Originally Posted By: kallog
Originally Posted By: ImagingGeek

The difference is one of ideal verse non-ideal gases. NASA has the classical (i.e. ideal gas-based) formula on their page, while I provided the one derived for non-ideal gases

Hang on, for a straight nozzle:
With ideal gasses the thrust is pressure thrust + an extra, nonzero term.
With non-ideal gasses the thrust is only pressure thrust, and the other term becomes exactly zero?

After seeing your comment I realize I explained things very, very badly. My apologies.

Ideal gases do not exist in the real world - they are a theoretical construct used as an approximation of real gases. For static gases - i.e. ones just sitting around, not moving - ideal gas calculations work very well. Once you start talking about a flowing gas, ideal gases begin to fall flat.

Ideal gases are almost exactly the same as real gases - ideal oxygen would have the same molecular weight as real oxygen, and thus carry the same kinetic energy at a given temperature as oxygen, etc. Where idea gases differ is the atoms/molecules of an ideal gas are are considered to have zero volume (as in they are infinitely small). Real gases of course are made of atoms/molecules, and those atoms/molecules have volume.

Having no volume has two dramatic implications. even at the atomic scale, friction is determined by contact area. By definition, a particle with zero volume also has zero surface area - hence, ideal gases do not experience friction. The second major implication is that atoms/molecules of ideal gases do not interact with each other - if they have no volume, they can never run into each other.

I'd also point out at this point that I stated something very badly - helium is not an ideal gas, but at low temperatures its inter-molecular interactions are so weak that it takes on some of the properties of an ideal gas.

So the Q now is how does the ideal gas differ compared to a real gas in pauls tank/pipe system or in a rocket engine.

The answer is both "no difference" and "big difference". Keep in mind the formula for force: m(dot)Ve + [Pt-Pe]Ae

The later half - [Pt-Pe]Ae - calculates the pressure force. Pressure is simply force per area - if pauls tank was sealed the entirety of the tank would experience the pressure all around. When you cut a hole in it things don't magically change - the pressure (and thus force) remains distributed around the whole tank - the only difference is the force normally applied to the part of the tank where we cut our hole is now applied to the outside environment. This force is the same, whether your talking about a real or ideal gas (ignoring any friction the real gas experiences leaving the tank).

m(dot)Ve however, is another matter. If you push on an ideal gas it moves as per F=ma. In the case of pauls tank/pipe apparatus we have ~6000N of force coming out of the tank due to [Pt-Pe]Ae. Keep in mind its particles have no volume - they don't collide with each other or interact with the tube. So the harder you push, or the longer you push, the faster they will go. Hence, pauls 6000N of force will continually accelerate an ideal gas - there is no opposing force to slow them down. Hence the m(dot)Ve, for an ideal gas without a divergent nozzle, will be non-zero.

Real gases are very different - like an idea gas they too will be accelerated by a force. However, when you push on them, they run into the gas atoms/molecules in front of them, resulting in a force pushing back. The harder you push the stronger this backwards force (termed back-pressure) becomes. Once the gas accelerates to the speed of sound (which is the acceleration imparted by [Pt-Pe]Ae), back-pressure will equal the pressure in the tank, which is why the gas no longer accelerates, and instead holds steady at the speed of sound. Keep in mind what that means - there is 6000N of force coming out of the tank, trying to push that gas forward. Since the gas is not accelerating, that means your back pressure is equal to that 6000N.

So with real gases F = [(m(dot)Ve) - Backpressure] +[Pt-Pe]Ae, and in a straight tube m(dot)Ve and Backpressure are the same, hence the total for that side of the equation is zero.

So lastly, how a divergent nozzle changes things
Picture the very end of pauls tube, now with a cone attached to it. At the very end of that tube you'll have the gas traveling at the speed of sound. Behind that gas is 6000N of force trying to push it forward, in front of that gas is 6000N of force trying to push it back.

As soon as the gas leaves the tube and enters the nozzle, things change. Backpressure is due to the compaction of gas by its own flow. As soon as you enter the cone, the gas spreads out to fill the cross-section of the cone, thus reducing backpressure. So now you have 6000N of thrust pushing the gas out of the tube, but less than 6000N of backpressure - the gas accelerates, past the speed of sound.

I won't both re-posting the formula I had posted earlier, but if you recall, it calculates Ve based on the change in pressure along the length of the nozzle. What that change in pressure represents is, in essence, the change in backpressure.
Originally Posted By: kallog

What if the gas is nearly ideal but not quite? There's a discontinuity which is really rare in the real world!

As mentioned above, there isn't really such a thing as a gas that acts like an ideal gas. The closest, as I said above, is super-cooled helium. I have no idea how that acts - I used to build my own rocket engines, but the only experience I have with cold helium is its use in cooling electron microscopes.

Originally Posted By: kallog

Your equation's site says Ve is exit velocity too, just like every other site. None of this "relative to some other speed" or "effective velocity". Actual velocity relative to the rocket.

Ve is the accepted term - doesn't mean its an ideal description of what is being measured. But, as outlined in the site I linked to (and the NASA articles I provided), Ve is proportional to the change in pressure through the nozzle.

Originally Posted By: kallog

I think there's a big mix-up with P symbols too. Your equation's site says:
Pe is exit pressure
Po is chamber pressure
Pa is ambient atmospheric pressure

There may be - I was trying to use pauls terms. Its quite possible I mixed them up as I was converting from one to the other.

Originally Posted By: kallog
But still we don't get zero exit velocity. I think any ratio of those pressures <> 1 for a tube shaped nozzle.

If I ever said Ve was zero that was in error. Keep in mind we're talking about net force produced by the engine. When you have a real gas there are three forces created by the engine - the pressure force, the m(dot)Ve force and the backpressure. When a system is choked - as in the gas is flowing at the speed of sound - the force created by m(dot)Ve and backpressure is the same, but in opposite direction. Hence the force due to Ve is zero, as it is countered by backpressure.

Mathematically, there are two ways of handling that. The accurate way is to calculate m(dot)Ve as per the formula I provided. The alternative way is to calculate the effective Ve - basically the amount of velocity not countered by backpressure. With choked flow, effective Ve is zero. I'd also add that the later way of calculating Ve is also what you do for real gases in a non-choked condition.

Originally Posted By: kallog

force; I used the one that is used by engineers to actually design and build rocket engines.

Why? This is all just a thought experiment to see if a closed tube can fly isn't it? There's no need for engineering. While we're at it why don't we just use that 90K helium and forget all this non-ideal business? Remember Paul never said it's burning fuel, it's just compressed gas coming out of a tank.

I agree - this started off as a thought experiment. Then I made the "mistake" of actully calcuating the forces involved in pauls device, as he described it. He whined that I had done it "wrong" (silly me, using the air he specified instead of an ideal gas...). And things snowballed to here.

We could do away with the whole force generated by pressurized air, and just accept there is a force. As I said (I think on page 3) - the magnitude of the force produced by the tank is irrelevant. What is relevant is the momentum produced will remain zero so long as that tank stays in a sealed pipe.

I have enjoyed "talking" with you though - so it wasn't entirely wasted.