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# ch06 - Problem 6.1[2 Given Velocity field Find Acceleration...

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Problem 6.1 [2] Given: Velocity field Find: Acceleration of particle and pressure gradient at (1,1) Solution: NOTE: Units of B are s -1 not ft -1 s -1 Basic equations For this flow u x y , ( ) A y 2 x 2 ( ) B x = v x y , ( ) 2 A x y B y + = a x u x u v y u + = A y 2 x 2 ( ) B x x A y 2 x 2 ( ) B x 2 A x y B y + ( ) y A y 2 x 2 ( ) B x + = a x B 2 A x + ( ) A x 2 B x + A y 2 + ( ) = a y u x v v y v + = A y 2 x 2 ( ) B x x 2 A x y B y + ( ) 2 A x y B y + ( ) y 2 A x y B y + ( ) + = a y B 2 A x + ( ) B y 2 A x y + ( ) 2 A y B x A x 2 y 2 ( ) + = Hence at (1,1) a x 1 2 1 1 + ( ) 1 s 1 1 2 1 1 + 1 1 2 + ( ) × ft s = a x 9 ft s 2 = a y 1 2 1 1 + ( ) 1 s 1 1 2 1 1 1 + ( ) × ft s 2 1 1 1 s 1 1 1 1 2 1 2 ( ) + × ft s = a y 7 ft s 2 = a a x 2 a y 2 + = θ atan a y a x = a 11.4 ft s 2 = θ 37.9 deg = For the pressure gradient x p ρ g x ρ a x = 2 slug ft 3 9 × ft s 2 lbf s 2 slug ft × = x p 18 lbf ft 2 ft = 0.125 psi ft = y p ρ g y ρ a y = 2 slug ft 3 32.2 7 ( ) × ft s 2 lbf s 2 slug ft × = y p 78.4 lbf ft 2 ft = 0.544 psi ft =

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