Given the following Full Load Currents (FLC):

- Phase AB = 38 A continuous
- Phase BC = 36 A continuous
- Phase CA = 31 A continuous

- A = 59.9 A
- B = 64.1 A
- C = 58.1 A

Thanks!

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Given the following Full Load Currents (FLC):

- Phase AB = 38 A continuous
- Phase BC = 36 A continuous
- Phase CA = 31 A continuous

- A = 59.9 A
- B = 64.1 A
- C = 58.1 A

Thanks!

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Your calcs need to be with 208 not 120. And your protecting the conductors feeding the other panel. So size at 1.25 percent of conductors. Screw the total load. Your ocpd cares about the conductor not what your putting on it.This is my first time calculating a breaker for a three-phase source.

I have a fuse panel (we’ll call it FP #1) supplying another fuse panel (FP #2) withthree phase power.

The output of FP #1 is fitted with a 50 A breaker. FP #2 is connected to 80 loads(unbalanced)which total the following:

Apparent Power: 12000 VA

Voltage: 120 V

Frequency: 60 Hz

Full Load Current (FLC):

I calculated the FLC for each phase to be:

- Phase AB = 38 A continuous
- Phase BC = 36 A continuous
- Phase CA = 31 A continuous

I feel like someone has put a 50 A breaker on FP #1 based on the maximum line-to-line current of FP #2 (38 A * 1.25 = 47.5). Am I correct in thinking that a three-phase breaker should be rated based on the average of the phase currents (59.9 + 64.1 + 58.1)/3 = 60.7A; 60.7A * 1.25 = 80 A breaker on a 80 A+ rated cable.

- A = 59.9 A
- B = 64.1 A
- C = 58.1 A

Thanks!

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Edit: Oh wait, I remember you now! Okay, we can work this out.. but I think you're in over your head. lol

That is what I would use for balanced loads; I'm under the impression that you can't use that for unbalanced loads?12000/(208x1.732) hint hint

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where do you live that you have three phase in residential

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Give out the data (homework) you used to calculate this and we can show you how to do it right.

For phase A:Give out the data (homework) you used to calculate this and we can show you how to do it right.

x=0.866*(AB + CA)

y=-0.5*AB + 0.5*CA

z=x + yj

|z|=(x+y) ^ 1/2

Current in Line 1 (Phase A) = 59.9A

This process is useful for unbalanced loads. For balanced loads I'd simply sum the currents, multiply by 3^1/2, and divide by 3. This value would then be used to find the circuit breaker rating (times FLC by 1.25 if continuous).

Hope this helps you understand what I'm trying to wrap my head around.

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And I thought I knew something. Now even apprentice calcs look like gibberish to meFor phase A:

x=0.866*(AB + CA)

y=-0.5*AB + 0.5*CA

z=x + yj

|z|=(x+y) ^ 1/2

Current in Line 1 (Phase A) = 59.9A

This process is useful for unbalanced loads. For balanced loads I'd simply sum the currents, multiply by 3^1/2, and divide by 3. This value would then be used to find the circuit breaker rating (times FLC by 1.25 if continuous).

Hope this helps you understand what I'm trying to wrap my head around.

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First disregarding nec:

look at the name plate flc on all 3 phase motors that current will be on all 3 legs.

Y208/120

Ex: 3phase 20a motor.

Ex: A 20a B 20a C 20a

Next add the flc for and loads across 2 poles that current will be on. AKA phase it is attached to.

Ex: Single phase 20 amp 208 volt motor. On phase a and b

Ex: 1000va heat strip on phases A and C. 1000/208=5a

Ex: A 45a B 40a C 25a

Last add any single phase loads attached to single phases.

Ex: 2400w microwave on phase C. 2400/120=20

Ex: A 45a B 40a C45a is the total current.

The nec states you must use 125% of the largest motor load (art 430) and any continuous loads.

Pro tip: When you are taking a test add another line to your line items label it 25% largest motor. That method makes it easier to check yourself and make changes.

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I would round to the 80-amp breaker instead of the 85-amp breaker as far as the calculations go. If I was installing #4, I would use the 85-amp breaker.

Given the following Full Load Currents (FLC):

I calculated the FLC for each phase to be:

- Phase AB = 38 A continuous
- Phase BC = 36 A continuous
- Phase CA = 31 A continuous

I'm wondering how three-phase breakers work. 64.1A x 1.25 = 80.125A. Does this mean a 85A breaker would be ideal for protecting this circuit as it covers our highest load (phase B)?

- A = 59.9 A
- B = 64.1 A
- C = 58.1 A

Thanks!

Kind of a round about way of saying yes.

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ship to www.mikeholt.com

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Why they need more trolls there?ship to www.mikeholt.com

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