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Monday, October 10, 2016
Question:
I have a question that involves 220.55 and how to calculate single phase loads, 120/208v connected to a three phase system. The code language gives us direction, “where two or more single-phase ranges are supplied by a 3-phase, 4-wire feeder of service, the total load shall be calculated on the basis of twice the maximum number connected between any two phases.” The problem is that the code language does not describe the method that is illustrated in Annex D 5(a). Using the same example of 10 ranges, the maximum number between any two phase legs is 4 x 2 = 8 ranges. The demand in 220.55 column ‘C’ for 8 ranges is 23 kw. According to 220.55 code language, 23 kw should be the maximum demand for 10 ranges connected to a three phase system. In keeping with 90.3, (“informative annexes are not part of the requirements of this code…”) is the method illustrated in Annex D 5(a), part of the process for finding the maximum demand for the 10 range scenario? Please verify if this is correct or not.
Shawn Haggin
A
Answer:
Hey Shawn thanks for your question. The example for range calculations in Informative Annex D5(a) is correct but it is based on 20 ranges. The calculation is based on twice the number of ranges that are connected to 2 of the phase conductors as stated in 220.55. It might be helpful to create a diagram that shows ranges connected A-B, B-C, and C-A which could result in 7 ranges A-B, 7 ranges B-C, and 6 ranges C-A totaling 20 ranges. The maximum number connected between any two phases is 7, twice that is 14 which is used in Table 220.55 resulting in a 29 KVA demand for those two conductors.
A simple method is to multiply the total number of ranges by 3 (3 phase system) then round up to the next larger number if there is any fraction, multiply that by 2 and use the result in Table 220.55.