Art. 430.24 calculation question

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royta

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I took the Mike Holt 25 question quiz HERE. I got a few more wrong than I would have liked, but afterwards I have been able to find the code that helps me arrive at the correct answers. I think I know why I got #10 wrong, but I'd like to verify.

Q10 The motor feeder conductor size for three 15hp, 208V, three-phase motors; three 3 hp, 208V single-phase motors; and three 1 hp, 120V, single-phase motors will be ________.

A. 2/0 AWG
B. 3/0 AWG
C. 4/0 AWG
D. 250 kcmil

I chose D but the correct answer is C. I used 430.24 but figured all three of the identical largest motors at 125% and the remaining motors at 100% and arrived at 277.35A which is 250 kcmil as per the 90C column of 310.16. If I only figure one the three identical largest motors at 125% and the remaining motors at 100% than I arrive at 254.25A which is 4/0 as per the 90C column of 310.16. So, I guess when a group of two or more identical motors are the largest, than I only figure one of them at 125% and the remaining motors at 100%. Is this correct? Thanks.

Roy
 
infinity said:
Why are you using the 90 degree C column for ampacity?
Click to expand...

I realize there are many variables missing in the question, but I just figured the answer would be based on using the smallest wire possible.
 
royta said:
I realize there are many variables missing in the question, but I just figured the answer would be based on using the smallest wire possible.
Click to expand...

Must be because if #4/0AWG is acceptable, then so is 250KCMIL. So, technically, your first answer was not wrong.

Remember the Code is minimum requirements, so in that case the #4/0AWG is the better answer, just not necesarily the only answer. Be a little cautious when answering code questions, because the answers may not necessarily be the best way to do something. It just simply meets minimum safety.
 
Royta,
Look at 110.14(C). I don't know of any terminations that are rated for 90 degree C. You will almost always be using the 60 or 75 degree column.
 
254.25A seems right to me,

(3) 15hp @ 208V 3ph = 46.2A ea, per table 430.250.
(3) 3hp @ 208V 1ph = 18.7A ea, per table 430.248.
(3) 1hp @ 120V 1ph = 16.0A ea, per table 430.248.

So calculation results:
125% * largest motor(only one motor, not all 3 largest motors) = 1.25*46.2 = 57.75A

15hp sizing = 2*46.2 + 57.75 = 150.15
3hp sizing = 3*18.7 = 56.1
1hp sizing = 3*16 = 48

Total 150.15 + 56.1 + 48 = 254.25

Any errors here?
 
I will have to correct my answer to 198.6A after taking more time...


150.2 + 32.4 + 16


Except that would make B the correct answer...?
 
wasasparky said:
I will have to correct my answer to 198.6A after taking more time...


150.2 + 32.4 + 16


Except that would make B the correct answer...?
Click to expand...


The amperage is 254.25. How are you arriving at your answers? What amperages are you using for these 9 motors?
 
royta said:
The amperage is 254.25.
Click to expand...


Are you sure?

Take all the loads and convert them to VA and add.
Convert the total VA to 3ph equivalent.
What do you get?


Hint: 18.7 + 18.7 does not equal 37.4 in this case...
 
wasasparky said:
Are you sure?

Take all the loads and convert them to VA and add.
Convert the total VA to 3ph equivalent.
What do you get?


Hint: 18.7 + 18.7 does not equal 37.4 in this case...
Click to expand...

OK, you're confusing me. Why convert to VA at all? Six of the nine motors are 1ph, so I'm not sure how or why you'd convert to VA, add the values, then convert the VA sum to 3ph. The question said nothing about balancing the load across the phases. GH_Vegas has it right on the money, which is how I did it too (I actually just used my Ugly's), but I made the mistake of multiplying all the identical largest motors by 125%, instead of only one of the largest motors.
 
royta said:
The amperage is 254.25. How are you arriving at your answers? What amperages are you using for these 9 motors?
Click to expand...


How are you adding them together? Remember you have 3 phase, single phase 208 and single phase 120 volt motors.
 
infinity said:
How are you adding them together? Remember you have 3 phase, single phase 208 and single phase 120 volt motors.
Click to expand...

Just like this:

GH_Vegas said:
254.25A seems right to me,

(3) 15hp @ 208V 3ph = 46.2A ea, per table 430.250.
(3) 3hp @ 208V 1ph = 18.7A ea, per table 430.248.
(3) 1hp @ 120V 1ph = 16.0A ea, per table 430.248.

So calculation results:
125% * largest motor(only one motor, not all 3 largest motors) = 1.25*46.2 = 57.75A

15hp sizing = 2*46.2 + 57.75 = 150.15
3hp sizing = 3*18.7 = 56.1
1hp sizing = 3*16 = 48

Total 150.15 + 56.1 + 48 = 254.25

Any errors here?
Click to expand...

Originally, I figured all three of the 15hp motors at 125%, but after getting the question wrong, I learned only to figure one of the motors at 125%. Better wrong now, then when the test really counts.
 
You still are not permitted to use the Table 310.16, 90C column though.

Circuit ratings over 100A generally imply 75C terminals [110.14(C)(1)(b)] and LV motors generally have 75C terminals too [110.4(C)(1)(a)(4)].

A computed load of 254A would still require 250kcmil. wasasparky's FIRST answer is the closest although I'd like to see his work.
 
rbalex said:
You still are not permitted to use the Table 310.16, 90C column though.

Circuit ratings over 100A generally imply 75C terminals [110.14(C)(1)(b)] and LV motors generally have 75C terminals too [110.4(C)(1)(a)(4)].

A computed load of 254A would still require 250kcmil. wasasparky's FIRST answer is the closest although I'd like to see his work.
Click to expand...

I'd like to see his work too, just so I can see how he arrived at his answer. However, I don't feel he has the right answer. At least not when using 430.24 and the Tables 430.248 through 430.250. Where does it say in the NEC that you can use something besides 430.24 for this type of scenario?

Regarding 250 kcmil, it appears the question was incorrectly written, and the test taker should appeal when it is marked incorrect. I chose 250 kcmil, but I did the calculation incorrectly, so it doesn't really count. :) I find it frustrating that Mike Holt's practice question is written incorrectly. There is only one correct answer, but it is obvious they are using the 90C column for the 4/0 answer. Now, when I sit for my Journeyman's exam for ID and UT, and am presented with an answer that coincides with the 90C column AND the 75C column, I won't feel secure with any answer I give. How nice for me. :rolleyes:
 
Royta,

When you appeal your answer and they ask you how do you add single phase

motors and 208v single phase and three phase motors together, you better

not say just line them up and add. Do a little more listening and a little less

badgering, then you won't have to worry about the right column to use when

you take the test. VA is the only way engineer's will do it.
 
You will never be able to use the 90C column for this particular type of question and I don't beleive Mike did.

The correct answer is very close to 205A - but not quite. The METHOD wasasparky apparently used in his second answer was fundamentally correct, but some of his "raw" data was wrong.

The only possible error I see in the question itself was not specifying that the single-phase loads were balanced; but I suspect the question uses 3-single phase motors of each type to imply balanced loads. I am assuming you have reported the questinon accurately.

One of the problems with mathematical models is they work so well most of the time that they often get us to overlook the underlying physics of a problem.
 
benaround said:
Royta,

When you appeal your answer and they ask you how do you add single phase

motors and 208v single phase and three phase motors together, you better

not say just line them up and add. Do a little more listening and a little less

badgering, then you won't have to worry about the right column to use when

you take the test. VA is the only way engineer's will do it.
Click to expand...



I agree, you cannot add apples, oranges and watermelons. I came up with about 200 amps. That would make the 4/0 answer incorrect.
 
Royta,

Let us say you have a three phase motor, name plate says 30a 3ph, ok, now

we have three 120v motors, name plate says 10a 1ph 120v. how many amps

will be on the system. 60a or 40a ? The way YOU added them up you got 60a

but now you can see it is only 40a. Try redoing your math and taking some

good advise from these guys who do it every day. good luck
 
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