Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Consider a liquid solid interface $z =\zeta(x,t)$ moving at constant speed $v$, for a two dimensional problem. Due to solidification interface is changing it position. For simplicity heat conductivities, heat capacity of liquid and solid are assumed to be the same. Heat conduction equation in the liquid and solid region is \begin{eqnarray} C\frac{\partial T}{\partial t} &=& k\nabla^2 T \nonumber \\ \end{eqnarray} Here, $k$, $C$ are heat conductivity and heat capacity.

Heat balance equation at interface is \begin{eqnarray} Lv &=& k(\nabla T_s - \nabla T_l)\cdot \hat n \nonumber \\ \end{eqnarray}

$L$ latent heat of solidification. $\hat n$ is unit normal to the interface and $v_n$ is velocity of interface in the normal direction. $T_l$ and $T_s$ are temperatures in the liquid and solid domains.

Boundary conditions are: \begin{eqnarray} T &=& T_{\infty} \hspace{1cm}z \to \infty \\ T &=& T_m \hspace{1cm} z = \zeta(x,t) \end{eqnarray}

I need to find Green's function for this problem. Green's function for heat diffusion of point source is

\begin{eqnarray} G = \frac{1}{{4\pi D(t-t_1)}}\exp\left(-\frac{(x-x_1)^2 +(z-z_1)^2}{4\pi D (t-t_1)}\right) \end{eqnarray} Where $D= k/C$

Then I proceeded as follows:

Since all the heat generated is at the interface $\zeta(x,t)$, \begin{eqnarray} T - T_{\infty} = \int^{t}_{0}dt_1 \int^{}_{}dx_1\frac{1}{{4\pi D(t-t_1)}}\exp\left(-\frac{(x-x_1)^2 +(z-z_1)^2}{4\pi D (t-t_1)}\right) \zeta(x,t) \end{eqnarray}

Research paper gives following equation, \begin{eqnarray} \frac{T-T_\infty}{L/C} &=& \int^t_0 dt_1 \int^{\infty}_{-\infty}dx_1 G[x-x_1, z-\zeta(x_1,t_1)+ v(t-t_1), t-t_1] \left( v + \frac{\partial \zeta}{\partial t}\right) \end{eqnarray}

where,\begin{equation} G(x,t) = \frac{1}{{4\pi D(t)}}\exp\left(-\frac{(x)^2 +(z+vt)^2}{4\pi D (t)}\right) \end{equation}

which is clearly different from what I assume. How can I arrive at this equation?

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.