I'm trying to get an estimation of the temperature that a Kanthal (A1) wire would get, when applied an amount of electrical power.

The wire is 30cm long. The diameter is 0.3mm (or there about). The electrical resistance from the specifications sheet says "1.45 Ω mm2/m" (I'm not sure how to read this). I'll actually leave you the spec sheet link: https://www.kanthal.com/en/products/material-datasheets/wire/resistance-heating-wire-and-resistance-wire/kanthal-a-1

In the specifications sheet they provide a temperature table, from which I kind of understood that the temperature is not going to be a linear function, but a rough linear estimation would be just fine. My ultimate goal is to calculate the minimum and maximum electrical power that I need to supply to the wire, in order to control it between 50°C and 100°C 100°C and 200°C.

Edit, as requested by @Transistor

I am unsure where to find a more appropriate spec sheet, in that page you can expand the sections to reveal the data. Anyway, I have also found these 2 pages:



Maybe they provide more useful information.

The "element" will be used for a foam wire cutter (thus, not really an element, just a bare wire). As far as I could find, foam can be cut at 200°C, but that may be too aggressive so I'd like to keep it a good amount below that temperature. I am going to make tests later, but right now I only need to "convert" W to T(°C) in order to calibrate an initial electrical circuit...

  • \$\begingroup\$ You haven't linked to the spec. sheet correctly. You seem to have linked to a page that may link to the spec. sheet. Can you fix the link? The temperature will stabilise when power in = heat lost but you have told us nothing about how the element will be used. There's an edit link under your question. Welcome to EE.SE. \$\endgroup\$
    – Transistor
    Jan 20, 2019 at 23:14
  • \$\begingroup\$ Hi, thank you for replying! I have added some info as you requested, hopefully it'll be of any use. Just so you know, I only need rough estimations, not proper engineering-level calculations, so you may skip any power loss calculations or things like that. My device is battery powered, so I need to know how many batteries to use, and also, the power limits that I should consider when designing a proper electrical circuit. \$\endgroup\$
    – T3STY
    Jan 20, 2019 at 23:43
  • 1
    \$\begingroup\$ If you know the voltage applied to the wire and its resistance then you can easily calculate the power dissipated in the wire. Assume that all of this gets converted to heat. However, your question is about temperature. This is a heat flow problem, not an electrical engineering problem. It depends on air flow, ambient temperature, heat flow through mounting hardware, etc. \$\endgroup\$ Jan 21, 2019 at 0:09
  • \$\begingroup\$ I would suggest you make a test rig that allows you to apply varying voltage/current to your wire (and hold it, hacksaw/Cshaped form or whatever) and simply start cutting a block of foam and turning the power up slowly until you have the cutting rate you want. The faster you cut and the more and colder airflow in the room, the more power will be required, so it is probably much much simpler to test than calculate. \$\endgroup\$
    – K H
    Jan 21, 2019 at 2:53
  • \$\begingroup\$ High Density Cartridge heater core winding wire calculation, & turns calculation egy formula. \$\endgroup\$ Dec 12, 2020 at 7:12

3 Answers 3


Basically, it turns out there is no simple way of finding an answer. However, I was wondering if I could find a function whoose curve passes through, or near, most of the temperature points from the spec sheet table... turns out, yes, there is such a function! I have actually found 2 functions of interest. The first is a logarithmic function (base 10):

y = 2672 * Log(x) - 2733


This function is the one I found gets nearest to all thermal points of the table, but definitely deviates a lot at the 20W and 27W marks.

The second is a linear function:

y = 61x - 621


This function is simpler, not involving a Log. However, it also deviates the most compared to the Log one, noticeable at the 22W, 27W and 35W marks. This is the one that I ended up using. I could then calculate that: 11.8W = 100°C; 13.5W = 200°C. A quick math gives that my project needs a minimum of 3 Li-Ion cells in series, capable of 1.1A continous discharge current (max needed for 200°C).

I guess that, if the thermal expansion table would've been taken into account, a better estimation would've resulted. For some estimations however, it's not worth the trouble.

Hopefully this will be useful for someone else. And thanks to anyone here who tried to help me.

Graph plotting provided by Desmos.com

  • \$\begingroup\$ It is not clear where the datapoints on your graph are coming from, but note that the final temperature of the wire will depend on ambient temperature, airflow and cutting rate, so you should build some margin into your design. If you took your datapoints from something like settling temp when operated inside a furnace or enclosed space, they won't be of use for this application. If you calculate the heat lost to the air at a given temperature by a wire of that length in the ambient temperature you expect, that will tell you what wattage the wire need dissipate to maintain that temp. \$\endgroup\$
    – K H
    Jan 21, 2019 at 21:27
  • \$\begingroup\$ If you really want to do this by calculation, you can use the math on this site to calculate the wattage you need for a given temperature, but it's quite involved. You may be able to copy their use of a simulator too. \$\endgroup\$
    – K H
    Jan 21, 2019 at 21:34
  • \$\begingroup\$ The plot comes from the first link I have posted, at the physical properties section. I know that environment, airflow and all those parameters are going to influence the outcome. As I have stated initially, I don't need an accurate result, just something to base my further work on. My project will require a PWM controller, battery charger circuit, an external power supply, MOSFETs and much more. Without any power usage estimation it's hard to choose and put all these together.Even though this estimation may turn out wrong, it'll be better than having nothing to start with. \$\endgroup\$
    – T3STY
    Jan 21, 2019 at 21:48
  • \$\begingroup\$ If you don't mind participating a little longer I think I have a way to do a rough estimate across your temperatures of interest with a spreadsheet. \$\endgroup\$
    – K H
    Jan 21, 2019 at 21:50
  • \$\begingroup\$ I think I'll try it out and We'll see how close the numbers end up. \$\endgroup\$
    – K H
    Jan 21, 2019 at 21:53

For the wire to maintain a temperature, all heat that is lost by the wire to the air around it must be replaced by the power consumption(thermal dissipation) of the wire, so:

Heat is transferred in 3 ways: Conduction, Radiation and convection. Conduction isn't too bad to calculate, and due to the low thermal conductivity of air at the temperatures in question, we may be able to neglect it. Radiation will represent the bulk of thermal energy that leaves the wire, and convection requires simulation, so we'll assume the room has still air and apply a small arbitrary factor to represent some acceleration of heat transfer caused by convection due to heating.

For thermal energy radiated by non-ideal radiators, we have ze Stefan-Boltzmann Law:

enter image description here


  • \$\sigma\$(baby sigma) = 5.6703 x 10\$^{-8}\$ W/m\$^2\$K\$^4\$

  • P is Power, in (W)atts

  • A is surface area, in (m\$^2\$)eters squared

  • e is the emissivity of the object (1 for a black body radiator, 0.7 for Kanthal)

  • T is absolute temperature, in (K)elvin

Okay! So it looks like this formula will give us a pretty good idea. I'll conjugate it for P:

\$P=e\sigma AT^4\$

And the page says if the surroundings aren't at absolute zero, we should use

\$P=e\sigma A(T^4-T_C^4)\$

Where \$T_C^4\$ is the temperature of the surrounding air, which radiates back.

And we'll need to calculate the surface area of the wire (exact dimensions would be better) based on the surface area of a cylinder 0.3m long and .0003m wide(not including ends):

\$A=\pi d * L = \pi*0.0003m*0.3m = 0.00028274334m^2\$

And also convert our temperatures of merit from Celsius to Absolute(K), so we add 273.15

\$T_{Element}=200^\circ C+273.15=473.15 K\$

\$T_C=30^\circ C+273.15=303.15 K\$

Ham it together and hope it works:

\$P=e\sigma AT^4=(0.7)(5.6703e^-8 W/m^2K^2)(0.0002827m^2)(473.15 K^4-303.15 K^4)=0.467678W\$

Hmmm... Waaay less than I thought it would be. I guess tomorrow I'll try adding conduction in. This may mean I'm wrong about radiation being stronger than conduction to still air. Are your certain your wire is only 0.3mm diameter?

I'll just post this in the mean time so I don't accidentally delete it.


Stefan-Boltzmann Law

Thermal conductivity of air at 25'C: 26.24 mW/(m K) Engineering Toolbox Calculator

Emissivity of fully oxidized Kanthal: 0.7 (Datasheet, P15)

  • \$\begingroup\$ Woah, definitely some engineering degree here! I don't have such preparation, and I trust that your calculations are correct :) Yes, I'm sure the wire diameter is 0.3mm, I can fit this wire 3 times side by side, in 1mm. To be more specific, the wire is made to be used for electronic cigarettes. \$\endgroup\$
    – T3STY
    Jan 22, 2019 at 19:21

Well not sure. You could get a potentiometer or a temperature control dial and wire it to the kanthal wire or plate at both ends (two poles) and then measure the temp with a higher temp oven thermometer, then you could mark the different temperatures on the dial. You could also pull out an electrical multimeter and measure the voltage and amps. If theres enough resistance in the kanthal wire then you can probably just splice a cord and plug right into the outlet with out blowing a fuse.

If you’re using a higher electrical voltage the wire may get hotter with a lower number of watts. Volts * Amps = Watts.

The more pins in a potentiometer then the more different electrical options. You can raise resistance in the pins to stop electricity from going threw in a larger amount, thry probably use a metal with higher resistance as the pins in the potentiometer connecting the two poles. You can lower resistance in the pins as well to allow more electricity to flow threw or just add more pins, but im not sure with a certain voltage or amps if it will go threw the pins with a higher resistance. Meaning that it wont let a higher voltage or higher amps pass threw the pin without heating it up and possibly frying it or slowly frying it over time.

Higher resistance means it lets less electricity flow threw or it flows slower or it gets hotter when trying to pull excessive or more electricity threw the pin or wire. Lower resistance means the electricity flows faster threw it, or more goes threw it or it goes threw without increasing temperature much. We use to say silver was better than copper but the copper wasn’t .999. Now i got this issue where i think i cant use copper fir dc voltage because i guess it was a copper craft eire that had iron in it because it created an electro magnet and pulled my key lanyard. So i assume the coper wasn’t pure and pure copper will work with dc, but with a laptop charging cable between the rectifier and laptop only aluminum wire would work and aluminum has more resistance than copper.

I just used carbon mixed in with copper or other metal to increase resistance. But theres s lot of metals you can combine to reach your desired resistance voltage or amps level to reach your desired number of watts.


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