Thermodynamics HW Solutions 903

Thermodynamics HW Solutions 903 - Chapter 11 Fundamentals...

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Chapter 11 Fundamentals of Thermal Radiation 11-85 A small surface emits radiation. The rate of radiation energy emitted through a band is to be determined. A = 2 cm 2 T = 600 K 40 ° 50 ° Assumptions Surface A emits diffusely as a blackbody. Analysis The rate of radiation emission from a surface per unit surface area in the direction ( θ , φ ) is given as φθθ θφ d d I dA Q d dE e e sin cos ) , ( = = & The total rate of radiation emission through the band between 40 ° and 50 ° can be expressed as 4 4 2 0 50 40 1736 . 0 ) 1736 . 0 ( ) 1736 . 0 ( sin cos ) , ( T T I d d I E b e σ π φθ θθ = = = = ∫∫ == since the blackbody radiation intensity is constant (
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