Equilibrium Temperature of Aluminium Foil

Initial estimate

P=eσA(T4-Tc4) (Stefan's Law)

P=power radiated by object (aluminium foil)
e=emissivity
σ=Stefan's constant
A=surface area
T=temperature of object
Tc =surrounding temperature

The approximation that the temperature of the void is 0K is then made for the initial estimate for the sake of ease, and thus the above becomes:

P=eσAT4

Additionally, we know that:

Pa=S0αAcosθ ,

where:
Pa=the power absorbed by the material
S0=the solar constant (=1.37kW/m2)
θ=angle of incidence
A=surface area
α=absorptivity .

The solar radiation is considered perpendicular to the surface of the foil, thus cosθ=1 , and so:

Pa=S0αA .

Ignoring the thickness of the material, and assuming that it absorbs radiation through one face and emits through two, and recognising that the equilibrium temperature will occur when the two power expressions are equal, we have:

2eσAT4=S0αA
T=[(S0α)/(2eσ)]1/4

σ=5.6704x10-8W/m2/K2
e≈0.031
α≈0.2

Thus,

T≈528.3677411K
T≈528K .

This is the initial estimate of the temperature of the aluminium foil. Although this is not the final value, it does provide a good guide to the temperature the foil will reach.

More accurately, one can obtain a value of the absorptivity from the coefficient of extinction.

α=4πk/λ k=6.7x10-2
λ=501nm
Thus, α=0.06722130389 (taking the depth of material into account, as otherwise one gets a value per unit length).

α+ρ+τ=1 (reflectivity, transmissivity and absorption)

ρ=0.923
Thus, τ=9.77869611x10-3

Thus, actually;
13.39681369W/Area transmitted
92.09318633W/Area absorbed

These values differ considerable from previously basically because of fewer approximations used, and because these values assume a surface with a much higher reflectivity, thus pushing the other values down.

Thus, the actual equilibrium temp of the aluminium will occur when the radiated power equals the power incident (that is being radiated by the glass), the power incident due to the sun itself.

Glass

For glass, the same process can be undertaken for radiation incident upon it and radiation it emits. However, there is no k value published in the text I am using at the range of wavelengths we are interested in. The values do become vanishingly small however, so it is reasonable to assume that all the radiation incident on the glass is either reflected or transmitted. With no value published for reflectivity either, it can be assumed that all of the radiation is therefore transmitted.

Thus, the equilibrium temperature of the glass will be:

(ΔT)2/R=PrG=eGσTG4

once conduction has been taken into account, and

PrA=92.09318633-(ΔT)2/R=eAσTA4

where, R=Δx/k
k=1.05

Δx=0.5mm or 1mm (as of yet unsure).

Solving simultaneously, we have:

TG=20.51021051K ≈20.5K if Δx=0.5mm
=17.2580396K ≈17.3K if Δx=1.0mm
TA=476.944736K ≈477K if Δx=0.5mm
=476.3333298K ≈476K if Δx=1.0mm

N.B. in all of the above, there is a factor of A (the area) in each equation,however I have omitted it as it will all cancel.

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