Can you provide the formula for calculating ampacity of conductors in air please.
Regards
Greg.
Can you provide the formula for calculating ampacity of conductors in air please.
Regards
Greg.
Do people use web search engine's anymore?
Oh yes, it is quite simple if you know about heat transfer.
I (in kiloamperes) = SQRT((TC-TA+Delta TD)/((RDC)(1+YC)(RCA))) found in Section 310.15(C) of the 2002 National Electrical Code.
RDC = resistance of one foot of conductor that is .001 in in diameter in microhms TC = conductor temperature in degrees C. TA = ambient temperature in degrees C. YC = component ac resistance resulting from skin effect and proximity effect RCA = effective thermal resistance between conductor and surrounding ambient in thermal ohm feet Delta TD = Dielectric loss Temperature Rise That is all there is to it. Just substitute in and calculate. For more information read the Neher McGrath paper published in 1957.
Also read Understanding the Neher-McGrath Calculation and the ampacity of conductors at:
Correction on the above;
RDC is the resistance of one foot of conductor for the given Circular Mil Area and is given by:
RDC = (1.02 *pc /cma)* ((tah+tc)/(tah+20))
RDC = ohms pc = circular mil ohms per foot of conductor at 20 degrees C. (10.371 ohms for 100% IACS copper, 17.002 ohms for 61% IACS aluminum) tah = absolute value of inferred temperature of zero resistance. (234.5 degrees C. for copper and 228.1 degrees C. for aluminum) cma = circular mil area of conductor from Chapter 9 Table 8 of NEC tc = conductor temperature in degrees C.
from: NM paper and from:
Thanks for that. Do you know where I can get a copy of the Neher-McGrath paper?
Greg.
----- Original Message ----- From: "Mr. Smith" Newsgroups: alt.engineering.electrical Sent: Tuesday, January 13, 2004 3:40 PM Subject: Re: Ampacity in air
Thanks for the info.
I am having problems calculating RCA. I know how to calculate Ri for the insulation but cannot seem to arrive at the same current value as per NEC table. The example given at
Can you shine some light on the subject.
Greg.
PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.