Object falling through highly compressed air question

[Danimal]

If you had a tube a couple of miles long filled with very compressed air, say 6,000 PSI, would an object you dropped in it fall very slowly? Even a heavy object like an anvil, how long would it take to drop?

 

[Halc]

Terminal velocity is ((2mg)/(ρAC))
mass, gravity/acceleration, ρ density of fluid, Area and C=drag coefficient

So 6000 psi is about 400 times the density of atmosphere at sea level, so the anvil falls at around 1/400th the speed. This formula doesn't take into account the buoyancy of the falling object. If it is low density and immune to compression by the fluid, eventually its weight reduces to zero and it doesn't fall at all. Hence I suspect the mg part of the equation (which equates to force in a vacuum) should be converted to weight, which already has buoyancy worked in.
So 2f/(ρAC) where f is weight, is how I would express it.

 

[Baluncore]

Even a heavy object like an anvil, how long would it take to drop?

Welcome to PF.

I could ask a similar question.
At what air pressure will an iron anvil float with neutral buoyancy ?
When the air has the same density as steel; 7.8 g/cc = 7800 kg/m³.

At sea level, air pressure is 14.7 psi, and the density is 1.225 kg/m³.
7800 / 1.225 = 6367.3 atm.
6367.3 * 14.7 psi = 93600. psi.

 

[phyzguy]

Another way to look at it is that at 6000 psi, air has a density of about 500 kg/m^3, so about 1/2 the density of water. So you can get a rough idea by asking how fast it will fall in water. An iron anvil will certainly fall in water, but much more slowly than in air. You can calculate how fast with the formula in Post #2. It is estimated that the Titanic took about 5 minutes to fall the 13,000 feet to the ocean floor, which works out to about 30 miles/hour.

 

[Chestermiller]

Another way to look at it is that at 6000 psi, air has a density of about 500 kg/m^3, so about 1/2 the density of water. So you can get a rough idea by asking how fast it will fall in water. An iron anvil will certainly fall in water, but much more slowly than in air. You can calculate how fast with the formula in Post #2. It is estimated that the Titanic took about 5 minutes to fall the 13,000 feet to the ocean floor, which works out to about 30 miles/hour.

What's the viscosity of air at 6000 psi, and how does that come in for the drag?

 

[phyzguy]

What's the viscosity of air at 6000 psi, and how does that come in for the drag?

Good question. If I understand correctly, the viscosity of air, even at 6000 psi, is approximately 50X less than the viscosity of water. So I think you are saying that my analogy with water is false, and an object will fall a lot faster through 6000psi air than through water. Is that correct? How do I take the viscosity into account when calculating the drag force?

 

[Chestermiller]

Good question. If I understand correctly, the viscosity of air, even at 6000 psi, is approximately 50X less than the viscosity of water. So I think you are saying that my analogy with water is false, and an object will fall a lot faster through 6000psi air than through water. Is that correct? How do I take the viscosity into account when calculating the drag force?

I don't know the exact viscosity (it is, of course, available or can be obtained from a corresponding states plot). If the body were a sphere, you could use the drag coefficient vs Reynolds number for a sphere.

 

[jbriggs444]

Terminal velocity is ((2mg)/(ρAC))
mass, gravity/acceleration, ρ density of fluid, Area and C=drag coefficient

So 6000 psi is about 400 times the density of atmosphere at sea level, so the anvil falls at around 1/400th the speed. This formula doesn't take into account the buoyancy of the falling object. If it is low density and immune to compression by the fluid, eventually its weight reduces to zero and it doesn't fall at all. Hence I suspect the mg part of the equation (which equates to force in a vacuum) should be converted to weight, which already has buoyancy worked in.
So 2f/(ρAC) where f is weight, is how I would express it.

That formula makes terminal velocity inversely proportional to fluid density. But if you look at an energy analysis and consider the energy imparted to the column of air displaced in one unit of time then if you fall twice as fast...

You displace twice the mass of air and impart twice the velocity, leading to eight times the energy
You fall twice as far.

So your energy expended per unit fall scales as the square of velocity.

Meanwhile, if you halve the density, the column masses half as much and the energy imparted per unit fall is halved.

So the energy expended per unit fall scales directly with the density.

Since the energy available per unit fall is fixed for a given mg, you'd expect terminal velocity to scale inversely with the square root of density.

And indeed, after working the back of the envelope, I visit Wikipedia and see:
$$\displaystyle V_{t}={\sqrt {\frac {2mg}{\rho AC_{d}}}}$$

 

[Halc]

Indeed, I forgot the square root part. And I was looking right at it.

The formula you quote (and which I should have quoted) still doesn't take buoyancy into account, as described in my post. Terminal velocity of a helium balloon would be downward, not upward, according to that formula.

 

[Danimal]

Wow, thank you everybody, this forum is a goldmine!