Question: Could a free falling cylindrical tank pass down through a tube/gate mechanism into a compressed air chamber? Specifically: Would the force of the compressed air in the compressed air chamber stop the falling cylinder from passing through a gate mechanism into the compressed air chamber? Description of concept: A specially designed cylinder is allowed to free fall in a guided manner straight down into a tube which is part of a gate mechanism at the top of a compressed air chamber. The tube is lined with compression rings so as to maintain a tight seal with the specially designed cylinder as the cylinder passes down through it. Half way down this tube sits a gate. The gate is triggered to open just before the falling cylinder reaches it. The cylinder continues to fall passing through the now open gate and on down the tube. The cylinder continues down the tube and drops out of the bottom of the tube into a large compressed air chamber. The gate mechanism in the tube is triggered to close just as the cylinder exists the tube and enters the compressed air chamber.

Specifics
Tank Cylinder: 2 ft diameter X 15 ft long
( 47.10 cubic ft)
Weight 2,645 lbs

Cross sectional area of cylinder 3.14 sq ft. (452 sq inches)
Tube cross sectional area just slightly larger than cylinder's to accommodate the compression rings.

Free fall distance of cylinder: 576 ft.

Speed of cylinder at tube/gate mechanism 192 ft/sec

Pressure of air in compressed air chamber 500 psi

Length of Tube through which the cylinder must pass to enter the air chamber = 30 ft. (15 ft above the gate and 15 ft below the gate.)

Size of compressed air chamber
30 ft X 60 ft(deep) X 30 ft = 54,000 cubic ft.

The cylindrical tank is designed with a rounded nose cone type base which hopefully gives it a more aerodynamic design for its free fall and its encounter with the compressed air in the tube that enters the comptessed air chamber.

The question here is: Could the momentum force of the free falling cylinder overcome the 500 psi air pressure resistance in the 30 ft of tube that the cylinder must pass through before it exists the tube and enters the compressed air chamber?

Consideration: Friction in the tube that the cylinder must pass through would be similar to the friction of a piston in a compression-ring lined cylinder of an automobile engine. I do not think that friction within the tube would be prohibitive.

Would appreciate help you could give me on this one.

[email protected] if more info is needed.

Rowland Reeves

If I understand the question properly, this cylinder is falling vertically at 192 ft/sec when it is suddenly exposed to 500 psi upward pressure, right? If you draw a force diagram for the cylinder it will have a downward force due to gravity of 2645 pounds, and an upward force due to the air pressure of 500psi x 452 sq in = 226,000 pounds. The acceleration that the cylinder experences, using F = ma, is:


a = \frac {(226000-2645)lb_f} {2645 lb_m} \cdot {32.2} \frac {lb_m ft} {lb_f s^2}\
= 2719 \frac {ft} {s^2}


This is an upward acceleration, equal to 84.4 times the acceleration due to gravity. The distance the cylinder travels downward before coming to a stop and reversing course is:

d = v^2/2a = \frac {(192 \frac {ft} {s})}^2 {2 \cdot 2719 \frac {ft} {s^2}} = 6.8 ft


At this point it reverses course and goes shooting upward, like out of a canon. So, no, the cylinder won't make it through the tube and into the pressurized vessel.