I'm looking at the example calculations in the NEC (2023) Handbook.
Here is the example question in 220.54:
Where does the 208 V in that last line of calculations come from?
Here is the example question in 220.55:
Where does the 240 A come from?
Additionally, and maybe I should post this in another part of this form, I think that there are some mistakes in this calculation example (220.55). I think that they need to write 11.6 kVA, rather than just VA. Also, I think that the 240 A should be 240 V.
Thank you in advance for any help. I would have formatted this more nicely, but I couldn't figure out how to include LaTeX on this post. Also, for some reason, when I pasted this text here from the LaTeX document I was working in, I got an error that took me over an hour to figure out. Turns out that I needed to cut the word 'system' out of the first question and paste it back in. Only then was I able to post this. Strange...
Here is the example question in 220.54:
Assuming a load of 10 single-phase dryers, each with a nameplate rating of 5500 W, is connected as evenly as possible to the 3-phase system (3 dryers connected between phases A and B, 3 dryers connected between phases B and C, and 4 dryers connected between phases A and C), what is the minimum calculated load?
Solution
Although no VA-rating is provided, the 5500 W may be treated as a VA-rating, according to 220.54 (kVA shall be considered equivalent to kW for loads calculated in this section). Since there is more than one dryer present on this 3-phase, 4-wire system, the total load shall be calculated on the basis of twice the maximum number connected between any two phases, which in this case is four.
4 dryers x 2 = 8 dryers
Refer to Table 220.54 using the calculated 8 dryers; the demand factor is 60%.
5500 VA/dryer x 8 dryers x 0.06 = 26400 VA
The dryers are connected between two phases; find the load on a single phase.
26400 VA / 2 phase loads = 13200 VA/phase load
A three phase system carries this load; find the load on the entire service.
13200 VA/phase load x 3 phase loads = 39600 VA
Find the actual load on each of the feeder conductors.
39600 VA / 208 V x sqrt (3) = 110 A
Where does the 208 V in that last line of calculations come from?
Here is the example question in 220.55:
Calculate the load for a single branch circuit that supplies the following cooking appliances:
- One counter-mounted cooking unit with rating of 8 kW
- One wall-mounted oven with rating of 7 kW
- A second wall-mounted oven with rating of 6 kW
Solution
Table 220.55 applies. Since there is one counter-mounted cooking unit and two wall-mounted ovens, Note 6 applies: add the nameplate rating of the individual appliances and treat this total as equivalent to one range.
8 kW + 7 kW + 6 kW = 21 kW
Now, use Note 1. Determine the amount of rating above 12 kW.
21 kW - 12 kW = 9 kW
According to Note 1, the maximum demand in Column C shall be increased 5 percent for each additional kilowatt of rating or major fraction thereof by which the rating of the individual ranges exceeds 12 kW.
5 %/kW x 9 kW = 45 %
One range corresponds to a maximum demand on Column C of 8 kW. The maximum demand is increased by the percentage found above.
8 kW x 145 % = 11.6 kW ~= 11.6 VA
Calculate the branch circuit load.
11,600 VA / 240 A = 48.3 A = 48 A
Where does the 240 A come from?
Additionally, and maybe I should post this in another part of this form, I think that there are some mistakes in this calculation example (220.55). I think that they need to write 11.6 kVA, rather than just VA. Also, I think that the 240 A should be 240 V.
Thank you in advance for any help. I would have formatted this more nicely, but I couldn't figure out how to include LaTeX on this post. Also, for some reason, when I pasted this text here from the LaTeX document I was working in, I got an error that took me over an hour to figure out. Turns out that I needed to cut the word 'system' out of the first question and paste it back in. Only then was I able to post this. Strange...