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Liquid Nitrogen Cannon, 1H11.30

Topic and Concept:

Location:

attachment:CannonEq04-400.jpg

Abstract:

A homemade aluminum cannon with a 3" bore diameter and bore length of nearly 34" is mounted on two 24" bicycle wheels and shoots either rubber stoppers or bottles filled with sand to show recoil. The only propellant is the pressure buildup from the liquid nitrogen transitioning from liquid to gas.

Equipment

Location

ID Number

Cannon

Floor Item: ME, South Wall

Equipment bag

Floor Item: ME, South Wall

Should be with Canon

Liquid N2

Important Setup Notes:

Setup and Procedure:

  1. Select the firing angle (angle of the cannon barrel relative to the ground). An angle of 45° will give the farthest range.
  2. Choose the desired projectile: soft stopper, hard stopper, or sand-filled bottle.
  3. Make sure the gas bleeder valve that is situated on the barrel is in the open position, (handle parallel to the barrel).
  4. Pour about 0.5 L of liquid nitrogen into the barrel.
  5. Using the rubber mallet, pound the projectile into the muzzle, tapered end first. The rubber stoppers can be pounded flush with muzzle.
  6. Close the gas bleeder valve. This will allow the nitrogen gas to build up which in turn causes the pressure buildup.
  7. Grab the gas bleeder valve assembly and tip the cannon muzzle down to let the liquid nitrogen to slosh up and down the bore to accelerate the evaporation of the liquid nitrogen.
  8. Made sure that the cannon is aimed at the mat on the wall

  9. You won't need to wait too long, about 30 seconds or so. Once enough pressure has accumulated within the barrel, the projectile will shoot out with an extreme force and velocity.

Cautions, Warnings, or Safety Concerns:

Discussion:

After the valve is closed, the barrel becomes air-tight and nitrogen gas (and therefore pressure) is free to build up. In general, the aluminum cannon is at STP (P = 1 atm or 101,325 Pa; T = 293 K or 68 °F or 20 °C), and the liquid nitrogen has a temperature of -196 °C (-321 °F or 77 K). There is quite a lot of heat available to cause the liquid nitrogen to boil rapidly and transition into gas. However, the amount of volume that nitrogen occupies when in a liquid state is miniscule compared to the volume it can occupy when in a gas state. This ratio is 1 to 694 at STP and is called the liquid to gas expansion ratio.

The pressure is given by the ideal gas law: P = n*R*T / V where P is the gas pressure, n is the number of moles of gas within the container, R is the gas constant, T is the temperature of the gas, and V is the volume of the container. We will be using R = 0.082 L  atm  K−1  mol−1 and assuming the ambient air is at STP, and that the liquid nitrogen rapidly comes to thermal equilibrium with the ambient air to standard temperature. One should note that this is an ideal case, and that the barrel will cool down rapidly thus reducing the local ambient temperature and therefore the pressure.

With this, the maximum possible pressure within the barrel can be calculated. The barrel has a length of 37.5" but the bore length is 34.8" or .884 m and a radius of 1.50" or .0381 m. However, when the stopper is pounded into the bore, the volume decreases. The thickness of the hard stopper is 1.50" or .0381 m. The thickness of the soft stopper is 1.625" or .0413 m. Let's use the soft stopper in the calculation. Then the effective bore length is 33.19" or 0.843 m. Thus the effective volume in the barrel is 3.84 L (assuming cylindrical geometry V = π * r2 * L and converting to liters)).

Just after the barrel is sealed off from the atmosphere, there is .5 L of liquid nitrogen and therefore 3.235 L of air inside. Thus there are [(1 atm)*(3.235 L)] / [(.082 atm L mol-1 K-1)*(293 K)] = .135 mol of air in the barrel. Liquid nitrogen has a density of 807 g L-1 therefore there are 404 g of liquid nitrogen in the bore. Nitrogen has a molar mass of 28 g mol-1 so there are 14.4 mol nitrogen within.

If all the liquid transitions to a gas, assuming there was enough friction to keep the projectile in place, the pressure would rise to Pmax = [(nair + nN2) * R * T] / V = [(14.65 mol)*(.082 atm L mol-1 K-1)*(293 K)] / (3.235 L) = 108.8 atm or 1,598.9 psi or 1102.4 N cm-2!!! It is difficult to determine the magnitude of the friction between the projectile and aluminum. However, it is apparent that 108.8 atm of pressure is more than enough. We can refine this number to get the projectile firing back pressure, if we estimate the evaporation rate of LN2 with the cannon. This can be done several ways. The easiest way is to just time how long a liter take to evaporation at STP. I'll work on this and add that info when I get it.

In any case when the projectile is shot, the cannon imparts a force on the projectile and the projectile imparts an equal but opposite force on the cannon as per Newton's third law. From the perspective of momentum conservation, the total amount of momentum in the cannon-projectile system is zero: Pprojectile = -Pcannon. The extent of the cannon's recoil is much smaller than that of the projectiles rage due to the large mass differences - made apparent by Newton's second law F = m * a.

Pertinent values:

attachment:Cannon16-250.jpg

attachment:Cannon03-250.jpg

attachment:Cannon08-250.jpg

attachment:Cannon07-250.jpg

attachment:Cannon12-250.jpg

attachment:CannonEq04-250.jpg

attachment:Eq01-250.jpg

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References:

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