What amount of ampacity (short circuit current) can an insulated #14 copper wire handle for a certain amount of time before there is damage to the insulation or the conductor itself? I'm looking for a Time vs Current curve that displays the information similar to the Time vs Current curves that manufacturers make for a circuit breaker to clear a fault. Example: At a short-cirucit current of 75 amps a typical 15A Cutler Hammer breaker should interrup the circuit current in approximately 6 to 7 seconds based on the Time vs Current curves for there specific breaker. But what I'm wondering, for what time duration could that #14 copper wire continue to heat up before the thermal limits of the insulation is damaged from the 75 amps? I'm sure wire manufacturer's have this information readily available. Please help. Thank you.
Overload on #14 AWG Copper Wire before Damaging the Insu
- Thread starter sweso
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- Location
- Mission Viejo, CA
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- Professional Electrical Engineer
Re: Overload on #14 AWG Copper Wire before Damaging the Insu
This website will give you a graph and the general formula. Take care though. True ambient conditions are rarely known.
http://www.okonite.com/engineering/short-circuit-currents.html
This website will give you a graph and the general formula. Take care though. True ambient conditions are rarely known.
http://www.okonite.com/engineering/short-circuit-currents.html
- Location
- Illinois
- Occupation
- retired electrician
Re: Overload on #14 AWG Copper Wire before Damaging the Insu
A formula used for calculating the withstand for insulated cables is that the cable can withstand 1 amp for every 43.2(not 100% sure of this number, but it is close) circular mils for 5 seconds. That would give you a 5 second withstand of 95 amps for the #14. You can use the formula I^2t to adjust for other times. 95^2 = 9025 X 5 = 45125 = the I^2t rating for #14. If the expected trip time was 3 seconds, you would divide the I^2t rating by 3 and take the square root of that number to find the 3 second withstand for #14. 45125/3 =15042. The square root of 15042 is 122. The 3 second withstand of #14 is 122 amps. If you have an OCPD that will clear the fault in 1 cycle (0.01667 seconds) the withstand of #14 is 1645 amps.
Don
A formula used for calculating the withstand for insulated cables is that the cable can withstand 1 amp for every 43.2(not 100% sure of this number, but it is close) circular mils for 5 seconds. That would give you a 5 second withstand of 95 amps for the #14. You can use the formula I^2t to adjust for other times. 95^2 = 9025 X 5 = 45125 = the I^2t rating for #14. If the expected trip time was 3 seconds, you would divide the I^2t rating by 3 and take the square root of that number to find the 3 second withstand for #14. 45125/3 =15042. The square root of 15042 is 122. The 3 second withstand of #14 is 122 amps. If you have an OCPD that will clear the fault in 1 cycle (0.01667 seconds) the withstand of #14 is 1645 amps.
Don
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