Help Understanding Ampacities and Parallels, table 310.15(B)(16)

[paulengr]

You assumed 75 C, why?? THHN will be 90 C. Inside the raceway the full cable temperature (90 C) would be allowed. At the terminals unless you use lugs rated to 90 C onto busbar you correctly assumed 75 C. That gives you the extra ampacity you are looking for.

 

[ramsy]

I have a 1000A/3phase feeder that's called out to be 4 sets of [(3)#300, 2/0g, in 3" EMT)], copper.

My question is, according to table 310.15(B)(16), (1)#300 in column 75C has ampacity rating of 285. Why must I have 4sets of (3)300s, a total of 12, if (4)#300s provides 1140A. I am clearly reading this table wrong. Please help me understand.

If 50% of load calc includes 220.18(A) motor loads, 125 Amps must be added for a total load of 1125 Amps.

Most contractor's have knuckled heads that can't stretch their ability beyond the adjustments sections & idiot tables of 310.15.

Among other things, engineers have less trouble bending their geeky minds into motor-load adjustments, per section 430.

That's why the geek's plans always over-rule knuckle heads in the field.

 

[ggunn]

Okay, with 2 wires each carrying 1000A in a single phase circuit, how many 'total amps' do you have?

And before you say the answer is the same, consider that at the same line-neutral voltage the 3 phase circuit is delivering more power with the 'same' current. This is why I don't like the idea of saying 'total amps'.

-Jon

It is the same. Power does not equal current; the same current delivers more power in three phases than in two. I don't say "total current", either; that implies you have to add something. I say "line current".

 

[ggunn]

For the same current at the same voltage, of course

Single phase: P = IV
Three phase: P = (sqrt3)IV

 

[JEFF MILLAR]

1000 Amp feeder , ok , 1000 A over current protection. Max. load on the. breaker 80 % . Load Amps equals 800 Amps . Feeder size copper. conductors equals 800 A x 1.25 equals 1000 A / 3 = 334 Amps. NEC. 310.15 , 75 C Column = 400 lcmil. My calculation makes 3 runs of conduit each with 3 - 1 / C # 400 kcmil + Eqpt. Grnd as per NEC 250.122. ( Each conduit has phase A , B and C ) and EG. Do you agree or do I need to be corrected ? (. I have assumed no ampactiy correction required or voltage loss consideration , for simplicity )

 

[infinity]

You should have said maximum continuous load on breaker is 800 amps. If the load is non-continuous then it can be 1000 amps.

 

[Carultch]

1000 Amp feeder , ok , 1000 A over current protection. Max. load on the. breaker 80 % . Load Amps equals 800 Amps . Feeder size copper. conductors equals 800 A x 1.25 equals 1000 A / 3 = 334 Amps. NEC. 310.15 , 75 C Column = 400 lcmil. My calculation makes 3 runs of conduit each with 3 - 1 / C # 400 kcmil + Eqpt. Grnd as per NEC 250.122. ( Each conduit has phase A , B and C ) and EG. Do you agree or do I need to be corrected ? (. I have assumed no ampactiy correction required or voltage loss consideration , for simplicity )

You are correct. 3 sets of 400 kcmil Cu at 75C terminations adds up to 1000A of ampacity.

If I were in the OP's position, and there were no other issues stopping me from doing so, that is what I would use, instead of 4 sets of 300 kcmil. 3 sets of 400 kcmil has roughly the same conductance as 4 sets of 300 kcmil, so I would expect that the voltage drop would still be about the same. Unless the skin effect gives you marginally less voltage drop that makes a difference.

When voltage drop governs the size, roughly speaking, you'll need about the same total kcmil per phase no matter how many sets in parallel you install. It's only a second order factor (the skin effect) that makes a difference. But when ampacity governs the size, because more surface area means more ability to release the heat, you get disproportionately more ampacity per kcmil by using multiple sets in parallel. So much that it has a practical advantage over a single large conductor per phase.

 

[JEFF MILLAR]

Give the one way length of the copper conductors and the size of the conductor and a maximum specified voltage loss of 3 % as an example it is easy to calculate the maximum length give these values.

 

[wwhitney]

Since he is running 4 conduits anyway, why not run (6) 1/0s in each conduit?

So with 4 conduits and 4 circuit conductors (on a wye system), any number of parallel sets can be used, and with N parallel sets, each conduit will have 3N CCCs. 1/0 wire is apparently 105.6 kcmils. Assuming 90C insulation, 1/0 Cu has a base ampacity for derating of 170A.

For 2 sets of 1/0, that's 211 kcmils of copper, and you get 0.8*2*170 = 272A of ampacity. While 250 kcmil has a 75C ampacity of only 255A.

For 3 sets of 1/0, that's 317 kcmils of copper, and you get 0.7*3*170 = 357A. While 350 kcmil at 75C is only 285A.

For 4 sets of 1/0, that's 422 kcmils, and you get 0.5*4*170 = 340A. You're better off using 3 sets of 1/0. 400 kcmil at 75C is 335A, slightly better at ampacity per unit area of copper, after derating.

For 5 sets of 1/0, that's 528 kcmils, and you get 0.5*5*170 = 425A. While 600 kcmil at 75C is only 420A.

For 6 sets of 1/0, that's 634 kcmils, and you get 0.5*6*170 = 510A. While 700 kcmil at 75C is only 460A.

The above only looks at minimizing copper quantity for a given ampacity. There's also the question of conduit fill, comparing the area of the insulated conductors. I looked at THHN-2 dimensions, and I got that area-wise, 2, 3, or 6 sets of 1/0 was always a win, but for 4 or 5 sets, the 1/0 takes up more area as far as conduit fill.

Cheers, Wayne

 

[infinity]

If I were in the OP's position, and there were no other issues stopping me from doing so, that is what I would use, instead of 4 sets of 300 kcmil. 3 sets of 400 kcmil has roughly the same conductance as 4 sets of 300 kcmil

I see many guys say use more sets but more sets usually cost more money when you factor in running additional raceways and more terminations on each end. Pulling 300's versus 400's isn't significantly different so there is no real cost change there.

 

[wwhitney]

I see many guys say use more sets but more sets usually cost more money when you factor in running additional raceways and more terminations on each end.

Agreed on the terminations, but the point of my previous post was that for 2 or 3 sets of 1/0, no need to add conduits (assuming you can properly partition the sets), the 1/0 sets win over the comparable single conductor even after derating for multiple CCCs. And their cross sectional area is less for conduit fill.

Cheers, Wayne