The radiant heat exchange between two infinite parallel surfaces per m2 at temperatures T1 and T2 may be given as,
qr = e s (T14 – T24)
qr = hr№ – T2)
hr = es (T12 + T22)(T1 + T2) = e(4sT) for T1 ffi T2
and e = — + — — 1, for two parallel surfaces
Є1 Є2
= e, for surface exposed to atmosphere
e1 and e2 are the emissivities of the two surfaces. When one of the surfaces is sky, Eq. (1.24a) becomes,
Qr = es (T14 — Tsky4) (1.25a)
The above equation may be rewritten as,
Qr = es (T14 — Ta4)+es(Ta4 — Tsky4) (1.25b)
or, Qr = hr (T1 — Ta)+ e AR (1.25c)
where AR = s[(Ta + 273)4-(Tsky + 273)4] is the difference between the long-wavelength radiation incident on the surface from sky and surroundings and the radiation emitted by a blackbody at ambient temperature. The reduction of Qr in the form of Eq. (1.25c) will enable one to find the exact closed form solution for T1. It may be noted here that this solution is based on the assumption that Ta and Tsky are constant.
