How to calculate the Max Amp-capacity of a wire?

[elpres05]

My 1900 Watt hair dryer has a seemingly thinner wiring compared to what my computer uses, almost twice as thick and power consumption no greater than 300 Watts. The thickness is however not of the covering over it, keep that in mind.

Whether it be 220 Volts or 110V, is there a way to measure the maximum electrical capacity of wire before melt down sequence?

Lets say a wire, cross sectional area 1cm^2, copper construction, how would i know its maximum electrical capacity?

Also, which is better...

1. Wire with X thickness, 1 piece like that on poles.
2. Wire with X thickness comprised of several thinner wires.

 

[DrPizza]

Yes.
oversimplification:
P=I^2*R losses to the wire (in reality, the wire heats up and as it does so, the resistance increases as well)

Resistance = resistivity*length/cross-sectional area.
To calculate current, include the resistance of the wire with the resistance of the device (blow dryer).

For meltdown, you'll have to consider the type of metal, thermal insulation, exposure to air, etc.
i.e. a lightbulb filament does just fine when there's no oxygen. Remove the glass, and it'll burn out quickly.
toaster elements: I have no clue what they're made out of.
extension cord: I'd be worried about the temperature at which the plastic melts, allowing the separate wires to come into contact, short circuit, increasing the current, and rapidly melting wire - somewhere (or tripping a breaker)

 

[Mark R]

The current rating of a wire depends on:

It's electrical resistance (and therefore it's cross sectional area)
It's surface area
It's maximum acceptable temperature (usually the max temp acceptable for the insulation)
The environment it's in (on it's own, installed under glassfibre insulation, installed in a big bundle with other power cables, hot environement, cold environment, ventilation, etc.)
Degree of overload capacity required (If you're right next to a transformer on 400 V 3phase, a short circuit might cause 10,000 A for 10 ms (when the breaker cuts the power) - this would be a more significant stress than if you were 100 yards away from the transformer on 100 V - perhaps 1,000 A for 10 ms). A high potential fault current will mean your 'normal' maximum would have to be reduced so that the cable could still withstand a fault, even after running at 'maximum' power for a considerable period.

Because surface area increases more slowly than cross sectional area - you need to increase the CSA more than you might otherwise expect.

E.g. 1.5 mm^2 would be satisfactory for about 15A (in free air, room temperature, PVC insulation), but you'd need 4 mm^2 for 30 A and 25 mm^2 for 120 A.

If you installed a cable, so that it ran under carpet, or under some foam insulation you would need to use a thicker cable (because the heat could't escape so easily). On those big transmission lines, acceptable temperature is limited by thermal expansion - the wires sag as they heat up - so maximum capacity has to be reduced in Summer, to avoid the cables sagging and hitting tree branches (In the 2003 blackouts, this happened in a sort of chain reaction. One power line hits a tree branch and trips out, the others have to take the load, heat up, lines sag and hit more tree branches, etc.)

There's not a lot of difference between solid conductors and stranded conductors, in terms of current capability. Normally, cable is sold based on its CSA - so the cables are designed to have a specific CSA. However, there is a difference in handling - solid core wire is very stiff, whereas stranded is flexible (but more expensive). You'd put solid core in a permanent installation, where the cable would never be touched - but you'd need stranded if the cable has to move.

 

[RossGr]

It is not hard just look up the ratings. Most likely your hair dryier used 14ga 14ga and your computer is 12ga. I am sure that you will find that your hair dryer is within spec. Most likely they use the minimium gauge possible to cut costs.

 

[TitanDiddly]

I have a chart for this at my dorm. I'll look it up for you when I get there.

 

[DrPizza]

Hey, thanks RossGr for that link...

I have a barn "next to" my house. From my breaker box in the basement, I'd estimate it's about 200 feet of wire to the barn. The entire barn is fed with one 12awg feed. I've recognized the need to rewire (I hate running a circular saw and all but turning out the lights in the barn). That site helps.

 

[pm]

Thanks for the link to the chart, RossGR. Very useful, I added it to my bookmarks.

I was amused to note, these numbers compared to some of my home projects. I've been wiring up hand-wound 3-phase brushless motors and I am so much higher than those numbers. I guess I can tolerate wire melt-down and I have better cooling than you'd get behind a sheetrock wall, but I'm running about 2700% more current through the wiring in the motors that I've been building. They do get very hot though... particularly the ones where I'm running 16A though a 25AWG wire.

 

[elpres05]

Originally posted by: RossGr
It is not hard just look up the ratings. Most likely your hair dryier used 14ga 14ga and your computer is 12ga. I am sure that you will find that your hair dryer is within spec. Most likely they use the minimium gauge possible to cut costs.

thanks for the chart.

 

[Mday]

Originally posted by: RossGr
It is not hard just look up the ratings. Most likely your hair dryier used 14ga 14ga and your computer is 12ga. I am sure that you will find that your hair dryer is within spec. Most likely they use the minimium gauge possible to cut costs.

that chart and website brings back memories lol... stupid college.

 

[RossGr]

An interesting artifact of cylindrical symmetry (as in a wire); as you add insulation you increase surface area faster then the radius increases. Since head loss is related directly to surface area, up to some critical radius you need to add insulation to cool a wire, thus increasing the current carrying capacity.

I cannot tell you what the critical radius is as it depends on the insulating material

 

[Mday]

Originally posted by: RossGr
An interesting artifact of cylindrical symmetry (as in a wire); as you add insulation you increase surface area faster then the radius increases. Since head loss is related directly to surface area, up to some critical radius you need to add insulation to cool a wire, thus increasing the current carrying capacity.

I cannot tell you what the critical radius is as it depends on the insulating material

typically you learn the hard way lol...

wires typically have the maximum temperature and voltage ratings printed on the insulation.

dont forget, voltage ratings are a function of insulation thickness and material. so it's not only the current you have to worry about when choosing wire.

 

[dkozloski]

All this is clearly presented in the National Electrical Code. Books can be found at any well stocked bookstore or electrical supply store.