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Discussion Starter · #1 ·
I have twenty four 8.1kW (at 208)1ph heaters to run a service for. This is continuos. The incoming power is 120/208 three phase. Can I get away with a 600 amp 3 ph service?
 

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I have twenty four 8.1kW (at 208)1ph heaters to run a service for. This is continuos. The incoming power is 120/208 three phase. Can I get away with a 600 amp 3 ph service?

That comes out to 930 amps if I did the math right.


220.51 Fixed Electric Space Heating. Fixed electric space-heating loads shall be calculated at 100 percent of the total connected load. However, in no case shall a feeder or service load current rating be less than the rating of the largest branch circuit supplied.
Exception:  Where reduced loading of the conductors results from units operating on duty-cycle, intermittently, or from all units not operating at the same time, the authority having jurisdiction may grant permission for feeder and service conductors to have an ampacity less than 100 percent, provided the conductors have an ampacity for the load so determined.
 

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Discussion Starter · #4 · (Edited)
HARRY304E said:
That comes out to 930 amps if I did the math right.

220.51 Fixed Electric Space Heating. Fixed electric space-heating loads shall be calculated at 100 percent of the total connected load. However, in no case shall a feeder or service load current rating be less than the rating of the largest branch circuit supplied.
Exception:  Where reduced loading of the conductors results from units operating on duty-cycle, intermittently, or from all units not operating at the same time, the authority having jurisdiction may grant permission for feeder and service conductors to have an ampacity less than 100 percent, provided the conductors have an ampacity for the load so determined.
But spread out over a three phase service can I get away with a smaller amp service? I should only have 547 amps on any one leg. Or am I confused.
 

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I think any phase will see a max of 16 * 8.1kw. So If each phase sees 129600 watts then divide by 360 (208 * 3 sq.rt) and you get 360 amps. So yes a 600 amp feeder should be more than adequate but I screw these up alot---
 

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Discussion Starter · #6 ·
Dennis Alwon said:
I think any phase will see a max of 16 * 8.1kw. So If each phase sees 129600 watts then divide by 360 (208 * 3 sq.rt) and you get 360 amps. So yes a 600 amp feeder should be more than adequate but I screw these up alot---
Thanks Dennis, I was thinking I would see 66% of the load on each leg. That's where I can up with just under 600 amps. I like your number better. I will have to pull out my books and double check, not that I doubt you.
 

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I calculated 539.6A at 208 3 phase.

Are you saying these heaters are on 100% of the time, 24/7? no thermostat?
I did not take that into consideration. How did you get 539.6 amps
 

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I think any phase will see a max of 16 * 8.1kw. So If each phase sees 129600 watts then divide by 360 (208 * 3 sq.rt) and you get 360 amps. So yes a 600 amp feeder should be more than adequate but I screw these up alot---
Yup,see I knew I messed it up...:no::laughing:
 

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Discussion Starter · #11 ·
Vintage Sounds said:
I calculated 539.6A at 208 3 phase.

Are you saying these heaters are on 100% of the time, 24/7? no thermostat?
No, they are storage heaters. They charge at night for up to 8 hrs. Off peak rate, under 5 cents a kilowatt hour. Only cheaper heat here is ground source heat pump.

That was mine number too. But that was off the top of my head
 

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Discussion Starter · #14 ·
HARRY304E said:
kVA x 1000
1.73 x E

24x8100=194,400/1.73x208=359.84=194,400/359.84= 540.24 amps:blink::laughing:

Okay Dennis how did you come up with 129,600 watts?

I'm missing a key point:)
I think because there are 8 heaters across each phase pair. So they overlap, that's how he got 16. If you draw it out it makes sense. The rest of his calcs, I don't know. But I trust him, besides I'll put his name on the calc sheet!
 

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I think because there are 8 heaters across each phase pair. So they overlap, that's how he got 16. If you draw it out it makes sense. The rest of his calcs, I don't know. But I trust him, besides I'll put his name on the calc sheet!
Yeah,that's the ticket...:laughing:
 

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I didn't see where he mentioned anything abut delta/wye
Correct me if i'm wrong but if it's a delta wouldn't it only be 208 off of one leg?:)
 

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Discussion Starter · #17 ·
ablyss said:
I didn't see where he mentioned anything abut delta/wye
Correct me if i'm wrong but if it's a delta wouldn't it only be 208 off of one leg?:)
It's a 120/208 wye.
 

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I get what Dennis got. 16 heaters on any one phase =360.16 amps. 220.51 says calcuate at 100%, so 400 amp service should do ?
 

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HARRY304E said:
kVA x 1000 1.73 x E 24x8100=194,400/1.73x208=359.84=194,400/359.84= 540.24 amps:blink::laughing: Okay Dennis how did you come up with 129,600 watts? I'm missing a key point:)
He's figuring each unit single phase " only using 2 legs" so each phase won't have the full load of all 24 heaters, only 16 max
 

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Sorry, but you will need 600 amp service. Each group of 8 heaters will draw 311.5 amps.... 8100 x 8 / 208= 311.5. Each node will be supplying 2 groups of heaters 120 degrees out of phase with each other... 311.5 x 1.73 = 538.9. Or just work it with total wattage as was shown earlier.
 
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