What has the most effect in determining heat transfer through radiation?

Answer 1

The radiative heat transfer from one surface to another is equal to the radiation entering the first surface from the other, minus the radiation leaving the first surface. For black bodies, i.e. perfect efficiency of radiation and absorption of impinging radiation:

Q = σAF(T14 - T24)

where

σ is the Stefan-Boltzmann constant

A is the area

F is the form factor

T1 is the temperature of the hotter surface (heat source)

T2 is the temperature of the cooler surface (heat sink)

From this equation we see the main factors affecting the heat transfer are the size/shape of the heat source and the temperatures of the two surfaces. Note also that a change in the temperature of the heat source has a greater effect on the heat transfer than equal changes in the heat sink - due to the temperatures being raised to the fourth power.

Reality is that there are no perfect "black bodies". Nothing radiates perfectly nor absorbs perfectly. We call these imperfect surfaces "grey bodies". For a grey body with only two surfaces the heat transfer is equal to:

Q = σ(T14 - T24)/(e1 + f + e2)

where

e1= (1-ε1)/(A1ε1)

e2= (1-ε2)/(A2ε2)

f = 1/(A1F12)

ε = thermal emissivity across the radiative spectrum

A1 = the area of the heat source

F12 = the form factor between the two surfaces.

Note that the temperature of the heat source usually still dominates.