Boundary Layer Analysis
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1 Boundary Layer Analysis ME 322 Lecture Slides, Winter 27 Gerald Recktenwald February 1, 27 Associate Professor, Mechanical and Materials Engineering Department Portland State niversity, Portland, Oregon,
2 Displacement Thickness (1) Streamline h h + δ * x = coordinate measured from the leading edge δ is the amount by which the streamline just outside the boundary layer is displaced. Boundary Layer Analysis: February 1, 27 page 1
3 Displacement Thickness (2) Apply mass conservation to the control volume Z ρ( V ˆn)dA = CS Streamline h h + δ * x = coordinate measured from the leading edge Z h ρbdy + ρbh + Z h+δ Z h+δ = h = ρu(y)bdy = ρu(y)bdy = Z h+δ u(y)dy ( ) Boundary Layer Analysis: February 1, 27 page 2
4 Displacement Thickness (3) Continue... add and subtract to the integrand on the right hand side of Equation ( ). h = Z h+δ ( + u(y))dy = (h + δ ) + Z h+δ (u(y) )dy Solve for δ δ = 1 Z h+δ ( u(y))dy = Z h+δ 1 u(y) «dy Boundary Layer Analysis: February 1, 27 page 3
5 Displacement Thickness (4) The preceding analysis shows tht δ = Z h+δ 1 u(y) «dy Since u(y) = = constant outside the boundary layer, the upper limit is arbitrary as long as h and h + δ are outside the boundary layer. So, we can change the upper limit of integration to δ = Z 1 u(y) «dy Boundary Layer Analysis: February 1, 27 page 4
6 Scale Analysis for Laminar Boundary Layers (1) Assume the boundary layer is thin, i.e. assume δ L 1 The continuity equation requires that v is small, i.e. v δ L The x direction momentum equation requires that δ L Re 1/2 L Therefore δ L will be small if Re L is large. Generally we take Re L 1 as the minimum Re L for a boundary layer to exist. Boundary Layer Analysis: February 1, 27 page 5
7 Boundary Layer Flow Regimes Laminar Transition Turbulent x = coordinate measured from the leading edge L = total length of the plate Re x = ρx µ Re L = ρl µ The critical Reynolds number for transition from laminar to turbulent flow is Re cr Boundary Layer Analysis: February 1, 27 page 6
8 Integral Analysis for Laminar Boundary Layers (1) Karman Boundary Layer Analysis: February 1, 27 page 7
9 Integral Analysis for Laminar Boundary Layers (2) Derive momentum integral for flat plate MYO, Equation (9.22), p 52. D(x) = ρb Z δ(x) u( u)dy (1) von Kàrmàn wrote equation (1) as D(x) = ρb 2 θ (2) where θ = is called the momentum thickness. Z δ u 1 u «dy (3) θ is a measure of total plate drag. Note that θ has dimensions of length. Boundary Layer Analysis: February 1, 27 page 8
10 Integral Analysis for Laminar Boundary Layers (3) Since the plate is parallel to the on-coming flow, the drag is only due to wall shear stress D(x) = b Z x τ w (x) dx (4) Take derivatives of equation (4) and (2) dd dx = bτ w (5) Boundary Layer Analysis: February 1, 27 page 9
11 Integral Analysis for Laminar Boundary Layers (4) Assume is constant and take derivative of equation (2) dd dx = ρb 2 dθ dx (6) Combine equation (5) and equation (6) τ w = ρ 2 dθ dx constant laminar or turbulent (7) Boundary Layer Analysis: February 1, 27 page 1
12 Integral Analysis for Laminar Boundary Layers (5) Summary so far. We have the von Kàrmàn integral momentum equation τ w = ρ 2 dθ dx (7) Equation (7) relates the local wall shear stress to the local momentum thickness. Both τ w and θ vary with position along the plate. Equation (7) is a tool for analysis of flat plate boundary layers. All we need to do is make assumptions for the profile shape, i.e., u «y = fcn, and equation (7) will δ allow us to calculate τ w (x), and from there, D(x) and D total Boundary Layer Analysis: February 1, 27 page 11
13 Integral Analysis for Laminar Boundary Layers (6) Apply von Kàrmàn parabolic profile: Assume 1 u = 2y δ y2 δ y/δ u/ Substitute into definition of θ θ = Z δ 2 y δ y2 δ 2! 1 2 y! δ + y2 δ 2 dy = 2 15 δ Boundary Layer Analysis: February 1, 27 page 12
14 Integral Analysis for Laminar Boundary Layers (7) Substitute parabolic profile into definition of τ w τ w = µ u y y= u y = 2 τ w = 2µ δ δ 2y δ «= u = 2 y y= δ Boundary Layer Analysis: February 1, 27 page 13
15 Integral Analysis for Laminar Boundary Layers (8) Put the pieces back into equation (7) τ w = ρ 2 dθ dx = 2µ δ = ρ 2 d dx «2 15 δ Rearrange where ν = µ/ρ Integrate to get or δdδ = 15 ν dx 1 2 δ2 = 15νx «δ ν 1/2 x = 5.5 = 5.5Re 1/2 x x Boundary Layer Analysis: February 1, 27 page 14
16 Integral Analysis for Laminar Boundary Layers (9) Now we know u/ = fcn(y/δ). From this velocity profile we can compute the wall shear stress τ w = µ u = 2µ = (2µ) (5.5xRe 1/2 x ) y y= δ Make dimensionless as c f c f = τ 2 (1/2)ρ 2 = r 8 15 Re 1/2 x =.73 Re 1/2 x Boundary Layer Analysis: February 1, 27 page 15
17 Integral Analysis for Laminar Boundary Layers (1) Recall definition of displacement thickness δ = Z δ 1 u «dy So the displacement thickness for parabolic profile is δ = Z δ 1 2 y! δ + y2 δ 2 dy = δ 3 or δ x = 1.83 Re 1/2 x Boundary Layer Analysis: February 1, 27 page 16
18 Integral Analysis for Laminar Boundary Layers (11) Summary of results from von Kàrmàn integral analysis Boundary layer thickness Momentum thickness δ x = 5.48 Re 1/2 x θ x =.73 Re 1/2 x Friction coefficient c f =.73 Re 1/2 x Boundary Layer Analysis: February 1, 27 page 17
19 Blasius Analytical Solution for Laminar Boundary Layers (1) Boundary Layer Analysis: February 1, 27 page 18
Practice Problems on Boundary Layers. Answer(s): D = 107 N D = 152 N. C. Wassgren, Purdue University Page 1 of 17 Last Updated: 2010 Nov 22
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