Basic Electrical questions

[halifax]

Ok some one help me out. I don't have my electrical hat on and its not making sense.

Standard house wire, #14 solid copper is rated at 15 amps (120V)

Granted it has some safety built into the number but that's about standard # 14-15 amps.

The same wire should, support 150 amps, mathematically.

15aX120v=1800 watts so ...... 1800w/12V=150a. No way would I try to put that much through a #14 wire, but the computations say it would work.
I might be a bit rusty on the math part, but I think I'm missing something obvious here.

Side note, I made a 12v portable alternator for the cabin, using a GM alternator. Its rated at 63 amp so I used 4 Gauge wire to and from it just to make sure. But it got me thinking, which is dangerous, about what would be the proper size wire for 63 amp.

What wrong with the picture?
Aubrey

[GuglioLS]

Hi Aubrey,

Great question - your wattage math is correct, the reason 14 Ga wire will carry 1800 watts @ 15 amps, 120 volts AC but not carry 1800 watts @ 12 volts, 150 amps DC is amp capacity (ampacity) of the wire. Basically it boils down to resistance as a function of voltage and amps. The higher the voltage the lower the effective resistance and power loss of the wire. As voltage and amps decrease from high voltage to low voltage, resistance of the same gauge wire effectively increases creating a loss of power and heat to build up, as heat builds up the resistance of wire further increases which reduces the amp carrying capacity further exacerbating the problem.

To help visualize whats going on - think of it this way... a long 1/4" ID hollow tube (analogous to 14 ga wire) will allow a certain viscosity fluid to flow at a certain rate at a certain pressure. The density (Viscosity) of the fluid being pushed through the tube is analogous to Amps. The pressure pushing the fluid through the tube is analogous to volts. As voltage decreases the density of the fluid must increase to meet the same power demand (watts). Increased density of the fluid (amps) meets much resistance on the walls of the 1/4" tube which eventually builds up heat and pressure beyond the capacity of the tube (Wire) bursting the tube (wire burns out). Increasing the size of the tube (wire) from 1/4" to 1" ID (analogous to going from 14 ga to 4 ga wire) has less resistance allowing the higher density fluid (150 amps) to be pushed through the tube at a lower pressure (lower voltage), to maintain the same amount of work available (Power in watts)

Hope that helps the wire size vs varying voltage * amps to carry the same amount of power conundrum.

As far as what ga wire to use for your 63 amp alternator, it depends on the length of wire, are you really using all 63 amps and what voltage drop you are willing to accept.

Here is a link to all the calculations:
DC wire size calculations

Or a basic wire size chart:
www.aaasolar.com/ProdLit/HelpfulInfo/06-WIRESIZE.PDF

Larry

[bracabric]

Larry, Thanks for that, the explanation was simple to follow and very informative. I have always (until now) been unsure of the definitions of Amps,watts,ohms,volts etc and your explanation has helped me a lot !
Dick
PS I am sorry I have not been contributing much of late, The b-lls aching bureaucracy we are still going through buying this new place is totally boring and uninteresting and seems endless. It's like punching foam rubber, you can't seem to connect with anything, still patience, one day at a time etc.... and I haven't gone away !!

[bradblazer]

Another way to look at it with those simple equations.

You know V = IR and P = VI. If you solve for the power generated in the wire which is the undesirable limiting factor:
P = I² R
Resistance for a given piece of wire is constant so the limit for that wire is some amount of current, say 15 amps for 14 gauge wire and the wire will not overheat. Volts do not matter.

The happy result is that you can transmit a lot of power over the over a small wire by increasing the voltage. (Power transmitted is different than power lost.) If the ratio of P/V is held constant, the current will be constant as will the power loss in the wire. There are of course electrical field and containment issues associated with high voltage that limit how much you can put through an extension cord. I practical terms, a given wire can transmit 10x as much power at 120V compared to 12v and twice as much power at 240V. vs 120V.

The error in your calculation above was that you incorrectly assumed the power transmitting capacity of the wire was constant regardless of voltage.


It's also important to know that the power loss in the wire is proportional to length so even though it will not overheat the wire (as long as it's not in a coil), running a long cord near it's capacity will result in wasted power and a low supplied voltage.


Brad

[halifax]

I knew it had to be something obvious, I just could not think of it. Forgot all about factoring the resistance, should have., could have, but didn't.

The little generator is rated at 63A but the power curve that came with it has a max of 90 amps at max rpm. I'll never get it to that point, matter of fact I've only gotten it to 40 amps. I always try to go over what I need for wires, just wanted to make sure. From the one chart, 4 gauge wire I'm using is good to 95 amps, it only 40" long. Right now it charges the batteries, and I have 6 gauge wire running from the batteries to inside the cabin on a 30 amp fuse. Sometime later I'll change out the wire going inside, put a breaker box in, and add more batteries for the capacity. Right now I'm good for what I use.

Thanks

Aubrey