range calculation for feeder

[yanert]

I am studying for a test and I am supposed to do a feeder calculation for ranges.

Can anybody give me the rule for the notes 1 and 2 for table 220.19.
The question is at the point where you increase 5 percent for each additional kilowatt of rating or major fraction thereof by which the rating of individual ranges exceeds 12 kw.

What is major fraction? Does that mean anything over 12? I have a questions that calls for:
12- 12.8 KW ranges
12- 12.5 KW ranges
24- 11.6 KW ranges

I come up with a number of 12.325
Is that a major fraction thereof? 12.325 is not over 12.5 so I don?t know if I should give the 5% addition or not.
The test questions I have are designed to be tough I think, so I am finding all kinds of little troubles with my answers!
If anybody is curious the question is this:
What is the feeder demand for 12 ? 12.8KW household ranges, 12 ? 12.5KW ranges and 24 ? 11.6KW ranges if connected to a three-phase service? The answer is 45-50KW and is supposed to come right from the table.
I keep coming up with 33.75 without the 5% or 35.4375 with the 5%
Is there something I might be missing with the wording of household range and then just ranges? Oh, and everything I have done I have never done a 3 phase calculation until the end. And that was to find the amperage after the sums of all the loads had been calculated.

 

[charlie b]

Re: range calculation for feeder

You have a total of 48 ranges. Go first to Note 2. Calculate an average. Keep in mind that you must count the 11.6 KW ranges as though they were 12 KW.

(12 times 12.8 KW) plus (12 times 12.5 KW) plus (24 times 12 KW), all divided by 48, equals a 12.325 KW average (I agree with you here).

So you treat this as though you had 48 ranges, each rated at 12.325 KW. That fits into the row marked 41 ? 50 ranges. The Column C value is ?25 KW plus 3/4 KW for each of the 48 ranges.? You then take 25 KW, and then add 3/4 times 48, and get 61 KW.

Finally, since 12.325 does not have a ?major fraction? of a KW in excess of 12, you do not add the 5% ?upcharge.? So the final load value is 61 KW.

To convert this to amps, you divide by the three phase line-to-line voltage, and divide again by the square root of three (about 1.732). But your question did not give us a voltage, only the fact that the system is three phase. So for illustration, I will use a value of 208 volts. Divide 61 KW by 208, and divide again by 1.732, and you get a load of 169 amps.

Do you have the book?s answer?

 

[yanert]

Re: range calculation for feeder

Hi Charlie, thanks for doing the calculation with me.
the answer is supposed to be 45-50 kw

 

[yanert]

Re: range calculation for feeder

Hi Charlie, I just looked at my figures again, I mistyped. I too got the the number 61 KW when I didnt apply the extra 5% So what is the trick about 45 to 50 KW for the answer? I have been going through a few different books studying problems, and its stating to get a bit frustrating with all the wrong answers and typos that are getting published!! I can spend a condiderabe amount of time on a problem, only to discover that for once it is not my calculations, but a typo in the text of the problem.

 

[charlie b]

Re: range calculation for feeder

When a question like this is posed, the answer will be a single number. It will not be a range of numbers. There is something out of whack in the way the question is worded or in the nature of the "book answer."

 

[yanert]

Re: range calculation for feeder

Hi charlie, I think he was just trying to get a range. He has done this before in some of the other calculations. But I agree, I wish he would ask for the exact number, that way I see if my calcs are exactly correct!

 

[yanert]

Re: range calculation for feeder

Hi charlie,
The final paragraph of 220.19 states: Where two or more single phase ranges are supplied by a 3-phase, 4 wire feeder or service, the total load shall be computed on the basis of twice the maximum number conected between any two phases. Not sure about how to do the calc yet, but I think that is what he is after. I just didnt see it because its on the next page of my softbound code book and I was not vigilant enough to read the entire text. It give you example in the annex, so maybe I can muddle through it. You know what they say, Give a monkey a typewriter, he will eventually pound out a word!
Thanks for you help!

 

[pattbaa]

Re: range calculation for feeder

It may be noted that when 208 volt loads are equally divided and distributed across a 3-phase 4-wire "Wye" system, the connected loads form a "Delta" arrangement. With each Phase-lead connected to two 100 amp ( 20KW/208) loads, the line-current in the Phase-lead is 100 X 1.73 = 173 amps.