Derating ???

[necnotevenclose]

Is it me or has the way you derate changed in the code from 2005 to 2008?

I was looking at NEC 2005 vs. NEC 2008 handbook for derating examples. I did not find a calculation in 2005, but what I found online on ecmweb.com gave examples of multiplying the derating factor. However, in 2008 handbook there is a calculation example (page 310-311) that divides the derating value making the amps of the wire more, thus possibly resulting in a bigger wire (pipe) size. Has anyone reviewed this? If you have, what is your take on how to calculate this?

Also, its my understanding that the code states for devices rated over 800A the conductors shall be equal to or greater than the rating of the OCPD. So if you have a 2000A device and you derate per 2005 (multiply) would you have to upsize your wire to meet 240.4C?

 

[charlie b]

However, in 2008 handbook there is a calculation example (page 310-311) that divides the derating value making the amps of the wire more, thus possibly resulting in a bigger wire (pipe) size.

You are mis-reading the example. It is not making the amps of the wire more, it is making you select a larger wire than would have been needed, if the only consideration were the load amps. That is what derating is all about.

You can do the same calculation two ways, and get the same answer.



Information used for both methods:

• The load is 175 amps.
• The derating factor is 0.82.

Method 1: Pick a conductor size, apply the derating factor, and see if it works. If it doesn't work, try the next larger conductor size.

• Try 4/0 aluminum XHHW-2. The 90C ampacity is 205. Multiply 205 times 0.82, and you get 168. That is too low, since we need 175.
• Try 250 MCM aluminum XHHW-2. The 90C ampacity is 230. Multiply 230 times 0.82, and you get 188. That is high enough, since all we needed was 175. So you select that conductor.

Method 2: Apply the derating factor to the load amps, and find a conductor that has that ampacity.

• This is what the handbook did. You need 175 amps. The derating factor multiplied by the tabulated ampacity must be at least 175.
• So you get the answer by dividing the load amps of 175 by the derating factor of 0.82, and you see that you need a conductor with a minimum ampacity of 213.
• Looking in the table, you see that 250 MCM aluminum XHHW-2 has an ampacity of 230. So you select that conductor.

Bottom line: You multiply the derating factor times the ampacity in the table, and you get the "derated ampacity."

 

[necnotevenclose]

thanks Charlie. My head is on straight now.

 

[mull982]

Information used for both methods:

• The load is 175 amps.
• The derating factor is 0.82.

Method 1: Pick a conductor size, apply the derating factor, and see if it works. If it doesn't work, try the next larger conductor size.

• Try 4/0 aluminum XHHW-2. The 90C ampacity is 205. Multiply 205 times 0.82, and you get 168. That is too low, since we need 175.
• Try 250 MCM aluminum XHHW-2. The 90C ampacity is 230. Multiply 230 times 0.82, and you get 188. That is high enough, since all we needed was 175. So you select that conductor.

Method 2: Apply the derating factor to the load amps, and find a conductor that has that ampacity.[/SIZE

It should be noted that when derating using 90deg C column the final cable size after derating must be equal to or greater than the cable in the 75deg ampacity rating for the origonal cable ampacity before derating.