Challenge question 2 in 2005 NEC prep book motor calcs.

[electricalperson]

The question states:

The motor feeder conductor size for 3 15 hp 208v 3 phase motors, 3 3 HP 208v 1 phase motors and 3 1 HP, 120v 1 phase motors will be_____

A 2/0 awg, B 3/0 AWG, C 4/0 AWG, D 250 MCM

15 HP = Largest motor load. 46.2 x 1.25 = 57.75

(46.2 x 1.25) + 46.2 + 46.2 + 18.7 + 18.7 + 18.7 + 16 + 16 +16 = 254.25

250 MCM is rated for 255 amps

my answer was D but the example in the books answer is C

i think the book is wrong since i have the answer key right infront of me and they only added in 1 1 HP motor.

 

[S'mise]

I didn't do the math but your probably right. I found the book loaded with mistakes

 

[Dennis Alwon]

I agree with the 15 hp motor being 46.2 but a 3 hp motor is 10.6.

 

[electricalperson]

I agree with the 15 hp motor being 46.2 but a 3 hp motor is 10.6.

the 3 hp 208v motor is single phase. table 430.248 18.7 amps

a 3 hp 208v 3 phase is 10.6

 

[electricalperson]

this is in the 2008 edition actually. not the 2005

 

[augie47]

Note the single phase motors, if properly connected, would only add single phase loads. For example your 1 hp 120v motors would add only 1 HP to each phase ( each phase would only be loaded 16 amps, not 16+16+16. The same is true for the 208 single phase.)
I think that gives you 204 amps for a 4/0

 

[Dennis Alwon]

the 3 hp 208v motor is single phase. table 430.248 18.7 amps

a 3 hp 208v 3 phase is 10.6

Oops. But Gus is correct.

 

[electricalperson]

Note the single phase motors, if properly connected, would only add single phase loads. For example your 1 hp 120v motors would add only 1 HP to each phase ( each phase would only be loaded 16 amps, not 16+16+16. The same is true for the 208 single phase.)
I think that gives you 204 amps for a 4/0

in the answer key it shows a table with L1 L2 L3. i did not understand why they made the table.

Article 430.24: Conductors supplying several motors, or motor(s) and other load(s) shall have an ampacity not less than 125% of the full load current rating of the highest rated motor plus the sum of the full load current ratings of all the other motors in the group as determined by 430.6(A) plus the ampacity required for the other loads.

the way i thought you do that is you add all the FLC's of the motors regardless if there single phase or 3 phase.

 

[augie47]

If you look at each line (phase) separately and you only has (1) 120v single phase motor it would only load the one phase to which it was connected.
In your case you have (3) 120v motors so if they were each connected to a phase you would only see 1 motor per phase.
(Motor 1 on Phase A, Motor 2 on B, 3 on 3.. so each phase sees only 16 amps not 3 x16)

 

[electricalperson]

If you look at each line (phase) separately and you only has (1) 120v single phase motor it would only load the one phase to which it was connected.
In your case you have (3) 120v motors so if they were each connected to a phase you would only see 1 motor per phase.
(Motor 1 on Phase A, Motor 2 on B, 3 on 3.. so each phase sees only 16 amps not 3 x16)

i understand a little bit better now. but the way article 430.24 is worded doesnt explain thats what you need to do. i dont fully understand this at all

 

[Dennis Alwon]

in the answer key it shows a table with L1 L2 L3. i did not understand why they made the table.

Article 430.24: Conductors supplying several motors, or motor(s) and other load(s) shall have an ampacity not less than 125% of the full load current rating of the highest rated motor plus the sum of the full load current ratings of all the other motors in the group as determined by 430.6(A) plus the ampacity required for the other loads.

the way i thought you do that is you add all the FLC's of the motors regardless if there single phase or 3 phase.

The feeder that carries the load of all these motor will be based on the load of the largest current or ampacity of a phase.
So for a 3 phase motor it is clearly divided up equally.
For the 3- 208 SP motor it would be AB CA BC . As you can see Phases A & B & C will divide the load so that the load is only on any phase twice. With 120 single phase it is obviously equal on all 3 phases so we add it once.

46.2 *1.25= 57.75
46.2*2= 92.4
Add 18.7 twice = 37.4
Add 16 once since the load is equal on all three phase = 16

57.75 (largest motor on all 3 phase at 125%+92.4(the other 2 3 phase motors) +37.4( the 208 sp motors)+16(the single phase 120V motor)= 203.55 amps

 

[Dennis Alwon]

Phase A.....Phase B.....Phase C
57.75 .......57.75........ 57.75
46.2 .........46.2..........46.2
46.2 .........46.2..........46.2

18.7..........18.7..........18.7
18.7..........18.7..........18.7

16.............16............16

Now add up any column

 

[electricalperson]

The feeder that carries the load of all these motor will be based on the load of the largest current or ampacity of a phase.
So for a 3 phase motor it is clearly divided up equally.
For the 3- 208 SP motor it would be AB CA BC . As you can see Phases A & B & C will divide the load so that the load is only on any phase twice. With 120 single phase it is obviously equal on all 3 phases so we add it once.

46.2 *1.25= 57.75
46.2*2= 92.4
Add 18.7 twice = 37.4
Add 16 once since the load is equal on all three phase = 16

57.75 (largest motor on all 3 phase at 125%+92.4(the other 2 3 phase motors) +37.4( the 208 sp motors)+16(the single phase 120V motor)= 203.55 amps

i just wrote out a table that was different than the one in the answer key. the answer key has L1 having 3, 18.7 amp loads, 1 16 amp load and obviously 3 46.2 amp loads

my new table has 3 46.2 amp loads, 2 18.7 amp loads and 1 16 amp load on each phase, L1 L2 L3. the table in the answer key was throwing me off

 

[Dennis Alwon]

i just wrote out a table that was different than the one in the answer key. the answer key has L1 having 3, 18.7 amp loads, 1 16 amp load and obviously 3 46.2 amp loads

my new table has 3 46.2 amp loads, 2 18.7 amp loads and 1 16 amp load on each phase, L1 L2 L3. the table in the answer key was throwing me off

The new table needs to have one motor at 125%-- the largest motor. Look at the chart I made above your last post.

 

[electricalperson]

each phase conductor will have a total of 192 amps on it when perfectly balanced. the table in the book was made wrong but the total at the bottom was 192.

i would take any of the balanced phase conductors L2 for example

(46.2 x 1.25) + 46.2 + 46.2 + 18.7 + 18.7 + 16 = 203.55 = 4/0 awg conductor

i understand better thanks to everyones help. i feel less stupid now

 

[david luchini]

I think 3/0 would be the correct answer, not 4/0.

46.2 + 46.2 + 46.2 + 32.4 (the 3 208V-1ph motors) + 16 (the 3 120V motors) = 187 Amps full load for all the motors. Add 25% of the largest motor (46.2x25%=11.6) and you get a minimum circuit ampacity of 198.6. That would be a 3/0 AWG (cu)