ce2two

[ce2two]

question :1999 master electricians study guide.what is the demand load in va for a single range with a 18kw rating ,my answer was 18kw times 5% equals ???? the code states above 12kw must multiply 5% ,where did i go wrong ..........they have 4 answers (1)8,450,(2)9,560#(3)10,100(4)10,400 none of these come close to my answer...........

 

[celtic]

It's not 18kw*5%...it's:

T220.19
Notes: 1. Over 12 kW through 27 kW ranges all of same rating.
For ranges individually rated more than 12 kW but not more than 27 kW, the maximum demand in Column A shall be increased 5 percent for each additional kilowatt of rating or major fraction thereof by which the rating of individual ranges exceeds 12 kW.

Try that...see what you get...let us know

 

[Mike Lang]

this might be confusing let me know if it helps

this might be confusing let me know if it helps

It says that over 12K you have to add 5% for each number over 12 . So if it is 18 Kw then the difference between 12 & 18 is 6 . Now multiply 6 x 5%= 30 %. Look in column C and the derating for 1 applicance over 12K is 8K. Now multiply 8Kw x .30= 2.4 Then the last step is to add the 2.4 to the 8Kw so the derated value is 10.4 Kw or 10400

 

[ce2two]

ce2two

thanks! nothing like chasing one"s tail ,then back the other the way ...... they should keep it simple (k.i..s.s.)