Question 3 of 5: Radiative/Convective Solidification of a Casting Top Layer

Question 3: Radiative/Convective Solidification of a Casting Top Layer (25 marks: (a) 3, (b) 2.5, (c) 3.5, (d) 4, (e) 6, (f) 6)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) $k_{Fe}$; (b) the flux expression; (c) $h_{total}$; (d) the solidified-layer thickness $Y$ where $Bi=0.1$; (e) $dY/dt$ under a uniform-temperature (Biot $\ll0.1$) layer; (f) the governing (unsolved) equation once conduction through the shell also matters.

Approach. (a) uses the Wiedemann–Franz–Lorenz law to convert the given electrical conductivity to a thermal conductivity, since no $k$ value is supplied directly; (b)–(c) linearize the combined radiative/convective boundary condition into a single surface coefficient; (d) applies the standard lumped-capacitance Biot criterion; (e) is an interface energy balance (latent heat released = heat lost from the top) valid while the layer itself is isothermal; (f) restates that same interface balance once the layer's own conduction resistance can no longer be neglected.

1. (a) Thermal conductivity via Wiedemann–Franz–Lorenz. For a free-electron conductor, $k=L_0\sigma_eT$ with Lorenz number $L_0=2.44\times10^{-8}\ \text{W}\cdot\Omega/\text{K}^2$: $$k=L_0\sigma_eT_m=(2.44\times10^{-8})(5\times10^5)(1800)=\boxed{22.0\ \text{W/m}\cdot\text{K}}$$ consistent with handbook values for iron near its melting point.
2. (b) Combined flux expression. With the environment black and the layer top grey (view factor 1, no reflected radiation returning), the net radiative flux is $\varepsilon\sigma_{SB}(T_s^4-T_\infty^4)$; adding Newton's-law convection: $$q=h(T_s-T_\infty)+\varepsilon\sigma_{SB}\left(T_s^4-T_\infty^4\right)$$
3. (c) Total heat transfer coefficient. With $T_\infty\ll T_s$, drop $T_\infty$ terms and factor out $T_s$: $q\approx h\,T_s+\varepsilon\sigma_{SB}T_s^4=T_s\big(h+\varepsilon\sigma_{SB}T_s^3\big)=h_{total}T_s$, evaluated at the layer-top temperature $T_s=T_m=1800$ K: $$h_{total}=h+\varepsilon\sigma_{SB}T_m^3=100+0.6(5.67\times10^{-8})(1800)^3=100+198.4=\boxed{298.4\ \text{W/m}^2\cdot\text{K}}$$
4. (d) Solidified-layer thickness at $Bi=0.1$. $Bi=h_{total}Y/k_{Fe}$: $$Y=\frac{Bi\cdot k_{Fe}}{h_{total}}=\frac{0.1(22.0)}{298.4}=\boxed{7.4\ \text{mm}}$$ Below this thickness the uniform-layer-temperature assumption of part (e) is valid; above it, conduction resistance in the layer (part f) becomes significant.
5. (e) Growth rate, uniform-layer limit. With the layer isothermal at $T_s=T_m$, the heat leaving the top, $q=h_{total}T_m$, must equal the rate of latent heat released as the solidification front advances (quasi-steady, no accumulation in the thin layer): $$\rho\,\Delta H_f\frac{dY}{dt}=h_{total}T_m\;\Rightarrow\;\frac{dY}{dt}=\frac{h_{total}T_m}{\rho\,\Delta H_f}=\frac{298.4(1800)}{7500(267{,}000)}=\boxed{2.68\times10^{-4}\ \text{m/s}\ (=0.268\ \text{mm/s})}$$
6. (f) Combined conduction + surface-loss equation (not solved). Once the solidified layer is thick enough that its own conduction resistance matters, the top surface temperature $T_s$ is no longer $T_m$; it is set by requiring the conductive flux through the layer to match the flux leaving the top surface, and the interface energy balance uses that same flux: $$\frac{k_{Fe}\left(T_m-T_s\right)}{Y}=h\left(T_s-T_\infty\right)+\varepsilon\sigma_{SB}\left(T_s^4-T_\infty^4\right)\qquad\text{(surface-flux match, implicitly defines }T_s(Y)\text{)}$$ $$\rho\,\Delta H_f\frac{dY}{dt}=\frac{k_{Fe}\left(T_m-T_s\right)}{Y}\qquad\text{(interface energy balance)}$$ These two coupled, nonlinear equations in $Y(t)$ and $T_s(t)$ define the growth law; they are not solved here, per the question's instruction.