Calculating wire size for a motor.

[kda3310]

(This is a hypothetical situation) I have two 40 horse motors 50 feet from the MCC. They will be a continuous load. They are running through a boiler room that is 120°F. They will be running in the same conduit. How do I calculate the wire size I need to run?

40H Motor is 52Amps
52 x 1.25 for continuous load is 65Amp
Now for having 6 current carrying conductors I have to multiply that by 80%. But do I multiply that by the 52 amps or by the 65 amps?
It will be the same question for the ambient temperature. Which amperage do I multiple by?

 

[augie47]

You apply the derating factors to the conductor based on the insulation. If you are using a 90° conductor you use the 90° rating.

 

[infinity]

You have a 52 amp load, to that you need to add 125% for a motor load and also factor in the ambient temperature of 120° F and 6 CCC's in a raceway.

 

[augie47]

FWIW, I think you will find you need a #3

 

[kda3310]

40H is 52A
80% for continuous load
80% for 6 current carrying conductors
82% for ambient temperature

Do you do the calculation like this?
52 x 1.25 = 65
65 x 1.25 = 81.25
81.25 x 1.219 = 99.043

Is the answer 99 amps?
I don't really need the answer. I need to know how to come up with the answer. Is this right?

 

[augie47]

looks like you are on target

 

[kda3310]

I guess where my confusion comes in is I would think you would do that calculation three individual times off of the 52 amps. Then just calculate the sum. Why would you multiply a multiplied number instead of just multiplying the base number three times individually?

52 x 1.25 = 65 - 52 = 13
52 x 1.25 = 65 - 52 = 13
52 x 1.219 = 63.388 - 52 = 11.388
52 + 13 + 13 + 11.388 = 89.388

Why couldn't I base my wire size off of 89 amps?

 

[augie47]

I mathematician would best answer your question but I believe the answer lies in the first 125% (of FLA)is to find the required ampacity
and the remaining math is to apply adjustment factors to that amapacity.

 

[infinity]

For the derating portion of the calculation you can use the 90° C ampacity of the conductor if it's a 90° C insulation type

 

[kda3310]

I mathematician would best answer your question but I believe the answer lies in the first 125% (of FLA)is to find the required ampacity
and the remaining math is to apply adjustment factors to that amapacity.

That would make sense in a way to me. But, this is how it is still working out in my mind. 40H motor = 52 amps. Do to it being a continuous loud the new base number is 65. So after I multiply for my continuous load factor my 65 amps is what I have to derate off of. Now if I multiply the derating two individual times and then calculate the sum I can still use a #4 wire and hypothetically same money. But is this still wrong?

52 x 1.25 = 65
65 x 1.25 = 81.25 - 65 = 16.25
65 x 1.219 = 79.235 - 65 = 14.235
65 + 16.25 + 14.235 = 95.485

95 amps

 

[infinity]

That would make sense in a way to me. But, this is how it is still working out in my mind. 40H motor = 52 amps. Do to it being a continuous loud the new base number is 65. So after I multiply for my continuous load factor my 65 amps is what I have to derate off of. Now if I multiply the derating two individual times and then calculate the sum I can still use a #4 wire and hypothetically same money. But is this still wrong?

52 x 1.25 = 65
65 x 1.25 = 81.25 - 65 = 16.25
65 x 1.219 = 79.235 - 65 = 14.235
65 + 16.25 + 14.235 = 95.485

95 amps

You have already added 25% for the motor load why are you adding an additional 25% for a continuous load?

 

[kda3310]

You have already added 25% for the motor load why are you adding an additional 25% for a continuous load?

If you recall my original statement I have to factor three issues. Continuous load, 6 current currying conductors and ambient temperature.

Continuous Load 1.25
6 CCC 1.25
Ambient Temp. 1.219

 

[infinity]

If you recall my original statement I have to factor three issues. Continuous load, 6 current currying conductors and ambient temperature.

Continuous Load 1.25
6 CCC 1.25
Ambient Temp. 1.219

Although the result is the same conductors for a motor circuit are sized at 125% of the nameplate HP rating and the ampacity in the corresponding table in Article 430. You do not need to add 25% for a continuous load.

 

[kda3310]

Although the result is the same conductors for a motor circuit are sized at 125% of the nameplate HP rating and the ampacity in the corresponding table in Article 430. You do not need to add 25% for a continuous load.

If the motor runs for three hours or longer you have to add in for continuous load. 2014 NEC 210.20A

 

[infinity]

If the motor runs for three hours or longer you have to add in for continuous load. 2014 NEC 210.20A

That section is for OCPD sizing not the conductors feeding a motor.

 

[kda3310]

That section is for OCPD sizing not the conductors feeding a motor.

OK, help me out here. What is the right answer then? How did you come up with that answer? Please spell it out for me. I see what you are saying but I am still no closer to the answer to my question or how to solve it.

 

[augie47]

My way:
Your motor has a Table ampacity of 52 amps. Code says your cable must have an ampacity = to 125% of that or 65 amps.
You have 6 ccc in your conduit requiring a derating to 80%, 65/.8= 81.25. You elected to use a 90° conductor in a 120°F enviornment so you have an addition derating of .82 (310.15(B)(2), 81.25/.82 = 99 maos so you need a 90° rated conductor good for 99 amps= #3

(You would double check the termination rating to comply with 110.14 but that's not an issue here as even the 60° rating of the #3 far exceeds the actual current)

 

[kwired]

If the motor runs for three hours or longer you have to add in for continuous load. 2014 NEC 210.20A

That section is for OCPD sizing not the conductors feeding a motor.

You get same end result, but for different reasons. Sizing motor conductors isn't done per art 210, it is done per art 430. 430.22 says you must have 125% motor full load current for conductor ampacity unless you have situation in subparts A thru G.

So your base is 65 amps before adjustments -

 

[kwired]

That would make sense in a way to me. But, this is how it is still working out in my mind. 40H motor = 52 amps. Do to it being a continuous loud the new base number is 65. So after I multiply for my continuous load factor my 65 amps is what I have to derate off of. Now if I multiply the derating two individual times and then calculate the sum I can still use a #4 wire and hypothetically same money. But is this still wrong?

52 x 1.25 = 65
65 x 1.25 = 81.25 - 65 = 16.25
65 x 1.219 = 79.235 - 65 = 14.235
65 + 16.25 + 14.235 = 95.485

95 amps

I was pretty sure that method wasn't right but didn't have time earlier to look into things to prove why. Here it is:

your minimum needed ampacity before any adjustments for this conductor is 65 amps - we have already established that

310.15(B)(2) requires an 82% adjustment for ambient temperature so 65/.82=79.27

310.15(B)(3) requires an 80% adjustment for number of conductors in the raceway - but what you were missing is in the header of the table: Percent of values in table 310.15(B)(16)....as adjusted for ambient temperature if necessary.

this means your adjustment for number of conductors in raceway applies to the 79.27 amps you get from ambient temp adjustment - 79.27/.8=99.08.

 

[kda3310]

Thank all of you for your help. I did learn this a long time ago with Mike Holt's programs. About 10 years ago. However, in my line of work doing electrical maintenance I don't use this very often. I love this forum.:thumbsup: