Consider 1D heat conduction on a rod of length L. Assume the rod has a known density $\rho$, specific heat $c_p$ , and thermal conductivity $k$, all of which are constant with respect to temperature. Suppose that the initial temperature of the rod is constant $T_0$. I want apply a constant heat flux $Q$ to the rod at one end, with the goal of heating this end of the rod to a desired temperature $T_{desired}$. Assume the other end of the rod has an insulated (zero heat flux) boundary condition.

Without explicitly solving the heat equation $\rho c_pu_t=ku_{xx}$, is there a simpler formula I can use to estimate how much time it takes for the heated end of the rod to rise by a a factor of $\Delta T = T_{desired}-T_0$? Is there a formula that only requires knowing $Q,\Delta T, L,\rho, c_p,$ and $k$ to estimate this time?


Well no not to any desired temperature, but you can estimate within engineering accuracy what the temperature rise will be by dimensional analysis. It is not an exact solution but will be accurate within an order of magnitude.

For this problem it is quite trivial but for more exotic equations (perhaps with heat transfer to the surroundings, or 2D etc.) dimensional analysis is a powerful tool which can provide insight into the dynamics of an equations which may otherwise be unsolvable (e.g. due to non-linearity, etc).

First, lets state the system we are dealing with: $$\partial_{t}T=a\partial_{x}^{2}T\quad T\left(x,0\right)=T_{0}$$ with thermal diffusion coefficient $a=k/\rho c_p$ and boundary conditions: $$-k\partial_{x}T\left(0,t\right)=Q\quad\partial_{x}T\left(L,t\right)=0$$

Now for the dimensional analysis we will determine characteristic time ($\Delta t$), length ($\Delta x$) and temperature ($\Delta T$) scales which will result in the above equations to only contain terms of $O(1)$ and smaller when they are non-dimensionalized.

We define the relevant dimensionless variables: $$\theta=\frac{T-T_{0}}{\Delta T}\quad\eta=\frac{x}{\Delta x}\quad\tau=\frac{t}{\Delta t} $$ Substituting these dimensionless variables in the equations yields: $$\partial_{\tau}\theta=\frac{a\Delta t}{\Delta x^{2}}\partial_{\eta}^{2}\theta\quad\theta\left(\eta,0\right)=0$$ $$-\partial_{\eta}\theta\left(0,\tau\right)=\frac{Q\Delta x}{k\Delta T}\quad\partial_{\eta}\theta\left(\frac{L}{\Delta x},\tau\right)=0$$

To obtain equations where all terms are $O(1)$ or less we define: $$\Delta x=L\quad\Delta t=\frac{L^{2}}{a}\quad\Delta T=\frac{QL}{k}$$

which result in the equations becoming: $$\partial_{\tau}\theta=\partial_{\eta}^{2}\theta\quad\theta\left(\eta,0\right)=0$$

$$-\partial_{\eta}\theta\left(0,\tau\right)=1\quad\partial_{\eta}\theta\left(1,\tau\right)=0$$ which are indeed $O(1)$ or smaller.

Having determined the characteristic scales of the system, it follows that it take approximately time $\Delta t = L^2/a$ to increase the temperature by $\Delta T = \frac{QL}{k}$. Hopefully, you find this useful.

  • $\begingroup$ Does this approximate the time required to raise the temperature of the entire rod by $\Delta T$? Or just the heated tip of the rod? $\endgroup$ – Paul Nov 14 '15 at 13:32
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    $\begingroup$ @Paul - It will be at the tip of the rod as the boundary condition is applied there. $\endgroup$ – nluigi Nov 14 '15 at 13:37
  • $\begingroup$ I would hope so, but the characteristic time here reflects the time it takes for heat to diffuse completely through the entire length $L$, right? $\endgroup$ – Paul Nov 14 '15 at 18:47
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    $\begingroup$ @Paul Correct, by the time heat has diffused along the rod, the temperature at the boundary will have increased by $\Delta T$. $\endgroup$ – nluigi Nov 14 '15 at 19:16

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