Thermodynamics HW Solutions 580

Thermodynamics HW Solutions 580 - s k hD Nu The heat...

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Chapter 7 External Forced Convection 7-51 A light bulb is cooled by a fan. The equilibrium temperature of the glass bulb is to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The light bulb is in spherical shape. 4 The local atmospheric pressure is 1 atm. Properties The properties of air at 1 atm pressure and the free stream temperature of 25 ° C are (Table A-15) 7296 . 0 Pr kg/m.s 10 181 . 2 kg/m.s 10 849 . 1 /s m 10 562 . 1 C W/m. 02551 . 0 5 C 100 @ , 5 2 5 - = × = × = × = ° = ° s k μ υ Lamp 100 W ε = 0.9 Air V = 2 m/s T = 25 ° C Analysis The Reynolds number is 4 2 5 10 280 . 1 /s m 10 562 . 1 m) m/s)(0.1 (2 Re × = × = υ = D V The proper relation for Nusselt number corresponding to this Reynolds number is [] [] 06 . 68 10 181 . 2 10 849 . 1 ) 7296 . 0 ( ) 10 280 . 1 ( 06 . 0 ) 10 280 . 1 ( 4 . 0 2 Pr Re 06 . 0 Re 4 . 0 2 4 / 1 5 5 4 . 0 3 / 2 4 5 . 0 4 4 / 1 4 . 0 3 / 2 5 . 0 = × × × + × + = μ μ + + = =
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Unformatted text preview: s k hD Nu The heat transfer coefficient is C . W/m 36 . 17 ) 06 . 68 ( m 1 . C W/m. 02551 . 2 = = = Nu D k h Noting that 90 % of electrical energy is converted to heat, & ( . )( Q = 0 90 100 W) = 90 W The bulb loses heat by both convection and radiation. The equilibrium temperature of the glass bulb can be determined by iteration, 2 2 2 m 0314 . ) m 1 . ( = = = D A s [ ] [ ] C 133.2 = = + + + = + = + = K 2 . 406 ) K 273 25 ( ) .K W/m 10 )(5.67 m (0.0314 ) 9 . ( K ) 273 25 ( ) m C)(0.0314 . W/m 36 . 17 ( W 90 ) ( ) ( 4 4 4 2-8 2 2 2 4 4 total s s s surr s s s s rad conv T T T T T A T T hA Q Q Q & & & 7-35...
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