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Thermodynamics HW Solutions 881

# Thermodynamics HW Solutions 881 - Chapter 11 Fundamentals...

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Chapter 11 Fundamentals of Thermal Radiation 11-42 The variation of reflectivity of a surface with wavelength is given. The average reflectivity, emissivity, and absorptivity of the surface are to be determined for two source temperatures. Analysis The average reflectivity of this surface for solar radiation ( = 5800 K) is determined to be T λμ μ λ T =→ = (. 3 0 978746 m)(5800 K) = 17400 mK f ρ λ λ , μ m 3 0.95 0.35 ρ ρρ λλ () (. ) ) (. ) ( . ) Tf T ff =+ = −− 10 2 12 11 1 0 35 0 978746 0 95 1 0 978746 0.362 Noting that this is an opaque surface, τ = 0 At = 5800 K: T 0.638 = = = ⎯→ = + 362 . 0 1 1 1 ρα Repeating calculations for radiation coming from surfaces at = 300 K, T μ λ T =⎯ ⎯= 3 0 0001685 m)(300 K) = 900 mK f 1 0.95 = + = ) 0001685 . 0 1 )( 95 . 0 ( ) 0001685 . 0 )( 35 . 0 ( ) ( T At = 300 K: T 0.05 = = = = + 95 . 0 1 1 1 and ε α == 0.05 The temperature of the aluminum plate is close to room temperature, and thus emissivity of the plate will be equal to its absorptivity at room temperature. That is, εε αα room s 005 0638 . . which makes it suitable as a solar collector. ( 0 and 1 room = = s for an ideal solar collector) 11-43
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