I will try and go through the calculation. This is not a solution however, because what I'll be providing you is ok for flat plates/walls, and the irregular geometry that you have will need some numerical tools to simulate and get an reasonably accurate estimate.
Assuming that you have a constant wall temperature of 70oC at the wall of the fiber wall, then the heat power (energy in the unit of time) $\dot{Q}$ ending up in the environment will be:
$$\dot{Q} = \frac{\Delta T_{tot}}{R_{tot}}$$
where:
The total thermal resistance can be found (for the flat wall case) as:
$$R = \frac{L}{kA} + \frac{1}{h_{\infty}A}$$
where:
- k is the Thermal Conductivity 230 (W/m K)
- $h_{\infty}$ is the convective coefficient of air side.
- L is the thickness of the aluminium
- A is the cross-sectional area of the heat exchange.
The reason you will need a numerical method (FEM or similar), is that the last two parameters (L and A) provide a problem, because:
- the thickness L of the aluminium is not constant
- the cross-sectional area A is not constant.
However for a simplified analysis you can calculate an average quantity for both L, and A and use that to get a ball park figure.
convective heat transfer coefficient of the process
The convective heat transfer coefficient is expressed in $\frac{W}{m^2.^oC}$. This heat coefficient is very much depended on the velocity. The easiest way (avoiding Nusselt and Prandtl number which are required for a detailed calculation) the easiest way would be to use the following equation (ref)):
$$h_\infty = 12.12 - 1.16 V + 11.6 \sqrt{V} \tag{eq.1}$$
where:
- $V$ is the velocity of air as it approaches the exchange surface.
- $h_\infty$ is the convectivity heat transfer coefficient in $\frac{W}{m^2. C}$
Make a note, that although in your case is zero, you can use a fan to provide faster air and thus increase the convective coefficient and thus the overall heat transfer. (that is why computers use fans).
quick calculation of required power per area
Assuming you have a 50 W source for the rod of $d_f=1.2 mm$ diameter, with a fiber length of $L_f=60mm$, then the exchange area of the rod would be:
$$A= \pi\cdot d_r \cdot L_f=226 mm^2$$
That means that per $m^2$ the energy is in the order of $ \frac{P}{A} = 221\frac{kW}{m^2}$ (which is quite a lot). You might need to increase the dimensions (also I don't know the fraction of the 50W that ends up on the heat sink, however even if its only 10%, you'd still have 22 $\frac{kW}{m^2}$). Just to give a measure of reference direct sun on a clear sky is about 1 $\frac{kW}{m^2}$.
assuming the fibre is not touching then the $R_{tot}$ would be modified as follows (in red is the additional quantity):
$$R =\color{red}{\frac{1}{h_{fr}A_{fr}}}+ \frac{L}{kA} + \frac{1}{h_{\infty}A}$$
where:
- $h_{fr}$ is the heat convectivity coefficient between the fibre rod wall and the hole
- $A_{fr} = \pi d_{fr} L_{fr}$ the exchange are between the fibre rod wall and the hole
- $d_{fr}$ the diameter of the fiber rod
- $L_{fr}$ the length of the fiber rod
