complete_solution2_2

complete_solution2_2 - FLUID MECHANICS II Assignment 2...

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Unformatted text preview: FLUID MECHANICS II Assignment 2 Question 3: Problem 4.2 ~ V = u i = V 1 + 2 x L i Only x component for acceleration, i.e. a y = and a z = . a x = u t + u u x + v u y + w u z = V 1 + 2 x L 2 V L a x = 2V 2 L 1 + 2 x L For L=6 in = 0.1524 m and V = 10 ft/s = 3.048 m/s, Du Dt = 2(3 . 048) 2 . 1524 1 + 2 x . 1524 = 121 . 92 1 + 2 x . 1524 where x is in m. At x = 0, Du Dt = 121 . 92 m/s 2 . ( 12 g s ) At x=L, Du Dt = 365 . 92 m/s 2 . ( 37 g s ) Question 4: Problem 4.3 Two-dimensional velocity field given by ~ V = ( x 2- y 2 + x ) i- (2 xy + y ) j Here the components of the acceleration vector are a x = Du Dt = u u x + v u y = ( x 2- y 2 + x )(2 x + 1) + (- 2 xy- y )(- 2 y ) a y = Dv Dt = u v x + v v y = ( x 2- y 2 + x )(- 2 y ) + (- 2 xy- y )(- 2 x- 1) At (x,y) = (1,2), we obtain a x = 18 [ m/s 2 ] a y = 26 [ m/s 2 ] b) At (x,y) = (1,2), the velocity field is given by ~ V =- 2 i- 6 j A unit vector along a 40 line would be...
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This note was uploaded on 09/14/2011 for the course ME 362 taught by Professor Ceciledevaud during the Winter '11 term at Waterloo.

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complete_solution2_2 - FLUID MECHANICS II Assignment 2...

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