A homogeneous half-space y>0 has thermal diffusivity α=1× 10^(-6) m^2/s and…

A homogeneous half-space $y>0$ has thermal diffusivity $\alpha=1\times 10^{-6}\,\mathrm{m}^2/\mathrm{s}$ and conductivity $k=2\,\mathrm{W}/(\mathrm{m}\cdot\mathrm{K})$. The surface temperature is the maintained harmonic \[ T(0,t)=T_0+10\cos(\omega t),\qquad \omega=2\times 10^{-4}\,\mathrm{s}^{-1}, \] and $T\to T_0$ as $y\to\infty$. Write $\theta:=T-T_0$ and $A=10\,\mathrm{K}$. The field satisfies the one-dimensional heat equation $\partial_t\theta=\alpha\partial_{yy}\theta$. 1. Define the thermal penetration depth $\delta=\sqrt{2\alpha/\omega}$. Prove $\delta=0.1\,\mathrm{m}$ and that the unique bounded time-periodic steady-state solution is \[ \theta(y,t)=10\,e^{-y/\delta}\cos(\omega t-y/\delta). \] 2. At the station $y=\delta$, prove that the local amplitude is $10/e$ and that the temperature lags the surface by $1\,\mathrm{rad}$. Prove that the depth at which the amplitude has fallen to $1\%$ of the surface value is \[ y_{1\%}=\delta\ln 100=0.460517\,\mathrm{m}. \] 3. The inward surface flux (heat entering the solid through $y=0$, per unit area) is $q=-(k\partial_y\theta)|_{y=0}$. Prove \[ q=\frac{kA}{\delta}\bigl(\cos(\omega t)-\sin(\omega t)\bigr) \] and that this waveform has amplitude $282.842712\,\mathrm{W}/\mathrm{m}^2$ and leads the surface temperature by $45^{\circ}$. 4. Audit both sentences: (i) “this is the Heat-course semi-infinite step, so $\theta=A\,\mathrm{erfc}(y/\sqrt{4\alpha t})$ and there is no finite $\delta$ or $45^{\circ}$ lead”; (ii) “the surface flux is $kA/\delta$ in phase with $\cos(\omega t)$, or the $1\%$ depth is $\delta$ itself.” Do not replace the periodic penetration field by a Graetz scaling, a matched-diffusivity triple analogy, or a coupled thermodiffusion slab.

Compute $2\alpha/\omega=2\cdot 10^{-6}/2\cdot 10^{-4}=10^{-2}$, so $\delta=0.1\,\mathrm{m}$. The real part of $A\exp\bigl(-(1+i)y/\delta+i\omega t\bigr)$ is the displayed cosine.