Explanation Given: The temperature of blackbody in flame is ( T f ) is 1700 K . The temperature of blackbody in moonlight is ( T m ) is 4000 K . Calculation: Calculate the peak spectral blackbody emissive power for flame ( E b f ) can be calculated using the relation. E b f = 3.74177 × 10 8 ( λ ) 5 [ exp ( 1.43878 × 10 4 / λ T f ) − 1 ] W / m 2 ⋅ μm E b f ( T f ) = 3.74177 × 10 8 T f 5 ( 2897.8 ) 5 [ exp ( 1.43878 × 10 4 / 2897.8 ) − 1 ] W / m 2 ⋅ μm = 1.278 × 10 − 11 T f 5 W / m 2 ⋅ μm = 1.278 × 10 − 11 ( 1700 ) 5 W / m 2 ⋅ μm = 1.81 × 10 5 W / m 2 ⋅ μm Calculate the peak spectral blackbody emissive power for moonlight ( E b m ) can be calculated using the relation

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Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala

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Chapter 21, Problem 19P

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The peak spectral blackbody emissive power for both lighting sources (match flame and moonlight).

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Chapter 21, Problem 18P

Chapter 21, Problem 20P

Chapter 21 Solutions

Fundamentals of Thermal-Fluid Sciences

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