derating wire for conditions of use - value > OCPD?

[david luchini]

If you used the method I proposed, "648A actual x 1.25 = 810A rated and 810A rated / 0.58 = 1396 A total ampacity required. 1396/4 runs = 349 amps per ckt. The minimum size conductor is 500 kcm at 380 amps is adequate. " and compare it to this method "rated load = 810 amps. Using 90C rating 500 kcm 430 amps x .58 = 249.4 amps x 4 = 997.6 the results are the same as far as conductor size required. However if you use the 1000 amp breaker the 500 kcm is not correct. I think the 2nd calculation is the way it is usually done but would the 1st calculation be incorrect? Any comments?

I'll give it a go. The first calculation is incorrect, because the adjusted/corrected ampacity need only be adequate for the load to be served, which is 648A. 648A/0.58 = 1117. 3 sets of 400mcm 90C would be adequate for the load (380A*3*0.58= 661A.)

However, as you noted, a conductor with an ampacity of 661A is not properly protected by a 1000A OCPD (or an 800A OCPD for that matter.)

 

[ggunn]

So when inverter experiences an overload or short circuit condition does it limit the current to it's max output or does it trigger some kind of overcurrent protection and shut down the output? Or is either scenario a possibility?

I am starting to see this is a source that doesn't exactly follow the same rules as other sources often do.

most of my inverter experience is with small units used to derive AC power from a 12 volt DC source, or with VFD's.

Yes, it's a different type of source. Most sources are constant voltage sources; they try to keep the voltage constant at the terminals and the current is inversely proportional to the load. A GT inverter behaves as a constant current source (it tries to keep current constant irrespective of voltage) whose voltage is clamped by the line voltage from the grid, which also gives it a virtually infinite sink to push current into, which in turn essentially turns it into a constant power source.

A grid tied inverter cannot experience an overload as long as it is connected to the grid and the grid is operational. In the event of either a short or an open circuit condition on its output conductors (or the grid going down), it shuts down immediately because its reference voltage goes away.

The difference between a GT inverter and the battery powered ones you are used to dealing with is that in your experiences the DC input is from a battery, which is a voltage source and the inverter can pull from it on demand. PV modules are current sources over most of their IV curve. Off grid PV systems use battery inverters as well; they run off the batteries and need a separate system to charge the batteries from the PV modules.

 

[bob]

I'll give it a go. The first calculation is incorrect, because the adjusted/corrected ampacity need only be adequate for the load to be served, which is 648A. 648A/0.58 = 1117. 3 sets of 400mcm 90C would be adequate for the load (380A*3*0.58= 661A.)

However, as you noted, a conductor with an ampacity of 661A is not properly protected by a 1000A OCPD (or an 800A OCPD for that matter.)

The OP indicated he was using a continuous load and the 1.25 factor should be included. Therefore we are back to 810 amps.

 

[david luchini]

The OP indicated he was using a continuous load and the 1.25 factor should be included. Therefore we are back to 810 amps.

That is incorrect. Common misconception, however. The load is 648A. The feeder conductor must have an ampacity not less than required to supply the load (648A) per 215.2(A)(1).

The minimum feeder conductor SIZE, before the application of any adjustment or correction factors, must have an allowable ampacity not less than the non-continuous load plus 125% of the continuous loads.
That is 810A would would make the minimum feeder conductor SIZE two sets of 600mcm or three sets of 300mcm, for example. (Also from 215.2(A)(1))

Of course, neither two sets of 600mcm or three sets of 300mcm would have sufficient ampacity for the 648A load when they are corrected for the ambient.

 

[bob]

OK. Minor multiplication at the wrong point results in major adjustment. Good job.

 

[Smart $]

OK. Minor multiplication at the wrong point results in major adjustment. Good job.

While I agree with David's numerical assessment, the correct applicable sections are 690.8(B)(1)&(2). FWIW, terminal temperature limitations are accounted for under overcurrent devices rather than conductor ampacities... irrelevant for the case at hand.

One area which hasn't been mentioned for this case is derating of the overcurrent device when used in an ambient greater than 40?C. See 690.8(B)(1)(c).