Clarification - Art. 630.12 - Calculation Example

[eager2learn]

So I'm reading the calculation example that is in art. 630.12 in the 2014 Handbook. I'm confused by the very last sentence given in the conclusion. Picture provided.

My question is in regard to the OCPD

I have a welder that has an input of 65 amps with a duty cycle of 60%

I took the 65 amp input and multiplied it by the duty cycle at 60% which is .81 giving me 52.65 amps which will use #8 thhn in the 90 degree category.

If I follow the calculation example in order to find out the OCPD best utilized for this application I end up with a OCPD that has a rating of 130a rated at 200% of the input amperage which is 65amps.

The clarification that I need is can I use a smaller OCPD that is less than 130 amps but equal to or greater than the initial 65amps at input.

I also provided a picture of the info on the actual welder

 

[jumper]

I took the 65 amp input and multiplied it by the duty cycle at 60% which is .81 giving me 52.65 amps which will use #8 thhn in the 90 degree category.

You cannot use 90C temp for final ampacity of conductors, derating and correction factors only because of termination limits. 110.14(C).

 

[eager2learn]

Thank you for the information, I appreciate it however I am still wondering about my original question in regards to OCPD

 

[jumper]

The clarification that I need is can I use a smaller OCPD that is less than 130 amps but equal to or greater than the initial 65amps at input

Yes.

 

[Smart $]

...I end up with a OCPD that has a rating of 130a rated at 200% of the input amperage which is 65amps.

The clarification that I need is can I use a smaller OCPD that is less than 130 amps but equal to or greater than the initial 65amps at input.
...

Thank you for the information, I appreciate it however I am still wondering about my original question in regards to OCPD

What do you think "not more than 200%" means?

 

[eager2learn]

Thank you.