Heat equation with heat loss to surroundings

Question

I am trying to model one dimensional heat diffusion within a uniform copper rod with insulated ends. In addition, I am including a term that models the heat loss to the surroundings from the lateral sides. This term would account for the heat losses (assuming it is Newtonian) of an infinitesimally small surface area on the copper rod. Here is the heat equation for a rod of arbitrary length L:

$\quad \quad u_t$=$\alpha^2$$u_{xx}$-$\beta$(u-$T_0$)

$\quad \quad u(x,0)=f(x)$

$\quad \quad u(0,t)_x=0$

$\quad \quad u(L,0)_x=0$

My question is with the lateral heat loss constant:

$\quad \quad -\beta$(u-$T_0$)

As this constant $\beta$ (heat transfer coefficient) should affect the rate at which heat is lost to the surroundings, how could I possibly determine an approximate value for this? Stated previously it accounts for the heat lost by an infinitesimally small lateral surface area on the copper rod, so what I did to approximate this was to get a 5 mm long copper round (same diameter and alloy as the rod) and determine the constant from Newton's law of cooling. After doing so I got $\ 2.61*10^{-3}$, however the thermal diffusivity ($\alpha^2$) of the copper alloy is $\ 1.12*10^{-4}$. From this it makes sense that the Newtonian losses will dominate the diffusion within the rod. However intuitively I think this is wrong as copper diffuses heat very slowly. Any ideas on how I could possibly determine this constant or if my approach is correct? Thank you all for your help.

Answer 1

$\beta$ is not a property of the rod alone. It depends on the rod's interaction with the rod's surroundings. Let us consider some of the possibilities:

1. The rod is surrounded by a fast moving fluid. This situation is called forced convention. $\beta$ will depend on the speed of the fluid as well as several fluid properties such as its viscosity and thermal conductivity. It also depends on the direction of the flow which could be longitudinal or transverse.

2. The rod is in a nominally stationary fluid. In this case, the heat transfer between rod and fluid will likely cause changes in fluid density which, under the influence of gravity, causes bouyancy which causes the fluid to move. This case is called natural convection.

3. The rod is inside a stationary solid. This is called conductive heat transfer. The solid material could have a thermal conductivity or it could be chosen because it is a thermal insulator.

4. The rod could be in vacuum. In this case, the heat transfer is due to radiative heat transfer. The magnitude of this will depend on the geometry and the effective thermal emissivities.

Until you pick your situation, you cannot choose a $\beta$.

Some formulas for heat transfer coefficient of cylinders in convective flow can be found here and here.