# Buoyancy force problem

I'm trying to design a kind of water valve with inexpensive materials as a first prototype. The water flow from the PVC pipe (1) reach the body of the valve and pass through an aluminum grid (3) to the water tank. When the water level goes up pushes the float closing the water intake at point (2). How can I calculate the buoyancy force needed to stop the water flow? And, what will be the mass of the float?

Let’s back to the basics; here I present the problem and some math that I been doing, I would like your opinion:

• Connected to the PVC pipe (2) I have a garden hose with a water flow pressure of, let's say... 49 kPa (I need to check this with a manometer), and I attached a 25 mm diameter and 0.5 m long PVC pipe. Let’s pretend that the float seals the other side of the PVC pipe, so I need to calculate the force generate the water flow pressure against the float.

Please take in consideration that I'm not a fluid mechanic expert.

When I open the garden hose, the PVC pipe starts to fill, so based on this situation: $$P_1+\dfrac{\rho gh_1+v^2\rho}{2}=P_2+\rho gh_2+\dfrac{v^2\rho}{2}$$

If I took the height of $P_1$ as the reference, $h=0$, and the diameter of the PVC pipe and the garden hose pipe are the same (25 mm), the water flow velocity at those points are equals:

$$P_1=P_2+\rho gh_2$$

So, if the garden hose pressure is 49 kPa:

$$\begin{gather} 49000\text{ kg}/\text{ms}^2=P_2+9.8\text{ m}/\text{s}^2\cdot 1000\text{ kg/m}^3\cdot 0.5\text{ m} \\ P_2=53900\text{ kg/ms}^2 \\ P_2=53.9\text{ kPa} \end{gather}$$

Ok, assuming this math is correct…now I have to calculate the force against the bottom of the PVC pipe at point 2:

$$P=F/A$$

In order to simplify this example, I took the diameter of the PVC pipe as the contact area.

$$\begin{gather}A=\pi r^2=\pi\cdot0.025^2=0.002\text{ m^2} \\ F=107.8\text{ N} \end{gather}$$

If the pressure of the water flow generates a force of 107.8 N, I need an opposite force with a higher value to counteract it. Is that correct?

My goal is to find a material (mass; area) that generate enough buoyancy force to stop the water flow through the valve and seal the water intake, and when the water level goes down, the float valve will let pass the water flow to continue to fill the water tank.

• Is that not exactly what the float valve in your toilet cistern does? The buoyancy force of the float is magnified by a long lever which applies the force necessary to close the valve. Dec 23, 2017 at 8:45
• I guess, in theory, this is possible. But for actual, you will be dealing with much more problems, e.g. dynamic impact of water in the float, the stability of floating body. I suggest do a spherical float to help you with this. Dec 27, 2017 at 2:14
• I tried with a ping pong ball, but the body valve volume was too big, so the ball didn't seal the water intake. I'm preparing a couple of tests to figure it out the mass value of the float that I need to cut the water flow. Maybe if I use a small valve diaphragm to seal the water intake...I don't know. Dec 27, 2017 at 11:22
• Given there is a gap between the float and it's body, won't the float move around and thus decrease the chances of it's notch passing through the mouth of the pipe for effective shutoff? Oct 28, 2021 at 7:03

To answer your question requires simple calculations of equilibrium. In order to close the valve, the external force must exceed internal force by a fraction.

Thinking logically, $$\rho = \frac {Force} {Area}$$ where $$\rho$$ is applied stress. This means that the volumetric value of the float directly affects the force it exerts against incoming water pressure. To reduce the volume of the float, it is placed on an arm because the fulcrum effect substantially multiplies the force exerted against incoming water pressure. This is why the ballcock is used in toilet cysterns so effectively. Were the ballcock arm half as long, the ball would be double in size... and therein less your design problem: float size.

The lever arm system is the most effective at keeping float size to a minimum.

In the problem as you have stated it, equilibrium is reached when the forces, and not the pressures, at the top and bottom of the stop cock are equal. Thus,

P1A1 + mg =ρghA2 where:

P1 is net of atmospheric pressure i.e. the gauge pressure. If the absolute pressure is used, then there's P2 = 1 atm on the RHS * A2

m is the weight of the cock h is the height of the water column in the tank above the cock (in the diagram it appears that the water surface and the bottom of the float are at the same level. This cannot be, as in such a case there is no hydrostatic pressure acting on the float bottom to counter the pressure from the top).

A2 is the cross sectional area of the cock bottom

A1 is the cross sectional area of the cock top $$\sum Fy = 0$$

$$F_1 + W_F = \gamma_wh_2A_F$$

$$h_2A_F = \dfrac{108 + m_f*g}{\gamma_w}$$

From here, you need to solve this problem by iteration. For the preliminary trials, I would assume $$W_F$$ is negligible. Once the optimum $$h_2$$ & $$A_F$$ is find, re-iterate the result with $$W_F = m_f*g = \rho Vg$$ (\$\rho = mass density of the material, V is the solid volume of the float).