In class, we learned that ”most” of the sun’s radiation happens in the band 0.2 ?...

In class, we learned that ”most” of the sun’s radiation happens in the band 0.2 ? ? ? 3µm, and Stephen asked what ”most” means. What proportion of solar radiation (for a black body sun) falls in the above wavelength range. Similarly, what proportion of the Earth’s radiation (black body Earth at 290K) occurs in the wavelength band 4 ? ? ? 40µm? What wavelength range contains the middle 95% of the sun’s radiation?

Solutions

Expert Solution

1. Given that T = 5778 K, (for a blackbody sun)

?₁T = 0.2 * 5778 = 1156 ?mK and ?₂T = 3 * 5778 = 17334 ?mK

the blackbody radiation functions are determined from Table 11-12 , (Introduction to Thermodynamics and Heat Transfer, Yunus A. Cengel)

f _?1 = 0.00173514 f _?2 = 0.97851368

Then the fraction of radiation emitted between these two wavelengths becomes

f _?2 - f _?1 = 0. 97851368– 0. 00173514= 0.9768 ? 97.68%

2. Given that T = 290 K, (for a blackbody earth)

?₁T = 4 * 290 = 1160 ?mK and ?₂T = 40 * 290 = 11600 ?mK

the blackbody radiation functions are determined from Table 11-12 , (Introduction to Thermodynamics and Heat Transfer, Yunus A. Cengel)

f _?1 = 0.001714 f _?2 = 0.9409868

Then the fraction of radiation emitted between these two wavelengths becomes

f _?2 - f _?1 = 0. 9409868– 0. 001714= 0.9393 ? 93.93%

3. Let the range for this 95% radiation be 0 ?m to ? ?m

thus, f _?2 - f _?1 = 0. 95 i.e., f _?2 – 0 = 0.95 i.e., f _?2 = 0.95

from the table 11-2, for 0.95

?₂T = 12488 ?mK

?₂*5778 = 12488

?₂ = 2.16 ?m

thus, 95% of the radiation has a range of 2.16 ?m

samet mamat answered 2 years ago

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