Ch7Fall2012

058:0160 Jianming Yang Fall 2012 Chapter 7 1 Chapter 7: Flow Past Immersed Bodies Boundary layer flows: External flows around streamlined bodies at high Re have viscous (shear and no-slip) effects confined close to the body surfaces and its wake, but are nearly inviscid far from the body. Bluff body flows: Separation. 058:0160 Jianming Yang Fall 2012 Chapter 7 2 1 Flat-Plate Momentum Integral Analysis & Laminar approximate solution Consider flow of a viscous fluid at high Re past a flat plate, i.e., flat plate fixed in a uniform stream of velocity 𝑈. Boundary-layer thickness arbitrarily defined by 𝑦 = 𝛿99% , where 𝑢 = 0.99𝑈. Streamlines outside 𝛿99% will deflect an amount 𝛿 ∗ (the displacement thickness). Thus the streamlines move outward from 𝑦 = ℎ at 𝑥 = 0 to 𝑦 = 𝛿 = ℎ + 𝛿 ∗ at 𝑥 = 𝑥1 . 058:0160 Jianming Yang Chapter 7 3 Fall 2012 Conservation of mass: ℎ ∫0 𝜌𝐕 ∙ 𝐧𝑏𝑑𝑦 = 0 = ℎ − ∫0 𝜌𝑈𝑏𝑑𝑦 + ℎ+𝛿 ∗ 𝜌𝑢𝑏𝑑𝑦 ∫0 Assuming incompressible flow (constant density), this relation simplifies to 𝑈ℎ = ℎ+𝛿 ∗ (𝑈 ∫0 + 𝑢 − 𝑈)𝑑𝑦 = 𝑈(ℎ + 𝛿 ∗) ℎ+𝛿 ∗ (𝑢 + ∫0 − 𝑈)𝑑𝑦 We get the definition of displacement thickness: ∗ 𝛿 = ℎ+𝛿 ∗ 𝑈−𝑢 ( ) 𝑑𝑦 ∫0 𝑈 = ℎ+𝛿 ∗ (1 ∫0 𝑢 − ) 𝑑𝑦 𝑈 𝛿 ∗ ( a function of 𝑥 only) is an important measure of effect of BL on external flow. Conservation of x-momentum: ∑ 𝐹𝑥 = −𝐷 = ℎ ∫0 𝜌𝑢𝐕 2 𝐷 = 𝜌𝑏𝑈 ℎ − ∙ 𝐧𝑏𝑑𝑦 = ℎ − ∫0 𝜌𝑈𝑈𝑏𝑑𝑦 ℎ+𝛿 ∗ 2 𝜌𝑏 ∫0 𝑢 𝑑𝑦 + ℎ+𝛿 ∗ 𝜌𝑢𝑢𝑏𝑑𝑦 ∫0 Fluid force on plate ℎ+𝛿 ∗ 𝑢 Again assuming constant density and using continuity: ℎ = ∫0 𝑑𝑦 𝑈 ∗ ∗ ℎ+𝛿 ∗ 2 𝑢 2 ℎ+𝛿 𝑢 2 ℎ+𝛿 𝑢 𝐷 = 𝜌𝑏𝑈 ∫0 𝑑𝑦 − 𝜌𝑏 ∫0 𝑢 𝑑𝑦 = 𝜌𝑏𝑈 ∫0 (1 − ) 𝑑𝑦 𝑈 𝑈 𝑈 ℎ+𝛿 ∗ 𝑢 𝐷 𝑢 = 𝜃 = (1 − ) 𝑑𝑦 ∫ 0 𝜌𝑏𝑈 2 𝑈 𝑈 𝑥 = ∫0 𝜏𝑤 𝑏𝑑𝑥 𝜃 is the momentum thickness (a function of 𝑥 only), an important measure of the drag. 058:0160 Jianming Yang Chapter 7 4 Fall 2012 𝐶𝐷 = 1 𝐷 2 2𝜌𝑏𝑈 𝑥 = 2𝜃 𝑥 𝐶𝑓 = 1 𝑥 = ∫0 𝑥 𝑑 𝜏𝑤 2 (𝑥𝐶𝐷 ) = 2 𝑑𝑥 𝑑𝜃 𝑑𝑥 𝜏𝑤 = = 1 𝑥 𝑏𝑑𝑥 = ∫0 𝐶𝑓 𝑑𝑥 1 𝑥 𝜌𝑏𝑈2 𝑑𝜃 𝑑𝑥 𝐶𝑓 2 𝑑𝜃 𝜌𝑈 2 𝑑𝑥 Special case: 2D momentum integral equation for 𝑝𝑥 = 0 058:0160 Jianming Yang Chapter 7 5 Fall 2012 Simple velocity profile approximations: 𝑢 = 𝑈 ( 𝑢(0) = 0 𝑢(𝛿) = 𝑈 B.C.: 𝑑𝑢 𝑑𝑦 2𝑦 𝛿 − 𝑦2 𝛿2 ) no slip matching with outer flow (𝑦 = 𝛿 ) = 0 Use velocity profile to get 𝐶𝑓 (𝛿 ) and 𝜃(𝛿 ) and then integrate momentum integral equation to get 𝛿 (𝑅𝑒𝑥 ) ∗ 𝛿 = 𝜃= = ℎ+𝛿 ∗ (1 ∫0 𝛿 2𝑦 ∫0 ( 𝛿 𝛿𝑑𝛿 = 𝑥 − ) 𝑑𝑦 = 𝑈 ℎ+𝛿 ∗ 𝑢 (1 ∫0 𝑈 𝜏𝑤 = 𝜇 𝛿 𝑢 − 𝑑𝑢 − ) 𝑑𝑦 = 𝑈 𝑦2 4𝑦 2 𝛿 𝛿2 2 − | 𝑑𝑦 𝑦=0 15𝜇 𝜌𝑈 𝑢 =𝜇 −1⁄2 𝐶𝐷 (𝑥 = 𝐿) = + 2𝑦 3 𝛿3 𝑥 2𝜃 𝐿 + 2𝑦 3 𝛿3 2𝑦 𝛿 − − 𝐶𝑓 = 1 + 𝑦2 𝛿2 = 30𝜇 𝜌𝑈 1𝛿 3𝑥 𝑦2 𝑦2 𝛿 𝛿 2 ) 𝑑𝑦 = (𝑦 − ) (1 − 2𝑦 + 𝛿 𝑦4 𝑦2 𝛿 𝜏𝑤 𝛿 4 ) 𝑑𝑦 = ( 2 2𝜌𝑈 𝛿 𝛿∗ − 𝛿 2𝑦 ∫0 ( 𝛿 2𝑈 𝛿2 = 𝑑𝑥 = √30𝑅𝑒𝑥 𝛿 ∫0 (1 𝑥 = 30 = 𝜇 𝜌𝑈𝑥 4𝜇 𝜌𝑈𝛿 =2 𝑥2 = ≈ 1.83𝑅𝑒𝑥 −1⁄2 − 𝑦2 𝛿2 3𝛿 𝑑𝜃 30 𝑥 3𝛿 0 1 2 )| = 𝛿 𝑦4 𝑦5 𝛿 4 𝑑𝛿 5𝛿 2 + = 𝑥2 𝑅𝑒𝑥 𝜃 2 𝛿 = 𝛿 3 ) 𝑑𝑦 5𝑦 3 𝑑𝑥 + 𝑦3 15 𝑥 3 − 𝛿 4 )| = 0 2 15 𝛿 15 𝑑𝑥 𝑅𝑒𝑥 = 𝜌𝑈𝑥 𝜇 ≈ 0.73𝑅𝑒𝑥 −1⁄2 ≈ 1.46𝑅𝑒𝐿 −1⁄2 theses estimates are 6% away from the exact solution. 058:0160 Jianming Yang Chapter 7 6 Fall 2012 2 Boundary layer approximations, equations and comments 2D steady Navier-Stokes equation, 𝜌 = const., neglect 𝑔 𝜕𝑢 𝜕𝑣 + =0 𝜕𝑥 𝜕𝑦 𝜕𝑢 𝜕𝑢 𝑢 𝑢 𝜕𝑥 𝜕𝑣 𝜕𝑥 +𝑣 +𝑣 𝜕𝑦 𝜕𝑣 𝜕𝑦 =− =− 1 𝜕𝑝 𝜌 𝜕𝑥 1 𝜕𝑝 𝜌 𝜕𝑦 +𝜐( +𝜐( 𝜕2 𝑢 𝜕𝑥 2 𝜕2 𝑣 𝜕𝑥 2 + + 𝜕2 𝑢 𝜕𝑦 2 𝜕2 𝑣 𝜕𝑦 2 ) ) Introduce non-dimensional variables that includes scales so all variables are of 𝑂(1): 𝑅𝑒 = 𝑈𝐿 𝜐 𝑥 𝑦 𝑢 𝑣 , 𝑥 ∗ = , 𝑦 ∗ = √𝑅𝑒, 𝑢∗ = , 𝑣 ∗ = √𝑅𝑒, 𝑝∗ = 𝐿 𝐿 𝑈 𝑈 𝑝−𝑝0 𝜌𝑈 2 The NS equations become 𝜕𝑢∗ 1 𝑅𝑒 + 𝜕𝑣 ∗ =0 𝜕𝑥 ∗ 𝜕𝑦 ∗ ∗ ∗ ∗ 𝜕𝑢 ∗ 𝜕𝑢 𝑢 +𝑣 𝜕𝑥 ∗ 𝜕𝑦 ∗ ∗ ∗ ∗ 𝜕𝑣 ∗ 𝜕𝑣 (𝑢 𝜕𝑥 ∗ +𝑣 𝜕𝑦 ∗ =− )=− 𝜕𝑝 𝜕𝑥 ∗ 𝜕𝑝 𝜕𝑦 ∗ + + 1 𝜕 2 𝑢∗ 𝑅𝑒 𝜕𝑥 ∗ 2 1 𝜕2 𝑣 ∗ 𝑅𝑒 2 𝜕𝑥 ∗ 2 + + 𝜕 2 𝑢∗ 𝜕𝑦 ∗ 2 1 𝜕2 𝑣 ∗ 𝑅𝑒 𝜕𝑦 ∗ 2 For large Re (BL assumptions) the red terms drop out and the BL equations are obtained. Therefore, y-momentum equation reduces to 𝜕𝑝⁄𝜕𝑦 = 0, i.e., 𝑝 = 𝑝(𝑥, 𝑡) External flow is presumed to be known and irrotational, then from the Bernoulli equqtion 𝑝 + 12𝜌𝑈𝑒2 = const. ⇒ 𝑑𝑝⁄𝑑𝑥 = −𝜌𝑈𝑒 𝑑𝑈𝑒 ⁄𝑑𝑥 058:0160 Jianming Yang Chapter 7 7 Fall 2012 2D Boundary Layer equations: 𝜕𝑢 𝜕𝑣 + =0 𝜕𝑥 𝜕𝑦 𝜕𝑢 𝜕𝑢 𝑢 𝜕𝑥 +𝑣 𝜕𝑦 = 𝑈𝑒 𝑑𝑈𝑒 𝑑𝑥 +𝜐 𝜕2 𝑢 𝜕𝑦 2 Note: (1) 𝑈(𝑥, 𝑦), 𝑝(𝑥, 𝑦) imposed on BL by the external flow. 𝜕2 (2) = 0, i.e. streamwise diffusion is neglected. 𝜕𝑥 2 (3) Due to (2), the equations are parabolic in 𝑥. Physically, this means all downstream influences are lost other than that contained in external flow. A marching solution is possible. matching (4) Boundary conditions inlet δ No slip: 𝑢(𝑥, 𝑦 = 0) = 𝑣 (𝑥, 𝑦 = 0) = 0 Solution by Inlet condition: 𝑢(𝑥0 , 𝑦) given at 𝑥0 marching y ( ) Matching condition: 𝑢 𝑥, 𝑦 = ∞ = 𝑈(𝑥) x (5) When applying the boundary layer equations X0 one must keep in mind the restrictions No slip imposed on them due to the basic BL assumptions. Therefore they are not applicable for thick BL or separated flows (although they can be used to estimate occurrence of separation). 058:0160 Jianming Yang Chapter 7 8 Fall 2012 (6) Curvilinear coordinates Although BL equations have been written in Cartesian coordinates, they apply to curved surfaces provided 𝛿 ≪ 𝑅 and 𝑥, 𝑦 are curvilinear coordinates measured along and normal to the surface, respectively. In such a system we would find under the BL assumptions: 𝜕𝑝⁄𝜕𝑦 = 𝜌𝑢2 ⁄𝑅. Assume 𝑢 is a linear function of 𝑦: 𝑢 = 𝑈𝑦⁄𝛿 : 𝑑𝑝 𝑑𝑦 = 𝜌𝑈 2 𝑦 2 𝑅𝛿 2 ⇒ 𝑝(𝛿 ) − 𝑝(0) ∝ 𝜌𝑈 2 𝛿 3𝑅 or ∆𝑝 𝜌𝑈 2 ∝ 𝛿 3𝑅 therefore, we require 𝛿 ≪ 𝑅 (7) Practical use of the BL theory For a given body geometry: (a) Inviscid theory gives 𝑝(𝑥) → integration gives 𝐿, 𝐷 = 0 (b) BL theory gives → 𝛿 ∗ (𝑥 ), 𝜏𝑤 (𝑥 ), 𝜃(𝑥 ),etc. and predicts separation if any (c) If separation present then no further information → must use inviscid models, BL equation in inverse mode, or NS equation. (d) If separation is absent, integration of 𝜏𝑤 (𝑥 ) → frictional resistance body + 𝛿 ∗ , inviscid theory gives → 𝑝(𝑥), can go back to (2) for more accurate BL calculation including viscous – inviscid interaction 058:0160 Jianming Yang Chapter 7 9 Fall 2012 (8) Separation and shear stress At the wall, 𝑢 = 𝑣 = 0 → 𝜕2 𝑢 𝜕𝑦 2 = 1 𝜕𝑝 𝜇 𝜕𝑥 𝜕𝑢 1st derivative 𝑢 gives 𝜏𝑤 → 𝜏𝑤 = 𝜇 2nd derivative 𝑢 depends on 𝜕𝑝 𝜕𝑥 𝜕𝑦 , 𝜏𝑤 = 0: separation 058:0160 Jianming Yang Chapter 7 10 Fall 2012 3 Laminar Boundary Layer - Similarity Solution 2D, steady, incompressible: method of reducing PDE to ODE by appropriate similarity transformation 𝜕𝑢 + 𝜕𝑣 =0 𝜕𝑥 𝜕𝑦 𝜕𝑢 𝜕𝑢 𝑢 𝜕𝑥 +𝑣 𝜕𝑦 =𝑈 𝑑𝑈 𝑑𝑥 +𝜐 𝜕2 𝑢 𝜕𝑦 2 BC: No slip: 𝑢(𝑥, 𝑦 = 0) = 𝑣 (𝑥, 𝑦 = 0) = 0 Matching condition: 𝑢(𝑥, 𝑦 = 𝛿 (𝑥 )) = 𝑈(𝑥) Inlet condition: 𝑢(𝑥0 , 𝑦) given at 𝑥0 For Similarity 𝑢(𝑥,𝑦) 𝑈(𝑥) = 𝑓( 𝑦 ) 𝑔(𝑥) expect 𝑔(𝑥 )related to 𝛿 (𝑥 ) Or in terms of stream function 𝛹: 𝑢 = For similarity 𝛹 = 𝑈(𝑥 )𝑔(𝑥 )𝑓(𝜂 ) 𝑢= 𝜕𝛹 𝜕𝑦 ,𝑣 = − 𝜂= 𝜕𝑥 = −( 𝑑𝑈 𝑑𝑥 𝑔𝑓 + 𝑈 𝑑𝑔 𝑑𝑥 𝑓 − 𝑈𝑓 ′ 𝑑𝑔 𝑑𝑥 𝜕𝛹 𝜕𝑥 𝑦 𝑔(𝑥) = 𝑈𝑓 ′ 𝜕𝑦 𝜕𝛹 𝑣=− 𝜕𝛹 𝜂) 058:0160 Jianming Yang Chapter 7 11 Fall 2012 𝑢(𝑥, 𝑦 = 0) = 0 BC: 𝑈(𝑥 )𝑓 ′ (0) = 0 ⇒ 𝑣 (𝑥, 𝑦 = 0) = 0 ⇒ − ( 𝑑𝑈 ⇒ 𝑑𝑥 𝑑𝑈 𝑑𝑥 𝑓 ′ (0) = 0 ⇒ 𝑔(𝑥 )𝑓 (0) + 𝑈(𝑥 ) 𝑔(𝑥 )𝑓 (0) + 𝑈(𝑥 ) 𝑑𝑔 𝑑𝑥 𝑑𝑔 𝑑𝑥 𝑓(0) − 𝑈(𝑥 )𝑓 ′ 𝑓 (0) = 0 𝑢(𝑥, 𝑦 = ∞) = 𝑈(𝑥 ) ⇒ 𝑈(𝑥 )𝑓 ′ (∞) = 𝑈(𝑥 ) Write boundary layer equations in terms of 𝛹 𝜕𝑢 𝑢 𝜕𝑥 +𝑣 𝜕𝑢 𝜕𝑦 =𝑈 𝑑𝑈 +𝜐 𝑑𝑥 𝜕2 𝑢 𝜕𝛹 𝜕2 𝛹 ⇒ 𝜕𝑦 2 𝜕𝑦 𝜕𝑦𝜕𝑥 − 𝜕𝑥 𝜕𝑦𝜕𝑦 𝑑𝑥 0) = 0 𝑓(0) = 0 ⇒ 𝑓 ′ (∞) = 1 ⇒ 𝜕𝛹 𝜕2 𝛹 𝑑𝑔 =𝑈 𝑑𝑈 𝑑𝑥 +𝜐 𝜕3 𝛹 𝜕𝑦𝜕𝑦𝜕𝑦 Substitute 𝜕2 𝛹 𝜕𝑦𝜕𝑦 = 𝑈𝑓 ′′ ⁄ 𝑔 𝜕3 𝛹 𝜕𝑦𝜕𝑦𝜕𝑦 = 𝑈𝑓 𝜕2 𝛹 ′′′ ⁄ 2 𝑔 𝜕𝑦𝜕𝑥 = 𝑑𝑈 𝑑𝑥 𝑓 ′ − 𝑈𝑓 ′′ 𝜂 𝑑𝑔 𝑑𝑥 ⁄𝑔 Assemble them together: 𝑑𝑈 ′ 𝑑𝑔 𝜂 𝑑𝑈 𝑑𝑔 𝑑𝑔 𝑈𝑓′′ ′ ′′ ′ (𝑈𝑓 ) ( 𝑓 − 𝑈𝑓 ) − ( 𝑔𝑓 + 𝑈 𝑓 − 𝑈𝑓 𝜂) ( ) 𝑑𝑥 𝑑𝑥 𝑔 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑔 𝑑𝑈 𝑑𝑈 𝑑𝑔 1 𝑑𝑈 1 2 𝑈 𝑓 ′ − 𝑈 𝑓𝑓 ′′ − 𝑈 2 𝑓𝑓 ′′ = 𝑈 + 𝜐𝑈𝑓 ′′′ 2 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑔 𝑑𝑥 𝑔 𝑑𝑈 ′ 2 𝑈 𝑑 (𝑔𝑈) 𝑑𝑈 ′′ ′′′ 1 𝑈 𝑓 𝑑𝑥 ′′′ 𝑓 + − 𝑔 𝑑𝑥 𝑔 𝑑 (𝑔𝑈) 𝑓𝑓 ′′ 𝜐 𝑑𝑥 𝑓𝑓 = 𝑈 + 𝑔2 𝑑𝑈 𝜐 𝑑𝑥 𝑑𝑥 + 𝜐𝑈𝑓 2 =𝑈 𝑑𝑈 𝑑𝑥 +𝜐( 𝑈𝑓′′′ 𝑔2 ) 𝑔2 (1 − 𝑓 ′ ) = 0 or 2 𝑓 ′′′ + 𝐶1 𝑓𝑓 ′′ + 𝐶2 (1 − 𝑓 ′ ) = 0 Where for similarity 𝐶1 and 𝐶2 are constants or functions of 𝜂 only. 058:0160 Jianming Yang Chapter 7 12 Fall 2012 The Blasius Solution for Flat-Plate Flow 𝑈 = const. 𝐶1 = 𝑔 𝑑 (𝑔𝑈) 𝜐 𝑑𝑥 ⇒ = 𝑑𝑈 =0 𝑑𝑥 2 𝑔 𝑑𝑈 𝑈 𝑑𝑔 𝜐 𝑑𝑥 1 + 𝜐 𝑑𝑥 ⇒ 𝑔= 𝑈 𝑑𝑔 𝜐 𝑑𝑥 𝐶2 = 𝑔 =0 𝜐 𝑑𝑥 2 𝑑(𝑔 ) 2𝐶1 𝜐 For simplicity, let 𝐶1 = , then 𝑔(𝑥 ) = √𝜐𝑥 ⁄𝑈 2 Blasius equations for Flat Plate Boundary Layer 1 𝑔2 𝑑𝑈 𝑑𝑥 = 𝑈 ⇒ 𝑔(𝑥 ) = ( 2𝐶1 𝜐𝑥 1⁄2 𝑈 ) 𝜂 = 𝑦√𝑈⁄(𝜐𝑥 ) 𝑓 ′′′ + 𝑓𝑓 ′′ = 0 2 ′ 𝑓(0) = 𝑓 (0) = 0 𝑓 ′ (∞) = 1 Third-order nonlinear ODE for 𝑓, solution can be obtained from numerical integration. 058:0160 Jianming Yang 𝑢 𝑈 Chapter 7 13 Fall 2012 = 0.99 when 𝜂 = 𝑦√ 𝑈 𝜐𝑥 ≈ 5.0 𝛿 𝑥 ∗ 𝛿 = ∞ ∫0 (1 ∞𝑢 𝜃 = ∫0 𝑈 𝑢 − ) 𝑑𝑦 = 𝑈 𝛿 ∫0 (1 = 𝑦 𝑥 𝑦 𝑥 ′) 𝑈𝑥 √ ≈ 𝜐 ≈ 5.0 𝑅𝑒𝑥 = 5.0 √𝑅𝑒𝑥 − 𝑓 𝑑 (𝜂√ ) ⇒ 𝑈 𝜐𝑥 (1 − ) 𝑑𝑦 = ∫0 𝑓 ′ (1 − 𝑓 ′ )𝑑 (𝜂√ ) 𝑈 𝜐 (Blasius, 1908) 𝜐𝑥 𝛿 𝑢 𝑈𝑥 𝑈 ⇒ 𝛿∗ 𝑥 𝜃 𝑥 ≈ ≈ 1.721 √𝑅𝑒𝑥 0.664 √𝑅𝑒𝑥 The ratio of displacement to momentum thickness, 𝐻 = 𝛿 ∗ ⁄𝜃 = 1.721 0.664 = 2.59 which is called the dimensionless-profile shape factor A large shape factor implies that boundary layer is about to separate. 𝜏𝑤 = 𝜇 𝐶𝑓 = 1 𝑑𝑢 | =𝜇 𝑑𝑦 𝑦=0 𝜏𝑤 0.664 2 2𝜌𝑈 𝐶𝐷 = 1 ≈ 𝐷 2 2𝜌𝑏𝑈 𝑥 𝑣= 𝑑𝑔 𝜂𝑓 ′ 𝑑𝑥 √𝑅𝑒𝑥 = − 2𝜃 𝑥 𝑑𝑔 𝑑𝑥 𝑈𝑓′′ (0) √𝜐𝑥⁄𝑈 = 𝜃 𝑥 = 2𝐶𝑓 ≈ 𝑓= 𝜂𝑓′ −𝑓 √2𝑅𝑒𝑥 1.328 √𝑅𝑒𝑥 ≪1 for 𝑅𝑒𝑥 ≫ 1 𝑔(𝑥 ) = √𝜐𝑥 ⁄𝑈 058:0160 Jianming Yang Chapter 7 14 Fall 2012 4 Momentum Integral Equation Exact solutions of the boundary-layer equations are possible only in simple cases. In more complicated problems, approximate methods satisfy only an integral of the boundary-layer equations across the layer thickness. When this integration is performed, the resulting ordinary differential equation involves the boundary layer’s displacement and momentum thicknesses, and its wall shear stress. 4.1 Momentum integral equation 𝑢 Momentum equation: 𝜕𝑢 𝜕𝑥 +𝑣 𝜕𝑢 𝜕𝑦 =− 1 𝜕𝑝 𝜌 𝜕𝑥 1 𝜕𝜏 + 𝜌 𝜕𝑦 The pressure gradient is evaluated form the outer potential flow using Bernoulli equation 𝑝 + 12𝜌𝑈 2 = const. 𝑢 𝜕𝑢 𝜕𝑥 +𝑣 𝜕𝑢 (𝑢 − 𝑈) ( (𝑢 − 𝜕𝑢 +𝑣 𝜕𝑥 1 𝜕𝜏 𝜌 𝜕𝑦 −𝑈 𝜕𝑦 𝜕𝑢 𝜕𝑥 𝜕𝑢 𝜕𝑦 + 𝑑𝑥 𝜕𝑣 𝜕𝑦 −𝑈 = −2𝑢 = 𝑑𝑈 𝜕 𝜕𝑥 𝜕𝑢 𝜕𝑥 − 1 𝜕𝜏 𝜌 𝜕𝑦 𝜕𝑢 )=𝑢 𝑑𝑈 𝑑𝑥 − −𝑣 = −𝜌𝑈 𝑑𝑥 𝑑𝑈 𝑑𝑥 =0 +𝑢 𝜕𝑥 1 𝜕𝜏 𝜌 𝜕𝑦 𝜕𝑢 𝜕𝑦 𝑑𝑝 ⇒ 𝜕𝑣 𝜕𝑦 −𝑈 ) + (𝑢 +𝑈 𝑑𝑈 𝑑𝑥 𝜕𝑢 𝜕𝑢 𝜕𝑥 +𝑢 𝜕𝑥 𝜕𝑣 −𝑢 (𝑢𝑈 − 𝑢2 ) + (𝑈 − 𝑢) 𝜕𝑦 𝑑𝑈 𝑑𝑥 −𝑈 𝜕𝑣 + 𝜕𝑦 −𝑈 𝜕𝑦 +𝑈 𝜕𝑣 𝜕𝑢 𝜕𝑥 𝜕 𝜕𝑦 =0 𝜕𝑢 𝜕𝑥 +𝑈 −𝑈 𝜕𝑣 𝜕𝑦 (𝑣𝑈 − 𝑣𝑢) 𝜕𝑣 𝜕𝑦 )=0 058:0160 Jianming Yang ∞ Chapter 7 15 Fall 2012 1 𝜕𝜏 1 ∫0 − 𝜌 𝜕𝑦 𝑑𝑦 = − 𝜌 (𝜏∞ − 𝜏𝑤 ) = = = ∞ 𝑑 𝑑𝑈 𝜏𝑤 𝜌 ∞ ∞ 𝜕 ∫ 𝑢(𝑈 − 𝑢)𝑑𝑦 + 𝑑𝑥 ∫0 (𝑈 − 𝑢)𝑑𝑦 + ∫0 𝑑𝑥 0 ∞ 𝑑 𝑑𝑈 𝜕𝑦 (𝑣𝑈 − 𝑣𝑢)𝑑𝑦 ∞ ∞ ( ) ( ) ( )| 𝑢 𝑈 − 𝑢 𝑑𝑦 + 𝑈 − 𝑢 𝑑𝑦 + 𝑣𝑈 − 𝑣𝑢 ∫ ∫ 0 𝑑𝑥 0 𝑑𝑥 0 𝜏𝑤 𝜌 𝜏𝑤 𝜌 𝜏𝑤 𝜌 = = = ∞ ∫ 𝑢(𝑈 𝑑𝑥 0 𝑑 − 𝑢)𝑑𝑦 + ∞𝑢 𝑑 𝑑𝑥 𝑑 𝑑𝑥 [𝑈 2 ∫0 𝑈 𝑢 ∞ ∫ (𝑈 𝑑𝑥 0 𝑑𝑈 𝑑𝑈 (1 − ) 𝑑𝑦] + (𝑈 2 𝜃) + 𝑈𝛿 𝑈 ∗ 𝑑𝑈 𝑑𝑥 − 𝑢)𝑑𝑦 ∞ 𝑢 𝑈 ∫0 (1 − ) 𝑑𝑦 𝑈 𝑑𝑥 von Karman boundary-layer momentum integral equation, which is valid for steady laminar boundary layers and for time-averaged flow in turbulent boundary layers. It is a single ordinary differential equation that relates three unknowns 𝜃, 𝛿 ∗ , and 𝜏𝑤 , so additional assumptions must be made or correlations provided to obtain solutions for these parameters. 𝜏𝑤 𝜌 𝜏𝑤 = 𝑈2 𝜌𝑈 2 𝜏𝑤 𝜌𝑈 2 = = 𝐶𝑓 2 𝐶𝑓 2 𝑑𝜃 𝑑𝑥 = = + 2𝑈𝜃 𝑑𝜃 𝑑𝑥 𝑑𝜃 𝑑𝑥 𝑑𝑈 𝑑𝑥 + 𝑈𝛿 ∗ + (2 + 𝑑𝑈 𝑑𝑥 𝛿 ∗ 𝜃 𝑑𝑈 𝜃 ) + (2 + 𝐻 ) 𝑈 𝑑𝑥 𝜃 𝑑𝑈 𝑈 𝑑𝑥 = 𝑈2 𝑑𝜃 𝑑𝑥 𝐻= + 𝑈(2𝜃 + 𝛿 ∗ ) 𝛿∗ 𝜃 𝑑𝑈 𝑑𝑥 058:0160 Jianming Yang Chapter 7 16 Fall 2012 Historically two approaches for solving the momentum integral equation for specified potential flow 𝑈(𝑥): 1. Guessed Profiles 2. Empirical Correlations Best approach is to use empirical correlations to get integral parameters (𝛿, 𝛿 ∗ , 𝜃, 𝐻, 𝐶𝑓 , 𝐶𝐷 ) after which use these to get velocity profile 𝑢/𝑈. 4.2 Thwaites Method Multiply momentum integral equation by 𝜏𝑤 𝜌 = 𝑑 𝑑𝑥 𝑑𝑈 𝑈𝛿 ∗ 𝑑𝑥 2 (𝑈 𝜃) + 𝜏𝑤 𝜃 ⇒ 𝜇𝑈 𝜃 𝜐𝑈 = 𝜃𝑈 𝑑𝜃 𝜐 𝑑𝑥 + 𝜃 2 𝑑𝑈 𝜐 𝑑𝑥 (2 + 𝐻 ) LHS and 𝐻 are dimensionless and can be correlated with pressure gradient parameter 𝜆 = 𝜃 2 𝑑𝑈 𝜐 𝑑𝑥 as shear and shape-factor correlations 𝜃𝑈 𝑑𝜃 Note 𝜐 𝑑𝑥 1 𝑑 2 𝑑𝑥 = 𝑈 𝜃2 ( ) 𝜐 Substitute above into momentum integral equation 1 𝑑 2 𝑑𝑥 𝑆(𝜆) = 𝑈 𝜃2 ( ) + 𝜆(2 + 𝐻(𝜆)) 𝜐 2[𝑆(𝜆) − (2 + 𝐻(𝜆))𝜆] = 𝑈 𝑑 𝑑𝑥 𝜃2 ( ) 𝜐 058:0160 Jianming Yang Chapter 7 17 Fall 2012 2[𝑆(𝜆) − (2 + 𝐻(𝜆))𝜆] = 𝑈 𝑑 𝜆 ) 𝑑𝑥 𝑑𝑈⁄𝑑𝑥 ( = 𝐹 (𝜆) 𝐹 (𝜆) = 0.45 − 6𝜆 based on AFD and EFD Define 𝑧 = 𝑈 𝑈 𝑑𝑧 𝑑𝑥 𝑑𝑧 𝑑𝑥 6 𝜃2 𝜐 so that 𝜆 = 𝑧 𝑑𝑈 𝑑𝑥 = 0.45 − 6𝜆 = 0.45 − 6𝑧 + 6𝑧 𝑑𝑈 𝑑𝑥 𝑑𝑈 𝑑𝑥 1 𝑑 = 0.45 𝑈 5 𝑑𝑥 𝑥 (𝑧𝑈 6 ) = 0.45 𝑧𝑈 = 0.45 ∫0 𝑈 5 𝑑𝑥 + 𝐶 𝜃 2 = 𝜃02 + 𝑥 5 ∫ 𝑈 𝑑𝑥 𝑈6 0 0.45𝜐 𝜃0 (𝑥 = 0) = 0 and 𝑈(𝑥) from potential flow solution Complete solution: 𝜆 = 𝜆(𝜃) = 𝜃 2 𝑑𝑈 𝜐 𝑑𝑥 , 𝜏𝑤 𝜃 𝜇𝑈 = 𝑆(𝜆) , 𝛿 ∗ = 𝜃𝐻(𝜆) Thwaites (1949), 𝑚 = −𝜆 Overall, the accuracy of Thwaites’ method is ±3% or so for favorable pressure gradients, and ±10% for adverse pressure gradients but perhaps slightly worse near boundary-layer separation. The great strength of Thwaites’ method is that it involves only one parameter (λ) and requires only a single integration. This simplicity makes it ideal for preliminary engineering calculations that are likely to be followed by more formal computations or experiments. 058:0160 Jianming Yang Fall 2012 Chapter 7 18 058:0160 Jianming Yang Fall 2012 Chapter 7 19 058:0160 Jianming Yang Chapter 7 20 Fall 2012 𝑦 𝑦3 𝛿 𝛿3 𝑢 = 𝑈 (2 − 2 + 𝑦4 𝛿4 ) 𝑢(0) = 0 no slip 𝑢(𝛿) = 𝑈 matching with outer flow B.C.: 𝛿 ∗ 3 4 2 4 5 ℎ+𝛿 𝛿 𝑢 𝑦 𝑦 𝑦 𝑦 𝑦 𝑦 3 𝛿 ∗ = ∫0 (1 − ) 𝑑𝑦 = ∫0 (1 − 2 + 2 3 − 4 ) 𝑑𝑦 = (𝑦 − + 3 − 4)| = 𝛿 𝑈 𝛿 𝛿 𝛿 𝛿 2𝛿 5𝛿 10 0 ∗ 3 4 3 4 ℎ+𝛿 𝑢 𝛿 𝑢 𝑦 𝑦 𝑦 𝑦 𝑦 𝑦 𝜃 = ∫0 (1 − ) 𝑑𝑦 = ∫0 (2 − 2 3 + 4 ) (1 − 2 + 2 3 − 4 ) 𝑑𝑦 𝑈 𝑈 𝛿 𝛿 𝛿 𝛿 𝛿 𝛿 3 4 2 4 5 4 6 7 𝛿 𝑦 𝑦 𝑦 𝑦 𝑦 𝑦 𝑦 𝑦 𝑦 𝑦5 𝑦7 𝑦8 = ∫0 (2 − 2 3 + 4 − 4 2 + 4 4 − 2 5 + 4 4 − 4 6 + 2 7 − 2 5 + 2 7 − 8 ) 𝑑𝑦 𝛿 𝛿 𝛿 𝛿 𝛿 𝛿 𝛿 𝛿 𝛿 𝛿 𝛿 𝛿 2 3 4 5 6 7 8 𝛿 𝑦 𝑦 𝑦 𝑦 𝑦 𝑦 𝑦 𝑦 = ∫0 (2 − 4 2 − 2 3 + 9 4 − 4 5 − 4 6 + 4 7 − 8 ) 𝑑𝑦 𝛿 𝛿 𝛿 𝛿 𝛿 𝛿 𝛿 𝛿 𝛿 𝑦2 4 𝑦3 1 𝑦4 9 𝑦5 2 𝑦6 4 𝑦7 1 𝑦8 1 𝑦9 37 =( − 𝛿 𝜏𝑤 = 𝜇 𝛿𝑑𝛿 = 𝛿 𝑥 3 𝛿2 𝑑𝑢 | 𝑑𝑦 𝑦=0 630𝜇 37𝜌𝑈 1260 =√ 𝐶𝑓 = − 2 𝛿3 = 𝑑𝑥 1 37 𝑅𝑒𝑥 1⁄2 𝜃 37 𝛿 𝑥 = 315 𝑥 ≈ + − − + − )| = 𝛿 3 𝛿5 7 𝛿6 2 𝛿7 9 𝛿8 0 315 2𝑈 𝑑𝜃 37 𝑑𝛿 𝜇 = 𝜌𝑈 2 = 𝜌𝑈 2 𝛿 𝑑𝑥 315 𝑑𝑥 1260𝜇 1260 𝜇 1260 1 𝛿2 = 𝑥= 𝑥2 = 𝑥2 37𝜌𝑈 37 𝜌𝑈𝑥 37 𝑅𝑒𝑥 ≈ 5 𝛿4 5.83 𝑅𝑒𝑥 1⁄2 0.685 𝑅𝑒𝑥 1⁄2 𝛿∗ 𝑥 = 𝐻= 3 𝛿 10 𝑥 𝛿∗ 𝜃 ≈ 1.751 𝑅𝑒𝑥 1⁄2 ≈ 2.554. 𝑅𝑒𝑥 = 𝜌𝑈𝑥 𝜇 058:0160 Jianming Yang Fall 2012 Chapter 7 21 5 Turbulent Boundary Layer 5.1 Transition to Turbulence Chapter 6 described the transition process as a succession of Tollmien-Schlichting waves, development of Λ - structures, vortex decay and formation of turbulent spots as preliminary stages to fully turbulent boundary-layer flow. The phenomena observed during the transition process are similar for the flat plate boundary layer and for the plane channel flow, as shown in the following figure based on measurements by M. Nishioka et al. (1975). Periodic initial perturbations were generated in the BL using an oscillating cord. For typical commercial surfaces transition occurs at 𝑅𝑒𝑥,𝑡𝑟 = 5 × 105 . However, one can delay the transition to 𝑅𝑒𝑥,𝑡𝑟 = 3 × 106 with care in polishing the wall. 058:0160 Jianming Yang Chapter 7 22 Fall 2012 5.2 Reynolds Averaged 2D Boundary Layer Equations 𝑢̃𝑖 = 𝑈𝑖 + 𝑢𝑖 𝑝̃ = 𝑃 + 𝑝 Substituting 𝑢̃𝑖 into continuity equation and taking the time average we obtain, 𝜕𝑈𝑗 𝜕𝑥𝑗 𝜕𝑢𝑗 =0 𝜕𝑥𝑗 =0 Similarly for the momentum equations and using continuity (neglecting 𝑔), 𝜕𝑈𝑖 𝜕𝑡 𝜕𝑈𝑖 + 𝑈𝑗 𝜕𝑥𝑗 =− 1 𝜕𝑃 𝜌 𝜕𝑥𝑖 + 1 𝜕 𝜌 𝜕𝑥𝑗 𝜏̅𝑖𝑗 = − 1 𝜕𝑃 𝜌 𝜕𝑥𝑖 + 1 𝜕 𝜌 𝜕𝑥𝑗 [𝜇 ( 𝜕𝑈𝑖 𝜕𝑥𝑗 + 𝜕𝑈𝑗 𝜕𝑥𝑖 ) − 𝜌𝑢 ̅̅̅̅̅] 𝑖 𝑢𝑗 a. 𝛿 (𝑥 ) ≪ 𝑥, which means 𝑉 ≪ 𝑈, 𝜕⁄𝜕𝑥 ≪ 𝜕⁄𝜕𝑦 b. mean flow structure is two-dimensional: 𝑊 = 0, 𝜕⁄𝜕𝑧 = 0 Note the mean lateral turbulence is actually not zero, 𝑤𝑤 ̅̅̅̅̅ ≠ 0, but its 𝑧 derivative is assumed to vanish. Then, we get the following BL equations for incompressible steady flow: Assume 𝜕𝑈 + 𝜕𝑉 𝜕𝑥 𝜕𝑦 𝜕𝑈 𝑈 𝜕𝑃 𝜕𝑦 𝜕𝑥 =0 +𝑉 ≈ −𝜌 𝜕𝑈 𝜕𝑦 ̅̅̅̅ 𝜕𝑣𝑣 𝜕𝑦 Continuity ≈ 𝑈𝑒 𝑑𝑈𝑒 𝑑𝑥 + 1 𝜕𝜏 𝜌 𝜕𝑦 x-momentum y-momentum Where 𝑈𝑒 is the free-stream velocity and 𝜏 = 𝜇 𝜕𝑈 𝜕𝑦 − 𝜌𝑢𝑣 ̅̅̅̅. 058:0160 Jianming Yang Chapter 7 23 Fall 2012 Note:  The equations are solved for the time averages 𝑈 and 𝑉  The shear stress now consists of two parts: 1. first part is due to the molecular exchange and is computed from the timeaveraged field as in the laminar case; 2. The second part appears additionally and is due to turbulent motions.  The additional term is new unknown for which a relation with the average field of the velocity must be constructed via a turbulence model. Integrate y- momentum equation across the boundary layer 𝑃 ≈ 𝑃𝑒 (𝑥 ) − 𝜌𝑣𝑣 ̅̅̅ So, unlike laminar BL, there is a slight variation of pressure across the turbulent BL due to velocity fluctuations normal to the wall, which is no more than 4% of the stream velocity and thus can be neglected. The Bernoulli relation is assumed to hold in the inviscid free-stream: 𝑑𝑃𝑒 𝑑𝑥 = −𝜌𝑈𝑒 𝑑𝑈𝑒 𝑑𝑥 Assume the free stream conditions, 𝑈𝑒 (𝑥 ) is known, the boundary conditions: No slip: 𝑈(𝑥, 0) = 𝑉 (𝑥, 0) = 0 Free stream matching: 𝑈(𝑥, 𝛿 ) = 𝑈𝑒 (𝑥 ) 058:0160 Jianming Yang Chapter 7 24 Fall 2012 5.3 Momentum Integral Equations The momentum integral equation has the identical form as the laminar-flow relation: 𝜏𝑤 𝜌𝑈𝑒2 = 𝐶𝑓 2 = 𝑑𝜃 𝑑𝑥 + (2 + 𝐻 ) 𝜃 𝑑𝑈𝑒 𝑈𝑒 𝑑𝑥 𝐻= 𝛿∗ 𝜃 For laminar flow: (𝐶𝑓 , 𝐻, 𝜃) are correlated in terms of simple parameter 𝜆 = 𝜃 2 𝑑𝑈𝑒 𝜐 𝑑𝑥 For Turbulent flow: (𝐶𝑓 , 𝐻, 𝜃) cannot be correlated in terms of a single parameter. Additional parameters and relationships are required that model the influence of the turbulent fluctuations. There are many possibilities all of which require a certain amount of empirical data. 5.4 Flat Plate Boundary Layer (Zero Pressure Gradient) 5.4.1 Log Law Analysis of Smooth Flat Plate Assume log-law can be used to approximate turbulent velocity profile and use to get 𝐶𝑓 = 𝐶𝑓 (𝛿 ) relationship 𝑈 𝑢∗ 1 𝑦𝑢∗ 𝜅 𝜐 = ln +𝐵 At 𝑦 = 𝛿 (edge of boundary layer) 𝑈𝑒 𝑢∗ 1 𝛿𝑢∗ 𝜅 𝜐 = ln +𝐵 where 𝜅 = 0.41 and 𝐵 = 5 058:0160 Jianming Yang Chapter 7 25 Fall 2012 However: 𝑈𝑒 𝑢∗ = 𝛿𝑢∗ 𝜐 𝑈𝑒 √𝜏𝑤 ⁄𝜌 = = 𝛿𝑈𝑒 𝑢∗ 𝜐 𝑈𝑒 1⁄2 2 1 ( ) 𝑈𝑒 √(12𝜌𝑈2𝑒 𝐶𝑓 )⁄𝜌 =( ) 𝐶𝑓 𝐶𝑓 1⁄2 = 𝑅𝑒𝛿 ( ) 2 𝐶𝑓 1⁄2 = ln [𝑅𝑒𝛿 ( ) 𝐶𝑓 1⁄2 2 𝜅 2 ]+𝐵 Skin friction law for turbulent flat-plate flow Following a suggestion of Prandtl, we can forget the complex log law and simply use a power-law approximation: ⁄ 𝐶𝑓 ≈ 0.02𝑅𝑒𝛿−1 6 Use 𝑈⁄𝑈𝑒 Profile to Get 𝜃, 𝐶𝑓 , 𝛿 , 𝛿 ∗ , and 𝐻 for Smooth Plate 𝜏𝑤 = 𝜌𝑈𝑒2 𝑑𝜃 𝑑𝑥 1 = 𝐶𝑓 𝜌𝑈𝑒2 2 𝐶𝑓 = 2 or ⁄6 LHS: From Log law or 𝐶𝑓 ≈ 0.02𝑅𝑒𝛿−1 RHS: Use 𝑈 𝑈𝑒 to get 𝑑𝜃 𝑑𝑥 𝑑𝜃 𝑑𝑥 058:0160 Jianming Yang Chapter 7 26 Fall 2012 Following another suggestion of Prandtl, the turbulent velocity profile can be approximated by a one-seventh-power law 𝑈 𝑈𝑒 𝜃= 𝛿 𝑈 ∫0 𝑈 (1 𝑒 𝐶𝑓 = 2 ⁄6 𝑅𝑒𝛿−1 𝑑𝜃 𝑑𝑥 − 𝑦 1⁄7 ≈( )  𝑈 𝛿 𝑦 1⁄7 ∫0 (𝛿 ) [1 𝛿 𝑈𝑒 ) 𝑑𝑦 = ⁄6 = 0.02𝑅𝑒𝛿−1 = 9.72 𝑑𝛿 𝑑𝑥 = =2 𝑑 ( 7 𝑦 1⁄7 −( ) 𝛿 ] 𝑑𝑦 = 7 72 𝛿 𝛿) 𝑑𝑥 72 𝑑 (𝑅𝑒 ) 𝑑 (𝑅𝑒𝛿 ) 9.72 ( 𝛿) −1⁄6 𝑑 𝑅𝑒𝑥 𝑅𝑒 𝛿 = 1 9.72 𝑑 (𝑅𝑒𝑥 ) Assuming that: 𝛿 = 0 at 𝑥 = 0 or 𝑅𝑒𝛿 = 0 at 𝑅𝑒𝑥 = 0: 𝑅𝑒𝛿 ≈ 0.16𝑅𝑒𝑥 6⁄7 or 𝛿 𝑥 ≈ 0.16 𝑅𝑒𝑥 1⁄7 Turbulent BL has almost linear growth rate which is much faster than laminar BL which 𝛿 5.0 is ≈ 1⁄2 . 𝑥 𝑅𝑒𝑥 Other properties: 𝐶𝑓 ≈ 𝐶𝐷 ≈ 0.027 𝑅𝑒𝑥 1⁄7 0.031 𝑅𝑒𝐿 1⁄7 𝜏𝑤,𝑡𝑢𝑟𝑏 ≈ 7 ∗ 1 = 𝐶𝑓 (𝐿) 𝛿 = 𝛿 6 8 0.0135𝜇1⁄7 𝜌6⁄7 𝑈𝑒 13⁄7 𝑥 1⁄7 𝐻= 𝛿∗ 𝜃 = 1.3 𝜏𝑤,𝑡𝑢𝑟𝑏 decreases slowly with 𝑥, increases with 𝜌 and 𝑈𝑒2 and insensitive to 𝜇 058:0160 Jianming Yang Chapter 7 27 Fall 2012 5.4.2 Influence of Roughness The influence of roughness can be analyzed in the manner as done for pipe flow i.e. 1 𝑢+ = ln 𝑦 + + 𝐵 − ∆𝐵(𝜖 + ) 𝜅 1 ∆𝐵(𝜖 + ) = ln(1 + 0.3𝜖 + ) 𝜅 i.e. rough wall velocity profile shifts downward by a constant amount ∆𝐵(𝜖 + ) which, increases with 𝜖 + = 𝜖𝑢∗ ⁄𝜐. A complete rough-wall analysis can be done using the composite log-law in a similar manner as done for a smooth wall, i.e., determine 𝐶𝑓 (𝛿 ) and 𝜃(𝛿 ) from and equate using momentum integral equation 𝐶𝑓 (𝛿 ) = 2 𝑑 𝑑𝑥 𝜃(𝛿 ). Then eliminate 𝛿 to get 𝐶𝑓 (𝑥, 𝜖 ⁄𝑥 ). However, analysis is complicated: solution is Fig. 7.6. For fully rough-flow a curve fit to the 𝐶𝑓 and 𝐶𝐷 equations is given by, 𝑥 −2.5 𝐶𝑓 = (2.87 + 1.58 log ) 𝜖 𝐿 −2.5 𝐶𝐷 = (1.89 + 1.62 log ) 𝜖 Again, shown on Fig. 7.6. along with transition region curves developed by Schlichting which depend on 𝑅𝑒𝑡𝑟𝑎𝑛𝑠 = 5 × 105 and 3 × 106 . 058:0160 Jianming Yang Chapter 7 28 Fall 2012 6 Boundary Layer with Pressure Gradient 𝑈 𝜕𝑈 𝜕𝑥 +𝑉 𝜕𝑈 𝜕𝑦 =− 1 𝜕𝑃 𝜌 𝜕𝑥 + 1 𝜕 𝜌 𝜕𝑦 (𝜇 𝜕𝑈 𝜕𝑦 − 𝜌𝑢𝑣 ̅̅̅̅) The pressure gradient term has a large influence on the solution. Especially, adverse pressure gradient (i.e. increasing pressure) can cause flow separation. Recall that the ymomentum equation subject to the boundary layer assumptions reduced to 𝜕𝑃 𝜕𝑦 = 0, i.e., 𝑃 = 𝑃𝑒 = const. across BL. That is, pressure (which drives BL equations) is given by external inviscid flow solution which in many cases is also irrotational. Even without solving the BL equations we can deduce information about the shape of the velocity profiles just by evaluating the BL equations at the wall (𝑦 = 0) 𝜇 𝜕2 𝑈 𝜕𝑦 2 = 𝜕𝑃𝑒 𝜕𝑃𝑒 𝜕𝑥 𝜕𝑥 = −𝜌𝑈𝑒 𝑑𝑈𝑒 𝑑𝑥 058:0160 Jianming Yang Chapter 7 29 Fall 2012 Thus the curvature of the velocity profile at the wall is related to the pressure gradient. Point of inflection: a point where a graph changes between concave upward and concave downward. The point of inflection is the location where second derivative of 𝑈 is zero, i.e. 𝜕2 𝑈 𝜕𝑦 2 =0 (a) favorable gradient: px<0, Ux>0, uyy<0 No point of inflection i.e. curvature is negative all across the BL and BL is very resistant to separation. Note uyy()<0 in order for u to merge smoothly with U. (b) zero gradient: px = Ux = uyy = 0 (c) weak adverse gradient: px>0, Ux<0, uyy>0 PI in flow, still no separation (d) critical adverse gradient: px>0, Ux<0, uyy>0, uy = 0 PI in flow, incipient separation (e) excessive adverse gradient: px>0, Ux<0, uyy>0, uy < 0 PI in flow, backflow near wall, i.e., separated flow region. Main flow breaks away or separates from the wall: large increase in drag and loss of performance: Hseparation = 3.5 laminar = 2.4 turbulent 058:0160 Jianming Yang Fall 2012 Chapter 7 30 058:0160 Jianming Yang Chapter 7 31 Fall 2012 7 Separation 7.1 What causes separation? The increasing downstream pressure slows down the wall flow and can make it go backward-flow separation. 𝑑𝑃⁄𝑑𝑥 > 0 adverse pressure gradient, flow separation may occur. 𝑑𝑃⁄𝑑𝑥 < 0 favorable gradient, flow separation can never occur Previous analysis of BL was valid before separation. 7.2 Separation Condition 𝜕𝑈 𝜏𝑤 = 𝜇 ( ) =0 𝜕𝑦 𝑦=0 Note: 1. Due to backflow close to the wall, a strong thickening of the BL takes place and BL mass is transported away into the outer flow 2. At the point of separation, the streamlines leave the wall at a certain angle. 058:0160 Jianming Yang Fall 2012 Chapter 7 32 7.3 Separation of Boundary Layer 1. D to E, pressure drop, pressure is transformed into kinetic energy. 2. From E to F, kinetic energy is transformed into pressure. 3. A fluid particle directly at the wall in the boundary layer is also acted upon by the same pressure distribution as in the outer flow (inviscid). 4. Due to the strong friction forces in the BL, a BL particle loses so much of its kinetic energy that is cannot manage to get over the “pressure gradient” from E to F. 5. The following figure shows the time sequence of this process: a. reversed motion begun at the trailing edge b. boundary layer has been thickened, and start of the reversed motion has moved forward considerably. c. and d. a large vortex formed from the backflow and then soon separates from the body. 058:0160 Jianming Yang Fall 2012 Chapter 7 33 058:0160 Jianming Yang Fall 2012 Chapter 7 34 058:0160 Jianming Yang Fall 2012 Chapter 7 35 058:0160 Jianming Yang Fall 2012 Chapter 7 36 058:0160 Jianming Yang Fall 2012 Chapter 7 37 8 Drag and Lift 8.1 Basic Considerations Characteristic area 𝐴, which may differ depending on the body shape: 1. Frontal area, the body as seen from the stream; suitable for thick, stubby bodies, such as spheres, cylinders, cars, trucks, missiles, projectiles, and torpedoes. 2. Planform area, the body area as seen from above; suitable for wide, flat bodies such as wings and hydrofoils. 3. Wetted area, customary for surface ships and barges. In using drag or other fluid force data, it is important to note what length and area are being used to scale the measured coefficients. Drag is decomposed into form and skin-friction contributions: 𝐶𝐷 = 1 1 2 2𝜌𝑈 𝐴 𝐶𝐿 = 1 1 2 2𝜌𝑈 𝐴 {∫𝑆 (𝑝 − 𝑝∞ )𝐧 ∙ 𝐢𝑑𝐴 + ∫𝑆 𝜏𝑤 𝐭 ∙ 𝐢𝑑𝐴} = 𝐶𝐷,𝑝 + 𝐶𝑓 {∫𝑆 (𝑝 − 𝑝∞ )𝐧 ∙ 𝐣𝑑𝐴 + ∫𝑆 𝜏𝑤 𝐭 ∙ 𝐣𝑑𝐴} 058:0160 Jianming Yang 𝑡 𝑐 𝑡 𝑐 Chapter 7 38 Fall 2012 ≪1 𝐶𝑓 ≫ 𝐶𝐷,𝑝 streamlined body ~1 𝐶𝐷,𝑝 ≫ 𝐶𝑓 bluff body 058:0160 Jianming Yang Fall 2012 Streamlining: One way to reduce the drag  reduce the flow separationreduce the pressure drag  increase the surface area  increase the friction drag  Trade-off relationship between pressure drag and friction drag Trade-off relationship between pressure drag and friction drag Benefit of streamlining: reducing vibration and noise Chapter 7 39 058:0160 Jianming Yang Fall 2012 8.2 Drag of 2-D Bodies Chapter 7 40 058:0160 Jianming Yang Fall 2012 Chapter 7 41 058:0160 Jianming Yang Fall 2012 Chapter 7 42 058:0160 Jianming Yang Fall 2012 Chapter 7 43 058:0160 Jianming Yang Chapter 7 44 Fall 2012 8.3 3D Bodies 𝑅𝑒 ≪ 1: ∇ ∙ 𝐮 = 0 and ∇𝑝 ≈ μ∇2 𝐮 𝐹sphere = 3𝜋𝜇𝑈𝑑 (Stokes) 𝐶𝐷 = 𝐹sphere 1 2 2𝜌𝑈 𝐴 = 24 𝜌𝑈𝑑⁄𝜇 = 24 𝑅𝑒𝑑 058:0160 Jianming Yang Fall 2012 Chapter 7 45 058:0160 Jianming Yang Fall 2012 Chapter 7 46 058:0160 Jianming Yang Fall 2012 Chapter 7 47 8.4 Effect of Compressibility on Drag 𝐶𝐷 = 𝐶𝐷 (𝑅𝑒, 𝑀𝑎) 𝑈 𝑀𝑎 = ∞ 𝑎 𝑎: speed of sound = rate at which infinitesimal disturbances are propagated from their source into undisturbed medium 𝑀𝑎 < 0.3 flow is incompressible, i.e.,   constant 𝑀𝑎 < 1 subsonic 𝑀𝑎  1 transonic (=1 sonic flow) 𝑀𝑎 > 1 supersonic 𝑀𝑎 >> 1 hypersonic 𝐶𝐷 increases for 𝑀𝑎  1 due to shock waves and wave drag 𝑀𝑎critical (sphere)  .6 𝑀𝑎critical (slender bodies)  1 For 𝑈 ≥ 𝑎: upstream flow is not warned of approaching disturbance which results in the formation of shock waves across which flow properties and streamlines change discontinuously 058:0160 Jianming Yang Fall 2012 Chapter 7 48 058:0160 Jianming Yang Fall 2012 Chapter 7 49 058:0160 Jianming Yang Fall 2012 Chapter 7 50

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