Thermodynamics HW Solutions 74

# Thermodynamics HW Solutions 74 - W/m(0.7 2 cond = ° − °...

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Chapter 1 Basics of Heat Transfer 1-137 The glass cover of a flat plate solar collector with specified inner and outer surface temperatures is considered. The fraction of heat lost from the glass cover by radiation is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values. 2 Thermal properties of the glass are constant. Properties The thermal conductivity of the glass is given to be k = 0.7 W/m ⋅° C. Analysis Under steady conditions, the rate of heat transfer through the glass by conduction is W 770 m 0.006 C ) 25 (28 ) m C)(2.2
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Unformatted text preview: W/m (0.7 2 cond = ° − ° ⋅ = Δ = L T kA Q & The rate of heat transfer from the glass by convection is 25 ° C 28 ° C L =0.6 cm A = 2.2 m 2 Q & Air, 15 ° C h =10 W/m 2 . ° C W 220 C ) 15 )(25 m C)(2.2 W/m (10 2 2 conv = ° − ° ⋅ = Δ = T hA Q & Under steady conditions, the heat transferred through the cover by conduction should be transferred from the outer surface by convection and radiation. That is, W 550 220 770 conv cond rad = − = − = Q Q Q & & & Then the fraction of heat transferred by radiation becomes 0.714 = = = 770 550 cond rad Q Q f & & (or 71.4%) 1-74...
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