Saturday 16 March 2019

fluid dynamics - How does a ball cause a splash? (With the relevant math)


Problem Statement:



Imagine a spherical ball is dropped from a height $h$, into a liquid. What is the maximum average height of the displaced water? For instance, although one particular drop of water might travel a high distance, the average displacement height of all the displaced water might be very low.


Approximation:


Let's make some assumptions. Let's assume that on impact, a fraction $\lambda$ of the ball's Kinetic Energy is transferred to the liquid. Let's also assume that all of the energy transferred will go into shooting water up. This should give the upper bound on how high the water "tower" will be.


The Kinetic Energy of the ball at impact with the surface of the liquid is,


$$KE=m \cdot g \cdot h$$


The fraction that is transferred into making the water tower is,


$$KE_t=m \cdot g \cdot \lambda \cdot h$$


When the formation of the water tower reaches its maximum height, the potential energy $PE$ must be equal to $KE_t$. We'll assume that the tower has cross-sectional area $A$, and integrate the $PE$ of each slice of the tower at a particular height to get the total $PE$.


$$PE=\int_0^{h_t} g \cdot l \cdot \rho \cdot A \ dl={{g \cdot \rho \cdot A \cdot {h_t}^2} \over 2}$$


Where $h_t$ is the height of the water tower, $\rho$ is the density of the liquid, and $g$ is gravity.



Set $KE_t=PE$ and get,


$$m \cdot g \cdot \lambda \cdot h={{g \cdot \rho \cdot A \cdot {h_t}^2} \over 2}$$


We'll use $A=\pi \cdot r^2$, which assumes the water tower has a radius equal to the ball. Solving for $h_t$, we note that the $g$ terms cancel, and we get.


$$h_t={1 \over r} \cdot \sqrt{{{2 \cdot \lambda \cdot m \cdot h} \over {\pi \cdot \rho}}}$$


Let's take $\lambda={1}$. let's use a water droplet. It has a radius of $0.5 \ cm$, a mass of $0.125 \ g$. Let's drop it into water from a height of $20 \ cm$.


We get,


$$h_t = 2.52... \ cm$$


Referring back to the original problem statement, we need only divide $h_t$ by two to get the average height of the displaced water.


Thus,


$$\mu_{h_t}=1.26... \ cm$$



Questions


First, I don't have access to materials other than water and a sink. So I can only test my approximation for water droplets dropped into water. Are there any results that point to flaws in my model?


Second, given this initial approximation, how does one go about adding in more complex phenomena? For instance, I've assumed a very ideal energy and water tower mass distribution. Is there any way you can go from this original approximation to a more useful one? Perhaps, there is an efficient way to conduct an experiment to get useful data?


Edit: This question inspired from another answer I wrote. I deleted the answer, and put the material here instead.




No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...