University:
Massachusetts Institute of Technology
Course:
2.29 | Numerical Marine Hydrodynamics (13.024)
Academic year:

2022
Views:

411

Pages:

63
Author:

Yasmin Shaffer

Two-Dimensional Steady Boundary Layer Equations z is horizontal direction along direction of main flow velocity u. Velocity at outer edge of boundary layer is called U or V or U or Ve. y is perpendicular to wall and velocity in this direction is v. The boundary layer begins, say, atz 0 and the boundary layer thickness is 8. δ «x. Because the boundary layer is thin, to leading order the pressure is constant through the thickness of the boundary layer, Also, vu, and Ju = 0. Edge of boundary layer X 8 V Outer flow velocity profile Wall shear stress (7) Plate Streamlines of boundary layer flow Boundary layer velocity profile FMBBL6A(1) Calculation of Turbulent Boundary Layer when Pressure Distribution is Known This result is approximate since the boundary layer thickness will alter the pressure distribution. The principal unknowns (quantities to be determined) are: 0(z) and d(z). An equivalent set of unknowns is 0(x) and H(z). There are two fundamental equations: do dx =-(H+2) O dUC Ue dx + 2 (1) 0 dH H dx O due 2CD+(H-1)d H 2 u dr (2) To be able to integrate the unknowns along the boundary layer, the deriv- atives of each of them are required: de/dz and dH/da. Equation 1 is in the desired form. To put equation 2 in the desired form, use the chain rule: dH dH dH dx dx dH (3) Empirical "closure relations" for H*(H) and dH/dH exist. Therefore we write the energy equation in the desired form as: dH dx H C /H+(-1)] 0 dH/dH 2 (4) FMISMLEA (2) To do the integrals numerically, we need a means of determining C, CD, H and dH/dH in terms of the principal quantities H and Re, where R = Uco/v. These empirical "closure relations" have been determined by as- sembling a large amount of experimental data. Laminar Closure Relations 0.76(H-4)²/H+1.515, Η < 4.0 H = 0.015(H4)2/H+1.515, H≥ 4.0 Cy = [0.03954[(7.4-H)/(H-1.0)] -0.134)/Re, H< 7.4 [0.044(1.0-1.4/(H-6)2-0.134/Ro H≥ 7.4 2CD H [0.00205(4H)5.5 +0.207)/Re, [-0.003(H-4.0)2/(1+0.02(H4)2)+0.207/Rg, H≥ 4.0 Η < 4.0 Turbulent Closure Relations H = {3+400/Re, Ra 400 Re≤ 400 Rez { Re, Re > 200 200, Re≤ 200 H< H H = { 1.505 +4/R+ (0.165-1.6/Ro), (HH)2 (0.007 H-He+(Ra)+0.015/H)+1.505+4.0/Re, H≥ Ho -(1.74+0.31H) C = 0.3-1.334 In(R) 2.3026 200=0.5C, 40/H-1+0.03 (1-1) H 3 Dominated by inviscid irrotational solution (V20) Boundary Conditions Ot + + ду ду (kinematic) {+[()+()+()]} +9 = constant (0) 1 2 (dynamic) Linearized Boundary Conditions Case of onset flow velocity of -iU. Now is the perturbation potential and the total potential is -Ux + ф. Oz20 Ot = θα Ot + 9 = 0 For steady flow with onset flow: -U θα U θα = 96 U²² Case of 2D waves and zero onset flow so is the total potential. Ot [84] Ot20 +96 = 0 Dispersion Relations for waves of ciruclar frequency w = 2f and wavenumber = k = 2/1 and zero onset flow. w² = gk deep water w² = gk tanh kh water of depth h

Week 11 Notes: Boundary Layers

Related Documents

Recommended Documents

Description

Related Documents

Free up your schedule!

Take 5 seconds to unlock