# Thread: who's got some time today

1. Senior Member
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Aug 2011
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## who's got some time today

prepping for exam. heres a question. 36 units apartment, with 15 kw ranges (seems far out , but i gotta try, i come up with a demand of 18 k from column c any input ? thanks ( 16 puls 2.4 from the 5 % over 12 kw dropping the fraction, givers 18

2. I get 59 kW. I think you misinterpreted the rule for above 26 ranges (i.e., 15 kW+ 1 kw for each range). I interpret that to mean you start with 15 kW, then add 36 kW because you have 36 ranges. That gives you 51 kW. Then you add 15% because each range is 3 kW over the 12 kW value, so you need to add 3 time 5 %. That brings another 7.65 kW to the party. The 0.65 is more than half, so I round up. Finally, 51 kW plus 8 kW gives the total of 59 kW.

But ignoring the math for a moment, don't you think it unreasonable that a total connected load of 540 kW (15 times 36) could be counted, after applying demand factors, as only 18 kW? If I had come up with that result, I would think I must have done something wrong.

3. Is this calc for a branch circuit, apt feeder or service?

4. Senior Member
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service demand

5. Senior Member
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that makes sense. *(59 kw) more sense. it's been years since i've had to do this stuff, i'm rusty.

6. Originally Posted by PEDRO ESCOVILLA
that makes sense. *(59 kw) more sense. it's been years since i've had to do this stuff, i'm rusty.
Don’t worry, most of us are!!!

7. Senior Member
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Originally Posted by PEDRO ESCOVILLA
service demand
Then you would need to know the square footage. Right? or are you just trying to get the range load?

8. Senior Member
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northern new jersey
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4000 amps should do it!

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