# How to calculate length needed for heating element (2mm Nickeline/Nichrome wire) to a specified power?

to start off, i dont really understand about electrical engineering. im a mechanical engineering student tasked to continue the project. i dont really understand electrical stuff beside the very basic stuff.

the data we have:

• wire resistance (Rn): 0.85 ohm/m
• voltage: 220 volt (actually 3 phase but i've been told that we'd use each phase separately, so i think its 220 v, but still not too sure about it)
• specified power output: 7000 watt

these are what i've calculated, but still not sure if there's anything wrong in it:

• Current = P/V = 7000/220 = 31.8 ampere
• Resistance required = P/I^2 = 7000/31.8^2 = 6.91 ohm
• Required length = R/Rn = 6.91/0.85 = 8.13 meter

the thing is, previous project uses up to 75 meters. a lot more that what my calculation shows, which is 8.13 meters.

so, did i calculate it wrong? if so, how do i properly calculate this?

did i calculate it wrong by calculating current based on required power output? shouldnt it be power input from power line which i still have no data yet, beside voltage and it being 3 phase?

another data i forgot to add:

• the target temperature is around 1000°C
• power input: 3 phase, 10 A, 220 V, 6600VA

• What are you heating up? Must it be NiCr wire? or can you a quartz heater or tungsten heater with forced air? Or is this just homework? Variables include hotspot temp max, thermal conductance/resistance, radiant and conducted heat transfer and temperature gradients Commented Apr 10, 2022 at 5:27
• its for electric furnace for heating metal that is going to be tested (for metallurgy class). idk what the wire name's is in english, tried to google it but result seems to be inconclusive. its called "Nikelin" here. here's a picture of it Commented Apr 10, 2022 at 5:58
• btw, id be glad if you can help me more for that stuff you mentioned. resistance is assumed to be constant so far. > Variables include hotspot temp max, thermal conductance/resistance, radiant and conducted heat transfer and temperature gradients Commented Apr 10, 2022 at 6:02
• I'll do. Its about 1000 degree Celsius. I didnt mention it cause it only mentioned by my prof in passing while telling me to read the paper published by previous student. and due to covid situation, their paper is a lot less than ideal. Commented Apr 10, 2022 at 8:36
• The resistance rises with termperature thus power is reduced. It can be used in parallel to lower resistance but must not exceed working temperature of wire with suitable insulation and PID temperature control. So you have insufficient info. Commented Apr 10, 2022 at 13:53

A second, probably more challenging, issue besides power sharing on the three mains phases is the maximum heating wire temperature.

You cannot dissipate any power you want per wire unit, it could overheat and eventually melt down. This is, by the way, how fuses work.

Working out the power to temperature relationship is a rather complex task involving radiation, conduction and convection in the specific boundary conditions like surrounding medium (air?), flow conditions, temperature, mounting methods and many other.

From an engineering point of view you'd better be given a number by the wire supplier assuming some kind of standard conditions.

On the other hand I love ballparking and try to toss something, given the (supposed) high temperature of the wire I'll stick on a radiation only model and neglect a few hundred K ambient temperature.

Let's take l meter of wire of diameter d and resistance per meter R ohm/m carrying I current, equilibrium is when dissipated power equals radiated one.

$$R\,l\,I^2=\sigma T^4 A$$ Where A is the outer surface area of wire $$R\,l\,I^2=\sigma T^4 l \,\pi\,d$$

And hence $$T=\left(\frac{R\,I^2}{\sigma\,\pi\,d}\right)^{\frac{1}{4}}$$

Which worked out with the initial datas

$$T=\left(\frac{0.85\,\Omega\mathrm{/m}\times(31.8\,\mathrm{ A})^2}{56.7\,\mathrm{ nW/m^2K^4}\times\pi\times 2\,\mathrm{mm}}\right)^{\frac{1}{4}}\approx 1500\,\mathrm{K}$$

rises a flag on the temperature being too high.

Anything else equal it comes out that temperature rise is proportional with the square root of current $$T \propto \sqrt{I}$$ so the three phase option with approximately 10 A could be reckoned to hit some 900 K.

I once more recall this is a very course estimate of wire surface temperature when anything surrounding is much cooler aiming to reckon max power per length unit. Furnace temperature is a far cry from this.

In any case I dare say this kind of engineers ballpark always gives very good clues.

• is this the rough estimates of how high the temperature will be if i used 31.8 A current? but yeah, as pointed by other comment, i did my calculation wrong and the current should be around 10 A. Commented Apr 10, 2022 at 8:26
• @Bramble yes, and 10A reckon is already covered in my answer, approx 900K Commented Apr 10, 2022 at 8:46
• btw, can you point out the formula if it would include conduction and convection? also, what is formula above called? i'd like to study about it more Commented Apr 10, 2022 at 8:49
• oh, just realized its just basic radiation power formula Commented Apr 10, 2022 at 9:11
• Just to make sure, in case you mentioned above, does the temperature will maxes out at that value for the current given? ~1500 K at 31.8 A and ~900 K at 10 A Commented Apr 10, 2022 at 16:41

Actually if you are using a 3-phase system with 220v on each phase you can use 3 wires (one from each phase to common), each line dissipating only 2333.3 watts each, (for a 7000w total). Recalculate the needed wire length for a 2333.3w line. So you would need 3 wires of that length. (The total of the 3 wires seems to be about 73m, and that's fairly close to the 75m length you mentioned).

Your original calculation assumes the power is being dissipated from a single phase 220v line at over 30A. A 30A single phase supply would require bulky wiring and large receptacles, and likely be near the limits for residential use. By using a 3 phase system each phase would only need about 10A, a much more reasonable value.

• yeah, using each phase separately is what ive been told to do, but im still not sure how to properly calculate it. so the formula would be 3P = VI for current calculation if its using 3 phase, i see thanks for the answer, it really helps Commented Apr 10, 2022 at 5:53
• wait a minute, i tried to lower the power but the wire seem to getting too long. could it be that there's 2 different kind of power here? P input from power line, and P output for what i desired. P input then used to calculate the current, and P output would be used to calculate the resistance (and wire length) required. Commented Apr 10, 2022 at 8:05
• Just determine the wire length needed for one phase, (which will be 1/3 the total required power). So using 3 wires (one on each phase) results in the total system of 3x the single phase power.
– Nedd
Commented Apr 10, 2022 at 8:10
• no, wait, the current should be already specified from power line, shouldnt it? then id just calculate the Resistance required with R = P/I^2, right? Commented Apr 10, 2022 at 8:11
• did i calculate it wrong by calculating current based on required power output? shouldnt it be power input from power line which i still have no data yet, beside voltage 220v and it being 3 phase? Commented Apr 10, 2022 at 8:16