Thursday, February 4, 2010

Reinforced Helium Balloons

Getting concrete to float on water is a common contest for first year engineering students. By molding the concrete into a boat shape, one can make it so that the mass of water displaced is greater than the mass of the concrete. When this happens, the buoyant force will allow the concrete to float on the surface of the water. Buoyancy is also what causes helium balloons to float in air. That said, I’d like to propose an even more challenging contest for advanced students: getting a concrete balloon to float in air.   How large of a concrete balloon would you need in order for it to float when you filled it with helium?

Consider a spherical balloon of radius R. The surface area A of the balloon is given by the formula,

A = 4 π R2.

Since concrete is brittle and likely to crack, the balloon will need to be fairly thick so it won’t leak. If we assume its thickness T is 1.0 cm, then we can approximate the total volume V of concrete as,

V ~ 4 π R2 T.

According to, the density ρ of concrete is 2400 kg/m3. From this we can compute the total mass M of the concrete,

M = 4 π R2 T ρ.

To get the balloon to float, we need the net mass displaced to be at least equal to the mass of the balloon. Put another way, the mass of air displaced by the balloon has to be equal to the mass of the concrete balloon plus the helium inside. The density of air is ρair =1.2 kg/m3 and the density of helium is ρHe= 0.1786 kg/m3. The total volume of the balloon is given by,

V = 4 π R3 / 3.

By using the buoyancy condition described above, we can set up an equation to find the radius of the balloon:

(Mass of Concrete) + (Mass of He) = (Mass of Displaced Air),
(4 π R2 T ρ) + (4 π R3 ρHe / 3) = (4 π R3 ρair / 3).

This equation can be solved for R:

R = 3 T ρ / (ρair - ρHe)
= 3 · (0.01 m) · (2400 kg/m3) / (1.2 kg/m3 – 0.1786 kg/m3)
= 70 m.

That’s a balloon 140 m in diameter or roughly one and a half football fields.

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