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# 3 phase calculation

#1
04-26-10, 06:30 PM
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3 phase calculation

I'm trying to calculate the amperage load for an oven that will have (4) 3000 watt heating elements(240 volts). My uncertainty is how to compute the amperage load of 4 elements spread over 3 phases.

If the load were balanced(12000 watts total) I think the load would be around 29 amps.

But since there are 4 elements divided over 3 phases that would place one element on A&B A&C B&C and the fourth on say A&B. So the A and B phases have 3 element connections each and the C phase only 2. How do you compute this load?

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#2
04-26-10, 07:38 PM
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If this is a school question I believe the number they are looking for, is the one you came up with, 29(28.9)amps. You might be over thinking the question.

Watts/(VoltsxPFx1.73) = Amps

If it was me, I would just look at the nameplate

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04-26-10, 08:56 PM
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Not a school question and there's no nameplate as the oven is custom built. The question is more for my curiosity than anything else.

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04-27-10, 05:11 AM
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I'm only guessing here since i can't find an exact formula, but I would think you would use the formula I posted above for the balanced load (9000 watts) and then add the unbalanced load using the single phase formula: P/E

Balanced load = 9000 watts
9000/(240x1x1.73) = 21.6

Unbalanced load = 3000 watts
3000/240 = 12.5

Total amps = 34.1 amps

These numbers will change if you connect it to a 208 volt system.

#5
04-27-10, 06:33 AM
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Thanks Tolyn, I was hoping there was a simple solution/calculation. In searching I did find a reference to Fortescue Transformation that was cited for use in an unbalanced 3 phase load. The calculations and theory within are beyond my capabilities and knowledge, but I'm thinking your calculation would be worst case so that may be sufficient.

Last edited by ednu99; 04-27-10 at 07:03 AM.
#6
04-27-10, 06:52 AM
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Originally Posted by ednu99
Not a school question and there's no nameplate as the oven is custom built. The question is more for my curiosity than anything else.
You may want to be careful with installing a custom piece without a nameplate. If you have a firecracker for an inspector, technically everything must have a UL or comperable listing or label. If you don't have that, they can force you to get it and it is not cheap. Not even close to cheap.

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04-27-10, 11:28 AM
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My calculation presumes the system available is a 3-phase , 4-wire, 120 / 208 volts "Wye" system ; if not , then you are connecting to a 3-wire "Delta" system.

The current for each element = 3000W / 208 V = 14.4

Three loads are connected "Delta" , with two elements connected to each Phase lead.
The three Phase currents = 14.4 amps X 1.73 = 25 . If you connect a 4th element to say Phase-leads A & B , then the current in these two Phase leads = 25 + 14.4 = 40 (approx ).

If classified as a "continuous load" , the ampacity of the Branch Circuit conductors = the line-current X 1.25

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04-27-10, 09:07 PM
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This is my figures that on 240 volt circuit

Phase A and Phase B you will have 6000 watts on that circuit while the B/C and C/A each will have 3000 watts so here the quick way I do the about the simauir way as Toyln did

9000/(240X1X1.73) = 21.6 amps that is balanced loads

Now add one more heating element in the mix

3000/240 = 12.5 amps

so for phase A and B you will get 34.1 amps while the other two phase will get 21.6 amps

Now the sticky part is sizing the OCPD { over current protection } if this is strictly heating elment only no fan motours or anything else it pretty straght foward it simple to do that with fuse ( 2X40 and 1X30 ) but with breakers it get tricky you will need 40 amp 3 pole breaker on this one.

If this is a commercal oven or custom made with unbalanced load I am pretty sure some of them will have fused disconenction switch to address this issue.

Merci,Marc

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04-28-10, 07:02 PM
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Happy to see I wasn't too far off!

Thanks Marc!

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