need help on calculation

[kendrichchan]

the feeder protection for one 25 hp,208v,three phase motor,and three 3hp,120v,single phase motor will be________ after balancing,note :use inverse time breakers.
one 25 hp 208v=74.8x1.25=59.84
three 3hp 120= 34ax3=102
inverse time breaker=2.5

59.8+102=161.8 x2.5=404a but answer is 225 amp
what step did I make mistake on?
thanks

 

[Dennis Alwon]

how did you multiply 74.8 amps times 1.25 and get 59.8 amps?

 

[Dennis Alwon]

Table 430.52 is for individual motors not feeders

 

[Dennis Alwon]

I am looking at 430.62(B). It seems that the OCP was figured this way. 78+34+34+34= 180 . 180*1.25= 225-- I did not think that was correct but this is not my forte.

 

[CONDUIT]

Using 430.52(c) 74.8 x 2.5 = 187 amps . 430.52(c)(1)exception (1) allows me to go up to next standard size of breaker. see 240.6(a). 200amp breaker. For sizing the feeder breaker you use the largest Branch circuit , short circuit, ground fault device for any motor in the group which is 200 amps and add the sum of the full load currents of the other motors. If you keep the system balance you should have only one single phase motor on each phase and the neutral conductor. 200 + 34 = 234 But section 430.62(a) does not give you permission to round up. It says "shall not exceed" The next standard size below 234 is 225 Amps.

 

[kendrichchan]

typo

typo

how did you multiply 74.8 amps times 1.25 and get 59.8 amps?

wrong calcution it should be93.5

 

[Little Bill]

I am looking at 430.62(B). It seems that the OCP was figured this way. 78+34+34+34= 180 . 180*1.25= 225-- I did not think that was correct but this is not my forte.

Dennis, I agree 430.52 doesn't apply here. I also agree in that I don't think the book answer is right. What I think they did was total up the motor amps and then multiplied by 1.25%. 74.8+34+34+34=176.8, 176.8x1.25=221, next size up would be 225. What should have been done was largest motor 74.8x1.25%=93.50
93.50+34+34+34=195.5, next size up 200A.
That's my story and I'm sticking to it!

 

[kwired]

Balance the three 120 volt motors across all three phases instead of adding them together as though they are on one phase and your result will be smaller.

3 hp @120 v = 34 x 1.73 = 58.8

25 hp @208 3ph = 78.2 x 1.25 = 97.8

97.8 + 58.8 = 156.6 Minimum feeder conductor ampacity.

It could easily be on a 400 amp breaker if it will not hold during starting.

 

[don_resqcapt19]

...

It could easily be on a 400 amp breaker if it will not hold during starting.

I don't see how you could use a 400 amp inverse time breaker on this feeder and comply with the code rules.

 

[kwired]

I don't see how you could use a 400 amp inverse time breaker on this feeder and comply with the code rules.

After recalculating I don't know either, but I do come up with a max possibility of 350 standard inverse time breaker.

I would guess that 200 amp will work in most cases.

 

[Smart $]

After recalculating I don't know either, but I do come up with a max possibility of 350 standard inverse time breaker.

I would guess that 200 amp will work in most cases.

CONDUIT's post is the correct method of determining [max] ITB rating—225A.

 

[kwired]

CONDUIT's post is the correct method of determining [max] ITB rating?225A.

I was using correct method but for some reason can't seem to plug in the right numbers even after two or three tries with different numbers. Maybe better go back to bed today

OK this time I come up with max of 333 = next lower standard device of 300 - if the 225 will not allow starting of the 25 hp motor.

 

[Smart $]

I was using correct method but for some reason can't seem to plug in the right numbers even after two or three tries with different numbers. Maybe better go back to bed today

OK this time I come up with max of 333 = next lower standard device of 300 - if the 225 will not allow starting of the 25 hp motor.

Well, if we get into using exceptions, namely 430.52(C)(3) Exception No. 1, and though OP did not state the motor design letter, using an ITB we can go up to 1700% of a Design B energy-efficient motor's full-load current if demonstrated by engineering evaluation...

74.8A ? 1700% = 1272A

Next higher standard size is 1600A...!!!

Without demonstration by engineering evaluation, the adjustment setting can be as high as 1100%...

74.8A ? 1100% = 823A :blink::blink::blink:

 

[Little Bill]

You guys are way over thinking this. Think back to when you were studying. This is a question from a text book and there are usually mistakes made. There is also info that is put in that has nothing to do with the actual question. The OP doesn't have an actual motor circuit to install. You may know this, but I think going deeper than the study question will confuse the student.