I Need Clarification on Ampacity, I Think (?)

[GarageTinker]

Background: Induction coil of 16AWG copper magnet wire(4.01600 Ohms/1000 Feet); 39 winds in coil which has an area of .00188m^2 and a total length of 23.79682 feet which gives me a total resistance of .09557Ohms. With the coil under the conditions (which can be varied) of a moving magnetic field; such that it produces a projected output voltage of 4.41069 volts at 46.15239Amps. Now, an AWG reference guide I have lists the wire as having a Current Carrying Ampacity (based on 1A/700mils) of 3.247. So does this mean the wire can only handle up to the 3.247Amps per the resistance of 1000 feet of the wire? Am I trying to shove about 15 times too much current through the wire based on this, or is this just the point that the wire will start to heat up and do nasty things? I understand that Ampacity is the amount of current a wire can handle (like how much water pressure a garden hose can handle before it bursts), but I'm stuck on how to figure it out so I can know what changes to make to the coil (ie lengthen or shorten) or to the magentic field conditions (stronger/ weaker or faster/slower rpms etc). I would greatly appreciate any input / clarification on this.

 

[Cliff_J]

Pressure in a water hose is not a handy physical metaphor to use here, its analogous to the voltage rating. I think it might be easier to think hydraulics and the gallons of fluid per time you are trying to force through the hose - if the hose is too small the friction will cause the fluid to heat up and if it gets too hot the hose will weaken and burst.

The amount of current a wire can handle depends on how much time you need the wire to handle the current and how much voltage drop is acceptable. If the timeframe is extremely short, like only a fraction of a second, that 16ga wire could easily flow 46A or more.

The length of the wire has no bearing on current capability, it only determines the resistance and thus the voltage drop.

The figure of 700 mils per amp is very conservative, there are charts that use 300 mils and some even 200 mils. With about 2500 mils in 16ga, that's 2500/700 = 3.6A, 2500/300 = 8.3A, and 2500/200 = 12.5A of continuous current capability, although the latter two will have a heating affect on the wire and now the ability of the wire to dissapate the heat becomes important and why a conservative rating ensures safety.

Also, as the temperature rises so does the resistance - well of most conductors used for wire, some substances go down (positive or negative temperature coefficient). So as the resistance rises, so does the voltage drop and thus somewhat limits the current flowing too.

The voltage drop is simple to calculate, its I*R so if you are trying to flow 46A through .1Ω you'll loose 4.6V. Since this is more voltage than your predictions, something sounds amiss.

I'll let someone with more experience with motor design handle your question on solutions to the problem.

 

[Michael_McGovern]

The voltage drop is simple to calculate, its I*R so if you are trying to flow 46A through .1Ω you'll loose 4.6V.

You meant 46V, right?

 

[berkeman]

You meant 46V, right?

No, I think he did the math right.

 

[Michael_McGovern]

No, I think he did the math right.

Oh, .1ohms! Didn't see the decimal place. That was stupid!

 

[berkeman]

No worries. Say, OP, how in the heck are you making such a big change in magnetic field? Zowies. What are you making?

 

[GarageTinker]

Pressure in a water hose is not a handy physical metaphor to use here, its analogous to the voltage rating. I think it might be easier to think hydraulics and the gallons of fluid per time you are trying to force through the hose - if the hose is too small the friction will cause the fluid to heat up and if it gets too hot the hose will weaken and burst.

I think I see what you're saying here.

The amount of current a wire can handle depends on how much time you need the wire to handle the current and how much voltage drop is acceptable. If the timeframe is extremely short, like only a fraction of a second, that 16ga wire could easily flow 46A or more.

The preferrable amount of time needed for the coil to be able to handle the current is 24/7.

The figure of 700 mils per amp is very conservative, there are charts that use 300 mils and some even 200 mils. With about 2500 mils in 16ga, that's 2500/700 = 3.6A, 2500/300 = 8.3A, and 2500/200 = 12.5A of continuous current capability, although the latter two will have a heating affect on the wire and now the ability of the wire to dissapate the heat becomes important and why a conservative rating ensures safety.

This what I'm primarily trying to figure out; how to mathematically tell when, under a certain set of variable parameters, a coil of copper Xgauge magnet wire will become too hot to be considered safe for the application.

Also, as the temperature rises so does the resistance - well of most conductors used for wire, some substances go down (positive or negative temperature coefficient). So as the resistance rises, so does the voltage drop and thus somewhat limits the current flowing too.
The voltage drop is simple to calculate, its I*R so if you are trying to flow 46A through .1Ω you'll loose 4.6V. Since this is more voltage than your predictions, something sounds amiss.

More than likely what's amiss is my calculations. I've set up an Xcel sheet to run the calculations to project the output volts and amps of a coil of wire, based on variables (coil geometrics, operating rpms and magnetic field strengths) that are hand-entered. I didn't even think to take voltage drop into account. Thanks for pointing that out. I know that regardless of what I do, the coil is going to heat up to some extent due to constant flow of current. I'm just trying to find the parameters that will give me a "happy medium" so to speak between acceptable power output and safe heat levels. The more I can work out mathematically, the less it will cost me in materials and experimentation time. The information I need to decide on how to make the coil (i.e.- how big in area, what length of what gauge of wire to use, rpms etc.) is either already available or calculable and I'd rather not have to "reinvent the wheel". Thank you for your help in this and for reminding me of that little matter of voltage drop.

 

[Robine]

The power (heat) generated by the wire resistance is I x I x R or more than 46 x 46 x .1...since resistance will increase as the wire heats...This is way too much power for your coil. ampacity may be given based on voltage loss or safe operating temperatures, whichever is limiting. In 12v dc systems for boats for example, a common electrical design criteria is to keep within 3% voltage drop in wiring so appliances get sufficient power. For diesel engine starters, a thousand amps may be drawn but only for a few seconds so steady state heating is not a concern. Another example is a light bulb which draws a surge of current at start up before the filament heats and as it does often blows from the high current and rapid temperature change.

 

[Averagesupernova]

The power (heat) generated by the wire resistance is I x I x R or more than 46 x 46 x .1...since resistance will increase as the wire heats...

Actually it will be less than 46 x 46 x .1. Remember ohms law? Current flowing depends on the resistance of the wire. Unless the magnetic field is strengthened in order to make up for the increased resistance of the wire, the power dissipated in the coil will go down as the resistance of the wire goes up. The temp of the coil will eventually balance out unless the coil becomes so hot that it melts open.

 

[GarageTinker]

Thank you for your input gentlemen. Right now the main thing I'm trying to figure out is how to properly formulate ampacity and voltage drop. I have reason to think that the information in the ampacity chart that I was initially using may be incorrect. It sights that the circular mils of the listed wire guages; which the ampacity is based on;is calculated as being the wire's dia. (in mils) squared. I'd have thought it would based on the area of the cross-section, PIr^2. Which, if either, is correct?