Panel Schedule Calculation

[transman2]

I'm trying to figure what size Main I need in my 3 phase panel. It will be a 3 phase 120/208 panel with 18 spaces. My loads will be 7 booths that pull around 18 amps each and will be fed with a 2 pole breaker. So if my total connected load is 52416 VA/ 208 x 1.73 = 145.66 amps. So the main would be 150 amps. Did I do this right or do I multiply by 125%. 182 amps so 200 amp panel. Not sure. Seems like I'm missing a step!

 

[augie47]

I didn't get the same number as you even considering the loads at 3 phase.
Your loads (2 pole breaker) are single phase so your (7) booths at 18 amps 208v = 26,208 va.
With 7 booths your loads won't be evenly distributed but a 100 amp panel should be sufficient (36,000 va) unless I made a mistake.

 

[transman2]

I didn't get the same number as you even considering the loads at 3 phase.
Your loads (2 pole breaker) are single phase so your (7) booths at 18 amps 208v = 26,208 va.
With 7 booths your loads won't be evenly distributed but a 100 amp panel should be sufficient (36,000 va) unless I made a mistake.

Is that figuring 18 amps for each booth?

 

[transman2]

Got it! Thanks!!!

 

[Smart $]

Is that figuring 18 amps for each booth?

It is.

7 circuits × 18A/circuit × 208V = 26,208VA​

But as Gus mentioned, your panel will be unbalanced because you cannot distribute 7 loads equally across three phases. Safest bet is to assume the next-greater-multiple-of-three, same-size loads.

9 circuits × 18A/circuit × 208 = 33,696VA
33,671VA ÷ 208V ÷ √3 = 93.6A​

You need to multiply the continuous portion of your load by 25% and add it to this figure.

Another way to do it, and achieve a less conservative value...

6 circuits × 18A/circuit × 208V = 22,464VA
22,464VA ÷ 208V ÷ √3 = 62.4A (balanced load)
62.4A + 18A (7th circuit) = 80.4A (unbalanced load)
​

Still need to add 25% of the continuous portion, though.

 

[Fnewman]

You can get pretty close by just filling out a 'theoretical' panel schedule and adding up the amps for each phase.

 

[GoldDigger]

The answers given above assume that the booths only use 208V equipment and so the 18A is at 208V.

If the booths actually have 120V loads and draw 18A at 120V from each of the two L wires to neutral it will work out to more power than the first calculation.
Not much more, but....
The panel schedule method will be exact for Line to Neutral loads, and conservative for Line to Line loads.

 

[Smart $]

...
The panel schedule method will be exact for Line to Neutral loads, and conservative for Line to Line loads.

It would be exact for line to line loads provided they are balanced and the same power factor.

 

[GoldDigger]

If you divide the line to line amps by 2 and add the magnitudes at each node, then multiply each sum by 120 you end up with the sum of the line amps times 240 instead of the line amps times 208.
Multiplying each line to line amperage by 208 and then adding the VA values is exact for power calculation.

Whether the three phase power formula then gives you the correct value for the line amps depends on the factors you mentioned.

 

[Smart $]

Multiplying each line to line amperage by 208 and then adding the VA values is exact for power calculation.

Whether the three phase power formula then gives you the correct value for the line amps depends on the factors you mentioned.

I am only referring to a panel schedule which uses VA or kVA as the row item values, where VA or kVA is equally divided to connected lines.

 

[GoldDigger]

I am only referring to a panel schedule which uses VA or kVA as the row item values, where VA or kVA is equally divided to connected lines.

And how do you calculate the VA values to enter as row items for a 12A line to line load on a two pole breaker in a 208Y/120 panel?
That is the essence of my question.
If you instead of working directly from the 12A you calculate 12 x 208 and then divide that between the rows, it is indeed exact (under the conditions you specified.)
How would you enter the VA value for a 12A line to neutral load? Presumably by multiplying 12 x 120 and entering it all on the one row. Also exact under the conditions specified.

 

[Smart $]

And how do you calculate the VA values to enter as row items for a 12A line to line load on a two pole breaker in a 208Y/120 panel?
That is the essence of my question.

... VA or kVA is equally divided to connected lines.

A 12A 208V load (2,496VA), connected line to line, say A to B, gets 1,248VA enter under Line A (commonly ØA) and 1,248VA entered under Line B (commonly ØB).

A 12A 120V load connected to Line A gets all 1,440VA entered under Line A (ØA).

A 3Ø load is connected to all three lines, so one-third in each of the three "Line" columns on the respective row.