I agree that momentum is conserved also , but its not like Im agreeing with you , as that is something that makes sence.

so let me get this straight - you agree with the fundamental physical principal which clearly states your "engine" is a physical impossibility, and yet you insist your "engine" will impart momentum...

A little schizoid, are we wink

yes the momentum is conserved in the moving pipe itself. nothing left the pipe.

Ergo, the pipe does not move.

Actually, the speed of sound is neither a constant, nor is it just the "speed that sound can travel at". The speed of sound varies with the physical makeup of the medium and its temperature.

so does g 9.8m/s^2
but we use it as a constant --- ACTUALY---
whats your point other than nic picking.

The point is the very part you dishonestly cropped - the speed of sound sets the maximum velocity a gas can accelerate itself to, using its own pressure as the source of that acceleration.

Nor is the speed of sound anywhere near as constant as the gravitational acceleration of earth - the speed of sound changes with the constituents of the gas and temperature.

if you will notice everything you are refering to is about rocket engines , im not using a rocket engine.

so all your links to rocket engines are null.

The physical principals are the same, regardless of the source of the pressure. The fact you expect otherwise defies not only the laws of physics,but basic logic as well.

a rocket engine uses a nozzle to direct the thrust

Nope, even your own sources refute that claim - the nozzle accelerates the gas, providing thrust beyond that of the pressure differential alone.

so just like the bottle rocket does not have a nozzle after the thrust passes the lip of the bottle , my pipe and tank anology also does not have a nozzle it
just has a 1 inch sq area tube , why do you keep insisting on using nozzles when there isnt a nozzle?

But I'm not insisting its a conical nozzle!!!! In fact, I've repetitively stated tat your nozzle is linear - as in a tube, exactly as you describe.

That is, after all, the entirety of the point I've been trying to make the past 10 or so posts - for Ve to have a non-zero value, the nozzle must have a divergent section. No divergent section, no change in the velocity of the gas escaping the "tube", thus Ve = 0.

I never said it had a nozzle , you did.
in fact at first you acknowledged it was just a tube.

Nothing, I simply used the word "nozzle" to differentiate that part of your contraption from the pipe and the tank. Too many tube-like objects.

As I've stated time and time again, your nozzle is tube-shaped; as in it has no divergent angle.

No divergent angle = no acceleration of the gas within the nozzle

No acceleration of the gas in the nozzle = no Ve

you included a formula to compute the nozzle thrust
and disreguarded the only thing that actually was there
the tube.

Wrong again. If you go back and check my calcs you'll see I was using the exact conditions you set in the formula - no divergence in the nozzle (i.e. it is a tube).

So basically you're complaining that I did the math right for the conditions you explained, but chose to use what I thought was a less confusing term for the particular part of your contraption.

ok , I want to allow the air to expand as it wants...I dont want the air to be directed towards the rear as you seem to want it to.

Wants and desires have no role in physics - physics obeys set, unalterable laws. In the case of the force generated when a gas moves from a region of higher pressure there are two factors involved - the pressure differential over the exit area; determined mathematically by [Pt-Pe]Ae, and the force generated by the gas as it expands in a divergent nozzle (if there is a divergent nozzle); determined mathematically by m*Ve.

As you said, and as I've agreed since the beginning, you do not have a divergent nozzle; simply a straight pipe. Under that condition, Ve is zero, leaving the sole source of force as the pressure differential - [Pt-Pe]Ae.

As for whether your pipe moves or not, that's a matter of the momentum in the system not the total force generated by the expulsion of air from the tube/nozzle.

Your contraption, nozzleless or nozzle containing, imparts momentum to the tank, which then transfers that momentum to the pipe. However, the same amount of momentum will be transferred to the air coming out of the pipe/nozzle, in the opposite direction. That momentum will be transferred to the opposite end of the pipe, where it will neutralize the momentum induced by the tank.

The nozzle doesn't change the momentum factor - all it does is more effectively transfer the pressure in the tank into momentum. But regardless, that momentum will be equal, but opposite, in the tank verses the air, and thus when the tank and air impart their momentum onto the pipe, the momentum cancels out. The only thing the nozzle does is determine how much momentum you "extract" from the pressurized gas. But regardless of your "conversion efficiency", the momentum imparted onto the tank will always be equal and opposite to the momentum imparted into the air.


will the pipe move?

is like saying will the bottle rocket go anywhere.

Completely different situations - the bottle rocket is part of an open system - its exhaust transferes its momentum into the environment, not back into the bottle rocket. If you were to seal that rocket into a pipe, the rocket would move - the pipe would not.

And the reason is simple - the water coming out of the back of the rocket will propel it forward in the pipe. The expelled water will be propelled backwards in the pipe. Both will have the same momentum. When the rocket hits the front of the pipe, it transfers that momentum to the front of the pipe. When the water hits the back of the pipe, it transfers its momentum to the back of the pipe. Since both the rocket, and the water, have the same momentum, the net momentum imparted on the pipe is zero.

Ergo, the pipe will not move.

And since you're stuck on the nozzle issue, bottle rockets are propelled by a liquid - water - not a gas. Because water is incompressible at water-bottle pressures, the water will not expand in volume when it leaves the "tube". Ergo, having a divergent nozzle will not improve performance, as there is no expanding gas to take advantage of.


But fill that bottle with compressed air alone, and a nozzle will give you great dividends:

But, and I emphasize this again, regardless of how efficiently or inefficiently you harness the expansion of that gas, the amount of momentum imparted onto the tank will always be equal, but opposite, to the amount of momentum imparted on the gas leaving that tank. And since your system is closed, the net effect will be no net change in momentum of your pipe.


Edited by ImagingGeek (06/09/10 01:14 AM)