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Problem 6.12 - p x p in ฯ x x a x x โŒ  โŽฎ โŒ d โ‹… โˆ’ =...

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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 U 20 m s = L 2 m = p in 50 kPa = ρ 900 kg m 3 = Here u x ( ) U e x L = u 0 ( ) 20 m s = u L ( ) 7.36 m s = The x component of acceleration is then a x x ( ) u x ( ) x u x ( ) = a x x ( ) U 2 e 2 x 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 2 x L = Integrating momentum
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Unformatted text preview: p x ( ) p in ฯ x x a x x ( ) โŒ  โŽฎ โŒก d โ‹… โˆ’ = p x ( ) p in U 2 ฯ โ‹… e 2 x โ‹… L โˆ’ 1 โˆ’ โŽ› โŽœ โŽ โŽž โŽ  โ‹… 2 โˆ’ = Hence p L ( ) p in U 2 ฯ โ‹… e 2 โˆ’ 1 โˆ’ ( ) โ‹… 2 โˆ’ = p L ( ) 206 kPa โ‹… = 0.5 1 1.5 2 50 100 150 200 x (m) dp/dx (kPa/m) 0.5 1 1.5 2 200 โˆ’ 150 โˆ’ 100 โˆ’ 50 โˆ’ x (m) ax (m/s2)...
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