Exam2_2010_solution - 57:020 Mechanics of Fluids and...

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Unformatted text preview: 57:020 Mechanics of Fluids and Transport Fall 2010 EXAM2 Solutions Problem 1: Momentum equation (Chapter 5) Information and assumptions • 1 = 760 kPa; 2 = = 101 kPa • 1 = 5 cm; 2 = 2 cm • ρ = 788.42 kg/m 3 for alcohol • No losses in the nozzle flow Find • Alcohol jet velocity V 2 and the force F required to hold the plate stationary Solution (a) As no losses in the nozzle flow, one can apply the Bernoulli equation between the sections 1 and 2 to find 2 . By knowing that 1 = ( 2 1 ⁄ ) 2 = ( 2 1 ⁄ ) 2 2 from the continuity equation and that 1 = 2 , the Bernoulli equation can be written as 1 + 1 2 ¡ 2 1 ¢ 2 2 £ 2 = 2 + 1 2 2 2 (+3 points) or 2 = ¤ 2( 1 − 2 ) [1 − ( 2 1 ⁄ ) 4 ] Thus, 2 = ¥ 2(760,000 Pa − 101,000 Pa) ¡ 788.42 kg m 3 ¢ [1 − (0.02 m 0.05 m ⁄ ) 4 ] = 41.42 m s ⁄ (+2 point) (b) For a control volume that encompasses the jet from section 2 and the plate, a momentum analysis gives − = − ̇ ( 2 ) + 0 = − 2 2 2 (+4 points) Thus, =...
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Exam2_2010_solution - 57:020 Mechanics of Fluids and...

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