Thermodynamics HW Solutions 887

# Thermodynamics HW Solutions 887 - ° = = d D solar G G G θ...

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Chapter 11 Fundamentals of Thermal Radiation 11-56C Due to its nearly horizontal orientation, windshield exchanges heat with the sky that is at very low temperature. Side windows on the other hand exchange heat with surrounding surfaces that are at relatively high temperature. 11-57C Because of different wavelengths of solar radiation and radiation originating from surrounding bodies, the surfaces usually have quite different absorptivities. Solar radiation is concentrated in the short wavelength region and the surfaces in the infrared region. 11-58 A surface is exposed to solar and sky radiation. The net rate of radiation heat transfer is to be determined. T sky = 280 K T s = 350 K α s = 0.85 ε = 0.5 G d = 400 W/m 2 θ G D =150 W/m 2 Properties The solar absorptivity and emissivity of the surface are given to α s = 0.85 and ε = 0.5. Analysis The total solar energy incident on the surface is 2 2 2 W/m 1 . 703 ) W/m 400 ( 30 cos ) W/m 350 ( cos = +
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Unformatted text preview: ° = + = d D solar G G G θ Then the net rate of radiation heat transfer in this case becomes [ ] surface) the (to K) 280 ( K) 350 ( ) K W/m 10 67 . 5 ( 5 . ) W/m 1 . 703 ( 85 . ) ( 4 4 4 2 8 2 4 4 , 2 W/m 347 = − ⋅ × − = − − = − sky s solar s rad net T T G q εσ & 11-59E A surface is exposed to solar and sky radiation. The equilibrium temperature of the surface is to be determined. Properties The solar absorptivity and emissivity of the surface are given to α s = 0.10 and ε = 0.8. Analysis The equilibrium temperature of the surface in this case is [ ] R 413.3 = − ⋅ × = − εσ = α = − εσ − α = − s s sky s solar s sky s solar s rad net T T T T G T T G q 4 4 4 2 8 2 4 4 4 4 , R) ( ) R Btu/h.ft 10 1714 . ( 8 . ) Btu/h.ft 400 ( 1 . ) ( ) ( & T sky = 0 R T s = ? s = 0.1 ε = 0.8 G solar = 400 Btu/h.ft 2 Insulation 11-25...
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