YesShouldn't the demand load then be 8.8kw? using note 1
12 + 2.2 = 14.2= 14Average the two using 12 as the smallest number...gives you an average of 17.... 5% for each kw over 12... If I did the math correct = 14
I still don't get it. Note #4... This is for branch circuit but I assume it would be the same for the demand loadLook at Example D6B in Annex D (Pg 799 in '17)
I'm confused now, is the answer 8800W or 10000W lolIf someone thinks note #4 doesn't apply for demand load then you would have to use each one individually. The 11.5 kw = 12 kw = 8kw demand from table
2.2 kw *80% = 1.76 =
8kw + 1.76 Kw = 9.76= 10 kw
IMO... it 8.8I'm confused now, is the answer 8800W or 10000W lol