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. 22 1 uuu p u u uv txy x x y ν ρ  ∂∂∂ ++= + +   3. 1 vvv p v v y x y + + ∂∂ BC’s w u h x u h x u h x v = + = = ± ) , ( 0 ) , ( 0 ) , ( Turning the crank: N 0 00 ( ) but 0 const Apply bc's 0 x v vv x xy y v v x v = += = =⇒= Now, momentum y : Since 0 = v , we have: 0( , ) ( ) p pxy px 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 LR p p over the length ,( ) Lp px = . Finally momentum x : N N N N 22 0 0 0 0 2 2 1 1 ,where steady x t uuu d p u u uv txy d x x y ud p yd x ν ρ µ µρ = =∂ = =   ∂∂∂ ++ = + +  ⇒= Observe that ) ( ) ( x g RHS y f LHS = = and L p p dx dp x g y f L R = = = = const const. ) ( ) ( For this problem, I’ll just use the gradient dx dp but realize this is specified by the end pressures.
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This note was uploaded on 11/06/2011 for the course AERO 100 taught by Professor Willcox during the Fall '03 term at MIT.

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

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