- Location
- Chapel Hill, NC
- Occupation
- Retired Electrical Contractor
So what is the feeder or service load for 2- 3 kva ovens and 1- 6kva cooktop?
Using Note 3 we get 2 * 3kva= 6kva. Demand for 2 units in Col. A Table 220.55 is 75%.
6kva * .75 = 4.5 kva
Also using Col. B for 1 unit at 6 kva we get 80% demand factor.
6kva * .8 = 4.8 kva
4.5 kva + 4.8 Kva = 9.3 kva
However if we run a branch circuit to feed all 3 units we can use note 4 and add all the units together and treat it as one range. That is 12 kva-- using Col. C we get 8 kva.
So in summary we can use 9.3kva for the feeder calculation but only use 8 kva for the branch circuit load. Do you agree? do you find that odd?
Using Note 3 we get 2 * 3kva= 6kva. Demand for 2 units in Col. A Table 220.55 is 75%.
6kva * .75 = 4.5 kva
Also using Col. B for 1 unit at 6 kva we get 80% demand factor.
6kva * .8 = 4.8 kva
4.5 kva + 4.8 Kva = 9.3 kva
However if we run a branch circuit to feed all 3 units we can use note 4 and add all the units together and treat it as one range. That is 12 kva-- using Col. C we get 8 kva.
So in summary we can use 9.3kva for the feeder calculation but only use 8 kva for the branch circuit load. Do you agree? do you find that odd?