Thermodynamics HW Solutions 621

# Thermodynamics HW Solutions 621 - Pr Re 037 2 3 1 8 5 3 1 8...

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Chapter 7 External Forced Convection 7-91 A car travels at a velocity of 60 km/h. The rate of heat transfer from the bottom surface of the hot automotive engine block is to be determined for two cases. Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Re cr = 5 × 10 5 . 3 Air is an ideal gas with constant properties. 4 The pressure of air is 1 atm. 5 The flow is turbulent over the entire surface because of the constant agitation of the engine block. 6 The bottom surface of the engine is a flat surface. Properties The properties of air at 1 atm and the film temperature of (T s + T )/2 = (75+5)/2 = 40 ° C are (Table A-15) L = 0.7 m 7255 . 0 Pr /s m 10 702 . 1 C W/m. 02662 . 0 2 5 - = × = ° = υ k Engine block Air V = 60 km/h T = 5 ° C T s = 75 ° C ε = 0.92 Analysis The Reynolds number is [] 5 2 5 10 855 . 6 /s m 10 702 . 1 m) (0.7 m/s ) 3600 / 1000 60 ( Re × = × × = υ = L V L T s = 10 ° C which is less than the critical Reynolds number. But we will assume turbulent flow because of the constant agitation of the engine block. C . W/m 97 . 58 ) 1551 ( m 7 . 0 C W/m. 02662 . 0 1551 ) 7255 . 0 ( ) 10 855 . 6 ( 037
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Unformatted text preview: . Pr Re 037 . 2 3 / 1 8 . 5 3 / 1 8 . ° = ° = = = × = = = Nu L k h k hL Nu L [ ] W 1734 = C 5)-(75 m) m)(0.7 (0.6 C) . W/m 97 . 58 ( ) ( 2 ° ° = − = ∞ s s conv T T hA Q & The heat loss by radiation is then determined from Stefan-Boltzman law to be [ ] W 181 K) 273 + (10 K) 273 + (75 ) .K W/m 10 (5.67 ) m 7 . )( m 6 . )( 92 . ( ) ( 4 4 4 2-8 4 4 = − × = − = surr s s rad T T A Q σε & Then the total rate of heat loss from the bottom surface of the engine block becomes W 1915 = + = + = 181 1734 rad conv total Q Q Q & & & The gunk will introduce an additional resistance to heat dissipation from the engine. The total heat transfer rate in this case can be calculated from W 1668 = m) 0.7 m 6 . )( C W/m. 3 ( ) m 002 . ( m)] m)(0.7 C)[(0.6 . W/m 97 . 58 ( 1 C 5)-(75 1 2 × ° + ° ° = + − = ∞ s s s kA L hA T T Q & The decrease in the heat transfer rate is 1734-1668 = 66 W 7-76...
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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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