Thermodynamics HW Solutions 662

Thermodynamics HW Solutions 662 - Chapter 8 Internal Forced...

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Chapter 8 Internal Forced Convection 8-49 A circuit board is cooled by passing cool helium gas through a channel drilled into the board. The maximum total power of the electronic components is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat flux at the top surface of the channel is uniform, and heat transfer through other surfaces is negligible. 3 The inner surfaces of the channel are smooth. 4 Helium is an ideal gas. 5 The pressure of helium in the channel is 1 atm. Properties The properties of helium at the estimated average temperature of 25 ° C are (Table A-16) 669 . 0 Pr C J/kg. 5193 /s m 10 233 . 1 C W/m. 1565 . 0 kg/m 1635 . 0 2 4 - 3 = ° = × = ° = = p C k υ ρ He 15 ° C 4 m/s T e Electronic components, 50 ° C L = 20 cm Air channel 0.2 cm × 14 cm Analysis The cross-sectional and heat transfer surface areas are 2 2 m 028 . 0 ) m 2 . 0 )( m 14 . 0 ( m 00028 . 0 ) m 14 . 0 )( m 002 . 0 ( = = = = s c A A To determine heat transfer coefficient, we need to first find the Reynolds number m 003944 . 0 m) 0.14 + m 002 . 0 ( 2 ) m 00028 . 0 ( 4 4 2 = = = P A D c h 9 . 127 /s m 10 233 . 1 m) 944 m/s)(0.003 (4 Re 2 4 = × = = h m D V which is less than 2300. Therefore, the flow is laminar and the thermal entry length is m 0.20 << m 0.01687 = m) 003944 . 0 )( 669 . 0 )( 9 . 127 ( 05 . 0 Pr Re 05 . 0 =
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This note was uploaded on 01/22/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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