Thermodynamics HW Solutions 528

Thermodynamics HW Solutions 528 - L A kL kA dy dT kA Q Q W...

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Chapter 6 Fundamentals of Convection 6-42 A shaft rotating in a bearing is considered. The power required to rotate the shaft is to be determined for different fluids in the gap. Assumptions 1 Steady operating conditions exist. 2 The fluid has constant properties. 3 Body forces such as gravity are negligible. Properties The properties of air, water, and oil at 40 ° C are (Tables A-15, A-9, A-13) Air: μ = 1.918 × 10 -5 N-s/m 2 Water: = 0.653 × 10 -3 N-s/m 2 Oil: = 0.212 N-s/m 2 Analysis A shaft rotating in a bearing can be approximated as parallel flow between two large plates with one plate moving and the other stationary. Therefore, we solve this problem considering such a flow with the plates separated by a L =0.5 mm thick fluid film similar to the problem given in Example 6-1. By simplifying and solving the continuity, momentum, and energy equations it is found in Example 6-1 that 12 m/s 5 cm 2500 rpm 10 cm () L A
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Unformatted text preview: L A kL kA dy dT kA Q Q W y L 2 2 1 2 2 2 2 mech V V V = = = = = = = &amp; &amp; &amp; First, the velocity and the surface area are m/s 545 . 6 s 60 min 1 ) rev/min 2500 )( m 0.05 ( = = = N D &amp; V 2 bearing m 01571 . ) m 10 . )( m 0.05 ( = = = DL A ( a ) Air: W 0.013 = = = m/s N 1 W 1 ) m 2(0.0005 m/s) 545 . 6 )( s/m N 10 918 . 1 ( ) m 01571 . ( 2 2 2 5 2 2 mech L A W V &amp; ( b ) Water: W 0.44 = = = = m/s N 1 W 1 ) m 2(0.0005 m/s) 545 . 6 )( s/m N 10 653 . ( ) m 01571 . ( 2 2 2 3 2 2 mech L A Q W V &amp; &amp; ( c ) Oil: W 142.7 = = = = m/s N 1 W 1 ) m 2(0.0005 m/s) 545 . 6 )( s/m N 212 . ( ) m 01571 . ( 2 2 2 2 2 mech L A Q W V &amp; &amp; 6-17...
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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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