Could someone help me on this. I have a 3phase 4wire circuit. All loads are 120Volts. Phase A=5Amps, B=10Amps, C=15Amps All wires are # 12THHN all 3 C/Bs are 20Amps each. How many amps are there on the nuetral? What is the fast way to figure this out in the field? What is the actual formula for this type of a circuit? Assume that all loads are continuos. Any help is appreciated.
3 Phase 4 Wire NUETRAL?
- Thread starter kiss
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ron
Senior Member
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- New York, 40.7514,-73.9925
Re: 3 Phase 4 Wire NUETRAL?
Halfway down on
http://www.mikeholt.com/technical.php?id=calculations/technicalcalculationformulas
[ November 05, 2004, 02:47 PM: Message edited by: ron ]
Halfway down on
http://www.mikeholt.com/technical.php?id=calculations/technicalcalculationformulas
[ November 05, 2004, 02:47 PM: Message edited by: ron ]
Re: 3 Phase 4 Wire NUETRAL?
I get 13.1 amps. The formula gets rather complex, but here is what I came up with:
I neutral = sqrt( (Ia + Ib * cos 120 + Ic * sin 240)^2 + (Ib * sin 120 + Ic * cos 240)^2)).
This formula basically divides the Ib and Ic currents into components that are at right angles to Ia.
One tip makes this calculation much easier:
Calculate each part one at a time, writing each number down as you go. For example, I get:
sqrt((5 - 5 - 13)^2 + (8.66-7.5)^2)
then: sqrt(-13^2+1.66^2)
(when you square -13, it becomes a + number. Don't just square the 13 and leave the negative sign).
then: sqrt(171) = 13.1 amps.
Also, make sure your calculator is set for degrees.
Steve
I get 13.1 amps. The formula gets rather complex, but here is what I came up with:
I neutral = sqrt( (Ia + Ib * cos 120 + Ic * sin 240)^2 + (Ib * sin 120 + Ic * cos 240)^2)).
This formula basically divides the Ib and Ic currents into components that are at right angles to Ia.
One tip makes this calculation much easier:
Calculate each part one at a time, writing each number down as you go. For example, I get:
sqrt((5 - 5 - 13)^2 + (8.66-7.5)^2)
then: sqrt(-13^2+1.66^2)
(when you square -13, it becomes a + number. Don't just square the 13 and leave the negative sign).
then: sqrt(171) = 13.1 amps.
Also, make sure your calculator is set for degrees.
Steve
Re: 3 Phase 4 Wire NUETRAL?
You can also go to this thread and substitute your numbers into the calculation in my post of September 21, 2004 11:30 AM.
Roger
You can also go to this thread and substitute your numbers into the calculation in my post of September 21, 2004 11:30 AM.
Roger
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- Illinois
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- retired electrician
Re: 3 Phase 4 Wire NUETRAL?
Steve,
I get 8.66 amps.
Don
Steve,
I get 8.66 amps.
Don
ron
Senior Member
- Location
- New York, 40.7514,-73.9925
Re: 3 Phase 4 Wire NUETRAL?
Kiss,
If it were straight resistive load, then you might guess it is less than the phase current. If there are harmonics, then it may exceed the phase current amount, depending on if you are using a true RMS meter or an averaging RMS meter.
Kiss,
If it were straight resistive load, then you might guess it is less than the phase current. If there are harmonics, then it may exceed the phase current amount, depending on if you are using a true RMS meter or an averaging RMS meter.
Re: 3 Phase 4 Wire NUETRAL?
I'm kind of rusty at these calculations, and I did it in kind of a hurry. I have the cos and sin switched on the Ic phases. The correct formula is:
I neutral = sqrt( (Ia + Ib * cos 120 + Ic * cos 240)^2 + (Ib * sin 120 + Ic * sin 240)^2)).
That gives 8.6 amps instead of the 13 I got.
Steve
I'm kind of rusty at these calculations, and I did it in kind of a hurry. I have the cos and sin switched on the Ic phases. The correct formula is:
I neutral = sqrt( (Ia + Ib * cos 120 + Ic * cos 240)^2 + (Ib * sin 120 + Ic * sin 240)^2)).
That gives 8.6 amps instead of the 13 I got.
Steve
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