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Problem 6.14

Problem 6.14 - Problem 6.14[Difficulty 3 Given Velocity...

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Problem 6.14 [Difficulty: 3] Given: Velocity field Find: The acceleration at several points; evaluate pressure gradient Solution: The given data is q 2 m 3 s m = K 1 m 3 s m = ρ 1000 kg m 3 = V r q 2 π r = V θ K 2 π r = The governing equations for this 2D flow are The total acceleration for this steady flow is then r - component a r V r r V r V θ r θ V r + V θ 2 r = a r q 2 K 2 + 4 π 2 r 3 = θ - component a θ V r r V θ V θ r θ V θ + V r V θ r + = a θ 0 = Evaluating at point (1,0) a r 0.127 m s 2 = a θ 0 = Evaluating at point (1, π /2) a r 0.127 m s 2 = a θ 0 = Evaluating at point (2,0) a r 0.0158 m s 2 = a θ 0 = From Eq. 6.3, pressure gradient is
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