Thermodynamics HW Solutions 523

# Thermodynamics HW Solutions 523 - Chapter 6 Fundamentals of...

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Chapter 6 Fundamentals of Convection 0 2 1 2 2 = μ = L y kL dy dT V ⎯→ 2 L y = Therefore, maximum temperature will occur at mid plane in the oil. The velocity and the surface area are m/s 425 . 9 s 60 min 1 ) rev/min 3000 )( m 0.06 ( = = = ππ n D V 2 m 0377 . 0 ) m 20 . 0 )( m 0.06 ( = π = π = bearing DL A The maximum temperature is C 53.3 ° = ° + ° = μ + = μ + = = m/s N 1 W 1 ) C W/m 8(0.17 m/s) 425 . 9 )( s/m N 05 . 0 ( C 50 8 ) 2 / ( 2 / 2 ) 2 / ( 2 2 2 0 2 2 2 0 max k T L L L L k T L T T V V ( b ) The rates of heat transfer are () W 419 = = = = = = m/s N 1 W 1 ) m 2(0.0002 m/s) 425 . 9 )( s/m N 05
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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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