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Chapter 9, Flow Over Immersed Body (External Flow) 1. Lift and drag concepts dFx = pdA cos + wdA sin dFy = -pdA sin + wdA cos Drag force, D = dFx Lift force, L = dFy Drag coefficient, CD = D/[(1/2)V2AD] Lift coefficient, CL = L/[(1/2)V2AL] 2. Boundary Layer Characteristics i) Flow over a flat plate Boundary layer thickness y= when u/U = 0.99 /L << 1, (Since v/u << 1) Displacement thickness, * bU* = b0 (U - u)dy * = 0 [1 - (u/U)]dy Momentum thickness, bU2 = b0 u(U - u)dy = 0 (u/U)[1 - (u/U)]dy Using the liniear momentum equation Fx = cs uVndA -b0 w dx = b0 [u2 -uU]dy 0 w dx = U20 (u/U)[1 - (u/U)]dy w = U2d/dx eq. Integrated linear momentum Skin friction drag, D = 0 w dx Friction coeffricient, Cf = w/[(1/2)U2] = 2d/dx Drag coeficient, CD = D/[(1/2)U2] = (1/)0 Cf dx If the velocity profile is known (Table 9.2), the shear stress at the wall can be found w = µdu/dy| y=0 Blasius’ solution is to solve the differential equation for u. However, if an approximate velocity profile is assumed, the momentum thickness, , can be found as a function of , the shear stress can be found from the integrated linear momentum equation: Assume velocity profile, u/U = 2(y/) - (y/)2 * = 0 [1 - (u/U)]dy */ = 01 [1 - 2(y/) + (y/)2]d(y/) = 1/3 = 0 (u/U)[1 - (u/U)]dy / = 01 [2(y/) - (y/)2][1 - 2(y/) + (y/)2]d(y/) = 2/15 w = µdu/dy| y=0 = 2µU/ Cf = w/[(1/2)U2] = [2µU/]/[(1/2)U2] = 2d/dx = 2d(2/15)/dx d = 15µdx/(U) /x = 5.48/(Rx)1/2, */x = 1.83/(Rx)1/2, Table 9.2 /x = Cf = 0.73/(Rx)1/2, CD = 1.46/(R)1/2 Laminar flow, Rx = Ux/µ < 5 x 105 Shear at the wall (Blasius’ solution) w = µdu/dy|y=0 = 0.332 U2/3 [µ/x] Friction coefficient Cf = w /[(1/2)U2] Friction drag coefficient (CD)f = (1/)o Cf dx = 1.328/R1/2 Turbulent flow, Rx = Ux/µ > 5 x 105 w = 0.0228U2/Rx1/5 Emperical Cf = w /[(1/2)U2] = 0.0576/Rx1/5 (CD)f = (1/)o Cf dx = 0.72/R1/5 ii) Effect of pressure gradient In addition to friction drag, there is pressure drag which can only be determined experimentally 3. Drag CD = (CD)f + (CD)p depending in Reynolds number and shape of the body A. Skin Friction Drag Shear stress at wall w = µdu/dy| y=0 Friction coefficient Cf = w/[(1/2)U2] Drag coefficient Cf dx CD = D/[(1/2)U2] = (1/)0 Boundary Layer Theory, solution for velocity a. Differential Method Continuity u/x + v/y = 0 Momentum uu/x + vu/y = -(1/)dp/dx + µ2u/x2 i) Flat Plate, dp/dx = 0 Blasius solution ii) Turbine Blade, dp/dx 0 Falkner and Skan solution b) Integral Method Displacement thickness * = 0 [1 - (u/U)]dy Momentum thickness = 0 (u/U)[1 - (u/U)]dy Shear stress at wall w = U2d/dx Friction coefficient Cf = w/[(1/2)U2] Drag coefficient Cf dx CD = D/[(1/2)U2] = (1/)0 Assuming a velovity profile u/U = a(y/) + b(y/)2 B. Pressure Drag Experimentally determined data are in Figs. 9-19, -20, -22, -23, -24 Ex.9-10 Fz = d(mV)/dt FB + D = W W = V FB = waterV D = (1/2)U2ACD CD is from Table 9-4, solve for U and ckeck Reynolds number Composite body drag CD = 0.16 + 0.0095Ni Ex. Autobomile, Table 9-5 Ni Applications Drag force, D = (1/2)U2ACD Moment (torque), Mo = 0 Power, = M Ex. 9-13 M = Ds(b + ds/2) + Dc(b/2) Ds = (1/2)U2(/4)ds2 CDs Dc = (1/2)U2bds CDc Rs = Uds/ = 2.24 x 107 CDs = 0.3 Rc = Udc/ = 8.41 x 106 CDc = 0.7 M = 3.64 x 105 ft-lb