Harmonic Distortion Limit and Neutral Cable Sizing

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Edelma_1 (Electrical)

(OP)

25 Jul 24 15:10

Hello Everyone,

I found out from IEC 60364-5-52 that for a multiphase circuit you can actually reduce the size of the neutral conductor than it`s active size if the harmonic distortion current is less than 15%. I would like to know by how many percent can one reduce the neutral.

Replies continue below

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RE: Harmonic Distortion Limit and Neutral Cable Sizing

waross (Electrical)25 Jul 24 15:50

While this has no standing in IEC land, it may be of interest:
The law in Canada:
(1) The neutral conductor shall have sufficient ampacity to carry the unbalanced load.
(2) The maximum unbalanced load shall be the maximum connected load between the neutral and any one
ungrounded conductor as determined by Section 8 but subject to the following:
(a) there shall be no reduction in the size of the neutral for that portion of the load that consists of
(i) electric-discharge lighting; or
(ii) non-linear loads supplied from a three-phase, 4-wire system; and
(b) except as required otherwise by Item (a), a demand factor of 70% shall be permitted to be applied
to that portion of the unbalanced load in excess of 200 A.
(3) The size of a service neutral shall be not smaller than the size of a neutral selected in accordance with
Subrule (1) and shall
(a) be not smaller than No. 10 AWG copper or No. 8 AWG aluminum; and
(b) be sized not smaller than a grounded conductor as required by Rule 10-204(2), except in service
entrance cable or where the service conductors are No. 10 AWG copper or No. 8 AWG aluminum.
(4) In determining the ampacity of an uninsulated neutral conductor run in a raceway, it shall be considered
to be insulated with insulation having a temperature rating not higher than that of the adjacent circuit
conductor

That said, it is common in Canada to have breaker panels supplied with a 200% neutral bus.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

RE: Harmonic Distortion Limit and Neutral Cable Sizing

Edelma_1 (Electrical)

(OP)

25 Jul 24 15:58

I actually read the standard half way, the full standards goes thus:

for multicore circuits the neutral conductor size can be smaller than the active size if the Harmonic distortion current is less than 15%.

When the Harmonics are between 15% to 33% the neutral conductor size shall be same size as the active conductor.

meanwhile the current THD should not exceed 20% of the fundamental frequency, in other words the neutral size should be same as conductor size for a nominal THD current of 20%.

RE: Harmonic Distortion Limit and Neutral Cable Sizing

che12345 (Electrical)26 Jul 24 00:36

Edelma_1 (Electrical)(OP)25 Jul 24 15:10
"..I found out from IEC 60364-5-52 that for a multiphase circuit you can actually reduce the size of the neutral conductor than it`s active size if the harmonic distortion current is less than 15%....I actually read the standard half way, the full standards goes thus:....I would like to know by how many percent can one reduce the neutral".
1. I am an octogenarian, the IEC 60364-5-52 (2001) I have in my collection could be outdated/superseded. Please refer to the latest.
2. Per 2001 publication , in brief ... 523.8.3 ... if the harmonic current is greater than 10% the neutral conductor shall not be smaller than the phase conductors...and the corresponding reduction factors for higher harmonic currents are given in annex D.
Che Kuan Yau (Singapore)

RE: Harmonic Distortion Limit and Neutral Cable Sizing

Parchie (Electrical)26 Jul 24 06:43

@EdeLma_1,
The formula I got is In/Ip = 3 X (per unit THD)/(sqrt(1+(per unit THD)^2) for circuits laden with 3rd harmonics. In = neutral current, and Ip = phase current.
I got 45% with 15% THD

RE: Harmonic Distortion Limit and Neutral Cable Sizing

Edelma_1 (Electrical)

(OP)

26 Jul 24 07:02

thanks Parchie,

asides using instruments to find out the THD, is it possible to get it by manual calculation first before proceeding to find voltage and current values of the harmonic orders?

RE: Harmonic Distortion Limit and Neutral Cable Sizing

che12345 (Electrical)27 Jul 24 04:48

@ Mr Parchie (Electrical)26 Jul 24 06:43
"...The formula I got is In/Ip = 3 X (per unit THD)/(sqrt(1+(per unit THD)^2) for circuits laden with 3rd harmonics. In = neutral current, and Ip = phase currentI got 45% with 15% THD"
1. I am skeptical of the simplistic formula on the In/Ip value =..... Reason: the magnitude of In and Ip (unless further qualified) do NOT indicate the harmonic content.
2. To illustrate, consider a Y with N system:
a) 3 de-similar linear loads connected across AN, BN , CN. There will be current in the Neutral. The neutral current is the fundamental, with harmonic = 0.
b) 3 de-similar non-linear loads connected across AN, BN , CN. There will be current in the Neutral. The neutral current is the fundamental, + the arithmetical sum of the third harmonic currents etc., where harmonic current would NOT = 0.
3. Logically, neither the In nor Ip indicates the value of the harmonic current in the Neutral line.
Che Kuan Yau (Singapore)

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