90Deg C cable sizing for Motor

[ventilator]

I am having trouble understanding how to apply 90deg C rating of a conductor to a motor circuit with 75deg termination.
I am dealing with a 550hp motor at 480v so 650 FLA.
We have 90deg C cable for the motor feeder. Using the derating factor given by the installtion site of 35deg C ambient and the 125% requirement for a motor feed the required cable ampacity calculates out as (650 * 1.25)/.96 = 846A. So my cable must be able to carry 846 amps which at 90deg means a parallel set of 500mcm copper.

Now I take into account the 75deg rating of the terminations so I look at 500mcm for 75deg and it is good for 380 amps. This is where I get really confused. The site engineer states that 500mcm are OK to use in this application because the amapcity required is 650a/.96=677 and parallel 500s are good for 730amps after derating for ambient.

Here are the questions I have with this situation:
1. How can you apply the 90deg derating factor of .96 when you are dealing with the 75deg rating (380a)? Shouldn't the ampacity be 650/.94?
2. Why does the 125% motor feeder rule from 430.22 not apply to the 75deg rating? Shouldn't the required ampacity be (650 * 1.25 / .94) = 776 which would bump you to a 600mcm parallel run?

I guess I dont' understand how the terminal rating fits into the feeder sizing, if the terminal is rated for 75deg doesn't that mean that it can only handle 380amps?

I know this is a long post so thank you in advance to anyone willing to get through it.

 

[bob]

The table giving the ampacity of cables is for a 30C environment. When you increase the temp you must reduce to cable rating. You can use the 90C rating when you are derating the cable for temp and number of cables in a conduit or other situations dictated by the code. In your case you would take the 90C rating x the temp reduction factor to get the ampacity rating of the cable. That would be 430 amps x 0.96 = 413 amps. The required ampacity is 650 x 1.25 = 812 amps/2 = 406 amps per phase. You are limited to 75C by the motor terminal rating. The 406 amps exceeds the 75C rating of 500 kcm which 380 amps. You will need to go to a larger conductor or increase the number of smaller cables to obtain this ampacity.

 

[david luchini]

I am having trouble understanding how to apply 90deg C rating of a conductor to a motor circuit with 75deg termination.
I am dealing with a 550hp motor at 480v so 650 FLA.
We have 90deg C cable for the motor feeder. Using the derating factor given by the installtion site of 35deg C ambient and the 125% requirement for a motor feed the required cable ampacity calculates out as (650 * 1.25)/.96 = 846A. So my cable must be able to carry 846 amps which at 90deg means a parallel set of 500mcm copper.

Per 430.22, you need a cable with an ampacity of not less than 125% of the FLA. This will be 813A (not 846A).

Now I take into account the 75deg rating of the terminations so I look at 500mcm for 75deg and it is good for 380 amps. This is where I get really confused. The site engineer states that 500mcm are OK to use in this application because the amapcity required is 650a/.96=677 and parallel 500s are good for 730amps after derating for ambient.

You are correct, per 110.14(C), the temperature rating associated with the ampacity of the conductor shall be selected and coordinated so as not to exeed the lowest temperature rating of any termination. In your case this is 75deg. To get the required ampacity (813), you need at least two sets of 600mcm: 420*2=840A. Two sets of 500mcm is too small (760A). But in your case, the correction for ambient, two sets of 600mcm (75deg) would be too small. 420*2*0.94 = 790A. You would need two sets of 700mcm, 75 deg conductors.

However, 110.14(C) also lets you use the 90 deg rating for ampacity adjustment, correction or both. So two sets of 600mcm, 90 deg would give you 475*2*0.96 = 912A. This ampacity is larger than the required 813, so it is OK. Two sets of 600mcm, 90 deg will work.

Here are the questions I have with this situation:
1. How can you apply the 90deg derating factor of .96 when you are dealing with the 75deg rating (380a)? Shouldn't the ampacity be 650/.94?

You would apply the 90deg factor to 90deg conductors, and the 75deg factor to 75deg conductors.

2. Why does the 125% motor feeder rule from 430.22 not apply to the 75deg rating? Shouldn't the required ampacity be (650 * 1.25 / .94) = 776 which would bump you to a 600mcm parallel run?

It does apply. You need a conductor with an ampacity of 813A.

I guess I dont' understand how the terminal rating fits into the feeder sizing, if the terminal is rated for 75deg doesn't that mean that it can only handle 380amps?

You are correct here. The 75deg terminals for two sets of 500mcm is only good for 760A. This is too small for the needed 813A. You need two sets of 600mcm.

 

[Smart $]

Adjustment and correction factors do not apply to the termination temperature limitations. Just remember the terminations are not in raceway or cable, which is what adjustment and correction factors apply to.

650A_FLC ? 125% = 812.5A minimum required conductor ampacity

2 sets minimum 406A each @75?C = 600kcmil
3 sets minimum 271A each @75?C = 300kcmil

 

[iceworm]

Slow poster - others already covered it.

Follow them and you got it

ice

 

[iceworm]

I am having trouble understanding how to apply 90deg C rating of a conductor to a motor circuit with 75deg termination. ...

One question: Is the switchboard with the motor feeder CB in the high ambient area?

That would change things a bit.

ice

 

[ventilator]

One question: Is the switchboard with the motor feeder CB in the high ambient area?

That would change things a bit.

ice

No, the feeder CB is in a climate controlled room. I don't know what the temp will be set to but certainly less than 35deg C.