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16100lectre28_cg

16100lectre28_cg - Assume steady =0 t L Assume > 1 h V =0 x...

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Assume steady 0 = t Assume 1 >> h L 0 V x = K Assume 2-D 0 , 0 = = z w Incompressible N-S equations: 1. 0 = + y v x u 2. 2 2 2 2 1 u u u p u u u v t x y x x y ν ρ + + = − + + 3. 2 2 2 2 1 v v v p v v u v t x y y x y ν ρ + + = − + + BC’s w u h x u h x u h x v = + = = ± ) , ( 0 ) , ( 0 ) , ( Turning the crank: N 0 0 0 ( ) but 0 const Apply bc's 0 x u v v v v x x y y v v x v = + = = = = = = Now, momentum y : Since 0 = v , we have: 0 ( , ) ( ) p p x y p x y = = y 2h x y=-h y= h p L p R u w
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Lecture 28 16.100 2002 2 Note: since the pressure does change from to L R p p over the length , ( ) L p p x = . Finally momentum x : N N N N 2 2 2 2 0 0 0 0 2 2 1 1 ,where steady x t u u u dp u u u v t x y dx x y u dp y dx ν ρ µ ν µ ρ = = = = + + = − + + = −
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