Originally Posted By: paul
the larger area did not reduce the velocity below the
340.29 m/s^2
the same mass is involved.

so the momentum I stated earlier is the same
the momentum is 99,557 kg-m/s

Absolutely, but this comes back to another common issue with you - you consistently confuse momentum and force.

Lets use simpler quantities to make the math more obvious - your tank is at 1000Pa (1kPa), and contains 1kg of air. For the sake of simplicity, the environment is a vacuum - 0Pa.

With your straight pipe that 1kg of air will be accelerated to the speed of sound - 340m/s. This means that after leaving the tank the air has a momentum of 1kg*340m/s = 340kg*m/2 - regardless of the size of tube you use.

However, the force produced will vary depending on the size of the tube. Take the case where you have a tube with a cross-section of 1m^2 verses 2m^2:

F = [Pt-Pe]Ae
If Ae = 1m^2, then F = [1000Pa-0Pa]1m^s = 1000N
If Ae = 2m^2, then F = [1000Pa-0Pa]2m^s = 2000N

Now what if we add a nozzle that has a divergence angle which doubles the speed of the air (i.e. accelerates it to 2X the speed of sound - 680m/s)?

Now your momentum doubles - 1kg*680m/s = 680kg*m/2 - regardless of the size of tube you use (so long as the nozzle is scaled appropriately).

Your force will also be larger:

The [Pt-Pe]Ae part will remain the same as above - i.e. a 1m^2 will impart 1000N of force. But we've also got the additional acceleration of the air to 2X the speed it would have if there was no nozzle. I've not included enough information here to calculate that, based on the above formula.

But I kinda cheated - the only time you would have a situation where a nozzle would double the speed (and thus double the momentum) of the air passing through it is when you have a nozzle designed such that it produces an amount of force equal to that of [Pt-Pe]Ae.

Ergo, the force produced by our nozzle is:
If Ae = 1m^2, then F = 2000N
If Ae = 2m^2, then F = 4000N

The above nozzle, BTW, is fairly inefficient. Most nozzles accelerate to 3-5X the speed of sound. There is a theoretical maximum you can extract, which depends mostly on the pressure and temperature of the gas exiting the nozzle. For a rocket this usually comes in around mach 6.4. It would be lower in your case, due to the relatively low temperatures involved.


Edited by ImagingGeek (06/09/10 09:11 PM)