Problem 5.68 - Problem 5.68[Difficulty 3 Given Find...

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Problem 5.68 [Difficulty: 3] Given: One-dimensional, incompressible flow through circular channel. Find: Solution: We will apply the particle acceleration definition to the velocity field  t V V V Dt V D a p Governing Equations: (Particle Accleration) CS CV A d V V d t 0 (Continuity equation) Assumptions: (1) Incompressible flow (2) One-dimensional flow Based on the above assumptions the continuity equation can provide the velocity at any location: uU A 1 A R 1 r 2 Now based on the geometry of the channel we can write rR 1 R 1 R 2 x L R 1 Δ R x L Therefore the flow speed is: R 1 2 R 1 Δ R x L 2 U 0 U 1 sin ω t () 1 Δ R R 1 x L 2 Based on the above assumptions the particle acceleration reduces to: i t u x u u a p ˆ
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Problem 5.68 - Problem 5.68[Difficulty 3 Given Find...

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