1 Phase Loads on 3 Phase system Calculations.

[JArch]

Yes, again. I have looked and can not find anything that pertains to what I need.

I am trying to calculate the wattage and load on a three phase system that consists of only single phase loads. All loads are resistive.
I have qty 6 units at 2.6 amps, 536 watts @208
qty 1 unit at 23.2 amps, 4815 watts @208
qty 1 unit at 28.9 amps, 6015 watts @208
qty 2 units at .43 amps, 89.44 watts @208
qty 1 unit at .23 amps, 47.84 watts @208.

The only set up that is making any sense so far is:

L1 L2 L3
268 268
268 268
268 268
268 268
268 268
268 268
2407.5 2407.5
3007.5 3007.5
3007.5 3007.5
89.44
89.44
47.84
______________________________
6445.72 6755 7087 = 20287.72

20287.72/208/1.732 = 56.3 Amps

I need to know amp draw and watt total. Not necessarily related to services in the NEC.
From other threads I've read, I kind of understand that you can't use the 1.732 on the unbalanced part. So do I take the difference from 6445.72 and 7087 to get 641.28 /120 for additional amps on that leg?

I would love someone to figure this out for me and then tell me how to do it on my own.
Thanks in advance!!

 

[david luchini]

I need to know amp draw and watt total.

Your watt total is simply the sum of all the watts of your loads, or 14,272.72 Watts.

The amp draw will depend on how the 11 single phase loads are configured. Assuming you are looking for the load to be as balanced as possible, lets say you connect the 23.2A load between A and B, the 28.9A load between B and C and the 9 small loads between C and A. So;

Iab=23.2
Ibc=28.9
Ica=16.69

The amp draw on each line is then Ia=Iab-Ica, Ib=Ibc-Iab, Ic=Ica-Ibc;

So;

Ia=34.7
Ib=45.2
Ic=39.9

 

[dkarst]

David's answer is correct of course but for the benefit of the original poster, remember these have to be added as phasors, i.e. paying attention to both magnitude and phase angle. Adding them arithmetically will give you an incorrect solution.

 

[JArch]

my total wattage is 20287.72. Assuming I balance as best I can, do I use the 1.732 correction factor on the entire amount? There is some conflicting information on the web. Do I divide by 1.732 or multiply by it?
20287.72/208/1.732 or 20287.72/208*1.732?

also, when you put your wattage in columns, do you split the wattage between the two phases or use the full amount? Splitting the watts between the phases seems to make more sense, mathmatically.

My original post messed up some how and doesn't show my balance correctly.

L1	L2	L3
268	268
	268	268
268		268
268	268
	268	268
268	268
2407.5	2407.5
	3007.5	3007.5
3007.5		3007.5
44.72	44.72
44.72	44.72
23.92	23.92
6600.36	6868.36	6819

 

[JArch]

Your watt total is simply the sum of all the watts of your loads, or 14,272.72 Watts.

The amp draw will depend on how the 11 single phase loads are configured. Assuming you are looking for the load to be as balanced as possible, lets say you connect the 23.2A load between A and B, the 28.9A load between B and C and the 9 small loads between C and A. So;

Iab=23.2
Ibc=28.9
Ica=16.69

The amp draw on each line is then Ia=Iab-Ica, Ib=Ibc-Iab, Ic=Ica-Ibc;

So;

Ia=34.7
Ib=45.2
Ic=39.9

This math does not make sense to me... 23.2-16.69=6.51 not 34.7. Can you please go through all of your steps?

 

[david luchini]

my total wattage is 20287.72.

Your original post indicates only one 6015 watt load, but you are using two 6015 watt load in your calculation. Which is correct?

Assuming I balance as best I can, do I use the 1.732 correction factor on the entire amount?

Yes.

There is some conflicting information on the web. Do I divide by 1.732 or multiply by it?
20287.72/208/1.732 or 20287.72/208*1.732?

Divide: 20287.72/208/1.732 or 20287.72/(208*1.732). With the loads nearly balanced this will give you a close approximation of the line currents, but not the exact line currents.

This math does not make sense to me... 23.2-16.69=6.51 not 34.7. Can you please go through all of your steps?

As Dennis mentioned, currents are phasors, with magnitude and angle. The angle between each of the currents is 120 degrees, so they can be written as:

Iab=23.2<0
Ibc=28.9<-120
Ica=16.69<-240

So;

Iab=23.2<0-16.69<-240,
.....=(23.2+j0)-(-8.35+j14.45)
.....=(31.55-j14.45)
.....=34.7<-24.6

 

[JArch]

Thank you!
Yes, my original post was wrong and you would have caught that when I listed the phases, had they posted correctly.
This makes sense to me now, thanks for spelling it out.
:thumbsup: