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PS4_sol_S08

# PS4_sol_S08 - Problem Set 4(2.006 Spring 08 Problem 1 Flow...

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Problem Set 4 (2.006, Spring 08) Problem 1 Flow past an infinite plate can be modeled using the steady two-dimensional form of the continuity and Navier-Stokes equations. 2 2 2 2 2 2 2 2 0 1 1 u x y u u P u u u x y x x y P u x y y x y υ υ ν ρ υ υ υ υ υ ν ρ + = + = − + + + = − + + From the problem statement we are told that the x-velocity, u , does not depend on the position x along the plate. Therefore, 0 u x = Using the continuity equation and the boundary condition for the y-velocity at the plate, we conclude that the y-velocity, υ , is constant and equal to the suction velocity, o υ . ) ( 0 x f y = = υ υ Apply the boundary condition: o υ υ = at all x. Since the y-velocity is constant the Navier-Stokes equation in the y-direction reduces to 0 = y P ) ( x f P = The pressure outside the boundary layer is uniform and equal to P atm . Thus, the pressure inside the boundary is P = P atm Plugging in the pressure, y-velocity and 0 u x = , the x- N.S. equation reduces to 2 2 y u y u o = ν υ 0 2 2 = y u y u o ν υ The general solution to the above differential equation is 2 1 exp ) ( C y C y u o + = ν υ The boundary conditions: 1) u(y →∞ ) = u o , and 2) u(y=0) = 0 allow us to determine C 1 and C 2 . C 2 = u o and C 1 = - u o = y u y u o o ν υ exp 1 ) ( b) The displacement thickness is defined as = δ δ 0 * ) ( 1 dy u y u o = = ν δ υ υ ν ν υ δ δ o o o dy y exp 1 exp 0 * The momentum thickness is defined as = δ θ 0 ) ( 1 ) ( dy u y u u y u o o

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= = ν δ υ ν δ υ υ ν ν υ ν υ θ δ o o o o o dy y y 2 exp 1 2 1 exp 1 exp 1 exp 0 We now have to find δ . At y = δ , u = u o . Therefore, δ→∞ , and we get o υ ν δ = * and o υ ν θ 2 = It is also acceptable to use the following definition: y = δ when u =0.99 u o . This gives us 99 . 0 exp 1 = ν δ υ o , and therefore, o υ ν δ 99 . 0 * = and o υ ν θ 49 . 0 c) The drag force on the plate is = = = = = bL u bL y u bL D o o y w ν υ μ μ τ 0 bL u o o υ ρ ( τ w is constant over x and b is the width of the plate into the page) You can also use the momentum equation to derive the drag force – it will give you the same answer.
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