Determine the launching speed of the object

[Kryptix]

So this is probably a very basic question. I'm studying EE so this is kind of out of my field as i have never taken this but I'm working on a project and it involves a pressurized container releasing air into a tube and launching something out the nozzle.

The problem I'm having is that I'm not entire sure how to determine the launching speed of the object given purely the velocity of the air exit the nozzle.

I've been doing research and some calculations that at ~54psia the air exiting the nozzle would be moving at mach 1 speeds (~300m/s) as long at the ratio between p/po is less than 0.528

I've been thinking of using the basic principle of conservation of momentum in order to determine the launching velocity of the solid leaving the nozzle, but I run into a problem. I'm not sure of the mass of air.

I've asked around as well and people have mentioned things such as control volume but have provided no explanation as to what it is or how to set it,(not to mention how to use it).


Any help on determining the exit velocity of a solid given the velocity of the air would be greatly appreciated.

Thanks again :)Other info: Diameter of Nozzle: 0.5 inch. PSI: 40. Atmospheric pressure: 1atm
Mass of solid being launched: 7 grams.

 

[mfb]

Do you have a sketch of the setup? In the limit of long nozzles, the mass will approximately reach the same speed as air without the projectile. For very short nozzles, speed is limited by the acceleration given by the pressure difference. Everything between is more complicated.

 

[Kryptix]

Do you have a sketch of the setup? In the limit of long nozzles, the mass will approximately reach the same speed as air without the projectile. For very short nozzles, speed is limited by the acceleration given by the pressure difference. Everything between is more complicated.

umm yeah, i have the sketch but i'd have to wait til i get home to get. For now I can tell you that the travel path of the air and the solid is about 2 ft.

 

[Kryptix]

I will post sketch later.