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Thermodynamics HW Solutions 548

# Thermodynamics HW Solutions 548 - surface because of the...

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Chapter 7 External Forced Convection 7-20 A car travels at a velocity of 80 km/h. The rate of heat transfer from the bottom surface of the hot automotive engine block is to be determined. 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 flow is turbulent over the entire surface because of the constant agitation of the engine block. T s = 80 ° C ε = 0.95 Air V = 80 km/h T = 20 ° C L = 0.8 m Engine block Properties The properties of air at 1 atm and the film temperature of (T s + T )/2 = (80+20)/2 =50 ° C are (Table A-15) 7228 . 0 Pr /s m 10 798 . 1 C W/m. 02735 . 0 2 5 - = × = υ ° = k Analysis Air flows parallel to the 0.4 m side. The Reynolds number in this case is 5 2 5 10 888 . 9 /s m 10 798 . 1 m) m/s](0.8 ) 3600 / 1000 80 [( Re × = × × = υ = L L V which is less than the critical Reynolds number. But the flow is assumed to be turbulent over the entire
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Unformatted text preview: surface because of the constant agitation of the engine block. Using the proper relations, the Nusselt number, the heat transfer coefficient, and the heat transfer rate are determined to be C . W/m 98 . 70 ) 2076 ( m 8 . C W/m. 02735 . 2076 ) 7228 . ( ) 10 888 . 9 ( 037 . Pr Re 037 . 2 3 / 1 8 . 5 3 / 1 8 . ° = ° = = = × = = = Nu L k h k hL Nu L W 1363 = C 20)-)(80 m C)(0.32 . W/m 98 . 70 ( ) ( m 0.32 = m) m)(0.4 8 . ( 2 2 2 ° ° = − = = = ∞ s s conv s T T hA Q wL A & The radiation heat transfer from the same surface is W 132 = × = − = ] K) 273 + (25-K) 273 + )[(80 .K W/m 10 )(5.67 m 32 . )( 95 . ( ) ( 4 4 4 2-8 2 4 4 surr s s rad T T A Q σε & Then the total rate of heat transfer from that surface becomes W 1495 = + = + = W ) 132 1363 ( rad conv total Q Q Q & & & 7-10...
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