ch734 - Chapter 7: Flow Past Immerses Bodies 7.1 7.2 7.3...

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Chapter 7: Flow Past Immerses Bodies 7.1 Reynolds Number and Geometry Effects 7.2 Momenum Integral Estimates 7.3 The Boundary Layer Equations 7.4 The Flat Plate Boundary Layer 7.5 Boundary Layers with Pressure Gradients 7.6 Experimental External Flows

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Momentum Integral Equation – Laminar Flow 2(y/ δ ) – (y/ δ ) 2 μ ∂ u/ y at y=0
Von Karman and Poulhausen derived momentum integral equation (approximation) which can be used for both laminar and turbulent flow (with and without pressure gradient)

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Von Karman and Polhausen devised a simplified method by satisfying only the boundary conditions of the boundary layer flow rather than satisfying Prandtl’s differential equations for each and every particle within the boundary layer.
τ w = μ∂ u/ y = [ μ U/ δ ]d(u/U)/d η Want to solve for δ in Laminar Flow Assume velocity profile: u/U = f(y/ δ = η ), similar profiles 1. u U e at y = δ ; 2. u/ y 0 at y = δ 3. u = 0 at y = 0

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LAMINAR FLOW – NO PRESSURE GRADIENT
(plate is 2% thick, Re x=L = 10,000; air bubbles in water) For flat plate with dp/dx = 0, dU/dx = 0

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Realize (like Blasius) that u/U similar for all x when plotted as a function of y/ δ . Substitutions: η = y/ δ ; so dy = δ d η η =0 when y=0; η =1 when y= δ Not f(x)
Strategy: assume velocity profile: u/U o = f( η ), obtain an expression for τ w as a function of δ , and solve for δ = f ( η )

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Laminar Flow Over a Flat Plate, dp/dx = 0 Assume velocity profile: u = a + by + cy 2 B.C. at y = 0 u = 0 so a = 0 at y = δ u = U so U = b δ + c δ 2 at y = δ u/ y = 0 = b + 2c δ so b = -2c δ U = -2c δ 2 + c δ 2 = -c δ 2 so c = -U/ δ 2 δ u = a + by + cy 2 = 0 + 2Uy/ δ – Uy 2 / δ 2 u/U = 2 η - η 2 Want to know τ w ( x ) and δ
Laminar Flow Over a Flat Plate, dp/dx = 0 u/U = 2 η - η 2 Can find τ w ( x ) and δ

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τ w = 2 μ U/ δ ; u/U = 2 η - η 2 2 η - 4 η 2 + 2 η 3 - η 2 +2 η 3 - η 4 Strategy: obtain an expression for τ w as a function of δ , and solve for δ (x)
2 μ U/( δρ U 2 ) = (d δ /dx) ( η 2 – (5/3) η 3 + η 4 – (1/5) η 5 )| 0 1 2 μ U/( δρ U 2 ) = (d δ /dx) (1 – 5/3 + 1 – 1/5) = (d δ /dx) (2/15) 15 μ dx = δρ U(d δ ) Assuming δ = 0 at x = 0, then c = 0 δ 2 /2 = 15 μ x/( ρ U) Strategy: obtain an expression for τ w as a function of δ , and solve for δ (x)

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