Alwayslearningelec
Senior Member
- Location
- NJ
- Occupation
- Estimator
200% N.
- Thread starter Alwayslearningelec
- Start date
- Status
- Location
- New Jersey
- Occupation
- Journeyman Electrician
If those are both the same feeder then it would require 5-#2/0 not 4, and 2 EGC's unless you're using the raceway as one EGC. Might not fit in a 2" either. What is the IG for?
zbang
Senior Member
- Location
- Roughly 5346 miles from Earls Court
Sounds like another case of 30 year old boilerplate designs. (Does anyone install double neutrals these days?)
xptpcrewx
Power System Engineer
- Location
- Las Vegas, Nevada, USA
- Occupation
- Licensed Electrical Engineer, Licensed Electrical Contractor, Certified Master Electrician
In all cases, follow the design documents. If there is a conflict, issue an RFI requesting clarification. Sometimes it might be as simple as not having the latest drawing. You aren’t allowed to just change stuff based on what you think (even if it’s wrong).
- Location
- New Jersey
- Occupation
- Journeyman Electrician
No, a 100 amp hot conductor and a 100 amp neutral conductor would be the same. A 100 amp hot conductor with a 200% neutral would be a 200 amp neutral conductor.
I see what your saying regarding the amps, but wouldn’t a 100% RATED neutral would be half size?
so if (say residential) needed 4/0 hot, a 100% rated neutral would be 2/0, correct?
a 200% would be a 4/0 neutral.
Does this make sense the way I’m seeing this?
Alwayslearningelec
Senior Member
- Location
- NJ
- Occupation
- Estimator
UPS
- Location
- New Jersey
- Occupation
- Journeyman Electrician
No, a 100% neutral would be equal in size to the ungrounded conductor. For a 120/240 volt system the neutral will carry the current imbalance of the two hot legs which could theoretically be 100% of the current in one hot leg.
Alwayslearningelec
Senior Member
- Location
- NJ
- Occupation
- Estimator
Why would it be 2 neutrals ? You stated below that a 200% neutral would be twice the ampacity of the hots.
- Location
- New Jersey
- Occupation
- Journeyman Electrician
You have:
#2/0 hot leg= 175 amp
#2/0 neutral= 175 amps or
#2/0 hot leg= 175 amp
(2)*#2/0 neutral = 350 amps (200%) or (1) 500 kcmil = 380 amps
Yes, I understand the math behind it. I’ve calculated triplens as high as 150% on some systems, but I’m in agreement with others triplen addition that high is hard to accomplish in the real world.
I guess I’m arguing semantics with myself.
you made the point for me.
synchro
Senior Member
- Location
- Chicago, IL
- Occupation
- EE
The worst case neutral current in a 3-phase circuit would be when the waveforms of the three L-N currents do not overlap each other in time at all, resulting in no cancellation of currents in the neutral. An example would be three equal nonlinear L-N loads that only draw current near the peaks of their L-N voltage, but but otherwise draw zero current. Then the mean square (i.e., average of the squared waveform) of the total neutral current will be 3 times the mean square current of each L-N load because their L-N currents do not overlap at any point in time. Therefore the RMS (root mean square) value of the neutral current will be √3 ≈ 1.73 times the L-N current. But apparently they've rounded this up to 2 times (i.e., 200%) for convenience.
Like others have said, this would be a situation that rarely occurs to this extent.
Like others have said, this would be a situation that rarely occurs to this extent.
wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
What's the maximum phase shift you can get from a linear load?
E.g. if I understand correctly, a pure capacitor load and a pure inductor load would each phase shift 90 degrees in opposite directions. So then if you have a resistor, an inductor and a capacitor arranged with each L-N on a 3 phase supply, the current waveforms could be only 30 degrees apart (60 degrees between the capacitor and the inductor). Which if they are all equal magnitude, would sum to 1 + √3 times that magnitude, instead of 0 for the case of 3 resistors.
But perhaps I've made a mistake in the physics or math. And I imagine in practice real loads would never have those characteristics.
Cheers, Wayne
E.g. if I understand correctly, a pure capacitor load and a pure inductor load would each phase shift 90 degrees in opposite directions. So then if you have a resistor, an inductor and a capacitor arranged with each L-N on a 3 phase supply, the current waveforms could be only 30 degrees apart (60 degrees between the capacitor and the inductor). Which if they are all equal magnitude, would sum to 1 + √3 times that magnitude, instead of 0 for the case of 3 resistors.
But perhaps I've made a mistake in the physics or math. And I imagine in practice real loads would never have those characteristics.
Cheers, Wayne
synchro
Senior Member
- Location
- Chicago, IL
- Occupation
- EE
That is true, but you obviously would be getting no real power from the L-N legs with the inductor and capacitor. So as you say, I don't think this would be found in practice.
...
You could have a phase shift between the load current and the applied voltage greater than 90 degrees, but then the real part of the impedance would likely be negative (i.e., a negative resistance). In such a case, the load would be supplying real power to the source instead of the other way around. It could still be a linear load but it would not be a passive load.
There are semiconductor devices such as Gunn diodes that have a negative resistance region (negative slope) in their I/V curve, but you have to apply a substantial DC bias current to get into this region and so you're not getting something for nothing.
- Location
- New Jersey
- Occupation
- Journeyman Electrician
I think that time has proven that the hype about 200% neutrals, k rated transformers and no MWBC's was just that, hype.
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