Thermodynamics HW Solutions 549

# Thermodynamics HW Solutions 549 - 10 899 . 1 /s m 10 896 ....

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Chapter 7 External Forced Convection 7-21 Air flows on both sides of a continuous sheet of plastic. The rate of heat transfer from the plastic sheet is to be determined. Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Re cr = 5 × 10 5 . 3 Radiation effects are negligible. 4 Air is an ideal gas with constant properties. Plastic sheet T s = 90 ° C Air V = 3 m/s T = 30 ° C 15 m/min Properties The properties of air at 1 atm and the film temperature of (T s + T )/2 = (90+30)/2 =60 ° C are (Table A-15) 7202 . 0 Pr /s m 10 896 . 1 C W/m. 02808 . 0 kg/m 059 . 1 2 5 - 3 = × = ° = = υ ρ k Analysis The width of the cooling section is first determined from m 0.5 = s) 2 ( m/s] ) 60 / 15 [( = Δ = t W V The Reynolds number is 5 2 5
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Unformatted text preview: 10 899 . 1 /s m 10 896 . 1 m) m/s)(1.2 (3 Re = = = L L V which is less than the critical Reynolds number. Thus the flow is laminar. Using the proper relation in laminar flow for Nusselt number, the average heat transfer coefficient and the heat transfer rate are determined to be C . W/m 07 . 6 ) 7 . 259 ( m 2 . 1 C W/m. 0282 . 7 . 259 ) 7202 . ( ) 10 899 . 1 ( 664 . Pr Re 664 . 2 3 / 1 5 . 5 3 / 1 5 . = = = = = = = Nu L k h k hL Nu L W 437 = C 30)-)(90 m C)(1.2 . W/m 07 . 6 ( ) ( m 1.2 = m) m)(0.5 2 . 1 ( 2 2 2 2 2 = = = = s s conv s T T hA Q LW A &amp; 7-11...
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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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