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:

• 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.

Voltage drop calculation:

• The single phase AC voltage drop is calculated as:

Vd1ϕ=IL(2Zc)/1000

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 don't have any links or facts to back this up at hand, but if I remember right I believe the difference in deratings between DC and AC is simply due to the nature of AC. Assuming unity of current and voltage phases, and speaking in terms of temperature derating, when the sine wave drops to 0V the wire isnt carrying current and not heating up and so has a chance to cool down giving it the ability to carry more current at a given temperature without breaking down vs. DC which is always flowing in the same direction and always heating the conductor. I know your question is in regards to voltage drop, but I do believe that is a function of temperature when it comes to derating. Either way I think the factor you are missing in your analysis is time, which has really has virtually no effect on a DC current through a wire but makes a great deal of difference in an AC current through the same/similar wire. You can't really directly compare VrmsAC to VDC in this case. While we can ignore certain properties for some equations, or calculate a relative approximation(ie.. Vrms), I don't think that's a property we can ignore in this particular case and maintain an accurate analysis.

I could be wrong on some of this, and I am sure someone will chime in and correct me if I am.

 

To me, it doesn't make sense that AC and DC should be different. The resistance of the wire doesn't change at power transmission frequencies.
A resistor doesn't have different power ratings for AC and DC.

 

That's why I'm asking Albert. I remember reading about the Edison/Tesla war over AC vs DC distribution and that AC won. Never remember derating 120VAC for distance and in an industrial setting, some runs were rather long. Maybe for the number of conductors in conduit and used THHN rated wire so not for temperature either? Too long ago and my memory is failing me these days. I did have to derate on my boats. In particular, I had a 24' cuddy cabin cruiser that over the years the helm station had become a rat's nest of wire from the previous owner's additions and modifications. After adding my marine VHF 25W radio, Color CRT Sonar, plug-in QBeam spotlight and Loran if anyone pushed the cigarette/cigar lighter in the voltage drop killed the Loran and it took quite a while for it to re-initialize and reconnect to the remote Loran signal transmitters. I ended up stripping the helm station wiring and running a ~15' 20A 8 gauge feeder from the battery to a fused distribution plate at the helm and rewiring all of the helm electrical systems to eliminate the voltage drop. It was being fed by a 12 gauge feeder wire that was undersized and not fused. Was obvious trying to play a cassette tape in the stereo system that there was voltage loss from the sound that came out even before I added my instrumentation. Everything worked properly after upgrading the wiring.

 

It seems I may have been working under a false understanding of AC vs. DC wire derating?

I didn't read all of the material, but I suspect it has something to do with electromigration and self-heat. We treated DC and AC currents differently for reliability purposes in microprocessor design. Metal lines with DC current were more impacted by electromigration. Lines with AC current were more impacted by conductor heating.

 

> 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

Well, looking at the chart 10.15 ohms/1000 ft is for AWG 20 wire.

Your 20A*100ft(2*10.15Ω/1000ft)/1000 has an extra factor of 1/1000 in it
and really is: 20A * 100ft * 2 * (10.15ohms/1000.) -> 40.6 volts drop

So instead of 120 volts the load will see 120-40 -> about 80 volts
and is unlikely to be very happy...

 

So instead of 120 volts the load will see 120-40 -> about 80 volts
and is unlikely to be very happy...

Something is not right, a 40V drop in 100' on a 20A 12AWG 120VAC circuit does not happen. If it did 120V appliance would indeed not be "happy" or even work properly. And while most house circuits are actually close to 100' (50' to & 50' from including elevation changes) you will not see 80V at the outlet, in fact, the mV drop is usually not noticed.

 

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:

DC 60V, Rload=60Ω --> long distance, Rwires=60Ω --> 50% losses of energy.
AC 60V, Rload=60Ω --> transformer 60/600V --> long distance, Rwires=60Ω --> transformer 600/6V --> 1% losses of energy.
To use transformers or not to use, - it is sense of Edison vs Tesla "electrical war".

 

Something is not right, a 40V drop in 100' on a 20A 12AWG 120VAC circuit does not happen.

I calculate the resistance of 100' of 12 AWG to be 0.16 ohms which would give a drop of about 3V at 20A. The appliance wouldn't care about the voltage drop.

 

From https://www.rapidtables.com/about/about.html


Well this says it is identical...
Yet the ABYC chart for 20A @ 100' says AWG8 and the NEC for 20A is AWG12? I think my confusion is partly due to the fact that a 20A circuit typically doesn't carry 20A but is rated to, so when measured you typically don't see the voltage drop for full current capacity? I think I'm getting too old for this...

 

Well this says it is identical...

It also says that 10 feet of 24AWG has a resistance of 2.5MΩ so I wouldn't put too much weight on that.

 

Hmmm... Missed that, didn't look at the bottom line. I always wonder as to provenance and accuracy of online "calculators" and have to take them at face value... I even went to the AWG site to get the Ω/meter value for 24AWG wire to plug into the calculator.

 

Look at the following details.

Sizing for AC circuits are usually in one-way lengths, the actual length of wire is 2x the one-way length.
Usually you size for <3% voltage drop at 80% of the breaker capacity.

With the body of the car being the return path you are mostly interested in the one-way drop.

Ambient temperature matters. The number of conductors in a raceway matters. The engine compartment temperature and cabin temperature will be hotter for cars and boats. A larger amount of conductors were bundled together in a vehicle.

In the old days, upping the voltage for longer distances was easy with a transformer. It's difficult to change 1000VDC to 10,000V DC.

A 10 MW Solar array field will have a high voltage DC bus. The inverters can locally adjust the power factor which is something you don;t get with AC. High voltage DC transmission lines to exist.

https://circuitglobe.com/hvdc-high-voltage-direct-current.html

technology has changed.

 

It also says that 10 feet of 24AWG has a resistance of 2.5MΩ so I wouldn't put too much weight on that.

The wire type (assume Aluminum or Copper) is a paameter that isn't filled out in the calculator.

 

but the ohm/meter for 24AWG copper is... weird...

 

no it's not.

 

but the ohm/meter for 24AWG copper is... weird...

Do they mean milli-ohms, perhaos??

 

Really, there are two separate sets of considerations with wire sizing and current ratings. The first consideration is voltage drop, which directly impacts efficiency, since every it of power turned into heat is wasted, at least considered wasted in most areas. The second consideration is the conductor heating, which is why there are different wire tables for different kinds of insulation, and conductors in air versus conductors in conduit. Heating leads to connection failures and also leads to insulation breakdown, both of them are problems.
What is interesting is a comparison of heating in wires carrying DC and AC. With the current specified as either RMS amps or average amps, the heating is about the same for AC and DC. Not exactlt the same, but quite close.

 

but the ohm/meter for 24AWG copper is... weird...

Do they mean milli-ohms, perhaos??

0.0841976Ω is in milliohms, taken directly from the AWG table.

Most of my career I dealt with 120/240/480 VAC THHN/TWWN power wiring in conduits and 120VAC control voltage with the same wire in conduit or cable in cable tray cables. Usually 25HP (or less) pump motors and controls integrated into my DCS systems. Some of those runs could be a few hundred feet and I don't remember ever derating except for maybe 1" or larger conduit carrying the maximum of #12 conductors and not very often.

Working on my boats was totally different, dealing with only 12VDC. I think that is where I got mislead as I never worked with 12VAC power distribution which is apparently very different.

 

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

ell, looking at the chart 10.15 ohms/1000 ft is for AWG 20 wire.

Your 20A*100ft(2*10.15Ω/1000ft)/1000 has an extra factor of 1/1000 in it
and really is: 20A * 100ft * 2 * (10.15ohms/1000.) -> 40.6 volts drop

From the table 1.588 ohms/1000 ft for 12 AWG wire.

1.588 ohms/1000' is 0.1588 ohms/100 feet; divide both sides by 10
There is 2 lengths so 2*0.1588 / 200 feet of actual wire, 100 foot run.

=0.3176 ohms/100' run; so 20*0.3176 = 6.352 V

6.352/120V = ~ 6%

At 240V it's 3%

It would not be a 20A breaker. A 20A breaker would be used for a 16A continuous load.