220.19

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luke warmwater

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In a residential load calculation, say we have a 16kW electric range. The demand rating for this range in the calculation would be 9600W. Code actually tells us how to do this in the notes, but does anyone know the easier alternate math method for getting the same answer?
 
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bphgravity

Senior Member
Location
Florida
Re: 220.19

16kw - 12 kw = 4kw x 5% = 20% + 8 = 9.6kw

I can't think of an easier way to teach this, and I know this is how Mr. Holt and others like Tom Henry show this calculation. I know electricians that have worked every calculation from 13kw to 27kw and wrote the value in the blank space within the column for fast and easy reference.

Ex: 13=8.4
14=8.8
15=9.2 and so on :confused:
 
roger

roger

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Staff member
Location
Fl
Occupation
Retired Electrician
Re: 220.19

Todd, I dont know how you're doing the math, but this is my method

Reduce to 8KW
mutiply by 120% = 5% of each KW over 12

8000 x 120% = 9600

This may not be what you're looking for.

Roger
 
roger

roger

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Staff member
Location
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Occupation
Retired Electrician
Re: 220.19

Todd, for the 8KW value yes. But this will change with each step as far as the column that might apply and the notes to 220.19

Example for two ranges of 16KW using column C we would reduce to 11KW

11000 x 120% = 13200
13200 - 11000 = 2200
2200 / 4 = 550

I guess if you could remember each constant this would be a way to do it, but I think I will stay with the 100% plus adder. :D

Roger
 
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