MAE210C: Fluid Mechanics III Spring Quarter sgls/mae210c 2013/ Solution II

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1 MAE210C: Fluid Mechanics III Spring Quarter sgls/mae210c 2013/ Solution II D 4.1 The equations are exactly the same as before, with the difference that the pressure in the fluid is now p and the dynamic boundary condition becomes p η = γ a ) a 2 η θθ + η zz, since the normal which goes out of the fluid) has changed direction. The solution to Laplace s equation bounded at infinity is ˆp = A n αr) it should really be α so use α > 0 to avoid that problem) and the only difference in the boundary conditions is the minus sign. Hence s 2 = γ α nα) a 3 ρ n α) 1 n2 α 2 ). Now α nα)/ n α) < 0 for all α so modes are unstable if and only if n = 0 and 0 < α < 1. The maximum dimensionless growth rate can be found e.g. using MATLAB) as sρa 3 /γ) 1/2 = at α = see Figure 1). alpha=0.01:0.01:1; % note that f is minus s^2 to use matlab s fminbnd command; plot -f f=inline -sqrt-alpha.*-besselk1,alpha))./besselk0,alpha).*1-alpha.^2)) ); plotalpha,-falpha)) xlabel \alpha ) a=fminbndf,0,1) -fa) D 6.10 i) Substituting into the continuity equation gives u = 2 1/2 k 2 + 1)k 1 XkC x C z 2 1/2 k 2 + 1)XC x C z = 0. ii) S z 0) = S z π) = S 2z 0) = S 2z π) = 0 so the boundary conditions are satisfied. iii) The nonlinear terms are uu x + wu z = k 1 k 2 + 1) 2 X 2 sin 2kx, uw x + ww z = k 2 + 1) 2 X 2 S 2z. The vorticity is q = 2 1/2 k 2 + 1) 2 k 1 XS x S z. The curl of the momentum equation takes the form q θ + u )q = σ t x + σ q. The nonlinear term u )q is in fact zero. We read off 2 1/2 k 2 + 1) 2 k 1 X t = σk 2 + 1) 3 k 2 2 1/2 Y k) 2 1/2 k 2 + 1) 2 k 1 X k 2 1). 1

2 α Figure 1: Nondimensional growth rate sρa 3 /γ) 1/2 for n = 0 and 0 < α < 1. Simplifying gives iv) The convective terms are dx dτ = σ Y + X). u θ = k 2 + 1) 4 k 2 XS x C z [ Y kc x )S z + YC x C z + 2 1/2 Z2C 2z )] = k 2 + 1) 4 k 2 XYS 2z + 2 3/2 ZXC x S z C 2z ). Note that S z C 2z = 1 2 S 3z S z ), the first term of which will not appear in the truncations used here. Identifying the C x S z and S 2z terms in turn gives and Simplifying gives k 2 + 1) 3 k 2 2 1/2 Y t + k 2 + 1) 4 k 2 2 3/2 ZX 1 2 ) = R2 1/2 k 2 + 1)X k 2 + 1) 3 k 2 2 1/2 Y k 2 1) k 2 + 1) 3 k 2 Z t + k 2 + 1) 4 k 2 XY = k 2 + 1) 3 k 2 Z 4). dy dτ ZX = Rk2 k 2 + 1) 3 X + Y, dz dτ + XY = 4 k 2 + 1) Z. 2

3 C Since and µ j are constants, U j can be written as the gradient of a velocity potential Φ j. The continuity equation then gives 2 Φ j = Uniform ascension means U j = 0, V). This corresponds to Φ j = Vy + C, where C is a function of time. For convenience, use the new variable z = y Vt so that the unperturbed boundary between the fluids is at z = 0. Then Φ j = Vz and equating potential and pressure from Darcy s law gives the second relation. 3. From Darcy s law and continuity, the perturbation potential and pressure are both harmonic functions with 2 φ j = 2 p j = 0 no linearization required). The conditions at infinity are decay for φ j and p j. The conditions at the front become kinematic and Darcy s law) φ 1 z = φ 2 z = Dh Dt, µ1 V 1g h + p 1 = + ρ 2g h + p 2 at z = hx, t). Linearizing about z = 0 leads to φ 1 z = φ 2 z = h t, µ1 V 1g h + p 1 = + ρ 2g h + p 2 at z = This form is appropriate since the coefficients of the governing equations do not depend on z or t; k is the wavenumber and σ is the growth rate. Writing all quantities as ˆf z) exp ikx + σt) gives d 2 ˆφ j dz 2 k2 ˆφ j = 0, d 2 ˆp j dz 2 k2 ˆp j = 0 as field equations. Solve to get ˆφ 1 = A 1 e k z, ˆφ 2 = A 2 e k z, ˆp 1 = B 1 e k z, ˆφ 2 = B 2 e k z. The conditions at the interface give k A 1 = k A 2 = σĥ, µ1 V 1g ĥ + B 1 = + ρ 2g ĥ + B 2, while Darcy s law corresponds to A j = µ j B j. Hence µ2 µ 1 ) V + ρ 2 ρ 1 )g ĥ = B 2 B 1 = µ 1 A 1 µ 2 A 2 = µ 1 + µ 2 ) σ k ĥ. Hence σ is as required, although with k to be precise. 5. These curves are straight lines. For ρ 1 = ρ 2, the slope is positive if µ 2 µ 1 )V > 0, i.e. there is instability if the front moves into the more viscous fluid. For ρ 1 = ρ 2, the curves are still straight lines but the condition of instability is more complicated. The divergence at large k is not physical. The model is no longer valid beyond wavenumbers of the order of the inverse of the viscous or surface tension length scales, depending which is smaller. 3

4 6. Instability when V > ρ 1 ρ 2 )g µ 2 µ 1 = 10 7 m s 1, which is small. For a speed of 1 cm/s, σ = 1 the gravitional term is negligible), corresponding to a time of t = σ 1 log 10 = 2.3 s. C Conservation of mass; conservation of momentum; continuity of tangential velocity; continuity of normal velocity and definition of front velocity. 2. The Euler equations are trivially satisfied since the variables are constant. The normal velocity is zero and the fourth jump condition shows that U 1 = U L. The third is trivially satisfied since the normal is in the same direction as the base flow. The two other jump conditions reduce to and are clearly satisfied. 3. Small perturbations satisfy Linearize 2.30) and 2.31): at z = 0. The jump conditions become at z = 0. These simplify to ρ 1 U 1 = ρ 2 U 2, P 1 + ρ 1 U 2 1 = P 2 + ρ 2 U 2 2, u j = 0, ρ j t u j + U j x u j ) = p j. n = 1, η y ) w n = t η. ρ 1 u 1 t η) = ρ 2 u 2 t η), p 1 + 2ρ 1 U 1 u 1 t η) = p 2 + 2ρ 2 U 2 u 2 t η), U 1 η y v 1 = U 2 η y v 2, u 1 t η = 0 u 1 = u 2 = t η, p 1 = p 2, U 1 η y + v 1 = U 2 η y + v 2, at z = This form is appropriate since the coefficients of the governing equations do not depend on Y or t; k is the wavenumber and σ is the growth rate. The amplitudes of the normal modes satisfy dû j dx + ik ˆv j = 0, ρ j σû j + U j x û j ) = d ˆp j dx, ρ jσ ˆv j + U j x ˆv j ) = ik ˆp j. 5. Now take k > 0 and σ > 0. The divergence of the momentum equation shows that pressure is harmonic, so we have p 1 = B 1 e kx and p 2 = B 2 e kx. The momentum equations in the x-direction give û 1 = A 1 e kx, û 2 = A 2 e kx + C 2 e σx/u 2, 4

5 with ρ 1 σ + ku 1 )A 1 = kb 1, ρ 2 σ ku 2 )A 2 = kb 2, The homogeneous solution for û 1 vanishes because of the boundary condition as x. The continuity equation gives ˆv 1 = ia 1 e kx, ˆv 2 = ia 2 e kx iσ ku 2 C 2 e σx/u The jump conditions become the four algebraic equations û 1 = û 2 = σ ˆη, ˆp 1 = ˆp 2, iku 1 ˆη + ˆv 1 = iku 2 ˆη + ˆv 2, at z = 0. The last equation can be replaced by iku 1 û 1 + σ ˆv 1 = iku 2 û 2 + σ ˆv 2, at z = 0. The quantities B 1 and B 2 can also be eliminated. The final result is three equations in three unknowns: A 1 = A 2 + C 2, ρ 1 σ + ku 1 )A 1 = ρ 2 σ ku 2 )A 2, iku 1 A 1 + iσa 1 = iku 2 A 2 + C 2 ) iσ This can be rewritten as the matrix system rσ + ku L ) σ rku L 0 σ + ku L σ rku L σ 2 r 1 k 1 UL 1 rku L A 2 + σ ) C 2. ku 2 A 1 A 2 C 2 = 0. In order for a non-trivial solution to exist, the determinant of this system must vanish. This leads to the cubic The first factor is uninteresting. 7. The solution to the quadratic is σ rku L )[1 + r)σ 2 + 2rkU L σ + r1 r)k 2 U 2 L ] = 0. r ± rr σ = ku 2 + r 1) L. 1 + r For r > 1, the square root is real. Hence the system is unstable since it has at least) one positive root. The divergence at large k is not physical: this analysis neglects surface tension and viscosity and below the larger of these two length scales the model is no longer valid. 5

UNIVERSITY OF EAST ANGLIA

UNIVERSITY OF EAST ANGLIA School of Mathematics May/June UG Examination 2007 2008 FLUIDS DYNAMICS WITH ADVANCED TOPICS Time allowed: 3 hours Attempt question ONE and FOUR other questions. Candidates must