Q2. What are the main reasons for the use of IR transparent covers?
IR transparent covers such as polyethylene to reduce convection are important to consider, though some aperture structures can reduce convection inputs naturally.
Q3. How does a high emittance roof paint achieve a net output power?
A known high emittance roof paint at 10° C below ambient, under a 45° aperture lined with shiny aluminium, can achieve a net output power near 135 Wm-2 under a clear sky.
Q4. How much is the net output of the black body at ambient?
At 45° apertures for surfaces approaching the ”ideal”, total output drops by 45 Wm-2 to yield a net rate of 55 Wm-2, while with no heat mirrorstheir net total heat pumping rate at 10° C below ambient is also close to 50 Wm-2.
Q5. What is the effect of the reflective surfaces on the output and input?
If present they reduce both output and input, which leads to a reduction factor for net output power near 0.90 if covers are thick enough for desired rigidity.
Q6. What is the absorptance of a sky window?
The peak absorptance value in known near “ideal” sky window emitters can reach around 0.90 so that their absorptance as in the sky window becomes as ~0.90.
Q7. What is the response for the rest of the Planck radiation spectrum?
The best response for the remainder of the Planck radiation spectrum surprisingly switches between two spectral extremes at a temperature which falls as the aperture gets smaller.
Q8. How much is the gain in total output power for a near ideal sky window system?
If however radiant blocking by heat mirrors is not used the near ideal sky window system becomes superior at 10° C below ambient, by a factor ~3 relative to the surface with es ~ 0.95.
Q9. How much cooling area can a vertical grid blocker achieve with a 45° aperture?
They can achieve around a 45° aperture with thin Al walls which do not waste much cooling area relative to the fully open surface.
Q10. What is the spectral dependence of the coating/substrate system?
A key aspect of cooling technologies based on exploitation of the sky window is the spectral properties of the coating/substrate system defined by their spectralemittance es(η,λ).
Q11. What is the effect of the heat mirrors on the output?
The resulting addition to incoming radiation with the mirrors at T = Ta is given by∆Pmirror = (σTa 4 )emirror cos 2 ηmax (7)so for ηmax = 45° and emirror = 0.1 the net reduction in output due to the heat mirrors being not perfect is 0.05(σTa4) which is 10% of the input from the atmosphere in the range 0° to 45°.
Q12. What is the effect of the aperture on the night sky?
Net thermal radiation cooling, from surfaces at sub-ambient temperatures, to the night sky is amplified if the aperture to the sky is partially blocked with heat mirrors.
Q13. What is the way to check the es of a window emitter?
Standard emittance data for practical sky window emitters is a useful quick quality check, with the better ones typically having es in the range 0.30 to 0.35.
Q14. What is the energy absorbed from the atmosphere?
Equation (1) gives the instantaneous net radiated power for an aperture with aconical opening to zenith angle ηmax and perfect heat mirrors,(1)with the second term the energy absorbed from the atmosphere since es(η,λ) = as(η,λ) the surface spectral absorptance.
Q15. What is the definite integral of equation 3?
ρsw(T) is the fraction of total energy under the Planck spectrum at temperature T within the range 7.9 µm to 13 µm and Isw(ηmax) is the definite integral of equation (3).
Q16. How much is the gain in total output power in using a 45° aperture?
Thus the estimated gain in total output power in using a 45° aperture and with es ~ 0.95 is around a factor of 8 relative to a surface with no heat mirror blocking, and a factor of 3 relative to either configuration of the near ideal sky window spectral system.
Q17. What is the effect of non-ideality of the heat mirrors?
The impact of non-ideality of the heat mirrors is easily estimated from an additionalintegral in equation (1) to cover the zenith aperture angles ηmax to π/2 with a constant low emittance value emirror for the metal mirrors in place of theatmospheric emittance.
Q18. How much is the gain with the aperture relative to no aperture?
The gain with the ideal emitter in using the aperture relative to no aperture is seen in figure 3 to be marginal down to 30° C below ambient.