# 240.4(C) Overcurrent Protection



## jar546 (Mar 29, 2011)

For devices rated over 800 Amperes.

Would this section apply to the main disconnect of a 1200A service or is this section specifically for OCPD's that are on the load side of the main disco?

What is your thought?

One of the guys that does plan review for my company wrote this up on a 1200A service and we are getting some resistance from the electrical engineer who designed it.


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## Bryan Holland (Mar 29, 2011)

If the 1200A overcurrent device is protecting conductors, then yes, the section would apply.  (The conductors being protected by the 1200A overcurrent device would also have to be rated for 1200A or greater.)

I am not sure that is the actual question you are asking???


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## jar546 (Mar 29, 2011)

Bryan, the argument was that 240 does not apply to services.


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## jar546 (Mar 29, 2011)

I have yet to find an engineer who takes into consideration 240.4© and Table 320.15(B)(2)(a) for 3 phase services over 800A.

4 sets of 350MCM copper does not cut it for a 1200A service


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## raider1 (Mar 29, 2011)

jar546 said:
			
		

> Bryan, the argument was that 240 does not apply to services.


If that is the case then why does 230.90(A) Exception #2 specify that fuses and circuit breakers with a rating or setting that complies with 240.4(B) or © and 240.6 shall be permitted.

That exception permits a 400 amp breaker to protect a 400 amp service that is supplied with 500 kcmil copper conductors.

The general rule for overcurrent protection sizing in 230.90(A) states that the overcurrent protective device have a rating or a setting not higher than the ampacity of the service conductors.

Chris


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## chris kennedy (Mar 29, 2011)

jar546 said:
			
		

> Bryan, the argument was that 240 does not apply to services.


Where does it say, "Other than service conductors"? Unless POCO owns the conductors on the line side, IMO 240.4 applies.



> 240.4 Protection of Conductors.Conductors, other than flexible cords, flexible cables, and fixture wires, shall be protected against overcurrent in accordance with their ampacities specified in 310.15, unless otherwise permitted or required in 240.4(A) through (G).





			
				jar546 said:
			
		

> I have yet to find an engineer who takes into consideration 240.4© and Table 320.15(B)(2)(a) for 3 phase services over 800A.


How would T310.15(B)(2) be a factor unless you have multiple sets of parallel conductors in one raceway?


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## jar546 (Mar 29, 2011)

raider1 said:
			
		

> If that is the case then why does 230.90(A) Exception #2 specify that fuses and circuit breakers with a rating or setting that complies with 240.4(B) or © and 240.6 shall be permitted.That exception permits a 400 amp breaker to protect a 400 amp service that is supplied with 500 kcmil copper conductors.
> 
> The general rule for overcurrent protection sizing in 230.90(A) states that the overcurrent protective device have a rating or a setting not higher than the ampacity of the service conductors.
> 
> Chris


EXACTLY what my reply was


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## jar546 (Mar 29, 2011)

chris kennedy said:
			
		

> Where does it say, "Other than service conductors"? Unless POCO owns the conductors on the line side, IMO 240.4 applies.How would T310.15(B)(2) be a factor unless you have multiple sets of parallel conductors in one raceway?


3phase with 4 current carrying conductors in each conduit.


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## raider1 (Mar 29, 2011)

jar546 said:
			
		

> 3phase with 4 current carrying conductors in each conduit.


Unless the neutal was carrying a large nonlinear load the neutral of a 4 wire 3 phase service would not be considered a current carrying conductor in accordance with 310.15(B)(2)(a) and 310.15(B)(4)©.

Chris


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## chris kennedy (Mar 29, 2011)

jar546 said:
			
		

> 3phase with 4 current carrying conductors in each conduit.


Then T310.15(B)(2) won't apply unless you can prove its a 310.15(B)(4)© install.


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## jar546 (Mar 29, 2011)

I'll buy 310.15(B)(4)(a) but not © unless they can prove it.

√((Ia²+Ib²+Ic²)-(Ia*Ib)-(Ib*Ic)-(Ia*Ic))


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## raider1 (Mar 30, 2011)

jar546 said:
			
		

> I'll buy 310.15(B)(4)(a) but not © unless they can prove it.√((Ia²+Ib²+Ic²)-(Ia*Ib)-(Ib*Ic)-(Ia*Ic))


The formula that you posted does not take into account harmonics. What you have posted is just the simple formula to find neutral current in a 3 phase circuit.

Without harmonics in the circuit the neutral of a 4 wire 3 phase circuit is not a current carrying conductor for deration purposes.

I would be amazed if I found a 3 phase service where the neutral had a large harmonic load that actually would make the neutral count as a current carrying conductor for derating purposes.

Chris


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