Thermodynamics HW Solutions 888

# Thermodynamics HW Solutions 888 - Chapter 11 Fundamentals...

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Chapter 11 Fundamentals of Thermal Radiation 11-60 Water is observed to have frozen one night while the air temperature is above freezing temperature. The effective sky temperature is to be determined. T = 4 ° C T sky = ? Water T s = 0 ° C ε = 0.8 Properties The emissivity of water is ε = 0.95 (Table A-21). Analysis Assuming the water temperature to be 0 ° C, the value of the effective sky temperature is determined from an energy balance on water to be hT T T T air surface s sky () ( −= εσ 44 ) and [ ] K 254.8 = ⎯→ × = ° ° ° sky sky T T 4 4 4 2 8 2 K) 273 ( ) K W/m 10 67 . 5 ( 95 . 0 ) C 0 C 4 )( C W/m 18 ( Therefore, the effective sky temperature must have been below 255 K. 11-61 The absorber plate of a solar collector is exposed to solar and sky radiation. The net rate of solar energy absorbed by the absorber plate is to be determined. Properties The solar absorptivity and emissivity of the
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## This note was uploaded on 01/24/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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