By Jon Delman

A few days ago I saw a science-related YouTube video and, in need
of a physics news topic, decided to watch the video. The video showed someone
playing in a pool of sand. The guy in the video had filled a hot-tub with sand,
and was pumping in air to the bottom of the tub, causing the sand to exhibit
fluid-like properties. In the picture below, a smaller version of the sand hot-

tub is shown. A hose is pumping gas in from underneath the bin, and the ping
pong balls in the sand appear remarkably similar to ping pong balls in a fluid
solution.

So I decided to figure out the physics behind this fluidized bed.
The air being pumped into the bin is compressed nitrogen (N

_{2}) gas, which causes the sand particles to hover in an equilibrium state. We know that for a particle in a fluid, there are two main forces acting on it: the buoyancy force and the weight force.
F

_{B}= (ρ_{fluid})(V_{displaced})g
F

_{w}= mg
For a particle of sand to exhibit fluid properties, it would have
to have a net force of 0 acting upon it. With too much air (F

_{B}), the sand particles would fly out of the bin, but without enough air pressure, the sand would not resemble a fluid.
ΣF = ma = 0

F

_{B}— F_{w}= 0
F

_{B}= F_{w}
(ρ

_{fluid})(V_{displaced})g = mg
(ρ

_{fluid})(V_{displaced}) = m
It is challenging to put estimated numbers into this equation,
though, because we are referring to the sand particle as both the fluid and the
object within the fluid. However, we can make some approximations about the
ping pong balls in the fluid. The following calculation is based on the
assumption that the ping-pong is “hovering” (negligible F

_{N}) in the sand.
The mass of a ping-pong ball (m) = 2.7 g = .0027 kg

The radius of a ping-pong ball (r) = 20 mm = .02 m

ΣF = ma = 0

F

_{B}— F_{w}= 0
F

_{B}= F_{w}
(ρ

_{fluid})(V_{displaced})g = mg
(ρ

_{fluid})(V_{displaced}) = m
(ρ)*(4/3 * π * .02

^{3}) = .0027 kg
ρ = 80.6 kg/m

^{3}
If 1 cup (.00024 m

^{3}) of normal sand has a mass of .36 kg, then the density of a normal bin full of sand is:
ρ = m/v

ρ = .36 / .00024 = 1500 kg/m

^{3}
1500 / 80.6

^{ }= 18.75; so the ρ of a normal cup of sand is 18.75 greater than the ρ of a fluidized cup of sand.
Thus, the volume of fluidized sand is roughly 18.75 times larger
than the volume of non-fluidized sand. An interesting next step to this problem
would be to figure out how much gas would need to be pumped into the bin of
sand each minute in order to fluidize the bin of sand.

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