# Formula to calculate wattage required to heat stainless steel?

Original Scope:

I am trying to make a heating element using a 304 stainless steel flat bar for simplicity. I'm choosing stainless steel instead of nichrome since it will be more durable.

If I know its electrical resistance is 0.72 x 10-6 Ω.m and I figure out the object's mass, how can I calculate the wattage required to bring that stainless steel bar to a specific temperature?

All I have been able to find so far are formulas for heating wire.

It seems like there should be a very simple formula to calculate this but in all of my searching online I can't seem to find it. If your answer is simply lookup the "so and so" formula that would be all the help I need.

What I've learned from your responses:

First, thanks everyone for your answers! From what many of you have said I now understand that this is a multifaceted problem that is quite complicated to solve in it's entirety. It seems that the best way to go about this is to do some testing and construct a temperature control circuit.

My Plan

My initial approach will be to supply the element with excessively more amperage then I expect it will require. Then, as it heats up I will monitor amp draw and temperature to get an estimate of amps required to reach desired temp. Then, based on that information I can find a power supply that is a good match. I'll also make a circuit that measure's temp and turns power on or off to maintain that temperature.

What is this for?

Some of you asked for more context about the purpose of this element. To many of you serious engineers this will seem silly but it's for a custom sword. The Cyberpunk 2077 Thermal Katana to be exact. One of my hobbies is to try and make real prototypes of technology found in science fiction. What makes this sword unique is that the edge is super heated. Why? No idea, it's a terrible design. Any metal that has reached a high enough temp to turn red will be useless. The blades design itself is terrible as well but I want to try and replicate it anyway. I have since changed my original element design to achieve this but the process will be the same. I want to make this sword still able to cut and yet have the super heated capability. So what I'll be doing now is using NiChrome flat ribbon wire on the flat of each side of the blade (ceramic barrier between 1095 blade and Nichrome element). This way I have a high carbon edge that begins the cut and the sides that apply the heat. There are many more details involved in the entire build that I have figured out but the element was the primary obstacle.

Just an FYI to anyone wondering, yes I am trying to make a functional-ish weapon, no I do not ever plan to use it for anything more than a display piece and the challenge of making it work. I am aware of the dangers of the voltage required for this and the intense heat output. There are a lot of technical and mechanical challenges as well but that is exactly the reason I want to do it. As safely as possible of course.

• Stefan–Boltzmann law. P = AeT^4. Assume e=0.95. At high temperatures, this dominates completely. The closer you get to room temperature, convection dominates instead. Aug 30, 2021 at 21:25
• Not simple. At all. You need the heat equation which is a partial differential equation and therefore tricky to solve in the best of cases because the temperature something equilibriates at depends on how much energy you put into it relative to how much it is losing to its surroundings. But how much it loses to its surroundings depends on the temperature difference between it and its surroundings. And then you have to throw in geometry, material properties, and circulation into the heat equation and it gets real messy. Aug 30, 2021 at 21:31
• Start by calculating the resistance of that bar, which determines the voltage and current for any given power. It's mass and the heat capacity will tell you the energy you need to supply to raise its mass to any temperature, then power tells you the the time it'll take to reach that temperature ... then you have to factor in cooling, which will (a) slow up the heating you just calculated, and (b) require constant power to keep it at temperature. Cooling is where the complexity happens. Frankly at that stage the simplest approach is to experiment and measure.
– user16324
Aug 30, 2021 at 22:08
• Your question is unanswerable since there are no time slew rate specs and thermal settling time can be > 5T Aug 31, 2021 at 4:15
• The outside temperature won't be stable, so the temperature of the rod won't be stable. Have you considered a PID-like controller on one or multiple thermic sensors (a couple of PT100(0) would suffice) to take care of the difference?
– Mast
Aug 31, 2021 at 9:26

There is no simple formula. In general you must include conduction, convection and radiation effects, the latter two of which are highly nonlinear and dependent on the temperatures involved and all are dependent on geometry. Crack open a book on Fluid Dynamics (air is a fluid) to get a feel of what you're up against on convection, for example.

Easiest thing is to mock it up and test it, or use rather complex software that you would have to verify anyway since there are many ways to go wrong. You may have enough intuition to guess roughly (within, say, 5:1) the power required from experience with heat sinks and incandescent bulbs, kilns etc. and test from there.

Depending on the temperature, stainless steel may not be more durable than NiCr, especially if moisture may be present. There's a reason why we use NiCr and not cheaper less brittle materials.

• “A reason why NiCr is used …” absolutely. Aug 30, 2021 at 21:52
• As having studied that (as part of my EE studies), I would always test on a real life example. You can model it so-so and there is still unexpected behaviour in some corners. Better models are trade secrets. Aug 30, 2021 at 22:55

The problem is that as soon as the bar gets hot, it will start to transfer heat to the surrounding environment. In order to calculate the temperature, you will need to know exactly how much heat it is transferring as well as how much heat you are adding. However, this depends on the ambient temperature, the airflow, and also the temperature and emissivity of objects within direct line of sight view of the bar.

In some specialized cases it may be possible to do the calculation. If the bar is heated rapidly in air to a peak temperature for a brief time, then allowed to cool, then you may be able to ignore heat transfer and just use the heat capacity of the bar. But as time goes on, if you want to hold it at a specific temperature, you will need to directly measure the bar temperature (maybe with a thermocouple) and control the flow of energy to the bar to maintain temperature.

For hot wire devices (hot wire foam cutters, hot wire vaping, etc) there are probably rules of thumb used by designers of those devices that you can use (so many Watts per meter or something).

Also, one reason for using nichrome instead of stainless is that nichrome does not have a large temperature coefficient. The resistance of a stainless bar will increase as it gets hotter. Around 4% per 10 degrees C, if I remember correctly. Many metals have similar positive temperature coefficients.

Warning: danger! Take appropriate safety measures before you set your house on fire!

If you look at nearly any heating element, you’ll notice that it is equipped with a thermostat, and uses hysteresis to maintain the target temperature (or more precisely, a temperature close to the target).

That’s because it’s difficult to just get to the target temperature with a fixed current and stay there. And for it to work, it would need to be an asymptotic function, meaning it would take forever to actually reach the target temperature.

So it’s a lot easier to “overshoot”. You heat the element as much as is practically possible, and once you get slightly above the target, you stop heating, until it reaches slightly below the target, and you start heating again, and so on (how much margin to have on either side of the target is another interesting topic).

So the question is not so much how much current do you need to reach a given temperature or what size element you need for that, but rather how fast you want to get there. Also, in general, you don’t want to heat the element per se, what you want is heat something else (water in a tank, air in your oven or in your home…). What you want to heat (nature, volume, initial temperature, isolation), how quickly you want to heat it, and how much you are willing to spend are some of the parameters which will dictate exactly how much current you should inject into the element.

Now if you look at most heating devices in your home, their main characteristic is usually the power they use (in watts). Values in the thousands of watts are common, though of course in some cases it may be much less (consider a soldering iron).

Find an appliance close to what you are trying to emulate and see its power rating. From that and the voltage you’ll apply, you can easily derive the resistance you need to achieve it.

For instance, if you want to reach 2400 W (warning, danger, this is a lot and requires proper safety measures and insulation) with 240 V, you need 10 A, which in turn tells you you need 24 ohms of resistance. Don’t forget a thermostat, though, otherwise your element will continue heating and reach much higher temperatures than your target (until something breaks or catches fire, most likely).

Of course, a 2400 W power source is something you want to take quite a few safety measures around, this is going to burn a lot of things! Don’t do this unless you understand what you are doing.

Also, remember that your heating element will transfer heat to most of the stuff it is in contact with: the medium (water or air, usually) into which it is inserted, anything that supports it, and, that’s easy to forget, the rest of the circuit. There’s no magical reason the copper wires on either end will not start heating. You need to have something in between which is a good electric conductor but a very bad heat conductor, otherwise things will end up badly.

• Is there any material you can think of that's a good electrical conductor but bad heat conductor? Sep 1, 2021 at 19:04
• @Cullub Much easier to find thermal conductors which aren't electrically conductive than the other way around. The qualities that make metals good electrical conductors also make them good thermal conductors. If there is a good electrical conductor which is a poor thermal conductor it is likely a ceramic. Sep 3, 2021 at 1:03

It's not a simple formula because it's not a simple problem. You'll see lots of posts on here that are along the lines of "why isn't there a simple way to determine how much current my wire can carry?", which is the same problem in reverse. It's because while it's easy to determine how much power will be dissipated in the material, it's tricky to determine what temperature change it translates into. Your bar's geometry, its orientation, its finish, the surrounding temperature, the altitude, and expected air circulation all factor into it, and I'm sure others can come up with more variables.

As a starting point, I suggest calculating the convective heat transfer coefficient for a bar of your geometry with the desired surface temperature, and adding the radiative heat loss as @winny suggests. Or, if the bar will be in intimate thermal contact with whatever it's heating, then the thermal resistance to the heated mass and its convective/radiative heat loss. Then you can use the bulk resistivity to get a (very rough) estimate for how much current you'll need and be prepared to either do some real-world testing or implement a closed-loop control scheme to achieve the desired temperature.

You would need to search in tables for Specific heat capacity of stainless steel. It's about 450 for 20C, but you need to find temperature characteristic. I suppose it is in EN norms.

Now you can calculate how much heat do you need to add to raise temperature.

$$\Delta Q = \frac{{c \cdot m \cdot \Delta T}}{\eta} \\\\ \Delta Q –\ heat \ in \ Watts,\\ m – mass \ of \ object,\\ \Delta T – temperature\ change.\\ \eta - efficiency \ of \ your \ circuit\\ \ c - Specific \ heat \ capacity$$

But you also have to subtract heat lose to enviroment.

As user Winny said you need Stefan–Boltzmann law for that.

$$\Phi = \sigma T^4 \\ \Phi - stream \ of \ heat [\frac{W}{m^2}]$$

It will give you some kind of estimate, but you need some measurements to make your formula.

Or just do experiments and your own table.

If you calculate the steady state wattage required to maintain a given temperature, applying that wattage constantly will cause the temperature to rise asymptotically towards said temperature. That said, using a controller with a thermocouple feedback is probably the best idea, because the convective heat transfer will vary wildly depending on orientation, and wind (swinging it around).

your sword will loose heat in 3 ways: Conduction, Convection, and Radiation. Conduction will be insignificant because air is a bad static heat conductor. Glowing red or orange will put you into the range where Radiation cannot be ignored in addition to natural convection.

I'll model the heated part of the sword as a thin flat plate, and only consider the wide surfaces for simplicity (not the thin edges). Let's say you want your sword at 1000C (a nice easy number that results in a pretty good orange) and the flat part has dimensions 1000x10 mm. https://www.engineeringtoolbox.com/ claims the emissivity for Nichrome is 0.7 ish (not an uncommon value for silvery metals). We will also assume the surroundings are far enough away to be a perfect absorber (emissivity 1.0) and are at room temp (20C). We then use Boltzman equations to calculate the energy loss in Watts:

q = heat transfer per unit time (W)
σ = 5.6703 10-8 (W/m2K4) - The Stefan-Boltzmann Constant
ε = emissivity coefficient of the object (one - 1 - for a black body)
Th = hot body absolute temperature (K)
Tc = cold surroundings absolute temperature (K)
Ah = area of the hot object (m2)

q = ε σ (Th4 - Tc4) Ah (source)
2079 watts = 0.7 * σ (12734 - 2934) * 2 (1*.01)

Calculating the convective heat transfer is much harder, but wikipedia has some equations to get started with. As winny suggested in the comments, at high temperatures, radiation will dominate anyway, so you may get away with just adding a "fudge factor" to accommodate convection heat loss.

As a real-world reference, the youtube channel Hacksmith Industries built a very similar concept, and according to the video it drew on the order of 10,000 - 20,000 watts.

• You seem to have used degreesC , not degrees K || and your Tc is 10 not 20 BUT that will make minimal difference when degrees K are used. Sep 2, 2021 at 11:26
• @RussellMcMahon yah, that's what I get for writing an answer without editing or double checking... Sep 3, 2021 at 0:35