Range Calc

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KP2

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Hello all, Happy New Year!!!

I wanted to check my work on this Range Calculation...

What is the Demand Factor for a house hold electric cooktop rated at 3.5 kw and a wall-mounted oven rated at 6 kw? ___________

Thanks
Kevin
 
Are you sizing a branch circuit, or doing a feeder or service calculation. What was your answer?
 
KP2 said:
Hello all, Happy New Year!!!

I wanted to check my work on this Range Calculation...

What is the Demand Factor for a house hold electric cooktop rated at 3.5 kw and a wall-mounted oven rated at 6 kw? ___________

Thanks
Kevin
Click to expand...

Show us your work and someone may help.
 
KP2 said:
Hello all, Happy New Year!!!

I wanted to check my work on this Range Calculation...

What is the Demand Factor for a house hold electric cooktop rated at 3.5 kw and a wall-mounted oven rated at 6 kw? ___________

Thanks
Kevin
Click to expand...

If he doesn't answer, I'll give it a shot.:)
 
Dennis Alwon said:
Go for it Bill tell us the answer. ;)
Click to expand...

Well, if he's sizing for a feeder or service load it would be 7.6kw. if for branch circuit, it would be 8kw if using note 4 of Table 220.55. If this is correct, could you use the lower 7.6kw service or feeder cal. for a branch circuit? Oh, and after re-reading the OP, if he just wants the demand factor, it is 80%.;)
 
Thanks for the replies.
The way I did the calc is, both units are in column B so 65% of the total is what I used.

6,000 + 3,500 = 9,500 * 65% = 6,175 va

The question is meant to be tricky because most think that the 3.5 kva unit falls in column A.

Also, a lot of us, including myself until you guys helped me ;), go straight to note 4.

As mentioned in this thread the question can seem unclear regarding the use of wording "demand factor".

So maybe the question should read, "What is the calculated load for the feeder ..."???

Any thoughts, and is the calculation correct this way?

Thanks Again
Kevin
 
KP2 said:
Thanks for the replies.
The way I did the calc is, both units are in column B so 65% of the total is what I used.

6,000 + 3,500 = 9,500 * 65% = 6,175 va

The question is meant to be tricky because most think that the 3.5 kva unit falls in column A.

Also, a lot of us, including myself until you guys helped me ;), go straight to note 4.

As mentioned in this thread the question can seem unclear regarding the use of wording "demand factor".

So maybe the question should read, "What is the calculated load for the feeder ..."???

Any thoughts, and is the calculation correct this way?

Thanks Again
Kevin
Click to expand...

I agree the 65% is the wtg.
 
KP2 said:
Thanks for the replies.
The way I did the calc is, both units are in column B so 65% of the total is what I used.

6,000 + 3,500 = 9,500 * 65% = 6,175 va

The question is meant to be tricky because most think that the 3.5 kva unit falls in column A.

Also, a lot of us, including myself until you guys helped me ;), go straight to note 4.

As mentioned in this thread the question can seem unclear regarding the use of wording "demand factor".

So maybe the question should read, "What is the calculated load for the feeder ..."???

Any thoughts, and is the calculation correct this way?

Thanks Again
Kevin
Click to expand...


I see I got it right the second time. The question didn't trick me, just fast response before putting brain in gear, ie:not going to bed when I should!:D
 
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