How to consider the dimension of a cylinder when computing its radiative budget?

Question

I am trying to derive a heat budget model to estimate the temprature of very small a stationary vertical cylinder (1mm) under environmental conditions. Imagine a wooden stick. Simply put, I have

R = H InSW * albedo + sigma * emissivity * Tcylinder^4 = rho_air * cp * (Tcylinder - Tair)/rH.

Where,

• R = sum of radiative components
• H = convective components (latent heat and conduction ignore)
• InSW = Incident solar radiation observed by a pyranometer
• albedo = albedo of the cylinder
• sigma = ste-boltz constant
• emissivity = emissivity of the cylinder to longwave IR
• Tcylinder = temperature cylinder
• rho_air = density of the air
• cp = heat capacity of the air
• rH = resistance to convection

This is the basis of a heat budget model, but how do I incorporate the fact that its a cylinder? I've managed to find information about how to deal with rH with the forced and natural convections, so what I'm more concerned about is the radiative components. Is it alright for me to leave them like this or do I need to start accounting for the geometry and its potential impact on the intercepted light and resistance to radiative transfer?

Thanks

Answer 1

You certainly need to take account of the shape of the object somehow.

As a simple example, consider a sphere of radius $r$ in space where everything except solar radiation can be ignored.

The cross section area of the sphere which is absorbing solar radiation is $\pi r^2$ but the surface area which is emitting it is $4 \pi r^2$. So the radiative heat balance equation looks something like $4 \pi r^2 T_{\text{sphere}}^4 = \pi r^2 T_{\text{sun}}^4$ and in the equilibrium state (assuming the sphere has high conductivity so the surface temperature is uniform) will be $T_{\text{sphere}}= T_{\text{sun}}/\sqrt 2$.

For a unsymmetrical body like a cylinder, the amount of radiation absorbed will depend on the orientation of the cylinder relative to the sun, but the amount emitted will not.