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Ampacity of Copper Conductors 9

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Mbrooke

Electrical
Nov 12, 2012
2,546
Anyone know why Table 310.15(B)17) lists 4/0 copper at 360*C but wire spec sheets list it at 480 amps? Same for other wire sizes... what is the actual ampacity? Am I correct to think NFPA70 assumes no wind while wire manufacturers assumes wind?


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You are comparing bare copper wire with insulated copper wire.
The Wire insulation provides thermal insulation as well as electrical insulation.
Note also that your 4/0, insulated wire has an ampacity of 405 Amps at 90 degrees C.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
I guess thats behind it. Any idea what bare copper is rated at 90 or 100*C?
 
The NEC current ratings in free air is based on a wind speed of [highlight #FCE94F]2 ft/sec[/highlight] (610 mm/sec).
[sub]It should be noted the following:
[ul]
[li]Insulation provides extra thermal resistance that makes this cable operate warmer than the bare conductor of the same size exposed to the same ambient conditions. Therefore, it should be expected that the rated Apmt of the insulated conductor < bare cable amp.[/li]
[li]For overhead applications, Insulated conductors are not limited by the insulation and sagging but for bare conductor only sagging is the limiting factor.[/li]
[li]The term standstill is used in some ampacity tables for ambient without wind. Also, the ampacity rating includes 961 watts/sq. foot sun, 0.5 coefficients of emissivity and absorption primarily for bare cables.[/sub]
[/li]
[/ul]
 
In the Canadian code, the equivalent table lists bare copper or cable with special insulation at:
110 degrees C, 455 Amps
125 degrees C,485 Amps
200 degrees C, 590 Amps
Note (3) These ampacities apply to bare wire or under special circumstances where the use of insulated conductors
having this temperature rating is acceptable.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
@Cuky:
Insulation provides extra thermal resistance that makes this cable operate warmer than the bare conductor of the same size exposed to the same ambient conditions. Therefore, it should be expected that the rated Apmt of the insulated conductor < bare cable amp.


Now this is getting interesting! [smarty] I originally assumed the opposite- that a covering would increase dissipation due to increase surface area not carrying electricity. Then I doubted my train of thought when I posted here and I would have agreed with this. However, a code expert pointed me to 310.15 B 21 which seems to actually show bare copper as having a lower current rating:

Clipboard01u_sg0cye.jpg


None the less it is still lower than most manufacturer data.

I looked at publicly available PJM documents which seem to use 2ft wind only for Emergency conditions and no wind for steady state operating though in theory this is a separate issue outside the scope of this thread but none the less makes me wonder where the true ampacity of 4/0 stands.
 
Thank you for your information Cuky.
I don't see our information as being contradictory, rather we are sharing information from different sources based on slightly different parameters.
Cuky said:
but none the less makes me wonder where the true ampacity of 4/0 stands.
That's easy.
The ampacity is whatever the AHJ says it is. grin

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
I think we can all agree on the AHJ having the last word.

Any idea at what temperature annealing takes place?
 
In NEC Annex B it is noted:
"For additional information concerning the application of these ampacities, see IEEE
STD 835-1994, Standard Power Cable Ampacity Tables."
Let’s take IEEE 835/1994 page 33:
0.6 to 5 kV Unshielded Single Conductor Extruded Dielectric Cable
90ºC - Copper Conductor - Concentric Stranded in Free Air - Single Isolated Cable
40ºC Air Ambient
In free air there are 4 possibility:
no sun no wind 434 A ; 2 Ft/s 542 A
full sun no wind 344 A ; 2 Ft/s 477 A
NEC Table 310.15(B)(17) 4/0 90oC 405 A 30oC air.
Table 310.15(B)(2)(a) Ambient Temperature Correction Factors Based on 30°C (86°F)
36-40oC Conductor temp. 90oC 0.91
That means according to NEC 4/0 copper conductor insulated for 90oC in 40oC air =
405*0.91= 368.55 A . No any condition meats IEEE 835/1994.
For ampacity calculation there are 2 possibilities :
according to Neher and McGrath[corrected according to IEEE 835/1994] or IEC 60287.

 
For ampacity calculation there are 2 possibilities :
according to Neher and McGrath[corrected according to IEEE 835/1994] or IEC 60287.
From NEC table 5
Type XHHW, XHHW-2, XHHN insulated core dia. 16.21 mm [0.638 inches]
From NEC table 8 dcopper= 13.41 mm[ 0.528”] 40oC AIR
IEEE Method 435 A NO SUN NO WIND
IEC Method 448 A NO SUN NO WIND

 
Alright, so I now want to ask: do I assume wind or no wind?
 
Calculated in "no sun no wind" condition for bare conductor the ampacity is lower than for covered case due to decreased heat dissipation by radiation: the radiation factor for covered is close to black [92%] but for bare decrease up to 23%.
Ventilation can amplify convection and thus compensate for the reduction in radiation.
How much? I will have to use the calculation method from IEEE 738 and that as soon as I can.

 
Seasonal ratings are bad enough. Adding time of day, day of year, and wind speed plus cloud cover is just asking for trouble unless you have a fully automated dynamic rating scheme and a way to enforce those ratings. No long term profit in squeezing out the last first cost dollar.
 
@ David, where would you put 4/0 under your seasonal ratings?

@7anoter4: Can you explain the theory behind radiation factor?
 
Calculated according IEEE 738/2007 IEEE Standard for Calculating the Current-Temperature of Bare Overhead Conductors for no sun, wind 2ft/s and ε[Emissivity (0.23 to 0.91)]=0.23 the ampacity of 4/0 copper [75oC conductor 25oC air]=487 A
 
I agree with davibeach and I did not introduce the sun exposure in my calculation.
 
487- close to that table. Any idea of the current at 100*C?
 
The heat evacuation from cable surface to air has two parts-convection and radiation.
The radiation depends on surface emissivity-ε it is the ratio between the black body emissivity
4.96 kcal/m^3.hr.(100oK)^4 and the this body emissivity q=vol.4.96*ε*(T/100)^4 where T it is the body temperature in Kelvin degrees and vol it is the body volume [m^3] q[= kcal/hr].

 
I owe you now.

Can you explain how surface emissivity works?
 
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