ME 509, Spring 2016, Final Exam, Solutions 05/03/2016 DON T BEGIN UNTIL YOU RE TOLD TO! Instructions: This exam is to be done independently in 120 minutes. You may use 2 pieces of letter-sized (8.5 11 ) paper with whatever you want on it (back and front, typed or handwritten or both). No other books, notes or electronic materials are to be used in preparing your answers, including calculators, phones, laptops, etc. Please make use of scratch paper for trial & error, then write down your solution steps legibly on the exam. If you need extra space, please attach an extra sheet of paper and don t work on the back of any of the pages. ME 509, Spring 2016, Final Page 1 of 8
Problem 1 21 pts. (= 7 3 pts.) True or False? (Also state why in only a few words.) (a) The Reynolds number for characteristic velocity U and length L is Re = U/ gl. Re = ϱul/µ. (b) ψ φ = 0 is a relation satisfied by the streamfunction ψ and potential φ for an ideal flow in the plane. T /F: ψ = ( x ψ, y ψ) = ( v, u), φ = ( x φ, y φ) = (u, v), hence ψ φ = vu + uv = 0. (c) All fluid flow fields v can be expressed in terms of a potential φ as v = φ. v = φ if and only if v = 0 (irrotational), which is a subset of all fluid flows. (d) The equations of low Reynolds number (Stokes) flow reduce to 2 ψ = 0. Equations of Stokes flow reduce to 4 ψ = 0, this could be true even if 2 ψ 0. (e) The boundary layer equations are valid for both small and large Re. In deriving the BL equations, we took the limit Re. (f) Ideal flows cannot satisfy all boundary conditions on a solid surface. Ideal flows can satisfy either v n = 0 or v t = 0. penetration (v n = 0). T /F: We require them to satisfy no (g) The equations of lubrication theory are an exact form of Navier Stokes. The lubrication approximation requires both h 0 /L 1 and Re small. We dropped many terms. The equations are approximate. ME 509, Spring 2016, Final Page 2 of 8
Problem 2 16 pts. (= 4 4 pts.) Consider the Cartesian velocity field v(x, y, z, t) = (cos y, 0, sin y). (a) Show that the vorticity vector ω of this given flow field is exactly equal to v. î ĵ ˆk ω = x y z cos y 0 sin y = î( y sin y z 0) ĵ( x sin y z cos y) + ˆk( x 0 y cos y) = î cos y + ˆk sin y = ( cos y, 0, sin y ) v. (b) The helicity (per unit volume) is defined as h = v ( v). Compute h for the given flow field v. (a) {}}{ h = v ( v) = v ω = v v = cos 2 y + sin 2 y = 1. ME 509, Spring 2016, Final Page 3 of 8
(c) Show that the given flow field v satisfies conservation of mass for an incompressible fluid in 3D. v = x cos y + y 0 + x sin y = 0 + 0 + 0 = 0 (d) Compute p(x, y, z, t) such that the given flow field v satisfies Euler s momentum equation(s) for inviscid incompressible flow in the absence of body forces, i.e., Dv/Dt = p. Dv Dt v +v v = ( cos y, 0, sin y ) 0 0 0 sin y 0 cos y = 0 }{{} t 0 0 0 =0 Hence, p = 0 p = const. (say, 0). ME 509, Spring 2016, Final Page 4 of 8
Problem 3 32 pts. (= 4 8 pts.) Consider ideal stagnation point flow in the plane for y > 0 and < x < in the presence of a source of strength Q at (x, y) = (0, 0). (a) Determine the potential function φ(x, y), in Cartesian coordinates, for the given flow. Could start with F (z) = Az 2 + Q log z, 2π then take the real part, or start directly with φ(x, y) = A(x 2 y 2 ) + Q 2π ln x 2 + y 2 = A(x 2 y 2 ) + Q 4π ln ( x 2 + y 2). (b) Determine the velocity profile v = ( u(x, y), v(x, y) ), in Cartesian coordinates, for the given flow. and u(x, y) x φ = 2Ax + v(x, y) y φ = 2Ay + Qx 2π (x 2 + y 2 ) Qy 2π (x 2 + y 2 ) ME 509, Spring 2016, Final Page 5 of 8
(c) This flow has exactly one stagnation point (x s, y s ). Find x s and y s as functions of the two parameters associated with flow you defined in (a). Verify that u(x s, y s ) = 0 and v(x s, y s ) = 0. [Hint: x s = 0; you re allowed to use this fact to speed up your calculation.] and Using hint, v(0, Q 4πA v(x s = 0, y s ) = 2Ay s + ) = 0 is easily verified. u(x s = 0, y s ) = 0 + 0 = 0 Q Q = 0 y s = 2πy s 4πA. (d) Could we model ideal stagnation point flow onto a bump using φ from (a)? Why or why not? [Hint: a sketch and a key fact about solid walls in ideal flows is enough.] Plot shows streamlines and equipotential lines. Black is the streamline ψ(x, y) 1.57075. We can identify it as a solid wall. Thus, φ can model ideal stagnation point flow onto a bump. ME 509, Spring 2016, Final Page 6 of 8
Problem 4 31 pts. (= 8 + 8 + 8 + 7 pts.) Consider steady flow, under the lubrication approximation [(h 0 /L) 2 Re 1], in a slot of non-uniform height h(x). Assuming the top wall is elastic, its deformation is proportional to the local fluid pressure, specifically: h(x) = h 0 +αp(x), where α is a proportionality constant with appropriate units. y = h(x) y h 0 x L (a) Starting from the conservation of linear momentum under the lubrication approximation, develop an expression for the horizontal (i.e., x-) velocity u(x, y), assuming no slip at y = 0 and y = h(x) = h 0 + αp(x). 0 = p x + µ 2 u y 2, 0 = p x p p(y). Solving first eq. (once again, plane Poiseuille flow), imposing BCs: u(x, y) = 1 dp y[y h(x)]. 2µ dx ME 509, Spring 2016, Final Page 7 of 8
(b) Given a constant flow rate Q (at any x) and using your result in (a), find an ODE for p(x) as a function of Q, µ, h 0 and α. [Hint: the ODE will be nonlinear in p.] Q = h(x) 0 u(x, y) dy = 1 [ dp y 3 2µ dx 3 h(x)y2 2 ] h(x) 0 = [h(x)]3 12µ which must hold at every x by conservation of mass. Then, using given h(x): 12µQ = [h 0 + αp(x)] 3 dp(x) dx. dp dx, (c) If correct, the ODE in (b) is separable and can be immediately integrated. Find Q as a (nonlinear) function of p(x) assuming the boundary condition p(x = L) = 0. 12µQdx = [h 0 + αp] 3 dp 12µQx + C = 1 4α [h 0 + αp(x)] 4. Using the BC p(l) = 0 gives C = 12µQL + h4 0 4α. Hence, Q = [h 0 + αp(x)] 4 h 4 0. 48αµ(L x) (d) Carefully take the limit as α 0 to recover p(x) in terms of Q, µ, L, x and h 0 for a (undeformed) slot of uniform height. Using the binomial theorem (or Pascal s triangle, or squaring the bracketed quantity twice in a row...): Q = [h 0 + αp(x)] 4 h 4 0 48αµ(L x) = h4 0 + 4h 3 0αp(x) + 6h 2 0α 2 [p(x)] 2 + 4h 0 α 3 [p(x)] 3 + α 4 [p(x)] 4 h 4 0, 48αµ(L x) Then canceling an α from top and bottom and taking α 0: Q = h3 0p(x) 12µ(L x) p(x) = 12µQ (L x). h 3 0 ME 509, Spring 2016, Final Page 8 of 8