load on 3 phase

[ohmhead]

Well i have a question balance load on 3 phase panel .


I was asked why so here it goes if you had 48 motors each at 11 amps 208 volts single phase no other load in panel .

So i took the 48 motors and equally put 16 on each phase A B C .

Next i took the total wattage of each phase with 16 motors attached 11 amps each and total them up of 16 on each A or B or C using one phase total on one leg .


Since there all the same thats what i used for my load i picked one phase and came up with around 178 amps .


I took the largest motor of the 16 and went 125 % of added that into load and picked my wire size .


I figured around a 200 amp panel and wire size ?


What is your way ?


I was shown to balance the load on each phase .


That ab bc ca rotation wise and that each motor only sees ab bc ca if iam wrong please let me know ?

 

[david luchini]

Well i have a question balance load on 3 phase panel .


I was asked why so here it goes if you had 48 motors each at 11 amps 208 volts single phase no other load in panel .

So i took the 48 motors and equally put 16 on each phase A B C .

Next i took the total wattage of each phase with 16 motors attached 11 amps each and total them up of 16 on each A or B or C using one phase total on one leg .


Since there all the same thats what i used for my load i picked one phase and came up with around 178 amps .


I took the largest motor of the 16 and went 125 % of added that into load and picked my wire size .


I figured around a 200 amp panel and wire size ?

Your 200A panel will be undersized. Each motor load is 11A*208V= 2288VA.

48 motors total will be: 48 * 2288VA = 109,824VA.

On a 208V, 3phase system: 109,824VA/208V/1.732=304.8 Amps.

Adding 25% to one motor would give you line currents of 307.2/307.2/304.8.

Where you took 11A time 16 motors per phase, you should also multiply by 1.732, since the motors are phase to phase connected.

 

[charlie b]

Where you took 11A time 16 motors per phase, you should also multiply by 1.732, since the motors are phase to phase connected.

I see that you and I both infer that the OP made the mistake of using the phase to neutral voltage of 120. That would give the OP's answer. The motors are connected across phase to phase, so the voltage is 1.732 times 120, or 208. I agree with your results.

 

[david luchini]

I see that you and I both infer that the OP made the mistake of using the phase to neutral voltage of 120. That would give the OP's answer. The motors are connected across phase to phase, so the voltage is 1.732 times 120, or 208. I agree with your results.

Interesting...I hadn't really considered that the mistake was using the phase to neutral voltage.

So i took the 48 motors and equally put 16 on each phase A B C .

The mistake, it seemed to me, was the concept of "connecting" the 208V motors to Phases A, B and C. In other words, 11 Amps per motor, 16 motors per phase = 176Amps per phase.

But of course, balancing the 208V motors would not give you 16 motors connected to each phase, it would give you 32 motors connected to each phase: 16 from A-B and 16 from A-C, etc. The balanced condition would give you 16 motors connected between each of the the phases.

 

[iwire]

What is your way ?

I would use the VA and not amps.

 

[ohmhead]

Thanks for the input were having a debate on how rotation works and why we add these up in that order of 16 to 32 .

And why ?

 

[david luchini]

Thanks for the input were having a debate on how rotation works and why we add these up in that order of 16 to 32 .

And why ?

I'm not sure I understand what you're asking about rotation.

With your 48 motors connected in a delta connection, the current on phase A will equal the load current from A-B, minus the load current from B-C, or Ia=Iab-Ica.

I think by rotation, you are referring to the phase angle between the load currents. If we assign 0 degrees to the load on A-B, then the load on B-C will be at -120 degrees, and the load on C-A will be at -240 degrees.

We know that the load current for each group of 16 motors is 176A (16*11) so we can assign load currents of Iab=176<0, Ibc=176<-120, Ica=176<-240. With Ia=Iab-Ica, we have Ia=(176<0)-(176<-240).

This solves to Ia=304.8<-30 Amps.

 

[kingpb]

I would use the VA and not amps.

Good answer!

When dealing with panels and panel loads, regardless of type, you should always use KVA.

Total KVA: (11 x 208V x 48) + (11 x 208V x 0.25) = 110.4KVA

The service is 3ph so: 110.4KVA/(208 x 1.732) = 306A/phase

Use 400A Rated Panelboard with a 400A main, done.