Heat Chap09-103 - Chapter 9 Natural Convection 9-103E The...

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Chapter 9 Natural Convection 9-103E The components of an electronic device located in a horizontal duct of rectangular cross section is cooled by forced air. The heat transfer from the outer surfaces of the duct by natural convection and the average temperature of the duct are 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 Radiation effects are negligible. 5 The thermal resistance of the duct is negligible. Properties The properties of air at 1 atm and the anticipated film temperature of ( T s + T )/2 = (120+80)/2 = 100 ° F are (Table A-15E) 1 - 2 3 R 001786 . 0 R ) 460 100 /( 1 / 1 726 . 0 Pr /s ft 10 1808 . 0 F Btu/h.ft. 01529 . 0 = + = = = × = ° = - f T k β υ Analysis ( a ) Using air properties at the average temperature of (85+100))/2 = 92.5 ° F and 1 atm for the forced air, the mass flow rate of air and the heat transfer rate by forced convection are determined to be lbm/min 581 . 1 ) /min ft 22 )( lbm/ft 07186 . 0 ( 3 3 = = = V m ρ Btu/h 1 . 342 F ) 85 100 )( F Btu/lbm. 2405 . 0 )( lbm/h 60 581 . 1 ( ) ( = ° - ° × = - = in out p forced T T C m Q Noting that radiation heat transfer is negligible, the rest of the 180 W heat generated must be dissipated by natural convection, Btu/h 272 = - × = - = 1 . 342 ) 412 . 3 180 ( forced total natural Q Q Q ( b ) We start the calculations by “guessing” the surface temperature to be 120 ° F for the evaluation of the properties and h . We will check the accuracy of this guess later and repeat the calculations if necessary. Horizontal top surface: The characteristic length is ft 2222 . 0 ft) 6/12 + ft 2(4 ft) ft)(6/12 4 ( = = = P A L s c . Then, 5 2 2 3 3 -1 2 2 3 10 604 . 5 ) 726 . 0 ( ) /s ft 10 1808 . 0 ( ) ft 2222 . 0 ( R) 80 120 )( R 001786 . 0 )( ft/s 2 . 32 ( Pr ) ( × = × - = - = - c s L T T g Ra 77 . 14 ) 10 604 . 5 ( 54 . 0 54 . 0 4 / 1 5 4 / 1 = × = = Ra Nu bottom top c top A A Nu L k h = = = ° = ° = = 2 2 ft 2 ) ft 12 / 6 )( ft 4 ( F . Btu/h.ft 016 . 1 ) 77 . 14 ( ft 2222 . 0 F Btu/h.ft. 01529 . 0 Horizontal bottom surface: The Nusselt number for this geometry and orientation can be determined from 387 . 7 ) 10 604 . 5 ( 27 . 0 27 . 0 4 / 1 5 4 / 1 = × = = Ra Nu F . Btu/h.ft 5082 . 0 ) 387 . 7 ( ft 2222 . 0 F Btu/h.ft. 01529 . 0 2 ° = ° = = Nu L k h c bottom Vertical side surfaces: The characteristic length in this case is the height of the duct, L c = L = 6 in. Then, 6 2 2 3 3 -1 2 2 3 10 383 . 6 ) 726 . 0 ( ) /s ft 10 1808 . 0 ( ) ft 5 . 0 ( R) 80 120 )( R 001786 . 0 )( ft/s 2 . 32 ( Pr ) ( × = × - = - = - L T T g Ra s 9-97 180 W L = 4 ft Air duct 6 in × 6 in Air 85 ° F 22 cfm 100 ° F Air 80 ° F
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Chapter 9 Natural Convection 57 . 27 726 . 0 492 . 0 1 ) 10 383 . 6 ( 387 . 0 825 . 0 Pr 492 . 0 1 Ra 387 . 0 825 . 0 2 27 / 8 16 / 9 6 / 1 6 2 27 / 8 16 / 9 6 / 1 = + × + = + + = Nu 2 2 ft 4 ) ft 5 . 0 )( ft 4 ( 2 F . Btu/h.ft 843 . 0 ) 57 . 27 ( ft 5 . 0 F Btu/h.ft. 01529 . 0 = = ° = ° = = side side A Nu L k h Then the total heat loss from the duct can be expressed as [( ) ( ) ( ) ]( ) Q Q Q Q hA hA hA T T total top bottom side top bottom side s = + + = + + - Substituting and solving for the surface temperature, F ) 80 ]( F Btu/h. ) 4 843 . 0 2 5082 . 0 2 016 . 1 [( Btu/h 272 ° - ° × + × + × = s T T s = 122.4 ° F which is sufficiently close to the assumed value of 120 ° F used in the evaluation of properties and h .
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This homework help was uploaded on 03/19/2008 for the course ME 410 taught by Professor Benard during the Spring '08 term at Michigan State University.

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Heat Chap09-103 - Chapter 9 Natural Convection 9-103E The...

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