I have a pipe with a ball inside it and a blowing air through it. Look this image:

enter image description here

The air is blowing in the direction of $F_a$. $F_a$ Is the air force and $P$ is the weight force.

What I want is something very very simple, not complex.

I want to assemble a formula of the height $h$ of this ball.

What I've done till here:

I know that the resultant force $F_r$ is: $$F_r = \frac{\partial^2h}{\partial t^2}.m_b$$ where $m_b$ is the mass of the ball. And I know that $F_r$ is also: $$F_r = F_a-P, then: \\ \frac{\partial^2h}{\partial t^2}.m_b = F_a-P$$ $P = m_b.g$, $g$ is the gravity acceleration. According to this, $F_a = A_b . \rho_w . D_{sphere}$. I made $A_b$ as the area of the center cross section of the sphere (the center circle). According to he link, $\rho_w = 0.00256 . v_w^2$, where $v_w$ is the speed of the wind inside the pipe. And I made $D_{sphere} = 0.47$, according to this. Then: $$\frac{\partial^2h}{\partial t^2}.m_b = A_b \times 0.00256.v_w^2 \times 0.47 - m_b.g$$

Here starts the problems... I have to build an approximation for the speed of the wind inside the pipe. This speed is not linear and It's in function of the actual height of the ball. Think in a fan blowing an air from bottom to top. The speed is gonna be high in the bottom and very low at the top.

What would be a nice formula to substitute $v_w$? I thought something like this:

$$v_w = \frac{k}{h^p}$$

Where $k$ is a constant (no problem at letting it just as a single constant) and $p$ is an index. What would be a good $p$?

The whole formula would be:

$$\frac{\partial^2h}{\partial t^2}.m_b = A_b \times 0.00256.\left(\frac{k}{h^p}\right)^2 \times 0.47 - m_b.g$$

What you guys think about the whole process for a simple approximation? Any absurd mistake?


Following the suggestions, I made some changes. I discarded the idea of $v_w = \frac{k}{h^p}$. Actually, this is the relative velocity between the air and the ball, so I made:

$$v_r = v_w - \frac{\partial h}{\partial t}$$

where $v_w$ is now a constant. I also decided to change the equation by this one from wikipedia. I also changed the air density at 25 deg to $1.1839 \ kg/m^3$.

So here is the new formula:

$$\frac{\partial^2h}{\partial t^2}.m_b = \frac{A_b}{2} \times 1.1839.\left(v_w - \frac{\partial h}{\partial t}\right)^2 \times 0.47 - m_b.g$$

Using Runge-Kutta, I made these graphics for different wind speeds:

enter image description here

Values used:

$$gravity = 9.8 \ m/s^2\\ mass = 0.01 \ kg\\ sphere \ diameter = 0.1 \ m\\$$

I think this is pretty reasonable...

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    $\begingroup$ You can't just make up a formula for $v_w$ as function of $h$! Why do you think the air speed will be high at the bottom and low at the top? Granted there will be a decrease in speed due to resistance, but it will be small. In general if we impose mass conservation, the speed in a cylinder of constant diameter must be the same at the bottom as at the top. Your air speed in the cylinder then follows simply from the imposed flowrate and the area of the cylinder it flows through. Now wether the speed in the drag equation should be the speed in the cylinder or around the ball is another question! $\endgroup$ – nluigi May 22 '16 at 22:40
  • $\begingroup$ That makes sense, there's no reason to the air speed decrease... So I can let $v_w$ be a constant, right? $\endgroup$ – João Paulo May 22 '16 at 22:45
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    $\begingroup$ "What I want is something very very simple, not complex." We can't always have what we want. I have tried Classic drag functions acting on the ball. They don't lead to a stationary ball. The only possible model here is one where pressure below the ball depends on $v_w$. But that dependency ($p=f(v_w)$) is hard to model theoretically. I think real world applications of this problem (ball flow meters) have been modelled empirically. $\endgroup$ – Gert May 23 '16 at 0:21
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    $\begingroup$ en.wikipedia.org/wiki/Thorpe_tube_flowmeter $\endgroup$ – Gert May 23 '16 at 15:50

If the pipe is cylindrical, there is no equilibrium height. All the air that goes in at the bottom must come out at the top, so the force balance will not depend on the height of the ball.

In a ball flow meter (rotameter), the pipe is slightly conical, so that the gap between the ball and the pipe increases as the ball rises.

If you want a very coarse estimate of the forces involved, you can use Bernoulli and make the additional assumption that due to turbulence, the kinetic energy gained in the gap between ball and cylinder is mostly (say a fraction $\alpha<1$) converted to heat. Start by assuming an incompressible fluid - the fact that air expands as the pressure drops makes it more difficult. If the cross section of the pipe is $A$, the volume flow rate is $Q$, the fluid density $\rho$, the cross section of the ball is $A'$, then the fluid velocity in the pipe is $v=Q/A$ and the fluid velocity in the gap is $v'=Q/(A-A')$. The net pressure difference would then be $\Delta p=\alpha\rho(v'^2-v^2)/2$, which acts on the area $A'$, resulting in a force $F=A'\Delta p$.

  • $\begingroup$ Very good answer. I'd worked out that with a cylindrical pipe there cannot possibly be an equilibrium height but couldn't get any further than that, until I saw a picture of a ball plus tapered cylinder. $\endgroup$ – Gert May 23 '16 at 16:06
  • $\begingroup$ I'll try this approach too and compare the results, I think that your model is nicer. I'll check this as the answer. Thank you! $\endgroup$ – João Paulo May 23 '16 at 16:43

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