9 Single Phase Ranges on Three Phase Feeder
- Thread starter jbwhite
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ranges
ranges
Heres how I would do the calculation for the demand load on the three phase feeder.
9 ranges 3 ranges on Phase "a -b" 3 ranges on phases "b - c"
3 ranges on phase "a - c"
The maximum number between any two phases is 3. 2 x 3 = six ranges.
Table 220.55 column "c" six ranges = 21,000 V.A. This is for two of the phases. To find the load for all three legs. I would undo what i did earlier by dividing by two, to get the v.a. per phase. 21,000/ 2 = 10,500 v.a. phase. Then multiply this by three to get the total. v.a. 10,500 x 3 = 31,500 v.a.
Then divide 31,500/ (208 x 1.732) = 87.5 amps or 88 amps
ranges
Heres how I would do the calculation for the demand load on the three phase feeder.
9 ranges 3 ranges on Phase "a -b" 3 ranges on phases "b - c"
3 ranges on phase "a - c"
The maximum number between any two phases is 3. 2 x 3 = six ranges.
Table 220.55 column "c" six ranges = 21,000 V.A. This is for two of the phases. To find the load for all three legs. I would undo what i did earlier by dividing by two, to get the v.a. per phase. 21,000/ 2 = 10,500 v.a. phase. Then multiply this by three to get the total. v.a. 10,500 x 3 = 31,500 v.a.
Then divide 31,500/ (208 x 1.732) = 87.5 amps or 88 amps
- Location
- Chapel Hill, NC
- Occupation
- Retired Electrical Contractor
Thank you JB-- but your book is correct--jbwhite said:
2* 3 Range max per phase =
6 ranges Table 220.55 = 21kw
21000/2= 10500
10500*3= 31500
31500/360= 87.5 amps
Dang I finally got it but I don't know if I ever would have worked backwards from Article 220.55 to table 220.55. I would have assumed its not a dwelling and never gone back. Thanks again for your patience. I am thick sometimes.
hardworkingstiff
Senior Member
- Location
- Wilmington, NC
Dennis Alwon said:
That's better than me, because I'm think all of the time.
...and exactly how are the test takers going to make the leap from 240V to 208V... :roll: ...everyone here has the benefit of knowing the "book" answer. I doubt very much the answer will be provided for a question on the test :grin:Dennis Alwon said:
Perhaps someone should contact the author and ask him/her how the answer was derived...
- Location
- Chapel Hill, NC
- Occupation
- Retired Electrical Contractor
Hey, it took me 2 days to come to understand this-- I just accepted the 208v but I agree it ought to be divided by 415 (240*1.73) rather than 360. This would change the answer to 76 amps. Wow I thought 88 was way to low. I am just thinking about 9 people cooking with the ovens going on all 9 ranges and some burners working also. Sounds mighty low to me in either caseSmart $ said:
I agree. Especially when one considers all ovens going full tilt is:Dennis Alwon said:
9 ? 10,000VA ? (240 ? √3) = 216.5A/Line
I'm picturing a "restaurant chef" school where they are being taught to expedite meals, and all the "burners" are fired up at the same time :grin:
(Yes, I'm aware that most likely such a school would have "gas" ranges and ovens... but I can picture anything I want :grin
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