Yea certainly looks non-zero to me.

He gave another equation for Ve in message 34741, and allowed Po=Pe, but I'm not sure you can do that. For any nozzle shape there'll be more pressure on the inside than the outside - and thus some gas expansion. Without a pressure difference there wouldn't be any flow.

But we'll see better when he returns.

Yaaaaaayyy - I get to explain this to someone who'll actually try and understand, instead of trying to force the science to fit their preconceptions!

When you have a pressurized gas being released by a tank there are two potential sources of thrust - the pressure pushing the gas out of the tank, and the expansion of that compressed gas once it leaves the tank. The former will be applied onto the tank/rocket/whatever regardless of the nozzle configuration, the later requires a properly designed nozzle to be harnessed.

The former is calculated using [Pt-Pe]Ae, the later using m*Ve. The total force produced is the sum of those two values.

Pressure thrust:The pressure thrust is determined by Fp = [Pt-Pe]Ae, where:

Fp = force produced by pressure

Pt = pressure in the tank/rocket engine/whatever

Pe = pressure of the environment outside of the tank

Ae = area of the connection between the tank and outside environment

The calculation is fairly obvious (IMO) - pressure is the measure of

force per unit area, produced in this case by a gas. So you determine the pressure difference between the tank and the environment, multiply that by the area connecting them, and you'll get the force produced by the pressure differential acting through the opening between the tank and the environment.

Paul complains that this calculation does not take into account mass (and by extension, he should complain it doesn't take into account the exhaust velocity). However, those values are not needed to calculate the pressure force for the following reasons:

1) Pressure is dependent on the amount of gas present (i.e. on its mass, or more specifically, on the moles of gas present). Ergo, by knowing the pressure you already "know the mass", and that quantity is part-and-parcel of the resulting pressure.

2) The movement of air from the tank to the environment is choked - a fancy way of saying it has a maximum possible velocity (the speed of sound). And this is the velocity the gas travels at - with the exception of any slowing due to friction in the nozzle area. Once again, this velocity is part-and-parcel of pressure (its determined solely by gas density).

So despite Pauls complaints to the contrary, [Pt-Pe]Ae calculates the force provided by differential pressure under ideal conditions (i.e. no friction, ideal gas).

Nozzle thrust:The second source of trust is the stream of gas once it leaves the tank. This stream of gas is under pressure, relative to the environment. This pressure exerts an expansive force, perpendicular to the direction of gas flow. A properly designed nozzle can capture this outwards force, thus "converting" it into additional thrust.

To do so you need a conical nozzle which will allow the gas to expand at a rate close to, but slightly less than, the rate it would expand unencumbered. This will allow the gas to be accelerated to the maximum speed (several times the speed of sound!), at which point the nozzle is impacted by that gas, thus imparting additional force to the nozzle, and thus the tank/rocket.

This value is determined by Fn = m*Ve, where:

Fn = force produced by nozzle

m = mass

flow through the nozzle (i.e. kg/s)

Ve = the exit velocity of the gas from the nozzle, relative to the velocity of the gas when it enters the nozzle.

The formula I provided earlier allows for the calculation of Ve, given a known pressure change along the length of the nozzle (determined by the divergence angle), and the characteristics of the gas (pressure, temperature, etc).

In the situation provided by Paul - no divergence, simply a straight tube connecting the tank to the environment - Ve will be zero. This is because the gas is not allowed to expand within the nozzle, thus no additional velocity is given to the gas, and thus no additional thrust is provided. Simply put, in Pauls situation the additional thrust that could be provided is lost in the form of the gas expanding behind the nozzle.

EDIT: I would add at this point that, at least in the case of rocket engines, this expansive force is where the majority of the thrust comes from. In the case of the sugar-potassium nitrate rockets I used to build, 75-80% of the thrust was due to the nozzle, meaning the nozzle "amplified" thrust, on average, 4-5X. People with better machining skills than I often would get increases in thrust of 6X or more, relative to the pressure thrust alone.

The Math:The total force produced by this tank of pressurized air is equal to the sum of the above two forces:

Ftotal = Fn + Fp = [m*Ve] + [Pt-Pe]Ae

In Pauls case Fn is zero, as there is no divergence of the nozzle and thus Ve is zero. Meanwhile, Fp is some non-zero value, thus:

Ftotal = Fn + Fp

Ftotal = 0 + Fp

Ftotal = Fp

Complications:There are some complications ignored in the above formula. There is the issue of in-engine friction, which reduces thrust. This can be largely ignored assuming proper engineering, but the true force of such a system will always be slightly less than calculated due to friction. Likewise, those formulas assume an ideal gas, and thus any non-ideal gas (i.e. all of them) will behave a little different than predicted - your total thrust tends to be calc'd right, but the force produced at any one time, the time it takes to empty the tank, etc, will be different than those calculated above.

All that said, Paul accidentally gave us a situation where the biggest complication has been eliminated - notably he has an engine expelling one phase (a gas). Solid rockets, as well as many liquid/gas rockets, produce two-phase flow (i.e. you'll have solid or liquid particulates in the gas stream). In these cases it is not as straight forward as illustrated above, as the force provided by the gaseous fraction components will only produce force due to their acceleration as they leave the engine. This gets quite complicated mathematically speaking, as the particulates can be accelerated twice - first as they go from the combustion chamber into the linear part of the nozzle, and again when they are in the divergent part of the nozzle. That later acceleration is particularly difficult to deal with, as much of the acceleration is perpendicular to the gas stream (and thus does not provide thrust), while at the same time there is some acceleration along the gas stream AND you have particulates impacting the nozzle itself.

Anyway, how does this show that momentum isn't conserved? Or how does it show that you can get sustained propulsion of a closed cylinder?

Exactly - that is where this whole thread started. Momentum is conserved in a closed system. Paul has described a closed system. Ergo, Pauls engine cannot produce a net thrust on his closed system...

...Paul pretends otherwise, by pretending the gas escaping the tank doesn't have momentum of its own, but ignoring reality doesn't change reality.

I realize the above was ridiculously long, but I hope I described things in a way people can understand.

Bryan