Calculation for ranges of unequal KVA

[Larry Bohn]

We had a calculation in class as follows: "What is the service demand for 5 8KW ranges, 5 10KW ranges, and 5 12 KW ranges"
A question arose when we calculated them, shown below
Column B
5 X 8KVA=40KVA X .45(demand factor) = 18 KVA
Note 2
raising the KW rating to 12KW for the five 10 KW, the calculation is
(5 X 12KW) + (5 X 12KW) = 120KW/10 ranges = 12KW each
demand for 10 12 KW ranges is 25KW
Adding the two demands you get 18KW +25KW = 43 KW

Can the total demand just be taken from column C since they are all under 12 KW? The demand would be 30KW instead of 43KW. I think so.

The question arose was when it was asked "what if one or more ranges were over 12 KW? Would we be required to calculate the demand using note 2 for the ranges over 8 3/4KW from column C, and the ranges below 8 3/4 KW from column B, and add them, as in the first example"? The demand factor is quite a bit larger calculated this way, even if only one range was rated at 14KW, instead of 12KW.

Opinions?

 

[david luchini]

Can the total demand just be taken from column C since they are all under 12 KW? The demand would be 30KW instead of 43KW. I think so.

I agree with 30kW.

 

[Larry Bohn]

I agree with 30kW.

What if one of the 12kW ranges was raised to 14kW? Would you be locked into using the original calculation that results in an answer of 43kW?
That seems like a big jump for the demand by simply adding an additional 2kW of possible load, even though the averaging of the ranges larger than 8 3/4kw allow the additional 2kW load to be ignored. Am I missing something?