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

Question

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

thanks in advance

Answer 1

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.

Answer 2

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.