Thermodynamics HW Solutions 739

Thermodynamics HW Solutions 739 - . )( m/s 81 . 9 ( Pr ) (...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 9 Natural Convection 9-44 The equilibrium temperature of a light glass bulb in a room is to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The local atmospheric pressure is 1 atm. 4 The light bulb is approximated as an 8-cm-diameter sphere. Properties The solution of this problem requires a trial-and-error approach since the determination of the Rayleigh number and thus the Nusselt number depends on the surface temperature which is unknown. We start the solution process by “guessing” the surface temperature to be 170 ° C for the evaluation of the properties and h . We will check the accuracy of this guess later and repeat the calculations if necessary. The properties of air at 1 atm and the anticipated film temperature of ( T s + T )/2 = (170+25)/2 = 97.5 ° C are (Table A-15) 1 - 2 5 K 002699 . 0 K ) 273 5 . 97 ( 1 1 7116 . 0 Pr /s m 10 279 . 2 C W/m. 03077 . 0 = + = = β = × = υ ° = f T k Analysis The characteristic length in this case is L c = D = 0.08 m. Then, 6 2 2 5 3 -1 2 2 3 10 694 . 2 ) 7116 . 0 ( ) /s m 10 279 . 2 ( ) m 08 . 0 )( K 25 170 )( K 002699
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . )( m/s 81 . 9 ( Pr ) ( = = = D T T g Ra s Air T = 25 C Lamp 60 W = 0.9 D = 8 cm Light, 6 W ( ) [ ] ( ) [ ] 42 . 20 7116 . / 469 . 1 ) 10 694 . 2 ( 589 . 2 Pr / 469 . 1 589 . 2 9 / 4 16 / 9 4 / 1 6 9 / 4 16 / 9 4 / 1 = + + = + + = Ra Nu Then 2 2 2 2 m 02011 . m) 08 . ( C . W/m 854 . 7 ) 42 . 20 ( m 08 . C W/m. 03077 . = = = = = = D A Nu D k h s Considering both natural convection and radiation, the total rate of heat loss can be written as ] ) K 273 25 ( ) 273 )[( .K W/m 10 67 . 5 )( m 02011 . )( 9 . ( C ) 25 )( m 02011 . )( C . W/m 854 . 7 ( W ) 60 90 . ( ) ( ) ( 4 4 4 2 8 2 2 2 4 4 + + + = + = s s surr s s s s T T T T A T T hA Q & Its solution is T s = 169.4 C which is sufficiently close to the value assumed in the evaluation of properties and h . Therefore, there is no need to repeat calculations. 9-40...
View Full Document

Ask a homework question - tutors are online