Consider a continuous stream of water flowing down from a tap. Since the water flow is continuous, by equation of continuity, the cross sectional area of the stream decreases. But what makes the water flow sideways, ie, which force is responsible for decreasing the area of cross-section?
[Physics] How does the cross section of a stream of falling water decrease
flowfluid dynamicssurface-tension
Best Answer
For an incompressible viscous fluid, the equation of continuity (mass conservation) for an axisymmetric deformation is given by$$\frac{1}{r}\frac{\partial (ur)}{\partial r}+\frac{\partial w}{\partial z}=0$$ where u is the radial velocity, w is the axial velocity, r is the radial coordinate, and z is the axial coordinate. The key components of the stress tensor for axisymmetric stretching of a cylinder of viscous fluid are given by: $$\sigma_r=-p+2\eta \frac{\partial u}{\partial r}$$ $$\sigma_z=-p+2\eta \frac{\partial w}{\partial z}$$ where p is a parameter with units of pressure that must be determined from the boundary conditions, $\eta$ is the fluid viscosity, and the sigmas are the stresses in the radial and axial directions, respectively.
From the continuity equation, it follows that if dw/dz is the (constant) axial rate of deformation for the cylinder, the radial velocity can be integrated to give:$$u=-\frac{r}{2}\frac{dw}{dz}$$ This equation indicates that, as the fluid is stretched axially, it contracts radially. The stress in the radial direction is equal to zero, so, the parameter p is given by:$$p=2\eta \frac{du}{dr}=-\eta \frac{dw}{dz}$$ If we substitute this into the axial stress equation, we obtain:$$\sigma_z=3\eta\frac{dw}{dz}$$ This is a well-know equation which indicates that the so-called elongational viscosity of a fluid is 3 times its shear viscosity.
In summary, according to this development, even through the overall stress in the radial direction is zero, the cylinder of viscous fluid still contracts radially when stretched axially in order to satisfy the continuity equation (provided that the fluid does not break up). The viscous compressive contribution to the radial stress holds the cylinder together. This all is totally analogous to the Poisson effect in stretching a solid (as a result of elastic forces).
Of course, in the case of molten polymers, the contribution of surface tension is typically negligible, unless the cylinder is of tiny diameter.