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| Assuming the ambient air is at STP (P = 1 atm T = 293 K) and that the liquid N,,2,, comes to thermal equilibrium with the ambient air to room temperature, the maximum possible pressure in the barrel can be calculated. The barrel has a length of 34.75" or .88 m and a radius of 1.5" or .039 m. Thus the barrel has a volume of 4.2 L (assuming cylindrical geometry V = π * r^2^ * L )). Just after the barrel is sealed off from the atmosphere, there is .5 L of liquid N,,2,, and therefore 3.7 L of air inside. Thus there are [(1 atm)*(3.7 L)] / [(.082 atm L mol^-1^ K^-1^)*(293 K)] = .15 mol of air in the barrel. Liquid N,,2,, has a density of 807 g L^-1^ therefore there are 404 g of N,,2,, in the barrel. N,,2,, has a molar mass of 28 g mol^-1^ so there are 14.4 mol N,,2,,. Once all the liquid transitions to a gas, assuming there was enough friction to keep the projectile in place, the pressure would rise to P,,max,, = [(n,,air,, + n,,N2,,) * R * T] / V = [(144 mol)*(.082 atm L mol^-1^ K^-1^)*(293 K)] / (4.2 L) = ~ 80 atm or 810 N cm^-2^!!! It is difficult to determine the magnitude of the friction between the projectile and aluminum. However, it is apparent that 80 atm of pressure is more than enough. | Assuming the ambient air is at STP (P = 1 atm, T = 293 K) and that the liquid N,,2,, comes to thermal equilibrium with the ambient air to room temperature, the maximum possible pressure in the barrel can be calculated. The barrel has a length of 34.75" or .88 m and a radius of 1.5" or .039 m. Thus the barrel has a volume of 4.2 L (assuming cylindrical geometry V = π * r^2^ * L )). Just after the barrel is sealed off from the atmosphere, there is .5 L of liquid N,,2,, and therefore 3.7 L of air inside. Thus there are [(1 atm)*(3.7 L)] / [(.082 atm L mol^-1^ K^-1^)*(293 K)] = .15 mol of air in the barrel. Liquid N,,2,, has a density of 807 g L^-1^ therefore there are 404 g of N,,2,, in the barrel. N,,2,, has a molar mass of 28 g mol^-1^ so there are 14.4 mol N,,2,,. Once all the liquid transitions to a gas, assuming there was enough friction to keep the projectile in place, the pressure would rise to P,,max,, = [(n,,air,, + n,,N2,,) * R * T] / V = [(144 mol)*(.082 atm L mol^-1^ K^-1^)*(293 K)] / (4.2 L) = ~ 80 atm, 1,175 psi or 810 N cm^-2^!!! It is difficult to determine the magnitude of the friction between the projectile and aluminum. However, it is apparent that 80 atm of pressure is more than enough. |
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[:MEEquipmentList: List of Mechanics Equipment & Supplies] |
[:Demonstrations:Lecture Demonstrations] |
Liquid Nitrogen Cannon, 1H11.30
Topic and Concept:
- Newton's Third Law, [:Newtons3RDLaw#Recoil: 1H11. Recoil]
Location:
Cabinet: [:MechanicsCabinet:Mechanic (ME)]
Floor Item: ME, South Wall
attachment:CannonEq04-400.jpg
Abstract:
A homemade aluminum cannon with a 3" bore diameter and nearly 34" bore length 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 |
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Cannon |
Floor Item: ME, South Wall |
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Equipment bag |
Floor Item: ME, South Wall |
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Liquid N2 |
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Important Setup Notes:
This demonstration requires about .5 L of liquid nitrogen per shot.
Be sure to leave at least 4 m of room behind the cannon for recoil.
Setup and Procedure:
- Select the firing angle(angle of the cannon barrel relative to the ground). An angle of 45° will give the farthest range.
- Choose the desired projectile: soft stopper, hard stopper, or sand-filled bottle.
- Make sure the valve, situated on the barrel, is in the open position (handle parallel to the barrel).
- Pour about 0.5 L of liquid nitrogen into the barrel.
- Using the hammer, pound the projectile into the barrel, shortest diameter side in.
- Without waiting too long (more than 30 seconds and the LN2 will boil off), close the valve when ready to shoot the projectile.
- The shot will occur soon - once enough pressure has accumulated inside the barrel.
Cautions, Warnings, or Safety Concerns:
Do Not fire cannon into audience.
- The rubber stopper is shot with enough force to severely dent drywall at a distant of 70 feet away from the cannon.
Discussion:
After the valve is closed, the barrel becomes an air-tight vessel. Since the aluminum that the liquid nitrogen is in contact with is at room temperature, there is a lot of heat available to the liquid N2 causing it to boil rapidly. The amount of space the liquid N2 occupies is miniscule compared to the volume it occupies as a gas. The pressure is given by the ideal gas law: P = n*R*T / V where P is pressure, n is the number of moles of gas are in the container, R is the gas constant, T is the temperature of the gas, and V is the volume of the container.
Assuming the ambient air is at STP (P = 1 atm, T = 293 K) and that the liquid N2 comes to thermal equilibrium with the ambient air to room temperature, the maximum possible pressure in the barrel can be calculated. The barrel has a length of 34.75" or .88 m and a radius of 1.5" or .039 m. Thus the barrel has a volume of 4.2 L (assuming cylindrical geometry V = π * r2 * L )). Just after the barrel is sealed off from the atmosphere, there is .5 L of liquid N2 and therefore 3.7 L of air inside. Thus there are [(1 atm)*(3.7 L)] / [(.082 atm L mol-1 K-1)*(293 K)] = .15 mol of air in the barrel. Liquid N2 has a density of 807 g L-1 therefore there are 404 g of N2 in the barrel. N2 has a molar mass of 28 g mol-1 so there are 14.4 mol N2. Once 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 = [(144 mol)*(.082 atm L mol-1 K-1)*(293 K)] / (4.2 L) = ~ 80 atm, 1,175 psi or 810 N cm-2!!! It is difficult to determine the magnitude of the friction between the projectile and aluminum. However, it is apparent that 80 atm of pressure is more than enough.
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.
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 |
Videos:
- Demonstration video to come...
[https://www.youtube.com/user/LectureDemostrations/videos?view=1 Lecture Demonstration's Youtube Channel]
References:
Concept for our design was obtained from, [http://www.physics.uci.edu/~demos/mechanics.html U.C.I]; [http://www.physics.uci.edu/~demos/pdf/mechanics/1h11.30-liquid_nitrogen_cannon.pdf UCI's liquid_nitrogen_cannon.pdf].
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