to burst any bubbles here, but IME the only time such an estimation came in handy for me is when doing demo or revision work... and even then, if you have a conduit with more than one neutral and ungrounded conductors outnumbering the neutrals, there are easier methods of determining for certain which ungrounded conductor matches which neutral, e.g. turning the ungrounded circuit conductors on and off while monitoring neutral conductor current. So the only time estimating is really helpful anymore is if management wants to stay up and running and meet the live work requirements.

 

[iceworm]

I'm still trying to figure out the original problem of the nerd sniping comic. "On this infinite grid of 1 ohm resistors, what is the equivalent resistance between the two marked nodes (that are a chess knight's move apart from one another)?"

Oh that one - consider this:
Infinite plane represents the earth. Once one gets by the ground rod to earth resistance, the resistance between any two points is zero.

KAPOW!! and another one bites the dust
That would be a nerd-snipe, not shooting down your thoughts

The worm

 

[iceworm]

...Not to burst any bubbles here, ....

Be encouraging. This is msu's first post.

 

[msuchoff]

Bunch of snarky guys you all can be.

Here's the proof:

Equation for Neutral based on addition of sinusoidals at angles 0, 120, -120

(1) Ia * sin(x +0) + Ib * sin(x +120) + Ic * sin(x -120) = N

When phase legs are same equal current X, neutral is 0

(2) X * sin(x +0) + X * sin(x +120) + X * sin (x -120) = 0

Subtract (2) from (1)

(3) (Ia-X) sin(x +0) + (Ib-X) sin(x +120) + (Ic -X) * sin(x -120) = N

Rank the Ia, Ib, Ic phase currents:
max = maximum value of (Ia, Ib, Ic)
min = minimum value of (Ia, Ib, Ic) (can be zero)
rem = the remaining phase leg (will be value >= 0 and <= max)

Substitute "min" into (3). This will cause one of the terms of (3) to cancel out.
Result of the substitution is:

(4) (max-min) * sin(x) + (mid-min) * sin(x +- 120) = N

Define the estimated neutral current to be max - min.

(5) estimate * sin(x) + (mid-min) * sin (x +- 120)

If mid-min = 0, (5) reduces to single phase and estimate is 100% accurate

If mid-min > 0, we have sum of two sinusoidals that can be solved using trig identity

(6) a * sin(x) + b * sin(x + p) = c * sin(x + q)
(7) c = sqrt( a^2 + b^2 + 2ab*cos(p))

From (5) phase angle "p" is +120 or -120 and cos(120) = cos(-120) = -.5.


(8) c = sqrt (a^2 + b^2 - a*b);

Inspection of (8) shows that "a" and "b" can be interchanged and result in same answer.

We set "a"=estimate and "b"=(mid-min) and substitute into (8) to get neutral current

(9) N = sqrt(estimate^2 + (mid-min) - estimate * (mid-min))

(9) can be solved but more informative to plot out results as (mid-min) varied from 0 to estimate.

From plot below we conclude:

• (mid-min) value causes measured neutral to be between 86%-100% of estimate
• estimate accuracy is +0% to +15%

 

[msuchoff]

Corrections for typos in proof

The variable "rem" should be "mid" as "mid" is used elsewhere

mid = the remaining phase leg (will be value >= 0 and <= max)

Equation (9) should be:

(9) N = sqrt(estimate^2 + (mid-min)^2 - estimate * (mid-min))