Calculating Water Pool Depth: Factors to Consider in Fluid Dynamics

[Daniel Ivanov]

TL;DR Summary

Hi there,
I was curious about the way how to calculate the pool's depth given that my spring board is 10 m above the water and mass dropped from it is 65 kg. I assume no air friction and rotationless motion in only vertical direction.

First of all, obviously, we need to get the velocity before entering the water from the energy conservation 1/2mv^2 = mgh -> v = sqrt(2*g*h). After entering the water, we still have a gravity force and also drag force. Since the velocity is low, I take a drag term as "-bv", where b is a constant that depends on both the material properties of the object and fluid, as well as the geometry of the object.
Now I consider second time energy, namely now the kinetic energy we had at the surface of the water is going to be spent on work W.

1/2 mv^2 - W = 0
Where W = F*d = (mg - bv)*d, d - depth and F written here is the net force acting on a body. Thus I got the following result:
d = mv^2 /2(mg - bv).
So the very first question was how one gets b parameter if it is not a simple sphere ? The second question: naively assuming that b = 1 [kg/s], for h = 10[m], m = 65[kg] and g = 9.8 m/s^2 I get that the depth should be approximately the same as spring board height (10.22[m]), what seems to me too large. According to different sources, it varies from 3.5 to 5 m.

My assumption what could be wrong is that I should consider more aspects of fluid dynamics in order to get a better result, but so far only b parameter is the only place I see it.

 

[Hill]

"W = F*d = (mg - bv)*d"
The v will change until mg=bv. After that, the body will keep descending without doing any work, "forever". It does not say anything about the depth of the pool.

 

[Lnewqban]

After entering the water, we still have a gravity force and also drag force.

Welcome, @Daniel Ivanov !

It seems that you are not considering buoyancy.
Please, see:
https://en.wikipedia.org/wiki/Buoyancy

Should we assume that you will calculate the pool's depth after you calculate the depth at which the body will reach zero velocity, plus a safety factor?

 

[Daniel Ivanov]

"W = F*d = (mg - bv)*d"
The v will change until mg=bv. After that, the body will keep descending without doing any work, "forever". It does not say anything about the depth of the pool.

Yes, I have read about it (terminal velocity). Then I think the second stage does not help me at all.

Welcome, @Daniel Ivanov !

It seems that you are not considering buoyancy.
Please, see:
https://en.wikipedia.org/wiki/Buoyancy

Should we assume that you will calculate the pool's depth after you calculate the depth at which the body will reach zero velocity, plus a safety factor?

Yes, I totally ignored it, thanks for your responce ! In general I thought to take it but wasn't sure how to do it properly.
Now I think that I need to answer a slightly different question: what is the distance that body would travel in water, and according to it I should define depth of the pool plus, as you've mentioned, safety factor.
Does it sound reasonable to you ? I found some relevant formulas here (http://hyperphysics.phy-astr.gsu.edu/hbase/lindrg.html), the lower part of the page.
The only one question that is still to be answered is the "b" parameter. I can of course assume that the falling body is a sphere and take it as 6п*eta*r (maybe I should do it actually for simplicity), but is there any experimental data measuring this parameter, when human body (or something close) is falling in the fluid ?

 

[Hill]

what is the distance that body would travel in water

I think that a major missing factor is effect of the impact on the body's velocity as it enters the water. This velocity will be different from velocity before the impact depending on the splash.

PS. For safety, perhaps you can assume that they are the same.

 

[erobz]

Yes, I have read about it (terminal velocity). Then I think the second stage does not help me at all.

Yes, I totally ignored it, thanks for your responce ! In general I thought to take it but wasn't sure how to do it properly.
Now I think that I need to answer a slightly different question: what is the distance that body would travel in water, and according to it I should define depth of the pool plus, as you've mentioned, safety factor.
Does it sound reasonable to you ? I found some relevant formulas here (http://hyperphysics.phy-astr.gsu.edu/hbase/lindrg.html), the lower part of the page.
The only one question that is still to be answered is the "b" parameter. I can of course assume that the falling body is a sphere and take it as 6п*eta*r (maybe I should do it actually for simplicity), but is there any experimental data measuring this parameter, when human body (or something close) is falling in the fluid ?

When you say body, do you literally mean "person" ( i.e. how deep will a person travel before beginning to ascend)? If it is a person, note that we typically spring off of a spring board.

 

[Daniel Ivanov]

When you say body, do you literally mean "person" ( i.e. how deep will a person travel before beginning to ascend)? If it is a person, note that we typically spring off of a spring board.

It's not important I think. Physically it barely changes something, for now I assumed that the "body" is just a sphere but in general of course I'm interested in a human's body to get a more "realistic result".

 

[erobz]

It's not important I think. Physically it barely changes something, for now I assumed that the "body" is just a sphere but in general of course I'm interested in a human's body to get a more "realistic result".

The idea is this "body" is buoyant though (like a person with a lungs full of air), and will tend to return to the surface. It wasn't clear what was being asked before.

 

[Daniel Ivanov]

The idea is this "body" is buoyant though (like a person with a lungs full of air), and will tend to return to the surface. It wasn't clear what was being asked before.

You're right, but the thing is that after jumping into the pool there are a few seconds when we go down and then up, so after literally touching a bottom of a pool I became curious.

 

[erobz]

You're right, but the thing is that after jumping into the pool there are a few seconds when we go down and then up, so after literally touching a bottom of a pool I became curious.

Well, the buoyant force is the key component in that analysis, and drag is probably better modeled as quadratic $F_d \propto v^2$. Can you write Newtons Second Law for the scenario?

 

[Daniel Ivanov]

Well, the buoyant force is the key component in that analysis, and drag is probably better modeled as quadratic $F_d \propto v^2$. Can you write Newtons Second Law for the scenario?

From what I know in fluid dynamics, v^2 is the case when the velocity of the object is high. Given that I'm dropping a sphere\jumping myself from 10 meters, the velocity is going to be not very large (~ 14m/s).
Newton's second law in my case looks like:
mv' + bv = -rho*g*V. According to different sources, -rho*g*V is the Buoyant force itself.
The result is:
v(t) = v_0*exp(-b*t/m) - rho*V*g/b. The result looks reasonable to me (ones we're in water, our initial velocity v_0 is decreasing and after significant amount of time the only one velocity v_t = rho*V*g/b = mg/b will be terminal velocity in upwards direction, due to a minus sign.
Can you find some problems in this approach ?

 

[erobz]

From what I know in fluid dynamics, v^2 is the case when the velocity of the object is high. Given that I'm dropping a sphere\jumping myself from 10 meters, the velocity is going to be not very large (~ 14m/s).

Reynolds number is used to characterize the drag regime, not velocity. $Re = \frac{VD}{\nu}$

Plugging in $V = 14 ~\text{m/s}, D = 0.3 ~\text{m}, \nu = 1.14 \cdot 10^{-6} ~\text{m}^2/ \text{s}$, gives a Reynolds number of $\approx 3.7 \cdot 10^6$. The regime where linear drag is dominant is for laminar flow ($Re < 2300$). If we use the same flow characteristics that is velocities less than $0.01~\text{m/s}$.

I would use quadratic drag since the body's speed is going to be significantly higher than that for the vast majority of the motion.

Newton's second law in my case looks like:
mv' + bv = -rho*g*V. According to different sources, -rho*g*V is the Buoyant force itself.
The result is:
v(t) = v_0*exp(-b*t/m) - rho*V*g/b. The result looks reasonable to me (ones we're in water, our initial velocity v_0 is decreasing and after significant amount of time the only one velocity v_t = rho*V*g/b = mg/b will be terminal velocity in upwards direction, due to a minus sign.
Can you find some problems in this approach ?

For one thing you are missing the force of weight.

 

[Daniel Ivanov]

Reynolds number is used to characterize the drag regime, not velocity. $Re = \frac{VD}{\nu}$

Plugging in $V = 14 ~\text{m/s}, D = 0.3 ~\text{m}, \nu = 1.14 \cdot 10^{-6} ~\text{m}^2/ \text{s}$, gives a Reynolds number of $\approx 3.7 \cdot 10^6$. The regime where linear drag is dominant is for laminar flow ($Re < 2300$).For one thing you are missing the force of weight.

Of course you're right regarding mg, missed it for some reason.
Regarding the Reynolds number I also agree, thanks for pointing it out. I will take a few moments to recalculate it

 

[erobz]

While you are doing that please take a moment to learn "latex" for formatting mathematics on the site.

Follow the link :LaTeX Guide

Most people have trouble with the delimiters, so pay attention to the section titled "Delimiting your Latex code"