Low voltage across low resistance for heating elements

Lets make this more real world. Carbon steel has a resistivity of 17 x 10-8 ohm meters. So let's consider a 1 meter coil of 16 gauge steel. 16 awg has an area of 1.29 sqmm or 0.00000129 sqm. The resistance should be ##\mathrm R = \frac {\rho \cdot l} A = \frac {17 \times 10^{-8} \times 1} {0.00000129} = ~0.1318 ~ \Omega\\I = \frac V R = \frac {0.001} {0.13178} = 7.6 mA\\P = VI = 0.001 \times 0.0076 = 7.6 \mu W##
Well that would not do much. I need 10 - 20 watts. So do I raise the voltage to get more current? Nope. Let's use a thicker and maybe shorter wire such that R = ##\frac {V^2} P = \frac {0.001^2} {10} = 0.1 \mu \Omega##

So how thick does the steel wire need to be at 1 meter length? ## A = \frac {\rho \cdot l} R = \frac {17 \times 10^{-8} \times 1} {0.0000001} = 1.7 sqm ## Wait, what? Ok. So clearly resistance is a big problem. Even if we shorten the length ten times to 4" we still need a 0.17 sqm conductor like a 0.4m x 0.4m x 0.1m piece of solid steel. Haha no way. Well what about copper then? It has a resistivity of only ##1.7 \times 10^{-8}## or 10 times less than steel. I think that just means the cross section would be .017 sqm. This would mean my copper heating element would have to be a 130mm x 130mm x 100mm piece of solid copper. Doable I guess but not very practical. Would be an interesting experiment except that the copper wire connecting to it would create way too much resistance.

I think what I am learning here is to start with a practical but low resistance conductor like copper or aluminium and determine the lowest voltage I can use with that to get my 10 watts. To continue with extremes I will consider a 10mm x 100mm x 100mm copper plate as the heating element .

There will still be the problem of the resistance in the secondary transformer winding and the wire to connect the transformer to the heating element. Unless they are refrigerated superconductors they are going to add a huge amount of resistance killing the current and power even if the transformer is a part of the heater and only a few inches away, but I will ignore that voltage divider problem for now. To be even more extreme I will solder the secondary winding chilled superconductor connector wires to the centers of the two faces of the plate so that the element length is only 10mm. So how much voltage do I need to get 10 watts of heat from that element?

## R = \frac {\rho \times l} A = \frac {1.7 \times 10^{-8} \times 0.010} {0.01} = 1.7 \times 10^{-8} = 170 \mu \Omega\\
P =\frac {V^2} R ~~ V^2 = PR ~~ V = \sqrt {PR} = \sqrt {10 \times 1.7 \times 10^{-8}} = 0.412 mV ##

So due to practical material concerns half a volt is about as low as I could go if I ignore the resistance of the other conductors in the circuit. Let's consider just the copper wire connected to the transformer winding. Let's use 4" and 4" of AWG 6 Copper wire for the hot and neutral of the secondary winding for a total of 8" or 0.2 meters. ## R = \frac {1.7 \times 10^{-8} \times 0.2} {0.0000133} = 0.256 \Omega ##

So even assuming a superconducting transformer this will basically kill the current and power in the circuit. In fact the 170 ##\mu \Omega## for the copper plate can be ignored as it will have almost no effect on the overall current and power. So using the new value for R how much voltage would be needed for 10w? I think it does not matter because almost all of the power would be dissipated in the 6 gauge wire. Those wires might melt before getting enough power to the heating element itself.

So I guess to go with a low voltage low resistance heating element one really has to consider both the wiring to the transformer and the secondary transformer winding as they seem to be the limiting factors to how low you can go instead of the element itself. Its all series resistance voltage divider considerations. Clearly you would want the transformer as close to the element as possible. I was also thinking of those electrically heated gloves and socks. I guess how low you can go with those depends a lot on the internal resistance of the batteries used as well as on the conductor sizes.

Maybe the first thing to do in designing a low voltage heater is to make sure that the resistance of the heating element is much greater than that of the sum of the secondary transformer winding (or battery resistance) and the wires leading to it. Not sure how much greater but maybe 10-20 times as a minimum. I will redo my design with this in mind. Now I want to take apart a soldering gun and measure the resistances there.

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I am currently experimenting with resistance wire to generate heat. I have 2 types here. One is insulated, the other isn't. The one that isn't insulated works just fine, but the insulated one gives a huge amount of smoke the first time I heat it and then the outside becomes completely black. After that the smoking stops.

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What is the use of that insulation if as soon as I put some current on the wire it smokes and turns black?

I measured the temperatures and I did not exceed 80 degrees Celcius. Voltage applied is around 10V and it draws a current of 2.5A.

Some info about the wire: Resistance: 10 Ohm / metre Diameter: 0.25mm Material: CuNi44

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