[試題] 103-2 李 雨 流體力學導論 期末考

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課程名稱︰ 流體力學導論 課程性質︰ 必修 課程教師︰ 李雨 開課學院: 工學院 開課系所︰ 應力所 考試日期(年月日)︰ 2015.06.24 考試時限(分鐘): 120 試題 : Fluid Mechanics Final Examination(open book) (1) (25%) Starting from the continuity and Navier-Stokes equations for an incompressible flow of Newtonian fluid. How do you make appropriate assumptions to simplify the fluid flow problems when the Reynolds number is small and when the Reynolds number is large. Write down the appropriate equations, and state the physical reasoning. (2) (25%) Find the relation for the pressure drop versus flow rate for the flow in a two-dimensional slot driven by the constant motion of the lower wall. The slot has length L and height d=d +Asin(2πx/L), where d , A, and 0 0 L are constants. (3) (25%) A tornado is sometimes represented by the superposition of a line vortex and a sink. Using that representation, derive an expression for the pressure in a tornado vs. distance from the center. (4) (25%) Consider the boundary layer flow over a flat plate with uniform suction as shown in Figure 1. (a) Show that the flow is nonsimilar, i.e., the problem cannot be solved by the similarity transformation method. (b) Solve the problem by using the integral method, and express the boundary layer thickness, the displacement thickness, the momentum thickness, and the shear stress at the wall in terms of the parameter, λ, which is defined as λ=(-v / U )√(U x/ν). 0 ∞ ∞ (Figure 1: http://imgur.com/LZkXjOA
) (5) (30%) You need to show me the detailed derivation in this problem! Consider the two-dimensional laminar jet as shown in Figure 2. The fluid is ejected from a slit with width b in a wall into infinite space filled with the same fluid. The mass and momentum discharge per unit depth are 2 Q =bρU and M=bρU , respectively. On setting b → 0 and U → ∞ m simultaneously such that M=constant, we have Q =M/U → ∞. Thus a constant m momentum source can be generated approximately by a weak mass source. Thus we can consider a jet as a flow generated by a constant momentum source, i.e., ∞ 2 ∫ρu dy = M = constant (1) -∞ for any x in the Cartesian coordinates shown in Figure 2. As the jet is a slender region of flow variation, and the pressure is uniform outside the jet, we may apply the boundary layer equations, ∂u ∂v ── + ── = 0, (2a) ∂x ∂y 2 and ∂u ∂u ∂u u ── + v ── = ν──── (2b) ∂x ∂y ∂y ∂y to describe the jet flow. These equations are solved subjected to the boundary conditions u = 0 at y → ∞ (3a) and ∂u ── = v = 0 at y = 0 (symmetric condition), (3b) ∂y together with integral constraint in equation (1). Equation (2a) is satisfied automatically by introducing the stream function, Ψ, such that ∂Ψ ∂Ψ u = ── and v = -──. (4) ∂y ∂x Equation (2b) then becomes 2 2 3 ∂Ψ ∂Ψ ∂Ψ ∂Ψ ∂Ψ ── ──── - ── ──── = ν──────, (5a) ∂y ∂y ∂x ∂x ∂y ∂y ∂y ∂y ∂y and equation (1) becomes ∞ ∂Ψ 2 ∫ρ(──) dy = M = constant (5b) -∞ ∂y Let x, δ and Q be the length scale in the x-direction, the length scale in y-direction and the scale for Ψ. Show that 2 2 ρν x 1/3 Mνx 1/3 Mx 1/3 δ~(────) and Q~(───) = ν(───) (6) M ρ ρνν by carrying out a scaling analysis of equations (5a) and (5b). Thus let Mνx 1/3 M 1/3 Ψ(x,y)=(───) f(η) with η=(────) y, (7) ρ ρννxx prove that f(η) is governed by 2 3f'''+ff'+f' = 0 (8) which is solved subject to boundary conditions f(0) = f''(0) = f'(∞) = 0. (9) Show that the solution is Cη f = (√2)C tanh(──). (10) 3√2 The constant C is determined by the integral constraint, equation (5b). Show that 9 1/3 C = (──) . 4√2 With (10), calculate the velocity components, u and v. Show that the results are 2 3M 1/3 2 u = (─────) sech ξ (11a) 32ρρνx and Mν 1/3 2 v = (────) (2ξsech ξ-tanhξ), (11b) 6ρxx 1/3 where ξ = η/(48) . Also show that Mν 1/3 v(x,y→∞) = -(────) , (12) 6ρxx which is not zero, and this phenomenon is called the entrainment. As there exists entrrainment at the outer edge of the jet, the total volume flow rate, Q , across the jet also varies with x. Show that x ∞ 9Mνx 1/3 ∞ 2 Q ≡ ∫u(x,y)dy = 2(────) ∫sech ξdξ, (13) x -∞ 2ρ 0 which increase with x. (Figure 2: http://imgur.com/mkGDGhM
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