Problem 5.48 - r e r R V V ˆ(Q.E.D Based on...

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Problem 5.48 [Difficulty: 2] Given: Incompressible, inviscid flow of air between parallel disks Find: a pr 3 10 3 m s 2   r e r R V V ˆ Solution: We will apply the conservation of mass and the definition of acceleration to the velocity.    0 1 1 t V z V V r rV r r z r Governing Equations: (Continuity Equation)  t V V V Dt V D a p (Particle acceleration) Assumptions: (1) Incompressible flow ( ρ is constant) (2) One-dimensional flow (velocity not a function of θ or z) (3) Flow is only in the r-direction (4) Steady flow (velocity is not a function of t) Based on the above assumptions, the continuity equation reduces to: 1 r r rV r  0 or rV r C Thus: V r C r should be the form of the solution. Now since at r = R: RV C it follows that: V r R r V or:  
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Unformatted text preview: r e r R V V ˆ (Q.E.D.) Based on assumptions (2) - (4), acceleration is radial only, and that acceleration is equal to: a pr V r r V r a pr V R r V R r 2 V 2 R R r 3 Therefore, at r = ri: a pr 15 m s 2 1 0.075 m 75 25 3 a pr 8.1 10 4 m s 2 Therefore, at r = R: a pr 15 m s 2 1 0.075 m 75 75 3 (a) simplified version of continuity equation valid in this flow field (b) show that the velocity is described by: (c) acceleration of a particle at r = r i , r = R...
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