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. 22 0 1 1 u xy uu P u u u x yx x y P u x yy x y υ υν ρ υυ ∂∂ += ⎛⎞ + + ⎜⎟ ⎝⎠ + + 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, , is constant and equal to the suction velocity, o . ) ( 0 x f y = = Apply the boundary condition: o = 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 0 2 2 = y u y u o The general solution to the above differential equation is 2 1 exp ) ( C y C y u o + = 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 exp 1 ) ( b) The displacement thickness is defined as = δ 0 * ) ( 1 dy u y u o = = o o o dy y exp 1 exp 0 * The momentum thickness is defined as = θ 0 ) ( 1 ) ( dy u y u u y u o o
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= = ν δ υ θ 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 = * and o 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 = o , and therefore, o 99 . 0 * = and o 49 . 0 c) The drag force on the plate is = = = = = bL u bL y u bL D o o y w μ τ 0 bL u o o ρ ( 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. Problem 2 Air at 20°C, 1 atm: ρ = 1.2 kg/m 3 , μ = 1.8 x 10 -5 Pa-s, ν = 1.5 x 10 -5 m 2 /s a) Since the logarithmic law is valid, B u r R u u + = κ * * ) ( ln 1 , where w u = * , 41 . 0 = and 0 . 5 = B At r = 0, u = s m u / 8 max = . Therefore, ( ) ( ) () 0 . 5 / 10 1.5 075 . 0
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This note was uploaded on 10/06/2011 for the course MECHANICAL 2.006 taught by Professor Blah during the Spring '08 term at MIT.

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

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