Droplet formation in quickly stretched liquid filaments

Abstract

Film splitting necessarily occurs in roll coating and unwanted droplets can occur at high machine speeds when the resulting filaments break up. To study this ‘misting’ problem, an apparatus was designed and built to simulate filament fluid mechanics. The device creates a filament by elongating a liquid bridge and stretches the filament at a high and constant rate of acceleration to mimic coating machine kinematics. Filament breakup was observed using a high-speed video camera and the images were analyzed to yield droplet size and number. With Newtonian fluids, a single droplet formed at Ohnesorge (Oh) numbers less than 0.1 and more and smaller droplets were produced at Oh numbers above 0.1. Associative polymer solutions, prepared as weakly elastic fluids to represent industrial coatings, produced even more and smaller droplets, but only for Ohnesorge numbers in the range of 0.01 to 0.1.

Appendix

The objective here is to find how filament velocity varies with time. As stated earlier, typically the paper speed is up to 1,500 m/min, the roll diameter is over 1 m, filaments break within the first 5 to 10° of rotation, and filament length at breakup is about 10 mm. Because of the latter factors, it is reasonable to assume that filaments are almost vertical at breakup. In Fig. 17, the rotation angle of the roller is labelled θ and the small angle between the filament and the vertical is α. The relevant trigonometric calculations are

$$ {\begin{array}{lll} {\kern-2.5pt}A^{\prime}B^{\prime}\!\cos \alpha\! &=&\! \left({2R\; {\sin \theta } / 2} \right)\left( {{\sin \theta } / 2} \right)\!=\!R\left( {1-\cos \theta } \right), \\[3pt] {\kern-2.5pt}A^{\prime}B^{\prime}\sin \alpha\! &=& \!R\theta -\left( {2R\; {\sin \theta } / 2} \right)\left( {{\cos \theta } / 2} \right)\!=\! R\left( {\theta -\sin \theta } \right), \\[3pt] {\kern18.6pt}\tan \alpha\! &=&\! {\left( {\theta -\sin \theta } \right)} / {\left( {1-\cos \theta } \right)} \\[3pt] &&\sim {\left( {{\theta ^3} / 6} \right)} {\left( {{\theta ^2} / {\left( {{\theta ^2} / 2} \right)}\!\sim \!\theta / 3}\right)} \\[3pt] &&{\kern-100pt}{\rm Since} \ \alpha \ {\rm is \ small}, \ \alpha \sim \theta /3. \\ \end{array}} $$

Finding the components of velocity in the direction of the filament and in the direction perpendicular to the filament makes it possible to calculate the stretching velocity V _E and the shearing velocity V _S (Fig. 18):

$$ {\kern-1pt}V_{\rm E} =V\sin \left( {\theta -\alpha } \right)+V \sin \left( \alpha \right) $$

and

$$ {\kern-1pt}V_{\rm S} =V \cos \left( \alpha \right)-V\cos \left( {\theta -\alpha } \right), $$

where V is the paper speed. Because θ and α are small, V _E ~Vθ, and V _S is negligible by comparison. Furthermore, θ =  Vt/R, where t is the time of stretching and R is the roll radius, so that $V_{\rm E}=V^{2}t$/R. Hence, it is found that a filament is stretched at a constant acceleration, equal to V ²/R. For the values given above for V (25 m/s) and R (0.5 m), the acceleration is about 1,000 m/s².

Fig. 17

Roll coating kinematics

Fig. 18

Velocity vector diagram