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VisFlow-Tut-SolnsQ1-10

# VisFlow-Tut-SolnsQ1-10 - ME2135 FLUID MECHANICS II Part 2...

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1 ME2135 FLUID MECHANICS II Part 2: Viscous Flow Tutorial Solutions 1. (a) The flow is unsteady because time t appears in the velocity components. The flow is 3-D because all three velocity components are non-zero. k a j a i a k dt dw j dt dv i dt du dt V d a z y x + + = + + = = where: ( ) ( ) ( ) x t x xz y t t tx x z u w y u v x u u t u dt du Dt Du a x 2 2 16 4 0 4 0 2 4 4 4 + = + + = + + + = = = ( ) ( ) ( ) y t ty xz t y t tx ty z v w y v v x v u t v dt dv Dt Dv a y 4 2 2 4 4 0 4 2 2 0 4 4 + = + + + = + + + = = = ( ) ( ) ( ) z x z x t x xz y t z tx z w w y w v x w u t w dt dw Dt Dw a z 2 2 16 16 4 4 0 2 4 4 0 + = + + + = + + + = = = ( ) ( ) ( ) k z x txz j y t ty i x t x dt V d a 2 4 2 16 16 4 4 16 4 + + + + + = = At ( ) ( ) 0 , 1 , 1 , , + = z y x : ( ) ( ) k j t t i t a 0 1 4 4 1 4 3 2 + + + + = Answer

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2 (b) Recalling that V = u i + v j + w k where: z c y b x a u 1 1 1 + + = z c y b x a v 2 2 2 + + = z c y b x a w 3 3 3 + + = For incompressible flow: 0 = V continuity or 0 = + + z w y v x u Substituting u , v and w into the continuity will yield: 0 3 2 1 = + + c b a Answer (c) Again applying the continuity condition: 0 = + + z w y v x u for incompressible flow in Cartesian coordinates Hence: ( ) 0 2 2 = + + z b y v ay ax x that is: ax y v y v ax 2 0 2 = = + Integrating partially with respect to y gives: ( ) ( ) t z x f axy t z y x v , , 2 , , , + = Answer
3 (d) A 2-D velocity field: ( ) ( ) j y x i x y x V + + = 2 2 2 x y x u + = 2 2 and ( ) y x v + = 2 { { j dt dv i dt du dt V d a y x a a + = = ( ) ( ) ( ) ( ) y y x x x y x y u v x u u t u Dt Du dt du a x 2 2 1 2 2 2 + + + = + + = = = ( ) ( ) ( ) y x x y x y v v x v u t v Dt Dv dt dv a y + + = + + = = = 2 2 2 2 At ( ) ( ) 2 , 1 , = y x : i a x 10 = and j a y 8 = Answer For steady, 2-D flow, the rate of change of pressure is: ( ) ( )( ) y y x x x y x y p v x p u t p Dt Dp dt dp 8 2 9 2 2 2 + + = + + = = At ( ) ( ) 2 , 1 , = y x : units 46 = Dt Dp Answer

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4 θ sin g θ θ dz ds s V streamline 2. Bernoulli’s Equation is for ideal fluids, that is, viscosity ν = 0 For ν = 0, the Navier-Stokes Equations become: 1 1 1 1 Bx B By Bz Du P f Dt x DV Dv P f P f Dt Dt y Du P f Dt z ρ ρ ρ ρ = Apply Euler’s Equation along a streamline in a steady, incompressible plane flow s V V = , where s = unit vector. θ sin = ds dz s s V V t V s V V t V Dt V D + = + = 0 (for steady flow) in s -direction: s P g s V V = ρ θ 1 sin s P s z g = ρ 1 that is: 0 1 = + + s z g s P s V V ρ or: 0 2 2 = + + gz P V s ρ or: = + + gz P V ρ 2 2 constant (Bernoulli’s Equation) (Euler’s Equation)
5 3. Let the flow be parallel to the x -axis. For a steady, 2-D, parallel flow (in Cartesian coordinate system): 2 2 y u dx dP = µ Since 0 = x u (from Continuity equation, that is, u is independent of x ): 2 2 2 2 dy u d y u = dx dP dy u d = 2 2 µ with the following boundary conditions: 0 = u at b y ± = (no-slip) 0 = dy du at 0 = y (symmetry of flow)

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