I have a 12 foot length of 14-gauge molybdenum wire that I have measured a resistance of 1.2 ohms. I would like to use it as a small heater coil.

How can I calculate the safe number of volts that I may give it without melting the wire?

Edit: I would like to operate this at a maximum of 1500C. Ideally it will be buried (protected) in refractory.

Edit 2: I read that molybdenum wire can be used as high temperature heating element, but I guess I just want to make sure that is indeed the case. I have variable voltage so I can start out low and raise it as needed.

NOTE: This is not the same low resistance MoSi2 heater elements I posted a question about yesterday - this is pure molybdenum 14 gauge wire.

  • \$\begingroup\$ It really depends on the external conditions, like ventilation, affecting the ambient temperature and thus the rate of the heat dissipation. \$\endgroup\$ – Eugene Sh. Oct 9 '15 at 15:42
  • \$\begingroup\$ What temperature do you want to heat it too, and approximately with what tolerance? \$\endgroup\$ – gbulmer Oct 9 '15 at 16:08
  • 1
    \$\begingroup\$ Why did you raise a new question when you could have established further answers on this: electronics.stackexchange.com/questions/194274/… \$\endgroup\$ – Andy aka Oct 9 '15 at 16:29
  • \$\begingroup\$ That is a completely different heating element element (low resistance) and question. \$\endgroup\$ – Jeshua Lacock Oct 9 '15 at 18:01
  • \$\begingroup\$ Please see edit. \$\endgroup\$ – Jeshua Lacock Oct 9 '15 at 18:03

You can't do this, simply by considering the electrical input power.

A constant-power heater (like most electric heater elements approximate) goes on rising in temperature until it's hot enough to drive the input power across the thermal resistance between the element and the rest of the world.

So the temperature of the element (which is what you care about because you don't want it to melt) is P x Rt, where P is the input power and Rt is the thermal resistance of the element to the environment.

So P and Rt are both equally important, and you need to care about them equally.

But to help you with some rules-of-thumb, a 1.2ohm resistor will dissipate V^2 / 1.2 Watts of power. i.e. 12V would be 120 watts.

I find thinking about incandescent light-bulbs is a good way to reason about this sort of amount of power, though I guess that's starting to mark me out as old.

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