Exam in Fluid Mechanics 5C1214

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1 Eam in Fluid Mechanics 5C1214 Final eam in course 5C / in Q24 Eaminer: Prof. Dan Henningson The point value of each question is given in parenthesis and you need more than 20 points to pass the course including the points obtained from the homework problems. Copies of appendi B from Kundu & Cohern can be used for the eam as well as a book of basic math formulas and a calculator. 1. (5 Consider the unsteady flow, u = u 0, v = kt, w = 0, where u 0 and k are positive constants. Show that the streamlines are straight lines, and sketch them at two different times. Also show that any fluid particle follows a parabolic path as time proceeds. 2. (5 For surface waves on water with finite depth h the phase speed is given by the relation c 2 = g tanh(kh. Compare the group velocity to the phase speed in the it of long and k short waves. Make a rough 2D sketch of a wave packet and eplain how it is moving. 3. Consider a thin layer of liquid in contact with a solid flat vertical wall. See figure 1. The wall is subject to a constant temperature gradient with increasing temperature in the downward direction. Due to a varying surface tension, as a result of the temperature gradient, a constant shear stress τ at the free surface is acting upward. At the same time, the gravitational acceleration g is acting downward. a (7 Assume constant layer thickness h and calculate the velocity field of the liquid layer. b (3 For what layer thickness h is the flu zero? z τ g h Figure 1: Liquid film on vertical wall driven by gravity and surface tension.

2 4. a (3 Show that u i u j = 1 (u j u j + ɛ ijk ω j u k 2 i b (4 Derive the vorticity equation starting with the Navier Stokes equation for incompressible flow. c (3 Show that the following relation holds for incompressible flow and discuss its implication for inviscid flow τ ij = µɛ ijk ω k 5. Consider the flow over an oscillating plate as illustrated in figure 2. y U(y = 0 = U cos(ωt Figure 2: Fluid flow driven by an oscillating plate. a (3 Motivate that the solution can be written as [u(y, t, 0, 0] and that it thus satisfies the equation u t = ν 2 u y 2 b (7 Show that u(y, t = Ue ky cos(ky ωt, where k = ω 2ν. 6. a (5 Derive the Reynolds average equation valid for turbulent flow. b (5 Assume that the Reynolds stress can be modeled as a turbulent viscosity and introduce this into the Reynolds equation and simplify. Good Luck! 2

3 Solutions to eam in Fluid mechanics 5C1214, Streamline (fi t: d ds = u = u 0 = u 0 s + a dy = u = kt ds y = kts + b Then we see that y( is a linear function, y = kt u 0 + d(t Particle path (fi 0 : ˆt = u = u 0 = u 0ˆt + 0 y ˆt = u = kˆt = 1 2 kˆt 2 + y 0 Then we see that y( is a parabola, y = 1 2 k ( y 0 u The wave number is defined as k = 2π/λ, where λ denotes the wave length of the wave. For long waves, go to the it k 0. For short waves, go to the it k. Phase speed: Short waves: Long waves: Group velocity: Short waves: Long waves: Longer waves travel faster. c = ω k = g k tanh(kh = tanh( tanh( 0 = 1 c = gh tanh(kh kh g k = 1 c = gh c g = dω dk = c ( 2kh sinh(2kh 0 sinh( = 0 c g = c 2 sinh( = 1 c g = c

4 3. a Assume steady, parallel flow ū = w(ē z and solve the z-momentum equation with boundary conditions: ν 2 w 2 g = 0, w(0 = 0, w (h = τ µ Integrate the equation once and introduce the second boundary condition: w = g ν + A, w gh (h = ν + A = τ A = 1 ( τ µ ν ρ gh Integrate the equation once more and introduce the first boundary condition: So we get: w = 1 g 2 ν 2 + A + B, w(0 = 0 B = 0 w = 1 g 2 ν ν ( τ ρ gh b The flu per unit width of the wall can be written h [ 1 g Q = w d = 0 6 ν ( ] h 1 τ 2 ν ρ gh 2 = 1 1 τ 0 2 ν ρ h2 1 g 3 Zero flu gives: h = 3 τ 2 ρg 4. a Begin with the second term on the right hand side: u m ɛ ijk ω j u k = ɛ ijk ɛ jlm (u m u k = (δ kl δ im δ km δ il u k l l ν h3 = u k u i k u k u k i u i u j u i u j = u j + ɛ ijk ω j u k u j = 1 i 2 b Navier Stokes momentum equation for incompressible flow: Use the epression from 4. a: u i t + u u i j = 1 p + ν 2 u i ρ i (u j u j + ɛ ijk ω j u k i u i t + 1 (u j u j + ɛ ijk ω j u k = 1 p + ν 2 u i 2 i ρ i Take the curl of the equation. The curl of a gradient vanish: ω i t + ɛ imn (ɛ njk ω j u k = ν 2 ω i m ω i t + ɛ nimɛ njk (ω j u k = ν 2 ω i m ω i t + (δ ijδ mk δ ik δ mj m (ω j u k = ν 2 ω i 4

5 ω i t + k (ω i u k (ω j u i = ν 2 ω i Use continuity. We end up with the vorticity equation: c Right hand side: Left hand side: Thus: µɛ ijk ω k = µɛ ijk ω i t + u ω i u i j = ω j + ν 2 ω i ( u m ɛ klm l µ(δ il δ jm δ im δ jl 2 u m l = µ τ ij = µɛ kij ɛ klm 2 u m l = ( 2 u j 2 u i i = µ ( ui + u j = µ 2 u i i τ ij = µɛ ijk ω k Irrotational flow is not subject to viscous forces. 5. a Consider the Navier Stokes equation in the direction: u t + uu + v u y + w u z = 1 ρ p + ν = µ 2 u i ( 2 u + 2 u 2 y + 2 u 2 z 2 With the current velocity field only terms with y derivatives will remain since there can be no change in the other directions, u t = ν 2 u y 2 ] b Make the ansatz: u = R [f(ye iω t = f(y cos(ω t. Insert into the equation: Introduce k = ω 2ν, We have ] u = R [Be yk(1+i e iwt iωf(ye iω t = νf (ye iωt f (y iω ν f(y = 0 with f(y = eλy gives λ 2 iω ν = 0 λ = ± iω ν = ± ω ν ( 1 + i 2 f(y = Ae yk(1+i + Be yk(1+i, f(y 0 as y A = 0 Boundary condition at y = 0 ] = R [Be ky e i(ωt ky = Be ky cos(ωt ky = Be ky cos(ky ωt At y = 0 we have u = U cos(ωt B = U 5

6 So we have, u(y, t = Ue ky cos(ky ωt, where k = ω 2ν 6. a Turbulent flow is inherently time dependent and chaotic. However, in applications one is not usually interested in knowing full the details of this flow, but rather satisfied with the influence of the turbulence on the averaged flow. For this purpose we define an ensemble average as 1 U i = u i = N N where each member of the ensemble u (n i flow. N n=1 u (n i is regarded as an independent realization of the We are now going to derive an equation governing the mean flow U i. Divide the total flow into an average and a fluctuating component u i as u i = u i + u i p = p + p and introduce into the Navier-Stokes equations. We find u i t + u i t + u j u i u i u i + u j + u j + u u i j = 1 ρ We take the average of this equation, using which gives u i = 0 u j u i = u i u j = 0 u i t + u u i j = 1 ρ u i = 0 i p i + ν 2 u i p i 1 ρ ( u i u j j p i + ν 2 u i + 2 u i where the average of the continuity equation also has been added. Now, let U i = u i, as above, and drop the. We find the Reynolds average equations where u i u j is the Reynolds stress. b Assume the Reynolds stress: U i t + U U i j = 1 P + ρ i U i = 0 i (τ ij u i u j u i u j = 2 3 Kδ ij 2ν T e ij The first term on the right hand side is introduced to give the correct value of the trace of u i u j. Here the deformation rate tensor is calculated based on the mean flow, i.e. 6

7 e ij = 1 2 ( Ui + U j 2 U r δ ij i 3 r where we have assumed incompressible flow. Introducing this into the Reynolds average equations gives U i t + U U i j = [ ( P ρ + 2 ] 3 k δ ij + 2 (ν + ν T e ij = 1 P + (ν + ν T 2 U i ρ i where the last equality assumes that ν T is constant, an approimation only true for very simple turbulent flow. It is introduced here only to point out the analogy between the molecular viscosity and the turbulent viscosity. 7

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