Exam Question: Feeder Size

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Adamjamma said:
Ok Jumper... used 6000 watts /240 = 25

plus 4000 watts /120 = 33.333
gave me 58.3

but... lessee something... How about...nope, math not working... You are dropping the amount of power for stove...

50 watts derated 80 percent equals 40 watts
4000 watts derated same way equals 26.67
equals 67 watts...

So.. if they took same thing I do for stoves.. 50 amp breaker before derating... then my math comes close to theirs but only if I de rate the house circuits of 80 percent as well...

If my math is right... that is...

But again, since minimum panel for residence is 100 amps if I remember as well, unlike Jamaica where I could put i a 70 amp... the rest of the question works out...
so question really a trick because you have the minimums you must have- 50 amp cooker circuit, and the minimum panel size...

Unless I really know nothing about the codes, which is totally possible with me trying to learn 3 codes at one time..lol
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Adam, you are mixing up a few things above. The 80% was a particular value from a chart for that range. You are just randomly trying to apply it and force the numbers, :happyno:

But hey, it was a good try!:D
 
so that leaves my calculations then at 73.7 amps... so still doing something wrong which I need to go find.. will figure it out in the morning..lol...
Either I am handling the 4000 watts wrong or am wrong on required circuits..lol...
but, it will probably be in my ugly book or my old notes some where about differences between UK, JA and USA..lol...
 
Adamjamma said:
so that leaves my calculations then at 73.7 amps... so still doing something wrong which I need to go find.. will figure it out in the morning..lol...
Either I am handling the 4000 watts wrong or am wrong on required circuits..lol...
but, it will probably be in my ugly book or my old notes some where about differences between UK, JA and USA..lol...
Click to expand...

The OP gave us limited info. No way to figure it, too many missing details, not worth it.

The answer is still #4 from the given info.

And yes, some of your math is wrong.
 
Ok.. still not right but... 4000 Watts given... three 15 amp circuits so 4500
appliance circuits times 2 at 20 amp for kitchen is 4800
bathroom required circuit 20 amp. 2400
laundry required circuit. 15 amp. 1800
total now..."................................................................................13500
Ok... subtract 3000 and then multiply by 0.35. 3675
add the 3000. 3000
add the range at 40 amps... 50 amps derated ... 9600
Total. 16275
Divide by 240 volts. 67.81 amps...



what what am I getting wrong..lol...
 
Adamjamma said:
Ok.. still not right but... 4000 Watts given... three 15 amp circuits so 4500
appliance circuits times 2 at 20 amp for kitchen is 4800
bathroom required circuit 20 amp. 2400
laundry required circuit. 15 amp. 1800
total now..."................................................................................13500
Ok... subtract 3000 and then multiply by 0.35. 3675
add the 3000. 3000
add the range at 40 amps... 50 amps derated ... 9600
Total. 16275
Divide by 240 volts. 67.81 amps...



what what am I getting wrong..lol...
Click to expand...

Assuming we have all the load info is what you are getting wrong.

OP says a net load of 4000W, meaningless.

Without details we cannot calculate. Do not waste your time.

And you are totaling incorrect values.
 
We do have a few things.. it is a residence so there are minimum circuits required that are calculated outside the 3va per sq ft... if the 4000 is the sqr ft.. va... then we can start there..but in some examples I am told to use that number straight, in other examples I am told that you then look at circuits using 1500 for a 15 amp circuit..yet I get 1800;...
so we then sub in 4500 for the 4000, and add the required kitchen circuits... 2 @ 20 amp
and the required laundry circuit @ 15 amps, and the required bathroom circuit at 20 amps...
subtract the first 3000... multiply the rest by 0.35 and then add back the original 3000...
unless I use the other calculation method of first ten thousand...lol.. that one lost me..but...
then we have the range.. but we have a standard of 50 amps but was told to debate that so 40 amps or 9600 volts..

sure i I am missing something... but looks like sun starting to come up and cannot find my old code book for USA.
 
Adamjamma said:
We do have a few things.. it is a residence so there are minimum circuits required that are calculated outside the 3va per sq ft... if the 4000 is the sqr ft.. va... then we can start there..but in some examples I am told to use that number straight, in other examples I am told that you then look at circuits using 1500 for a 15 amp circuit..yet I get 1800;...
so we then sub in 4500 for the 4000, and add the required kitchen circuits... 2 @ 20 amp
and the required laundry circuit @ 15 amps, and the required bathroom circuit at 20 amps...
subtract the first 3000... multiply the rest by 0.35 and then add back the original 3000...
unless I use the other calculation method of first ten thousand...lol.. that one lost me..but...
then we have the range.. but we have a standard of 50 amps but was told to debate that so 40 amps or 9600 volts..

sure i I am missing something... but looks like sun starting to come up and cannot find my old code book for USA.
Click to expand...

:happyno:

Too much to explain now, but no, your values are incorrect.

You are not using Art.220 correctly per NEC.
 
Adam, calculate this please.

Example D1(a) One-Family Dwelling
The dwelling has a floor area of 1500 ft2. Appliances are a 12-kW range and a 5.5-kW, 240-V dryer.

Answer is 78A.

Incorrect values and you will never get the correct answer. Need an NEC to get those values. See my point.
 
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jumper said:
Adam, calculate this please.

Example D1(a) One-Family Dwelling
The dwelling has a floor area of 1500 ft2. Appliances are a 12-kW range and a 5.5-kW, 240-V dryer.

Answer is 78A.

Incorrect values and you will never get the correct answer. Need an NEC to get those values. See my point.
Click to expand...
Ok.. watched two video... one of them boots something..lol...

pulled out my ugly guide from my testers box... got confused again by the two column stuff..but

on your example here I get 77.5 amps so you round to 78 amps...
on the one that started this headache, keeping me up til 4 am trying to do badly..lol... I get 40.52 amps with given info... but taking the dryer calc into consideration and saying that if you have a laundry load you must have a dryer circuit... in case the person plugs a dryer in..lol... but not in code that I read, just common sense to me... I add the minimum of 5000 and bring the calculation to 61.35 amps..lol...
still missing info but.. better math to the examples..
of course, neither load has water heaters, ac, heating, or other loads in them such as microwaves...
 
Now.. we don’t use the optional calculation here because load is less than 100 amps but.. if we did, your example would calculate as 22500 but then take first 10,000, .40 of 12500, giving total of 62.5 amps?? Or am I doing that one wrong?
 
Adamjamma said:
Ok.. watched two video... one of them boots something..lol...

pulled out my ugly guide from my testers box... got confused again by the two column stuff..but

on your example here I get 77.5 amps so you round to 78 amps...
on the one that started this headache, keeping me up til 4 am trying to do badly..lol... I get 40.52 amps with given info... but taking the dryer calc into consideration and saying that if you have a laundry load you must have a dryer circuit... in case the person plugs a dryer in..lol... but not in code that I read, just common sense to me... I add the minimum of 5000 and bring the calculation to 61.35 amps..lol...
still missing info but.. better math to the examples..
of course, neither load has water heaters, ac, heating, or other loads in them such as microwaves...
Click to expand...

Forget the OPs question, it is too badly written.

If you were able to come up with 77.5 correctly then call it good, fyi though, fractions of .5 or less may be rounded down per NEC.

In doing NEC calcs, never assume anything but the given info or what is the minimum requirements. Pretty much holds true for any NEC test question.
 
Adamjamma said:
Now.. we don’t use the optional calculation here because load is less than 100 amps but.. if we did, your example would calculate as 22500 but then take first 10,000, .40 of 12500, giving total of 62.5 amps?? Or am I doing that one wrong?
Click to expand...

IDK, I did not do the math.

Better to get good examples with proper info and start new thread.
 
david luchini said:
I'd be curious how they come up with a load of 68.3A.
Click to expand...


6000VA for the dryer

8000VA * 0.8 = 6400VA for the range load (NEC Table 220.55)

Net load = general load + dryer + range = 4000VA + 6000VA + 6400VA = 16400VA

net load current = 16400VA/240V = 68.3A

The question comes from PPIs book by Camara; Engineering Power Reference Manual
 
TheSmoothestCriminal said:
6000VA for the dryer

8000VA * 0.8 = 6400VA for the range load (NEC Table 220.55)

Net load = general load + dryer + range = 4000VA + 6000VA + 6400VA = 16400VA

net load current = 16400VA/240V = 68.3A

The question comes from PPIs book by Camara; Engineering Power Reference Manual
Click to expand...

I'm wondering if your original question missed some information...It mentioned a 6kw range, not an 8kw range, and didn't mention a dryer at all.

68.3A appears correct for that information, and as previously noted, 230.79 requires a 100A min. diconnecting means and 310.15(B)(7) permits #4 conductors.
 
TheSmoothestCriminal said:
6000VA for the dryer

8000VA * 0.8 = 6400VA for the range load (NEC Table 220.55)

Net load = general load + dryer + range = 4000VA + 6000VA + 6400VA = 16400VA

net load current = 16400VA/240V = 68.3A

The question comes from PPIs book by Camara; Engineering Power Reference Manual
Click to expand...
[Quote ]
"A single-family dwelling unit has a net load of 4000VA. A 6kW range will also be included in the design. The dwelling unit is supplied by a 120/240V, three wire, single phase copper feeder of type THW. According to the NEC, what size feeder is required? - [/Quote]

No Dryer mentioned- Just saying-
 
TheSmoothestCriminal said:
6000VA for the dryer

8000VA * 0.8 = 6400VA for the range load (NEC Table 220.55)

Net load = general load + dryer + range = 4000VA + 6000VA + 6400VA = 16400VA

net load current = 16400VA/240V = 68.3A

The question comes from PPIs book by Camara; Engineering Power Reference Manual
Click to expand...

Your OP:

A single-family dwelling unit has a net load of 4000VA. A 6kW range will also be included in the design

Your numbers and descriptions do not match.
 
david luchini said:
I'm wondering if your original question missed some information...It mentioned a 6kw range, not an 8kw range, and didn't mention a dryer at all.

68.3A appears correct for that information, and as previously noted, 230.79 requires a 100A min. diconnecting means and 310.15(B)(7) permits #4 conductors.
Click to expand...


Youre right, i missed that part when i was typing it off of my paper copy of the question. Sorry!
 
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