0

Some time ago i saw this question Why does water falling slowly from a tap bend inwards?

Which explains why the water flowing from a tap bend inwards while falling, using an inviscid flow model. However all calculations there assume steady flow, so that the continuity equation and bernoulli can be used in the usual forms.

But i don't see how can a free fall flow ever can be steady. I know it is the case, but i struggle to understand how a free fall flow would transition from unsteady to steady.

I imagine, for instance, a large volume of water with the exit closed at the bottom. At a initial moment, the bottom exit opens, and the water starts moving from rest. Since the fluid is now unconstrained, the pressure reduces to zero (or simply atmospheric pressure) and the water falls due to gravity, but every single point of the water should fall with the same acceleration g, therefore there is no relative motion between the fluid parts, so it should fall as an unique body. The flow is unsteady of course and would simply be that $\frac{\partial v}{\partial t} = g$.

How would this flow ever become steady?

Qmechanic
  • 220,844
Klaus3
  • 482

2 Answers2

1

If the acceleration of the fluid particles is expressed as $\frac{dv}{dt}=g$, we can use the chain rule to write $$\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=g$$where x is the distance measured downward from the tank exit. Integrating this with respect to x yields: $$\frac{v^2}{2}=\frac{v_0^2}{2}+gx$$where $v_0$ is the exit velocity from the tank. This shows that, from the perspective of a stationary observer, the downward velocity is a function only of x and not time. This is the definition of a steady flow.

Chet Miller
  • 35,124
-1

The reason slowly falling water contracts as it falls (up to a point at least) is that the center of the stream is falling faster than the periphery of the stream. This tends to draw the sides in. When viscous dissipation erases this velocity difference, the stream stops contracting.

NOTE ADDED: This model presupposes a velocity profile that exists across the diameter of the jet at its point of origin, and viscous coupling via the shear gradient across that cross-section which would act to accelerate the circumference of the jet while decelerating its faster-moving core- all while the jet itself is being pulled out in length by gravity. I am going to pose a separate question on this topic to determine if this effect is real and if not, I will delete my answer to this question. -NN

niels nielsen
  • 99,024