Calculating Total amps across 3 phase using 4 2 phase loads

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cloudfire181

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Hi was wondering if someone could point me toward a link or help solve this problem i'm having in finding total current being used across the 3 phases. So here is the problem I'm using 4 heater sets, each heater set has 2 heater in parallel with 208 Volts running across and the watts per heater set is 2000 Watts or 1000 Watts per heater. The amps I calculated per heater was 4.81 Amps or 9.62 Amps for the set I rounded up to 10 Amps, with 2 heater sets on AB, 1 on BC, and 1 AC. Now I working with a 30 Amp 3 pole breaker I was wondering if it would trip the breaker because there was 30 amps on phase A, Phase B, and 20 Amps on Phase C. Or I'm I wrong in thinking that there is 30 amps on A and B and 20 on C. Any help on this would be great.
 
cloudfire181 said:
Hi was wondering if someone could point me toward a link or help solve this problem i'm having in finding total current being used across the 3 phases. So here is the problem I'm using 4 heater sets, each heater set has 2 heater in parallel with 208 Volts running across and the watts per heater set is 2000 Watts or 1000 Watts per heater. The amps I calculated per heater was 4.81 Amps or 9.62 Amps for the set I rounded up to 10 Amps, with 2 heater sets on AB, 1 on BC, and 1 AC. Now I working with a 30 Amp 3 pole breaker I was wondering if it would trip the breaker because there was 30 amps on phase A, Phase B, and 20 Amps on Phase C. Or I'm I wrong in thinking that there is 30 amps on A and B and 20 on C. Any help on this would be great.
Click to expand...

You have 4 heater that = 10 amps each(rounded). If you connect 3 of the heaters AB, AC, BC you would have 10 amps on each phase.
You can then add the 4 heater to any of the 3 phases. Lets Say you add it to AB. Then AB will have and additional 10 amps.
Load would be approx. AB 20, AC 10, BC 10. That is assuming all of the heaters are on at the same time.
 
Vectors being what they are, 20A + 10A does not equal 30A, so to that extent you are okay.
But if you have to consider these heaters as continuous loads you could still have an NEC problem.
I have not done the detailed math.
 
Using "simple" math, you have 1080 volt-amps/heater/line. Lines with 3 heaters connected will be 3240 VA, one with 2 heaters is 2160 VA. Divide by 120 you get 27A and 18A respectively.

Not 100% accurate, but close enough for NEC work. :D
 
Another approach, not 100% accurate but good as a "sanity check," would be to assume (just for a moment) that the load was balanced. You have 8000 VA of load, and if balanced on a 208 volt, 3-phase system, the current would be about 22 amps. You could add a 25% "margin of safety" factor on top of that, and still be under your desired 30 amp limit.
 
Using my current and phase reference
Ia into N -24 ang 200
Ib into N 14.4 ang 30
Ic out of N -9.6 ang 0

when summed = 0 or balanced

V ref
a-b 208 ang 0 ... 4 heaters
b-c 208 ang 120 ...2 heaters
c-a 208 ang -120 ...2 heaters

I converted each heater to a 43.3 + 0j Z
 
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Since 3 of the 2kw heaters will be on different phases, are purely resistive and balanced, could one treat them collectively as a single 3ph load?:

6000W/208/1/732 = 16.7A

and add the 4th (9.6A/ph, say across AB phase) giving 26.6/16.7/16.7A across the breaker?

Even so, fixed heating is a continuous load so you'd be limited to 24A draw on a 30A breaker, yes? and that load cannot be averaged across all 3 phases. imo, breaker wont trip but it would be overloaded per NEC rules.
 
Smart $ said:
Using "simple" math, you have 1080 volt-amps/heater/line. Lines with 3 heaters connected will be 3240 VA, one with 2 heaters is 2160 VA. Divide by 120 you get 27A and 18A respectively.
...
Click to expand...
Just realized there's an error in my math. Should be 1040 volt-amps/heater/line, which makes the result 26.0A and 17.3A respectively.

Using vector math, with 4th heater connected AB:
A=26.5A@-349.1° (or 10.9°)
B=26.5A@-130.9°
C=17.3A@-240.0°
 
JFletcher said:
Since 3 of the 2kw heaters will be on different phases, are purely resistive and balanced, could one treat them collectively as a single 3ph load?:

6000W/208/1/732 = 16.7A

and add the 4th (9.6A/ph, say across AB phase) giving 26.6/16.7/16.7A across the breaker?

Even so, fixed heating is a continuous load so you'd be limited to 24A draw on a 30A breaker, yes? and that load cannot be averaged across all 3 phases. imo, breaker wont trip but it would be overloaded per NEC rules.
Click to expand...

I'm more agreeable with this numbers. I did my own calcualtions and got 25.44A on two of the lines and a 16.65 on one of the three lines. Yes, a 30A breaker would be just fine!:D
 
Gentlemen:

The rule 'Power=1.732*V*I ' is applicable in a 3 phase system for balanced loads only.

It is not applicable here.
 
Sahib said:
Gentlemen:

The rule 'Power=1.732*V*I ' is applicable in a 3 phase system for balanced loads only.

It is not applicable here.
Click to expand...
But if you look at a hypothetical balanced load which is larger than the real load and use that rule, the real current will equal to or smaller than that hypothetical current.
 
Sahib said:
smart$:
It is a 208V load and not 120V load as the OP stated ' using 4 2 phase loads'........
Click to expand...
Technically NOT 2 phase loads. Accurately described they are four (4) line-to-line loads, or four (4) 208V 1Ø loads.

As to using 120V instead of 208V, we are taking advantage of the panel schedule method of VA load divided by line connections. The result is divided by the line-to-neutral voltage.

Total load for 3 heaters:
3(qty)×10A×208V=6240VA

Balanced panel schedule method:
6240VA÷3(lines)=2080VA/line
2080VA÷120V/line=17.33A

Traditional 3Ø load calculation, and why the panel schedule method is a fair aproximation:
6240VA÷208V÷sqrt(3)=17.33A
6240VA÷[120V×sqrt(3)]÷sqrt(3)=17.33A
6240VA÷120V÷3=17.33A
 
Here's mine
sorry but I scribble
converted the loads to a Z
1000/208 = 4.8 A
208/4.8 = 43.3 ohm or 43.3 + 0 j
please check, won't be the first time I made a math or set-up error
won't be the last either lol
 
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