Wire sizing & 240.4(B) next size up rule

Status
Not open for further replies.

SunFish

NABCEP Certified
Location
ID
Occupation
Sr. PV Systems Design Engineer
I am confused by the next size up rule (240.4(B)). Lets say I have 6 current carrying conductors (THWN-2 - 90 deg C wire) in a raceway and my cont. currents are 38 A. The wiring is all indoors so no temp. adjustments are needed.

1) Cont. currents x 1.25 = 47.5 A so I would select a 50 A breaker - This would require a #8

2) Cont. currents / 0.8 (for more than three conductors in raceway) = 47.5 A - This also requires a min #8

So by 690.8 (B) both items 1 & 2 require a #8 minimum

Also, checking my 75 deg C terminals #8 is rated at 50 A so my 50 A breaker still protects my terminals

So everything should be ok to use #8 THWN-2/THHN according to 690.8(B)(2)

Ok here is where I am confused. If I multiply #8 at 55 A (90 deg C rating) by 0.8 (6 conductors in raceway) this comes out as 44 A. This is greater than my continuous currents as allowed by 690.8(B)(2) and this should be ok. But wouldn't 240.4 say that the maximum breaker size I can use to protect this wire at these conditions of use is the next size up 45 A breaker? What am I missing here?

Am I ok to use a 50 A breaker?
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
I am confused by the next size up rule (240.4(B)). Lets say I have 6 current carrying conductors (THWN-2 - 90 deg C wire) in a raceway and my cont. currents are 38 A. The wiring is all indoors so no temp. adjustments are needed.

1) Cont. currents x 1.25 = 47.5 A so I would select a 50 A breaker - This would require a #8

2) Cont. currents / 0.8 (for more than three conductors in raceway) = 47.5 A - This also requires a min #8

So by 690.8 (B) both items 1 & 2 require a #8 minimum

Also, checking my 75 deg C terminals #8 is rated at 50 A so my 50 A breaker still protects my terminals

So everything should be ok to use #8 THWN-2/THHN according to 690.8(B)(2)

Ok here is where I am confused. If I multiply #8 at 55 A (90 deg C rating) by 0.8 (6 conductors in raceway) this comes out as 44 A. This is greater than my continuous currents as allowed by 690.8(B)(2) and this should be ok. But wouldn't 240.4 say that the maximum breaker size I can use to protect this wire at these conditions of use is the next size up 45 A breaker? What am I missing here?

Am I ok to use a 50 A breaker?
No. If you use a 50A breaker then your worst case derated conductor ampacity must be more than 45A. Go to a #6. This happens occasionally.
 

Carultch

Senior Member
Location
Massachusetts
No. If you use a 50A breaker then your worst case derated conductor ampacity must be more than 45A. Go to a #6. This happens occasionally.

If the equipment on both sides of the circuit are "listed and marked otherwise" for 75C, which is very likely, you can use a 50A breaker and protect a #8 Cu wire. Assuming negligible derate factors, if applicable.
 

Smart $

Esteemed Member
Location
Ohio
If the equipment on both sides of the circuit are "listed and marked otherwise" for 75C, which is very likely, you can use a 50A breaker and protect a #8 Cu wire. Assuming negligible derate factors, if applicable.
But the derating factor is not negligible for this installation.

The derated conductor ampacity is less than 46A, the minimum derated ampacity permitted to be protected with a 50A breaker.
 
Status
Not open for further replies.
Top