Originally Posted By: paul
this is not a thrust that is the result of combustible gasses , it is simply air that is compressed.

But Paul, it is the same in either case.

Combustion in a confined space produces compressed gas, exactly as you would have in your tank. The physics is the same afterwards - that pressurized air will have potential energy which cam be extracted in the two ways described above.

There are only two ways in which the two systems differ:
1) The pressure dynamics over time are different for a rocket, as the pressure in the "tank" in continually replaced by combustion products, and
2) The two-phase flow issue, as described in my earlier post

Originally Posted By: paul

I would like to see your results from the following

and your explaination of it.

Already provided, in detail, in my first post today. But to clarify:

If by 'm' you mean simple mass, and V is velocity, than the above formula calculates the momentum of the air exiting your tank. In this case your F should be a P, as F is the symbol for force, not momentum.

If m = mass flow (kg/sec) and V equals the exhaust velocity (i.e. Ve), that it describes the force produced by an divergent nozzle, as described in my last post.

Exhaust velocity is not the same as velocity, but rather is the increase in velocity due to expansion in the nozzle. No expansion, no increase. No increase, Ve = 0.

In simplest terms, Ve = Vx-Vi, where
Ve = exhaust velocity,
Vx = velocity of the gas when it exits the nozzle
Vi = velocity of the gas when it enters the nozzle

For a non-divergent nozzle, as in what you described, there is no acceleration along the length of the nozzle, and ergo Vx = Vi. As such:

Ve = Vx-Vi; Vx=V1, therefore
Ve = Vx-Vx
Ve = 0

The formula I provided earlier for Ve allows for the calculation of Ve without having to actually measure these velocities directly. This is because these velocities will be equal to the speed of sound (Vi) and the speed induced by the pressure, over the length of the nozzle. In the post where I first presented the formula for Ve I went through the critical part of this calc - notably that Ve is a multiple of the fraction of pressure "consumed" by the divergent nozzle. This is strictly dependent on the difference in the pressure of the exhaust when it leaves the tank [Pt] verses when it leaves the nozzle [Pe], as the ratio between these two determines how much of the potential expansion energy in the compressed gas stream is being "captured".

In the case of your straight tube connecting the tank to the outside world there is no divergence. Thus there is no decrease in the pressure of the air stream within your nozzle. Ergo, Pe=Pt, and therefore the part of the formula posted earlier which is:

Fraction of pressure consumed = 1-[Pe/Pt]
= 1-[Pt/Pt]
= 1- [1]
= 0

Since the entirety of the Ve formula is multiplied by that value, Ve itself will be zero.

Originally Posted By: paul

although it is a absolute meaningless part of the formula
according to you , do you honestly think that there will be no mass comming out of the tube therefore there will be zero velocity.

I never said that, and it is clear from this comment you did not read the very post you quote-mined.

The force produced due to the pressure-driven movement of air is determined by [Pt-Pe]Ae. Because pressures are used, both the mass and gas velocity are intrinsic to the measurement, and as such you do not need to calculate them as separate entities.

What you are trying to do is force the m*Ve to your preconceptions, without calculating m*Ve properly. m*Ve is ***not*** the mass flowing through the nozzle times its speed - that doesn't even calculate a force, but rather a momentum. A momentum is not a force, and thus you cannot simply add the momentum you're calculating to the force calc'd by [Pt-Pe]*Ae