robyne
Member
- Location
- kamiah, idaho
Just prepping for a masters exam and want to get this right. Appreciate some clarification if you can figure out what I have typed in 
Thanks!
Master exam #4 from the 2008 calcs prep book, #4:
20-800 sq.ft. apartments
20-12 kw ranges
40- small appliance circuits
20 - laundry ckts.
Size the Neutral conductor to ______ amperes. 230/115v single phase.
Answer according to book: 20x 800 sq. ft. x 3 va = 48000va
small app. & laundry ckts. 60 x 1500va = 90000va
138000va
Table 220.42 Demand :
First 3000va @ 100% = 3000va
Next 117000va @ 35% = 40950va
Final 18000va @ 25% = 4500va
Table 220.55 Column C : 35kw x .70% = 24500va
Up to this point everything seems good, but I have a problem with adding it to the 48450va and then taking an additional demand deduction as shown below:
...its added onto the the 48,450 which equals 72,950va, and then he further says, 72,950va/230v = 317amps
220.61: First 200a @ 100% and the next 117A @ 70% = 82A for a total of 282 A on the neutral"
As I read 220.61(b) it says " A service or feeder supplying the following loads shall be permitted to have an additional d.f. of 70% applied to the amount in 220.61b1 OR portion of the amoutn int 220.61b2 determined by the basic calculation:
1) a feeder or service supplying household electric ranges, w.m. ovens, c.m. cooking units, & elect. dryers, where the max. unbal. ld. has been determined in accordance with T220.55 and T220.54
2) That portion of the unbal. ld. in excess of 200a where the feeder or servaice is supplied from .... etc...
I would figure that the 48,540va \ 230 would be 211a - first 200 at 100% and the 11A at 70% making it 208A and the Ranges at the 35,000KW x's the 70% and divided by 230v would be 107A and the two added together would be 315A. Is Tom right (hes a pretty smart guy, but sometimes I run into clerical errors in his books, Mikes too--- no one is perfect! Or would I use the second method when testing?
Thanks!Master exam #4 from the 2008 calcs prep book, #4:
20-800 sq.ft. apartments
20-12 kw ranges
40- small appliance circuits
20 - laundry ckts.
Size the Neutral conductor to ______ amperes. 230/115v single phase.
Answer according to book: 20x 800 sq. ft. x 3 va = 48000va
small app. & laundry ckts. 60 x 1500va = 90000va
138000va
Table 220.42 Demand :
First 3000va @ 100% = 3000va
Next 117000va @ 35% = 40950va
Final 18000va @ 25% = 4500va
48450va
Table 220.55 Column C : 35kw x .70% = 24500va
Up to this point everything seems good, but I have a problem with adding it to the 48450va and then taking an additional demand deduction as shown below:
...its added onto the the 48,450 which equals 72,950va, and then he further says, 72,950va/230v = 317amps
220.61: First 200a @ 100% and the next 117A @ 70% = 82A for a total of 282 A on the neutral"
As I read 220.61(b) it says " A service or feeder supplying the following loads shall be permitted to have an additional d.f. of 70% applied to the amount in 220.61b1 OR portion of the amoutn int 220.61b2 determined by the basic calculation:
1) a feeder or service supplying household electric ranges, w.m. ovens, c.m. cooking units, & elect. dryers, where the max. unbal. ld. has been determined in accordance with T220.55 and T220.54
2) That portion of the unbal. ld. in excess of 200a where the feeder or servaice is supplied from .... etc...
I would figure that the 48,540va \ 230 would be 211a - first 200 at 100% and the 11A at 70% making it 208A and the Ranges at the 35,000KW x's the 70% and divided by 230v would be 107A and the two added together would be 315A. Is Tom right (hes a pretty smart guy, but sometimes I run into clerical errors in his books, Mikes too--- no one is perfect! Or would I use the second method when testing?