Problem 6.12 - p x ( ) p in ρ x x a x x ( ) ⌠ ⎮ ⌡ d...

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Problem 6.12 [Difficulty: 2] Given: Velocity field Find: Expression for acceleration and pressure gradient; plot; evaluate pressure at outlet Solution: Basic equations Given data U2 0 m s = L2 m = p in 50 kPa = ρ 900 kg m 3 = Here ux () Ue x L = u0 () 2 0 m s = uL ( ) 7.36 m s = The x component of acceleration is then a x x () ux () x ux () = a x x () U 2 e 2x L L = The x momentum becomes ρ u x u d d ρ a a = x p d d = The pressure gradient is then dp dx ρ L U 2 e 2x L = Integrating momentum
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Unformatted text preview: p x ( ) p in ρ x x a x x ( ) ⌠ ⎮ ⌡ d ⋅ − = p x ( ) p in U 2 ρ ⋅ e 2 x ⋅ L − 1 − ⎛ ⎜ ⎝ ⎞ ⎠ ⋅ 2 − = Hence p L ( ) p in U 2 ρ ⋅ e 2 − 1 − ( ) ⋅ 2 − = p L ( ) 206 kPa ⋅ = 0.5 1 1.5 2 50 100 150 200 x (m) dp/dx (kPa/m) 0.5 1 1.5 2 200 − 150 − 100 − 50 − x (m) ax (m/s2)...
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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Problem 6.12 - p x ( ) p in ρ x x a x x ( ) ⌠ ⎮ ⌡ d...

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