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Problem 2.34 - Problem 2.34[Difficulty 3 Given Velocity...

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Problem 2.34 [Difficulty: 3] Given: Velocity field Find: Equation for streamline through point (2.5); coordinates of particle at t = 2 s that was at (0,4) at t = 0; coordinates of particle at t = 3 s that was at (1,4.25) at t = 1 s; compare pathline, streamline, streakline Solution: Governing equations: For streamlines v u dy dx = For pathlines u p dx dt = v p dy dt = Assumption: 2D flow Given data a 2 = m s b 1 = 1 s x 0 2 = y 0 5 = x 1 = x x = For streamlines v u dy dx = b x a = So, separating variables a b dy x dx = Integrating a b y y 0 ( ) 1 2 x 2 x 0 2 = The solution is then y y 0 b 2 a x 2 x 0 2 + x 2 4 4 + = = Hence for pathlines u p dx dt = a = v p dy dt = b x = Hence dx a dt = dy b x dt = Integrating x x 0 a t t 0 ( ) = dy b x 0 a t t 0 ( ) + dt = y y 0 b x 0 t t 0 ( ) a 2 t 2 t 0 2 + a t 0 t t 0 ( ) = The pathlines are x x 0 a t t 0 ( ) + = y y 0 b x 0 t t 0 ( ) a 2 t 2 t 0 2 + a t 0 t t 0 ( ) + =
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For a particle that was at x 0 0 = m, y 0 4 = m at t 0 0 = s, at time t 2 = s we find the position is x x 0 a t t 0 ( ) + 4 = = m y y 0 b x 0 t t 0 ( ) a 2 t 2 t 0 2
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