It seems I may have been working under a false understanding of AC vs. DC wire derating? Going back to the Edison vs Tesla "electrical war" on comparison given was:
I'll admit I don't understand the physics behind it but by comparing the DC derating chart from the American Boating and Yacht Council which I have used for many years now.
Versus the National Electrical Code
Derating:
where I is the load current, L is the distance, and Zc is the cable impedance in Ω/km.
So let's limit this to single phase, no temperature derating, no derating for multiple conductors. Using 120VAC, 100', 20A,
From the AWG
So 100' AWG 12 wire with 10.15Ω/1000 ft. on a 20A circuit breaker
Vd1ϕ=IL(2Zc)/1000 = 20A*100ft(2*10.15Ω/1000ft)/1000= 40.3mV
Which is far less than the derating for DC?
So I am obviously missing something here... I understand that @ Vrms, volts are volts no matter whether AC or DC, and amps are amps. What am I missing here?
I'll admit I don't understand the physics behind it but by comparing the DC derating chart from the American Boating and Yacht Council which I have used for many years now.
Versus the National Electrical Code
Derating:
- No derating is currently applied to the current ratings tables 310.15(B)(16) and 310.15(B)(17)..
- It is assumed that the maximum ambient temperature is 30°C and the maximum ground temperature id 20°C. For higher temperatures, a derating will have to be applied according to NEC.
- It is assumed that there is only one three current carrying conductors in a raceway, cable or buried. And that the raceways, cables and buried conductors are spaced according to NEC to prenet derating.
- To apply a manual derating, divide the load by the derating factor from NEC, and enter the new load value in the calculator.
- The single phase AC voltage drop is calculated as:
where I is the load current, L is the distance, and Zc is the cable impedance in Ω/km.
So let's limit this to single phase, no temperature derating, no derating for multiple conductors. Using 120VAC, 100', 20A,
From the AWG
So 100' AWG 12 wire with 10.15Ω/1000 ft. on a 20A circuit breaker
Vd1ϕ=IL(2Zc)/1000 = 20A*100ft(2*10.15Ω/1000ft)/1000= 40.3mV
Which is far less than the derating for DC?
So I am obviously missing something here... I understand that @ Vrms, volts are volts no matter whether AC or DC, and amps are amps. What am I missing here?
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