Temperature generated by resistance wire?

AI Thread Summary
The discussion revolves around calculating the temperature generated by a 9.42-inch resistance wire rated at 4.701 ohms/foot, powered by a 12VAC, 600mA source. Participants highlight that the power supply may not provide sufficient current to achieve desired heating levels, suggesting that measuring actual output is crucial. The importance of heat loss and wire resistance at elevated temperatures is acknowledged, with a recommendation to experiment with different transformers for optimal results. Additionally, using a power-regulating knob is proposed to adjust voltage and current effectively. Overall, achieving consistent heating for the intended application requires careful consideration of power supply capabilities and heat transfer principles.
John Galt
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Temperature generated by resistance wire??

I've got a problem I need some help on.

I have 9.42 inches of a semi-coiled (the coils are horizantal, not vertical like wrapping wire around a pencil) design resistance wire rated at 4.701 ohms/foot. I have a 12vac power source with a 600ma rating. Is it possible to calculate what temperature is generated by the wire? Let's consider the increased resistance of the wire with increased temperature as negligible.

Thanks all you smarties :biggrin:
 
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John Galt said:
I've got a problem I need some help on.

I have 9.42 inches of a semi-coiled (the coils are horizantal, not vertical like wrapping wire around a pencil) design resistance wire rated at 4.701 ohms/foot. I have a 12vac power source with a 600ma rating. Is it possible to calculate what temperature is generated by the wire? Let's consider the increased resistance of the wire with increased temperature as negligible.

Thanks all you smarties :biggrin:
I'm assuming the form of winding is irrelevant to the problem since you know its resistance is 3.96 ohms. Before trying to calculate temperature, you have a power supply problem; you won't come close to supplying enough current to drive this fully. If you have a good supply that current-limits cleanly, and assuming your 12Vac is rms, then the power generated is 12V*0.6A=7.2W. If it's just a transformer or something like that, then the output voltage will sag and you don't know how a priori how much power you'll get. Measuring it would be the best bet.

Is the wire temperature really important, or are you trying to heat something up?
 
The power supply is from a 12vac transformer rated at .6A. The transformer is just one of those 'plug into the wall' types. I'm embedding the resistance wire in ceramic to make a heating element. The temperature I'm shooting for would be enough to light a cigarette upon contact with the clay. The manufacturer of the resistance wire says that about half the amps are needed to heat the same length of wire to a specific temperature if it is coiled rather than strait, which is the only reason I mentioned the shape of the wire. How much should I expect the voltage to sag? I wasn't expecting that.

The accuracy tolerance of the temperature needed is very high. I don't need any specific temperature, just something hot and consistent.

Thanks for the speedy reply :)
 
Hmm, I'm not happy designing a cigarette lighter that's going to give you lung cancer You could figure out your transformer's internal source impedance to know what it will do with that load, but it's not worth the trouble, it won't do the job.

You're going to need to put time in on some technical work here, are you willing to read up on heat transfer, etc.? For instance, cherry red is about 1500F. You need to estimate/calculate the heat losses of your device to figure out the heat input needed to reach that surface temp. The dominant losses are a) radiation and b) conduction to air and to the mount, whatever it is. Also the temperature dependence of wire resistance will not be negligible.

Alternately, build your lighter, borrow a big power supply with a knob, and turn it 'til you get what you want. Or light your stogies with a match...
 
Oops, I forgot that inductance results in a coiled wire with current going through it, hence the voltage sag.

I like your idea of the power regulating knob. I didn't want to purchase a voltmeter or ammeter, but it's possible to find one for cheap I suppose.

Edit* I understand that heat losses occur and that in the temperatures I am working with the resistance will only increase by a factor of 3% or so, making it negligible considering my high accuracy tolerance. I suppose experimenting with a couple different voltage and ampere rated transformers would be best. This project isn't rocket science...lol, so it must not be treated as such.
 
Last edited:
John Galt said:
Oops, I forgot that inductance results in a coiled wire with current going through it, hence the voltage sag.

I like your idea of the power regulating knob. I didn't want to purchase a voltmeter or ammeter, but it's possible to find one for cheap I suppose.

Edit* I understand that heat losses occur and that in the temperatures I am working with the resistance will only increase by a factor of 3% or so, making it negligible considering my high accuracy tolerance. I suppose experimenting with a couple different voltage and ampere rated transformers would be best. This project isn't rocket science...lol, so it must not be treated as such.
Inductance shouldn't be a problem at low frequency (60 Hz). I Googled nichrome wire (I'm guessing that's what you have) and found this web site. 28AWG matches your resistance/foot. Looks like about 3A will get you a dull red, about 3-1/2A for a nice cherry red, in a straight line (probably in still air), about half that for a coil. For a really tight coil like a cigarette lighter you'll probably need even less. So 1 to 1.5A across your 4 ohm wire is 4 to 6 V.
If the website is accurate, then a 5V supply that can source a couple of amps will do the job for you.
 
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