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complete_solution2_2

# complete_solution2_2 - FLUID MECHANICS II Assignment 2...

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FLUID MECHANICS II Assignment 2 Question 3: Problem 4.2 ~ V = u ˆ i = V 0 1 + 2 x L ˆ i Only x component for acceleration, i.e. a y = 0 and a z = 0 . a x = ∂u ∂t + u ∂u ∂x + v ∂u ∂y + w ∂u ∂z = V 0 1 + 2 x L ¶‚ 2 V 0 L a x = 2V 0 2 L 1 + 2 x L For L=6 in = 0.1524 m and V 0 = 10 ft/s = 3.048 m/s, Du Dt = 2(3 . 048) 2 0 . 1524 1 + 2 x 0 . 1524 = 121 . 92 1 + 2 x 0 . 1524 where x is in m. At x = 0, Du Dt = 121 . 92 m/s 2 . ( 12 g 0 s ) At x=L, Du Dt = 365 . 92 m/s 2 . ( 37 g 0 s ) Question 4: Problem 4.3 Two-dimensional velocity field given by ~ V = ( x 2 - y 2 + x ) ˆ i - (2 xy + y ) ˆ 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 ˆ i - 6 ˆ j

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A unit vector along a 40 line would be ˆ n = cos (40 ) ˆ i + sin (40 ) ˆ j Then the velocity component along a 40 line V 40 = ~ V . ˆ n = ( - 2 ˆ i - 6 ˆ j ) . ( cos (40 ) ˆ i + sin (40 ) ˆ j ) = 5 . 39 m/s
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complete_solution2_2 - FLUID MECHANICS II Assignment 2...

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