line to line loads

[john d]

How to properly size a 3phase panel with all singe phase loads. I have several elect. heater that draws 3664 watts 480v line to line. how do i find the actual amp draw to add to the total amps for each phase to size a panel.

 

[charlie b]

First of all, don't add amps. Convert everything (i.e., each load, one by one) to KVA, do your best to equalize the load among the three phases, add up all the KVA, and convert back to amps as the last step.

 

[LarryFine]

Don't forget to add some headroom for less-than-ideal balancing. The highest individual phase's load must still be accomodated.

 

[john d]

I am still a bit comfused, in my question i get 8 amps i think that is wrong and then do i put 8 [or the correct answer ]in column A and B [or whatever 2 phases i use]

 

[Smart $]

I am still a bit comfused, in my question i get 8 amps i think that is wrong and then do i put 8 [or the correct answer ]in column A and B [or whatever 2 phases i use]

You put one-half 1? line-to-line load's kVA in panel schedule columns.

In your case of a 3,664W heater, VA equals watts. Converted to kVA would be 3.664kVA... So you put 1.83 in each of the two columns a load is connected to... or multiples thereof if connecting more than one heater per circuit.

 

[charlie b]

. . . in my question I get 8 amps (I think that is wrong). . . .

3664 divided by 480 is about 7.6, so 8 is a close enough answer.

. . . and then do i put 8 . . . in column A and B. . . .

No. That is my earlier point. You don't put amps into any column. Never try to add amps to amps. It very seldom gives you the right answer.

You already have the load in terms of power. You enter the KVA values into the two columns, as Smart$ has already described. After you enter all the loads, add the KVA for each phase, to see if they are close to being evenly spread out. Make some changes, if needed, to get the loads closely balanced.

Then you add up all the KVA in all three phases. That gives you the total KVA for the panel. Multiply by 1000 to get total VA. Finally, you divide the total VA by 480 volts, and divide again by 1.732 (the square root of 3). That will give you the amps. At this point, since you will have already balanced the phases, it will be the same amps on each of the three phases. But do not try to add the three to get a total amps.

Example: Suppose your total KVA was 65.5 (65,500 VA). You divide that by 480, and get 136.5. Divide that by 1.732, and get 78.8. That tells me you have about 79 amps on phase A, 79 amps on phase B, 79 amps on phase C, and a "total" (I do not like using that word in this context) of 79 amps (not, Not, NOT 237 amps).

 

[jumper]

You put one-half 1? line-to-line load's kVA in panel schedule columns.

In your case of a 3,664W heater, VA equals watts. Converted to kVA would be 3.664kVA... So you put 1.83 in each of the two columns a load is connected to... or multiples thereof if connecting more than one heater per circuit.

Where does 1.83 come from?

 

[charlie b]

Where does 1.83 come from?

It is half of 3.66. For a single phase, 480 volt load that is connected between phases A and B, for example, half of the KVA load should be assigned to phase A, and the other half to phase B.

 

[jumper]

It is half of 3.66. For a single phase, 480 volt load that is connected between phases A and B, for example, half of the KVA load should be assigned to phase A, and the other half to phase B.

Got it. thanks. I was working the problem out in VA and not KVA and missed his decimal point.