Nonlinear instability of a viscous liquid jet
Résumé
We investigate the weakly nonlinear temporal instability of an axisymmetric Newtonian liquid jet. Early nonlinear studies on the capillary instability of inviscid liquid jets were carried up to the third order contributions to the jet deformation and showed the nonlinear interaction between different modes. A recent study on the weakly nonlinear instability of planar Newtonian liquid sheets revealed the role of the liquid viscosity in the sheet stability behavior and showed a complicated influence [1]. Here, the instability of a liquid jet is examined as the axisymmetric counterpart of the sheet, in search for corresponding insight into the role of the liquid viscosity in the jet instability mechanism. Copyright line will be provided by the publisher The weakly nonlinear instability of an incompressible, Newtonian axisymmetric liquid jet is studied as an alternative to and for comparison with numerical simulations such as in [2]. The problem is formulated in cylindrical coordinates. Body forces are not accounted for, since the Froude number is large. The variables and equations of motion are non-dimensionalized with the undeformed jet radius a, the capillary timescale (ρa 3 /σ) 1/2 and the pressure σ/a where ρ is the liquid density and σ the liquid-air surface tension. The jet surface is described as a place where r(z, t) = 1 + η(z, t), where η is the deformation against the undisturbed cylindrical shape. The initial surface disturbance is assumed to be purely sinusoidal with amplitude η 0 and wavenumber k. With this hypothesis, mass conservation leads to the expression η(z, 0) = η 0 cos kz + (1 − η 2 0 /2) 1/2 − 1 for the initial deformation [3]. As usual in weakly nonlinear analysis, the initial deformation is supposed to be small, so that η 0 ≪ 1. This allows the velocity components, the pressure field and the deformed interface shape to be represented by series expansions with respect to η 0. The expansion for the axial flow velocity, as an example, reads u z = u z1 η 0 + u z2 η 2 0 +. .. . Introducing the expansions into the equations of motion and into the boundary conditions, we obtain equations determining the linear problem in terms of the quantities with subscript 1, and the second-order problem by quantities with subscript 2. The related equations are linear in the quantities of the respective order, but non-linear in the quantities of the next lower order. The first-order contribution to the jet surface deformation is formulated as η 1 = ˆ η 1 exp(ikz − α 1 t), where α 1 is a complex frequency andˆηandˆ andˆη 1 is the first-order initial amplitude fixed to 1/4 by the initial condition. Representing the first-order velocities as derivatives of a disturbance stream function ψ, and taking the curl of the first-order momentum equation, the fourth-order partial differential equation 1 Oh ∂ ∂t − E 2 E 2 ψ = 0 for the disturbance stream function is obtained, where E 2 = r∂/∂r((1/r)∂/∂r) + ∂ 2 /∂z 2. Its solution reads ψ(r, z, t) = [C 1 rI 1 (kr) + C 3 rI 1 (lr)] exp(ikz − α 1 t) := ψ 1 + ψ 2 , where I 1 is the first-order modified Bessel function of the first kind. We have discarded the modified Bessel functions of the second kind for regularity of the solution on the jet axis. We have furthermore defined l 2 = k 2 − α 1 /Oh and the Ohnesorge number Oh = µ/(σaρ) 1/2 , the characteristic parameter distinguishing the viscous from the inviscid case, with the liquid dynamic viscosity µ. The flow field is determined by spatial derivatives of the stream function, and the two integration constants C 1 and C 3 are determined by the kinematic and dynamic zero-shear stress boundary conditions. The normal-stress boundary condition yields the dispersion relation α 2 1 − 2α 1 k 2 Oh 1 − I 1 (k) I 0 (k) 1 k + 2kl k 2 + l 2 I 0 (l) I 1 (l) − 1 l = k 1 − k 2 I 1 (k) I 0 (k) l 2 − k 2 l 2 + k 2 governing the linear temporal stability behaviour of the jet, which was first presented by Weber [4]. For Oh → 0, Rayleigh's solution is retrieved [5]. The second-order continuity and momentum equations read 1 r ∂ ∂r (ru r2) + ∂u z2 ∂z = 0 ∂u r2 ∂t + u r1 ∂u r1 ∂r + u z1 ∂u r1 ∂z = − ∂p 2 ∂r + Oh ∂ ∂r 1 r ∂ ∂r (ru r2) + ∂ 2 u r2 ∂z 2 ∂u z2 ∂t + u r1 ∂u z1 ∂r + u z1 ∂u z1 ∂z = − ∂p 2 ∂z + Oh 1 r ∂ ∂r r ∂u z2 ∂r + ∂ 2 u z2 ∂z 2