# Problem 5 – Hint 4

To begin with, notice that there is a very useful small parameter, the ratio $\delta/L$, where $\delta=\delta (t)$ denotes the heat penetration depth perpendicular to the pipe that grows diffusively in time. The diffusive growth law $\delta\sim\sqrt{Dt}$ can be obtained by relating the heat flux, in its one-dimensional version expressed as $\kappa\partial T/\partial x$, to the growth rate of the accumulated heat. Here $D$ denotes the diffusivity. Strictly speaking, $\delta$ is also a function of the coordinate $z$ along the pipe, because in the close-to-the-inlet regions of the pipe, the heat starts diffusing radially earlier than in the farther regions. However, as it appears, the propagation along the pipe slows down, and as a result, the time during which radial diffusion has taken place is on the same order of magnitude for almost all the points along the pipe, except in the immediate neighborhood of the heat front.

Since $\delta\ll L$, in the soil, the radial temperature gradient is much bigger than the gradient along the pipe, and the latter can be neglected. Furthermore, in the close-to-pipe regions where the distance from the axis of the pipe $s$ is approximately smaller than $\delta$, the heat density is almost constant, and hence, we can use a quasi-stationary heat balance equation: for any small soil region, the heat flux carries in as much flux as it carries out. This equation allows us to find the quasi-stationary temperature profile in the close-to-pipe region. The reason why we can use a quasi-stationary equation is as follows: The process can usually be considered quasi-stationary if the characteristic time of the process is much longer than the time required to reach equilibrium, for thermodynamic processes, the so-called thermalization time. For diffusive processes, the thermalization time is estimated as $a^2/D$, where $a$ is the characteristic scale; in our case, $a$ is the distance of an observation point from the pipe, and one can easily see that for $a<\delta$, the thermalization time is smaller than the observation time $t$.

The steps towards a solution are:
(i) Find an estimate for the radius $\delta$ of the sausage-like region where the soil is heated, as a function of time;
(ii) Find the radial temperature profile, i.e., the dependence $T=T(s)$ for $s$ approximately smaller than $\delta$;
(iii) Find an estimate for the average temperature of soil in that sausage-like region, as a function of time;
(iv) Equate the total heat that has been brought in by the hot water flowing in the pipe, to the heat accumulated in the sausage-like region of length $L$ and radius $\delta$ — this is an equation for finding $L$.

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