exam1_solutions - 2.58 Spring 2006 Midterm 1 Solutions...

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Unformatted text preview: 2.58 Spring 2006 Midterm 1 Solutions Question 1. The peak wavelength of the solar radiation is around 500 nm. For a particle with a diameter d less than 50 nm, Rayleigh scattering is a good approximation. The absorption efficiency of the small particle is given by 2 2 m − 1 4 π d ( 2 + 0.1 i ) − 1 0.133 π d (1) = 4 x Im = Im = 2 Q a m 2 + 2 λ ( 2 + 0.1 i ) + 2 λ The solar irradiance that reaches the particle is 2 2 R (2) G λ = s e b λ ( T s ) = R s C 1 d d se se 5 λ e xp C 2 − 1 λ T s where R s is the radius of the sun, d se the distance between the sun and the earth, and C 1 and C 2 are constants. (a) The total solar energy absorbed by the particle is 2 ∞ π d q = G Q d λ (3) a 4 λ a Plug Eqs. (1) and (2) into Eq. (3) to yield 2 2 3 q = 0.133 π d R s T 5 ∞ C 1 (4) a 0 4 d se s ( λ T s ) 6 exp C 2 − 1 d ( λ T...
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This note was uploaded on 11/27/2011 for the course CHEMICAL E 20.410j taught by Professor Rogerd.kamm during the Spring '03 term at MIT.

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exam1_solutions - 2.58 Spring 2006 Midterm 1 Solutions...

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