# How does the cross section of a stream of falling water decrease?

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?

• For the principle of continuity shouldn't the initial momentum of the liquid remain constant? But when the water falls down because of gravity, there is a force acting on it Jan 23, 2016 at 10:11
• Does it have something to do with the air drag ?? This is what I think :- The outer edges of flowing water experience the air drag and hence moves slower as compared to the one in the middle of the flow and so it (water along the edges) falls a lesser distance as compared to the one in middle and hence creating the narrow shape we see and if allowed to fall more and more it becomes narrower and ultimately unstable .... Aug 4, 2022 at 8:10

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.

• Can you please write an answer which does not make use of tensors(if it is possible)? I and many other high-school students have not yet studied tensors but this example is often shown in books as an application of continuity equation. Nov 11, 2017 at 5:27
• @Chet Miller hey Miller .. do u think air drag also matters here ? Check the comment I wrote below the question.. Aug 4, 2022 at 8:13
• Air drag can have an effect on the shape, although it is typically secondary. Aug 4, 2022 at 10:59

It is due to a matter of fact that same mass and hence same volume of water has to flow through a given cross section.This is termed as law of continuity in fluid mechanics. It is a known fact that velocity of a falling object increases with height, and hence to satisfy the law of continuity the cross section has to decrease.(to keep the volume constant).And so higher the velocity of fluid flow lesser the cross section becomes.

• Your answer does not address which force is responsible for the contraction. Jan 23, 2016 at 22:44

For the sake of the argument, assume that the stream of water is 10 cm tall. The water drops at the bottom of the stream have been falling for a longer period of time than the water drops 1 cm above them, and this is true for all the water drops in the falling stream. Because of this, the water drops at the bottom of the stream are moving at a higher velocity than the water drops 1 cm above them, because the water drops that are lower in the stream have been accelerating for a longer time. Since the flow rate is constant, the stream must get thinner in cross-sectional area as it falls, in order to satisfy the continuity equation.

• Your answer does not address which force in responsible for the contraction. You only prove that continuity is satisfied, but that is not really what is asked here. Jan 23, 2016 at 22:45
• The acceleration due to gravity causes the thinning. When the stream gets thin enough, surface tension breaks the stream up into drops. Jan 24, 2016 at 14:49
• Gravity acceleration is downwards, it can't cause the water particles to accelerate horizontally (which is necessary for the radius to decrease). I think the horizonal change is caused by attraction between the particles Oct 28, 2023 at 16:41

If you want to see this in terms of force you can see it as below.

Consider the point when the water has just started falling(take it as a reference cross section), obviously at this point the atmospheric pressure and pressure from inside the cross section in outward direction will be in equilibrium. Also at this moment the water has not reached its max compressibility yet.

As i suppose you already know that as velocity increases pressure decreases (this same phenomenon is used for air particles to fly you know what(in accordance with Bernoulli theorum)). Due to this fact as velocity of falling water increases the outward pressure exerted by it decreases and thus due to atmospheric pressure the cross section decreases until it reaches its max compressibility.

Consider 3 particles moving in a line, one after another, in a tube (say) with uniform speed '$$v$$', maintaining a constant distance '$$d$$' between successive particles; at time $$t=0$$, the first particle is at the end of tube, after escaping from which, the particles accelerate (with acceleration '$$a$$').

Now, after a time $$\delta t = \frac{d}{v}$$, the 2nd particle reaches the end of the tube, with the third particle a distance '$$d$$' behind it; but the first particle is ahead of the second particle by: $$D \: = d \: + \: \frac{ad^2}{2v^2}$$

Hence, the spacing between the first two particles has increased. By the time the 3rd particle comes out, the second particle will be ahead of it by a separation '$$D$$', whereas the first one will be ahead of the second one by a separation even greater than '$$D$$'. Hence, the particles will be sparsed-out.

Extend the argument to a bunch of particles (of the fluid) moving together. As the fluid accelerates, the separation between nearby particles increases. This would start to decrease the fluid pressure, and the force due to atmospheric pressure would push the particles inwards, until the pressure in fluid equals the atmospheric pressure.

Hence, the tube of flow narrows down.