PV Systems & Applying NEC 705.12

Status
Not open for further replies.
cdcengineer

cdcengineer

Senior Member
I have been posed a new one for me. The PV installer has designed an 8,280 W system to be beck-fed froma signle phase 208V inverter into a 208Y/120V distribution panel. The unbalanced interconnection (NEC 705.100) is not a concern, but the issue is with 705.12(D)(2).

We typically see residential and/or commercial systems where we tie in a breaker and add the sum of the back-feed breaker and the MOCP device. This resultant value cannot exceed 120% of the busbar or conductor. However, with a 45A/2P back-feed breaker and a 200A/3 MOCP, theoretically this is not acceptable due to 705.12 - allowed 1.2 x 200 = 240A versus proposed 200 + 45 = 245A.

I believe this article should read that the sum of the load (kW or kVA) rating of the circuitsd supplying a busbar or conductor shall not exceed that of the busbar.

The existing busbar is rated for 200A, 3-phase or 200 x 208 x SQRT 3 = ~ 72,053 VA x 1.2 (Per NEC 705.12(D)(2) ~ 86,464 VA, the 200A/3P main is rated for ~ 72,053 VA and the back-feed breaker is 45 x 208 ~ 9,360 VA. Thus 72,053 + 9,360 = 81,413 VA / 208 x SQRT 3 = 81,413 VA.

Anyone have any experience with something like this, or have an opinion on the interpretation of 705.12(D)(2)? Feedback, comments and input are greatly appreciated.
 
120426-0857 EDT

My comment only applies to theory.

Consider a simple 120 V single phase panel with one neutral and one hot bus bar. From the power company feed the top end of the bus bars, and from the local PV system feed the bottom end of the bars.

Put a single 100 A load at the bus bars mid points. With no PV supply there is 100 A flowing in the upper half of the bus bars.

Add 50 A at the bottom from the PV, and there is only 50 A flowing thru the full length of the bus bars.

Last supply 100 A from the PV and only the lower half of the bus bars have current flow.

Total power dissipated in the bus bars in the three cases are:
1 ---- 1002 * R / 2 = 5000 * R
2 ----- 502 * R / 1 = 2500 * R
3 ---- 1002 * R / 2 = 5000 * R

Thus, total power dissipation is never worse than for no PV attachment. And power dissipation per unit length of the heated part of the bus bar is 4 times larger for case 1 or 3 vs 2.

.
 
cdcengineer said:
...

The existing busbar is rated for 200A, 3-phase or 200 x 208 x SQRT 3 = ~ 72,053 VA x 1.2 (Per NEC 705.12(D)(2) ~ 86,464 VA, the 200A/3P main is rated for ~ 72,053 VA and the back-feed breaker is 45 x 208 ~ 9,360 VA. Thus 72,053 + 9,360 = 81,413 VA / 208 x SQRT 3 = 81,413 VA.

...
Click to expand...
The way I see it, you are trying to apportion the single phase inverter current to all three busbars, rather than the two that it affects... :happysad:
 
Smart $ - that's a fair statement. I wasn't looking at it like that.

Even though it's single phase, it could output up to ~ 45A on A & B and if the main OCP device has significant load, than we could overload the 2 phases.
 
cdcengineer said:
Smart $ - that's a fair statement. I wasn't looking at it like that.

Even though it's single phase, it could output up to ~ 45A on A & B and if the main OCP device has significant load, than we could overload the 2 phases.
Click to expand...
On the other side of the coin, you could say that with a power factor of 1 for both the utility and inverter currents, the inverter current will be out of phase ?30?. The sum of these currents at magnitudes of 200A and 45A is 239A@ر5.4?.
 
But with that said, why doesn't the NEC require that we look at the sum of the currents, not the sum of the ampere rating of the OCP? Looks like a needed code revision for 2014.
 
Status
Not open for further replies.
Top