Recall Differential Form of Heat Transfer Equation:
k\vec \nabla^2T + \cancel{\dot S_v} $$
Therefore:
\alpha \frac{\partial^2 T}{\partial x^2} $$
where:
$$ \alpha = \frac{k}{\rho c_p} $$
Initial Condition:
At $t = 0$, $T = T_i \quad \forall x\ge 0$
Boundary Conditions:
At $x = 0$, $T=T_s \quad \forall t\ge 0^+$
At $x\to \infty$, $T\to T_i \quad \forall t\ge 0^+$
<aside> đŸ’¡ Over time, the temperature profile is simply the same shape but ‘stretched’ in the x-direction - SELF-SIMILARITY
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For:
\alpha \frac{\partial^2 T}{\partial x^2} $$
Normalise the variables:
$$ T^* = \frac{T}{T_c}\\ t^* = \frac{t}{t_c}\\ x^* = \frac{x}{x_c} $$
Therefore:
\alpha \frac{T_c}{x_c^2}\frac{\partial^2T^*}{\partial x^{*2}}\\
\alpha \frac{1}{x_c^2}\frac{\partial^2T^*}{\partial x^{*2}} $$
Choose characteristic quantities to be same order of magnitude as real quantities. Then:
$$ T^* \thicksim \, t^* \thicksim \, x^* \thicksim \,0(1) \qquad \text{Order of magnitude similar as 1}\\ \therefore \frac{\partial T^}{\partial t^} \thicksim \, \frac{\partial^2T^*}{\partial x^{*2}} \thicksim \, 0(1) $$
\alpha \frac{1}{x_c^2}\frac{\partial^2T^*}{\partial x^{*2}} \quad \to \quad \frac{1}{t_c}\thicksim \, \frac{\alpha}{x_c^2} \quad \to \quad x_c \thicksim \, \sqrt{\alpha t_c} $$
Units: $[m]$