Resistive heating - calculating how much a wire will heat up before it melts

When you pass a current through a wire, for a fraction of a second (a time $Delta t$, say) almost all the electrical work ($I^2 RDelta t$) goes into internal thermal energy in the wire (that is, as you say, $mcDelta T$). But as the wire gets hotter more and more of the energy being supplied is lost as heat to the surroundings. Let's suppose that the wire never reaches its melting point. Instead it will reach a maximum temperature, when it gives off heat at the same rate that energy is being supplied electrically.

How do we calculate this maximum temperature? If the wire gets red hot, or not far short of it, much of the heat – my limited researches suggest more than 80% for temperatures over 600°C – will be lost as infrared radiation. In this case we have $$I^2 R=alphasigma 2pi r l(T^4-T_0^4)$$ The right hand side is the net rate of heat loss by radiation. $alpha$ is a number between 1 and 0, dependent on the nature of the surface of the wire. Putting $alpha=1$ isn't usually too bad an approximation. $sigma (text{the Stefan-Boltzmann constant})=5.67times10^{-8} text{Wm}^{-2} text{K}{^{-4}}$, $2pi rl$ is the surface area of the wire, $T$ is its temperature and $T_0$ is room temperature. $T_0^4$ may well be negligible compared with $T^4$. Remember to use kelvin temperatures.

Hence you can find a theoretical value for the wire's final temperature, $T$. If it works out to be more than the wire's melting point – then we predict that the wire will melt!

Note that this is all rather approximate. $alpha$ is difficult to put a precise figure on, and we have neglected to consider the heat loss rate by convection (which is not easy to estimate, as it depends on whether the wire is vertical or horizontal and on the presence of surrounding objects).

If you consider a wire made of the special alloy constantan, you remove one complication. Constantan (aka Eureka) is is designed so that its resistivity changes very little with temperature. Its resistivity is 0.49 $mu Omega text m$ and its melting point is 1210 °C = 1483 K. [Using these figures I found that a constantan wire of diameter 0.50 mm would need a current of 13 A to reach its melting point. Not unreasonable?]

This is a great opportunity to teach your students about performing an energy balance, about various heat transfer modes, about solving differential equations (or approximating them with finite-difference equations), and about model simplification.

For any small section of the wire (length $dx$, cross-sectional area $A$), we can perform an energy balance with the following components:

• Volumetric heat generation: $J^2 rA,dx$, where $J=I/A$ is the current density and $r$ is the wire material resistivity, a material property.

• Conductive heat transfer: $kAfrac{partial^2 T(x)}{partial x^2}dx$, as derived here, where $k$ is the thermal conductivity of the wire, a material property, and $T$ is the temperature in that wire section. This term implies that conduction tends to eliminate curvature (the second derivative) in temperature distributions.

• Convective heat loss: $-hP[T(x)-T_infty]dx$, where $h$ is the convective coefficient, characterizing the natural or forced airflow past the wire, $P$ is the perimeter length (e.g., pi times the diameter), and $T_infty$ is the ambient temperature.

• Radiative heat loss: $-sigmaepsilon P[T(x)^4-T_infty^4]dx$, where $sigma=5.67times 10^{-8},mathrm{W,m^{-2},K^{-4}}$ is the Stefan-Boltzmann constant and $epsilon$ is the emissivity of the wire surface.

Add these together and divide by the section volume to obtain the general heat equation:

$$kfrac{partial^2 T(x)}{partial x^2}+J^2r-frac{hP}{A}[T(x)-T_infty]-frac{sigmaepsilon P}{A} [T(x)^4-T_infty^4]=rho cfrac{partial T(x)}{partial t},$$

where $rho$ is the wire material density and $c$ is its specific heat capacity. The term on the right indicates that the result of the energy balance on the left (whether positive or negative) goes to change the wire temperature over time $t$.

This equation can be hard to solve. Let's try some simplifying assumptions. If the wire is very long and thin, then we might consider conduction along its length to be unimportant (this is equivalent to assuming that it remains at the same temperature, something you can evaluate by looking at the glow when it gets very hot). Furthermore, at very high temperatures, the $T(x)^4$-containing radiation term dominates over the $T(x)$-containing convection term. Plus, let's change differentials to finite differences to obtain some general scaling laws:

$$J^2r-frac{sigmaepsilon P}{A} [T(x)^4-T_infty^4]=rho cfrac{Delta T}{Delta t}.$$ From this, we can, for example, estimate the time required for melting, which you noted was of interest to you:

$$tapproxfrac{rho c T_m}{J^2r-sigmaepsilon P T_m^4/A}.$$

Here, I've assumed that the ambient or room temperature $T_infty$ is small compared to the melting temperature $T_m$ of the wire material. Does this make sense? Try plugging in your specific conditions!