Electronic – Formula for heat wire Kanthal or NiCr Watt consume

Short answer: no.

50% longer answer: yes.

But seriously, how hot the wire will get for a given power input will depend on how well it's insulated. If the insulation is perfect, then there is theoretically no limit to how hot you can make it. Realistically, either at some temperature the heat radiated to the environment equals the power input, and it stops getting hotter, or the wire melts.

I'm not entirely sure what "4.7mW at 47% efficiency" means. 47% of what? Is 4.7mW the power output, or input of the power supply? All that's really relevant in the power output of your power supply, because the power input to the wire will equal exactly this, and 100% of this electrical power will be converted to (what else?) heat.

Anyhow, let's say 4.7mW was the power you put into the wire. How power translates to temperature rise for a particular component is usually expressed as an absolute thermal resistance in units \$K/W\$, or equivalently, \$^\circ C/W\$. It expresses how much temperature rises per unit power input.

So let's say you want a temperature of \$750^\circ F \approx 400^\circ C\$, and ambient temperature is \$20^\circ C\$, thus you need to raise the temperature \$ 400^\circ C - 20^\circ C = 380^\circ C\$, and you need to do it with \$4.7mW\$. The thermal resistance must then be:

$$ \frac{380^\circ C}{4.7mW} \approx 80800 ^\circ C / W $$

or more. That's not physically impossible, but it's some pretty serious insulation. For comparison, a typical thermal resistance to ambient for a typical TO-220 package, which is approximately the same size as you describe, is \$34 ^\circ C / W\$. This table of thermal resistances of common packages from Linear Technology may be helpful to put those numbers in perspective.

You also asked "how long", and again the answer is complex, and will depend on the environment and insulation. The ultimate speed of heating will be limited by heat capacity of your heating element and the rate of energy input (power). This will be offset by the rate at which heat is lost to the environment. But, having already established that you need some pretty incredible insulation to achieve the target temperature in your case, we can skip the rigorous analysis and intuitively know that the high temperature is reached by integrating the power over a long time to accumulate enough energy, and say simply the time required to reach the target temperature is "a long while".

I don't quite follow your calculations. But if you are doing I^2 *R as the power, and assuming the same max power for each wire.. then that's what I would have done.

However I went here and it looks like I*R is about constant. (?)
(I had to plot it.) Still looks linear.

Maybe someone can tell us both why.

Edit: I*R dependence. (Thanks Spehro, it was suddenly obvious on the drive home.) No matter the thermal loss mechanism (convection, radiation..) It will go as the area of the wire. 2 * pi *r * l (r - radius and l - length), so bigger wire will need more heat to get to a given temperature. (more later)